SSC CGL 2024 Tier-I · 13 September 2024, Shift 1 · Quantitative Aptitude · Question 15 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.

Simplify the following expression. [{70 − 60÷ 2 of 6} + (20− 10)÷ 10] + 600÷ 25 of (2 ×3)× (2 of 3) + 2 of 5.

(a)100
(b)99
(c)90
(d)110

Answer

Why

Correct — A. In the convention SSC's keys use, of is resolved before ÷, so read every x of y as a bracketed product.

First bracket:
2 of 6 = 12, so 60 ÷ 12 = 5, and 70 − 5 = 65
(20 − 10) ÷ 10 = 10 ÷ 10 = 1
[65 + 1] = 66

Second block:
25 of (2 × 3) = 25 of 6 = 150, so 600 ÷ 150 = 4
2 of 3 = 6, so 4 × 6 = 24

Last term: 2 of 5 = 10
66 + 24 + 10 = 100 → option (a)

Why the others are wrong

(b)
99 — 99 is the total with the (20 − 10) ÷ 10 term dropped or read as zero. It is worth exactly 1, so the first bracket falls from 66 to 65.
(c)
90 — 90 is the total with the trailing 2 of 5 = 10 left off. It sits after a multiplication and is the easiest term on the line to lose.
(d)
110 — 110 needs one block to come out ten too high. All three are pinned: the bracket is 66, the ÷ and × chain is 24, the last term is 10.

Concept

Strict BODMAS has no separate rank for the word of. Indian school practice, and SSC's keys with it, treat x of y as a bracket — evaluate it before any division sitting to its left.

You can see that the key relies on it. Read 60 ÷ 2 of 6 left to right as (60 ÷ 2) × 6 = 180 and the first bracket turns negative, and the whole line lands nowhere near any printed option.

So the working order here is: resolve each of, then brackets inside out, then ÷ and × left to right, then + and −.

The expression is long rather than hard. Splitting it into the three blocks the plus signs already mark — the square bracket, the 600 ÷ … chain and the final 2 of 5 — turns it into three small sums that can be checked separately.

Key facts

SSC's keys resolve of before division, so 60 ÷ 2 of 6 is 60 ÷ 12 = 5.
The square bracket evaluates to [{70 − 5} + 1] = 66.
600 ÷ (25 of 6) × (2 of 3) = 4 × 6 = 24.
Adding the final 2 of 5 = 10 gives 66 + 24 + 10 = 100.

Study next

The VBODMAS order and where of sits inside itNested bracket simplification worked from the innermost outwardsSimplification lines that mix decimals with of, such as 21 − 4.9 ÷ 7 + 3.9 × 0.4

Common traps

Reading 60 ÷ 2 of 6 left to right as 30 × 6
Dropping the (20 − 10) ÷ 10 term, which contributes 1
Forgetting the trailing 2 of 5 once the long chain is done

SSC prints one of these as a long single line and rewards a candidate who splits it at the plus signs rather than reading it end to end.

The same of convention is needed at Quant Q.19 of the 26 Sep 2024, 12:30 shift and at Quant Q.23 of the 25 Sep 2024, 12:30 shift, which sets 5 + 3 of (25 − 2 × 10) inside a bracket.

Related PYQs

No directly related past PYQ was found.