Question Detail

A is three times as old as B. C was twice-as old as A four years ago. In four years' time, A will be 31. What are the present ages of B and C ?

9, 46
9, 50
10, 46
10, 50

Answer: Option B

Explanation:

As from the question we have,
A = 3B --- (i)
c - 4 = 2(A+4) ---(ii)

Also, A + 4 = 31
or A = 31-4 = 27

Putting A = 27 in (i) we get
B = 9
Putting A = 27 in (ii) we get
C = 50

1. I have a few sweets to be distributed. If I keep 2, 3 or 4 in a pack, I am left with one sweet. If I keep 5 in a pack, I am left with none. What is the minimum number of sweets I have to pack and distribute ?

Answer: Option A

Explanation:

The required number will be such that it leaves a reminder of 1 when divided by 2, 3 or 4 and no reminder when divided by 5. Such a number is 25 among options

2. A father is now three times as old as his son. Five years back, he was four times as old as his son. The age of the son is :

Answer: Option B

Explanation:

Let son's age be x then father's age is 3x
Five years ago, father's age = 3x-5 and son's age x-5
So, 3x-5 = 4(x-5)
=> 3x-5 = 4x-20
=> x = 15

3. In a class, there are 18 boys who are over 160 cm tall. If these constitute three-fourths of the boys and the total number of boys is two-thirds of the total number of students in the class, what is the number of girls in the class ?

Answer: Option B

Explanation:

Let the number of boys be x.
Then, 3/4 x = 18
=> x = 24

If total number of students is y, then
2/3 y = 24
=> y = 36

Number of girls in class = y - x
= 36 - 24 = 12

4. A, B, C and D play a game of cards. A says to B. "If I give you 8 cards, you will have as many as C has and I shall have 3 less than what C has. Also if I take 8 cards from C, I shall have twice as many as D has." If B and D together have 50 cards, how many cards have A got?

Answer: Option D

Explanation:

If we read given question carefully, we get,
B+8 = C ---(i)
A-8 = C-3 ---(ii)
A+6 = 2D ---(iii)
B+D = 50 ---(iv)

Putting C = A -5 from (ii) into (i), we have :
B+8 = A-5 or A-B = 13 ---(v)

Putting D = 50-B from (iv) and (iii), we have :

A+6 = 100-2B or A+2B = 94 ---(vi)

Solving (v) and (vi), we get
B = 27 and A = 40

5. Find the least number which leaves a reminder of 3 when divided by 5, 6, 7 and 8, but leaves no reminder when divided by 9 ?

1683
1692
1598
1458

Answer: Option A

Explanation:

First of all, calculate LCM of 5, 6, 7 and 8 which is 840.
Suppose Number is K
then 840 * K + 3 is divisible by 9.
If K = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9

If K = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9

So 1683 is correct answer.