Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 251-260 of 600 questions.
If \(x=5+\sqrt{2}\), what type of number is (x-5)?
C. अपरिमेयIrrational
\(x-5=\sqrt{2}\) है जो अपरिमेय है। पहले परिमेय पदों को हटाकर सरल करें। / \(x-5=\sqrt{2}\), which is irrational. First remove rational terms and simplify.
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If \(x=\sqrt{12}+2\sqrt{3}\), what will be the simplified form of (x)?
D. \(4\sqrt{3}\)
\(\sqrt{12}=2\sqrt{3}\) इसलिए कुल \(4\sqrt{3}\) मिलता है। समान मूलों को जोड़ने से पहले सरल करें। / \(\sqrt{12}=2\sqrt{3}\) so the total is \(4\sqrt{3}\). Simplify before adding like radicals.
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Which is the simplified form of \(\sqrt{48}+\sqrt{27}\)?
A. \(7\sqrt{3}\)
\(\sqrt{48}=4\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए योग \(7\sqrt{3}\) है। समान मूलों को जोड़ने से पहले सरल करें। / \(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the sum is \(7\sqrt{3}\). Simplify before adding like radicals.
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What is the value of (\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\))?
A. (9)
यह \(a^2-b^2\) रूप है इसलिए मान (15-6=9) है। संयुग्मी गुणन में मूल हट जाता है। / This is the \(a^2-b^2\) form so the value is (15-6=9). Conjugate multiplication removes radicals.
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What is the value of (\(\sqrt{19}\)2+8)?
B. (27)
(\(\sqrt{19}\)2=19), इसलिए मान (27) है। वर्ग करने पर वर्गमूल हट जाता है। / (\(\sqrt{19}\)2=19), so the value is (27). Squaring removes the square root.
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Which is the rationalised form of \(\frac{4}{\sqrt{3}+1}\)?
C. \(2\sqrt{3}-2\)
संयुग्मी से गुणा करने पर (\frac{4\(\sqrt{3}-1\)}{2}=2\sqrt{3}-2) मिलता है। हर को परिमेय बनाते समय पूरा भिन्न सरल करें। / Multiplying by the conjugate gives (\frac{4\(\sqrt{3}-1\)}{2}=2\sqrt{3}-2). Simplify the whole fraction after rationalising.
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What is obtained by rationalising \(\frac{1}{\sqrt{6}+1}\)?
C. \(\frac{\sqrt{6}-1}{5}\)
संयुग्मी से गुणा करने पर हर (6-1=5) बनता है। इसलिए परिमेयकृत रूप \(\frac{\sqrt{6}-1}{5}\) है। / Multiplying by the conjugate makes the denominator (6-1=5). So the rationalised form is \(\frac{\sqrt{6}-1}{5}\).
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If \(a=\sqrt{8}-\sqrt{2}\), what is (a) equal to?
B. \(\sqrt{2}\)
\(\sqrt{8}=2\sqrt{2}\) इसलिए \(2\sqrt{2}-\sqrt{2}=\sqrt{2}\) है। समान मूलों के गुणांक घटाएं। / \(\sqrt{8}=2\sqrt{2}\) so \(2\sqrt{2}-\sqrt{2}=\sqrt{2}\). Subtract coefficients of like radicals.
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If \(a=3\sqrt{2}+4\) and \(b=3\sqrt{2}-4\), what is (a+b)?
B. \(6\sqrt{2}\)
स्थिर पद कट जाते हैं और \(3\sqrt{2}+3\sqrt{2}=6\sqrt{2}\) मिलता है। समान मूलों को जोड़ें। / The constant terms cancel and \(3\sqrt{2}+3\sqrt{2}=6\sqrt{2}\). Add like radicals.
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What is the value of (\(5+\sqrt{3}\)\(5-\sqrt{3}\))?
B. (22)
यह \(a^2-b^2\) है इसलिए मान (25-3=22) है। संयुग्मी गुणन में मूल हट जाता है। / This is \(a^2-b^2\), so the value is (25-3=22). Conjugate multiplication removes the radical.
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