CSAT Solved Papers/ 2022/Q57

2022 CSAT — Q57

Quant Arithmetic & numeracy 2.5 marks Medium

$24$ men and $12$ women can do a piece of work in $30$ days. In how many days can $12$ men and $24$ women do the same piece of work?

A 30 days
B More than 30 days
C Less than 30 days or more than 30 days
D Data is inadequate to draw any conclusion Answer

Worked rationale

Let a man’s daily rate be $m$ and a woman’s be $w$ . The work is

W = (24m + 12w)\cdot 30.

The second crew’s time is

d = \frac{W}{12m + 24w} = \frac{30(24m + 12w)}{12m + 24w} = 30\cdot\frac{2m + w}{m + 2w}.

This depends on the ratio $m : w$ , which is never given:

if $m = w$ : $d = 30 \cdot \tfrac{3}{3} = 30$ days;
if $m > w$ (men faster): $\tfrac{2m+w}{m+2w} > 1$ , so $d > 30$ ;
if $m < w$ : $d < 30$ .

Without knowing whether men or women work faster, $d$ cannot be pinned — it can be $=30$ , $>30$ , or $<30$ .

Answer: (d) Data is inadequate to draw any conclusion.

Why the other options miss

A

assumes what isn’t given: silently takes $m = w$ (equal efficiency), which the problem never states.
B

guesses the direction: assumes men are faster, so swapping toward women slows the work — only true under an unstated assumption.
C

misses a case: rules out the $m=w$ tie, wrongly excluding exactly $30$ .

Specialist insight

The crew swaps $12$ men $\leftrightarrow$ $12$ women; the new time is $30\cdot\tfrac{2m+w}{m+2w}$ , whose value hinges entirely on the unknown ratio $m:w$ . Because all three outcomes ( $=30$ , $>30$ , $<30$ ) are attainable, the honest answer is inadequate data — and note that option (c) is wrong precisely because it forgets the $m=w$ tie. This is the classic “looks symmetric, isn’t determined” trap.

The trap, in one line

$d = 30\cdot\tfrac{2m+w}{m+2w}$ depends on the unknown $m{:}w$ — can be $<,=,>30$ $\Rightarrow$ data inadequate $=$ (d).

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