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Quantitative Aptitude

GATE CSE 2024 Set 1 · Q2

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If two distinct non-zero real variables xx and yy are such that (x+y)(x + y) is proportional to (x−y)(x - y) then the value of xy\frac{x}{y}

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Official answer: D) is a constant

Quantitative Aptitude

Given that (x+y)(x + y) is directly proportional to (x−y)(x - y):

(x+y)=k(x−y)(x + y) = k(x - y)

where kk is a non-zero proportionality constant (k≠1k \neq 1, since y≠0y \neq 0).

Rearranging the terms:

x+y=kx−kyx + y = kx - ky
y+ky=kx−xy + ky = kx - x
y(1+k)=x(k−1)y(1 + k) = x(k - 1)

Dividing both sides by y(k−1)y(k - 1) (since xx and yy are distinct, x−y≠0  ⟹  k≠1x - y \neq 0 \implies k \neq 1):

xy=1+kk−1\frac{x}{y} = \frac{1 + k}{k - 1}

Since kk is a fixed constant, 1+kk−1\frac{1 + k}{k - 1} is also a fixed constant.

Therefore, the value of xy\frac{x}{y} is a constant.

Hence, D is the correct option.

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This Quantitative Aptitude previous year question is Q2 of the GATE CSE 2024 Set 1 paper. Find more GATE CSE PYQs on Quantitative Aptitude in General Aptitude, or browse all GATE CSE PYQs.

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Quantitative Aptitude in GATE CSE

Quantitative Aptitude has 53 GATE CSE PYQs across 13 papers from 2011 to 2026. 26 of them are 1-mark MCQs like this one, and GATE CSE 2024 Set 1 has 5.

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GateAI solution cardGATE CSE 2024 Set 1 · Q2 · MCQ

If two distinct non-zero real variables x and y are such that (x + y) is proportional to (x - y) then the value of (x)/(y)

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.