CSAT Solved Papers/ 2022/Q47

2022 CSAT — Q47

Quant Arithmetic & numeracy 2.5 marks Hard

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 Answer
D 300

Worked rationale

Let $A$ ‘s count be the pivot. Then $B = \tfrac{A}{2},\ C = \tfrac{A}{3},\ D = \tfrac{A}{4},\ E = \tfrac{A}{5}, \ F = \tfrac{A}{6}$ , all integers, so $A$ must be a multiple of $\operatorname{lcm}(2,3,4,5,6) = 60$ .

Write $A = 60k$ . Then

B = 30k,\ C = 20k,\ D = 15k,\ E = 12k,\ F = 10k.

Each must be even. The only one at risk is $D = 15k$ (odd $\times k$ ), which is even iff $k$ is even. Smallest even $k$ is $k = 2$ :

A=120,\ B=60,\ C=40,\ D=30,\ E=24,\ F=20\ (\text{all even}).

Total $= (60 + 30 + 20 + 15 + 12 + 10)\,k = 147k = 147 \times 2 = 294$ .

Answer: (c) 294.

Why the other options miss

A

stops at “whole numbers,” forgets “even”: takes $k = 1$ (total $147$ ), ignoring that $D = 15$ is odd and breaks the all-even requirement.
B

an arithmetic slip: rounds or mis-sums the share coefficients near $147$ .
D

overshoots the minimum: jumps to $A$ as a multiple of $120$ unnecessarily, past the true smallest total.

Specialist insight

Two layers: $A = \operatorname{lcm}(2,\dots,6)\cdot k = 60k$ makes every share an integer; then the parity constraint bites only on the odd-coefficient share $D = 15k$ , forcing $k$ even. The minimum is $k = 2$ , not $k = 1$ — the $147$ trap catches anyone who stops at integrality and forgets the even-number requirement.

The trap, in one line

$A = 60k$ , but $D = 15k$ even forces $k = 2$ ; total $= 147 \times 2 = 294 \Rightarrow$ (c).

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