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

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Operations on real numbers and the laws of exponents Practice Questions

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Question 501/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

यदि \(x\neq0\) हो, तो (\left\(\frac{2x^{-3}}{x^{2}}\right\)^{-2}\cdot x^{-4}) का सरल रूप क्या है?

If \(x\neq0\), what is the simplified form of (\left\(\frac{2x^{-3}}{x^{2}}\right\)^{-2}\cdot x^{-4})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{x^{6}}{4}\)

Explanation

Simple Explanation

अंदर \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\) है, इसलिए (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4})। परीक्षा में पहले अंदर की घातें घटाएं। / Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}). In exams, subtract the inner exponents first.

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Question 502/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

यदि \(a\neq0\) और \(\frac{a^{2p+1}\cdot a^{p-3}}{a^{p+4}}=a^{6}\), तो (p) का मान क्या है?

If \(a\neq0\) and \(\frac{a^{2p+1}\cdot a^{p-3}}{a^{p+4}}=a^{6}\), what is the value of (p)?

Explanation opens after your attempt
Correct Answer

C. (6)

Explanation

Simple Explanation

कुल घात ((2p+1)+(p-3)-(p+4)=2p-6) है। (2p-6=6) से (p=6) मिलता है। / The total exponent is ((2p+1)+(p-3)-(p+4)=2p-6). From (2p-6=6), we get (p=6).

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Question 503/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

\(\frac{3^{8}\cdot27^{-1}\cdot81^{2}}{9^{5}}\) का सरल मान क्या है?

What is the simplified value of \(\frac{3^{8}\cdot27^{-1}\cdot81^{2}}{9^{5}}\)?

Explanation opens after your attempt
Correct Answer

B. \(3^{2}\)

Explanation

Simple Explanation

सभी पदों को आधार (3) में लिखने पर कुल घात (8-3+8-10=3) नहीं बल्कि (3) है। इसलिए सही मान \(3^{3}\) है और विकल्पों में \(3^{3}\) चुनना चाहिए। / Writing all terms with base (3), the total exponent is (8-3+8-10=3). Therefore, the value is \(3^{3}\), so choose the option \(3^{3}\).

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Question 504/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

यदि \(u=\sqrt{13}+\sqrt{5}\) और \(v=\sqrt{13}-\sqrt{5}\), तो \(\frac{u^{2}-v^{2}}{uv}\) का मान क्या है?

If \(u=\sqrt{13}+\sqrt{5}\) and \(v=\sqrt{13}-\sqrt{5}\), what is the value of \(\frac{u^{2}-v^{2}}{uv}\)?

Explanation opens after your attempt
Correct Answer

B. \(2\sqrt{65}\)

Explanation

Simple Explanation

(u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{5}\cdot2\sqrt{13}=4\sqrt{65}) और (uv=8)। इसलिए मान \(\frac{\sqrt{65}}{2}\) है। / Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{5}\cdot2\sqrt{13}=4\sqrt{65}) and (uv=8). Hence the value is \(\frac{\sqrt{65}}{2}\).

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Question 505/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

\(\frac{1}{4-\sqrt{15}}+\frac{1}{4+\sqrt{15}}\) का मान क्या है?

What is the value of \(\frac{1}{4-\sqrt{15}}+\frac{1}{4+\sqrt{15}}\)?

Explanation opens after your attempt
Correct Answer

A. (8)

Explanation

Simple Explanation

हरों का गुणनफल (16-15=1) है और अंश (8) बनता है। परीक्षा में संयुग्म हरों को साथ जोड़ना तेज तरीका है। / The product of the denominators is (16-15=1), and the numerator becomes (8). In exams, adding conjugate denominators together is a fast method.

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Question 506/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

यदि \(4^{x+1}-4^{x}=192\), तो (x) का मान क्या है?

If \(4^{x+1}-4^{x}=192\), what is the value of (x)?

Explanation opens after your attempt
Correct Answer

B. (3)

Explanation

Simple Explanation

\(4^{x+1}-4^{x}=4\cdot4^{x}-4^{x}=3\cdot4^{x}=192\), इसलिए \(4^{x}=64\)। \(64=4^{3}\), इसलिए (x=3)। / Here \(4^{x+1}-4^{x}=4\cdot4^{x}-4^{x}=3\cdot4^{x}=192\), so \(4^{x}=64\). Since \(64=4^{3}\), (x=3).

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Question 507/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

(\left\(\frac{m^{-4}n^{3}}{m^{2}n^{-5}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{m^{-4}n^{3}}{m^{2}n^{-5}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

B. \(m^{6}n^{-8}\)

Explanation

Simple Explanation

अंदर \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\) है। (-1) घात लेने पर \(m^{6}n^{-8}\) मिलता है। / Inside, \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\). Raising to (-1) gives \(m^{6}n^{-8}\).

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Question 508/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

(\left\(\frac{64}{125}\right\)^{-\frac{2}{3}}) का मान क्या है?

What is the value of (\left\(\frac{64}{125}\right\)^{-\frac{2}{3}})?

Explanation opens after your attempt
Correct Answer

B. \(\frac{25}{16}\)

Explanation

Simple Explanation

(\left\(\frac{64}{125}\right\)^{\frac{1}{3}}=\frac{4}{5}), इसलिए (\left\(\frac{64}{125}\right\)^{-\frac{2}{3}}=\left\(\frac{4}{5}\right\)^{-2}=\frac{25}{16})। परीक्षा में पहले घनमूल निकालें। / Since (\left\(\frac{64}{125}\right\)^{\frac{1}{3}}=\frac{4}{5}), (\left\(\frac{64}{125}\right\)^{-\frac{2}{3}}=\left\(\frac{4}{5}\right\)^{-2}=\frac{25}{16}). In exams, take the cube root first.

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Question 509/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

यदि \(x-\frac{1}{x}=6\), तो \(x^{2}+\frac{1}{x^{2}}\) का मान क्या है?

If \(x-\frac{1}{x}=6\), what is the value of \(x^{2}+\frac{1}{x^{2}}\)?

Explanation opens after your attempt
Correct Answer

C. (38)

Explanation

Simple Explanation

(\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2) होता है। इसलिए \(36=x^{2}+\frac{1}{x^{2}}-2\) और मान (38) है। / We use (\left\(x-\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}-2). Thus \(36=x^{2}+\frac{1}{x^{2}}-2\), so the value is (38).

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Question 510/600 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 44

\(\frac{11^{5}\cdot121^{-2}}{1331^{-1}}\) का सरल मान क्या है?

What is the simplified value of \(\frac{11^{5}\cdot121^{-2}}{1331^{-1}}\)?

Explanation opens after your attempt
Correct Answer

C. \(11^{4}\)

Explanation

Simple Explanation

\(121^{-2}=11^{-4}\) और \(1331^{-1}=11^{-3}\), इसलिए \(\frac{11^{5}\cdot11^{-4}}{11^{-3}}=11^{4}\)। परीक्षा में ऋणात्मक घात से भाग करते समय घात जुड़ती है। / Here \(121^{-2}=11^{-4}\) and \(1331^{-1}=11^{-3}\), so \(\frac{11^{5}\cdot11^{-4}}{11^{-3}}=11^{4}\). In exams, division by a negative power adds the exponent.

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