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SSC CHSL 15 Nov 2025 Shift 1 Paper

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Question : 64 of 100

Marks: +1, -0

If

a^{2}+b^{2}+c^{2}=240

ab-bc-ca=64

, then find the value of

(a+b-c)^{2}

.

368
321
340
388

Solution:

Concept:

Use the algebraic identity to expand

(a+b-c)^2

and substitute the given values.

Explanation:

We know:

(a+b-c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca

.

Factor the last three terms:

2ab - 2bc - 2ca = 2(ab - bc - ca)

.

Given

a^2 + b^2 + c^2 = 240

and

ab - bc - ca = 64

.

So

(a+b-c)^2 = 240 + 2 \times 64

.

Compute:

2 \times 64 = 128

. Then

240 + 128 = 368

.

Answer:

368

, i.e., option A (368).

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