GATE 2025 EC GA Q.7: solution and answer

GATE 2025 · ECGA Q.7 · 2 marks · MCQ

A stick of length one meter is broken at two locations at distances of $b_{1}$ and $b_{2}$ from the origin (0), as shown in the figure. Note that $0 < b_{1} < b_{2} < 1$ . Which one of the following is NOT a necessary condition for forming a triangle using the three pieces?

Note: All lengths are in meter. The figure shown is representative.

GATE 2025 EC GA question 7: A horizontal line segment representing the stick, with short vertical tick marks at its left end, at two interior points… — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
$b_{1} < 0.5$
(B)
$b_{2} > 0.5$
(C)
$b_{2} < b_{1} + 0.5$
(D)
$b_{1} + b_{2} < 1$

Answer and solution

ANSWER

(D) $b_{1} + b_{2} < 1$ is NOT a necessary condition. The pieces form a triangle exactly when each is shorter than $0.5 m$ . Those three conditions are options (A), (B) and (C), and a triangle can form even when $b_{1} + b_{2} > 1$ (for example $b_{1} = 0.45$ , $b_{2} = 0.7$ ).

Official answer: DSource: Official answer key, GATE 2025 EC (IIT Roorkee)

The three pieces have lengths $b_{1}$ , $b_{2} - b_{1}$ and $1 - b_{2}$ . All three are positive because $0 < b_{1} < b_{2} < 1$ , and they add up to $1 m$ .
Triangle inequality: each side must be shorter than the sum of the other two. The three sides add up to 1, so the other two sides of any side $x$ add up to $1 - x$ . The condition is then $x < 1 - x$ , which means $x < 0.5$ . So a triangle can be formed exactly when every piece is shorter than $0.5 m$ . Each of these three conditions is necessary, and together they are sufficient.
First piece: $b_{1} < 0.5$ . This is option (A), so (A) is necessary.
Second piece: $b_{2} - b_{1} < 0.5 \Rightarrow b_{2} < b_{1} + 0.5$ . This is option (C), so (C) is necessary.
Third piece: $1 - b_{2} < 0.5 \Rightarrow b_{2} > 0.5$ . This is option (B), so (B) is necessary.
Option (D) is not one of these conditions, and one counter-example shows it is not necessary. Take $b_{1} = 0.45$ , $b_{2} = 0.7$ . The pieces are $0.45$ , $0.25$ and $0.30 m$ , all shorter than $0.5 m$ , so a triangle forms. Yet $b_{1} + b_{2} = 1.15 > 1$ , so (D) fails even though the triangle exists.
Therefore the condition that is NOT necessary is (D) $b_{1} + b_{2} < 1$ .

KEY IDEA

When three lengths add up to $L$ , the triangle inequality reduces to one simple test: every side must be shorter than $L /2$ . Here $L = 1$ , so each of $b_{1}$ , $b_{2} - b_{1}$ and $1 - b_{2}$ must be less than $0.5$ . Write out these three inequalities and match them to the options. To show that a condition is not necessary, one counter-example is enough: a valid triangle that breaks that condition.

WATCH OUT FOR

Confusing 'necessary' with 'sufficient'. Each of (A), (B) and (C) alone is not enough to guarantee a triangle, so students sometimes wrongly mark one of them as the answer. The question asks which condition a triangle does not REQUIRE. Another trap is assuming $b_{1} + b_{2} < 1$ from the figure, or from a symmetry argument that puts both cuts near the middle. The figure is only representative. Cuts at $0.45$ and $0.7$ give a valid triangle with $b_{1} + b_{2} > 1$ .

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 22 Geometry questions, with notes

GATE 2025 EC GA Q.7: solution and answer

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