(Ans). (C)
(Question). X can finish a work in 18 days. Y can finish the same work in 15 days. Yworked for 10 days and left the job. how many days does X alone need to finish the remaining work?
Options:
(A) 3
(B) 5
(C) 6
(D) 8
(E) 9
(CoT). Work done by X in 1 day = 1/18
Work done by Y in 1 day = 1/15
Work done by Y in 10 days = 10/15 = 2/3
Remaining work = 1 2/3 = 1/3
Number of days in which X can finish the remaining work = (1/3) / (1/18) = 6

(Ans). (B)
(Question). Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes?
Options:
(A) 6/18
(B) 6/11
(C) 6/68
(D) 6/19
(E) 6/12
(CoT). Part filled by (A + B + C) in 3 minutes = 3(1/30 + 1/20 + 1/10) = 11/20
Part filled by C in 3 minutes = 3/10
Required ratio = 3/10 * 20/11 = 6/11

(Ans). (E)
(Question). Let us say that a table tennis tournament was going on with knock out terms which means the one who loses the match is out of the tournament. 88 players took part in that tournament. How many matches were played?
Options:
(A) 90 matches
(B) 95 matches
(C) 99 matches
(D) 85 matches
(E) 87 matches
(CoT). E 87 matches. The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 87 matches were played.

(Ans). (C)
(Question).
he population of a city is 150,000. If it increases at the rate of 6% per annum, then what will be its population 2 years hence?
Options:
(A) 1697787
(B) 1697723
(C) 168540
(D) 169728
(E) 169718
(CoT). Population after n years = P [1 + (R/100)]2
Population after 2 years = 150000 * [1 + 6/100]2 = (150000 x 53/50 x 53/50) = 168540