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.
What is the simplified form of (\frac{\(3^2\)4}{35\cdot3^{-2}})?
A. \(3^5\)
पहले (\(3^2\)4=38) और हर की घात (5-2=3) है। इसलिए कुल घात (8-3=5) होती है। / First (\(3^2\)4=38) and the denominator exponent is (5-2=3). Therefore the total exponent is (8-3=5).
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Simplify: (\(3^2\)3 \div 34) is equal to which value?
A. \(,3^2,\)
घात के नियम से (\(3^2\)3=36) और \(3^6 \div 3^4=3^2\) होता है। परीक्षा में पहले power of power का नियम लगाएं फिर division का। / By exponent laws, (\(3^2\)3=36) and \(3^6 \div 3^4=3^2\). In exams, apply power of a power first and then the division law.
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What is the value of \(\frac{4^3\cdot2^{-1}}{8}\)?
B. (4)
(43=\(2^2\)3=26), \(2^{-1}\) और \(8=2^3\) है। कुल घात (6-1-3=2) है इसलिए मान \(2^2=4\) है। / (43=\(2^2\)3=26), \(2^{-1}\), and \(8=2^3\). The total exponent is (6-1-3=2), so the value is \(2^2=4\).
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If \(\dfrac{2^5 \times 8}{4^2}\) is simplified using exponents, what is its value?
A. (,16,)
यहां \(8=2^3\) और (42=\(2^2\)2=24), इसलिए \(\dfrac{2^5 \times 2^3}{2^4}=2^4=16\)। परीक्षा में सभी संख्याओं को समान आधार में बदलना उपयोगी होता है। / Here \(8=2^3\) and (42=\(2^2\)2=24), so \(\dfrac{2^5 \times 2^3}{2^4}=2^4=16\). In exams, converting numbers to the same base is useful.
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If \(x\neq0\), what is the simplified form of \(\frac{x^7\cdot x^{-3}}{x^2}\)?
A. \(x^2\)
समान आधार (x) की कुल घात (7-3-2=2) है। इसलिए सरल रूप \(x^2\) है। / The total exponent of the same base (x) is (7-3-2=2). Hence the simplified form is \(x^2\).
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What is the correct expansion of (\(2a^2b^3\)2)?
A. \(4a^4b^6\)
बाहरी घात गुणांक और सभी घातों पर लगती है। इसलिए (\(2a^2b^3\)2=4a-4b-6) है। / The outside exponent applies to the coefficient and all powers. Thus (\(2a^2b^3\)2=4a-4b-6).
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What is the simplified form of \(\frac{18x^5y^3}{6x^2y}\) if \(x\neq0\) and \(y\neq0\)?
A. \(3x^3y^2\)
गुणांक \(\frac{18}{6}=3\), \(x^{5-2}=x^3\) और \(y^{3-1}=y^2\) है। इसलिए उत्तर \(3x^3y^2\) है। / The coefficient is \(\frac{18}{6}=3\), \(x^{5-2}=x^3\), and \(y^{3-1}=y^2\). So the answer is \(3x^3y^2\).
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What is the value of (\left\(\frac{1}{5}\right\)^{-2}+\left\(\frac{1}{3}\right\)^{-2})?
B. (34)
(\left\(\frac{1}{5}\right\)^{-2}=25) और (\left\(\frac{1}{3}\right\)^{-2}=9) है। इसलिए योग (34) है। / (\left\(\frac{1}{5}\right\)^{-2}=25) and (\left\(\frac{1}{3}\right\)^{-2}=9). Therefore the sum is (34).
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What is the value of ((0.04)^{-1})?
B. (25)
\(0.04=\frac{4}{100}=\frac{1}{25}\) है। इसलिए ((0.04)^{-1}=25) होता है। / \(0.04=\frac{4}{100}=\frac{1}{25}\). Therefore ((0.04)^{-1}=25).
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If (a=5), what is the value of \(a^2-a^{-1}\)?
A. \(\frac{124}{5}\)
\(5^2=25\) और \(5^{-1}=\frac{1}{5}\) है। इसलिए \(25-\frac{1}{5}=\frac{124}{5}\) है। / \(5^2=25\) and \(5^{-1}=\frac{1}{5}\). Thus \(25-\frac{1}{5}=\frac{124}{5}\).
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