Concept-wise Practice

hard_simplification MCQ Questions for Class 10

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

Practice Questions

1 questions tagged with hard_simplification.

(\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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