STD-5 PRADNYA 2012 PAPER SOLUTIONS

1a:

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.

Answer:

A=2, B=4, C=9, D=5

        723

X       54

      2892

+  36150

    39042

Observations:

Step 1:

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

Step 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

1b: 

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

=4905-728

=4177

2a:

Let the quotient be Y

141Y +134 = 143Y+98

36=2Y

Y=18

So, the number is  141Y+134= 2672

2b: 

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

3a: 

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.

3b:

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.

4a:

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.

4b:

BC=BP+PC      and  PC=2BP

BC=3BP

similarly DC=PC=2BP

Now perimeter=2(BC+DC)

70=2(5BP)

BP=7

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

5a:

 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

5b:

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

6a:

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

6b:

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.

7a:

First convert mixed fractions into decimal fractions in the denominator and we get

=  = ½

7b:

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

 

8a:

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

15,42,22,39,29,36,36,33,43

8b:

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

9a:

This is a continued fraction problem and in such a problem one should start solving the them from the inner most fraction.

Step 1:

Step 2: =

Step 3:

Step 4:

Step 5:

9b:

The number of triangles are  44

The number of squares are  11

10a:

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

11x13x17=2431 and

13x17x19=4199

Thus the primes are 11,13,17,19

10b:

Simplifying terms of the equation:

0.5x0.05x4=0.1

Now the equation is:

0.1 + .75 x - 5.25 + 2 = 3/5

Simplifying further:

0.1 - 5.25 + 2 = - 3.15

3 /5 = 0.6

Therefore our equation is:

0.75x-3.15=0.6

0.75x = 3.75

x = 5