STD-5 PRADNYA 2012 PAPER SOLUTIONS
We are of the opinion that there is a printing mistake in the question.
The final product should have been 3C0BA and not 3C9BA as given in the question.
A=2, B=4, C=9, D=5
In the ten’s place of multiplication: 3x5=15 with 1 as carry over. So D=5.
So 3615 / 5 = 723 is the first number. So A = 2
In the unit’s place of multiplication: 3xB is giving unit’s place A=2. This is possible only when 3x4=12. So B=4
To find sum of all two digit numbers which are not divisible by 7, we will subtract the sum of all two digit numbers divisible by 7 from sum of all two digit numbers.
=(10+11+....+98+99) - (14+21+....+98)
=45(10+99) - 13x56
Let the quotient be Y
141Y +134 = 143Y+98
So, the number is 141Y+134= 2672
Let Z minutes be required to solve 1 question of the other subjects. Then, 2Z is the time needed for each question on maths.
So to solve 40 questions of other subjects will need 40Z minutes; and the 10 maths questions will take another 20Z minutes.
Thus, total time required is 60Z and that is given to be 180 min.
60Z = 180 min
Hence Z=3 min
Therefore, the time taken to solve the maths questions =20Z=20x3=60 min= 1 hour
Numerators are of the form (prime)2 + 1, and denominator is always 3 less than numerator.
So, after using 72+1 = 50 to get 50/47, the next four primes 11,13,17,19 will continue the sequence with 122/119, 170/167, 290/287 and 382/379.
Adding the fractions we get
⅕ + ⅕ + ⅛ + ⅛ + ¼ = 9/10
Hence the remaining 1/10th of the total fruits is comprised of the 100 custard apples. That means there are a total of 100x10=1000 fruits.
Multiplying 4 terms that is 7x7x7x7, we get a unit’s place digit as 1. This will happen for every group of 4 sevens.
But 153 = 38 x 4 + 1; so in 153 there are 38 groups of 4 each (whose answer will have 1 in the unit’s place), and there is one more 7 left to be multiplied.
Hence the final answer is 1x7=7.
BC=BP+PC and PC=2BP
Thus, AQ=BQ=BP=7 and DC=14 and BC=21
Hence sum of areas above will be 189.875 and area(rectangleABCD)=14x21=294
Now subtracting them we get the area (= 104.125
Constructing any one of the following possible rhombuses is a valid solution:
i) Square of side 5
ii) 2 Equilateral triangles of side 5, sharing one common side
Let vessel A with full of liquid is 8 gm
then vessel B’s full capacity is X/4 = 2 gm
vessel c”s full capapcity is X/8 = 1 gm
weight of liquid in C = 200-65=135 gm
Hence weight of liquid in A will be 135x8=1080
Weight of empty vessel A = weight of vessel with liquid - weight of liquid
= 1500 - 1080
= 420 gm
Let A be the required two digit number
We are given that 7xA + 7 = 7x(A+1) is a multiple of 13.
So 7x(A+1) = 13 x B for some B
Here, LHS is divisible by 7, so B on RHS is also divisible by 7.
Trying B = 7, we get:
7x(A+1) = 7 x 13 = 91
So A + 1 = 13, hence A=12 is the first answer.
Apply same method by taking all the multiples of 7 for B and you get possible answers for A as below
12, 25, 38, 51, 64, 77, 90
The 1st tap can empty ¼ th tank in 1 hour time
The 2nd tap can empty ⅛ th tank in 1 hour time
The 3rd tap can fill ½ the tank in 1 hour time
½ - ¼ - ⅛ = ⅛ th tank gets filled every hour.
hence the tank will be full in 8 hours.
First convert mixed fractions into decimal fractions in the denominator and we get
= = ½
Shamrao kept ½ for himself and half of the remaining that is ¼ th for wife.
So the remaining property is 1 - ½ - ¼ = ¼
Out of this ¼ th he gave 3/16 th of it to his sons and the leftover part that is 4/16 - 3/16 = 1/16 was split into 2 parts and each daughter gets one part and that is 1/32
So, 1/32 (1 part in 32) corresponds to Rs.50000/- , then the total property of 32 parts will be equal to 50000x32=160000
There are 2 different series mixed together:
The alternate numbers forms one series
15,22,29.. increasing by 7 with each term, so next two terms are 36,43
The other series is 42,39,36 decreasing by 3 with each term, so the next term is 33
when you mix them up you get the answer
Adding the fractions we get
The remaining are failed students and that is which corresponds to 1240.
Hence 31 parts are equal to 1240; so each part will be 1240 / 31 = 40
So totally the 77 parts will be 77x40=3080
This is a continued fraction problem and in such a problem one should start solving the them from the inner most fraction.
Step 2: =
The number of triangles are 44
The number of squares are 11
2431 is more than cube of 10 but less than cube of 20
Also 2431 is much close to 1000
Also observe unit’s place of the product 2431 is 1 and that of 4199 is 9
All these gives us some hints
and hence starting from primes above 10 we get
Thus the primes are 11,13,17,19
Simplifying terms of the equation:
Now the equation is:
0.1 + .75 x - 5.25 + 2 = 3/5
0.1 - 5.25 + 2 = - 3.15
3 /5 = 0.6
Therefore our equation is:
0.75x = 3.75
x = 5