SSC CGL 2024 Tier-I · 17 September 2024, Shift 2 · Quantitative Aptitude · Question 3 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 6 September 2026.

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For what value of k, the system of equations 4x + 12y + 36 = 0 and 5x + ky + 45 = 0 has an infinite number of solutions?

(a)22
(b)20
(c)25
(d)15

Answer

Why

Correct — D. For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the solutions are infinite when all three coefficient ratios agree.

a₁/a₂ = 4/5
c₁/c₂ = 36/45 = 4/5 — the constants already match

So the y-coefficients must match too:
12/k = 4/5
4k = 60 → k = 15 → option (d)

Check: 5x + 15y + 45 = 0 is exactly 1.25 × (4x + 12y + 36 = 0), so the two lines coincide.

Why the others are wrong

(a)
22 — k = 22 gives 12/22 = 6/11, about 0.55, well short of 0.8. Put back in, 5x + 22y + 45 = 0 is not a multiple of the first equation.
(b)
20 — k = 20 gives 12/20 = 3/5. The x-ratio and the constant ratio would agree while the y-ratio did not, so the lines cross at one point instead of coinciding.
(c)
25 — k = 25 gives 12/25 = 0.48, far from 0.8. A k that large would need the first equation to carry 20y rather than 12y.

Concept

Two linear equations in x and y are two lines. Compare the ratios a₁/a₂, b₁/b₂ and c₁/c₂.

All three equal → the equations are multiples of each other, the lines coincide, and there are infinitely many solutions.

a₁/a₂ = b₁/b₂ but c₁/c₂ different → the lines are parallel and distinct, so there is no solution.

a₁/a₂ ≠ b₁/b₂ → the lines meet once and there is exactly one solution.

Both equations must be in the same arrangement, ax + by + c = 0, before the ratios are compared.

Moving the constant to the right in only one of them flips its sign and the c-ratio test then misfires.

Key facts

Infinitely many solutions require a₁/a₂ = b₁/b₂ = c₁/c₂.
Equal a and b ratios with a different c ratio give parallel distinct lines and no solution.
Unequal a and b ratios give one intersection point and a unique solution.
Here 4/5 = 12/15 = 36/45, so 5x + 15y + 45 = 0 is 1.25 times 4x + 12y + 36 = 0.

Study next

The no-solution condition and how it differs from the coincident-lines conditionSolving a pair of linear equations by cross-multiplicationPairs where the unknown sits in the constant term as well as in a coefficient

Common traps

Checking only a₁/a₂ = b₁/b₂ and forgetting the constant ratio, which is the no-solution case
Inverting the ratio, solving 12/k = 4/5 as 12 × 4/5 = 9.6 instead of 12 × 5/4 = 15
Reading infinite number of solutions as no solution under time pressure

SSC keeps the stem identical and moves the unknown around the pair.

The same condition with K inside the y-coefficient runs at 24 Sep 2024, 12:30, Quant Q.11, on x + 2Ky − 8 = 0 and 2x − y − 16 = 0.

Related PYQs

No directly related past PYQ was found.