Reasoning: Total number of votes polled = (1136 + 7636 + 11628) = 20400
So, Required percentage = 11628/20400 * 100 = 57%
Ans: (C)
Question: 3 candidates in an election and received 1136, 7636 and 11628 votes respectively. What % of the total votes did the winningcandidate got in that election?
Options:
(A) 45%
(B) 47%
(C) 57%
(D) 48%
(E) 50%
Rationale: Given Exp. = (12)3 x 64 = (12)3 x 64 = (12)2 x 62 = (72)2 = 5184
432 12 x 62
The answer: (E)
Question: Find the value of y from (12)^3 x 6^4 ÷ 432 = y?
Options:
(A) 2234
(B) 4350
(C) 4560
(D) 5132
(E) y=5184
Reasoning: P(1 + 20/100)n > 2P or (6/5)n > 2
Now, (6/5 * 6/5 * 6/5 * 6/5) > 2. So, n = 4 years.
Ans: (B)
Question: The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is?
Options:
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9) = 5/12
The answer: (B)
Question: Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is
Options:
(A) 5/54
(B) 5/12
(C) 5/17
(D) 5/18
(E) 5/14
Reasoning: The intersection for two = 45 - 12 - 22 = 11 play both games.
Ans: (C)
Question: 45 pupil,out of them 12 in debate only and 22 in singing only.then how many in both?
Options:
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
Reasoning: Must be once on heads and twice on tails
1/2*1/2*1/2 = 1/8
Ans: (D)

Question: When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land once on heads and twice tails?
Options:
(A) 1/3
(B) 1/6
(C) 1/4
(D) 1/8
(E) 1/5
Reasoning: Total number of votes polled = (1136 + 7636 + 11628) = 20400
So, Required percentage = 11628/20400 * 100 = 57%
Ans: (C)
Question: 3 candidates in an election and received 1136, 7636 and 11628 votes respectively. What % of the total votes did the winningcandidate got in that election?
Options:
(A) 45%
(B) 47%
(C) 57%
(D) 48%
(E) 50%
Rationale: Given Exp. = (12)3 x 64 = (12)3 x 64 = (12)2 x 62 = (72)2 = 5184
432 12 x 62
The answer: (E)
Question: Find the value of y from (12)^3 x 6^4 ÷ 432 = y?
Options:
(A) 2234
(B) 4350
(C) 4560
(D) 5132
(E) y=5184
Reasoning: P(1 + 20/100)n > 2P or (6/5)n > 2
Now, (6/5 * 6/5 * 6/5 * 6/5) > 2. So, n = 4 years.
Ans: (B)
Question: The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is?
Options:
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9) = 5/12
The answer: (B)
Question: Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is
Options:
(A) 5/54
(B) 5/12
(C) 5/17
(D) 5/18
(E) 5/14
Reasoning: The intersection for two = 45 - 12 - 22 = 11 play both games.
Ans: (C)
Question: 45 pupil,out of them 12 in debate only and 22 in singing only.then how many in both?
Options:
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
Reasoning: Must be once on heads and twice on tails
1/2*1/2*1/2 = 1/8
Ans: (D)

user
Question: When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land once on heads and twice tails?
Options:
(A) 1/3
(B) 1/6
(C) 1/4
(D) 1/8
(E) 1/5
assistant
Reasoning: Total number of votes polled = (1136 + 7636 + 11628) = 20400
So, Required percentage = 11628/20400 * 100 = 57%
Ans: (C)
Question: 3 candidates in an election and received 1136, 7636 and 11628 votes respectively. What % of the total votes did the winningcandidate got in that election?
Options:
(A) 45%
(B) 47%
(C) 57%
(D) 48%
(E) 50%
Rationale: Given Exp. = (12)3 x 64 = (12)3 x 64 = (12)2 x 62 = (72)2 = 5184
432 12 x 62
The answer: (E)
Question: Find the value of y from (12)^3 x 6^4 ÷ 432 = y?
Options:
(A) 2234
(B) 4350
(C) 4560
(D) 5132
(E) y=5184
Reasoning: P(1 + 20/100)n > 2P or (6/5)n > 2
Now, (6/5 * 6/5 * 6/5 * 6/5) > 2. So, n = 4 years.
Ans: (B)
Question: The least number of complete years in which a sum of money put out at 20% C.I. will be more than doubled is?
Options:
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9) = 5/12
The answer: (B)
Question: Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is
Options:
(A) 5/54
(B) 5/12
(C) 5/17
(D) 5/18
(E) 5/14
Reasoning: The intersection for two = 45 - 12 - 22 = 11 play both games.
Ans: (C)
Question: 45 pupil,out of them 12 in debate only and 22 in singing only.then how many in both?
Options:
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
Reasoning: Must be once on heads and twice on tails
1/2*1/2*1/2 = 1/8
Ans: (D)

Question: When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land once on heads and twice tails?
Options:
(A) 1/3
(B) 1/6
(C) 1/4
(D) 1/8
(E) 1/5