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SSC CHSL 14 Nov 2025 Shift 1 Paper

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Question : 70 of 100

Marks: +1, -0

Given that

343\times (\sqrt{7})^{n}=49^{4}

, find n.

8
10
9
12

Solution:

Concept:

Express all numbers as powers of the same base (7) to compare exponents.

Explanation:

343 = 7^{3}

and

49 = 7^{2}

. Also

\sqrt{7} = 7^{1/2}

.

So the equation becomes:

7^{3} \times (7^{1/2})^{n} = (7^{2})^{4}

\Rightarrow 7^{3} \times 7^{n/2} = 7^{8}

Using

a^{m} \times a^{n} = a^{m+n}

, we get

7^{3 + n/2} = 7^{8}

.

Since bases are equal, equate exponents:

3 + n/2 = 8

.

Solve:

n/2 = 5 \Rightarrow n = 10

.

Answer:

n = 10

(Option B).

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