Rationale: The number of zeroes at the end of 100! will be less than the number of zeroes at the end of 200!
Hence it would be sufficient to calculate the number of zeroes at the end of 100!
Number of zeroes = [100/5] + [100/25] + [100/125] = 20 + 4 + 0 = 24
The answer: (E)
Question: How many zeroes are there at the end of the number N, if N = 100! + 200! ?
Options:
(A) 73
(B) 49
(C) 20
(D) 48
(E) 24
Reasoning: Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is = C/C = (12 * 11 * 10 * 9)/(15 * 14 * 13 * 12) = 33/91
Ans: (B)
Question: A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If four marbles are picked at random, what is the probability that none is blue?
Options:
(A) 17/91
(B) 33/91
(C) 51/91
(D) 65/91
(E) None of these
Rationale: ? = 9568 6548 1024 = 1996.
The answer: (B)
Question: What should come in place of the question mark (?) in the following equation?
9568 6548 1024 = ?
Options:
(A) 2086
(B) 1996
(C) 2293
(D) 1896
(E) None of these
Reasoning: 10224. This is the only option with last two digits divisible by 4
Ans: (B)
Question: Which of the following numbers is divisible by 4?
Options:
(A) 27767
(B) 10224
(C) 26778
(D) 26679
(E) 26711
Rationale: P = (P*6*R)/100
R = 16 2/3%
The answer: (B)
Question: In what time a sum of money double itself at 6% per annum simple interest?
Options:
(A) 33 1/8%
(B) 16 2/3%
(C) 33 7/3%
(D) 32 1/3%
(E) 23 1/3%
Rationale: He has has ten patterns of chairs and eight patterns of tables
Hence, A chair can be arranged in 10 ways and
A table can be arranged in 10 ways
Hence one chair and one table can be arranged in 10 x 10 ways = 100 ways
The answer: (E)
Question: An design manager has 10patterns of chairs & 10patterns of tables. In how many ways can he make a pair of table & chair?
Options:
(A) 20
(B) 40
(C) 80
(D) 90
(E) 100