यदि \(s=\sqrt{242}+\sqrt{128}\) है तो \(\frac{s}{\sqrt{2}}\) का मान क्या है?

If \(s=\sqrt{242}+\sqrt{128}\), what is the value of \(\frac{s}{\sqrt{2}}\)?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

B. (19)

Step 1

Concept

\(\sqrt{242}=11\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so \(s=19\sqrt{2}\). Dividing gives (19).

Step 2

Why this answer is correct

The correct answer is B. (19). \(\sqrt{242}=11\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so \(s=19\sqrt{2}\). Dividing gives (19).

Step 3

Exam Tip

\(\sqrt{242}=11\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\), इसलिए \(s=19\sqrt{2}\) है। भाग देने पर (19) मिलता है।

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FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(s=\sqrt{242}+\sqrt{128}\) है तो \(\frac{s}{\sqrt{2}}\) का मान क्या है? / If \(s=\sqrt{242}+\sqrt{128}\), what is the value of \(\frac{s}{\sqrt{2}}\)?

Correct Answer: B. (19). Explanation: \(\sqrt{242}=11\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\), इसलिए \(s=19\sqrt{2}\) है। भाग देने पर (19) मिलता है। / \(\sqrt{242}=11\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so \(s=19\sqrt{2}\). Dividing gives (19).

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{242}=11\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so \(s=19\sqrt{2}\). Dividing gives (19).

What exam hint can help solve this Mathematics question?

\(\sqrt{242}=11\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\), इसलिए \(s=19\sqrt{2}\) है। भाग देने पर (19) मिलता है।