Hard Mathematics Polynomials Class 10 Level 25

कौन सा विकल्प \(\sqrt{12}+\sqrt{27}+\sqrt{75}-\sqrt{48}\) का सरल रूप है?

Which option is the simplified form of \(\sqrt{12}+\sqrt{27}+\sqrt{75}-\sqrt{48}\)?

Explanation opens after your attempt
Correct Answer

A. \(6\sqrt{3}\)

Step 1

Concept

It is \(2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\). Add like radical terms.

Step 2

Why this answer is correct

The correct answer is A. \(6\sqrt{3}\). It is \(2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\). Add like radical terms.

Step 3

Exam Tip

यह \(2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\) है। समान जड़ वाले पद जोड़ें।

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FAQs

Mathematics Answer, Explanation and Revision Hints

कौन सा विकल्प \(\sqrt{12}+\sqrt{27}+\sqrt{75}-\sqrt{48}\) का सरल रूप है? / Which option is the simplified form of \(\sqrt{12}+\sqrt{27}+\sqrt{75}-\sqrt{48}\)?

Correct Answer: A. \(6\sqrt{3}\). Explanation: यह \(2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\) है। समान जड़ वाले पद जोड़ें। / It is \(2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\). Add like radical terms.

Which concept should I revise for this Mathematics MCQ?

It is \(2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\). Add like radical terms.

What exam hint can help solve this Mathematics question?

यह \(2\sqrt{3}+3\sqrt{3}+5\sqrt{3}-4\sqrt{3}=6\sqrt{3}\) है। समान जड़ वाले पद जोड़ें।

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