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IBPS Clerk Prelims Model Paper 10

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Question : 3 of 100

Marks: +1, -0

[(5\sqrt{7}+\sqrt{7}+(4\sqrt{7}+8\sqrt{7})]-(19)^2

= ?

143
$72\sqrt{7}$
134
$70\sqrt{7}$
None of these

Solution:

Concept:

Simplify like radicals, then apply multiplication and subtraction using BODMAS.

Explanation:

Combine the first group:

5\sqrt{7} + \sqrt{7} = 6\sqrt{7}

.

Combine the second group:

4\sqrt{7} + 8\sqrt{7} = 12\sqrt{7}

.

Multiply them:

6\sqrt{7} \times 12\sqrt{7} = (6 \times 12) \times (\sqrt{7} \times \sqrt{7}) = 72 \times 7 = 504

.

Calculate

(19)^2 = 361

.

Subtract:

504 - 361 = 143

.

Answer:

143

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