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 D

Explanation:

Let the number of keepers be x.
Then, total number of feet = 2*50+4*45+4*8+2x
= 2x + 312

Total number of heads = 50+45+8+x = 103 + x
Now, (2x+312) = (103+x)+224
=> x = 15

Answer: Option D

Explanation:

Let number of horses = number of men = x.

Then, number of legs = 4x + 2 x (x/2) = 5x.

So, 5X = 70 or x = 14.

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

Answer: Option B

Explanation:

Generally we may commit mistake of dividing 1200/15.
But out of 16 persons there is one captain.
so, it will be 1200/16 = 75