Reasoning: Hint: y^2-5 (y is an integer) is never multiple of 3 (try to prove this), hence 30 is out.
Ans: (B)
Question: For some integer y, y^2 - 5 is divisible by all of the following EXCEPT
Options:
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41
Rationale: There are altogether three arrangements, which are we can select two members each from team A and B, each from team A and C, and each from team B and C. For each arrangement, there are altogether 25 possibilities (5*5) since each team has 5 players. So there are a total of 75 possibilities (25*3).
The answer: (C)
Question: There are 3 teams each with 5 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?
Options:
(A) 50
(B) 23
(C) 75
(D) 28
(E) 45
Reasoning: Average of the first 'n' natural numbers = (n + 1)/2
We get (19 + 1)/2 = 10
Ans: (C)
Question: Find the average of the first 19 natural numbers?
Options:
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
Rationale: The simple logic is to divide total balls into three sets. Weigh any two sets against each other(these two must contain equal number of balls). From this we can identify which set contains the heavy ball. Divide this set into three and repeat until you find the heavier ball. Under this logic, the minimum number of weighings required turns out to be the smallest integer greater than or equal to log(n), where n is the total number of balls and the base of logarithm is 3. Or simply [log(n)/log(3)] with any base. Here, n = 6561. log 6561 / log 3 = 8
The answer: (B)
Question: There are 6561 balls out of them 1 is heavy.Find the min. no. of times the balls have to be weighed for finding out the haevy ball.
Options:
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Reasoning: Average of 20 numbers = 0
Sum of 20 numbers = (0 * 20) = 0
It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (-a).
Ans: (D)
Question: The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
Options:
(A) 22
(B) 28
(C) 29
(D) 20
(E) 21
Reasoning: Total distance = 8 Km
Total time = 50 Km
Time taken to cover the distence of 3 Km = 50*2/5 = 20 min = 1/3 hours
Remaining distance = 8 - 3 = 5 Km
Required speed = 5/1/3 = 15 Kmph
Ans: (B)
Question: Avinash covers a distance of 8km in 50 minutes. If he covers 3km distance in 2/5th of time then, What speed should he maintain to cover the remaining distance in the remaining time?
Options:
(A) 10 Kmph
(B) 15 Kmph
(C) 25 Kmph
(D) 27 Kmph
(E) 30 Kmph