SSC CGL 2024 Tier-I · 26 September 2024, Shift 2 · Quantitative Aptitude · Question 20 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.

In ΔABC, DE ∥ BC and 5AE = 3EC. If AB = 6.4 units, then the value of DB (in units) is:

(a)3.2
(b)5
(c)4
(d)2.4

Answer

Why

Correct — C. The stem reaches you as an image of one line of text: in ΔABC, DE ∥ BC and 5AE = 3EC, with AB = 6.4 units, and DB is asked for.

Rule: DE ∥ BC ⇒ AD⁄DB = AE⁄EC — the Basic Proportionality Theorem.

5AE = 3EC → AE⁄EC = 3⁄5
So AD⁄DB = 3⁄5, and D cuts AB in the ratio 3 : 5
DB = 6.4 × 5⁄(3 + 5) = 6.4 × 5⁄8

DB = 4 units → option (c), with AD = 2.4 making up the 6.4

Why the others are wrong

(a)
3.2 — 3.2 units is half of 6.4 — the answer if AE = EC. But 5AE = 3EC makes AE the smaller piece, so D cannot sit at the midpoint of AB.
(b)
5 — 5 units is the coefficient lifted straight out of '5AE', not a length. AD and DB have to add to 6.4, and DB = 5 would leave AD = 1.4, nowhere near 3⁄5 of 5.
(d)
2.4 — 2.4 units is AD, the piece next to A, from 6.4 × 3⁄8. The question asks for DB, the piece next to B — the two segments have been swapped.

Concept

A line drawn parallel to one side of a triangle cuts the other two sides in the same ratio. That is the Basic Proportionality (Thales) Theorem.

DE ∥ BC therefore forces AD⁄DB = AE⁄EC, and the ratio on AC transfers whole to AB. You never need AE and EC themselves, only how they compare.

Turning the stated equation into a ratio is the step most often fumbled: 5AE = 3EC means AE⁄EC = 3⁄5, because the larger multiplier sits with the smaller segment.

No diagram is attached — the row's figure is the sentence itself, set as a picture. D lies on AB and E on AC, the arrangement in which DE ∥ BC, AD + DB = AB and AE + EC = AC all hold together.

Key facts

Basic Proportionality Theorem: a line parallel to one side of a triangle divides the other two sides in the same ratio.
5AE = 3EC rearranges to AE⁄EC = 3⁄5, so the bigger coefficient marks the smaller segment.
With AD : DB = 3 : 5 and AB = 6.4, the parts are AD = 2.4 and DB = 4.
The converse holds too: if AD⁄DB = AE⁄EC, then DE must be parallel to BC.

Study next

The converse of the Basic Proportionality Theorem, used to prove two lines parallelThe midpoint theorem as the special case AD = DBΔADE ~ ΔABC and the ratio of the two areas

Common traps

Flipping 5AE = 3EC into AE⁄EC = 5⁄3 instead of 3⁄5.
Using AD⁄AB = 3⁄5 when the theorem gives AD⁄DB = 3⁄5, which needs the parts summed to 8 first.
Computing AD = 2.4 correctly and then answering it, when DB was asked.

SSC states the split as an equation such as 5AE = 3EC and lets the ratio conversion do the real work.

The same parallel-line proportionality turns up as the midpoint case at 25 Sep 2024, 16:00, Quant Q.25, and as a similar-triangle side ratio at 10 Sep 2024, 12:30, Quant Q.5.

Related PYQs

No directly related past PYQ was found.