Rationale: If students are A, B, C, D, E and F; we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get 5C2 = 10 ways of doing it.
The answer: (C)
Question: A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included in a particular student?
Options:
(A) 6
(B) 8
(C) 10
(D) 12
(E) 15
Reasoning: Given Exp. = (12)3 x 64 = (12)3 x 64 = (12)2 x 62 = (72)2 = 5184
432 12 x 62
Ans: (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) 5184
Rationale: Average of the first 'n' natural numbers = (n + 1)/2
We get (23 + 1)/2 = 12
The answer: (E)
Question: Find the average of the first 23 natural numbers?
Options:
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
Reasoning: Required time = (6)/3 + (6)/6 = 3 hours.
Ans: (E)
Question: A man walks at a speed of 3 km/hr and runs at a speed of 6 km/hr. How much time will the man require to cover a distance of 12 km, if he completes half of the distance, i.e., (6) km on foot and the other half by running?
Options:
(A) 2.6
(B) 2.2
(C) 2.5
(D) 2
(E) 3
Rationale: Sum of 10 odd no. = 100
Average = 100/10 = 10
The answer: (B)
Question: The average of first 10 odd numbers is?
Options:
(A) 44
(B) 10
(C) 99
(D) 77
(E) 62
Rationale: Option 'E'
The total number of arrangements is
3P3 = 3! = 6
The answer: (E)
Question: In how many different ways can the letters of the word "AIM" be rearrangement?
Options:
(A) 4
(B) 2
(C) 7
(D) 9
(E) 6