["User: Working alone at their respective constant rates, a can complete a task in ‘ a'days and b in ‘ b'days. they take turns in doing the task with each working 2 days at a time. if a starts they finish the task in exactly 12 days. if b starts, they take half a day more. how long does it take to complete the task if they both work together ?\nAssistant: Lete the work when both work together = 6.\nUser: Can you solve this step by step?\\Assistant: Sure.\nWork done by ab in a day = xy respectively.\nWhen a starts: number of days when a works = 6 number of days when b works = 6 → 6 x + 6 y = 1 when b starts: number of days when a works = 6.5 number of days when b works = 5.5 → 6.5 x + 5.5 y = 1.\nSolving the above two equations for xy x = 1/12 y = 1/12 → total work done by ab in a day = 1/12 + 1/12 = 2/12 = 1/6 → number of days to comp.\nLete the work when both work together = 6.", "User: A company that ships boxes to a total of 12 distribution centers uses color coding to identify each center. if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors, what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter. ).\nAssistant: Let's look at 5 we get 5 single color codes and out 5 color.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet's start with 4 minimum number of colors.\nSo single color code we can make 4 nos.\nNow if we need to make 2 color combination out of 4 we can do.\nSo in 4 !/2 ! * 2 ! or 4 * 3/2 or 6.\nSo total we can make 4 + 6 = 10 color combinations but we have 12 boxes.\nLet's look at 5 we get 5 single color codes and out 5 color.", 'User: A small company is planning to rent either computer a or computer b to print customer mailing lists. both computer a and computer b must be rented on an hourly basis. the rental fee is based only on the amount of time the computer is turned on. it will cost 40 percent more per hour to rent computer a than to rent computer b. computer b would, however, require 20 hours more than computer a to do the job. if either computer a, or computer b were rented the total cost to rent the computer would be $ 850.00. what would be the approximate hourly charge to rent computer b ?. Can you show me your work?\nAssistant: Pa = price of a pb = price of b ta = time for a to comp.\nLete the job tb = time for b to comp.\nLete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 850 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28/0.4 = 70 pb = 850/70 ~ 12.14 c.']

['User: Two goods trains each 500 m long are running in opposite directions on parallel tracks. their speeds are 45 km / hr and 75 km / hr respectively. find the time taken by the slower train to pass the driver of the faster one ?. Can you show me your work?\nAssistant: Relative speed = 45 + 75 = 120 km/hr.\n120 * 5/18 = 100/3 m/sec.\nDistance covered = 500 + 500 = 1000 m.\nRequired time = 1000 * 3/100 = 30 sec.', 'User: Two goods trains each 500 m long are running in opposite directions on parallel tracks. their speeds are 70 km / hr and 30 km / hr respectively. find the time taken by the slower train to pass the driver of the faster one ?\nAssistant: Required time = 1000 * 9/250 = 36 sec.\nUser: Can you solve this step by step?\\Assistant: Sure.\nRelative speed = 70 + 30 = 100 km/hr.\n100 * 5/18 = 250/9 m/sec.\nDistance covered = 500 + 500 = 1000 m.\nRequired time = 1000 * 9/250 = 36 sec.', 'User: Two goods trains each 500 m long are running in opposite directions on parallel tracks. their speeds are 20 km / hr and 30 km / hr respectively. find the time taken by the slower train to pass the driver of the faster one ?\nAssistant: Required time = 1000 * 9/125 = 72 sec.\nUser: Can you solve this step by step?\\Assistant: Sure.\nRelative speed = 20 + 30 = 50 km/hr.\n50 * 5/18 = 125/9 m/sec.\nDistance covered = 500 + 500 = 1000 m.\nRequired time = 1000 * 9/125 = 72 sec.']

['User: P and q can complete a work in 15 days and 10 days respectively. they started the work together and then q left after 2 days. p alone completed the remaining work. the work was finished in - - - days.. Can you show me your work?\nAssistant: Explanation: work done by p in 1 day = 1/15 work done by q in 1 day = 1/10 work done by p and q in 1 day = 1/15 + 1/10 = 1/6 work done by p and q in 2 days = 2 × ( 1/6 ) = 1/3 remaining work = 1 – 1/3 = 2/3 time taken by p to comp.\nLete the remaining work 2/3 = ( 2/3 )/( 1/15 ) = 10 days total time taken = 2 + 10 = 12 days.', "User: 12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days, all of them stopped working. how many women should be put on the job to complete the remaining work, if it so to be completed in 3 days ?. And explain please.\nAssistant: 1 man's 1 day work = 1/48.\n1 woman's 1 day work = 1/60.\n6 men's 2 day's work = 6/48 * 2 = 1/4.\nRemaining work = ( 1 - 1/4 ) = 3/4 now, 1/60 work is done in 1 day by 1 woman.\nSo, 3/4 work will be done in 3 days by ( 60 * 3/4 * 1/3 ) = 15 women.", "User: 12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for 2 days, all of them stopped working. how many women should be put on the job to complete the remaining work, if it is to be completed in 3 days.. Let's think step by step.\nAssistant: One man's one day's work = 1/48 one woman's one day's work = 1/60 6 men's 2 day's work = ( ( 6/48 ) x 2 ) = 1/4 remaining work = 3/4 now, 1/60 work s done in 1 day by 1 woman.\nThus, 3/4 work will be done in 3 days by ( 60 x ( 3/4 ) x ( 1/3 ) ) = 15 women."]

['User: Sarah is driving to the airport. after driving at 25 miles per hour for one hour, she realizes that if she continues at that same average rate she will be an hour late for her flight. she then travels 50 miles per hour for the rest of the trip, and arrives 30 minutes before her flight departs. how many miles did she drive in all ?. And explain please.\nAssistant: After driving at 25 miles per hourfor one hour, this distance left to cover is d - 25.\nSay this distance is x miles.\nNow, we know that the difference in time between covering this distance at 25 miles per hour and 50 miles per hour is 1 + 1/2 = 3/2 hours.\nSo, we have that x/25 - x/50 = 3/2.\n2 x/50 - x/50 = 3/2.\nX/50 = 3/2.\nx = 75.\nTotal distance = x + 25 = 100 miles.', "User: Tom traveled the entire 60 miles trip. if he did the first 12 miles of at a constant rate 24 miles per hour and the remaining trip of at a constant rate 48 miles per hour, what is the his average speed, in miles per hour ? Let's think step by step.\nAssistant: Avg speed = total distance/total time = ( d 1 + d 2 )/( t 1 + t 2 ) = ( 12 + 48 )/( ( 12/24 ) + ( 48/48 ) ) = 60 * 2/3 = 40 mph d.", "User: Tom traveled the entire 80 miles trip. if he did the first 30 miles of at a constant rate 30 miles per hour and the remaining trip of at a constant rate 50 miles per hour, what is the his average speed, in miles per hour ? Let's think step by step.\nAssistant: Avg speed = total distance/total time = ( d 1 + d 2 )/( t 1 + t 2 ) = ( 30 + 50 )/( ( 30/30 ) + ( 50/50 ) ) = 80/2 = 40 mph c."]

['User: On dividing 17698 by a certain number, we get 89 as quotient and 14 as remainder. what is the divisor ?\nAssistant: R = ( dividend ) - ( remainder )/quotient ( 17698 - 14 )/89 = 199.\nUser: Can you solve this step by step?\\Assistant: Sure.\nDivi.\nSor * quotient + remainder = dividend divi.\nSor = ( dividend ) - ( remainder )/quotient ( 17698 - 14 )/89 = 199.', 'User: On dividing 12401 by a certain number, we get 76 as quotient and 13 as remainder. what is the divisor ?. And explain please.\nAssistant: Divi.\nSor * quotient + remainder = dividend divi.\nSor = ( dividend ) - ( remainder )/quotient ( 12401 - 13 )/76 = 163.', 'User: What is the smallest integer that is the sum of 2 distinct prime integers, each greater than 70 ?\nAssistant: Immediate two prime numbers after 70 are - 71 and 73 sum = 71 + 73 = 144.\nUser: Can you solve this step by step?\\Assistant: Sure.\nA prime number ( or a prime ) is a natural number greater than 1 that has no positive divi.\nSors other than 1 and itself.\nHere are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc.\nImmediate two prime numbers after 70 are - 71 and 73 sum = 71 + 73 = 144.']

["User: How many positive integers less than 5,000 are there in which the sum of the digits equals 5 ?. Can you show me your work?\nAssistant: Basically, the question asks how many 4 digit numbers ( including those in the form 0 xxx, 00 xx, and 000 x ) have digits which add up to 5.\nThink about the question this way: we know that there is a total of 5 to be spread among the 4 digits, we just have to determine the number of ways it can be spread.\nLet x represent a sum of 1, and | represent a seperator between two digits.\nAs a result, we will have 5 x's ( digits add up to the 5 ), and 3 |'s ( 3 digit seperators ).\nSo, for example: xx | x | x | x = 2111 | | xxx | xx = 0032 etc.\nThere are 8 c 3 ways to determine where to place the separators.", "User: How many positive integers less than 8,000 are there in which the sum of the digits equals 5 ?. Help me solve this step by step.\nAssistant: Basically, the question asks how many 4 digit numbers ( including those in the form 0 xxx, 00 xx, and 000 x ) have digits which add up to 5.\nThink about the question this way: we know that there is a total of 5 to be spread among the 4 digits, we just have to determine the number of ways it can be spread.\nLet x represent a sum of 1, and | represent a seperator between two digits.\nAs a result, we will have 5 x's ( digits add up to the 5 ), and 3 |'s ( 3 digit seperators ).\nSo, for example: xx | x | x | x = 2111 | | xxx | xx = 0032 etc.\nThere are 8 c 3 ways to determine where to place the separators.", "User: How many positive integers less than 7,000 are there in which the sum of the digits equals 5 ?. Help me solve this step by step.\nAssistant: Basically, the question asks how many 4 digit numbers ( including those in the form 0 xxx, 00 xx, and 000 x ) have digits which add up to 5.\nThink about the question this way: we know that there is a total of 5 to be spread among the 4 digits, we just have to determine the number of ways it can be spread.\nLet x represent a sum of 1, and | represent a seperator between two digits.\nAs a result, we will have 5 x's ( digits add up to the 5 ), and 3 |'s ( 3 digit seperators ).\nSo, for example: xx | x | x | x = 2111 | | xxx | xx = 0032 etc.\nThere are 8 c 3 ways to determine where to place the separators."]

["User: The value of ( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.5 * 7.57 ) is close to. Let's think step by step.\nAssistant: ( 34.31 * 0.473 * 1.567 )/( 0.0673 * 23.5 * 7.57 ) = 25.4303/11.972 = 2.12.", "User: The value of ( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to. Let's think step by step.\nAssistant: ( 34.31 * 0.473 * 1.567 )/( 0.0673 * 23.25 * 7.57 ) = 25.4303/11.845 = 2.15.", 'User: The closest approximation of a ( 69.28 × 0.004 ) / 0.03 is.\nAssistant: , 6.9 < 6.928 × 1.33 < 13.8 9.2 is the only.\nUser: Can you solve this step by step?\\Assistant: Sure.\nA ( 69.28 × 0.004 )/0.03 1.\n0.004 = 4 × 10 ^ ( - 3 ) 2.\n0.03 = 3 × 10 ^ ( - 2 ) 3.\n( a × b )/c = a × ( b/c ) 4.\n0.004/0.03 = 4 × 10 ^ ( - 3 )/( 3 × 10 ^ ( - 2 ) ) = 4 × 10 ^ ( - 3 - ( - 2 ) )/3 = 4 × 10 ^ ( - 1 )/3 = ( 4/3 ) × 10 ^ ( - 1 ) = 1.333 × 10 ^ ( - 1 ).\nTherefore, ( 69.28 × 0.004 )/0.03 = 69.28 × ( 0.004/0.03 ) = 69.28 × 1.33 × 10 ^ ( - 1 ) = 69.28 × 1.33/10 = 6.928 * 1.33 now, 7 × 2 = 14 7 × 1 = 7 or better: 6.9 × 1 = 6.9 6.9 × 2 = 13.8.\nHence, 6.9 < 6.928 × 1.33 < 13.8 9.2 is the only.']

["User: A man rows his boat 90 km downstream and 45 km upstream, taking 2 1 / 2 hours each time. find the speed of the stream ? Let's think step by step.\nAssistant: Speed downstream = d/t = 90/( 2 1/2 ) = 36 kmph speed upstream = d/t = 45/( 2 1/2 ) = 18 kmph the speed of the stream = ( 36 - 18 )/2 = 9 kmph.", 'User: A man can row 3.6 km / hr in still water. it takes him twice as long to row upstream as to row downstream. what is the rate of the current ?\nAssistant: U = b/3 = 1.2 km/hr.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSpeed of boat in still water ( b ) = 3.6 km/hr.\nSpeed of boat with stream ( down stream ), d = b + u speed of boat against stream ( up stream ), u = b – u it is given upstream time is twice to that of down stream.\nDownstream speed is twice to that of upstream.\nSo b + u = 2 ( b – u ).\nu = b/3 = 1.2 km/hr.', "User: A man can row 3.3 km / hr in still water. it takes him twice as long to row upstream as to row downstream. what is the rate of the current ?. Let's think step by step.\nAssistant: Speed of boat in still water ( b ) = 3.3 km/hr.\nSpeed of boat with stream ( down stream ), d = b + u speed of boat against stream ( up stream ), u = b – u it is given upstream time is twice to that of down stream.\nDownstream speed is twice to that of upstream.\nSo b + u = 2 ( b – u ).\nu = b/3 = 1.1 km/hr."]

['User: Carl is facing very difficult financial times and can only pay the interest on a $ 30,000 loan he has taken. the bank charges him a quarterly compound rate of 10 %. what is the approximate interest he pays annually ?\nAssistant: Had it been a simple quarterly rate, we would have just found 4 * 10 % of 30,000 = $ 12000 as our.\nUser: Can you solve this step by step?\\Assistant: Sure.\nUsually, you are given the annual rate of interest and it is mentioned that it is annual rate.\nThe bank charges him a quarterly compounded annual rate of 40 %.\nSo the bank charges 10 % every quarter and compounds it in the next quarter.\nHad it been a simple quarterly rate, we would have just found 4 * 10 % of 30,000 = $ 12000 as our.', 'User: The amount of an investment will double in approximately 70 / p years, where p is the percent interest, compounded annually. if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest, compounded annually, what will be the approximate total value of the investment when thelma is ready to retire 28 years later ?. Can you show me your work?\nAssistant: The amount of an investment will double in approximately 70/p years, where p is the percent interest, compounded annually.\nIf thelma invests $ 40,000 in a long - term cd that pays 5 percent interest, compounded annually, what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70/p years.\nTherefore, the investment gets doubled in 70/5 = every 14 years.\nAfter 28 years, the investment will get doubled 28/14 = 2 times.\nSo the amount invested will get doubled thrice.\nSo, 40000 * 2 = 80000 80000 * 2 = 160000.', 'User: Carl is facing very difficult financial times and can only pay the interest on a $ 20,000 loan he has taken. the bank charges him a quarterly compound rate of 5 %. what is the approximate interest he pays annually ?\nAssistant: Had it been a simple quarterly rate, we would have just found 4 * 5 % of 20,000 = $ 4000 as our.\nUser: Can you solve this step by step?\\Assistant: Sure.\nUsually, you are given the annual rate of interest and it is mentioned that it is annual rate.\nThe bank charges him a quarterly compounded annual rate of 20 %.\nSo the bank charges 5 % every quarter and compounds it in the next quarter.\nHad it been a simple quarterly rate, we would have just found 4 * 5 % of 20,000 = $ 4000 as our.']

['User: The ratio between the length and the breadth of a rectangular park is 3 : 2. if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 6 minutes, then the area of the park ( in sq. m ) is. And explain please.\nAssistant: Perimeter = distance covered in 6 min.\n= 12000 x 6 m = 1000 m.\n60.\nLet length = 3 x metres and breadth = 2 x metres.\nThen, 2 ( 3 x + 2 x ) = 1000 or x = 100.\nLength = 300 m and breadth = 200 m.\nArea = ( 300 x 200 ) m 2 = 60000 m 2.', 'User: The ratio between the length and the breadth of a rectangular park is 3 : 1. if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 8 min, then the area of the park ( in sq. m ) is ?\nAssistant: Area = ( 600 x 200 ) m 2 = 120000 m e.\nUser: Can you solve this step by step?\\Assistant: Sure.\nPerimeter = distance covered in 8 min.\n= 12000 x 8 m = 1600 m.\n60.\nLet length = 3 x metres and breadth = 1 x metres.\nThen, 2 ( 3 x + 1 x ) = 1600 or x = 200.\nLength = 600 m and breadth = 200 m.\nArea = ( 600 x 200 ) m 2 = 120000 m e.', 'User: The ratio between the length and the breadth of a rectangular park is 1 : 3. if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 4 min, then the area of the park ( in sq. m ) is ?. Help me solve this step by step.\nAssistant: Perimeter = distance covered in 4 min.\n= ( 12000 x 4 )/60 m = 800 m.\nLet length = 1 x metres and breadth = 3 x metres.\nThen, 2 ( 1 x + 3 x ) = 800 or x = 100.\nLength = 100 m and breadth = 300 m.\nArea = ( 100 x 300 ) m 2 = 30000 m e.']

['User: A batsman makes a score of 82 runs in the 17 th inning and thus increases his averages by 3. what is his average after 17 th inning ?\nAssistant: X = 34.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 82 = total runs scored in 17 innings.\n16 ( x - 3 ) + 82 = 17 x.\n16 x - 48 + 82 = 17 x.\nx = 34.', 'User: A batsman makes a score of 56 runs in the 17 th inning and thus increases his averages by 3. what is his average after 17 th inning ?. Can you show me your work?\nAssistant: Let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 56 = total runs scored in 17 innings.\n16 ( x - 3 ) + 56 = 17 x.\n16 x - 48 + 56 = 17 x.\nx = 8.', 'User: A batsman makes a score of 87 runs in the 17 th inning and thus increases his averages by 3. what is his average after 17 th inning ?\nAssistant: X = 39.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 87 = total runs scored in 17 innings.\n16 ( x - 3 ) + 87 = 17 x.\n16 x - 48 + 87 = 17 x.\nx = 39.']

['User: A student scored an average of 85 marks in 3 subjects : physics, chemistry and mathematics. if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70, what are the marks in physics ?. Can you show me your work?\nAssistant: Given m + p + c = 85 * 3 = 255.\n( 1 ) m + p = 90 * 2 = 180.\n( 2 ) p + c = 70 * 2 = 140.\n( 3 ) where m, p and c are marks obtained by the student in mathematics, physics and chemistry.\np = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 255 = 65.', "User: A student scored an average of 75 marks in 3 subjects : physics, chemistry and mathematics. if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70, what are the marks in physics ?. Let's think step by step.\nAssistant: Given m + p + c = 75 * 3 = 225.\n( 1 ) m + p = 90 * 2 = 180.\n( 2 ) p + c = 70 * 2 = 140.\n( 3 ) where m, p and c are marks obtained by the student in mathematics, physics and chemistry.\np = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 225 = 95.", 'User: A student scored an average of 80 marks in 3 subjects : physics, chemistry and mathematics. if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70, what are the marks in physics ?\nAssistant: P = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 240 = 80.\nUser: Can you solve this step by step?\\Assistant: Sure.\nGiven m + p + c = 80 * 3 = 240.\n( 1 ) m + p = 90 * 2 = 180.\n( 2 ) p + c = 70 * 2 = 140.\n( 3 ) where m, p and c are marks obtained by the student in mathematics, physics and chemistry.\np = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 240 = 80.']

['User: Farm tax is levied on the 75 % of the cultivated land. the tax department collected total $ 3840 through the farm tax from the village of mr. william. mr. william paid only $ 480 as farm tax. the percentage of total land of mr. willam over the total taxable land of the village is :.\nAssistant: But the question asks ratio between his total land to total cultivated land.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThis will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village.\nThat leads to ( 480/3840 ) x 100 = 12.5 % in percentage terms.\nBut the question asks ratio between his total land to total cultivated land.', 'User: Farm tax is levied on the 90 % of the cultivated land. the tax department collected total $ 3840 through the farm tax from the village of mr. william. mr. william paid only $ 480 as farm tax. the percentage of total land of mr. willam over the total taxable land of the village is :.\nAssistant: But the question asks ratio between his total land to total cultivated land.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThis will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village.\nThat leads to ( 480/3840 ) x 100 = 12.5 % in percentage terms.\nBut the question asks ratio between his total land to total cultivated land.', "User: Farm tax is levied on the 60 % of the cultivated land. the tax department collected total $ 3840 through the farm tax from the village of mr. willam. mr. willam paid only $ 500 as farm tax. the percentage of total land of mr. willam over the total taxable land of the village is :. Help me solve this step by step.\nAssistant: Only trick n this question is to ignore 60 % information as farm tax is levied uniformly in the village and that includes mr william's land what percentage of tax mr william paid ? this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village.\nThat leads to ( 500/3840 ) x 100 = 13.02083 % in percentage terms.\nBut the question asks ratio between his total land to total cultivated land."]

['User: A, b, c subscribe rs. 50,000 for a business. if a subscribes rs. 4000 more than b and b rs. 5000 more than c, out of a total profit of rs. 39,000, what will be the amount a receives ?\nAssistant: X = 36000/3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a, b and c = 21000: 17000: 12000 = 21: 17: 12 share of a = total profit × 21/50 = 39000 × 21/50 = 16380.\nUser: Can you solve this step by step?\\Assistant: Sure.\nTotal amount invested = 50000 assume that investment of c = x.\nThen investment of b = 5000 + x, investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000.\n3 x + 14000 = 50000.\n3 x = 50000 – 14000 = 36000.\nx = 36000/3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a, b and c = 21000: 17000: 12000 = 21: 17: 12 share of a = total profit × 21/50 = 39000 × 21/50 = 16380.', 'User: A, b, c subscribe rs. 50000 for a business. if a subscribes rs. 4000 more than b and b rs. 5000 more than c, out of a total profit of rs. 36000, what will be the amount a receives ?. And explain please.\nAssistant: Total amount invested = 50000 assume that investment of c = x.\nThen investment of b = 5000 + x, investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000.\n3 x + 14000 = 50000.\n3 x = 50000 – 14000 = 36000.\nx = 36000/3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a, b and c = 21000: 17000: 12000 = 21: 17: 12 share of a = total profit × 21/50 = 36000 × 21/50 = 15,120.', 'User: A, b, c subscribe rs. 50,000 for a business. if a subscribes rs. 4000 more than b and b rs. 5000 more than c, out of a total profit of rs. 36,000, what will be the amount a receives ?. Can you show me your work?\nAssistant: Total amount invested = 50000 assume that investment of c = x.\nThen investment of b = 5000 + x, investment of a = 4000 + 5000 + x = 9000 + x x + 5000 + x + 9000 + x = 50000.\n3 x + 14000 = 50000.\n3 x = 50000 – 14000 = 36000.\nx = 36000/3 = 12000 investment of c = x = 12000 investment of b = 5000 + x = 17000 investment of a = 9000 + x = 21000 ratio of the investment of a, b and c = 21000: 17000: 12000 = 21: 17: 12 share of a = total profit × 21/50 = 36000 × 21/50 = 15,120.']

['User: A leak in the bottom of a tank can empty the full tank in 8 hours. an inlet pipe fills water at the rate of 6 litres a minute. when the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. how many litres does the cistern hold ?. And explain please.\nAssistant: Work done by the in.\nLet in 1 hour = ( 1/8 - 1/12 ) = 1/24.\nWork done by the in.\nLet in 1 min.\n= ( 1/24 × 1/60 ) = 1/1440.\nVolume of 1/1440 part = 6 litres.\nTherefore, volume of whole = ( 1440 × 6 ) ‹ = › 8640 litres.', "User: Two pipes a and b can separately fill a tank in 10 and 15 minutes respectively. a third pipe c can drain off 45 liters of water per minute. if all the pipes are opened, the tank can be filled in 15 minutes. what is the capacity of the tank ? Let's think step by step.\nAssistant: 1/10 + 1/15 - 1/x = 1/15 x = 10 10 * 45 = 450.", "User: A leak in the bottom of a tank can empty the full tank in 6 hours. an inlet pipe fills water at the rate of 2.5 liters per minute. when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours. the capacity of the tank is ? Let's think step by step.\nAssistant: 1/x - 1/6 = - 1/8 x = 24 hrs 24 * 60 * 2.5 = 3600."]

14_Geometry problems involving intersecting lines on a coordinate plane and finding the side length of a cube with given surface area___

['User: The two lines y = x and x = - 6 intersect on the coordinate plane. if z represents the area of the figure formed by the intersecting lines and the x - axis, what is the side length of a cube whose surface area is equal to 6 z ?\nAssistant: Clearly s = √ 18 = 3 √ 2, or.\nUser: Can you solve this step by step?\\Assistant: Sure.\n800 score official.\n: the first step to solving this problem is to actually graph the two lines.\nThe lines intersect at the point ( - 6, - 6 ) and form a right triangle whose base length and height are both equal to 4.\nAs you know, the area of a triangle is equal to one half the product of its base length and height: a = ( 1/2 ) bh = ( 1/2 ) ( 6 × 6 ) = 18.\nSo z = 18.\nThe next step requires us to find the length of a side of a cube that has a face area equal to 18.\nAs you know the 6 faces of a cube are squares.\nSo, we can reduce the problem to finding the length of the side of a square that has an area of 18.\nSince the area of a square is equal to s ², where s is the length of one of its side, we can write and solve the equation s ² = 18.\nClearly s = √ 18 = 3 √ 2, or.', 'User: The two lines y = x and x = - 4 intersect on the coordinate plane. if z represents the area of the figure formed by the intersecting lines and the x - axis, what is the side length t of a cube whose surface area is equal to 6 z ?. Help me solve this step by step.\nAssistant: 800 score official.\n: the first step to solving this problem is to actually graph the two lines.\nThe lines intersect at the point ( - 4, - 4 ) and form a right triangle whose base length and height are both equal to 4.\nAs you know, the area of a triangle is equal to one half the product of its base length and height: a = ( 1/2 ) bh = ( 1/2 ) ( 4 × 4 ) = 8.\nSo z = 8.\nThe next step requires us to find the length of a side of a cube that has a face area equal to 8.\nAs you know the 6 faces of a cube are squares.\nSo, we can reduce the problem to finding the length of the side of a square that has an area of 8.\nSince the area of a square is equal to s ², where s is the length of one of its side, we can write and solve the equation s ² = 8.\nClearly s = √ 8 = 2 √ 2, or.', 'User: The two lines y = x and x = - 4 intersect on the coordinate plane. if z represents the area of the figure formed by the intersecting lines and the x - axis, what is the side length r of a cube whose surface area is equal to 6 z ?. Help me solve this step by step.\nAssistant: 800 score official.\n: the first step to solving this problem is to actually graph the two lines.\nThe lines intersect at the point ( - 4, - 4 ) and form a right triangle whose base length and height are both equal to 4.\nAs you know, the area of a triangle is equal to one half the product of its base length and height: a = ( 1/2 ) bh = ( 1/2 ) ( 4 × 4 ) = 8.\nSo z = 8.\nThe next step requires us to find the length of a side of a cube that has a face area equal to 8.\nAs you know the 6 faces of a cube are squares.\nSo, we can reduce the problem to finding the length of the side of a square that has an area of 8.\nSince the area of a square is equal to s ², where s is the length of one of its side, we can write and solve the equation s ² = 8.\nClearly s = √ 8 = 2 √ 2, or.']

['User: The ages of two person differ by 20 years. if 5 years ago, the elder one be 5 times as old as the younger one, their present ages ( in years ) are respectively. Can you show me your work?\nAssistant: Let their ages be x and ( x + 20 ) years.\nThen, 5 ( x - 5 ) = ( x + 20 - 5 ).\n4 x = 40.\nx = 10 their present ages are 30 years and 10 year.', "User: Ages of two persons differ by 22 years. if 6 year ago, the elder one be 3 times as old the younger one, find their present age. Let's think step by step.\nAssistant: Let the age of younger person is x, then elder person age is ( x + 22 ).\n3 ( x - 6 ) = ( x + 22 - 6 ) [ 6 years before ].\n3 x - 18 = x + 16.\nx = 17.\nSo other person age is x + 22 = 39.", 'User: Ages of two persons differ by 16 years. if 6 year ago, the elder one be 3 times as old the younger one, find their present age. Help me solve this step by step.\nAssistant: Let the age of younger person is x, then elder person age is ( x + 16 ).\n3 ( x - 6 ) = ( x + 16 - 6 ) [ 6 years before ].\n3 x - 18 = x + 10.\nx = 14.\nSo other person age is x + 16 = 30.']

["User: Riya and priya set on a journey. riya moves eastward at a speed of 20 kmph and priya moves westward at a speed of 30 kmph. how far will be priya from riya after 30 minutes. Let's think step by step.\nAssistant: Total eastward distance = 20 kmph * 1/2 hr = 10 km total westward distance = 30 kmph * 1/2 hr = 15 km total distn betn them = 10 + 15 = 25 km.", "User: Riya and priya set on a journey. riya moves eastward at a speed of 30 kmph and priya moves westward at a speed of 50 kmph. how far will be priya from riya after 45 minutes. Let's think step by step.\nAssistant: Total eastward distance = 30 kmph * 3/4 hr = 22.50 km total westward distance = 50 kmph * 3/4 hr = 37.5 km total distn betn them = 22.5 + 37.5 = 60 km.", "User: Riya and priya set on a journey. riya moves eastward at a speed of 24 kmph and priya moves westward at a speed of 35 kmph. how far will be priya from riya after 45 minutes. Let's think step by step.\nAssistant: Total eastward distance = 24 kmph * 3/4 hr = 18 km total westward distance = 35 kmph * 3/4 hr = 26 km total distn betn them = 18 + 26 = 44 km."]

['User: The reciprocal of the hcf and lcm of two are 1 / 13 and 1 / 312. if one of the number is 24 then other no. is.\nAssistant: , other = 13 * 312/24 = 169.\nUser: Can you solve this step by step?\\Assistant: Sure.\nReciprocal of the hcf and lcm of two are 1/13 and 1/312.\nSo, hcf = 13, lcm = 312 lcm * hcf = product of two numbers = a * b.\nb = lcm * hcf/a.\nSo, other = 13 * 312/24 = 169.', 'User: The reciprocal of the hcf and lcm of two are 1 / 16 and 1 / 312. if one of the number is 24 then other no. is. Help me solve this step by step.\nAssistant: Reciprocal of the hcf and lcm of two are 1/16 and 1/312.\nSo, hcf = 16, lcm = 312 lcm * hcf = product of two numbers = a * b.\nb = lcm * hcf/a.\nSo, other = 16 * 312/24 = 208.', 'User: The reciprocal of the hcf and lcm of two are 1 / 12 and 1 / 312. if one of the number is 24 then other no. is. Can you show me your work?\nAssistant: Reciprocal of the hcf and lcm of two are 1/12 and 1/312.\nSo, hcf = 12, lcm = 312 lcm * hcf = product of two numbers = a * b.\nb = lcm * hcf/a.\nSo, other = 12 * 312/24 = 156.']

["User: Tough and tricky questions : word problems. a salesman's income consists of commission and base salary. his weekly income totals over the past 7 weeks have been $ 406, $ 413, $ 420, $ 436, $ 395, $ 410, $ 360. what must his average ( arithmetic mean ) income over the next two weeks be to decrease his average weekly income to $ 400 over the 9 - week period ?\nAssistant: Dividing $ 760 by 2 will give us the amount he needs to earn on average over the next 2 weeks: $ 760/2 = $ 380.\nUser: Can you solve this step by step?\\Assistant: Sure.\nOfficial.\n: ( b ) first, we need to add up the wages over the past 7 weeks: $ 406 + $ 413 + $ 420 + $ 436 + $ 395 + $ 410 + $ 360 = $ 2840.\nTo average $ 400 over 9 weeks, the salesman would need to earn: $ 400 × 9 = $ 3600.\nSubtract $ 2840 from $ 3600 to determine how much he would need to earn, in total, over the next 2 weeks to average $ 400 for the 9 weeks: $ 3600 – $ 2840 = $ 760.\nDividing $ 760 by 2 will give us the amount he needs to earn on average over the next 2 weeks: $ 760/2 = $ 380.", "User: Tough and tricky questions : word problems. a salesman's income consists of commission and base salary. his weekly income totals over the past 5 weeks have been $ 406, $ 413, $ 420, $ 436 and $ 395. what must his average ( arithmetic mean ) income over the next 5 weeks be to increase his average weekly income to $ 500 over the 10 - week period ?\nAssistant: Dividing $ 2930 by 5 will give us the amount he needs to earn on average over the next 5 weeks: $ 2930/5 = $ 586.\nUser: Can you solve this step by step?\\Assistant: Sure.\nOfficial.\n: ( b ) first, we need to add up the wages over the past 5 weeks: $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070.\nTo average $ 500 over 10 weeks, the salesman would need to earn: $ 500 × 10 = $ 5000.\nSubtract $ 2070 from $ 5000 to determine how much he would need to earn, in total, over the next 5 weeks to average $ 500 for the 10 weeks: $ 5000 – $ 2070 = $ 2930.\nDividing $ 2930 by 5 will give us the amount he needs to earn on average over the next 5 weeks: $ 2930/5 = $ 586.", "User: Tough and tricky questions : word problems. a salesman's income consists of commission and base salary. his weekly income totals over the past 5 weeks have been $ 406, $ 413, $ 420, $ 436 and $ 395. what must his average ( arithmetic mean ) income over the next two weeks be to decrease his average weekly income to $ 400 over the 7 - week period ?\nAssistant: Dividing $ 730 by 2 will give us the amount he needs to earn on average over the next 2 weeks: $ 730/2 = $ 365.\nUser: Can you solve this step by step?\\Assistant: Sure.\nOfficial.\n: ( c ) first, we need to add up the wages over the past 5 weeks: $ 406 + $ 413 + $ 420 + $ 436 + $ 395 = $ 2070.\nTo average $ 400 over 7 weeks, the salesman would need to earn: $ 400 × 7 = $ 2800.\nSubtract $ 2070 from $ 2800 to determine how much he would need to earn, in total, over the next 2 weeks to average $ 400 for the 7 weeks: $ 2800 – $ 2070 = $ 730.\nDividing $ 730 by 2 will give us the amount he needs to earn on average over the next 2 weeks: $ 730/2 = $ 365."]

['User: There are 56 lights which are functional and each is controlled by a separate on / off switch. two children x and y start playing with the switches. x starts by pressing every third switch till he reaches the end. y, thereafter, presses every fifth switch till he too reaches the end. if all switches were in off position at the beggining, how many lights are switched on by the end of this operation ?\nAssistant: Number of switches that are turned on = ( 18 - 6 ) + ( 11 - 6 ) = 17.\nUser: Can you solve this step by step?\\Assistant: Sure.\nEditing my.\n: number of switches = 56 number of switches turned on by x: 3, 6.\n54 = 18 number of switches turned on by y: 5, 10.\n55 = 11 few switches are turned on by x and later turned off by y: lcm ( 35 ) = 15 x = 15, 30.\n90 = 6.\nSubtract the above 6 switches from both x and y as they are turned off.\nNumber of switches that are turned on = ( 18 - 6 ) + ( 11 - 6 ) = 17.', "User: A certain university will select 1 of 7 candidates eligible to fill a position in the mathematics department and 3 of 10 candidates eligible to fill 3 identical positions in the computer science department. if none of the candidates is eligible for a position in both departments, how many different sets of 4 candidates are there to fill the 4 positions ?\nAssistant: We simplify instead.\nUser: Can you solve this step by step?\\Assistant: Sure.\n1 of 7 will be chosen for the math 3 of 10 will be chosen for the computer none of the 4 chosen people can be in more than one departments.\nWe can choose any of the 7 candidates for the math dep.\nWhich gives as 7 selections.\nWe can choose 3 of the 10 candidates for the computer dep.\nWhich gives us 3 selections and 7 rejections.\nSo, the way to find how many different selections of 2 candidates we can have for the computer dep.\nWe do: 10 !/3 ! * 7 ! = 120 we are multiplying our individual selections: 7 * 120 = 840 in the bolded part, we do n't have to multiply all of the numbers, as those in 8 ! are included in 10 !.\nSo we simplify instead.", "User: A certain university will select 1 of 6 candidates eligible to fill a position in the mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the computer science department. if none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions ?\nAssistant: We simplify instead.\nUser: Can you solve this step by step?\\Assistant: Sure.\n1 of 6 will be chosen for the math 2 of 10 will be chosen for the computer none of the 3 chosen people can be in more than one departments.\nWe can choose any of the 6 candidates for the math dep.\nWhich gives as 6 selections.\nWe can choose 2 of the 10 candidates for the computer dep.\nWhich gives us 2 selections and 8 rejections.\nSo, the way to find how many different selections of 2 candidates we can have for the computer dep.\nWe do: 10 !/2 ! * 8 ! = ( 9 * 10 )/2 = 90/2 = 45.\nWe are multiplying our individual selections: 6 * 45 = 270 in the bolded part, we do n't have to multiply all of the numbers, as those in 8 ! are included in 10 !.\nSo we simplify instead."]

['User: A theater box office sold an average ( arithmetic mean ) of 64 tickets per staff member to a particular movie. among the daytime staff, the average number sold per member was 80, and among the evening staff, the average number sold was 60. if there are no other employees, what was the ratio of the number of daytime staff members to the number of evening staff members ?. Help me solve this step by step.\nAssistant: Deviation from the mean for the daytime staff = 80 - 64 = 16.\nDeviation from the mean for the evening staff = 64 - 60 = 4.\nThus, the ratio of the number of daytime staff members to the number of evening staff members is 4: 16 = 1: 4.', "User: Among all sales staff at listco corporation, college graduates and those without college degrees are equally represented. each sales staff member is either a level - 1 or level - 2 employee. level - 1 college graduates account for 15 % of listco's sales staff. listco employs 90 level - 1 employees, 30 of whom are college graduates. how many sales staff members without college degrees are level - 2 employees ?\nAssistant: No c and no l 2 will be 100 - 60 = 40.\nUser: Can you solve this step by step?\\Assistant: Sure.\nI'm going in on this one.\nLet's say that we have the following.\nSo we know that l 1 = 90 and that c and l 1 = 0.15 x, we should set up a double set matrix btw but anyways, i'm just explaining the point with this problem.\nNow we are told that 0.15 x = 30.\nTherefore the grand total is 200.\nNow we know that l 2 is 200 - 90 = 110.\nWe\nAlso learn that c and no c are equally represented.\nThus 100 each.\nTherefore no c and no l 2 will be 100 - 60 = 40.", "User: Among all sales staff at listco corporation, college graduates and those without college degrees are equally represented. each sales staff member is either a level - 1 or level - 2 employee. level - 1 college graduates account for 15 % of listco's sales staff. listco employs 60 level - 1 employees, 30 of whom are college graduates. how many sales staff members without college degrees are level - 2 employees ?\nAssistant: No c and no l 2 will be 100 - 30 = 70.\nUser: Can you solve this step by step?\\Assistant: Sure.\nI'm going in on this one.\nLet's say that we have the following.\nSo we know that l 1 = 60 and that c and l 1 = 0.15 x, we should set up a double set matrix btw but anyways, i'm just explaining the point with this problem.\nNow we are told that 0.15 x = 30.\nTherefore the grand total is 200.\nNow we know that l 2 is 200 - 60 = 140.\nWe\nAlso learn that c and no c are equally represented.\nThus 100 each.\nTherefore no c and no l 2 will be 100 - 30 = 70."]

['User: Sari and ken climb up a mountain. at night, they camp together. on the day they are supposed to reach the summit, sari wakes up at 05 : 00 and starts climbing at a constant pace. ken starts climbing only at 07 : 00, when sari is already 700 meters ahead of him. nevertheless, ken climbs at a constant pace of 500 meters per hour, and reaches the summit before sari. if sari is 50 meters behind ken when he reaches the summit, at what time did ken reach the summit ?. Can you show me your work?\nAssistant: Both sari and ken climb in the same direction.\nSpeed of sari = 700/2 = 350 meters/hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters/hr at 8: 00, distance between ken and sari is 700 meters.\nKen needs to cover this and another 50 meters.\nTime he will take = total distance to be covered/relative speed = ( 700 + 50 )/( 500 - 350 ) = 5 hrs starting from 7: 00, in 5 hrs, the time will be 12: 00.', 'User: Sari and ken climb up a mountain. at night, they camp together. on the day they are supposed to reach the summit, sari wakes up at 06 : 00 and starts climbing at a constant pace. ken starts climbing only at 08 : 00, when sari is already 600 meters ahead of him. nevertheless, ken climbs at a constant pace of 500 meters per hour, and reaches the summit before sari. if sari is 50 meters behind ken when he reaches the summit, at what time did ken reach the summit ?. And explain please.\nAssistant: Both sari and ken climb in the same direction.\nSpeed of sari = 600/2 = 300 meters/hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters/hr at 8: 00, distance between ken and sari is 600 meters.\nKen needs to cover this and another 50 meters.\nTime he will take = total distance to be covered/relative speed = ( 600 + 50 )/( 500 - 300 ) = 3.25 hrs starting from 8: 00, in 5 hrs, the time will be 11: 15.', 'User: Sari and ken climb up a mountain. at night, they camp together. on the day they are supposed to reach the summit, sari wakes up at 06 : 00 and starts climbing at a constant pace. ken starts climbing only at 08 : 00, when sari is already 800 meters ahead of him. nevertheless, ken climbs at a constant pace of 500 meters per hour, and reaches the summit before sari. if sari is 50 meters behind ken when he reaches the summit, at what time did ken reach the summit ?\nAssistant: Time he will take = total distance to be covered/relative speed = ( 800 + 50 )/( 500 - 400 ) = 8.5 hrs starting from 8: 00, in 8.5 hrs, the time will be 16: 30.\nUser: Can you solve this step by step?\\Assistant: Sure.\nBoth sari and ken climb in the same direction.\nSpeed of sari = 800/2 = 400 meters/hr ( since she covers 800 meters in 2 hrs ) speed of ken = 500 meters/hr at 8: 00, distance between ken and sari is 800 meters.\nKen needs to cover this and another 50 meters.\nTime he will take = total distance to be covered/relative speed = ( 800 + 50 )/( 500 - 400 ) = 8.5 hrs starting from 8: 00, in 8.5 hrs, the time will be 16: 30.']

['User: If a / b = 1 / 3, b / c = 2, c / d = 1 / 2, d / e = 3 and e / f = 1 / 8, then what is the value of abc / def ?\nAssistant: Abc/def = ( 2 * 6 * 3 )/( 6 * 2 * 16 ) = 3/16.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSay a = 2.\nThen: a/b = 1/3.\nb = 6.\nB/c = 2.\nc = 3.\nC/d = 1/2.\nd = 6.\nD/e = 3.\ne = 2.\nE/f = 1/8.\nf = 16.\nAbc/def = ( 2 * 6 * 3 )/( 6 * 2 * 16 ) = 3/16.', "User: If a / b = 1 / 2, b / c = 5, c / d = 2 / 3, d / e = 1 / 6 and e / f = 2 / 3, then what is the value of abc / def ?. Let's think step by step.\nAssistant: Say a = 2.\nThen: a/b = 1/2.\nb = 4.\nB/c = 5.\nc = 4/5.\nC/d = 2/3.\nd = 6/5.\nD/e = 1/6.\ne = 36/5.\nE/f = 2/3.\nf = 54/2.\nAbc/def = ( 2 * 4 * 4/5 )/( 6/5 * 36/5 * 54/5 ) = 50/729.", 'User: If 1 tic equals 5 tacs and 2 tacs equal 8 tocs, what is the ratio of one tic to one toc ?. And explain please.\nAssistant: Tic = 5 * tac and 2 * tac = 8 * toc.\n2 * tic = 10 * tac and 10 * tac = 40 * toc.\n2 * tic = 40 * toc.\nTic/toc = 40/2 = 20/1.']

['User: A bag contains 4 white marbles and 4 black marbles. if each of 4 girls and 4 boys randomly selects and keeps a marble, what is the probability that all of the girls select the same colored marble ?\nAssistant: , total favorable outcomes/total outcomes = 2/70 = 1/35 a.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFirst, total ways to select for all boys and girls, i.\ne 8 !/( 4 ! * 4 ! ) = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1/4 * 3 * 2 * 1 * 4 * 3 * 2 * 1 = 70 then there are one two way girls can have all same colors, either white or black.\nThe number of ways in which 4 girls can select 4 white balls = 4 c 4 = 1 the number of ways in which 4 girls can select 4 black balls = 4 c 4 = 1.\nTherefore, total favorable outcomes/total outcomes = 2/70 = 1/35 a.', 'User: A bag contains 2 white marbles and 2 black marbles. if each of 2 girls and 2 boys randomly selects and keeps a marble, what is the probability that all of the girls select the same colored marble ?\nAssistant: , total favorable outcomes/total outcomes = 2/6 = 1/3 c.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFirst, total ways to select for all boys and girls, i.\ne 4 !/( 2 ! * 2 ! ) = 4 * 3 * 2 * 1/2 * 1 * 2 * 1 = 6 then there are one two way girls can have all same colors, either white or black.\nThe number of ways in which 2 girls can select 2 white balls = 2 c 2 = 1 the number of ways in which 2 girls can select 2 black balls = 2 c 2 = 1.\nTherefore, total favorable outcomes/total outcomes = 2/6 = 1/3 c.', 'User: A bag contains 3 white marbles and 3 black marbles. if each of 3 girls and 3 boys randomly selects and keeps a marble, what is the probability that all of the girls select the same colored marble ?. And explain please.\nAssistant: First, total ways to select for all boys and girls, i.\ne 6 !/( 3 ! * 3 ! ) = 6 * 5 * 4 * 3 * 2 * 1/3 * 2 * 1 * 3 * 2 * 1 = 20 then there are one two way girls can have all same colors, either white or black.\nThe number of ways in which 3 girls can select 3 white balls = 3 c 3 = 1 the number of ways in which 3 girls can select 3 black balls = 3 c 3 = 1.\nTherefore, total favorable outcomes/total outcomes = 2/20 = 1/10 b.']

['User: A positive number x is multiplied by 7, and this product is then divided by 3. if the positive square root of the result of these two operations equals x, what is the value of x ?. Help me solve this step by step.\nAssistant: Sq rt ( 7 x/3 ) = x.\n7 x/3 = x ^ 2.\nx = 7/3.', 'User: A positive number x is multiplied by 8, and this product is then divided by 3. if the positive square root of the result of these two operations equals x, what is the value of x ?. And explain please.\nAssistant: Sq rt ( 8 x/3 ) = x.\n8 x/3 = x ^ 2.\nx = 8/3.', "User: If x ^ 2 + 1 / x ^ 2 = 9, what is the value of x ^ 4 + 1 / x ^ 4 ?\nAssistant: , ( x ² + 1/x ² ) ( x ² + 1/x ² ) = 81 expand to get: x ⁴ + 1 + 1 + 1/x ⁴ = 81 simplify: x ⁴ + 1/x ⁴ = 79.\nUser: Can you solve this step by step?\\Assistant: Sure.\nImportant: I notice that if we square x ², we get x ⁴, and if we square 1/x ², we get 1/x ⁴.\nLet's see what happens if we take the equation x ² + 1/x ² = 9 andsquareboth sides: ( x ² + 1/x ² ) ² = 81.\nSo, ( x ² + 1/x ² ) ( x ² + 1/x ² ) = 81 expand to get: x ⁴ + 1 + 1 + 1/x ⁴ = 81 simplify: x ⁴ + 1/x ⁴ = 79."]

["User: At an upscale fast - food restaurant, shin can buy 3 burgers, 7 shakes, and one cola for $ 120. at the same place it would cost $ 160.50 for 4 burgers, 10 shakes, and one cola. how much would it cost for a meal of one burger, one shake, and one cola ?. Can you show me your work?\nAssistant: Let's suppose that the price of a burger is bb, of a shake - ss and that of a cola is cc.\nWe can then construct these equations: 3 b + 7 s + c = 120 4 b + 10 s + c = 160.5 subtracting the first equation from the second gives us b + 3 s = 40.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 39.\nIn any case, there is no necessity to know each item's price, just the sum.", "User: At an upscale fast - food restaurant, shin can buy 3 burgers, 7 shakes, and one cola for $ 130. at the same place it would cost $ 164.50 for 4 burgers, 10 shakes, and one cola. how much would it cost for a meal of one burger, one shake, and one cola ?\nAssistant: In any case, there is no necessity to know each item's price, just the sum.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet's suppose that the price of a burger is bb, of a shake - ss and that of a cola is cc.\nWe can then construct these equations: 3 b + 7 s + c = 130 4 b + 10 s + c = 164.5 subtracting the first equation from the second gives us b + 3 s = 34.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 61.\nIn any case, there is no necessity to know each item's price, just the sum.", "User: At an upscale fast - food restaurant, shin can buy 3 burgers, 7 shakes, and one cola for $ 120. at the same place it would cost $ 162.50 for 4 burgers, 10 shakes, and one cola. how much would it cost for a meal of one burger, one shake, and one cola ?. And explain please.\nAssistant: Let's suppose that the price of a burger is bb, of a shake - ss and that of a cola is cc.\nWe can then construct these equations: 3 b + 7 s + c = 120 4 b + 10 s + c = 162.5 subtracting the first equation from the second gives us b + 3 s = 42.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 35.\nIn any case, there is no necessity to know each item's price, just the sum."]

["User: Shannon and maxine work in the same building and leave work at the same time. shannon lives due north of work and maxine lives due south. the distance between maxine's house and shannon's house is 60 miles. if they both drive home at the rate 2 r miles per hour, maxine arrives home 60 minutes after shannon. if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour, shannon arrives home 2 hours before maxine. how far does maxine live from work ?\nAssistant: We get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 48.\nUser: Can you solve this step by step?\\Assistant: Sure.\nNice question + 1 we have that x/24 - ( 60 - x )/2 r = 40\nAlso x/r - ( 60 - x )/2 r = 120.\nSo we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 48.", "User: Shannon and maxine work in the same building and leave work at the same time. shannon lives due north of work and maxine lives due south. the distance between maxine's house and shannon's house is 50 miles. if they both drive home at the rate 2 r miles per hour, maxine arrives home 40 minutes after shannon. if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour, shannon arrives home 2 hours before maxine. how far does maxine live from work ?. And explain please.\nAssistant: Nice question + 1 we have that x/24 - ( 60 - x )/2 r = 40\nAlso x/r - ( 60 - x )/2 r = 120.\nSo we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 46.", "User: Shannon and maxine work in the same building and leave work at the same time. shannon lives due north of work and maxine lives due south. the distance between maxine's house and shannon's house is 40 miles. if they both drive home at the rate 2 r miles per hour, maxine arrives home 40 minutes after shannon. if maxine rider her bike home at the rate of r per hour and shannon still drives at a rate of 2 r miles per hour, shannon arrives home 2 hours before maxine. how far does maxine live from work ?\nAssistant: We get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 38.\nUser: Can you solve this step by step?\\Assistant: Sure.\nNice question + 1 we have that x/24 - ( 60 - x )/2 r = 40\nAlso x/r - ( 60 - x )/2 r = 120.\nSo we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 38."]

["User: A card is drawn from a pack of 52 cards. the probability of ge ƫ ng a queen of club or a king of heart is. Let's think step by step.\nAssistant: Explanation: total number of cases = 52 favourable cases = 2 probability = 2/56 = 1/26.", 'User: A card is drawn from a pack of 52 cards. the probability of getting a queen club or a king of heart ?\nAssistant: N ( e ) = 2 p ( e ) = n ( e )/n ( s ) = 2/52 = 1/26.\nUser: Can you solve this step by step?\\Assistant: Sure.\nN ( s ) = 52.\nLet e = event of getting a queen of club or a king of heart.\nn ( e ) = 2 p ( e ) = n ( e )/n ( s ) = 2/52 = 1/26.', 'User: A card is drawn from a pack of 52 cards. the probability of getting a queen of club or a king of heart is :. And explain please.\nAssistant: Here, n ( s ) = 52.\nLet e = event of getting a queen of club or a king of heart.\nThen, n ( e ) = 2.\np ( e ) = n ( e )/n ( s ) = 2/52 = 1/26.']

["User: The ratio of ages of aman, bren, and charlie are in the ratio 5 : 8 : 7 respectively. if 8 years ago, the sum of their ages was 76, what will be the age of bren 10 years from now ?\nAssistant: = 40 + 10 = 50.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the present ages of aman, bren, and charlie be 5 x, 8 x and 7 x respectively.\n5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of bren = 8 * 5 = 40 bren's age 10 years.\nHence = 40 + 10 = 50.", "User: The total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year, less the age of the tree in years ( rounded down to the nearest integer ). during its 5 th year, the plum tree grew 50 plums. if this trend continues, how many plums will it grow during its 7 th year ? Let's think step by step.\nAssistant: 1 st year: 0 - 1 ( age ), we take age = 0 ( as the question says that we have to ( rounded down to the nearest integer ) ) 2 ndyear: 1 - 2 ( age ), we take age = 1 3 rd year: 2 - 3 ( age ), we take age = 2 4 th year: 3 - 4 ( age ), we take age = 3 5 th year: 4 - 5 ( age ), we take age = 4 6 th year: 5 - 6 ( age ), we take age = 5 7 th year: 6 - 7 ( age ), we take age = 6.", "User: The total number of plums that grow during each year on a certain plum tree is equal to the number of plums that grew during the previous year, less the age of the tree in years ( rounded down to the nearest integer ). during its 8 th year, the plum tree grew 50 plums. if this trend continues, how many plums will it grow during its 6 th year ?. Let's think step by step.\nAssistant: 1 st year: 0 - 1 ( age ), we take age = 0 ( as the question says that we have to ( rounded down to the nearest integer ) ) 2 ndyear: 1 - 2 ( age ), we take age = 1 3 rd year: 2 - 3 ( age ), we take age = 2 4 th year: 3 - 4 ( age ), we take age = 3 5 th year: 4 - 5 ( age ), we take age = 4 6 th year: 5 - 6 ( age ), we take age = 5 7 th year: 6 - 7 ( age ), we take age = 6 8 th year: 7 - 8 ( age ), we take age = 7.\nThus for the 3 rd year = 50, 4 th year = 50 - 3 = 47 5 th year = 47 - 4 = 43 6 th year = 43 - 5 = 38."]

['User: The ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17. if ram has rs. 735, krishan has ?. Can you show me your work?\nAssistant: Ram: gopal = 7: 17 = 49: 119 gopal: krishan = 7: 17 = 119: 289 ram: gopal: krishan = 49: 119: 289 ram: krishan = 49: 289.\nThus, 49: 289 = 735: n & there n = 289 x 735/49 = rs.\n4335.', "User: The ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17. if ram has rs. 637, krishan has ?. Let's think step by step.\nAssistant: Ram: gopal = 7: 17 = 49: 119 gopal: krishan = 7: 17 = 119: 289 ram: gopal: krishan = 49: 119: 289 ram: krishan = 49: 289.\nThus, 49: 289 = 637: n & there n = 289 x 637/49 = rs.\n3757.", 'User: The ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17. if ram has rs. 539, krishan has ?\nAssistant: 3179.\nUser: Can you solve this step by step?\\Assistant: Sure.\nRam: gopal = 7: 17 = 49: 119 gopal: krishan = 7: 17 = 119: 289 ram: gopal: krishan = 49: 119: 289 ram: krishan = 49: 289.\nThus, 49: 289 = 539: n & there n = 289 x 539/49 = rs.\n3179.']

["User: 20 - 8 percent of the programmers in a startup company weigh 200 pounds or more. 20 - 5 percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less. what percent of the programmers in the startup company weigh between 100 and 200 pounds ?\nAssistant: , the required % is [ ( 3/4 ) ∗ 72/100 ] ∗ 100 = 54 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nInitially 72 % and 28 % split 80 % is further divided as 25 % and 75 % q is asking about that 75 %.\nLet total be'100'then that 75 % is ( 3/4 ) ∗ 72.\nSo, the required % is [ ( 3/4 ) ∗ 72/100 ] ∗ 100 = 54 %.", "User: 20 percent of the programmers in a startup company weigh 200 pounds or more. 20 - 5 percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less. what percent of the programmers in the startup company weigh between 100 and 200 pounds ?\nAssistant: , the required % is [ ( 3/4 ) ∗ 80/100 ] ∗ 100 = 60 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nInitially 80 % and 20 % split 80 % is further divided as 25 % and 75 % q is asking about that 75 %.\nLet total be'100'then that 75 % is ( 3/4 ) ∗ 80.\nSo, the required % is [ ( 3/4 ) ∗ 80/100 ] ∗ 100 = 60 %.", "User: Sixteen percent of the programmers in a startup company weigh 200 pounds or more. 20 - 5 percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less. what percent of the programmers in the startup company weigh between 100 and 200 pounds ?. Can you show me your work?\nAssistant: Initially 84 % and 16 % split 80 % is further divided as 25 % and 75 % q is asking about that 75 %.\nLet total be'100'then that 75 % is ( 3/4 ) ∗ 84.\nSo, the required % is [ ( 3/4 ) ∗ 84/100 ] ∗ 100 = 63 %."]

['User: If a square mirror has a 20 - inch diagonal, what is the approximate perimeter w of the mirror, in inches ?\nAssistant: We can eliminate.\nUser: Can you solve this step by step?\\Assistant: Sure.\nIf you draw the square and diagonal inside the square u can see square becomes part of two triangles opposite to each other.\nAnd we know the property of the triangle, addition of two sides of triangle must be greater than its diagonal in order to comp.\nLete the triangle.\nAnd each side must be less than 20 and perimeter w must be less than 80.\nSo we can eliminate.', 'User: If a square mirror has a 20 - inch diagonal, what is the approximate perimeter r of the mirror, in inches ?\nAssistant: We can eliminate.\nUser: Can you solve this step by step?\\Assistant: Sure.\nIf you draw the square and diagonal inside the square u can see square becomes part of two triangles opposite to each other.\nAnd we know the property of the triangle, addition of two sides of triangle must be greater than its diagonal in order to comp.\nLete the triangle.\nAnd each side must be less than 20 and perimeter r must be less than 80.\nSo we can eliminate.', "User: The size of a flat - screen television is given as the length of the screen's diagonal. how many square inches greater is the screen of a square 24 - inch flat - screen television than a square 17 - inch flat - screen television ?\nAssistant: E = 143.5 a.\nUser: Can you solve this step by step?\\Assistant: Sure.\nIf we take a square with side length x and draw a diagonal, we get two i.\nSosceles right triangles.\nIf we focus on one such right triangle, we see that the legs have length x.\nSquare 24 - inch flat - screen television the diagonal ( hypotenuse ) = 24.\nSo, we can apply the pythagorean theorem to get x ² + x ² = 24 ² simplify: 2 x ² = 24 ² divide both sides by 2 to get: x ² = 24 ²/2 since the area of the square = x ², we can see that the area of this square is 24 ²/2 square 17 - inch flat - screen television the diagonal ( hypotenuse ) = 17.\nSo, we can apply the pythagorean theorem to get x ² + x ² = 17 ² simplify: 2 x ² = 17 ² divide both sides by 2 to get: x ² = 17 ²/2 since the area of the square = x ², we can see that the area of this square is 17 ²/2 difference in areas = 24 ²/2 - 17 ²/2 = 288 - 144.5 i.\ne = 143.5 a."]

['User: A doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds. if the typical dosage is 2 cubic centimeters per 10 pounds of the body weight, by what percent was the prescribed dosage lesser than the typical dosage ?. Can you show me your work?\nAssistant: Typical dosage per 10 pound of the body weight = 2 c.\nc typical dosage per 120 pound of the body weight = 2 * ( 120/10 ) = 2 * 12 = 24 c.\nc dosage prescribed by doctor for 120 pound patient = 18 c.\nc % prescribed dosage greater than the typical dosage = ( 18 - 24/16 ) * 100 % = ( - 6/16 ) * 100 % = - 37.5 %.', 'User: A dessert recipe calls for 50 % melted chocolate and 50 % raspberry puree to make a particular sauce. a chef accidentally makes 15 cups of the sauce with 40 % melted chocolate and 60 % raspberry puree instead. how many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50 % of each ?\nAssistant: Lving x = 2.5.\nUser: Can you solve this step by step?\\Assistant: Sure.\nWe have 15 cups of sauce with 4040 choc and 6060 rasb cups of choc = 0.4 ∗ 15 = 6 cups of rasb = 0.6 ∗ 15 = 9\nLet say we removed x cup of original mix and replaced with x cups of choc.\nTherefore final number of cups of choc = 6 − 0.4 x + x now this number of cup should be 50 % of total = 15/2 = 7.5.\nTherefore 6 − 0.4 x + x = 7.5 on.\nSolving x = 2.5.', 'User: A dessert recipe calls for 50 % melted chocolate and 50 % raspberry puree to make a particular sauce. a chef accidentally makes 15 cups of the sauce with 30 % melted chocolate and 70 % raspberry puree instead. how many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50 % of each ?\nAssistant: We need to mix 5 * 15/7 = 12.5 cups of 40 % chocolate sauce with 1 * 15/7 = 2.14 cups of pure chocolate sauce.\nUser: Can you solve this step by step?\\Assistant: Sure.\nYes, we assume that the mix is homogeneous.\nOtherwise, we will not be able to solve the question.\nYou have 15 cups of sauce with 30 % chocolate.\nYou\nAlso have unlimited amount of pure chocolate sauce.\nNow you need to mix these two in such a way that you get total 15 cups of sauce with 50 % chocolate.\nUsing scale method: w 1/w 2 = ( 100 - 50 )/( 50 - 30 ) = 5/2 w 1 - amount of 30 % chocolate sauce w 2 - amount of pure chocolate sauce.\nSo for every 5/2 cups of 30 % chocolate sauce, we need 2 cup of pure chocolate sauce.\nThis will give us 7 cups of 50 % chocolate sauce.\nBut we need 15 cups of 50 % chocolate sauce.\nSo we need to mix 5 * 15/7 = 12.5 cups of 40 % chocolate sauce with 1 * 15/7 = 2.14 cups of pure chocolate sauce.']

["User: The length of a room is 5.5 m and width is 3.75 m. what is the cost of paying the floor by slabs at the rate of $ 400 per sq. metre.. Let's think step by step.\nAssistant: Area = 5.5 × 3.75 sq metre.\nCost for 1 sq metre.\n= $ 400.\nHence, total cost = 5.5 × 3.75 × 400 = $ 8250 a.", 'User: The lenght of a room is 5.5 m and width is 4 m. find the cost of paving the floor by slabs at the rate of rs. 750 per sq. metre.. And explain please.\nAssistant: Area of the floor = ( 5.5 ã — 4 ) m 2 = 22 m 2.\nCost of paving = rs.\n( 750 ã — 22 ) = rs.\n16500.', "User: The length of a room is 5.5 m and width is 3.75 m. find the cost of paving the floor by slabs at the rate of rs. 400 per sq. metre.. Let's think step by step.\nAssistant: Area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2.\n∴ cost of paving = rs.\n( 400 × 20.625 ) = 8250."]

["User: Because he's taxed by his home planet, mork pays a tax rate of 10 % on his income, while mindy pays a rate of 20 % on hers. if mindy earned 3 times as much as mork did, what was their combined tax rate ?\nAssistant: Combined tax % will be 70/100 + 300 = 15.5 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSay morks income is - 100.\nSo tax paid will be 10 say mindys income is 3 * 100 = 300.\nSo tax paid is 20 % * 300 = 60 total tax paid = 10 + 60 = 70.\nCombined tax % will be 70/100 + 300 = 15.5 %.", 'User: Farm tax is levied on the 50 % of the cultivated land. the tax department collected total $ 3840 through the farm tax from the village of mr. william. mr. william paid only $ 480 as farm tax. the percentage of total land of mr. willam over the total taxable land of the village is :. Help me solve this step by step.\nAssistant: This will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village.\nThat leads to ( 480/3840 ) x 100 = 12.5 % in percentage terms.\nBut the question asks ratio between his total land to total cultivated land.', "User: Farm tax is levied on the 50 % of the cultivated land. the tax department collected total $ 3840 through the farm tax from the village of mr. willam. mr. willam paid only $ 480 as farm tax. the percentage of total land of mr. willam over the total taxable land of the village is :.\nAssistant: But the question asks ratio between his total land to total cultivated land.\nUser: Can you solve this step by step?\\Assistant: Sure.\nOnly trick n this question is to ignore 50 % information as farm tax is levied uniformly in the village and that includes mr william's land what percentage of tax mr william paid ? this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village.\nThat leads to ( 480/3840 ) x 100 = 12.5 % in percentage terms.\nBut the question asks ratio between his total land to total cultivated land."]

['User: A particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 %. during the new year season, it further marked up its prices by 25 % of the original retail price. in february, the store then offered a discount of 18 %. what was its profit on the items sold in february ?\nAssistant: Total profit = 23 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nAssume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.82 * 150 x = 123 x.\nHence total profit = 23 %.', 'User: If xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers, how much will it cost to buy 45000 sheets of xerox paper ?. And explain please.\nAssistant: 30 sec approach.\nSolve it using approximation 45000 sheet at full price, 5 cent = 2250 45000 sheet at max discount price, 4 cent = 2000 your.', 'User: If xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers, how much will it cost to buy 15000 sheets of xerox paper ?. Help me solve this step by step.\nAssistant: 30 sec approach.\nSolve it using approximation 15000 sheet at full price, 5 cent = 750 15000 sheet at max discount price, 4 cent = 600 your.']

['User: If the sides of a triangle are 26 cm, 24 cm and 12 cm, what is its area ?. Help me solve this step by step.\nAssistant: The triangle with sides 26 cm, 24 cm and 12 cm is right angled, where the hypotenuse is 26 cm.\nArea of the triangle = 1/2 * 24 * 12 = 144 cm 2.', 'User: If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area ?\nAssistant: Area of the triangle = 1/2 * 24 * 10 = 120 cm 2.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThe triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.\nArea of the triangle = 1/2 * 24 * 10 = 120 cm 2.', 'User: If the sides of a triangle are 26 cm, 18 cm and 10 cm, what is its area ?\nAssistant: Area of the triangle = 1/2 * 18 * 10 = 90 cm 2.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThe triangle with sides 26 cm, 18 cm and 10 cm is right angled, where the hypotenuse is 26 cm.\nArea of the triangle = 1/2 * 18 * 10 = 90 cm 2.']

['User: Of all the homes on gotham street, 1 / 3 are termite - ridden, and 2 / 5 of these are collapsing. what fraction of the homes are termite - ridden, but not collapsing ?\nAssistant: Required ratio = 3/15 = 1/5.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet total homes be 15 termite ridden = 1/3 ( 15 ) = 5 termite ridden and collapsing = 2/5 ( 5 ) = 2.\nThus homes that are termite ridden, but not collapsing = 5 - 2 = 3.\nThus required ratio = 3/15 = 1/5.', "User: Of all the homes on gotham street, 1 / 3 are termite - ridden, and 1 / 4 of these are collapsing. what fraction of the homes are termite - ridden, but not collapsing ? Let's think step by step.\nAssistant: The fraction of homes which are termite - infested but not collapsing is 3/4 * 1/3 = 1/4.", "User: Of all the homes on gotham street, 1 / 3 are termite - ridden, and 7 / 10 of these are collapsing. what fraction of the homes are termite - ridden, but not collapsing ? Let's think step by step.\nAssistant: The fraction of homes which are termite - infested but not collapsing is 3/10 * 1/3 = 1/10."]

['User: By selling an article at rs. 800, a shopkeeper makes a profit of 25 %. at what price should he sell the article so as to make a loss of 20 % ?\nAssistant: 512.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSp = 800 profit = 25 % cp = ( sp ) * [ 100/( 100 + p ) ] = 800 * [ 100/125 ] = 640 loss = 25 % = 25 % of 640 = rs.\n128 sp = cp - loss = 640 - 128 = rs.\n512.', 'User: By selling an article at rs. 1000, a shopkeeper makes a profit of 25 %. at what price should he sell the article so as to make a loss of 25 % ?\nAssistant: 600.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSp = 1000 profit = 25 % cp = ( sp ) * [ 100/( 100 + p ) ] = 1000 * [ 100/125 ] = 800 loss = 25 % = 25 % of 800 = rs.\n200 sp = cp - loss = 800 - 200 = rs.\n600.', 'User: By selling an article at rs. 800, a shopkeeper makes a profit of 25 %. at what price should he sell the article so as to make a loss of 25 % ?\nAssistant: 480.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSp = 800 profit = 25 % cp = ( sp ) * [ 100/( 100 + p ) ] = 800 * [ 100/125 ] = 640 loss = 25 % = 25 % of 640 = rs.\n160 sp = cp - loss = 640 - 160 = rs.\n480.']

['User: We have boxes colored either red or blue. in each blue box there is a fixed number of blueberries. in each red box there is a fixed number of strawberries. if we dispose of one blue box for one additional red box, the total number of berries would increase by 10, and the difference between the total number of strawberries and the total number of blueberries would increase by 50. each blue box contains how many blueberries ?\nAssistant: X + ( x + 10 ) = 50 x = 20.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet x be the number of blueberries in each blue box.\nThen there are x + 10 strawberries in each red box.\nx + ( x + 10 ) = 50 x = 20.', 'User: We have boxes colored either red or blue. in each blue box there is a fixed number of blueberries. in each red box there is a fixed number of strawberries. if we dispose of one blue box for one additional red box, the total number of berries would increase by 20, and the difference between the total number of strawberries and the total number of blueberries would increase by 80. each blue box contains how many blueberries ?. Help me solve this step by step.\nAssistant: Let x be the number of blueberries in each blue box.\nThen there are x + 20 strawberries in each red box.\nx + ( x + 20 ) = 80 x = 30.', 'User: We have boxes colored either red or blue. in each blue box there is a fixed number of blueberries. in each red box there is a fixed number of strawberries. if we dispose of one blue box for one additional red box, the total number of berries would increase by 15, and the difference between the total number of strawberries and the total number of blueberries would increase by 87. each blue box contains how many blueberries ?. Can you show me your work?\nAssistant: Let x be the number of blueberries in each blue box.\nThen there are x + 15 strawberries in each red box.\nx + ( x + 15 ) = 87 x = 36.']

['User: What two - digit number is less than the sum of the square of its digits by 12 and exceeds their doubled product by 5 ?\nAssistant: The two numbers can be 15 or 99.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the digits be x and y.\nThe number would be 10 x + y.\nWe are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 12.\nThus 2 xy + 5 = x ^ 2 + y ^ 2 - 12 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9.\nThus the two numbers can be 15 or 99.', "User: What two - digit number is less than the sum of the square of its digits by 13 and exceeds their doubled product by 5 ?. Let's think step by step.\nAssistant: Let the digits be x and y.\nThe number would be 10 x + y.\nWe are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 13.\nThus 2 xy + 5 = x ^ 2 + y ^ 2 - 13 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9.\nThus the two numbers can be 15 or 99.", 'User: What two - digit number is less than the sum of the square of its digits by 15 and exceeds their doubled product by 5 ?\nAssistant: The two numbers can be 15 or 73.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the digits be x and y.\nThe number would be 10 x + y.\nWe are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 15.\nThus 2 xy + 5 = x ^ 2 + y ^ 2 - 15 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9.\nThus the two numbers can be 15 or 73.']

["User: Country x imported approximately $ 1.44 billion of goods in 1996. if country x imported $ 488 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year, by how much would country xs 1997 imports exceed those of 1996 ? Let's think step by step.\nAssistant: Convert units to millions as.", "User: 8 % people of a village in sri lanka died by bombardment, 15 % of the remainder left the village on account of fear. if now the population is reduced to 3553, how much was it in the beginning ? Let's think step by step.\nAssistant: X * ( 92/100 ) * ( 85/100 ) = 3553 x = 4543.", "User: 5 % people of a village in sri lanka died by bombardment, 15 % of the remainder left the village on account of fear. if now the population is reduced to 3553, how much was it in the beginning ? Let's think step by step.\nAssistant: X * ( 95/100 ) * ( 85/100 ) = 3553 x = 4400."]

["User: Of the female students at barkely university, 5 / 6 are on the honor roll. of the male students, 2 / 3 are on the honor roll. if 2 / 5 of the students are female, what fraction of all the students are on the honor roll ? Let's think step by step.\nAssistant: Let the total students be 30 given 2/5 of the students are females = 12 then males = 3/5 = 18 5/6 of the females are on honor roll = 10 males on the honor roll = 2/3 = 12 total students on honor roll = 12 + 10 = 22 fraction = 22/30 d.", "User: Of the female students at barkely university, 3 / 5 are on the honor roll. of the male students, 2 / 5 are on the honor roll. if 5 / 9 of the students are female, what fraction of all the students are on the honor roll ? Let's think step by step.\nAssistant: Let the total students be 45 given 5/9 of the students are females = 25 then males = 4/9 = 20 3/5 of the females are on honor roll = 15 males on the honor roll = 2/5 = 8 total students on honor roll = 15 + 8 = 23 fraction = 23/45 c.", "User: Of the female students at barkely university, 1 / 6 are on the honor roll. of the male students, 2 / 5 are on the honor roll. if 3 / 5 of the students are female, what fraction of all the students are on the honor roll ? Let's think step by step.\nAssistant: Let the total students be 100 given 3/5 of the students are females = 60 then males = 2/5 = 40 1/6 of the females are on honor roll = 10 males on the honor roll = 2/5 = 16 total students on honor roll = 10 + 16 = 26 fraction = 26/100 c."]

['User: In the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30. in the sport formulation, the ratio of flavoring to corn syrup is three times as great as in the standard formulation, and the ratio of flavoring to water is half that of the standard formulation. if a large bottle of thesportformulation contains 5 ounces of corn syrup, how many ounces of water does it contain ?. Help me solve this step by step.\nAssistant: Standard: fl: corn s: water = 1: 12: 30 sport: fl: corn s: water = 3: 12: 180 this simplifies to 1: 4: 60 if the large bottle has a capacity of x ounces, then 4 x/65 = 5.\nSo, x = 325/4 ounces.\nWater = ( 60/65 ) * ( 325/4 ) = 60 * 3/4 = 75 ounces.', 'User: In the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30. in the sport formulation, the ratio of flavoring to corn syrup is three times as great as in the standard formulation, and the ratio of flavoring to water is half that of the standard formulation. if a large bottle of the sport formulation contains 1 ounces of corn syrup, how many ounces of water does it contain ?\nAssistant: Water = ( 60/65 ) * ( 65/4 ) = 15 ounces.\nUser: Can you solve this step by step?\\Assistant: Sure.\nStandard: fl: corn s: water = 1: 12: 30 sport: fl: corn s: water = 3: 12: 180 this simplifies to 1: 4: 60 if the large bottle has a capacity of x ounces, then 4 x/65 = 1.\nSo, x = 65/4 ounces.\nWater = ( 60/65 ) * ( 65/4 ) = 15 ounces.', 'User: In the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30. in the sport formulation, the ratio of flavoring to corn syrup is 3 times as great as in the standard formulation, and the ratio of flavoring to water is half that of the standard formulation. if a large bottle of the sport formulation contains 2 ounces of corn syrup, how many ounces of water does it contain ?\nAssistant: Water = ( 60/65 ) * ( 65/2 ) = = 30 ounces.\nUser: Can you solve this step by step?\\Assistant: Sure.\nStandard: fl: corn s: water = 1: 12: 30 sport: fl: corn s: water = 3: 12: 180 this simplifies to 1: 4: 60 if the large bottle has a capacity of x ounces, then 4 x/65 = 2.\nSo, x = 32.5 ounces.\nWater = ( 60/65 ) * ( 65/2 ) = = 30 ounces.']

['User: An empty fuel tank with a capacity of 204 gallons was filled partially with fuel a and then to capacity with fuel b. fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume. if the full fuel tank contains 30 gallons of ethanol, how many gallons of fuel a were added ?. Help me solve this step by step.\nAssistant: Say there are a gallons of fuel a in the tank, then there would be 204 - a gallons of fuel b.\nThe amount of ethanol in a gallons of fuel a is 0.12 a.\nThe amount of ethanol in 204 - a gallons of fuel b is 0.16 ( 204 - a ).\nSince the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 204 - a ) = 30.\na = 66.', 'User: An empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b. fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume. if the full fuel tank contains 22 gallons of ethanol, how many gallons of fuel a were added ?\nAssistant: A = 250.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSay there are a gallons of fuel a in the tank, then there would be 200 - a gallons of fuel b.\nThe amount of ethanol in a gallons of fuel a is 0.12 a.\nThe amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ).\nSince the total amount of ethanol is 22 gallons then 0.12 a + 0.16 ( 200 - a ) = 22.\na = 250.', 'User: An empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b. fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume. if the full fuel tank contains 20 gallons of ethanol, how many gallons of fuel a were added ?\nAssistant: A = 300.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSay there are a gallons of fuel a in the tank, then there would be 200 - a gallons of fuel b.\nThe amount of ethanol in a gallons of fuel a is 0.12 a.\nThe amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ).\nSince the total amount of ethanol is 20 gallons then 0.12 a + 0.16 ( 200 - a ) = 20.\na = 300.']

['User: At veridux corporation, there are 180 employees. of these, 90 are female, and the rest are males. there are a total of 40 managers, and the rest of the employees are associates. if there are a total of 135 male associates, how many female managers are there ?\nAssistant: Now, in the â € œ associate â € \x9d row, 135 + e = 140, which me.\nUser: Can you solve this step by step?\\Assistant: Sure.\nWell, first.\nLet â € ™ s take care of the â € œ totals â € \x9d.\nThe numbers in the â € œ totals â € \x9d row must add up.\nIf 90 are females, the other 180 â € “ 90 = 90 must be males.\nSimilarly, the numbers in the â € œ totals â € \x9d column must add up.\nIf 40 are managers, then the other 180 â € “ 40 = 140 must be associates.\nNow, in the â € œ associate â € \x9d row, 135 + e = 140, which me.', 'User: Find the probability that a leap year selected at random will have 53 sundays.\nAssistant: The probability of that is 2/7.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThere are 366 days in a leap year: 52 weeks and 2 more days.\nSo, 52 sundays and 2 days.\nThese 2 days can be: { mon, tue }, { tue, wed }, { wed, thu }, { thu, fri }, { fri, sat }, { sat, sun } and { sun, mon } ( 7 cases ).\nIn order to have 53 sundays we should have either { sat, sun } or { sun, mon } case.\nThe probability of that is 2/7.', 'User: Find the probability that a leap year selected at random will have 53 mondays.\nAssistant: The probability of that is 2/7.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThere are 366 days in a leap year: 52 weeks and 2 more days.\nSo, 52 mondays and 2 days.\nThese 2 days can be: { mon, tue }, { tue, wed }, { wed, thu }, { thu, fri }, { fri, sat }, { sat, sun } and { sun, mon } ( 7 cases ).\nIn order to have 53 mondays we should have either { mon, tuesday } or { sun, mon } case.\nThe probability of that is 2/7.']

["User: Working alone at its constant rate, machine a produces x boxes in 10 minutes and working alone at its constant rate, machine b produces 2 x boxes in 5 minutes. how many minutes does it take machines a and b, working simultaneously at their respective constant rates, to produce 6 x boxes ?\nAssistant: Let's first find the combined rate of machine a and b rate of machine a = x/10 min + rate of machine b = 4 x/10 = 5 x/10 now combine time = combine work needs to be done/combine rate = 6 x/5 x * 10 = 12 min.\nUser: Can you solve this step by step?\\Assistant: Sure.\nRate = work/time given rate of machine a = x/10 min machine b produces 2 x boxes in 5 min.\nHence, machine b produces 4 x boxes in 10 min.\nRate of machine b = 4 x/10 we need tofind the combined time that machines a and b, working simultaneouslytakeat their respective constant rates.\nLet's first find the combined rate of machine a and b rate of machine a = x/10 min + rate of machine b = 4 x/10 = 5 x/10 now combine time = combine work needs to be done/combine rate = 6 x/5 x * 10 = 12 min.", "User: Working alone at its constant rate, machine a produces x boxes in 10 minutes and working alone at its constant rate, machine b produces 2 x boxes in 5 minutes. how many minutes does it take machines a and b, working simultaneously at their respective constant rates, to produce 10 x boxes ?. And explain please.\nAssistant: Rate = work/time given rate of machine a = x/10 min machine b produces 2 x boxes in 5 min.\nHence, machine b produces 4 x boxes in 10 min.\nRate of machine b = 4 x/10 we need tofind the combined time that machines a and b, working simultaneouslytakeat their respective constant rates.\nLet's first find the combined rate of machine a and b rate of machine a = x/10 min + rate of machine b = 4 x/10 = 5 x/10 now combine time = combine work needs to be done/combine rate = 10 x/5 x * 10 = 6 min.", "User: Working alone at its constant rate, machine a produces x boxes in 10 minutes and working alone at its constant rate, machine b produces 3 x boxes in 5 minutes. how many minutes does it take machines a and b, working simultaneously at their respective constant rates, to produce 3 x boxes ?. And explain please.\nAssistant: Rate = work/time given rate of machine a = x/10 min machine b produces 3 x boxes in 5 min.\nHence, machine b produces 4 x boxes in 10 min.\nRate of machine b = 6 x/10 we need tofind the combined time that machines a and b, working simultaneouslytakeat their respective constant rates.\nLet's first find the combined rate of machine a and b rate of machine a = x/10 min + rate of machine b = 6 x/10 = 7 x/10 now combine time = combine work needs to be done/combine rate = 3 x/7 x * 10 = 4.2 min."]

['User: The average expenditure of a labourer for 6 months was 85 and he fell into debt. in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30. his monthly income i. Can you show me your work?\nAssistant: Income of 6 months = ( 6 × 85 ) – debt = 510 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt.\n∴ income of 10 months = 780 average monthly income = 780 ÷ 10 = 78.', "User: The average monthly salary of 8 workers and one supervisor in a factory was 430. @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430. @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430. whenthesupervisor, whosesalarywas 430. when the supervisor, whose salary was 870 per month, retired, a new person was appointed and then the average salary of 9 people was $ $ 420 per month. the salary of the new supervisor is :. And explain please.\nAssistant: Explanation: total salary of 8 workers and supervi.\nSor together = 9 ã — 430 = 3870 now total salary of 8 workers = 3870 â ˆ'870 = 3000 total salary of 9 workers including the new supervi.\nSor = 9 ã — 420 = 3780 salary of the new supervi.\nSor = 3780 â ˆ'3000 = 780.", "User: The average monthly salary of 8 workers and one supervisor in a factory was 430. @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430. @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430. whenthesupervisor, whosesalarywas 430. when the supervisor, whose salary was 870 per month, retired, a new person was appointed and then the average salary of 9 people was $ $ 430 per month. the salary of the new supervisor is :. And explain please.\nAssistant: Explanation: total salary of 8 workers and supervi.\nSor together = 9 ã — 430 = 3870 now total salary of 8 workers = 3870 â ˆ'870 = 3000 total salary of 9 workers including the new supervi.\nSor = 9 ã — 430 = 3870 salary of the new supervi.\nSor = 3870 â ˆ'3000 = 870."]

['User: The average weight of a group of boys is 35 kg. after a boy of weight 40 kg joins the group, the average weight of the group goes up by 1 kg. find the number of boys in the group originally ?. Can you show me your work?\nAssistant: Let the number off boys in the group originally be x.\nTotal weight of the boys = 35 x after the boy weighing 40 kg joins the group, total weight of boys = 35 x + 40.\nSo 35 x + 40 = 36 ( x + 1 ).\nx = 5.', 'User: The average weight of a group of boys is 30 kg. after a boy of weight 40 kg joins the group, the average weight of the group goes up by 1 kg. find the number of boys in the group originally ?. And explain please.\nAssistant: Let the number off boys in the group originally be x.\nTotal weight of the boys = 30 x after the boy weighing 40 kg joins the group, total weight of boys = 30 x + 40.\nSo 30 x + 40 = 31 ( x + 1 ).\nx = 9.', 'User: The average weight of a group of boys is 30 kg. after a boy of weight 35 kg joins the group, the average weight of the group goes up by 1 kg. find the number of boys in the group originally ?\nAssistant: X = 4.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the number off boys in the group originally be x.\nTotal weight of the boys = 30 x after the boy weighing 35 kg joins the group, total weight of boys = 30 x + 35.\nSo 30 x + 35 + 31 ( x + 1 ).\nx = 4.']

['User: If x is real, find the maximum value of the expression - 2 x ^ 2 + 11 x + 13.\nAssistant: X = 2.75 the maximum value is - 2 ( 2.75 ) ^ 2 + 11 ( 2.75 ) + 13 = 28.125.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThis is an equation of a downward facing parabola.\nThe maximum value is the top point of the parabola.\n- 2 x ^ 2 + 11 x + 13 = ( - 2 x + 13 ) ( x + 1 ) the roots are 13/2 and - 1.\nThe maximum value must be when x is halfway between these two points.\nx = 2.75 the maximum value is - 2 ( 2.75 ) ^ 2 + 11 ( 2.75 ) + 13 = 28.125.', 'User: Y = x ^ 2 + bx + 64 cuts the x axis at ( h, 0 ) and ( k, 0 ). if h and k are integers, what is the least value of b ?. Help me solve this step by step.\nAssistant: As the curve cuts the x - axis at ( h, 0 ) and ( k, 0 ).\nTherefore h, k are the roots of the quadratic equation.\nFor the quadratic equation is in the form ofax ^ 2 + bx + c = 0, the product of the roots = c/a = 64/1 = 256 and the sum of the roots = - b/a = - b 64 can be expressed as product of two numbers in the following ways: 1 * 64 2 * 32 4 * 16 the sum of the roots is maximum when the roots are 1 and 64 and the maximum sum is 1 + 64 = 65.\nThe least value possible for b is therefore - 65.', "User: Y = x ^ 2 + bx + 512 cuts the x axis at ( h, 0 ) and ( k, 0 ). if h and k are integers, what is the least value of b ?. Let's think step by step.\nAssistant: As the curve cuts the x - axis at ( h, 0 ) and ( k, 0 ).\nTherefore h, k are the roots of the quadratic equation.\nFor the quadratic equation is in the form of ax ^ 2 + bx + c = 0, the product of the roots = c/a = 512/1 = 256 and the sum of the roots = - b/a = - b 512 can be expressed as product of two numbers in the following ways: 1 * 512 2 * 256 4 * 128 8 * 64 16 * 32 the sum of the roots is maximum when the roots are 1 and 256 and the maximum sum is 1 + 512 = 513.\nThe least value possible for b is therefore - 513."]

["User: Fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 20 a pair. during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans. if the sum of the two discounts rates is 22 percent, what is the discount rate on pony jeans ? Let's think step by step.\nAssistant: X discount on pony je.", "User: Fox jeans regularly sell for $ 10 a pair and pony jeans regularly sell for $ 18 a pair. during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans. if the sum of the two discounts rates is 22 percent, what is the discount rate on pony jeans ? Let's think step by step.\nAssistant: X discount on pony je.", "User: Fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair. during a sale these regular unit prices are discounted at different rates so that a total of $ 9 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans. if the sum of the two discounts rates is 33 percent, what is the discount rate on pony jeans ? Let's think step by step.\nAssistant: X discount on pony je."]

["User: A luxury liner, queen marry ii, is transporting several cats as well as the crew ( sailors, a cook, and captain ) to a nearby port. altogether, these passengers have 15 heads and 42 legs. how many cats does the ship host ?\nAssistant: Sa's ( 2 ) + 2 + 2 + cats * 4 = 42 sa's * 2 + cats * 4 = 38 or sa's + cats * 2 = 19 or 13 - cats + cat * 2 = 19 then cats = 6 b.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSa's + co + ca + cats = 15.\nSa's + 1 + 1 + cats = 15 or sa's + cats = 13.\nSa's ( 2 ) + 2 + 2 + cats * 4 = 42 sa's * 2 + cats * 4 = 38 or sa's + cats * 2 = 19 or 13 - cats + cat * 2 = 19 then cats = 6 b.", "User: A luxury liner, queen marry ii, is transporting several cats as well as the crew ( sailors, a cook, and one - legged captain ) to a nearby port. altogether, these passengers have 16 heads and 45 legs. how many cats does the ship host ?\nAssistant: Sa's ( 2 ) + 2 + 1 + cats * 4 = 45 sa's * 2 + cats * 4 = 42 or sa's + cats * 2 = 21 or 14 - cats + cat * 2 = 21 then cats = 7 e.\nUser: Can you solve this step by step?\\Assistant: Sure.\nA's + co + ca + cats = 16.\nSa's + 1 + 1 + cats = 16 or sa's + cats = 14.\nSa's ( 2 ) + 2 + 1 + cats * 4 = 45 sa's * 2 + cats * 4 = 42 or sa's + cats * 2 = 21 or 14 - cats + cat * 2 = 21 then cats = 7 e.", "User: A luxury liner, queen marry ii, is transporting several cats as well as the crew ( sailors, a cook, and one - legged captain ) to a nearby port. altogether, these passengers have 16 heads and 41 legs. how many cats does the ship host ?\nAssistant: Sa's ( 2 ) + 2 + 1 + cats * 4 = 41 sa's * 2 + cats * 4 = 38 or sa's + cats * 2 = 19 or 14 - cats + cat * 2 = 19 then cats = 5 a.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSa's + co + ca + cats = 16.\nSa's + 1 + 1 + cats = 16 or sa's + cats = 14.\nSa's ( 2 ) + 2 + 1 + cats * 4 = 41 sa's * 2 + cats * 4 = 38 or sa's + cats * 2 = 19 or 14 - cats + cat * 2 = 19 then cats = 5 a."]

["User: At what price must an article costing rs. 47.50 be marked in order that after deducting 6 % from the list price. it may be sold at a profit of 25 % on the cost price ? Let's think step by step.\nAssistant: Cp = 47.50 sp = 47.50 * ( 125/100 ) = 59.375 mp * ( 94/100 ) = 59.375 mp = 63.16.", "User: At what price must an article costing rs. 47.50 be marked in order that after deducting 9 % from the list price. it may be sold at a profit of 25 % on the cost price ? Let's think step by step.\nAssistant: Cp = 47.50 sp = 47.50 * ( 125/100 ) = 59.375 mp * ( 91/100 ) = 59.375 mp = 65.25.", "User: At what price must an article costing rs. 47.50 be marked in order that after deducting 8 % from the list price. it may be sold at a profit of 25 % on the cost price ? Let's think step by step.\nAssistant: Cp = 47.50 sp = 47.50 * ( 125/100 ) = 59.375 mp * ( 92/100 ) = 59.375 mp = 64.54."]

['User: A man invested rs. 14400 in rs. 100 shares of a company at 20 % premium. if his company declares 5 % dividend at the end of the year, then how much does he get ?\nAssistant: 600.\nUser: Can you solve this step by step?\\Assistant: Sure.\nExplanation: number of shares = = 120.\nFace value = rs.\n( 100 x 120 ) = rs.\n12000.\nAnnual income = = rs.\n600.', "User: A certain company expects quarterly earnings of $ 0.80 per share of stock, half of which will be distributed as dividends to shareholders while the rest will be used for research and development. if earnings are greater than expected, shareholders will receive an additional $ 0.04 per share for each additional $ 0.10 of per share earnings. if quarterly earnings are $ 1.10 per share, what will be the dividend paid to a person who owns 400 shares of the company's stock ?. Can you show me your work?\nAssistant: Eps actual > eps expected.\nEach gets and additional.\n12 per share 52 * 400.\n$ 208.", "User: A certain company expects quarterly earnings of $ 0.80 per share of stock, half of which will be distributed as dividends to shareholders while the rest will be used for research and development. if earnings are greater than expected, shareholders will receive an additional $ 0.04 per share for each additional $ 0.10 of per share earnings. if quarterly earnings are $ 1.10 per share, what will be the dividend paid to a person who owns 600 shares of the company's stock ?\nAssistant: $ 312.\nUser: Can you solve this step by step?\\Assistant: Sure.\nEps actual > eps expected.\nEach gets and additional.\n12 per share 52 * 600.\n$ 312."]

["User: George baked a total of 50 pizzas for 5 straight days, beginning on saturday. he baked 3 / 5 of the pizzas the first day, and 3 / 5 of the remaining pizzas the second day. if each successive day he baked fewer pizzas than the previous day, what is the maximum number of pizzas he could have baked on wednesday ?. Can you show me your work?\nAssistant: 3/5 of the 50 pizzas cooked on saturday = 30 pizzas 3/5 of the remaining pizzas on sunday = 12 pizzas we're left with ( 50 - 30 - 12 ) = 8 pizzas for the remaining 3 days.\nThe prompt tells us that each day has fewer pizzas than the day before it.\nSo we ca n't have duplicate numbers.\nm t w 5 2 1 = 8 w = 1 e.", "User: George baked a total of 75 pizzas for 7 straight days, beginning on saturday. he baked 3 / 5 of the pizzas the first day, and 3 / 5 of the remaining pizzas the second day. if each successive day he baked fewer pizzas than the previous day, what is the maximum number of pizzas he could have baked on wednesday ?. Can you show me your work?\nAssistant: 3/5 of the 75 pizzas cooked on saturday = 45 pizzas 3/5 of the remaining pizzas on sunday = 18 pizzas we're left with ( 75 - 45 - 18 ) = 12 pizzas for the remaining 5 days.\nThe prompt tells us that each day has fewer pizzas than the day before it.\nSo we ca n't have duplicate numbers.\nm t w th f 5 4 2 1 0 = 12 w = 2 c.", "User: George baked a total of 150 pizzas for 7 straight days, beginning on saturday. he baked 3 / 5 of the pizzas the first day, and 3 / 5 of the remaining pizzas the second day. if each successive day he baked fewer pizzas than the previous day, what is the maximum number of pizzas he could have baked on wednesday ?. Can you show me your work?\nAssistant: 3/5 of the 150 pizzas cooked on saturday = 90 pizzas 3/5 of the remaining pizzas on sunday = 36 pizzas we're left with ( 150 - 90 - 36 ) = 24 pizzas for the remaining 5 days.\nThe prompt tells us that each day has fewer pizzas than the day before it.\nSo we ca n't have duplicate numbers.\nm t w th f 8 7 6 2 1 = 24 w = 6 b."]

['User: Obra drove 180 π meters along a circular track. if the area enclosed by the circular track on which she drove is 57,600 π square meters, what percentage of the circular track did obra drive ?\nAssistant: , part of circumference covered = 180 π/480 π = 37.5 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nArea enclosed by the circular track on which she drove is 57,600 π square meters.\nSo, π ( r ^ 2 ) = 57,600 π.\n( r ^ 2 ) = 57,600.\nr = 240 circumference of the circular track = 2 π r = 480 π.\nTherefore, part of circumference covered = 180 π/480 π = 37.5 %.', 'User: A dog is tied to a tree by a long nylon cord. if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items, and the dog ran approximately 30 feet, what was the approximate length of the nylon cord w, in feet ?. Help me solve this step by step.\nAssistant: Because the cord was extended to its full length at all items, the dog ran along a semi - circular path, from north to south.\nThe circumference of a full circle is 2 * pi * r, but since we only care about the length of half the circle, the semi - circle path is pi * r.\nw = pi * r = 30.\nRound pi = 3, then r = 10.\nChord is about 10 feet long.', 'User: A dog is tied to a tree by a long nylon cord. if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items, and the dog ran approximately 30 feet, what was the approximate length of the nylon cord e, in feet ?\nAssistant: Chord is about 10 feet long.\nUser: Can you solve this step by step?\\Assistant: Sure.\nBecause the cord was extended to its full length at all items, the dog ran along a semi - circular path, from north to south.\nThe circumference of a full circle is 2 * pi * r, but since we only care about the length of half the circle, the semi - circle path is pi * r.\ne = pi * r = 30.\nRound pi = 3, then r = 10.\nChord is about 10 feet long.']

['User: A reduction of 5 % in the price of oil enables a house wife to obtain 10 kgs more for rs. 5000, what is the reduced price for kg ?. And explain please.\nAssistant: 5000 * ( 5/100 ) = 250.\n- 10 ?\n- 1.\nRs.\n25.', 'User: A furniture store owner decided to drop the price of her recliners by 20 % to spur business. by the end of the week she had sold 60 % more recliners. what is the percentage increase of the gross ?\nAssistant: % increase = 2800 * 100/10000 = 28 % d is the.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSay a recliner is actually worth $ 100 if she sells 100 recliners then she earns $ 10000 after the discount of 20 %, she will earn $ 80 per recliner and she sells 60 % more ie.\n160 recliners.\nHence her sales tields 160 * 80 = $ 12800 increase in sales = 12800 - 10000 = $ 2800.\nSo % increase = 2800 * 100/10000 = 28 % d is the.', 'User: A furniture store owner decided to drop the price of her recliners by 20 % to spur business. by the end of the week she had sold 40 % more recliners. what is the percentage increase of the gross ?\nAssistant: % increase = 1120 * 100/10000 = 11.20 % c is the.\nUser: Can you solve this step by step?\\Assistant: Sure.\nSay a recliner is actually worth $ 100 if she sells 100 recliners then she earns $ 10000 after the discount of 20 %, she will earn $ 80 per recliner and she sells 40 % more ie.\n140 recliners.\nHence her sales tields 140 * 80 = $ 11200 increase in sales = 11200 - 10000 = $ 1120.\nSo % increase = 1120 * 100/10000 = 11.20 % c is the.']

['User: The avg weight of a, b & c is 84 kg. if d joins the group, the avg weight of the group becomes 80 kg. if another man e who weights is 6 kg more than d replaces a, then the avgof b, c, d & e becomes 79 kg. what is the weight of a ?\nAssistant: ( I ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 74 + 4 = 78 d.\nUser: Can you solve this step by step?\\Assistant: Sure.\nA + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320.\n- ( I ).\nSo, d = 68 & e = 68 + 6 = 74 b + c + d + e = 79 * 4 = 316.\n( ii ) from eq.\n( I ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 74 + 4 = 78 d.', 'User: The avg weight of a, b & c is 84 kg. if d joins the group, the avg weight of the group becomes 80 kg. if another man e who weights is 3 kg more than d replaces a, then the avgof b, c, d & e becomes 79 kg. what is the weight of a ?\nAssistant: ( I ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 71 + 4 = 75 c.\nUser: Can you solve this step by step?\\Assistant: Sure.\nA + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320.\n- ( I ).\nSo, d = 68 & e = 68 + 3 = 71 b + c + d + e = 79 * 4 = 316.\n( ii ) from eq.\n( I ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 71 + 4 = 75 c.', 'User: The avg weight of a, b & c is 84 kg. if d joins the group, the avg weight of the group becomes 80 kg. if another man e who weights is 5 kg more than d replaces a, then the avgof b, c, d & e becomes 79 kg. what is the weight of a ?. And explain please.\nAssistant: A + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320.\n- ( I ).\nSo, d = 68 & e = 68 + 5 = 73 b + c + d + e = 79 * 4 = 316.\n( ii ) from eq.\n( I ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 73 + 4 = 77 c.']

['User: A certain country is divided into 6 provinces. each province consists entirely of progressives and traditionalists. if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 9 the total number of progressives in the entire country, what fraction of the country is traditionalist ?\nAssistant: The total population is p + 2 p/3 = 5 p/3 p/( 5 p/3 ) = 3/5.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet p be the number of progressives in the country as a whole.\nIn each province, the number of traditionalists is p/9 the total number of traditionalists is 6 p/9 = 2 p/3.\nThe total population is p + 2 p/3 = 5 p/3 p/( 5 p/3 ) = 3/5.', 'User: A certain country is divided into 8 provinces. each province consists entirely of progressives and traditionalists. if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 12 the total number of progressives in the entire country, what fraction of the country is traditionalist ?\nAssistant: The total population is p + 2 p/3 = 5 p/3 p/( 5 p/3 ) = 3/5.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet p be the number of progressives in the country as a whole.\nIn each province, the number of traditionalists is p/12 the total number of traditionalists is 8 p/12 = 2 p/3.\nThe total population is p + 2 p/3 = 5 p/3 p/( 5 p/3 ) = 3/5.', 'User: At an international conference, “ red ” world countries and “ blue ” world countries are the only participants. the ratio of “ red ” world participants to “ blue ” world participants is 5 : 5. if one - third of “ red ” world participants are left - handed and two - thirds of “ blue ” world participants are left - handed, then what is the fraction of the participants who are left - handed ?\nAssistant: 2/3 * 5 x = 10 x/3 blue left handed fraction of participants who are left handed = total left handed/total participants = ( red left handed + blue left handed )/total participants = ( 5 x/3 + 10 x/3 )/( 5 x + 5 x ) = ( 15 x/30 x ) = 1/2.\nUser: Can you solve this step by step?\\Assistant: Sure.\nRed: blue = 5: 5.\nLet red = 5 x and blue = 5 x 1/3 of red are left handed.\n1/3 * 5 x = 5 x/3 red left handed 2/3 of blue are left handed.\n2/3 * 5 x = 10 x/3 blue left handed fraction of participants who are left handed = total left handed/total participants = ( red left handed + blue left handed )/total participants = ( 5 x/3 + 10 x/3 )/( 5 x + 5 x ) = ( 15 x/30 x ) = 1/2.']

['User: Seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 % fescue. if a mixture of x and y contains 30 % ryegrass, what percent of the weight of the mixture is x ?\nAssistant: The weight of mixture x as a percentage of the weight of the mixture = ( weight of x/weight of mixture ) * 100 % = ( 100/3 )/100 * 100 % = 33 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nAssuming the weight of the mixture to be 100 g * *, then the weight of ryegrass in the mixture would be 30 g.\nAlso, assume the weight mixture x used in the mixture is xg, then the weight of mixture y used in the mixture would be ( 100 - x ) g.\nSo we can now equate the parts of the ryegrass in the mixture as: 0.4 x + 0.25 ( 100 - x ) = 30 0.4 x + 25 - 0.25 x = 30 0.15 x = 5 x = 5/0.15 = 500/15 = 100/3.\nSo the weight of mixture x as a percentage of the weight of the mixture = ( weight of x/weight of mixture ) * 100 % = ( 100/3 )/100 * 100 % = 33 %.', "User: When processing flower - nectar into honey bees'extract, a considerable amount of water gets reduced. how much flower - nectar must be processed to yield 1 kg of honey, if nectar contains 50 % water, and the honey obtained from this nectar contains 30 % water ?\nAssistant: Amount of flower - nectar needed = ( 0.70/0.51 ) kg = 1.4 kgs.\nUser: Can you solve this step by step?\\Assistant: Sure.\nExplanation: flower - nectar contains 50 % of non - water part.\nIn honey this non - water part constitutes 70 % ( 100 - 30 ).\nTherefore 0.5 x amount of flower - nectar = 0.70 x amount of honey = 0.70 x 1 kg.\nTherefore amount of flower - nectar needed = ( 0.70/0.51 ) kg = 1.4 kgs.", "User: When processing flower - nectar into honey bees'extract, a considerable amount of water gets reduced. how much flower - nectar must be processed to yield 1 kg of honey, if nectar contains 50 % water, and the honey obtained from this nectar contains 25 % water ?. Help me solve this step by step.\nAssistant: Explanation: flower - nectar contains 50 % of non - water part.\nIn honey this non - water part constitutes 75 % ( 100 - 25 ).\nTherefore 0.5 x amount of flower - nectar = 0.75 x amount of honey = 0.75 x 1 kg.\nTherefore amount of flower - nectar needed = ( 0.75/0.51 ) kg = 1.5 kgs."]

['User: 60 percent of movie theatres in town x have 3 screens or less. 20 % of those theatres sell an average of more than $ 200 worth of popcorn per showing. 56 percent of all the movie theatres in town x sell $ 300 or less of popcorn per showing. what percent of all the stores on the street have 4 or more screens and sell an average of more than $ 300 worth of popcorn per day ?\nAssistant: 18 ( 44 - 12 ) must be those with four screens or more b is the.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLets take numbers here.\nAssume that the total number of movie theaters in the town = 100 then number of movie theaters with 3 screens or less = 60.\nNumber of movie theaters with 4 screens or more = 40 movie theaters with 3 screens or less selling popcorn at more than $ 200 = 20 % of 60 = 12 number of movie theaters selling popcorn at $ 300 or less = 56.\nNumber of movie theaters selling popcorn at more than $ 300 = 100 - 56 = 44 of these 44 theaters, 12 are those with 3 screens or less.\nTherefore 18 ( 44 - 12 ) must be those with four screens or more b is the.', "User: A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. these book are occasionally loaned out through an inter - library program. if, by the end of the month, 70 percent of books that were loaned out are returned and there are 63 books in the special collection at that time, how many books of the special collection were loaned out during that month ?. Let's think step by step.\nAssistant: I did n't understand how did we get 100 ? total = 75 books.\n65 % of books that were loaned out are returned.\n100 % - 70 % = 30 % of books that were loaned out are not returned.\nNow, there are 63 books.\nThus 76 - 63 = 12 books are not returned.\n{ loaned out } * 0.30 = 7.\n{ loaned out } = 40.", "User: A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. these book are occasionally loaned out through an inter - library program. if, by the end of the month, 70 percent of books that were loaned out are returned and there are 60 books in the special collection at that time, how many books of the special collection were loaned out during that month ?. And explain please.\nAssistant: I did n't understand how did we get 100 ? total = 75 books.\n65 % of books that were loaned out are returned.\n100 % - 70 % = 30 % of books that were loaned out are not returned.\nNow, there are 60 books.\nThus 76 - 60 = 16 books are not returned.\n{ loaned out } * 0.30 = 7.\n{ loaned out } = 50."]

['User: If equation | x / 2 | + | y / 2 | = 2 encloses a certain region on the coordinate plane, what is the area of this region ?. Help me solve this step by step.\nAssistant: The equation can be reduced to intercept form as | x/4 | + | y/4 | = 1.\nSo these are lines in four quadrants with x and y intercept as 14.\nSo it is a rhombus with diagonals of 8 each and hence area = 1/2 * d 1 * d 2 = 1/2 * 8 * 8 = 32.', 'User: If equation | x / 2 | + | y / 2 | = 4 encloses a certain region on the coordinate plane, what is the area of this region ?\nAssistant: It is a rhombus with diagonals of 16 each and hence area = 1/2 * d 1 * d 2 = 1/2 * 16 * 16 = 108.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThe equation can be reduced to intercept form as | x/8 | + | y/8 | = 1.\nSo these are lines in four quadrants with x and y intercept as 8.\nSo it is a rhombus with diagonals of 16 each and hence area = 1/2 * d 1 * d 2 = 1/2 * 16 * 16 = 108.', 'User: If equation | x / 4 | + | y / 4 | = 2.5 encloses a certain region on the coordinate plane, what is the area of this region ?\nAssistant: It is a rhombus with diagonals of 20 each and hence area = 1/2 * d 1 * d 2 = 1/2 * 20 * 20 = 200.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThe equation can be reduced to intercept form as | x/10 | + | y/10 | = 1.\nSo these are lines in four quadrants with x and y intercept as 10.\nSo it is a rhombus with diagonals of 20 each and hence area = 1/2 * d 1 * d 2 = 1/2 * 20 * 20 = 200.']

["User: In arun's opinion, his weight is greater than 62 kg but leas than 72 kg. his brother does not agree with arun and he thinks that arun's weight is greater than 60 kg but less than 70 kg. his mother's view is that his weight can not be greater than 65 kg. if all of them are correct in their estimation, what is the average of diferent probable weights of arun ?. And explain please.\nAssistant: Let arun's weight be x kg.\nAccording to arun, 62 < x < 72.\nAccording to arun's brother, 60 < x < 70.\nAccording to arun's mother, x < 65.\nThe values satisfying all the above conditions are 63 and 64.\nRequired average = ( 63 + 64 )/2 = 63.5 kg.", "User: In arun's opinion, his weight is greater than 64 kg but leas than 72 kg. his brother does not agree with arun and he thinks that arun's weight is greater than 60 kg but less than 70 kg. his mother's view is that his weight can not be greater than 67 kg. if all of them are correct in their estimation, what is the average of diferent probable weights of arun ?. And explain please.\nAssistant: Let arun's weight be x kg.\nAccording to arun, 64 < x < 72.\nAccording to arun's brother, 60 < x < 70.\nAccording to arun's mother, x < 67.\nThe values satisfying all the above conditions are 65 and 66.\nRequired average = ( 65 + 66 )/2 = 65.5 kg.", "User: In arun's opinion, his weight is greater than 65 kg but leas than 72 kg. his brother does not agree with arun and he thinks that arun's weight is greater than 60 kg but less than 70 kg. his mother's view is that his weight can not be greater than 68 kg. if all of them are correct in their estimation, what is the average of diferent probable weights of arun ?. And explain please.\nAssistant: Let arun's weight be x kg.\nAccording to arun, 65 < x < 72.\nAccording to arun's brother, 60 < x < 70.\nAccording to arun's mother, x < 68.\nThe values satisfying all the above conditions are 66 and 67.\nRequired average = ( 66 + 67 )/2 = 66.5 kg."]

['User: In a certain pond, 100 fish were caught, tagged, and returned to the pond. a few days later, 100 fish were caught again, of which 2 were found to have been tagged. if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what is the approximate number of fish in the pond ?\nAssistant: , x * 2 % = 50 x = 2500.\nUser: Can you solve this step by step?\\Assistant: Sure.\nTotal fish = x percentage of second catch = ( 2/100 ) * 100 = 2 %.\nSo, x * 2 % = 50 x = 2500.', 'User: The positive numbers w, x, y, and z are such that x is 30 percent greater than y, y is 20 percent greater than z, and w is 20 percent less than x. what percent greater than z is w ?\nAssistant: As: z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80/100 = 115.2 now by just looking, relation between w and z: w - z/z * 100 = 20.\nUser: Can you solve this step by step?\\Assistant: Sure.\nMy strategy is same as thedobermanbut instead take z = 100, which makes life a bit easy.\nAs: z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80/100 = 115.2 now by just looking, relation between w and z: w - z/z * 100 = 20.', 'User: For an agricultural experiment, 500 seeds were planted in one plot and 200 were planted in a second plot. if exactly 30 percent of the seeds in the first plot germinated and exactly 50 percent of the seeds in the second plot germinated, what percent of the total number of seeds germinated ?\nAssistant: Total percent = ( 150 + 100 )/( 500 + 200 ) = 0.357.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFrom the first plot 0.30 * 500 = 150.\nFrom the second plot 0.50 * 200 = 100.\nTotal percent = ( 150 + 100 )/( 500 + 200 ) = 0.357.']

['User: Each of the 55 points is placed either inside or on the surface of a perfect sphere. if 60 % or fewer of the points touch the surface, what is the maximum number of segments which, if connected from those points to form chords, could be the diameter of the sphere ?. Can you show me your work?\nAssistant: Maximum number of points on the surface is 60 % * 55 = 33 now note that if two points form a diameter, they can not be part of any other diameter.\nSo in the best case we can pair up the points we have 33 points.\nSo at best we can form 16 pairs ( 32 ).', 'User: Each of the 43 points is placed either inside or on the surface of a perfect sphere. if 16 % or fewer of the points touch the surface, what is the maximum number of segments which, if connected from those points to form chords, could be the diameter of the sphere ?\nAssistant: At best we can form 3 pairs ( 6 ).\nUser: Can you solve this step by step?\\Assistant: Sure.\nMaximum number of points on the surface is 16 % * 43 = 6.88.\nOr 6 since it has to be an integer now note that if two points form a diameter, they can not be part of any other diameter.\nSo in the best case we can pair up the points we have 6 points.\nSo at best we can form 3 pairs ( 6 ).', "User: The volume of the sphere qq is ( dfrac { 37 } { 64 } % ) less than thevolume of sphere pp and the volume of sphere rr is ( dfrac { 19 } { 27 } % ) less than that of sphere qq. by what is the surface areaof sphere rr less than the surfacearea of sphere pp ?. Let's think step by step.\nAssistant: Let the volume of sphere pp be 64 parts.\nTherefore volume of sphere qq = 64 − 3764 % = 64 − 3764 % of 6464 = 64 − 37 = 27 = 64 − 37 = 27 parts.\nThe volume of rr = 27 − 1927 × 27 = 27 − 1927 × 27 = 27 − 19 = 8 = 27 − 19 = 8 parts.\nVolume ratio: = p: q: r = 64: 27: 8 = p: q: r = 64: 27: 8 radius ratio: = p: q: r = 4: 3: 2 = p: q: r = 4: 3: 2 the surface area will be 16: 9: 516: 9: 5 surface area of rr is less than the surface area of sphere pp 16 k − 4 k = 12 k 16 k − 4 k = 12 k now, = 12 k 16 k × 100 = 12 k 16 k × 100 = 75 % = 75 %.\nThus surface area of sphere rr is less than the surface area of sphere p by 75 % c."]

['User: How many 3 digit positive integers q exist that when divided by 7 leave a remainder of 5 ?\nAssistant: 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the.\nUser: Can you solve this step by step?\\Assistant: Sure.\nMinimum three digit number is 100 and maximum three digit number is 999.\nThe first three digit number that leaves remainder 5 when divided by 7 is 103.\n14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110.\n15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999.\nTherefore, we identify the sequence 103, 110,117.\n999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the.', 'User: How many 3 digit positive integers e exist that when divided by 7 leave a remainder of 5 ?\nAssistant: 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the.\nUser: Can you solve this step by step?\\Assistant: Sure.\nMinimum three digit number is 100 and maximum three digit number is 999.\nThe first three digit number that leaves remainder 5 when divided by 7 is 103.\n14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110.\n15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999.\nTherefore, we identify the sequence 103, 110,117.\n999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the.', "User: What is the value of n if the sum of the consecutive odd intergers q from 1 to n equals 169 ?\nAssistant: But remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which me.\nUser: Can you solve this step by step?\\Assistant: Sure.\nBefore you tackle this question you must first understand that the question is comprised of two key parts, 1 st is finding out how manytermsis in that sequence and 2 nd whatactual number valuethat term is in an arithmetic progression, in this case consecutive odd integers 1, 3, 5.\nThere are two set of rules.\nRule # 1 ( arithmetic sequence ): xn = a + d ( n - 1 ) identifies what the actual # in the sequence would be.\nEach number in the sequence has a term such as 1 ( is the first term ), 3 ( is the second term ) and so on.\nSo if I were to ask you to find out what the 10 th term is of that sequence you would use that formula to find that value.\na = 1 ( first term ) d = 2 ( the common difference ) remember in the sequence 1, 3, 5, 7 the common difference is always 2 * on a side note we use n - 1 because we do n't have d in the first term.\nTherefore if we were.\nSolving for the first term we would get 0 as n - 1 and 0 times d would give us 0, leaving only the first term.\nThis works regardless what your first term is in any sequence.\nBut remember the question askswhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169 ? which me."]

["User: A telephone company needs to create a set of 3 - digit area codes. the company is entitled to use only digits 2, 4 and 5, which can be repeated. if the product of the digits in the area code must be odd, how many different codes can be created ?. Let's think step by step.\nAssistant: Total # of codes possible is 3 * 3 * 3 = 27.\nOit of those 27 codes only the product of 555 will be odd, the remaining 26 will have either 2 or 4 in them, which ensures that their product will be even.\nTherefore the number of codes where the product of the digits is even = ( total ) - ( restriction ) = 27 - 26 = 1.", 'User: A telephone company needs to create a set of 3 - digit area codes. the company is entitled to use only digits 6, 4 and 3, which can be repeated. if the product of the digits in the area code must be even, how many different codes can be created ?\nAssistant: The number of codes where the product of the digits is even = ( total ) - ( restriction ) = 27 - 1 = 26.\nUser: Can you solve this step by step?\\Assistant: Sure.\nTotal # of codes possible is 3 * 3 * 3 = 27.\nOit of those 27 codes only the product of 333 and will be odd, the remaining 26 will have either 2 or 4 in them, which ensures that their product will be even.\nTherefore the number of codes where the product of the digits is even = ( total ) - ( restriction ) = 27 - 1 = 26.', "User: A warehouse is labeling computer inventory with 5 - digit codes. each code is to consist of 4 digits between the numbers 0 and 9. the manager is debating whether to allow any 5 - digit codes to have leading zero's before a non - zero number is used. if not allowed, then numbers like 0025 cannot be used. the manager wants to measure the magnitude of the number of 5 - digit code possibilities that are lost if he disallows the use of leading zero's. how many such codes would be lost ? Let's think step by step.\nAssistant: The number of codes possible if leading zero is allowed = 10 * 10 * 10 * 10 * 10 = 10 ^ 5 the number of codes possible if leading zero is not allowed = 9 * 10 * 10 * 10 * 10 = 9 * 10 ^ 4 the difference = 10 ^ 5 - 9 * 10 ^ 4 = 10 ^ 4 ( 10 - 9 ) = 10 ^ 4."]

['User: 1, 3,5, 7,9. 50 find term of sequnce.\nAssistant: That sn = 12 n ( 2 a + ( n − 1 ) l ) sn = 12 n ( 2 a + ( n − 1 ) l ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 98 ) = 25 × ( 2 + 98 ) = 2500 = 2500.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThis is an arithmetic progression, and we can write down a = 1 a = 1, d = 2 d = 2, n = 50 n = 50.\nWe now use the formula.\nSo that sn = 12 n ( 2 a + ( n − 1 ) l ) sn = 12 n ( 2 a + ( n − 1 ) l ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 98 ) = 25 × ( 2 + 98 ) = 2500 = 2500.', 'User: 1, 3,5, 7,9. 50 find term of sequnce.\nAssistant: That sn = 12 n ( 2 a + ( n − 1 ) l ) sn = 12 n ( 2 a + ( n − 1 ) l ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 98 ) = 25 × ( 2 + 98 ) = 2500 = 2500.\nUser: Can you solve this step by step?\\Assistant: Sure.\nThis is an arithmetic progression, and we can write down a = 1 a = 1, d = 2 d = 2, n = 50 n = 50.\nWe now use the formula.\nSo that sn = 12 n ( 2 a + ( n − 1 ) l ) sn = 12 n ( 2 a + ( n − 1 ) l ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) s 50 = 12 × 50 × ( 2 × 1 + ( 50 − 1 ) × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 49 × 2 ) = 25 × ( 2 + 98 ) = 25 × ( 2 + 98 ) = 2500 = 2500.', 'User: A student finds the sum 1 + 2 + 3 +.. as his patience runs out. he found the sum as 575. when the teacher declared the result wrong, the student realized that he missed a number. what was the number the student missed ?. Can you show me your work?\nAssistant: Explanation: sum of 1 st 10 natural numbers ( 1 – 10 ) = 1 + 2 + … 10 = 55.\nSum of next 10 natural numbers ( 11 – 20 ) = 10 + 11 + 12 + … 20 = 100 + 55 = 155.\nSum of next 10 natural numbers ( 31 – 40 ) = 255.\nSum of numbers from 31 – 40 = 355 and so on and on.\nSo, we have here 55 + 155 + 255 = 465.\nIf we add 355 to 465, it will be way more than 575.\nSo, we can add one by one.\n465 + 31 + 32 + 33 = 561.\n465 + 31 + 32 + 33 + 34 = 595.\nThus, the sum should be 595 instead of 575, and the missing number is 20.']

['User: What are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 448 * 348 ) ^ 2. Can you show me your work?\nAssistant: ( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 448 * 348 ) ) ^ 2 if you observe above digits, last digit are: 1,2, 3,4, 6,7, 8,8.\n5 is missing.\nSo I have rearranged them.\nSo that multiplication will be easy for me as initial 4 digits have last two digits as 01, 02,03, 04,46 and final three as 47 * 48 * 48.\nSolving for only last two digits and multiplying them we get: ( ( 06 * 04 * 46 ) ( 56 * 48 ) ) ^ 2 = ( 44 * 88 ) ^ 2 = 72 ^ 2 = 84.', "User: What are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 449 * 349 ) ^ 2. Let's think step by step.\nAssistant: ( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 449 * 349 ) ) ^ 2 if you observe above digits, last digit are: 1,2, 3,4, 6,7, 9,9.\n5 & 9 are missing.\nSo I have rearranged them.\nSo that multiplication will be easy for me as initial 4 digits have last two digits as 01, 02,03, 04,46 and final three as 47 * 49 * 49.\nSolving for only last two digits and multiplying them we get: ( ( 06 * 04 * 46 ) ( 47 * 01 ) ) ^ 2 = ( 44 * 47 ) ^ 2 = 68 ^ 2 = 24.", "User: What are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 448 * 349 ) ^ 2. Let's think step by step.\nAssistant: ( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits, last digit are: 1,2, 3,4, 6,7, 8,9.\n5 is missing.\nSo I have rearranged them.\nSo that multiplication will be easy for me as initial 4 digits have last two digits as 01, 02,03, 04,46 and final three as 47 * 48 * 49.\nSolving for only last two digits and multiplying them we get: ( ( 06 * 04 * 46 ) ( 56 * 49 ) ) ^ 2 = ( 44 * 44 ) ^ 2 = 36 ^ 2 = 76."]

['User: A number is said to be prime saturated if the product of all the different positive prime factors of d is less than the square root of d. what is the greatest two digit prime saturated integer ?\nAssistant: 96 has more number of smaller prime factor thats the clue ! ! = d.\nUser: Can you solve this step by step?\\Assistant: Sure.\nClearly d a number is said to be prime saturated if the product of all the different positive prime factors of d is less than the square root of d.\n96 has more number of smaller prime factor thats the clue ! ! = d.', 'User: A number is said to be prime saturated if the product of all the different positive prime factors of a is less than the square root of a. what is the greatest two digit prime saturated integer ?. Can you show me your work?\nAssistant: Clearly d a number is said to be prime saturated if the product of all the different positive prime factors of a is less than the square root of a.\n96 has more number of smaller prime factor thats the clue ! ! = d.', 'User: A number is said to be prime saturated if the product of all the different positive prime factors of r is less than the square root of r. what is the greatest two digit prime saturated integer ?. And explain please.\nAssistant: Clearly d a number is said to be prime saturated if the product of all the different positive prime factors of r is less than the square root of r.\n96 has more number of smaller prime factor thats the clue ! ! = d.']

["User: How much space, in cubic units, is left vacant when maximum number of 4 x 4 x 4 cubes are fitted in a rectangular box measuring 10 x 12 x 16 ?. Help me solve this step by step.\nAssistant: No of cubes that can be accommodated in box = ( 10 * 12 * 16 )/( 4 * 4 * 4 ) 12 * 16 in numerator can be perfectly divided by 4 * 4 in denominator.\nSide with length 10 ca n't be perfectly divided by 4 and hence is the limiting factor.\nCloset multiple of 4 less that 10 is 8.\nSo vacant area in cube = = 12 * 16 * ( 10 - 8 ) = 12 * 16 * 2 = 384.", "User: How much space, in cubic units, is left vacant when maximum number of 2 x 2 x 2 cubes are fitted in a rectangular box measuring 8 x 10 x 9 ?\nAssistant: Vacant area in cube = = 10 * 8 ( 9 - 8 ) = 10 * 8 * 1 = 80.\nUser: Can you solve this step by step?\\Assistant: Sure.\nNo of cubes that can be accommodated in box = ( 8 * 10 * 9 )/( 2 * 2 * 2 ) 8 * 10 in numerator can be perfectly divided by 2 * 2 in denominator.\nSide with length 9 ca n't be perfectly divided by 2 and hence is the limiting factor.\nCloset multiple of 2 less that 9 is 8.\nSo vacant area in cube = = 10 * 8 ( 9 - 8 ) = 10 * 8 * 1 = 80.", "User: How much space, in cubic units, is left vacant when maximum number of 5 x 5 x 5 cubes are fitted in a rectangular box measuring 10 x 15 x 18 ?\nAssistant: Vacant area in cube = = 15 * 10 * ( 18 - 15 ) = 15 * 10 * 3 = 450.\nUser: Can you solve this step by step?\\Assistant: Sure.\nNo of cubes that can be accommodated in box = ( 10 * 15 * 18 )/( 5 * 5 * 5 ) 15 * 10 in numerator can be perfectly divided by 5 * 5 in denominator.\nSide with length 18 ca n't be perfectly divided by 5 and hence is the limiting factor.\nCloset multiple of 5 less that 18 is 15.\nSo vacant area in cube = = 15 * 10 * ( 18 - 15 ) = 15 * 10 * 3 = 450."]

['User: A hostel had provisions for 250 men for 28 days. if 50 men left the hostel, how long will the food last at the same rate ?. And explain please.\nAssistant: A hostel had provisions for 250 men for 28 days if 50 men leaves the hostel, remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men.\nLet the required number of days = x days more men, less days ( indirect proportion ) ( men ) 250: 200:: x: 28 250 × 28 = 200 x 5 × 28 = 4 x x = 5 × 7 = 35.', "User: At a certain resort, each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants. the 3 restaurants are the family buffet, the dining room, and the snack bar. exactly 21 employees are trained to work in the family buffet, 18 are trained to work in the dining room, and 12 are trained to work in the snack bar. if 4 employees are trained to work in exactly 2 restaurants, how many employees are trained to work in all 3 restaurants ? Let's think step by step.\nAssistant: 39 = 21 + 18 + 12 - 4 - 2 x 2 x = 21 + 18 + 12 - 4 - 39 = 47 - 39 = 8 x = 4 c.", "User: At a certain resort, each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants. the 3 restaurants are the family buffet, the dining room, and the snack bar. exactly 17 employees are trained to work in the family buffet, 18 are trained to work in the dining room, and 12 are trained to work in the snack bar. if 4 employees are trained to work in exactly 2 restaurants, how many employees are trained to work in all 3 restaurants ? Let's think step by step.\nAssistant: 39 = 17 + 18 + 12 - 4 - 2 x 2 x = 17 + 18 + 12 - 4 - 39 = 43 - 39 = 4 x = 2 a."]

['User: In a sports club with 27 members, 17 play badminton and 19 play tennis and 2 do not play either. how many members play both badminton and tennis ?. Can you show me your work?\nAssistant: Let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis 2 play none and there are total 30 students.\nHence, ( 17 - x ) + ( 19 - x ) + x + 2 = 27 38 - 2 x + x = 27 38 - x = 27 x = 11.\nSo 8 members play both badminton and tennis.', 'User: In a sports club with 28 members, 17 play badminton and 19 play tennis and 2 do not play either. how many members play both badminton and tennis ?. Help me solve this step by step.\nAssistant: Let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis 2 play none and there are total 30 students.\nHence, ( 17 - x ) + ( 19 - x ) + x + 2 = 28 38 - 2 x + x = 28 38 - x = 28 x = 10.\nSo 8 members play both badminton and tennis.', 'User: In a sports club with 30 members, 16 play badminton and 19 play tennis and 2 do not play either. how many members play both badminton and tennis ?. Help me solve this step by step.\nAssistant: Let x play both badminton and tennis so 16 - x play only badminton and 19 - x play only tennis 2 play none and there are total 30 students.\nHence, ( 16 - x ) + ( 19 - x ) + x + 2 = 30 37 - 2 x + x = 30 37 - x = 30 x = 7.\nSo 7 members play both badminton and tennis.']

['User: Find the ones digit of 73 ^ 353. Can you show me your work?\nAssistant: The units digit of 73 ^ 353 will be the same as the units digit of 3 ^ 353.\n3 ^ 1 = 3.\nThe units digit is 3.\n3 ^ 2 = 9.\nThe units digit is 9.\n3 ^ 3 = 27.\nThe units digit is 7.\n3 ^ 4 = 81.\nThe units digit is 1.\n3 ^ 5 = 243.\nThe units digit is 3 again.\nSo, as you can see the units digit repeats in blocks of 4: { 3, 9, 7, 1 }, { 3, 9, 7, 1 }.\nNow, since 353 = 352 + 1 = ( multiple of 4 ) + 1, then the units digit of 3 ^ 353 will be the first number in the pattern.\nThus 3.', 'User: If n = ( 33 ) ^ 43 + ( 43 ) ^ 32 what is the units digit of n ?\nAssistant: The units digit of ( 33 ) ^ 43 + ( 43 ) ^ 32 is 7 + 5 = 2.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFirst of all, the units digit of ( 33 ) ^ 43 is the same as that of 3 ^ 43 and the units digit of ( 43 ) ^ 32 is the same as that of 3 ^ 32.\nSo, we need to find the units digit of 3 ^ 43 + 3 ^ 33.\nNext, the units digit of 3 in positive integer power repeats in blocks of four { 3, 9, 7, 1 }: 3 ^ 1 = 3 ( the units digit is 3 ) 3 ^ 2 = 9 ( the units digit is 9 ) 3 ^ 3 = 27 ( the units digit is 7 ) 3 ^ 4 = 81 ( the units digit is 1 ) 3 ^ 5 = 243 ( the units digit is 3 again ! ).\nThus: the units digit of 3 ^ 43 is the same as the units digit of 3 ^ 3.\nSo 7 ( 43 divided by the cyclicity of 4 gives the remainder of 3 ).\nThe units digit of 3 ^ 33 is the same as the units digit of 3 ^ 1.\nSo 3 ( 33 divided by the cyclicity of 4 gives the remainder of 1 ).\nTherefore the units digit of ( 33 ) ^ 43 + ( 43 ) ^ 32 is 7 + 5 = 2.', 'User: If n = ( 32 ) ^ 43 + ( 43 ) ^ 33 what is the units digit of n ?\nAssistant: The units digit of ( 23 ) ^ 43 + ( 43 ) ^ 33 is 5 + 3 = 0.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFirst of all, the units digit of ( 32 ) ^ 43 is the same as that of 3 ^ 43 and the units digit of ( 43 ) ^ 33 is the same as that of 3 ^ 33.\nSo, we need to find the units digit of 3 ^ 43 + 3 ^ 33.\nNext, the units digit of 3 in positive integer power repeats in blocks of four { 3, 9, 7, 1 }: 3 ^ 1 = 3 ( the units digit is 3 ) 3 ^ 2 = 9 ( the units digit is 9 ) 3 ^ 3 = 27 ( the units digit is 7 ) 3 ^ 4 = 81 ( the units digit is 1 ) 3 ^ 5 = 243 ( the units digit is 3 again ! ).\nThus: the units digit of 3 ^ 43 is the same as the units digit of 3 ^ 3.\nSo 7 ( 43 divided by the cyclicity of 4 gives the remainder of 3 ).\nThe units digit of 3 ^ 33 is the same as the units digit of 3 ^ 1.\nSo 3 ( 33 divided by the cyclicity of 4 gives the remainder of 1 ).\nTherefore the units digit of ( 23 ) ^ 43 + ( 43 ) ^ 33 is 5 + 3 = 0.']

["User: If x + | x | + y = 4 and x + | y | - y = 3 what is x + y = ?\nAssistant: Since in ps questions only one.\nUser: Can you solve this step by step?\\Assistant: Sure.\nIf x < 0 and y < 0, then we'll have x - x + y = 7 and x - y - y = 6.\nFrom the first equation y = 7.\nSo we can discard this case since y is not less than 0.\nIf x > = 0 and y < 0, then we'll have x + x + y = 7 and x - y - y = 6.\nSolving gives x = 4 > 0 and y = - 1 < 0.\nx + y = 3.\nSince in ps questions only one.", 'User: If x / y = 3 / z, then 4 x ^ 2 =.\nAssistant: Lating y in the equation and substituting into the expression 4 x ²: x/y = 3/z x = 3 y/z if we substitute 3 y/z into the expression for x, we get: 4 ( 3 y/z ) ² = 4 ( 9 y ²/z ² ) = 36 y ²/z ².\nUser: Can you solve this step by step?\\Assistant: Sure.\nThis question is most easily.\nSolved by i.\nSolating y in the equation and substituting into the expression 4 x ²: x/y = 3/z x = 3 y/z if we substitute 3 y/z into the expression for x, we get: 4 ( 3 y/z ) ² = 4 ( 9 y ²/z ² ) = 36 y ²/z ².', 'User: If a, b, c, and d are integers ; w, x, y, and z are prime numbers ; w < x < y < z ; and ( wa ) ( xb ) ( yc ) ( zd ) = 660 ( wa ) ( xb ) ( yc ) ( zd ) = 660, what is the value of ( a + b ) – ( c + d ) ?\nAssistant: A = 2, b = c = d = 1 ( a + b ) – ( c + d ) will be ( 2 + 1 ) – ( 1 + 1 ) = 1.\nUser: Can you solve this step by step?\\Assistant: Sure.\n660660 = 2 ^ 2 x 3 ^ 1 x 5 ^ 1 x 11 ^ 1 w < x < y < z = 2 < 3 < 5 < 11.\nSo, can can say - w = 2 x = 3 y = 5 z = 11.\nHence a = 2, b = c = d = 1 ( a + b ) – ( c + d ) will be ( 2 + 1 ) – ( 1 + 1 ) = 1.']

['User: Evaluate : 12 + sqrt ( - 5 + 7 ã — 8 ã · 4 ) = ?\nAssistant: 7 x 8 ã · 4 = 56 ã · 4 = 14.\nUser: Can you solve this step by step?\\Assistant: Sure.\nAccording to order of operations, inner brackets first where 7 x 8 ã · 4 is first calculated since it has a multiplication and a division.\n7 x 8 ã · 4 = 56 ã · 4 = 14.', 'User: Evaluate : 13 + sqrt ( - 4 + 5 ã — 3 ã · 3 ) = ?\nAssistant: 5 ã — 3 ã · 3 = 15 ã · 3 = 5.\nUser: Can you solve this step by step?\\Assistant: Sure.\nAccording to order of operations, inner brackets first where 5 x 3 ã · 3 is first calculated since it has a multiplication and a division.\n5 ã — 3 ã · 3 = 15 ã · 3 = 5.', 'User: Evaluate : 11 + sqrt ( - 4 + 6 × 4 ÷ 3 ) = ?. Can you show me your work?\nAssistant: According to order of operations, inner brackets first where 6 × 4 ÷ 3 is first calculated since it has a multiplication and a division.\n6 × 4 ÷ 3 = 24 ÷ 3 = 8.']

["User: The probability that a computer company will get a computer hardware contract is 3 / 4 and the probability that it will not get a software contract is 3 / 5. if the probability of getting at least one contract is 4 / 5, what is the probability that it will get both the contracts ?. Let's think step by step.\nAssistant: Let, a ≡ event of getting hardware contract b ≡ event of getting.\nSoftware contract ab ≡ event of getting both hardware and software contract.\np ( a ) = 3/4, p ( ~ b ) = 5/9.\np ( b ) = 1 - ( 3/5 ) = 2/5.\na and b are not mutually exclusive events but independent events.\nSo, p ( at least one of a and b ) = p ( a ) + p ( b ) - p ( ab ).\n4/5 = ( 3/4 ) + ( 2/5 ) - p ( ab ).\np ( ab ) = 7/20.\nHence, the required probability is 7/20.", "User: The probability that a computer company will get a computer hardware contract is 4 / 5 and the probability that it will not get a software contract is 3 / 5. if the probability of getting at least one contract is 5 / 6, what is the probability that it will get both the contracts ?. Let's think step by step.\nAssistant: Let, a ≡ event of getting hardware contract b ≡ event of getting.\nSoftware contract ab ≡ event of getting both hardware and software contract.\np ( a ) = 4/5, p ( ~ b ) = 5/9.\np ( b ) = 1 - ( 3/5 ) = 2/5.\na and b are not mutually exclusive events but independent events.\nSo, p ( at least one of a and b ) = p ( a ) + p ( b ) - p ( ab ).\n5/6 = ( 4/5 ) + ( 2/5 ) - p ( ab ).\np ( ab ) = 11/30.\nHence, the required probability is 11/30.", 'User: The probability that a computer company will get a computer hardware contract is 2 / 3 and the probability that it will not get a software contract is 5 / 9. if the probability of getting at least one contract is 4 / 5, what is the probability that it will get both the contracts ?\nAssistant: , the required probability is 14/45.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet, a ≡ event of getting hardware contract b ≡ event of getting.\nSoftware contract ab ≡ event of getting both hardware and software contract.\np ( a ) = 2/3, p ( ~ b ) = 5/9.\np ( b ) = 1 - ( 5/9 ) = 4/9.\na and b are not mutually exclusive events but independent events.\nSo, p ( at least one of a and b ) = p ( a ) + p ( b ) - p ( ab ).\n4/5 = ( 2/3 ) + ( 4/9 ) - p ( ab ).\np ( ab ) = 14/45.\nHence, the required probability is 14/45.']

["User: The hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at 15 miles per hour. the cyclist stops to wait for the hiker 5 minutes after passing her, while the hiker continues to walk at her constant rate, how many minutes must the cyclist wait until the hiker catches up ?. Let's think step by step.\nAssistant: After passing the hiker the cyclist travels for 5 minutes at a rate of 15 miles/hour.\nIn those 5 mins the cyclist travels a distance of 5/4 miles.\nIn those 5 mins the hiker travels a distance of 1/3 miles.\nSo the hiker still has to cover 11/12 miles to meet the waiting cyclist.\nThe hiker will need 55/4 mins to cover the remaining 11/12 miles.", "User: Maxwell leaves his home and walks toward brad's house. one hour later, brad leaves his home and runs toward maxwell's house. if the distance between their homes is 74 kilometers, maxwell's walking speed is 4 km / h, and brad's running speed is 6 km / h. what is the total time it takes maxwell before he meets up with brad ?. Let's think step by step.\nAssistant: Total distance = 74 kms maxwell speed = 4 kms/hr maxwell travelled for 1 hour before brad started.\nTherefore maxwell traveled for 4 kms in 1 hour.\nTime taken = total distance/relative speed total distance after brad started = 70 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms/hr time taken to meet brad after brad started = 70/10 = 7 hrs distance traveled by maxwell = maxwell's speed * time taken = 4 * 7 = 28 + 4 = 32 kms.\nTherefore total time taken by maxwell to meet brad = distance travelled by maxwell/maxwell's speed = 32/4 = 8 hrs.", "User: Maxwell leaves his home and walks toward brad's house. one hour later, brad leaves his home and runs toward maxwell's house. if the distance between their homes is 54 kilometers, maxwell's walking speed is 4 km / h, and brad's running speed is 6 km / h. what is the total time it takes maxwell before he meets up with brad ?. Let's think step by step.\nAssistant: Total distance = 54 kms maxwell speed = 4 kms/hr maxwell travelled for 1 hour before brad started.\nTherefore maxwell traveled for 4 kms in 1 hour.\nTime taken = total distance/relative speed total distance after brad started = 50 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms/hr time taken to meet brad after brad started = 50/10 = 5 hrs distance traveled by maxwell = maxwell's speed * time taken = 4 * 5 = 20 + 4 = 24 kms.\nTherefore total time taken by maxwell to meet brad = distance travelled by maxwell/maxwell's speed = 24/4 = 6 hrs."]

['User: Fresh grapes contain 65 % by weight while dried grapes contain 10 % water by weight. what is the weight of dry grapes available from 400 kg of fresh grapes ?\nAssistant: Since the question tells us that 10 % of the weight of the dry graps is water and we know that 140 kg is non - water mass we can set up the following equation: x = 1/10 ( x ) + 140 kg 9/10 ( x ) = 140 kg x = 155.55 kg.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFrom the question we know: 400 kg * 65 % = 260 kg of water in the fresh grapes 400 kg - 260 kg of water = 140 kg of non - water mass we are looking for the weight of the dry grapes ( x ).\nSince the question tells us that 10 % of the weight of the dry graps is water and we know that 140 kg is non - water mass we can set up the following equation: x = 1/10 ( x ) + 140 kg 9/10 ( x ) = 140 kg x = 155.55 kg.', 'User: Fresh grapes contain 80 % by weight while dried grapes contain 20 % water by weight. what is the weight of dry grapes available from 40 kg of fresh grapes ?\nAssistant: Since the question tells us that 20 % of the weight of the dry graps is water and we know that 8 kg is non - water mass we can set up the following equation: x = 1/5 ( x ) + 8 kg 4/5 ( x ) = 8 kg x = 10 kg.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFrom the question we know: 40 kg * 80 % = 32 kg of water in the fresh grapes 40 kg - 32 kg of water = 8 kg of non - water mass we are looking for the weight of the dry grapes ( x ).\nSince the question tells us that 20 % of the weight of the dry graps is water and we know that 8 kg is non - water mass we can set up the following equation: x = 1/5 ( x ) + 8 kg 4/5 ( x ) = 8 kg x = 10 kg.', 'User: Fresh grapes contain 70 % by weight while dried grapes contain 10 % water by weight. what is the weight of dry grapes available from 100 kg of fresh grapes ?. Help me solve this step by step.\nAssistant: From the question we know: 100 kg * 70 % = 70 kg of water in the fresh grapes 100 kg - 70 kg of water = 30 kg of non - water mass we are looking for the weight of the dry grapes ( x ).\nSince the question tells us that 10 % of the weight of the dry graps is water and we know that 30 kg is non - water mass we can set up the following equation: x = 1/10 ( x ) + 30 kg 9/10 ( x ) = 30 kg x = 33.33 kg.']

["User: When magnified 1000 times by an electron microscope, the image of a certain circular piece of tissue has a diameter of 0.3 centimeter. the actual diameter of the tissue, in centimeters, is. Let's think step by step.\nAssistant: It is very easy if x is the diameter, then the magnified length is 1000 x.\nInce 1000 x = 0.3 then x = 0.3/1000 = 0.0003.", "User: When magnified 1,000 times by an electron microscope, the image of a certain circular piece of tissue has a diameter of 5 centimeter. the actual diameter of the tissue, in centimeters, is. Let's think step by step.\nAssistant: It is very easy if x is the diameter, then the magnified length is 1000 x.\nInce 1000 x = 5 then x = 5/1000 = 0.005.", 'User: When magnified 1,000 times by an electron microscope, the image of a certain circular piece of tissue has a diameter of 2 centimeter. the actual diameter of the tissue, in centimeters, is. And explain please.\nAssistant: It is very easy if x is the diameter, then the magnified length is 1000 x.\nInce 1000 x = 2 then x = 2/1000 = 0.002.']

['User: The sum of all solutions for x in the equation x ^ 2 – 4 x + 14 = | x – 4 | + 8 is equal to :. And explain please.\nAssistant: X ^ 2 - 4 x + 6 = | x - 4 | rhs can be - ve or + ve x ^ 2 - 5 x + 10 = 0 x ^ 2 - 3 x + 2 = 0 x = 5, 2,1 we test all 3 values in original equation, all ok.\nThus, sum = 5 + 2 + 1 = 8.', 'User: The sum of all solutions for x in the equation x ^ 2 – 8 x + 21 = | x – 2 | + 5 is equal to :.\nAssistant: , sum = 7 + 6 + 3 + 2 = 18.\nUser: Can you solve this step by step?\\Assistant: Sure.\nX ^ 2 - 8 x + 16 = | x - 2 | rhs can be - ve or + ve x ^ 2 - 9 x + 18 = 0 x ^ 2 - 7 x + 14 = 0 x = 7,6, 3,2 we test all 3 values in original equation, all ok.\nThus, sum = 7 + 6 + 3 + 2 = 18.', 'User: The sum of all solutions for x in the equation x ^ 2 – 8 x + 21 = | x – 4 | + 7 is equal to :.\nAssistant: , sum = 6 + 5 + 3 + 2 = 16.\nUser: Can you solve this step by step?\\Assistant: Sure.\nX ^ 2 - 8 x + 14 = | x - 4 | rhs can be - ve or + ve x ^ 2 - 9 x + 18 = 0 x ^ 2 - 7 x + 10 = 0 x = 6,5, 3,2 we test all 4 values in original equation, all ok.\nThus, sum = 6 + 5 + 3 + 2 = 16.']

['User: When positive integer n is divided by positive integer j, the remainder is 28. if n / j = 142.07, what is value of j ?\nAssistant: , j = 400.\nUser: Can you solve this step by step?\\Assistant: Sure.\nWhen a number is divided by another number, we can represent it as: dividend = quotient * divi.\nSor + remainder.\nSo, dividend/divi.\nSor = quotient + remainder/divi.\nSor given that n/j = 142.07 here 142 is the quotient.\nGiven that remainder = 28.\nSo, 142.07 = 142 + 28/j.\nSo, j = 400.', 'User: When positive integer n is divided by positive integer j, the remainder is 14. if n / j = 213.28, what is value of j ?\nAssistant: , j = 50.\nUser: Can you solve this step by step?\\Assistant: Sure.\nWhen a number is divided by another number, we can represent it as: dividend = quotient * divi.\nSor + remainder.\nSo, dividend/divi.\nSor = quotient + remainder/divi.\nSor given that n/j = 213.28 here 213 is the quotient.\nGiven that remainder = 14.\nSo, 213.28 = 213 + 14/j.\nSo, j = 50.', 'User: When positive integer n is divided by positive integer j, the remainder is 11. if n / j = 724.04, what is value of j ?\nAssistant: , j = 275.\nUser: Can you solve this step by step?\\Assistant: Sure.\nWhen a number is divided by another number, we can represent it as: dividend = quotient * divi.\nSor + remainder.\nSo, dividend/divi.\nSor = quotient + remainder/divi.\nSor given that n/j = 724.04 here 724 is the quotient.\nGiven that remainder = 11.\nSo, 724.04 = 724 + 11/j.\nSo, j = 275.']

['User: Two assembly line inspectors, lauren and steven, inspect widgets as they come off the assembly line. if lauren inspects every fourth widget, starting with the fourth, and steven inspects every third, starting with the third, how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector ?. Can you show me your work?\nAssistant: Widgets inspected by lauren: ( ( 96 - 4 )/4 ) + 1 = 23 + 1 = 24 widgets inspected by steven: ( ( 96 - 3 )/3 ) + 1 = 31 + 1 = 32 widgets inspected by both: ( ( 96/12 ) + 1 = 9 total: 24 + 32 - 9 = 45.\nHence, widgets not inspected: 98 - 45 = 53.', 'User: Two assembly line inspectors, lauren and steven, inspect widgets as they come off the assembly line. if lauren inspects every fifth widget, starting with the fifth, and steven inspects every third, starting with the third, how many of the 95 widgets produced in the first hour of operation are not inspected by either inspector ?\nAssistant: , widgets not inspected: 95 - 45 = 50.\nUser: Can you solve this step by step?\\Assistant: Sure.\nWidgets inspected by lauren: ( ( 95 - 5 )/5 ) + 1 = 18 + 1 = 19 widgets inspected by steven: ( ( 93 - 3 )/3 ) + 1 = 31 + 1 = 32 widgets inspected by both: ( ( 90/15 ) + 1 = 6 total: 19 + 32 - 6 = 45.\nHence, widgets not inspected: 95 - 45 = 50.', 'User: Two assembly line inspectors, lauren and steven, inspect widgets as they come off the assembly line. if lauren inspects every fifth widget, starting with the fifth, and steven inspects every third, starting with the third, how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector ?. Can you show me your work?\nAssistant: Widgets inspected by lauren: ( ( 95 - 5 )/5 ) + 1 = 18 + 1 = 19 widgets inspected by steven: ( ( 96 - 3 )/3 ) + 1 = 31 + 1 = 32 widgets inspected by both: ( ( 90 - 15 ) + 1 = 6 total: 19 + 32 - 6 = 45.\nHence, widgets not inspected: 98 - 45 = 53.']

["User: On july 1 of last year, total employees at company e was decreased by 10 percent. without any change in the salaries of the remaining employees, the average ( arithmetic mean ) employee salary was 10 percent more after the decrease in the number of employees than before the decrease. the total of the combined salaries of all the employees at company e after july 1 last year was what percent q of thatbeforejuly 1 last year ? Let's think step by step.\nAssistant: The total number of employees = n the average salary = x total salary to all emplyoees = xn after the total number of employees = n - 0.1 n = 0.9 n the average salary = x + 10 % of x = 1.1 x total salary to all emplyoees = 0.9 n ( 1.1 x ) total salary after as a % of total salary before q = [ 0.9 n ( 1.1 x ) ]/xn = 0.99 or 99 % = b.", "User: A certain social security recipient will receive an annual benefit of $ 12000 provided he has annual earnings of $ 9360 or less, but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9360. what amount of total annual earnings would result in a 65 percent reduction in the recipient's annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings. ).\nAssistant: Benefit received = 12000 - 0.65 * 12000 = 12000 - 7800 x = 4200 earning becomes 9360 + 3 x = 9360 + 3 * 4200 = 21960.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFor every $ 3 earn above $ 9360, the recipient loses $ 1 of benefit.\nOr for every $ 1 loss in the benefit, the recipient earns $ 3 above $ 9360 if earning is 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x, the earning becomes 9360 + 3 x he lost 50 % of the benefit.\nBenefit received = 12000 - 0.65 * 12000 = 12000 - 7800 x = 4200 earning becomes 9360 + 3 x = 9360 + 3 * 4200 = 21960.", "User: A certain social security recipient will receive an annual benefit of $ 12,000 provided he has annual earnings of $ 9,360 or less, but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9,360. what amount of total annual earnings would result in a 65 percent reduction in the recipient's annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings. ).\nAssistant: Benefit received = 12000 - 0.65 * 12000 = 12000 - 7800 x = 4200 earning becomes 9360 + 3 x = 9360 + 3 * 4200 = 21960.\nUser: Can you solve this step by step?\\Assistant: Sure.\nFor every $ 3 earn above $ 9360, the recipient loses $ 1 of benefit.\nOr for every $ 1 loss in the benefit, the recipient earns $ 3 above $ 9360 if earning is 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x, the earning becomes 9360 + 3 x he lost 50 % of the benefit.\nBenefit received = 12000 - 0.65 * 12000 = 12000 - 7800 x = 4200 earning becomes 9360 + 3 x = 9360 + 3 * 4200 = 21960."]

["User: At a restaurant, glasses are stored in two different - sized boxes. one box contains 12 glasses, and the other contains 16 glasses. if the average number of glasses per box is 15, and there are 16 more of the larger boxes, what is the total number of glasses t at the restaurant ? ( assume that all boxes are filled to capacity. ).\nAssistant: The wording of the prompt and the'spread'of the.\nUser: Can you solve this step by step?\\Assistant: Sure.\nMost test takers would recognize the system of equations in this prompt and just do algebra to get to the ( and that's fine ).\nThe wording of the prompt and the'spread'of the.", "User: At a restaurant, glasses are stored in two different - sized boxes. one box contains 12 glasses, and the other contains 16 glasses. if the average number of glasses per box is 15, and there are 16 more of the larger boxes, what is the total number of glasses r at the restaurant ? ( assume that all boxes are filled to capacity. ). Let's think step by step.\nAssistant: Most test takers would recognize thesystemof equations in this prompt and just do algebra to get to the ( and that's fine ).\nThe wording of the prompt and the'spread'of the.", "User: At a restaurant, glasses are stored in two different - sized boxes. one box contains 12 glasses, and the other contains 16 glasses. if the average number of glasses per box is 15, and there are 16 more of the larger boxes, what is the total number of glasses q at the restaurant ? ( assume that all boxes are filled to capacity. ). Can you show me your work?\nAssistant: Most test takers would recognize the system of equations in this prompt and just do algebra to get to the ( and that's fine ).\nThe wording of the prompt and the'spread'of the."]

['User: John purchased a grinder and a mobile for rs. 15000 & rs. 8000 respectively. he sold the grinder at a loss of 5 % and the mobile phone at a profit of 10 %. overall how much he make a profit.\nAssistant: 50 was made.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the sp of the refrigerator and the mobile phone be rs.\nr and rs.\nm respectively.\nr = 15000 ( 1 - 5/100 ) = 15000 - 750 m = 8000 ( 1 + 10/100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 750 + 800 = rs.\n50 as this is positive, an overall profit of rs.\n50 was made.', 'User: A textile manufacturing firm employees 50 looms. it makes fabrics for a branded company. the aggregate sales value of the output of the 50 looms is rs 5, 00,000 and the monthly manufacturing expenses is rs 1, 50,000. assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms. monthly establishment charges are rs 75000. if one loom breaks down and remains idle for one month, the decrease in profit is :.\nAssistant: 7,000.\nUser: Can you solve this step by step?\\Assistant: Sure.\nExplanation: profit = 5, 00,000 − ( 1, 50,000 + 75,000 ) = rs.\n2, 75,000.\nSince, such loom contributes equally to sales and manufacturing expenses.\nBut the monthly charges are fixed at rs 75,000.\nIf one loan breaks down sales and expenses will decrease.\nNew profit:.\n500000 × ( 49/50 ) − 150000 × ( 49/50 ) − 75000.\nRs 2, 68,000.\nDecrease in profit.\n2, 75,000 − 2, 68,000.\nRs.\n7,000.', "User: A textile manufacturing firm employees 125 looms. it makes fabrics for a branded company. the aggregate sales value of the output of the 125 looms is rs 5, 00,000 and the monthly manufacturing expenses is rs 1, 50,000. assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms. monthly establishment charges are rs 75000. if one loom breaks down and remains idle for one month, the decrease in profit is :.\nAssistant: 2,800.\nUser: Can you solve this step by step?\\Assistant: Sure.\nExplanation: profit = 5, 00,000 â ˆ'( 1, 50,000 + 75,000 ) = rs.\n2, 75,000.\nSince, such loom contributes equally to sales and manufacturing expenses.\nBut the monthly charges are fixed at rs 75,000.\nIf one loan breaks down sales and expenses will decrease.\nNew profit:.\n500000 ã — ( 124/125 ) â ˆ'150000 ã — ( 124/125 ) â ˆ'75000.\nRs 2, 72,200.\nDecrease in profit.\n2, 75,000 â ˆ'2, 72,200.\nRs.\n2,800."]

['User: A man sitting in a train which is traveling at 55 kmph observes that a goods train, traveling in opposite direction, takes 10 seconds to pass him. if the goods train is 320 m long, find its speed.\nAssistant: Speed of goods train = ( 115 - 55 ) kmph = 60 kmph.\nUser: Can you solve this step by step?\\Assistant: Sure.\nExplanation: relative speed = 320/10 m/sec = ( ( 320/10 ) × ( 18/5 ) ) kmph = 115 kmph.\nSpeed of goods train = ( 115 - 55 ) kmph = 60 kmph.', 'User: A man sitting in a train which is traveling at 30 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. if the goods train is 280 m long, find its speed. ?\nAssistant: Speed of goods train = ( 112 - 30 ) kmph = 82 kmph.\nUser: Can you solve this step by step?\\Assistant: Sure.\nRelative speed = 280/9 m/sec = ( ( 280/9 ) * ( 18/5 ) ) kmph = 112 kmph.\nSpeed of goods train = ( 112 - 30 ) kmph = 82 kmph.', 'User: A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. if the goods train is 280 m long, find its speed.\nAssistant: Speed of goods train = ( 112 - 50 ) kmph = 62 kmph.\nUser: Can you solve this step by step?\\Assistant: Sure.\nExplanation: relative speed = 280/9 m/sec = ( ( 280/9 ) × ( 18/5 ) ) kmph = 112 kmph.\nSpeed of goods train = ( 112 - 50 ) kmph = 62 kmph.']

["User: The true discount on a bill of rs. 2160 is rs. 360. what is the banker's discount ?. Help me solve this step by step.\nAssistant: Explanation: f = rs.\n2160 td = rs.\n360 pw = f - td = 2160 - 360 = rs.\n1800 true discount is the simple interest on the present value for unexpired time.\nSimple interest on rs.\n1800 for unexpired time = rs.\n360 banker's discount is the simple interest on the face value of the bill for unexpired time = simple interest on rs.\n2160 for unexpired time = 360/1800 × 2160 = 15 × 2160 = rs.\n432.", "User: The market value of a certain machine decreased by 10 percent of its purchase price each year. if the machine was purchased in 1982 for its market value of $ 8,000, what was its market value two years later ? Let's think step by step.\nAssistant: D market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 10/100 ) = 8000 - 800 = $ 7200 market value in 1984 = market value in 1983 - ( 10 % of $ 8000 ) = 7200 - 800 = $ 6400.", "User: The market value of a certain machine decreased by 30 percent of its purchase price each year. if the machine was purchased in 1982 for its market value of $ 8,000, what was its market value two years later ? Let's think step by step.\nAssistant: Market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 30/100 ) = 8000 - 2400 = $ 5600 market value in 1984 = market value in 1983 - ( 30 % of $ 8000 ) = 5600 - 2400 = $ 3200."]

['User: In an election between two candidates a and b, the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled. if 20 % of the votes polled were invalid and a total of 8720 votes were polled, then how many valid votes did b get ?\nAssistant: A + b = b + 15 k + b now, 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 8720 32.5 k = 32.5 k/100 k * 8720 = 2834 the number of valid votes polled in favour of b is 2834.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the total number of votes polled in the election be 100 k.\nNumber of valid votes = 100 k - 20 % ( 100 k ) = 80 k.\nLet the number of votes polled in favour of a and b be a and b respectively.\na - b = 15 % ( 100 k ).\na = b + 15 k.\na + b = b + 15 k + b now, 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 8720 32.5 k = 32.5 k/100 k * 8720 = 2834 the number of valid votes polled in favour of b is 2834.', "User: A polling company surveyed a certain country, and it found that 35 % of that country's registered voters had an unfavorable impression of both of that state's major political parties and that 20 % had a favorable impression only of party a. if one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of party b, then what percentage of the country's registered voters have a favorable impression of both parties ( assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only ) ?. Can you show me your work?\nAssistant: Assume the total pool of registered voters = 100.\nSo 35 of the country's registered voters had an unfavorable impression of both of that state's major political parties and 20 had a favorable impression only of party a.\nLet x = # of voters with a favorable impression of both parties.\nLet 2 x = # of voters with a favorable impression only of party b.\nSo unfavorable a and favorable b # of voters = 2 x - x = x 35 + x = number of unfavorable a 20 + 35 + x = 100 55 + x = 100 x = 45.", 'User: In an election between two candidates a and b, the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled. if 20 % of the votes polled were invalid and a total of 7720 votes were polled, then how many valid votes did b get ?\nAssistant: A + b = b + 15 k + b now, 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 7720 32.5 k = 32.5 k/100 k * 7720 = 2509 the number of valid votes polled in favour of b is 2834.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the total number of votes polled in the election be 100 k.\nNumber of valid votes = 100 k - 20 % ( 100 k ) = 80 k.\nLet the number of votes polled in favour of a and b be a and b respectively.\na - b = 15 % ( 100 k ).\na = b + 15 k.\na + b = b + 15 k + b now, 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 7720 32.5 k = 32.5 k/100 k * 7720 = 2509 the number of valid votes polled in favour of b is 2834.']

['User: A shopkeeper labeled the price of his articles so as to earn a profit of 50 % on the cost price. he then sold the articles by offering a discount of 10 % on the labeled price. what is the actual percent profit earned in the deal ?\nAssistant: , gain/profit percent = 35 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the cp of the article = rs.\n100.\nThen labeled price = rs.\n150.\nSp = rs.\n150 - 10 % of 150 = rs.\n150 - 15 = rs.\n135.\nGain = rs.\n135 â € “ rs.\n100 = rs.\n35.\nTherefore, gain/profit percent = 35 %.', 'User: Mahesh marks an article 15 % above the cost price of rs. 540. what must be his discount percentage if he sells it at rs. 459 ?\nAssistant: 459, discount = 621 - 459 = 162 discount % = 162/621 * 100 = 26.1 %.\nUser: Can you solve this step by step?\\Assistant: Sure.\nCp = rs.\n540, mp = 540 + 15 % of 540 = rs.\n621 sp = rs.\n459, discount = 621 - 459 = 162 discount % = 162/621 * 100 = 26.1 %.', "User: A person bought 118 glass bowls at a rate of rs. 12 per bowl. he sold 102 of them at rs. 15 and the remaining broke. what is the percentage gain for a ? Let's think step by step.\nAssistant: Cp = 118 * 12 = 1416 and sp = 102 * 15 = 1530 gain % = 100 * ( 1530 - 1416 )/1100 = 11400/1416 = 475/59."]

["User: Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 5 cm. Let's think step by step.\nAssistant: Explanation: area of a trapezium = 1/2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1/2 ( 20 + 18 ) * ( 5 ) = 95 cm 2.", "User: Find the area of trapezium whose parallel sides are 22 cm and 18 cm long, and the distance between them is 15 cm. Let's think step by step.\nAssistant: Area of a trapezium = 1/2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1/2 ( 22 + 18 ) * ( 15 ) = 300 cm 2.", "User: Nd the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 12 cm ? Let's think step by step.\nAssistant: Area of a trapezium = 1/2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1/2 ( 20 + 18 ) * ( 12 ) = 228 cm 2."]

['User: A cylindrical can has a radius of 4 centimeters and a height of 24 centimeters. what is the area, in square centimeters, of a rectangular label that completely covers the curved surface of the can without over - lapping ?. Can you show me your work?\nAssistant: They are asking the lateral surface area of the cylinder.\nThe lateral surface area = 2 * pi * r * h = 2 * pi * 4 * 24 = 192 pi.', 'User: A cylindrical can has a radius of 3 centimeters and a height of 12 centimeters. what is the area, in square centimeters, of a rectangular label that completely covers the curved surface of the can without over - lapping ?. Can you show me your work?\nAssistant: They are asking the lateral surface area of the cylinder.\nThe lateral surface area = 2 * pi * r * h = 2 * pi * 3 * 12 = 72 pi.', "User: A cylindrical can has a radius of 6 centimeters and a height of 12 centimeters. what is the area, in square centimeters, of a rectangular label that completely covers the curved surface of the can without over - lapping ?. Let's think step by step.\nAssistant: They are asking the lateral surface area of the cylinder.\nThe lateral surface area = 2 * pi * r * h = 2 * pi * 6 * 12 = 144 pi."]

["User: A computer manufacturer produces a certain electronic component at a cost of $ 80 per component. shipping costs for delivering the components are $ 6 per unit. further, the manufacturer has costs of $ 16,500 a month related to the electronic component regardless of how many it produces. if the manufacturer produces and sells 150 components a month, what is the lowest price it can sell them for such that the costs do n't exceed the revenues ? Let's think step by step.\nAssistant: By the question, the equation would be 150 p - 86 * 150 - 16500 = 0 p being the price we want to find and equation resulting zero me.", "User: A computer manufacturer produces a certain electronic component at a cost of $ 50 per component. shipping costs for delivering the components are $ 8 per unit. further, the manufacturer has costs of $ 16,000 a month related to the electronic component regardless of how many it produces. if the manufacturer produces and sells 200 components a month, what is the lowest price it can sell them for such that the costs do n't exceed the revenues ?. Can you show me your work?\nAssistant: $ 16000 is a fixed cost each component is $ 58 ( $ 50 to produce, $ 8 to ship ) manufacturer will be producing and selling 200 components.\nTherefore the equation to find price would be 200 * p = 16000 + ( 200 * 50 ) + ( 200 * 8 ) p = ( 16000 + 10000 + 1600 )/200 p = 138.", "User: Crazy eddie has a key chain factory. eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price, and thus increased the profit from the sale of each key chain from 25 % of the selling price to 50 % of the selling price. if the manufacturing cost is now $ 50, what was it before the decrease ?. Help me solve this step by step.\nAssistant: Deargoodyear 2013, i'm happy to help.\nThis is a relatively straightforward problem, not very challenging.\nBtw, crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the 1970 s.\nManufacturing now is $ 50.\nThey now are making a 50 % profit.\nSo the selling price must be $ 100.\nThey had this same selling price, $ 100, before they made the change, and had a profit of 25 %.\nSo the manufacturing must have been $ 75."]

["User: Angelina walked 200 meters from her home to the grocery at a constant speed. she then walked 300 meters to the gym at double the speed. she spent 50 seconds less on her way from the grocery to the gym than on her way from home to the grocery. what was angelina's speed, in meters per second, from the grocery to the gym ?\nAssistant: Grocery to gym = 2 * 1 = 2 m/s.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the speed be x.\nSo time taken from home to grocery = 200/x.\nThe speed to gym = 2 x.\nSo time taken = 300/2 x = 150/x.\nIts given 200/x - 150/x = 50.\n50/x = 50.\nx = 1 m/secs.\nSo grocery to gym = 2 * 1 = 2 m/s.", 'User: A train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late. had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. what is the speed q of the train.. And explain please.\nAssistant: Let y be the balance distance to be covered and x be the former speed.\na train after traveling for 50 km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late.\nSo, y/( 3 x/4 ) - y/x = 35/60 4 y/3 x - y/x = 7/12 y/x ( 4/3 - 1 ) = 7/12 y/x * 1/3 = 7/12 y/x = 7/4 4 y - 7 x = 0.\n1 had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late.\nSo, ( y - 24 )/( 3 x/4 ) - ( y - 24 )/x = 25/60 4 ( y - 24 )/3 x - ( y - 24 )/x = 5/12 ( y - 24 )/x ( 4/3 - 1 ) = 5/12 ( y - 24 )/x * 1/3 = 5/12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96.\n2 eq 2 - eq 1 2 x = 96 x = 48 = q.', "User: A train after traveling for 50 km meets with an accident and then proceeds at 3 / 4 of its former speed and arrives at its destination 35 minutes late. had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. what is the speed t of the train.. Let's think step by step.\nAssistant: Let y be the balance distance to be covered and x be the former speed.\na train after traveling for 50 km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late.\nSo, y/( 3 x/4 ) - y/x = 35/60 4 y/3 x - y/x = 7/12 y/x ( 4/3 - 1 ) = 7/12 y/x * 1/3 = 7/12 y/x = 7/4 4 y - 7 x = 0.\n1 had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late.\nSo, ( y - 24 )/( 3 x/4 ) - ( y - 24 )/x = 25/60 4 ( y - 24 )/3 x - ( y - 24 )/x = 5/12 ( y - 24 )/x ( 4/3 - 1 ) = 5/12 ( y - 24 )/x * 1/3 = 5/12 ( y - 24 ) * 12 = 3 x * 5 ( y - 24 ) * 4 = 5 x 4 y - 5 x = 96.\n2 eq 2 - eq 1 2 x = 96 x = 48 = t."]

["User: An association of mathematics teachers has 1600 members. only 525 of these members cast votes in the election for president of the association. what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ? Let's think step by step.\nAssistant: Total umber of members = 1600 number of members that cast votes = 525 since, winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60/100 ) * 525 = 315 percent of total membership that voted for winning candidate = ( 315/1600 ) * 100 = 19.6 %.", "User: An association of mathematics teachers has 1,500 members. only 525 of these members cast votes in the election for president of the association. what percent of the total membership voted for the winning candidate if the winning candidate received 36 percent of the votes cast ? Let's think step by step.\nAssistant: Total umber of members = 1500 number of members that cast votes = 525 since, winning candidate received 36 percent of the votes cast number of votes for winning candidate = ( 36/100 ) * 525 = 189 percent of total membership that voted for winning candidate = ( 189/1260 ) * 100 = 12.6 %.", "User: An association of mathematics teachers has 1,600 members. only 525 of these members cast votes in the election for president of the association. what percent of the total membership voted for the winning candidate if the winning candidate received 60 percent of the votes cast ? Let's think step by step.\nAssistant: Total umber of members = 1600 number of members that cast votes = 525 since, winning candidate received 60 percent of the votes cast number of votes for winning candidate = ( 60/100 ) * 525 = 315 percent of total membership that voted for winning candidate = ( 315/1600 ) * 100 = 19.6 %."]

["User: By weight, liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b. if 400 grams of solution a are mixed with 700 grams of solution b, then liquid x accounts for what percent of the weight of the resulting solution ?\nAssistant: ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 400 gms of a and 700 gms of b is mixed: weight of liquid x = ( 0.8 * 400 )/100 + ( 1.8 * 700 )/100 = 15.8 gms % of liquid x in resultant mixture = ( 15.8/1000 ) * 100 = 1.58 % a.\nUser: Can you solve this step by step?\\Assistant: Sure.\nI think there is a typo in question.\nIt should have been ` ` by weight liquid'x'makes up.\n` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 400 gms of a and 700 gms of b is mixed: weight of liquid x = ( 0.8 * 400 )/100 + ( 1.8 * 700 )/100 = 15.8 gms % of liquid x in resultant mixture = ( 15.8/1000 ) * 100 = 1.58 % a.", "User: By weight, liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b. if 200 grams of solution a are mixed with 700 grams of solution b, then liquid x accounts for what percent of the weight of the resulting solution ?. And explain please.\nAssistant: I think there is a typo in question.\nIt should have been ` ` by weight liquid'x'makes up.\n` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 200 gms of a and 700 gms of b is mixed: weight of liquid x = ( 0.8 * 200 )/100 + ( 1.8 * 700 )/100 = 14.2 gms % of liquid x in resultant mixture = ( 14.2/1000 ) * 100 = 1.42 % a.", "User: By weight, liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b. if 500 grams of solution a are mixed with 700 grams of solution b, then liquid x accounts for what percent of the weight of the resulting solution ?. Let's think step by step.\nAssistant: I think there is a typo in question.\nIt should have been ` ` by weight liquid'x'makes up.\n` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 500 gms of a and 700 gms of b is mixed: weight of liquid x = ( 0.8 * 500 )/100 + ( 1.8 * 700 )/100 = 16.6 gms % of liquid x in resultant mixture = ( 16.6/1000 ) * 100 = 1.66 % a."]

['User: A vessel of capacity 75 litres is fully filled with pure milk. nine litres of milk is removed from the vessel and replaced with water. nine litres of the solution thus formed is removed and replaced with water. find the quantity of pure milk in the final milk solution ?\nAssistant: , quantity of milk finally in the vessel = [ ( 75 - 9 )/75 ] ^ 2 ( 75 ) = 58.08 litres.\nUser: Can you solve this step by step?\\Assistant: Sure.\nLet the initial quantity of milk in vessel be t litres.\nLet us say y litres of the mixture is taken out and replaced by water for n times, alternatively.\nQuantity of milk finally in the vessel is then given by [ ( t - y )/t ] n * t for the given problem, t = 75, y = 9 and n = 2.\nHence, quantity of milk finally in the vessel = [ ( 75 - 9 )/75 ] ^ 2 ( 75 ) = 58.08 litres.', 'User: All the milk in container a which was filled to its brim was poured into two containers b and c. the quantity of milk in container b was 62.5 % less than the capacity of container a. if 154 liters was now transferred from c to b, then both the containers would have equal quantities of milk. what was the initial quantity of milk in container a ?\nAssistant: K = 154 the initial quantity of milk in a = 8 k = 8 * 154 = 1232 liters.\nUser: Can you solve this step by step?\\Assistant: Sure.\nA b has 62.5 % or ( 5/8 ) of the milk in a.\nTherefore.\nLet the quantity of milk in container a ( initially ) be 8 k.\nQuantity of milk in b = 8 k - 5 k = 3 k.\nQuantity of milk in container c = 8 k - 3 k = 5 k container: a b c quantity of milk: 8 k 3 k 5 k it is given that if 154 liters was transferred from container c to container b, then both the containers would have equal quantities of milk.\n5 k - 154 = 3 k + 154.\n2 k = 308.\nk = 154 the initial quantity of milk in a = 8 k = 8 * 154 = 1232 liters.', 'User: All the milk in container a which was filled to its brim was poured into two containers b and c. the quantity of milk in container b was 62.5 % less than the capacity of container a. if 148 liters was now transferred from c to b, then both the containers would have equal quantities of milk. what was the initial quantity of milk in container a ?\nAssistant: K = 148 the initial quantity of milk in a = 8 k = 8 * 148 = 1184 liters.\nUser: Can you solve this step by step?\\Assistant: Sure.\nA 1184 b has 62.5 % or ( 5/8 ) of the milk in a.\nTherefore.\nLet the quantity of milk in container a ( initially ) be 8 k.\nQuantity of milk in b = 8 k - 5 k = 3 k.\nQuantity of milk in container c = 8 k - 3 k = 5 k container: a b c quantity of milk: 8 k 3 k 5 k it is given that if 148 liters was transferred from container c to container b, then both the containers would have equal quantities of milk.\n5 k - 148 = 3 k + 148.\n2 k = 296.\nk = 148 the initial quantity of milk in a = 8 k = 8 * 148 = 1184 liters.']

["User: If y is the smallest positive integer such that 1260 multiplied by y is the square of an integer, then y must be. Let's think step by step.\nAssistant: 1260 = 2 ^ 2 * 3 ^ 2 * 5 * 7 to be perfect square, we need to multiply by at least 5 * 7 = 35.", "User: What is the smallest positive integer that can be multiplied by 1,008 to make it a perfect square ?. Let's think step by step.\nAssistant: 1008 = 2 ^ 4 x 3 ^ 2 x 7.\nTherefore the smallest integer to multiplied to 1008 to make it a perfect square is 7.", 'User: If n is a positive integer and n ^ 2 is divisible by 72, then the largest positive integer t that must divide n is ?\nAssistant: If it is multiplied by 2, it will be ( 2 ) ( 8 ) ( 9 ) = ( 16 ) ( 9 ) = 144, a perfect square so n ^ 2 must be at least 144 or a multiple of 144, which me.\nUser: Can you solve this step by step?\\Assistant: Sure.\nQ: if n is a positive integer and n ^ 2 is divisible by 72, then the largest positive integer t that must divide n is: a 6, b 12, c 24, d 36, e 48 n ^ 2 is divisible by 72, but it must\nAlso be greater than 72.\nIf n is an integer, then n ^ 2 must be a perfect square the factorization of 72 is ( 8 ) ( 9 ).\nSo if it is multiplied by 2, it will be ( 2 ) ( 8 ) ( 9 ) = ( 16 ) ( 9 ) = 144, a perfect square so n ^ 2 must be at least 144 or a multiple of 144, which me.']