SSC CGL 2024 Tier-I · 12 September 2024, Shift 2 · Quantitative Aptitude · Question 4 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 6 September 2026.

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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 ΔDEF, the bisector of ∠ D intersects side EF at point N. If DE = 36 cm, DF = 40 cm and EF = 38 cm, then the length (in cm) of NF is ________.

(a)18
(b)20
(c)12
(d)16

Answer

Why

Correct — B. The internal bisector from D cuts EF in the ratio of the two sides that meet at D.

Angle bisector theorem: EN⁄NF = DE⁄DF
= 36⁄40 = 9⁄10

So EF is split into 9 parts and 10 parts, 19 parts in all:
one part = 38⁄19 = 2 cm
NF = 10 × 2 = 20 cm → option (b)

Sanity check: NF is the piece next to the longer arm DF (40 cm), so it must be the bigger half of EF — anything under 19 cm is wrong before you compute.

Why the others are wrong

(a)
18 — 18 cm is EN, the other piece of EF. The ratio 9 : 10 runs EN : NF, so the 9 parts land on the E side — the trap is attaching the ratio to the wrong segment.
(c)
12 — 12 would split EF as 26 : 12, a ratio of 13 : 6. The bisector forces 36 : 40 = 9 : 10, which 13 : 6 is nothing like.
(d)
16 — 16 splits EF as 22 : 16, that is 11 : 8. Only the split 18 : 20 reduces to the 9 : 10 the two arms at D demand.

Concept

The internal angle bisector theorem: the bisector of an angle divides the opposite side into segments proportional to the two sides containing that angle.

In ΔDEF the bisector from D meets EF at N, so EN⁄NF = DE⁄DF = 36⁄40.

The segment touching the longer arm is the longer segment, which gives you a free check: DF is 40 and DE is 36, so NF > EN and NF must exceed half of 38.

The third side, 38 cm, fixes only the total to be divided — it plays no part in the ratio itself.

No figure is supplied and none is needed, but the lettering matters: N lies on EF, so the two segments are EN and NF, and the stem asks for the one adjacent to F. Sketch the triangle before writing the ratio and the wrong-segment error disappears.

Key facts

Angle bisector theorem: the bisector from D divides EF so that EN⁄NF = DE⁄DF.
Here EN : NF = 36 : 40 = 9 : 10 with EF = 38, so the pieces are 18 cm and 20 cm.
The segment adjacent to the longer of the two arms is always the longer segment.
The external bisector of an angle divides the opposite side externally in that same ratio.

Study next

The length of the bisector itself, whose square is DE × DF − EN × NFThe incentre as the point where the three internal bisectors meetThe basic proportionality (Thales) theorem, where a parallel line cuts the sides in the same ratio

Common traps

Giving EN, 18 cm, when the stem asked for NF.
Dividing 38 in the ratio 40 : 36 the wrong way round.
Assuming the bisector also bisects EF, which is true only when DE = DF.

Bisectors return in a different guise at 9 Sep 2024, 16:00, Quant Q.13, where the bisectors of ∠Q and ∠R meet inside the triangle at O and ∠QOR = 107° fixes ∠P through ∠QOR = 90° + ∠P⁄2, giving 34°. Read the stem for which bisector fact it wants — the side-ratio one here, the incentre-angle one there.

Related PYQs

No directly related past PYQ was found.