Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 3/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
Options:
(A) 63/280
(B) 1/280
(C) 3/56
(D) 3/35
(E) 7/40

P = P(A wins, B wins, C loses) + P(A wins, B loses, C wins) + P(A loses, B wins, C wins) = 1/5*3/8*4/7 + 1/5*5/8*3/7 + 4/5*3/8*3/7 = 63/280. So the answer is: (A).

5,9,16,28,48,__
Options:
(A) 79
(B) 80
(C) 81
(D) 82
(E) 83

(9+5) +2 = 16
(16+9) +3 = 28
(28+16)+4 = 48
(48+28)+5 = 81 So the answer is: (C).

The average amount with a group of seven numbers is Rs. 30. If the newly joined member has Rs. 60 with him, what was the average amount with the group before his joining the group?
Options:
(A) s. 25.6
(B) s. 25
(C) s. 16.6
(D) s. 26
(E) s. 25.6

Total members in the group = 7
Average amount = Rs. 30
Total amount with them = 7 * 30 = Rs. 210
One number has Rs. 60. So, the amount with remaining 6 people = 210 - 60 = Rs. 150
The average amount with them = 150/6 = Rs. 25 So the answer is: (B).

70 is increased by 50%. Find the final number.
Options:
(A) 100
(B) 110
(C) 105
(D) 30
(E) 160
Final number = Initial number + 50%(original number) = 70 + 50%(70) = 70 + 35 = 105. So the answer is: (C).