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UPSC Prelims 2009 Q105: Science & Technology – Quantitative Aptitude

UPSC Prelims 2009, General Studies Paper I · Science & Technology

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?
  1. (a) 4
  2. (b) 5
  3. (c) 6
  4. (d) 7

Answer: (d) 7

Explanation

Which value of n cannot distribute 240 balls under the given rule

Boxes B1 to Bn hold a decreasing sequence of balls, each 4 more than the next, summing to 240.

StepReasoning
SetupLet the last box Bn hold x balls; the sequence is x, x+4, x+8, ..., up to n terms
Sum formulaSum = n·x + 4·(0+1+...+(n−1)) = n·x + 2n(n−1) = 240
Test n=44x + 24 = 240 → x = 54 (valid, positive integer)
Test n=55x + 40 = 240 → x = 40 (valid)
Test n=66x + 60 = 240 → x = 30 (valid)
Test n=77x + 84 = 240 → x = 156/7 = 22.28... (NOT a whole number — invalid)

Why n=7 fails:

PointDetail
Box contents must be whole numbersn=7 gives a non-integer value for the smallest box
n=4, 5, 6All give valid whole-number solutions

Answer: (d) 7.

Why each option is right or wrong

Key point

Testing each value of n in the equation n·x + 2n(n−1) = 240, only n=7 fails to produce a whole-number box count — making it the impossible value.

Remember

240 balls, boxes decreasing by 4 each: n=7 is the value that does NOT give a whole-number solution.

Common mistakes

Exam tip

For 'decreasing sequence totaling X' problems, test each candidate n in the sum formula and check whether the result is a valid whole number.

Sources

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