SSC CGL 2024 Tier-I · 12 September 2024, Shift 1 · Quantitative Aptitude · Question 11 of 99

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.

The average of 12 numbers is 48. The average of the first 5 numbers is 45 and the average of next 4 numbers is 52. If the 10 th number is 10 less than the 11 th number and is 5 more than the 12 th number, then the average of the 11 th and 12 th numbers is:

(a)48.5
(b)47.5
(c)46.5
(d)50.5

Answer

Why

Correct — A. Convert every average into a total first.

Sum of all 12 = 12 × 48 = 576
Sum of first 5 = 5 × 45 = 225
Sum of next 4 = 4 × 52 = 208
Sum of first 9 = 225 + 208 = 433

So the 10th, 11th and 12th together = 576 − 433 = 143.

Let the 10th be x. "10 less than the 11th" makes the 11th x + 10, and "5 more than the 12th" makes the 12th x − 5.

x + (x + 10) + (x − 5) = 143
3x + 5 = 143
x = 46

So the 11th is 56 and the 12th is 41, and their average = (56 + 41)⁄2 = 97⁄2 = 48.5 → option (a).

Why the others are wrong

(b)
47.5 — 47.5 puts the 10th number at 45, since the average of the 11th and 12th is always x + 2.5. But 3(45) + 5 = 140, three short of the 143 the first nine leave.
(c)
46.5 — 46.5 puts the 10th number at 44, and 3(44) + 5 = 137 — six short of 143. It also sits temptingly close to x = 46 itself.
(d)
50.5 — 50.5 puts the 10th number at 48, and 3(48) + 5 = 149, six more than the 143 available.

Concept

Two ideas are chained here. The first is that sum = average × count, which turns three averages into three totals and lets you subtract.

The first 5 and the next 4 account for nine numbers, so 576 − 433 = 143 is the total of the three that are left.

The second is translating the verbal relations into one variable. Naming the 10th number x is the efficient choice, because both relations are stated against it: the 11th becomes x + 10 and the 12th becomes x − 5.

Three unknowns then reduce to one equation, 3x + 5 = 143.

A useful shortcut once x is named: the average of the 11th and 12th is ((x + 10) + (x − 5))⁄2 = x + 2.5. You can test any option straight against that.

Key facts

Sum = average × count, so 12 numbers averaging 48 total 576.
The first 5 and the next 4 cover nine of the twelve numbers, leaving three unaccounted for.
If the 10th number is x, the 11th is x + 10 and the 12th is x − 5, so the average of the last two is x + 2.5.

Study next

Weighted averages of groups of unequal sizeThe assumed-mean (deviation) method for averagesTranslating "A is k less than B" into an equation without flipping the sign

Common traps

Averaging the averages — (45 + 52)⁄2 is meaningless because the two groups have different sizes.
Reversing the relation and making the 11th number x − 10.
Solving correctly to x = 46 and then answering with the 10th number instead of the average asked for.

SSC starts from a grand total that must be split, then hides the last few terms behind a verbal relation you have to translate.

Also asked 11 Sep 2024, 16:00, Quant Q.9, where a class of 30 splits into 12 and 18, and 9 Sep 2024, 12:30, Quant Q.9, where two sixteen-day halves of a 31-day month overlap on the day being asked for.

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