Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If (\left\(x^{4}y^{-3}\right\)^{k}=x^{16}y^{-12}), what is the value of (k)?
C. (4)
बाएँ पक्ष में घातें (4k) और (-3k) हैं। (4k=16) और (-3k=-12) दोनों से (k=4) मिलता है। / The left side has exponents (4k) and (-3k). Both (4k=16) and (-3k=-12) give (k=4).
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What is the simplified form of \(\sqrt[3]{343a^{15}b^{12}}\)?
A. \(7a^{5}b^{4}\)
\(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), और \(\sqrt[3]{b^{12}}=b^{4}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें। / We have \(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), and \(\sqrt[3]{b^{12}}=b^{4}\). In exams, divide exponents by (3) under a cube root.
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Which option is the simplified form of \(\frac{x^{12}-4096}{x^{6}-64}\), where \(x^{6}\neq64\)?
B. \(x^{6}+64\)
(x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\))। समान गुणनखंड कटने पर \(x^{6}+64\) मिलता है। / Since (x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\)), cancelling the common factor gives \(x^{6}+64\).
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If \(p=8-\sqrt{63}\), what is the value of \(\frac{1}{p}-p\)?
A. \(2\sqrt{63}\)
\(\frac{1}{8-\sqrt{63}}=8+\sqrt{63}\), क्योंकि (64-63=1) है। इसलिए \(\frac{1}{p}-p=2\sqrt{63}\)। / Since \(\frac{1}{8-\sqrt{63}}=8+\sqrt{63}\), because (64-63=1). Therefore, \(\frac{1}{p}-p=2\sqrt{63}\).
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What is the value of \(\frac{5^{-2}+5^{-3}}{5^{-4}}\)?
A. (30)
\(5^{-2}+5^{-3}=\frac{1}{25}+\frac{1}{125}=\frac{6}{125}\) और \(5^{-4}=\frac{1}{625}\)। भाग देने पर (30) मिलता है। / Here \(5^{-2}+5^{-3}=\frac{1}{25}+\frac{1}{125}=\frac{6}{125}\), and \(5^{-4}=\frac{1}{625}\). Division gives (30).
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If \(x=\sqrt{11}-\sqrt{6}\), what is the value of \(x^{2}+2\sqrt{66}\)?
C. (17)
\(x^{2}=11+6-2\sqrt{66}=17-2\sqrt{66}\)। इसलिए \(x^{2}+2\sqrt{66}=17\)। / Since \(x^{2}=11+6-2\sqrt{66}=17-2\sqrt{66}\), \(x^{2}+2\sqrt{66}=17\).
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What is the simplified form of (\left\(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}\right\)^{-1})?
A. \(9r^{6}s^{-8}\)
अंदर \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\) है। (-1) घात लेने पर \(9r^{6}s^{-8}\) मिलता है। / Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).
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If \(16^{x}=1024\) and \(32^{y}=1024\), what is the value of (x+y)?
A. \(\frac{9}{2}\)
\(1024=2^{10}\), \(16^{x}=2^{4x}\) से \(x=\frac{5}{2}\), और \(32^{y}=2^{5y}\) से (y=2)। इसलिए योग \(\frac{9}{2}\) है। / Since \(1024=2^{10}\), \(16^{x}=2^{4x}\) gives \(x=\frac{5}{2}\), and \(32^{y}=2^{5y}\) gives (y=2). Hence the sum is \(\frac{9}{2}\).
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What is the value of (\left\(\sqrt{29}+\sqrt{20}\right\)\left\(\sqrt{29}-\sqrt{20}\right\)-3^{2})?
A. (0)
संयुग्म गुणनफल (29-20=9) है और \(3^{2}=9\)। इसलिए अंतर (0) है। / The conjugate product is (29-20=9), and \(3^{2}=9\). Hence the difference is (0).
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What is the simplified form of \(\frac{24^{3}}{2^{6}\cdot3^{2}}\)?
B. (6)
(24^{3}=\(2^{3}\cdot3\)^{3}=2^{9}\cdot3^{3})। भाग देने पर \(2^{3}\cdot3=24\) मिलता है, इसलिए विकल्पों में सही मान नहीं है। / Since (24^{3}=\(2^{3}\cdot3\)^{3}=2^{9}\cdot3^{3}), division leaves \(2^{3}\cdot3=24\), so the correct value is not among the options.
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