SSC CGL 2024 Tier-I · 25 September 2024, Shift 1 · Quantitative Aptitude · Question 21 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.

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Find the values of 'a' and 'b' for which the system of equations 3x+y=3 and (a-b)x+ (a+b)y=3a+b-3 has infinite solutions.

Answer

Why

Correct — D. Infinite solutions means the two equations describe the same line, so all three coefficient ratios must agree.

For a₁x + b₁y = c₁ and a₂x + b₂y = c₂: a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂
Here: 3⁄(a − b) = 1⁄(a + b) = 3⁄(3a + b − 3)

First two: 3(a + b) = a − b → 2a = −4b → a = −2b
First and third: 3a + b − 3 = 3(a + b) → b − 3 = 3b → b = −3⁄2
Then a = −2 × (−3⁄2) = 3

Check: a − b = 4.5, a + b = 1.5 and 3a + b − 3 = 4.5, so all three ratios are 2⁄3.
a = 3 with b = −3⁄2 is option (d).

Why the others are wrong

(a)
Option (a) shows a = −3⁄2, b = 2 — the two values swapped. Then a − b = −3.5 while a + b = 0.5, so 3⁄(a − b) is negative and 1⁄(a + b) = 2. The ratios do not even share a sign.
(b)
Option (b) shows a = 2 with the correct b = −3⁄2. Then a − b = 3.5 and a + b = 0.5, giving 3⁄3.5 = 6⁄7 against 1⁄0.5 = 2 — the x and y ratios disagree.
(c)
Option (c) shows a = 3 with b = −2⁄3, the fraction turned over. Then 3⁄(a − b) = 3 ⁄ (11⁄3) = 9⁄11 while 1⁄(a + b) = 3⁄7, so the first equality already fails.

Concept

A pair of linear equations in two unknowns falls into exactly three cases, decided by the coefficient ratios.

a₁⁄a₂ ≠ b₁⁄b₂ — one solution, the lines cross.
a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂ — no solution, the lines are parallel.
a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂ — infinitely many, the lines coincide.

Infinite solutions is the strictest case: it demands both equalities. That is why two unknowns can be pinned down here — the ratio condition is really two equations in a and b.

Put the second equation into the form a₂x + b₂y = c₂ before taking any ratio. The constant term is 3a + b − 3, which carries the unknowns as well — treat it as c₂, not as something to be simplified away.

Key facts

Infinite solutions require a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂, all three ratios equal.
Equal first two ratios with a different third ratio means no solution, not infinitely many.
For a = 3 and b = −3⁄2 the common ratio here is 2⁄3.
The pair that satisfies both conditions is a = 3 and b = −3⁄2.

Study next

The no-solution condition a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂, and how one changed constant flips the caseReading the three cases off a graph as intersecting, parallel and coincident linesDeterminant form, where infinite solutions need D = Dx = Dy = 0

Common traps

Using only a₁⁄a₂ = b₁⁄b₂ and stopping, which cannot separate parallel from coincident
Reading 3a + b − 3 as belonging to the left-hand side rather than as the constant
Finding a = −2b from the first pair and then guessing which option fits, instead of using the third ratio

The unknowns are hidden inside the coefficients, so the ratio condition becomes a small simultaneous system in a and b.

The same condition is asked at 12 Sep 2024, 12:30, Quant Q.25, and at 12 Sep 2024, 16:00, Quant Q.4, where the answer wanted is a + b rather than the pair.

Related PYQs

No directly related past PYQ was found.