q: There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 16 other amateurs, what is the total number of chess games possible to be played in the tournament?
Options:
(A) 10
(B) 20
(C) 40
(D) 60
(E) 120
a: Stream of consciousness: Each person is one participant of 4 games. So there are in all 16*5 = 80 instances of one participant games. But each game has 2 participants so total number of games = 80/2 = 40 The answer is (C).
--
q: A person spends 1/3rd of the money with him on clothes, 1/5th of the remaining on food and 1/4th of the remaining on travel. Now, he is left with Rs 600. How much did he have with him in the beginning?
Options:
(A) s 200
(B) s 1500
(C) s 300
(D) s 450
(E) s 550
a: Stream of consciousness: Suppose the amount in the beginning was Rs 'x'
Money spent on clothes = Rs 1x/3 Balance = Rs 2x/3
Money spent on food = 1/5 of 2x/3 = Rs 2x/15
Balance = 2x/3 - 2x/15 = Rs 8x/15
Money spent on travel = 1/4 of 8x/15 = Rs 2x/15 = 8x/15 - 2x/15 = 6x/15 = Rs2x/5
Therefore 2x/5 = 100 = 1500 The answer is (B).
--
q: The average of five numbers is 41. The average of first two numbers is 32 and the average of last two numbers is 38. What is the middle number?
Options:
(A) 59
(B) 64
(C) 61
(D) 72
(E) 65
a:
Stream of consciousness: The total of five numbers = 5X41 = 205
The total of first 2 and last 2 numbers is = 2 X 32+2 X 38 = 140
So, the middle number is (205 - 140 ) = 65 The answer is (E).
--