A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?
- (a) 147
- (b) 150
- (c) 294 ✓ UPSC's answer
- (d) 300
Why the answer is (c)
• Let the number of pens received by F be $f$. Then A gets $6f$, B gets $3f$, C gets $2f$, D gets $1.5f$, E gets $1.2f$, and F gets $f$.
• For all counts to be integers, $f$ must be a multiple of 2 (to make $1.5f$ and $1.2f$ integers). Let $f = 2k$.
• The counts become: A=$12k$, B=$6k$, C=$4k$, D=$3k$, E=$2.4k$, F=$2k$. For E to be an integer, $2.4k$ must be an integer, so $k$ must be a multiple of 5. Let $k=5m$.
• The counts become: A=$60m$, B=$30m$, C=$20m$, D=$15m$, E=$12m$, F=$10m$.
• The problem requires each child to receive an **even** number of pens. A, B, C, E, and F are already even for any integer $m$. D receives $15m$, which is even only if $m$ is even.
• The minimum positive even value for $m$ is 2. Substituting $m=2$: A=120, B=60, C=40, D=30, E=24, F=20.
• Total pens = $120 + 60 + 40 + 30 + 24 + 20 = 294$.
Why the other options are wrong
- (a) 147
- 147 is the total if the 'even number' constraint is ignored and the minimum integer solution ($m=1$) is used, but D would receive 15 pens (odd).
- (b) 150
- 150 is not a valid total for any integer solution that satisfies the ratio constraints and integer pen counts.
- (d) 300
- 300 is not the minimum; it corresponds to a non-minimal or incorrect scaling of the base ratio.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.