UPSC Prelims 2009 · General Studies
Question 105: Basic Numeracy
There are 240 balls and n number of boxes B1, B2, B3, … , Bn. The balls are to be placed in the boxes such that B1 should contain 4 balls more than B2, B2 should contain 4 balls more than B3, and so on. Which one of the following cannot be the possible value of n?
AnswerThe answer is (d).
Let the last box hold a balls. The boxes then hold a, a+4, a+8, ... up to a + 4(n-1), so the total is na + 2n(n-1) = 240. For n = 4, a = 54; for n = 5, a = 40; for n = 6, a = 30. For n = 7, 7a = 240 - 84 = 156, which is not a multiple of 7. So n = 7 is not possible.
Why each option is right or wrong
- (a) n = 4 works with 54, 58, 62, 66 balls.
- (b) n = 5 works with 40, 44, 48, 52, 56 balls.
- (c) n = 6 works with 30, 34, 38, 42, 46, 50 balls.
- (d) Right: n = 7 would need a = 156/7, not a whole number.
Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2009, General Studies. UPSC's own paper isn't online; the words are from a published compilation. The paper on upsc.gov.in.