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 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

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 B

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