SSC CGL 2024 Tier-I · 17 September 2024, Shift 1 · Quantitative Aptitude · Question 11 of 98

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.

In a collection of rare coins, there is one gold coin for every four non-gold coins. If 20 more gold coins are added to the collection, the ratio of the number of gold coins to that of non-gold coins will be 2:3. The total number of coins in the collection will now become ________.

(a)80
(b)48
(c)60
(d)100

Answer

Why

Correct — A. One gold for every four non-gold fixes gold : non-gold = 1 : 4, so write gold = x and non-gold = 4x. Only the gold side changes.

After the 20 arrive: (x + 20) : 4x = 2 : 3
Cross-multiply: 3(x + 20) = 2(4x)
3x + 60 = 8x → 5x = 60 → x = 12

Gold now 12 + 20 = 32, non-gold 4 × 12 = 48.
Total = 32 + 48 = 80 → option (a).

Why the others are wrong

(b)
48 — 48 is the non-gold count on its own, 4 × 12. It never changed, but the question asks for the whole collection, which also carries the 32 gold coins.
(c)
60 — 60 is the collection before the coins were added — 12 gold and 48 non-gold. The word now in the last line puts the count after the addition.
(d)
100 — 100 is 20 more than the correct total: the 20 new coins counted a second time, on top of the 32 gold that already include them. Once x = 12 is solved, nothing further is added.

Concept

A ratio is a shared multiplier, not a pair of counts. 1 : 4 means x and 4x, and the single unknown x carries the whole collection.

When a fixed number is added to one side, that side becomes x + 20 while the other stays at 4x. The new ratio then gives one linear equation in one unknown — which is why these items are solved by cross-multiplication, not by trial.

The final ask is the total, so the value of x has to be pushed back through both sides before you answer.

'One gold coin for every four non-gold coins' is 1 : 4, not 1 in every 4. Gold is a fifth of the original collection, not a quarter.

Key facts

'One A for every four B' is the ratio A : B = 1 : 4, so a collection of 5x coins holds x gold.
Adding coins of one kind changes that kind's count and the total, and leaves the other kind untouched.
Here x = 12, so the collection began with 12 gold and 48 non-gold, a total of 60.

Study next

Ratio problems where a quantity is removed instead of addedMixtures, where water is poured in to shift a milk-water ratioCompounding two ratios that share a middle term, A : B with B : C

Common traps

Reading 'one for every four' as one in every four, which makes gold a quarter of the whole.
Stopping at x = 12 and answering with a count of gold coins.
Letting the non-gold side change as well, when only gold coins were added.

SSC states the starting ratio in words, changes one side by a fixed number, and then asks for the new total rather than the new ratio — so the last step is the one most often dropped.

Adding to one side of a ratio is also asked 23 Sep 2024, 09:00, Quant Q.20, where water is poured into a 32 l milk-water mixture. Chaining two symbolic ratios into a : b : c is asked 17 Sep 2024, 09:00, Quant Q.3, on A : B = 6 : 8 with B : C = 7 : 10.

Related PYQs

No directly related past PYQ was found.