SSC CGL 2025 Tier-I · 15 September 2025, Shift 2 · Quantitative Aptitude · Question 5 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

What is the average of all numbers between 1000 and 1300 that are divisible by 25?

(a)1150
(b)1200
(c)1575
(d)1500

Answer

Why

Correct — A. Multiples of 25 are evenly spaced, so their average is (first + last) ÷ 2.

Strictly between: first = 1025, last = 1275
Average: (1025 + 1275) ÷ 2 = 2300 ÷ 2 = 1150

Including 1000 and 1300: (1000 + 1300) ÷ 2 = 1150
Either reading gives 1150 → option (a)

Why the others are wrong

(b)
1200 — 1200 is not the centre of 1025 to 1275. The list is symmetric about 1150: 1025 and 1275, 1050 and 1250, and so on, each pair averaging 1150.
(c)
1575 — 1575 is above 1300, the top of the range. Every number averaged is at most 1300, so their average cannot exceed 1300.
(d)
1500 — 1500 is larger than every term. The biggest multiple of 25 in range is 1275, or 1300 if the end is counted, and an average never exceeds the largest value averaged.

Concept

The multiples of 25 form an arithmetic progression: 1025, 1050, …, 1275, rising by 25 each time. In any AP the average is (first term + last term) ÷ 2, because terms pair off from the two ends with equal sums.

Here every pair, 1025 + 1275, 1050 + 1250 and so on, sums to 2300. The number of terms is never needed.

'Between' can mean with or without the end points, and both 1000 and 1300 are multiples of 25. Here it does not matter: the two sit symmetrically about 1150, so adding or dropping them leaves the average unchanged.

Key facts

Average of an arithmetic progression = (first term + last term) ÷ 2.
From 1025 to 1275 there are (1275 − 1025) ÷ 25 + 1 = 11 multiples of 25.
An average always lies between the smallest and largest numbers averaged.

Study next

Sum of an AP: number of terms × (first + last) ÷ 2Counting the multiples of a number in a range, such as multiples of 25 from 1025 to 1275Average of the first n natural numbers: (n + 1) ÷ 2

Common traps

Treating 'between' as a problem when both readings give 1150. It changes a sum or a count, not this average.
Taking 1200 as the round middle of a list that actually runs from 1025 to 1275.

The same (first + last) ÷ 2 step answers 12 Sep 2025, 09:00, Quant Q.7: multiples of 11 between 100 and 250 run from 110 to 242, average 176.

12 Sep 2025, 16:00, Quant Q.10 does it with 13 between 300 and 450: 312 to 442, average 377.

Related PYQs

No directly related past PYQ was found.

Topic: Averages — all 53 SSC CGL questions on it, every solved shift.