CSAT Solved Papers/ 2025/Q45

2025 CSAT — Q45

Quant Arithmetic & numeracy 2.5 marks Hard

A set $(X)$ of $20$ pipes can fill $70\%$ of a tank in $14$ minutes. Another set $(Y)$ of $10$ pipes fills $\tfrac{3}{8}$ th of the tank in $6$ minutes. A third set $(Z)$ of $16$ pipes can empty half of the tank in $20$ minutes. If half of the pipes of set $X$ are closed and only half of the pipes of set $Y$ are open, and all pipes of the set $(Z)$ are open, then how long will it take to fill $50\%$ of the tank?

A 8 minutes
B 10 minutes
C 12 minutes
D 16 minutes Answer

Worked rationale

Get per-pipe rates (tank per minute), then scale by how many run.

Set X: $20$ pipes fill $0.70$ in $14$ min $\Rightarrow$ $20$ -pipe rate $= \tfrac{0.70}{14} = 0.05$ ; per pipe $= 0.0025$ .
Set Y: $10$ pipes fill $\tfrac38$ in $6$ min $\Rightarrow$ $10$ -pipe rate $= \tfrac{3/8}{6} = \tfrac{1}{16} = 0.0625$ ; per pipe $= 0.00625$ .
Set Z (empties): $16$ pipes empty $0.5$ in $20$ min $\Rightarrow$ $16$ -pipe rate $= \tfrac{0.5}{20} = 0.025$ (draining).

Now apply the scenario — half of X, half of Y, all of Z:

\underbrace{10 \times 0.0025}_{X:\,0.025} \;+\; \underbrace{5 \times 0.00625}_{Y:\,0.03125} \;-\; \underbrace{0.025}_{Z, \text{ all }16} \;=\; 0.03125 \text{ tank/min (net fill).}

The X-fill exactly cancels the Z-drain, leaving the half-Y rate. Time to fill $50\% = 0.5$ :

\frac{0.5}{0.03125} = 16 \text{ minutes.}

Answer: (d) 16 minutes.

Why the other options miss

A

a count slip: forgets to halve $Y$ (uses all $10$ Y-pipes $= 0.0625$ ), doubling the net rate and halving the time.
B

the drain treated as a fill: treats $Z$ as filling (adds it) or omits $Z$ entirely, inflating the net rate.
C

a scaling slip: mis-scales one set (e.g. keeps all $20$ X-pipes), landing on a plausible middle value.

Specialist insight

The discipline that makes multi-set pipe problems tractable: reduce everything to a per-pipe rate first, then multiply by the active count and sign (fill $+$ , empty $-$ ). The numbers are engineered so the active X-fill ( $0.025$ ) exactly cancels the full Z-drain ( $0.025$ ) — spotting that cancellation collapses the whole problem to “how long does half of Y take?” The deadly slips are arithmetic of count: forgetting to halve a set (a) or mis-signing the drain (b). Keep a tidy table of {set, per-pipe rate, active count, sign} and the net rate writes itself.

The trap, in one line

Halve X and Y, keep all of draining Z (sign $-$ ): X-fill cancels Z-drain, net $= 0.03125$ /min, so $0.5/0.03125 = 16$ min.

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