SSC CGL 2025 Tier-I · 13 September 2025, Shift 2 · Quantitative Aptitude · Question 1 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 13 September 2026.

Where this paper and its answers come from

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 value of 4 + 44 + 444 + 4444 + 44444?

(a)49380
(b)42300
(c)41300
(d)45000

Answer

Why

Correct — A. Add the terms one at a time and keep a running total.

4 + 44 = 48
48 + 444 = 492
492 + 4444 = 4936
4936 + 44444 = 49380 → option (a)

Check: every term is 4 times a string of 1s, so the sum is 4 × (1 + 11 + 111 + 1111 + 11111) = 4 × 12345 = 49380.

Why the others are wrong

(b)
42300 — It is below 48,888, which is 44444 + 4444, the two largest terms alone. A sum of five positive terms cannot be smaller than two of them added together.
(c)
41300 — It ends in 00, but the units column is five 4s = 20 and the tens column is four 4s plus the carried 2 = 18, so the total must end in 80.
(d)
45000 — A round figure that falls short: 44444 + 4444 already makes 48,888 before 4, 44 and 444 are added.

Concept

Each term is 4 times a repunit, a number written only with 1s: 44 = 4 × 11, 444 = 4 × 111, and so on.

The repunits 1 + 11 + 111 + 1111 + 11111 add to 12345. Their units column holds five 1s, the tens four, the hundreds three, the thousands two and the ten-thousands one, and no column carries.

So the whole sum is 4 × 12345 = 49380.

Column addition gives the same total, and the last two digits alone separate the options.

Units: five 4s = 20 → write 0, carry 2
Tens: four 4s + 2 = 18 → write 8, carry 1
Hundreds: three 4s + 1 = 13 → write 3, carry 1
Thousands: two 4s + 1 = 9, then ten-thousands: 4

Option (a) is the one choice ending in 80.

Key facts

d + dd + ddd + … = d × (1 + 11 + 111 + …), so 4 + 44 + 444 + 4444 + 44444 = 4 × 12345.
For up to nine terms, 1 + 11 + 111 + … writes the digits 1, 2, 3 … in order: 1 + 11 + 111 = 123.
A total of positive terms is always larger than any part of it: here 44444 + 4444 = 48,888 is a floor.

Study next

Decimal versions such as 0.4 + 0.44 + 0.444, which factor the same way: 4 × (0.1 + 0.11 + 0.111)Last-digit and last-two-digit checks to eliminate options before a full calculationColumn addition with carries for long sums

Common traps

Dropping the carry from the units column: the tens column is four 4s plus 2 = 18, and without the 2 it reads 16, turning the tens digit into 6.
Choosing the nearest round option by estimate without checking a lower bound such as 44444 + 4444.

Here the sum is printed in full and the options sit thousands apart, so a floor from the two largest terms, or the last two digits, settles it before the full addition.

Related PYQs

No directly related past PYQ was found.

Topic: Basic algebraic identities of School Algebra & Elementary surds — all 182 SSC CGL questions on it, every solved shift.