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Mathematics Operations on real numbers and the laws of exponents MCQ Questions for Class 10

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Question 491/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

\(\frac{12^{4}}{2^{5}\cdot3^{3}}\) का सरल रूप क्या है?

What is the simplified form of \(\frac{12^{4}}{2^{5}\cdot3^{3}}\)?

Explanation opens after your attempt
Correct Answer

A. \(2^{3}\cdot3\)

Explanation

Simple Explanation

(12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), इसलिए भाग देने पर \(2^{3}\cdot3\) बचता है। परीक्षा में पहले अभाज्य गुणनखंड करें। / Since (12^{4}=\(2^{2}\cdot3\)^{4}=2^{8}\cdot3^{4}), division leaves \(2^{3}\cdot3\). In exams, prime-factorize first.

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Question 492/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

यदि \(s=2+\sqrt{7}\), तो \(s^{2}-\frac{1}{s^{2}}\) का मान क्या है?

If \(s=2+\sqrt{7}\), what is the value of \(s^{2}-\frac{1}{s^{2}}\)?

Explanation opens after your attempt
Correct Answer

A. \(8\sqrt{7}\)

Explanation

Simple Explanation

\(\frac{1}{s}=\sqrt{7}-2\), इसलिए \(s-\frac{1}{s}=4\) और \(s+\frac{1}{s}=2\sqrt{7}\)। अतः \(s^{2}-\frac{1}{s^{2}}=8\sqrt{7}\)। / Here \(\frac{1}{s}=\sqrt{7}-2\), so \(s-\frac{1}{s}=4\) and \(s+\frac{1}{s}=2\sqrt{7}\). Thus \(s^{2}-\frac{1}{s^{2}}=8\sqrt{7}\).

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Question 493/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

(\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{-\frac{1}{3}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{-\frac{1}{3}})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{2xy^{2}}{3}\)

Explanation

Simple Explanation

(\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), इसलिए \(-\frac{1}{3}\) घात देने पर उसका व्युत्क्रम \(\frac{2xy^{2}}{3}\) है। परीक्षा में भिन्न घात के बाद ऋणात्मक संकेत को व्युत्क्रम मानें। / We get (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.

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Question 494/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

यदि \(x^{2}-\frac{1}{x^{2}}=24\) और \(x-\frac{1}{x}=4\), तो \(x+\frac{1}{x}\) का मान क्या है?

If \(x^{2}-\frac{1}{x^{2}}=24\) and \(x-\frac{1}{x}=4\), what is the value of \(x+\frac{1}{x}\)?

Explanation opens after your attempt
Correct Answer

A. (6)

Explanation

Simple Explanation

(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)), इसलिए (24=4\left\(x+\frac{1}{x}\right\))। परीक्षा में वर्गों के अंतर की पहचान लगाएं। / Since (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)), (24=4\left\(x+\frac{1}{x}\right\)). In exams, use the difference of squares identity.

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Question 495/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

(\frac{\(2a^{-1}+3a^{-1}\)}{5a^{-2}}) का सरल रूप क्या है, जहाँ \(a\neq0\)?

What is the simplified form of (\frac{\(2a^{-1}+3a^{-1}\)}{5a^{-2}}), where \(a\neq0\)?

Explanation opens after your attempt
Correct Answer

A. (a)

Explanation

Simple Explanation

ऊपर \(2a^{-1}+3a^{-1}=5a^{-1}\) है। इसलिए \(\frac{5a^{-1}}{5a^{-2}}=a^{1}=a\)। / The numerator is \(2a^{-1}+3a^{-1}=5a^{-1}\). Hence \(\frac{5a^{-1}}{5a^{-2}}=a^{1}=a\).

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Question 496/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

यदि \(\sqrt{x}=3\sqrt{2}\), तो \(x^{\frac{3}{2}}\) का मान क्या है?

If \(\sqrt{x}=3\sqrt{2}\), what is the value of \(x^{\frac{3}{2}}\)?

Explanation opens after your attempt
Correct Answer

A. \(54\sqrt{2}\)

Explanation

Simple Explanation

\(\sqrt{x}=3\sqrt{2}\) से (x=18), और \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें। / From \(\sqrt{x}=3\sqrt{2}\), (x=18), and \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).

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Question 497/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

(\left\(\frac{3}{5}\right\)^{-2}+\left\(\frac{5}{3}\right\)^{-2}) का मान क्या है?

What is the value of (\left\(\frac{3}{5}\right\)^{-2}+\left\(\frac{5}{3}\right\)^{-2})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{706}{225}\)

Explanation

Simple Explanation

(\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) और (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), इसलिए योग \(\frac{625+81}{225}=\frac{706}{225}\)। परीक्षा में ऋणात्मक घात पर भिन्न उलटें। / Here (\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) and (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.

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Question 498/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

यदि (\left\(3^{x}\right\)^{2}\cdot3^{x-1}=729), तो (x) का मान क्या है?

If (\left\(3^{x}\right\)^{2}\cdot3^{x-1}=729), what is the value of (x)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{7}{3}\)

Explanation

Simple Explanation

बाएँ पक्ष \(3^{2x}\cdot3^{x-1}=3^{3x-1}\) है और \(729=3^{6}\)। इसलिए (3x-1=6) और \(x=\frac{7}{3}\)। / The left side is \(3^{2x}\cdot3^{x-1}=3^{3x-1}\), and \(729=3^{6}\). Hence (3x-1=6) and \(x=\frac{7}{3}\).

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Question 499/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

\(\frac{\sqrt{147}-2\sqrt{12}+3\sqrt{27}}{\sqrt{3}}\) का मान क्या है?

What is the value of \(\frac{\sqrt{147}-2\sqrt{12}+3\sqrt{27}}{\sqrt{3}}\)?

Explanation opens after your attempt
Correct Answer

A. (16)

Explanation

Simple Explanation

\(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), और \(3\sqrt{27}=9\sqrt{3}\), इसलिए अंश \(12\sqrt{3}\) नहीं बल्कि \(7\sqrt{3}-4\sqrt{3}+9\sqrt{3}=12\sqrt{3}\) है। अतः मान (12) होना चाहिए। / Here \(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), and \(3\sqrt{27}=9\sqrt{3}\), so the numerator is \(12\sqrt{3}\). Therefore, the value should be (12).

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Question 500/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

यदि \(\frac{1}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}\) और (a>b>0), तो (a-b) का मान क्या है?

If \(\frac{1}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}\) and (a>b>0), what is the value of (a-b)?

Explanation opens after your attempt
Correct Answer

A. (1)

Explanation

Simple Explanation

दोनों पक्षों को \(\sqrt{a}+\sqrt{b}\) से गुणा करने पर (1=\(\sqrt{a}-\sqrt{b}\)\(\sqrt{a}+\sqrt{b}\)=a-b)। परीक्षा में संयुग्म गुणनफल सीधे लगाएं। / Multiplying both sides by \(\sqrt{a}+\sqrt{b}\), we get (1=\(\sqrt{a}-\sqrt{b}\)\(\sqrt{a}+\sqrt{b}\)=a-b). In exams, apply the conjugate product directly.

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