Concept-wise Practice

monomials MCQ Questions for Class 10

monomials se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

27 questions tagged with monomials.

यदि (\left\(3x^{-2}y^{3}\right\)^{2}\cdot\left\(9x^{4}y^{-1}\right\)^{-1}) को \(cx^{r}y^{s}\) लिखा जाए, तो (c+r+s) का मान क्या है?

If (\left\(3x^{-2}y^{3}\right\)^{2}\cdot\left\(9x^{4}y^{-1}\right\)^{-1}) is written as \(cx^{r}y^{s}\), what is the value of (c+r+s)?

Explanation opens after your attempt
Correct Answer

B. (2)

Step 1

Concept

The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus (c=1), (r=-8), (s=7), and (c+r+s=0).

Step 2

Why this answer is correct

The correct answer is B. (2). The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus (c=1), (r=-8), (s=7), and (c+r+s=0).

Step 3

Exam Tip

अभिव्यक्ति \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\) है। इसलिए (c=1), (r=-8), (s=7), और (c+r+s=0) होता है।

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(\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{-\frac{1}{3}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{-\frac{1}{3}})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{4x^{3}y^{4}}{5}\)

Step 1

Concept

We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{4x^{3}y^{4}}{5}\). We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).

Step 3

Exam Tip

(\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{4x^{3}y^{4}}{5}\) मिलता है।

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(\left\(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

A. \(9r^{6}s^{-8}\)

Step 1

Concept

Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).

Step 2

Why this answer is correct

The correct answer is A. \(9r^{6}s^{-8}\). Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).

Step 3

Exam Tip

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।

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\(\sqrt[3]{343a^{15}b^{12}}\) का सरल रूप क्या है?

What is the simplified form of \(\sqrt[3]{343a^{15}b^{12}}\)?

Explanation opens after your attempt
Correct Answer

A. \(7a^{5}b^{4}\)

Step 1

Concept

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.

Step 2

Why this answer is correct

The correct answer is A. \(7a^{5}b^{4}\). 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.

Step 3

Exam Tip

\(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), और \(\sqrt[3]{b^{12}}=b^{4}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।

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यदि (\left\(x^{4}y^{-3}\right\)^{k}=x^{16}y^{-12}), तो (k) का मान क्या है?

If (\left\(x^{4}y^{-3}\right\)^{k}=x^{16}y^{-12}), what is the value of (k)?

Explanation opens after your attempt
Correct Answer

C. (4)

Step 1

Concept

The left side has exponents (4k) and (-3k). Both (4k=16) and (-3k=-12) give (k=4).

Step 2

Why this answer is correct

The correct answer is C. (4). The left side has exponents (4k) and (-3k). Both (4k=16) and (-3k=-12) give (k=4).

Step 3

Exam Tip

बाएँ पक्ष में घातें (4k) और (-3k) हैं। (4k=16) और (-3k=-12) दोनों से (k=4) मिलता है।

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

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

Explanation opens after your attempt
Correct Answer

A. \(p^{8}q^{-12}\)

Step 1

Concept

Inside, (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}). Raising to (-2) gives \(p^{8}q^{-12}\).

Step 2

Why this answer is correct

The correct answer is A. \(p^{8}q^{-12}\). Inside, (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}). Raising to (-2) gives \(p^{8}q^{-12}\).

Step 3

Exam Tip

अंदर (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}) है। (-2) घात देने पर \(p^{8}q^{-12}\) मिलता है।

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यदि (\left\(2x^{-1}y^{2}\right\)^{3}\cdot\left\(4x^{2}y^{-1}\right\)^{-1}) को \(cx^{r}y^{s}\) लिखा जाए, तो (c+r+s) का मान क्या है?

If (\left\(2x^{-1}y^{2}\right\)^{3}\cdot\left\(4x^{2}y^{-1}\right\)^{-1}) is written as \(cx^{r}y^{s}\), what is the value of (c+r+s)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{17}{4}\)

Step 1

Concept

The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence (c+r+s=2-5+7=4).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{17}{4}\). The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence (c+r+s=2-5+7=4).

Step 3

Exam Tip

अभिव्यक्ति \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=;2x^{-5}y^{7}\) है। इसलिए (c+r+s=2-5+7=4) है।

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(\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{-\frac{1}{3}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{-\frac{1}{3}})?

Explanation opens after your attempt
Correct Answer

A. \(\frac{3x^{2}y^{3}}{4}\)

Step 1

Concept

We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{3x^{2}y^{3}}{4}\). We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).

Step 3

Exam Tip

(\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{3x^{2}y^{3}}{4}\) मिलता है।

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(\left\(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

A. \(7r^{5}s^{-6}\)

Step 1

Concept

Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).

Step 2

Why this answer is correct

The correct answer is A. \(7r^{5}s^{-6}\). Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).

Step 3

Exam Tip

अंदर \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\) है। (-1) घात लेने पर \(7r^{5}s^{-6}\) मिलता है।

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\(\sqrt[3]{216a^{12}b^{9}}\) का सरल रूप क्या है?

What is the simplified form of \(\sqrt[3]{216a^{12}b^{9}}\)?

Explanation opens after your attempt
Correct Answer

A. \(6a^{4}b^{3}\)

Step 1

Concept

We have \(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), and \(\sqrt[3]{b^{9}}=b^{3}\). In exams, divide exponents by (3) under a cube root.

Step 2

Why this answer is correct

The correct answer is A. \(6a^{4}b^{3}\). We have \(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), and \(\sqrt[3]{b^{9}}=b^{3}\). In exams, divide exponents by (3) under a cube root.

Step 3

Exam Tip

\(\sqrt[3]{216}=6\), \(\sqrt[3]{a^{12}}=a^{4}\), और \(\sqrt[3]{b^{9}}=b^{3}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।

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यदि (\left\(x^{-3}y^{2}\right\)^{k}=x^{-12}y^{8}), तो (k) का मान क्या है?

If (\left\(x^{-3}y^{2}\right\)^{k}=x^{-12}y^{8}), what is the value of (k)?

Explanation opens after your attempt
Correct Answer

C. (4)

Step 1

Concept

The left side has exponents (-3k) and (2k). Both (-3k=-12) and (2k=8) give (k=4).

Step 2

Why this answer is correct

The correct answer is C. (4). The left side has exponents (-3k) and (2k). Both (-3k=-12) and (2k=8) give (k=4).

Step 3

Exam Tip

बाएँ पक्ष में घातें (-3k) और (2k) हैं। (-3k=-12) और (2k=8) दोनों से (k=4) मिलता है।

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

Step 1

Concept

Inside, \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\). Raising to (-1) gives \(m^{6}n^{-8}\).

Step 2

Why this answer is correct

The correct answer is B. \(m^{6}n^{-8}\). Inside, \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\). Raising to (-1) gives \(m^{6}n^{-8}\).

Step 3

Exam Tip

अंदर \(m^{-4-2}n^{3-(-5)}=m^{-6}n^{8}\) है। (-1) घात लेने पर \(m^{6}n^{-8}\) मिलता है।

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

Step 1

Concept

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.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{2xy^{2}}{3}\). 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.

Step 3

Exam Tip

(\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।

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(\left\(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}\right\)^{-1}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}\right\)^{-1})?

Explanation opens after your attempt
Correct Answer

A. \(5m^{6}n^{-4}\)

Step 1

Concept

Inside, \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), so raising to (-1) gives \(5m^{6}n^{-4}\). In exams, do not forget to invert the coefficient too.

Step 2

Why this answer is correct

The correct answer is A. \(5m^{6}n^{-4}\). Inside, \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), so raising to (-1) gives \(5m^{6}n^{-4}\). In exams, do not forget to invert the coefficient too.

Step 3

Exam Tip

अंदर \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), इसलिए (-1) घात लेने पर \(5m^{6}n^{-4}\) है। परीक्षा में गुणांक भी उलटना न भूलें।

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\(\sqrt[3]{125a^{9}b^{6}}\) का सरल रूप क्या है?

What is the simplified form of \(\sqrt[3]{125a^{9}b^{6}}\)?

Explanation opens after your attempt
Correct Answer

A. \(5a^{3}b^{2}\)

Step 1

Concept

We have \(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), and \(\sqrt[3]{b^{6}}=b^{2}\). In exams, divide exponents by (3) under a cube root.

Step 2

Why this answer is correct

The correct answer is A. \(5a^{3}b^{2}\). We have \(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), and \(\sqrt[3]{b^{6}}=b^{2}\). In exams, divide exponents by (3) under a cube root.

Step 3

Exam Tip

\(\sqrt[3]{125}=5\), \(\sqrt[3]{a^{9}}=a^{3}\), और \(\sqrt[3]{b^{6}}=b^{2}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें।

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यदि (\left\(x^{2}y^{-1}\right\)^{k}=x^{10}y^{-5}), तो (k) का मान क्या है?

If (\left\(x^{2}y^{-1}\right\)^{k}=x^{10}y^{-5}), what is the value of (k)?

Explanation opens after your attempt
Correct Answer

C. (5)

Step 1

Concept

The left side has exponents (2k) and (-k). Both (2k=10) and (-k=-5) give (k=5).

Step 2

Why this answer is correct

The correct answer is C. (5). The left side has exponents (2k) and (-k). Both (2k=10) and (-k=-5) give (k=5).

Step 3

Exam Tip

बाएँ पक्ष में घातें (2k) और (-k) हैं। (2k=10) और (-k=-5) दोनों से (k=5) मिलता है।

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(\left\(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}\right\)^{2}\cdot\frac{y^{12}}{x^{4}}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}\right\)^{2}\cdot\frac{y^{12}}{x^{4}})?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Inside, \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), and its square is \(4x^{8}y^{-12}\). Multiplying by \(\frac{y^{12}}{x^{4}}\) gives \(4x^{4}\).

Step 2

Why this answer is correct

The correct answer is A. \(4x^{4}\). Inside, \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), and its square is \(4x^{8}y^{-12}\). Multiplying by \(\frac{y^{12}}{x^{4}}\) gives \(4x^{4}\).

Step 3

Exam Tip

अंदर \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), इसका वर्ग \(4x^{8}y^{-12}\) है। फिर \(\frac{y^{12}}{x^{4}}\) से गुणा करने पर \(4x^{4}\) मिलता है।

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(\left\(\frac{a^{-3}b^{2}}{a^{2}b^{-4}}\right\)^{-2}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{a^{-3}b^{2}}{a^{2}b^{-4}}\right\)^{-2})?

Explanation opens after your attempt
Correct Answer

A. \(a^{10}b^{-12}\)

Step 1

Concept

Inside, \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\), so raising to (-2) gives \(a^{10}b^{-12}\). In exams, a negative outer power changes the signs of both exponents.

Step 2

Why this answer is correct

The correct answer is A. \(a^{10}b^{-12}\). Inside, \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\), so raising to (-2) gives \(a^{10}b^{-12}\). In exams, a negative outer power changes the signs of both exponents.

Step 3

Exam Tip

अंदर \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\) है, इसलिए (-2) घात देने पर \(a^{10}b^{-12}\) मिलता है। परीक्षा में ऋणात्मक घात पर दोनों घातों के चिह्न बदलते हैं।

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(\left\(\frac{3x^{-2}}{y^{-1}}\right\)^{3}\cdot\frac{y^{2}}{27}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{3x^{-2}}{y^{-1}}\right\)^{3}\cdot\frac{y^{2}}{27})?

Explanation opens after your attempt
Correct Answer

A. \(x^{-6}y^{5}\)

Step 1

Concept

\(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), its cube is \(27x^{-6}y^{3}\), and multiplying by \(\frac{y^{2}}{27}\) gives \(x^{-6}y^{5}\). In exams, turn division by a negative power into multiplication.

Step 2

Why this answer is correct

The correct answer is A. \(x^{-6}y^{5}\). \(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), its cube is \(27x^{-6}y^{3}\), and multiplying by \(\frac{y^{2}}{27}\) gives \(x^{-6}y^{5}\). In exams, turn division by a negative power into multiplication.

Step 3

Exam Tip

\(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), इसका घन \(27x^{-6}y^{3}\) है, फिर \(\frac{y^{2}}{27}\) से गुणा करने पर \(x^{-6}y^{5}\) मिलता है। परीक्षा में भाग को ऋणात्मक घात से गुणा में बदलें।

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यदि (\left\(x^{2}y^{3}\right\)^{n}=x^{8}y^{12}), तो (n) का मान क्या है?

If (\left\(x^{2}y^{3}\right\)^{n}=x^{8}y^{12}), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

C. (4)

Step 1

Concept

The left side has exponents (2n) and (3n), so (2n=8) and (3n=12), giving (n=4). In exams, match exponents of both variables.

Step 2

Why this answer is correct

The correct answer is C. (4). The left side has exponents (2n) and (3n), so (2n=8) and (3n=12), giving (n=4). In exams, match exponents of both variables.

Step 3

Exam Tip

बाएँ पक्ष में घातें (2n) और (3n) हैं, इसलिए (2n=8) और (3n=12) से (n=4)। परीक्षा में दोनों चर की घात मिलाकर जांचें।

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(\left\(\frac{a^{3}b^{-2}}{a^{-1}b^{2}}\right\)^{2}) का सरल रूप क्या है?

What is the simplified form of (\left\(\frac{a^{3}b^{-2}}{a^{-1}b^{2}}\right\)^{2})?

Explanation opens after your attempt
Correct Answer

A. \(a^{8}b^{-8}\)

Step 1

Concept

Inside, \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), and squaring gives \(a^{8}b^{-8}\). In exams, watch the sign when subtracting negative exponents.

Step 2

Why this answer is correct

The correct answer is A. \(a^{8}b^{-8}\). Inside, \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), and squaring gives \(a^{8}b^{-8}\). In exams, watch the sign when subtracting negative exponents.

Step 3

Exam Tip

अंदर \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), इसलिए वर्ग करने पर \(a^{8}b^{-8}\) है। परीक्षा में ऋणात्मक घात घटाते समय चिह्न पर ध्यान दें।

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किस विकल्प में (\left\(ab^{-2}\right\)^{3}\cdot a^{-1}b^{5}) का सही सरल रूप है?

Which option gives the correct simplified form of (\left\(ab^{-2}\right\)^{3}\cdot a^{-1}b^{5})?

Explanation opens after your attempt
Correct Answer

A. \(a^{2}b^{-1}\)

Step 1

Concept

We have (\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.

Step 2

Why this answer is correct

The correct answer is A. \(a^{2}b^{-1}\). We have (\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.

Step 3

Exam Tip

(\left\(ab^{-2}\right\)^{3}=a^{3}b^{-6}), फिर \(a^{-1}b^{5}\) से गुणा करने पर \(a^{2}b^{-1}\) मिलता है। परीक्षा में हर चर की घात अलग-अलग जोड़ें।

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(\left\(\frac{4x^{2}y^{-3}}{2x^{-1}y}\right\)^{-2}) का सरल रूप क्या है, जहाँ \(x\neq0\) और \(y\neq0\)?

What is the simplified form of (\left\(\frac{4x^{2}y^{-3}}{2x^{-1}y}\right\)^{-2}), where \(x\neq0\) and \(y\neq0\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{y^{8}}{4x^{6}}\)

Step 1

Concept

Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{y^{8}}{4x^{6}}\). Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.

Step 3

Exam Tip

अंदर \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), इसलिए घात (-2) देने पर \(\frac{y^{8}}{4x^{6}}\) मिलता है। परीक्षा में पहले कोष्ठक के अंदर सरल करें।

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यदि \(x \neq 0\), तो (\dfrac{(2x)3\(3x^{-2}\)}{12x^{-1}}) का सरल रूप क्या है?

If \(x \neq 0\), what is the simplified form of (\dfrac{(2x)3\(3x^{-2}\)}{12x^{-1}})?

Explanation opens after your attempt
Correct Answer

A. \(,2x^2,\)

Step 1

Concept

The numerator is ((2x)3\(3x^{-2}\)=8x-3\cdot 3x^{-2}=24x), and \(\dfrac{24x}{12x^{-1}}=2x^2\). In exams, simplify both coefficient and variable parts.

Step 2

Why this answer is correct

The correct answer is A. \(,2x^2,\). The numerator is ((2x)3\(3x^{-2}\)=8x-3\cdot 3x^{-2}=24x), and \(\dfrac{24x}{12x^{-1}}=2x^2\). In exams, simplify both coefficient and variable parts.

Step 3

Exam Tip

ऊपर ((2x)3\(3x^{-2}\)=8x-3\cdot 3x^{-2}=24x), और \(\dfrac{24x}{12x^{-1}}=2x^2\)। परीक्षा में coefficient और variable दोनों सरल करें।

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(\(-4a^2b^{-3}\)\(3a^{-1}b^5\)) का गुणनफल क्या है?

What is the product of (\(-4a^2b^{-3}\)\(3a^{-1}b^5\))?

Explanation opens after your attempt
Correct Answer

A. \(,-12ab^2,\)

Step 1

Concept

The product of coefficients (-4) and (3) is (-12), and \(a^{2-1}b^{-3+5}=ab^2\). In exams, handle coefficients and exponents separately.

Step 2

Why this answer is correct

The correct answer is A. \(,-12ab^2,\). The product of coefficients (-4) and (3) is (-12), and \(a^{2-1}b^{-3+5}=ab^2\). In exams, handle coefficients and exponents separately.

Step 3

Exam Tip

गुणांक (-4) और (3) का गुणनफल (-12) है, और \(a^{2-1}b^{-3+5}=ab^2\) है। परीक्षा में गुणांक और घातांक अलग-अलग संभालें।

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यदि \(a \neq 0\) और \(b \neq 0\), तो \(\dfrac{6a^3b^2}{2ab^{-1}}\) का सरल रूप क्या है?

If \(a \neq 0\) and \(b \neq 0\), what is the simplified form of \(\dfrac{6a^3b^2}{2ab^{-1}}\)?

Explanation opens after your attempt
Correct Answer

A. \(,3a^2b^3,\)

Step 1

Concept

The coefficient is \(\dfrac{6}{2}=3\), \(a^{3-1}=a^2\), and \(b^{2-(-1)}=b^3\). In exams, the sign changes when subtracting a negative exponent.

Step 2

Why this answer is correct

The correct answer is A. \(,3a^2b^3,\). The coefficient is \(\dfrac{6}{2}=3\), \(a^{3-1}=a^2\), and \(b^{2-(-1)}=b^3\). In exams, the sign changes when subtracting a negative exponent.

Step 3

Exam Tip

गुणांक \(\dfrac{6}{2}=3\), \(a^{3-1}=a^2\) और \(b^{2-(-1)}=b^3\) है। परीक्षा में हर के ऋणात्मक घातांक को घटाते समय sign बदलता है।

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(\(2x^2y\)\(-3xy^2\)) का गुणनफल क्या है?

What is the product of (\(2x^2y\)\(-3xy^2\))?

Explanation opens after your attempt
Correct Answer

A. \(,-6x^3y^3,\)

Step 1

Concept

The product of coefficients (2) and (-3) is (-6), and powers of like variables are added. In exams, watch both the sign and the exponents carefully.

Step 2

Why this answer is correct

The correct answer is A. \(,-6x^3y^3,\). The product of coefficients (2) and (-3) is (-6), and powers of like variables are added. In exams, watch both the sign and the exponents carefully.

Step 3

Exam Tip

गुणांक (2) और (-3) का गुणनफल (-6) है, और समान चरों की घातें जुड़ती हैं। परीक्षा में sign और exponents दोनों ध्यान से देखें।

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