SSC CGL 2024 Tier-I · 25 September 2024, Shift 3 · Quantitative Aptitude · Question 8 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

‘x’ varies directly with the cube of ‘y’, and inversely with the square of ‘z’. If x = 1⁄36 when y = 2 and z = 3, then what is the value of 800x when y = 3 and z = 5?

Answer

Why

Correct — C. The stem figure says x varies directly with the cube of y and inversely with the square of z, with x = 1⁄36 when y = 2 and z = 3, and asks for 800x when y = 3 and z = 5.

Write the joint variation with one constant:
x = k·y³ ⁄ z²

Fix k from the given point:
1⁄36 = k × 8⁄9, so k = 9 ⁄ (36 × 8) = 1⁄32

Apply it at y = 3, z = 5:
x = (1⁄32) × 27⁄25 = 27⁄800
800x = 800 × 27⁄800 = 27 → option (c)

Why the others are wrong

(a)
800⁄9 is not reachable from this data. With k = 1⁄32 the value is x = 27⁄800 exactly, so 800x = 27, a whole number — the 800 in the question cancels the 800 in x rather than surviving into the answer.
(b)
9⁄800 is a value of x, not of 800x. It is what you get by reading y³ as y², since (1⁄32) × 9⁄25 = 9⁄800, and even that slip would give 9 once multiplied by 800 as the question asks.
(d)
9 is the cube read as a square. Keeping k = 1⁄32 but substituting y² = 9 instead of y³ = 27 gives 800x = 9. The stem says the cube of y, so 3³ = 27 belongs in the numerator.

Concept

Joint variation collapses into one equation. Directly with the cube of y puts y³ in the numerator, inversely with the square of z puts z² in the denominator, and a single constant k absorbs everything else.

x = k·y³ ⁄ z²

The given data point exists only to fix k, and k is then reused unchanged. Here 1⁄36 = k × 2³⁄3² = 8k⁄9, so k = 1⁄32.

At the second point x = (1⁄32)(27⁄25) = 27⁄800.

The 800 is not decoration. SSC asks for 800x rather than x precisely because 800 is the denominator the working lands on, turning an awkward fraction into the integer 27.

Read the last line of the stem before you start solving — the quantity asked for is often one step past the one you are computing.

Key facts

Varies directly with the cube of y means x contains y³ as a factor.
Varies inversely with the square of z means x contains 1⁄z² as a factor.
Combining them, x = k·y³ ⁄ z², and 1⁄36 = k × 8⁄9 fixes k at 1⁄32.
At y = 3 and z = 5 that gives x = 27⁄800, so 800x = 27.

Study next

Setting up direct, inverse and joint variation from the sentence that states itVariation where the constant carries a unit, as in cost varying as the square of weightRatio form of variation, x₁ ⁄ x₂ = (y₁ ⁄ y₂)³ × (z₂ ⁄ z₁)², which skips k altogether

Common traps

Answering x = 27⁄800 when the question asks for 800x.
Swapping the powers and using y² or z³.
Recomputing k from the second data point, which is the unknown being solved for.

SSC states the variation in words, hides one power inside cube or square, then asks for a multiple of x so the answer comes out whole. Variation is also set at 17 Sep 2024, 16:00, Quant Q.18, at 19 Sep 2024, 09:00, Quant Q.11 and at 09 Sep 2024, 12:30, Quant Q.22.

Related PYQs

No directly related past PYQ was found.