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 ?

Answer: Option B

Explanation:

I. matches of first player with other 5 players
II. matches of second player with 4 players other than the first player
III. matches of third player with 3 players other than the first player and second player.
IV. matches of fourth player with 2 players other than the first player, second player and third player.
V. matches of fifth player with 1 player other than the first player, second player, third player and fourth player.

So total matches will be 5+4+3+2+1 = 15

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

Answer: Option D

Explanation:

Let the number of bananas in the second bunch be x.
Then, the number of bananas in the first bunch
= x + 1/4 x
= 5/4 x
So,
5/4 x - x = 3
=> 5x - 4x = 12
x = 12

Number of bananas in first bunch
= 5/4 * 12 = 15

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

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