SSC CGL 2024 Tier-I · 11 September 2024, Shift 3 · 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.

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.

Naman bought few apples for ₹720 from a shop. He negotiated the price and the shopkeeper reduced it by ₹2 per apple. Due to this Naman could buy four more apples than what he had bought earlier. How many apples did he originally buy?

(a)36
(b)44
(c)40
(d)48

Answer

Why

Correct — A. Let n be the number of apples he first bought. The spend is fixed at ₹720, so the price per apple is 720⁄n before and 720⁄(n + 4) after.

The price fell by ₹2:
720⁄n − 720⁄(n + 4) = 2

720(n + 4 − n) = 2n(n + 4)
2880 = 2n² + 8n
n² + 4n − 1440 = 0

(n + 40)(n − 36) = 0, and a negative count is impossible, so n = 36 → option (a).

Check: 720⁄36 = ₹20 and 720⁄40 = ₹18, a fall of exactly ₹2.

Why the others are wrong

(b)
44 — 44 does not give a ₹2 fall. 720⁄44 is about ₹16.36 and 720⁄48 is ₹15, a drop of roughly ₹1.36. The stem fixes the drop at ₹2 exactly.
(c)
40 — 40 is the number he ended up with, not the number he started with. 720⁄40 = ₹18 is the reduced price, and the question asks how many he originally bought.
(d)
48 — 48 makes the apples too cheap to move ₹2. 720⁄48 = ₹15 and 720⁄52 is about ₹13.85, a fall near ₹1.15. The larger n gets, the smaller the possible drop.

Concept

This is the fixed-budget family: total spend stays the same while price per unit and quantity move in opposite directions.

Write price as total ÷ quantity and the condition becomes a difference of two reciprocals, 720⁄n − 720⁄(n + 4) = 2. Clearing the denominators always produces a quadratic.

The numbers are picked so that the quadratic factorises. Here n² + 4n − 1440 splits as (n + 40)(n − 36), and the negative root is thrown away because a count cannot be negative.

Back-solving from the options is faster than the algebra if you check the right thing: divide 720 by the option and by option + 4, and keep the pair that differs by exactly ₹2.

Key facts

With spend fixed at ₹720, the price per apple is 720⁄n, so a price change is a difference of reciprocals.
n² + 4n − 1440 = 0 factorises as (n + 40)(n − 36).
Originally: 36 apples at ₹20 each, costing ₹720.
After the ₹2 cut: 40 apples at ₹18 each, still ₹720.

Study next

Fixed-budget problems where a price change alters the quantity boughtForming and factorising a quadratic from a difference of reciprocalsFaulty-weight variants of unit-price arithmetic

Common traps

Solving for the new count and answering 40, which is the value after the reduction.
Writing the difference the other way round as 720⁄(n + 4) − 720⁄n = 2, which has no positive root.
Testing options by price alone without checking that the gap is exactly ₹2.

The story changes and the algebra does not: spend fixed, price per unit changed, quantity moved. Read the stem to see which of the two counts is being asked for.

A neighbouring variant sits at 19 Sep 2024, 09:00, Quant Q.1, where a vendor cuts ₹10 per kg to ₹7.2 per kg and also weighs 900 g as a kilogram, so both the price and the unit shift at once.

Related PYQs

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