CSAT Solved Papers/ 2025/Q17

2025 CSAT — Q17

Quant Arithmetic & numeracy 2.5 marks Medium

The petrol price shot up by 10%10\% as a result of the hike in crude oil prices. The price of petrol before the hike was ₹ 9090 per litre. A person travels 22002200 km every month and his car gives a mileage of 1616 km per litre. By how many km should he reduce his travel if he wants to maintain his expenditure at the previous level?

  1. A 180 km
  2. B 200 km Answer
  3. C 220 km
  4. D 240 km

Worked rationale

Expenditure == (price per litre) ×\times (litres used), and litres used == distance // mileage. Hold the rupee budget fixed and let the distance fall.

Old monthly fuel =220016=137.5= \dfrac{2200}{16} = 137.5 litres; old cost =137.5×90=12375= 137.5 \times 90 = 12375 (in ₹).

New price =90×1.10=99= 90 \times 1.10 = 99 per litre. To keep the cost at ₹ 1237512375, the litres he can now buy are

1237599=125 litres ⇒ distance=125×16=2000 km.\frac{12375}{99} = 125 \text{ litres} \ \Rightarrow\ \text{distance} = 125 \times 16 = 2000 \text{ km.}

Reduction =2200−2000=200= 2200 - 2000 = 200 km.

Shortcut: a 10%10\% price rise means he can afford 11.1=1011\tfrac{1}{1.1} = \tfrac{10}{11} of the old distance, i.e. 1011×2200=2000\tfrac{10}{11}\times 2200 = 2000 km — same 200200 km cut, no rupee figures needed.

Answer: (b) 200 km.

Why the other options miss

  • A
    wrong base for the cut: cuts distance by 10%10\% of the wrong quantity or applies the reduction to litres (137.5×0.1≈13.75137.5 \times 0.1 \approx 13.75 L →\to a too-small km figure).
  • C
    wrong base for the cut: reduces by a flat 10%10\% of 22002200 (=220= 220), treating the distance cut as 10%10\% of the old distance rather than the 111\tfrac1{11} that actually holds the budget.
  • D
    an arithmetic slip: over-corrects (e.g. uses 12%12\% or mis-divides 12375/9912375/99), inflating the cut.

Specialist insight

The clean lens is “budget fixed, so quantity scales by the reciprocal of the price ratio.” A +10%+10\% price means affordable quantity becomes 11.1=1011\tfrac{1}{1.1} = \tfrac{10}{11} of before — not 90%90\%. That single distinction kills the deadliest distractor (c) 220220, which is just ”10%10\% of 22002200.” The reciprocal route (1011×2200=2000\tfrac{10}{11}\times 2200 = 2000) avoids every rupee computation and is far faster under the clock. Mileage and price are both fixed, so they only set the conversion; the live variable is distance, and it moves as 11+r\tfrac{1}{1+r}.

The trap, in one line

A +10%+10\% price lets you afford 1011\tfrac{10}{11} of the distance (cut =200= 200 km), not 90%90\% (which gives the planted 220220).

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