CSAT Solved Papers/ 2021/Q68

2021 CSAT — Q68

Quant Arithmetic & numeracy 2.5 marks Easy

A cubical vessel of side 11 m is filled completely with water. How many millilitres of water is contained in it (neglect thickness of the vessel)?

  1. A 1000
  2. B 10000
  3. C 100000
  4. D 1000000 Answer

Worked rationale

Volume of the cube =(1 m)3=1 m3= (1\text{ m})^3 = 1\text{ m}^3. Convert through the metric ladder:

1 m3=1000 litres,1 litre=1000 mL,1\text{ m}^3 = 1000\text{ litres}, \qquad 1\text{ litre} = 1000\text{ mL},

so

1 m3=1000×1000=1,000,000 mL.1\text{ m}^3 = 1000 \times 1000 = 1{,}000{,}000\text{ mL}.

Answer: (d) 1000000.

Why the other options miss

  • A
    stopped one unit early: halts at litres (1 m3=10001\text{ m}^3 = 1000 L) and forgets the litre→\tomL step.
  • B
    a power-of-ten slip: uses one wrong factor of ten in the conversion chain.
  • C
    off by a single factor of ten: e.g. treats 1 L1\text{ L} as 100100 mL.

Specialist insight

The only skill tested is the two-step metric chain: 1 m3=1000 L1\text{ m}^3 = 1000\text{ L} and 1 L=1000 mL1\text{ L} = 1000\text{ mL}, multiplying to 10610^6 mL. Cubic-to-litre is where students drop a power of ten — anchor on 1 m3=10001\text{ m}^3 = 1000 L (a litre is a cubic decimetre, and there are 10310^3 of them in a cubic metre), then the litre→\tomL factor of 10001000 finishes it. No geometry beyond “side 11 gives volume 11.”

The trap, in one line

1 m3=1000 L=1,000,000 mL⇒1\text{ m}^3 = 1000\text{ L} = 1{,}000{,}000\text{ mL} \Rightarrow (d).

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