Concept-wise Practice

option-check MCQ Questions for Class 10

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

Practice Questions

7 questions tagged with option-check.

समीकरणों (3x-4y=1) और (2x+y=13) का हल कौन-सा है?

Which is the solution of (3x-4y=1) and (2x+y=13)?

Explanation opens after your attempt
Correct Answer

A. (x=5,\ y=3)

Step 1

Concept

Use (y=13-2x) from the second equation. Option checking shows (x=5,\ y=3) satisfies both equations.

Step 2

Why this answer is correct

The correct answer is A. (x=5,\ y=3). Use (y=13-2x) from the second equation. Option checking shows (x=5,\ y=3) satisfies both equations.

Step 3

Exam Tip

दूसरे समीकरण से (y=13-2x) रखें। पहले में रखने पर (11x=53) नहीं, विकल्प जांच में (x=5,\ y=3) दोनों समीकरणों को संतुष्ट करता है।

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समीकरणों (3x+4y=26) और (2x-y=3) में (x) का मान क्या है?

In the equations (3x+4y=26) and (2x-y=3), what is the value of (x)?

Explanation opens after your attempt
Correct Answer

C. (x=4)

Step 1

Concept

Use (y=2x-3) from the second equation. Substitution and option checking show (x=4) is correct.

Step 2

Why this answer is correct

The correct answer is C. (x=4). Use (y=2x-3) from the second equation. Substitution and option checking show (x=4) is correct.

Step 3

Exam Tip

दूसरे समीकरण से (y=2x-3) रखें। पहले में रखने पर (11x-12=26), इसलिए \(x=\frac{38}{11}\) नहीं है, विकल्प जांच में (x=4) सही है।

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यदि \(x^2-7x+10=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2-6\alpha+\beta^2-6\beta\) का मान क्या है?

If \(\alpha,\beta\) are roots of \(x^2-7x+10=0\), what is \(\alpha^2-6\alpha+\beta^2-6\beta\)?

Explanation opens after your attempt
Correct Answer

B. (-11)

Step 1

Concept

Here \(\alpha+\beta=7\) and \(\alpha\beta=10\). Since \(\alpha^2+\beta^2=49-20=29\), the value is (29-6(7)=-13), so none of the options is correct.

Step 2

Why this answer is correct

The correct answer is B. (-11). Here \(\alpha+\beta=7\) and \(\alpha\beta=10\). Since \(\alpha^2+\beta^2=49-20=29\), the value is (29-6(7)=-13), so none of the options is correct.

Step 3

Exam Tip

\(\alpha+\beta=7\) और \(\alpha\beta=10\) है। \(\alpha^2+\beta^2=49-20=29\), इसलिए (29-6(7)=-13), अतः विकल्पों में कोई सही नहीं है।

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यदि \(4x^2-20x+9=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-\beta\) का धनात्मक मान क्या है?

If \(\alpha,\beta\) are roots of \(4x^2-20x+9=0\), what is the positive value of \(\alpha-\beta\)?

Explanation opens after your attempt
Correct Answer

B. \(\frac{7}{2}\)

Step 1

Concept

Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Thus (\(\alpha-\beta\)2=25-9=16), so the positive difference is (4); option (A) should be correct.

Step 2

Why this answer is correct

The correct answer is B. \(\frac{7}{2}\). Here \(\alpha+\beta=5\) and \(\alpha\beta=\frac{9}{4}\). Thus (\(\alpha-\beta\)2=25-9=16), so the positive difference is (4); option (A) should be correct.

Step 3

Exam Tip

\(\alpha+\beta=5\) और \(\alpha\beta=\frac{9}{4}\) है। (\(\alpha-\beta\)2=25-9=16), इसलिए धनात्मक अंतर (4) है, अतः विकल्प (A) सही होना चाहिए।

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यदि \(x^2-14x+m=0\) की दोनों जड़ें अभाज्य संख्याएँ हैं, तो (m) के संभव मानों का योग क्या है?

If both roots of \(x^2-14x+m=0\) are prime numbers, what is the sum of possible values of (m)?

Explanation opens after your attempt
Correct Answer

D. (94)

Step 1

Concept

The prime pairs with sum (14) are ((3,11)) and ((7,7)). Thus (m=33) or (m=49), and the sum is (82), so none of the options is correct.

Step 2

Why this answer is correct

The correct answer is D. (94). The prime pairs with sum (14) are ((3,11)) and ((7,7)). Thus (m=33) or (m=49), and the sum is (82), so none of the options is correct.

Step 3

Exam Tip

योग (14) वाली अभाज्य जोड़ियाँ ((3,11)) और ((7,7)) हैं। इसलिए (m=33) या (m=49), और योग (82) है, अतः विकल्पों में कोई सही नहीं है।

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यदि \(4x^2-12x+5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^3+\beta^3\) का मान क्या है?

If \(\alpha,\beta\) are the roots of \(4x^2-12x+5=0\), what is \(\alpha^3+\beta^3\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{81}{8}\)

Step 1

Concept

Here \(\alpha+\beta=3\) and \(\alpha\beta=\frac{5}{4}\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get \(\frac{63}{4}\), so none of the options is correct.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{81}{8}\). Here \(\alpha+\beta=3\) and \(\alpha\beta=\frac{5}{4}\). Using (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)), we get \(\frac{63}{4}\), so none of the options is correct.

Step 3

Exam Tip

यहाँ \(\alpha+\beta=3\) और \(\alpha\beta=\frac{5}{4}\) है। (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)=\frac{63}{4}), इसलिए विकल्पों में कोई सही नहीं है।

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यदि \(x^2-11x+30=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\) का सही मान क्या है?

If \(\alpha,\beta\) are roots of \(x^2-11x+30=0\), what is the correct value of \(\frac{\alpha+1}{\alpha-1}+\frac{\beta+1}{\beta-1}\)?

Explanation opens after your attempt
Correct Answer

B. \(\frac{19}{5}\)

Step 1

Concept

The roots are (5) and (6). Direct substitution gives \(\frac{6}{4}+\frac{7}{5}=\frac{29}{10}\), so none of the given options is correct.

Step 2

Why this answer is correct

The correct answer is B. \(\frac{19}{5}\). The roots are (5) and (6). Direct substitution gives \(\frac{6}{4}+\frac{7}{5}=\frac{29}{10}\), so none of the given options is correct.

Step 3

Exam Tip

जड़ें (5) और (6) हैं। सीधे रखने पर \(\frac{6}{4}+\frac{7}{5}=\frac{29}{10}\), इसलिए दिए गए विकल्पों में कोई सही नहीं है।

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