Concept-wise Practice

option-audit MCQ Questions for Class 10

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

Practice Questions

15 questions tagged with option-audit.

Question 1/15 Expert Mathematics Chapter 2: Polynomials Irrational numbers and real numbers Class 10 Level 27

यदि \(\alpha+\beta=10\) और \(\alpha\beta=21\), तो कौन सा संयुग्मी अपरिमेय युग्म संभव है?

If \(\alpha+\beta=10\) and \(\alpha\beta=21\), which conjugate irrational pair is possible?

Explanation opens after your attempt
Correct Answer

B. \(5+\sqrt{5}\) और \(5-\sqrt{5}\)\(5+\sqrt{5}\) and \(5-\sqrt{5}\)

Step 1

Concept

The pair \(5+\sqrt{5}\) and \(5-\sqrt{5}\) has sum (10) and product (20) so it also fails. The pair (5+2) and (5-2) would be rational so none of the given options fits.

Step 2

Why this answer is correct

The correct answer is B. \(5+\sqrt{5}\) और \(5-\sqrt{5}\) / \(5+\sqrt{5}\) and \(5-\sqrt{5}\). The pair \(5+\sqrt{5}\) and \(5-\sqrt{5}\) has sum (10) and product (20) so it also fails. The pair (5+2) and (5-2) would be rational so none of the given options fits.

Step 3

Exam Tip

\(5+\sqrt{5}\) और \(5-\sqrt{5}\) का योग (10) और गुणनफल (25-5=20) है इसलिए यह भी नहीं है। सही युग्म (5+2) और (5-2) परिमेय होगा इसलिए दिए विकल्पों में कोई नहीं।

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Question 2/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 20

(0.01875) को सरलतम भिन्न में लिखने पर हर का अभाज्य गुणनखंडन क्या होगा?

When (0.01875) is written in lowest fraction form, what is the prime factorisation of the denominator?

Explanation opens after your attempt
Correct Answer

A. \(2^4\cdot 5\)

Step 1

Concept

\(0.01875=\frac{1875}{100000}=\frac{3}{160}\), and \(160=2^5\cdot 5\). The correct prime factorisation is \(2^5\cdot 5\), so complete the calculation before choosing.

Step 2

Why this answer is correct

The correct answer is A. \(2^4\cdot 5\). \(0.01875=\frac{1875}{100000}=\frac{3}{160}\), and \(160=2^5\cdot 5\). The correct prime factorisation is \(2^5\cdot 5\), so complete the calculation before choosing.

Step 3

Exam Tip

\(0.01875=\frac{1875}{100000}=\frac{3}{160}\) और \(160=2^5\cdot 5\) है। सही अभाज्य रूप \(2^5\cdot 5\) है इसलिए गणना पूरी करके विकल्प चुनें।

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Question 3/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 20

\(0.\overline{045}\) को सरलतम भिन्न में लिखने पर हर क्या होगा?

What is the denominator when \(0.\overline{045}\) is written in lowest fraction form?

Explanation opens after your attempt
Correct Answer

A. (37)

Step 1

Concept

\(0.\overline{045}=\frac{45}{999}=\frac{5}{111}\), so the denominator is (111). An initial zero inside the repeating block is also counted as a digit.

Step 2

Why this answer is correct

The correct answer is A. (37). \(0.\overline{045}=\frac{45}{999}=\frac{5}{111}\), so the denominator is (111). An initial zero inside the repeating block is also counted as a digit.

Step 3

Exam Tip

\(0.\overline{045}=\frac{45}{999}=\frac{5}{111}\) है इसलिए हर (111) है। आवर्ती भाग में आरंभिक शून्य को भी अंक माना जाता है।

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Question 4/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 20

\(\frac{13}{2^3\cdot 5^7}\) को \(\frac{N}{10^7}\) के रूप में लिखने पर (N) क्या होगा?

If \(\frac{13}{2^3\cdot 5^7}\) is written as \(\frac{N}{10^7}\), what is (N)?

Explanation opens after your attempt
Correct Answer

A. (104)

Step 1

Concept

Since \(10^7=2^7\cdot 5^7\), the denominator lacks \(2^4\). Thus \(N=13\cdot 16=208\), so the correct value is not listed.

Step 2

Why this answer is correct

The correct answer is A. (104). Since \(10^7=2^7\cdot 5^7\), the denominator lacks \(2^4\). Thus \(N=13\cdot 16=208\), so the correct value is not listed.

Step 3

Exam Tip

\(10^7=2^7\cdot 5^7\) है इसलिए हर में \(2^4\) की कमी है। \(N=13\cdot 16=208\) होगा इसलिए दिए विकल्पों में सही मान नहीं है।

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Question 5/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 20

\(0.00\overline{63}\) को सरलतम भिन्न में लिखने पर हर क्या होगा?

What is the denominator when \(0.00\overline{63}\) is written as a fraction in lowest form?

Explanation opens after your attempt
Correct Answer

A. (110)

Step 1

Concept

\(0.00\overline{63}=\frac{63}{9900}=\frac{7}{1100}\), so the denominator is (1100). In recurring decimals, the first denominator formed may not be final.

Step 2

Why this answer is correct

The correct answer is A. (110). \(0.00\overline{63}=\frac{63}{9900}=\frac{7}{1100}\), so the denominator is (1100). In recurring decimals, the first denominator formed may not be final.

Step 3

Exam Tip

\(0.00\overline{63}=\frac{63}{9900}=\frac{7}{1100}\) है इसलिए हर (1100) है। आवर्ती दशमलव में पहले बना हर हमेशा अंतिम हर नहीं होता।

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Question 6/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 19

(0.0375) को सरलतम भिन्न में लिखने पर हर का अभाज्य गुणनखंडन क्या होगा?

When (0.0375) is written in lowest fraction form, what is the prime factorisation of the denominator?

Explanation opens after your attempt
Correct Answer

A. \(2^3\cdot 5\)

Step 1

Concept

\(0.0375=\frac{375}{10000}=\frac{3}{80}\), and \(80=2^4\cdot 5\). The correct prime factorisation is \(2^4\cdot 5\).

Step 2

Why this answer is correct

The correct answer is A. \(2^3\cdot 5\). \(0.0375=\frac{375}{10000}=\frac{3}{80}\), and \(80=2^4\cdot 5\). The correct prime factorisation is \(2^4\cdot 5\).

Step 3

Exam Tip

\(0.0375=\frac{375}{10000}=\frac{3}{80}\) और \(80=2^4\cdot 5\)। सही अभाज्य रूप \(2^4\cdot 5\) है।

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Question 7/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 19

कौन-सी भिन्न असांत आवर्ती दशमलव देगी?

Which fraction will give a non-terminating recurring decimal?

Explanation opens after your attempt
Correct Answer

C. \(\frac{49}{2\cdot 5^2\cdot 7^2}\)

Step 1

Concept

In \(\frac{49}{2\cdot 5^2\cdot 7^2}\), \(49=7^2\) cancels completely, so it terminates. For a non-terminating recurring decimal, a factor other than (2) and (5) must remain in the reduced denominator.

Step 2

Why this answer is correct

The correct answer is C. \(\frac{49}{2\cdot 5^2\cdot 7^2}\). In \(\frac{49}{2\cdot 5^2\cdot 7^2}\), \(49=7^2\) cancels completely, so it terminates. For a non-terminating recurring decimal, a factor other than (2) and (5) must remain in the reduced denominator.

Step 3

Exam Tip

\(\frac{49}{2\cdot 5^2\cdot 7^2}\) में \(49=7^2\) पूरा कट जाता है, इसलिए यह सांत है। सही असांत आवर्ती के लिए सरलतम हर में (2) और (5) के अलावा कोई गुणनखंड बचना चाहिए।

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Question 8/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 19

किस विकल्प में दी गई भिन्न असांत आवर्ती दशमलव देगी?

Which option will give a non-terminating recurring decimal?

Explanation opens after your attempt
Correct Answer

A. \(\frac{121}{2^2\cdot 5^3\cdot 11}\)

Step 1

Concept

In the first option, \(121=11^2\) cancels the denominator's (11), leaving only (2) and (5) in the denominator, so it terminates. No option is non-terminating here, so the options need rechecking.

Step 2

Why this answer is correct

The correct answer is A. \(\frac{121}{2^2\cdot 5^3\cdot 11}\). In the first option, \(121=11^2\) cancels the denominator's (11), leaving only (2) and (5) in the denominator, so it terminates. No option is non-terminating here, so the options need rechecking.

Step 3

Exam Tip

पहले विकल्प में \(121=11^2\) से एक (11) कटेगा पर दूसरा (11) अंश में रहेगा और हर में केवल (2), (5) बचेंगे, इसलिए यह सांत है। सही असांत विकल्प नहीं बनता, इसलिए ऐसे प्रश्न में विकल्पों की दोबारा जाँच जरूरी है।

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Question 9/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 19

(0.00072) को सरलतम भिन्न में लिखने पर हर क्या होगा?

What is the denominator when (0.00072) is written as a fraction in lowest form?

Explanation opens after your attempt
Correct Answer

A. (1250)

Step 1

Concept

\(0.00072=\frac{72}{100000}\), and reducing by (8) gives \(\frac{9}{12500}\). So the correct denominator is (12500); check the common factor carefully in small decimals.

Step 2

Why this answer is correct

The correct answer is A. (1250). \(0.00072=\frac{72}{100000}\), and reducing by (8) gives \(\frac{9}{12500}\). So the correct denominator is (12500); check the common factor carefully in small decimals.

Step 3

Exam Tip

\(0.00072=\frac{72}{100000}\) और (72) से सरल करने पर \(\frac{9}{12500}\) मिलता है। इसलिए सही हर (12500) है, छोटे दशमलवों में महत्तम सामान्य गुणनखंड ध्यान से देखें।

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Question 10/15 Expert Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 19

\(0.3\overline{18}\) को सरलतम भिन्न \(\frac{p}{q}\) में लिखने पर (q) क्या होगा?

When \(0.3\overline{18}\) is written as \(\frac{p}{q}\) in lowest form, what is (q)?

Explanation opens after your attempt
Correct Answer

A. (110)

Step 1

Concept

Taking \(x=0.31818\ldots\), subtracting (10x) from (1000x) gives \(\frac{315}{990}=\frac{7}{22}\). The reduced denominator is (22), so none of the listed denominators is correct.

Step 2

Why this answer is correct

The correct answer is A. (110). Taking \(x=0.31818\ldots\), subtracting (10x) from (1000x) gives \(\frac{315}{990}=\frac{7}{22}\). The reduced denominator is (22), so none of the listed denominators is correct.

Step 3

Exam Tip

\(x=0.31818\ldots\) लेने पर (10x) और (1000x) घटाने से \(\frac{315}{990}=\frac{7}{22}\) मिलता है। सरलतम हर (22) है, इसलिए दिए विकल्पों में सही हर नहीं है।

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Question 11/15 Hard Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 21

किसी सरलतम भिन्न का दशमलव ठीक (2) स्थानों पर समाप्त होता है। इनमें से कौन-सा हर संभव नहीं है?

A reduced fraction terminates exactly after (2) decimal places. Which denominator is not possible?

Explanation opens after your attempt
Correct Answer

D. (50)

Step 1

Concept

For exactly (2) places, the larger exponent must be (2).

Step 2

Why this answer is correct

\(4=2^2\), \(20=2^2\cdot 5\), and \(25=5^2\) give exactly (2) places. \(50=2\cdot 5^2\) also gives exactly (2) places, so none of the listed choices is impossible.

Step 3

Exam Tip

If all options seem possible, check the question or options for an error. चरण 1: ठीक (2) स्थानों के लिए बड़ी घात (2) होनी चाहिए। चरण 2: \(4=2^2\), \(20=2^2\cdot 5\), और \(25=5^2\) ठीक (2) स्थान देते हैं। \(50=2\cdot 5^2\) भी ठीक (2) स्थान देता है, इसलिए दिए गए विकल्पों में कोई असंभव नहीं है। चरण 3: जब सभी विकल्प संभव लगें, तो प्रश्न या विकल्पों में त्रुटि जाँचें।

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Question 12/15 Hard Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 21

किस सरलतम हर से ठीक (4) दशमलव स्थान मिलेंगे?

Which reduced denominator will give exactly (4) decimal places?

Explanation opens after your attempt
Correct Answer

C. (625)

Step 1

Concept

For exactly (4) decimal places, the larger power of (2) or (5) in the reduced denominator must be (4).

Step 2

Why this answer is correct

\(625=5^4\), so it gives exactly (4) places. \(80=2^4\cdot 5\) also gives (4) places, so the choices would need checking if only one answer is expected.

Step 3

Exam Tip

Factorise all options in such questions. चरण 1: ठीक (4) दशमलव स्थानों के लिए सरलतम हर में (2) या (5) की बड़ी घात (4) होनी चाहिए। चरण 2: \(625=5^4\), इसलिए यह ठीक (4) स्थान देगा। \(80=2^4\cdot 5\) भी (4) स्थान देता है, पर एक से अधिक सही होने पर विकल्प जाँचनी होगी। चरण 3: ऐसे प्रश्नों में सभी विकल्पों की घातें निकालें।

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Question 13/15 Hard Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 21

(0.0075) को सरलतम भिन्न में लिखने पर हर का अभाज्य गुणनखंडन क्या होगा?

When (0.0075) is written as a fraction in lowest form, what is the prime factorisation of the denominator?

Explanation opens after your attempt
Correct Answer

B. \(2^3\cdot 5\)

Step 1

Concept

\(0.0075=\frac{75}{10000}\).

Step 2

Why this answer is correct

Reducing gives \(\frac{3}{400}\), and \(400=2^4\cdot 5^2\). This factorisation is not present in the listed choices, so the options have an error.

Step 3

Exam Tip

Do not choose an option before writing the final denominator in prime factor form. चरण 1: \(0.0075=\frac{75}{10000}\) है। चरण 2: सरल करने पर \(\frac{3}{400}\) मिलता है और \(400=2^4\cdot 5^2\)। यहाँ दिए विकल्पों में यह नहीं है, इसलिए सही विकल्पों की जाँच में त्रुटि दिखती है। चरण 3: अंतिम हर को अभाज्य रूप में लिखे बिना विकल्प न चुनें।

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Question 14/15 Hard Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 19

\(\frac{44}{2^3\cdot 5\cdot 11}\) का दशमलव प्रसार कैसा होगा?

What type of decimal expansion will \(\frac{44}{2^3\cdot 5\cdot 11}\) have?

Explanation opens after your attempt
Correct Answer

A. सांत और (2) दशमलव स्थानTerminating with (2) decimal places

Step 1

Concept

\(44=2^2\cdot 11\).

Step 2

Why this answer is correct

After cancellation, the denominator becomes \(2\cdot 5=10\). So the decimal terminates after (1) place. Since that exact statement is not listed, the given options contain an issue.

Step 3

Exam Tip

Complete your calculation before trusting the options. चरण 1: \(44=2^2\cdot 11\) है। चरण 2: कटौती के बाद हर \(2\cdot 5\) बचेगा, जो (10) है। इसलिए दशमलव (1) स्थान पर समाप्त होगा। दिए गए विकल्पों में यह बात सीधे नहीं है, इसलिए सबसे निकट भी गलत होगा। चरण 3: विकल्पों से पहले अपनी गणना पूरी करें।

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Question 15/15 Medium Mathematics Chapter 1: Real Numbers 7: Decimal expansion of rational numbers Class 10 Level 19

किस संख्या से \(\frac{5}{18}\) को गुणा करने पर प्राप्त भिन्न का दशमलव प्रसार समाप्त होगा?

By which number should \(\frac{5}{18}\) be multiplied so that the resulting fraction has a terminating decimal expansion?

Explanation opens after your attempt
Correct Answer

B. (3)

Step 1

Concept

\(18=2\times3^2\), so \(3^2\) in the denominator is the obstacle.

Step 2

Why this answer is correct

Multiplying by (3) gives \(\frac{15}{18}=\frac{5}{6}\), which still does not terminate.

Step 3

Exam Tip

Exam tip: The correct least multiplier should be (9), so none of the given options is suitable. चरण 1: \(18=2\times3^2\) है, इसलिए हर में \(3^2\) बाधा है। चरण 2: \(\frac{5}{18}\times3=\frac{15}{18}=\frac{5}{6}\) अभी भी समाप्त नहीं है, इसलिए (3) पर्याप्त नहीं है। चरण 3: परीक्षा सुझाव: सही न्यूनतम गुणक (9) होना चाहिए, इसलिए दिए गए विकल्पों में कोई उपयुक्त विकल्प नहीं है।

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