CSAT Solved Papers/ 2025/Q10

2025 CSAT — Q10

Quant Arithmetic & numeracy 2.5 marks Medium

The price (p)(p) of a commodity is first increased by k%k\%; then decreased by k%k\%; again increased by k%k\%; and again decreased by k%k\%. If the new price is qq, then what is the relation between pp and qq?

  1. A p(104−k2)2=qimes108p(10^4 - k^2)^2 = q imes 10^8 Answer
  2. B p(104−k2)2=qimes104p(10^4 - k^2)^2 = q imes 10^4
  3. C p(104−k2)=qimes104p(10^4 - k^2) = q imes 10^4
  4. D p(104−k2)=qimes108p(10^4 - k^2) = q imes 10^8

Worked rationale

Percentage changes multiply, they do not add. A +k%+k\% then −k%-k\% pair multiplies the price by

(1+k100)(1−k100)=1−k2104=104−k2104.\left(1 + \tfrac{k}{100}\right)\left(1 - \tfrac{k}{100}\right) = 1 - \frac{k^2}{10^4} = \frac{10^4 - k^2}{10^4}.

There are two such up-down pairs, so

q=p(104−k2104)2=p (104−k2)2108.q = p\left(\frac{10^4 - k^2}{10^4}\right)^2 = p\,\frac{(10^4 - k^2)^2}{10^8}.

Clear the denominator: q×108=p(104−k2)2q \times 10^8 = p(10^4 - k^2)^2.

Answer: (a) p(104−k2)2=q×108p(10^4 - k^2)^2 = q \times 10^8.

Why the other options miss

  • B
    an arithmetic slip: squares the numerator (104−k2)2(10^4-k^2)^2 correctly but forgets to square the denominator 10410^4, landing on 10410^4 instead of 10810^8.
  • C
    counted one pair, not two: applies the up-down multiplier once, ignoring that the cycle runs twice.
  • D
    mismatched numerator and denominator: one factor of (104−k2)(10^4-k^2) but a 10810^8 scale — an inconsistent mix of the one-pair numerator with the two-pair denominator.

Specialist insight

The single most-tested percentage trap is treating +k%+k\% and −k%-k\% as cancelling. They do not: the net of one pair is a loss of k2104\tfrac{k^2}{10^4} (a difference of squares), and the price never returns to pp. The exam-fast route is to write the multiplier (1−k2104)\left(1-\tfrac{k^2}{10^4}\right) per pair and count the pairs (here two), then square. Keep the bookkeeping symbolic — squaring 104−k2104\tfrac{10^4-k^2}{10^4} both top and bottom gives 10810^8 in the denominator, the line that separates (a) from every distractor. Two disciplines win: multiply don’t add, and square the whole fraction.

The trap, in one line

+k%+k\% then −k%-k\% does not cancel — each pair multiplies by 104−k2104\tfrac{10^4-k^2}{10^4}, and squaring the fraction puts 10810^8 (not 10410^4) on the right.

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