SSC CGL 2024 Tier-I · 9 September 2024, Shift 1 · Quantitative Aptitude · Question 9 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.

R pays ₹100 to P with ₹5, ₹2 and ₹1 coins. The total number of coins used for paying are 40. What is the number of coins of denomination ₹5 in the payment?

(a)13
(b)17
(c)16
(d)18

Answer

Why

Correct — A. Two equations, then a bound that removes three options.

Let f, t and o be the numbers of ₹5, ₹2 and ₹1 coins.
Coins: f + t + o = 40
Value: 5f + 2t + o = 100

Subtract the first from the second — the ₹1 coins drop out:
4f + t = 60

A coin count cannot be negative, so t = 60 − 4f ≥ 0 gives f ≤ 15.
That rules out 16, 17 and 18 immediately.

Check f = 13: t = 60 − 52 = 8 and o = 40 − 13 − 8 = 19.
Value 5(13) + 2(8) + 1(19) = 65 + 16 + 19 = 100, using 13 + 8 + 19 = 40 coins.

So the number of ₹5 coins is 13 — option (a).

Why the others are wrong

(b)
17 — 4f + t = 60 forces t = 60 − 68 = −8 two-rupee coins, which is impossible. Every value above 15 breaks the same bound.
(c)
16 — Sixteen ₹5 coins come to ₹80, leaving ₹20 to be paid with the other 24 coins — but 24 coins of ₹2 and ₹1 are worth at least ₹24 between them.
(d)
18 — Eighteen ₹5 coins already come to ₹90, leaving ₹10 to be made from the remaining 22 coins, whose minimum possible value is ₹22.

Concept

Three unknowns and two equations is normally underdetermined. What closes the gap here is that the unknowns are counts — whole numbers, none of them negative.

Subtracting the coin-count equation from the value equation clears the ₹1 coins altogether and leaves the single relation 4f + t = 60.

From there t ≥ 0 caps f at 15, and only one of the four options survives.

The two equations by themselves do not pin f down: f = 12, 13 and 14 each give a payment that uses all three denominations — (12, 12, 16), (13, 8, 19) and (14, 4, 22). It is the option list that fixes the answer at 13, so read the options before hunting for a unique solution.

Key facts

Subtracting the count equation from the value equation eliminates the ₹1 coins and leaves 4f + t = 60.
Because t cannot be negative, the number of ₹5 coins is at most 15.
13 five-rupee coins, 8 two-rupee coins and 19 one-rupee coins make ₹100 in exactly 40 coins.

Study next

Linear equations in whole numbers where a non-negativity bound does the selectingTwo-denomination coin problems solved by the ratio of the coin valuesEliminating a variable by subtracting a count equation from a value equation

Common traps

Hunting for a unique algebraic solution when the system has several and the options decide
Forgetting that the ₹1 coins carry value too, and treating the payment as if only fives and twos mattered

The stem hides a bound inside a story. The algebra is a single subtraction, and the real work is noticing that three of the four options make a coin count negative before any solving starts.

Related PYQs

No directly related past PYQ was found.