UPSC Prelims 2023 · CSAT
Question 36: Basic Numeracy
A box contains 14 black balls, 20 blue balls, 26 green balls, 28 yellow balls, 38 red balls and 54 white balls. Consider the following statements :
- 1.The smallest number n such that any n balls drawn from the box randomly must contain one full group of at least one colour is 175.
- 2.The smallest number m such that any m balls drawn from the box randomly must contain at least one ball of each colour is 167.
Which of the above statements is/are correct ?
AnswerUPSC’s answer is (c).
Total balls: 14 + 20 + 26 + 28 + 38 + 54 = 180. Statement 1: in the worst case you take one fewer than each group: 13 + 19 + 25 + 27 + 37 + 53 = 174; the next ball completes a group, so n = 175. Statement 2: you can draw the most balls while still missing a colour by leaving out the smallest group, black: 180 - 14 = 166. So m = 167. Both statements are correct.
- 1. Worst case 174 balls with no full group, so 175 guarantees one.
- 2. Worst case 180 - 14 = 166 balls can miss black, so 167 guarantees every colour.
How you could have got this
- If you were sure statement 1 is correct, (b) and (d) go: each of them leaves 1 out.
What you needed to know: Worst case 174 balls with no full group, so 175 guarantees one.
- Then, if you were sure statement 2 is correct, (a) goes: it leaves 2 out.
What you needed to know: Worst case 180 - 14 = 166 balls can miss black, so 167 guarantees every colour.
Only (c) is left. This one needs every statement.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) Wrong: statement 2 is also correct (167).
- (b) Wrong: statement 1 is also correct (175).
- (c) Right: n = 175 and m = 167.
- (d) Wrong: both statements are correct.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2023, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.