SSC CGL 2024 Tier-I · 13 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.

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.

The solution of the following equations is: 5x + 4y - 8z = 1 7x - 9y + z = - 1 2x + 3y - 4z = 1

(a)x = 1,y = 1,z = - 1
(b)x = 1,y = 1,z = 1
(c)x = - 1,y = 1,z = 1
(d)x = 1,y = - 1,z = 1

Answer

Why

Correct — B. Each option fixes all three unknowns, so this is a checking job, not a solving job — and a solution must satisfy all three equations.

Test x = 1, y = 1, z = 1:
5(1) + 4(1) − 8(1) = 5 + 4 − 8 = 1 ✓
7(1) − 9(1) + (1) = 7 − 9 + 1 = −1 ✓
2(1) + 3(1) − 4(1) = 2 + 3 − 4 = 1 ✓

All three right-hand sides are met, so option (b) is the solution.

Why the others are wrong

(a)
x = 1,y = 1,z = - 1 — Flips only the sign of z, and the first equation catches it at once: 5 + 4 − 8(−1) = 5 + 4 + 8 = 17, against a required 1.
(c)
x = - 1,y = 1,z = 1 — With x = −1 the first equation gives 5(−1) + 4 − 8 = −9, not the 1 the paper states. One failed equation is enough to discard a triple.
(d)
x = 1,y = - 1,z = 1 — Fails the second equation: with y = −1 the −9y term turns positive, so 7 + 9 + 1 = 17 where the paper needs −1.

Concept

A triple is a solution only if it satisfies every equation in the system, which means one failed check is enough to reject it.

That is what makes an MCQ system like this a verification task. Full elimination on three unknowns takes several minutes; substituting the four triples into the first equation alone separates them here, because only one of the four returns 1.

Solve properly only when the options do not fix every unknown — when they give a single variable, or an expression such as x + y + z.

All four triples are built from just 1 and −1, so none can be ruled out by inspection. Each has to be pushed through an equation, and the first equation is the cheapest filter because its coefficients are the largest.

Key facts

A candidate solution must satisfy every equation in the system, so failing any one of them rejects it.
For x = 1, y = 1, z = 1 the three left sides evaluate to 1, −1 and 1, matching the stated right sides.
Substituting the options is valid only when each option fixes every unknown in the system.

Study next

Solving three linear equations in three unknowns by eliminationConsistent, inconsistent and dependent systems — when no unique solution existsUsing back-substitution as a deliberate time-saver on multiple-choice systems

Common traps

Checking only the first equation and stopping, which can pass a triple that fails a later one.
Sign slips on the −8z and −9y terms, where a negative value for the variable turns the whole term positive.
Starting elimination out of habit when the options already carry complete solutions.

SSC prints the entire solution inside each option here, which converts the item into four quick substitutions. Quant Q.10 of the 13 Sep 2024, 09:00 paper rewards the same habit — the ₹240 spent on 7 drinks can be tested option by option instead of solved.

Related PYQs

No directly related past PYQ was found.