Concept-wise Practice

large-number MCQ Questions for Class 10

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

Practice Questions

35 questions tagged with large-number.

Question 1/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 524880 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 524880?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^8\times5\)

Step 1

Concept

Write \(524880=104976\times5\).

Step 2

Why this answer is correct

\(104976=16\times6561=2^4\times3^8\).

Step 3

Exam Tip

Therefore, \(524880=2^4\times3^8\times5\). चरण 1: \(524880=104976\times5\) लिखें। चरण 2: \(104976=16\times6561=2^4\times3^8\)। चरण 3: इसलिए \(524880=2^4\times3^8\times5\) है।

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Question 2/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 1000188 का अभाज्य गुणनखंडन क्या है?

What is the prime factorisation of 1000188?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3^6\times7^3\)

Step 1

Concept

Write \(1000188=4\times250047\).

Step 2

Why this answer is correct

\(4=2^2\) and \(250047=3^6\times7^3\), so the prime form is \(2^2\times3^6\times7^3\).

Step 3

Exam Tip

Treat 729 and 343 as powers of prime bases and keep prime bases in the final answer. चरण 1: \(1000188=4\times250047\) लिखें। चरण 2: \(4=2^2\) और \(250047=3^6\times7^3\), इसलिए अभाज्य रूप \(2^2\times3^6\times7^3\) है। चरण 3: 729 और 343 संयुक्त घातों के रूप में समझें, अंतिम उत्तर में अभाज्य आधार रखें।

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Question 3/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 443520 का सही अभाज्य गुणनखंडन कौन सा है?

Which is the correct prime factorisation of 443520?

Explanation opens after your attempt
Correct Answer

A. \(2^7\times3^2\times5\times7\times11\)

Step 1

Concept

Write \(443520=40320\times11\).

Step 2

Why this answer is correct

Since \(40320=2^7\times3^2\times5\times7\), the complete prime form is \(2^7\times3^2\times5\times7\times11\).

Step 3

Exam Tip

40320 is composite, so it should not remain in the final form. चरण 1: \(443520=40320\times11\) लिखें। चरण 2: \(40320=2^7\times3^2\times5\times7\), इसलिए पूरा अभाज्य रूप \(2^7\times3^2\times5\times7\times11\) है। चरण 3: 40320 संयुक्त है, इसलिए अंतिम रूप में नहीं रहना चाहिए।

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Question 4/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 262440 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 262440?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^8\times5\)

Step 1

Concept

Write \(262440=52488\times5\).

Step 2

Why this answer is correct

\(52488=8\times6561=2^3\times3^8\).

Step 3

Exam Tip

Therefore, \(262440=2^3\times3^8\times5\). चरण 1: \(262440=52488\times5\) लिखें। चरण 2: \(52488=8\times6561=2^3\times3^8\)। चरण 3: इसलिए \(262440=2^3\times3^8\times5\) है।

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Question 5/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 500094 का अभाज्य गुणनखंडन क्या है?

What is the prime factorisation of 500094?

Explanation opens after your attempt
Correct Answer

A. \(2\times3^3\times7^3\times13\)

Step 1

Concept

Write \(500094=2\times250047\).

Step 2

Why this answer is correct

\(250047=3^3\times7^3\times13\), so the full form is \(2\times3^3\times7^3\times13\).

Step 3

Exam Tip

Composite bases should not remain in final prime form. चरण 1: \(500094=2\times250047\) लिखें। चरण 2: \(250047=3^3\times7^3\times13\), इसलिए पूरा रूप \(2\times3^3\times7^3\times13\) है। चरण 3: अंतिम अभाज्य रूप में संयुक्त आधार नहीं रहने चाहिए।

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Question 6/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 221760 का सही अभाज्य गुणनखंडन कौन सा है?

Which is the correct prime factorisation of 221760?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^2\times5\times7\times11\)

Step 1

Concept

Write \(221760=20160\times11\).

Step 2

Why this answer is correct

Since \(20160=2^6\times3^2\times5\times7\), the complete prime form is \(2^6\times3^2\times5\times7\times11\).

Step 3

Exam Tip

20160 is composite, so do not keep it in the final answer. चरण 1: \(221760=20160\times11\) लिखें। चरण 2: \(20160=2^6\times3^2\times5\times7\), इसलिए पूरा अभाज्य रूप \(2^6\times3^2\times5\times7\times11\) है। चरण 3: 20160 संयुक्त है, इसलिए उसे अंतिम उत्तर में न रखें।

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Question 7/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 4

संख्या 131220 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 131220?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3^8\times5\)

Step 1

Concept

Write \(131220=26244\times5\).

Step 2

Why this answer is correct

\(26244=4\times6561=2^2\times3^8\).

Step 3

Exam Tip

Therefore, \(131220=2^2\times3^8\times5\). चरण 1: \(131220=26244\times5\) लिखें। चरण 2: \(26244=4\times6561=2^2\times3^8\)। चरण 3: इसलिए \(131220=2^2\times3^8\times5\) है।

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Question 8/35 Expert Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 4

संख्या 110880 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 110880?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^2\times5\times7\times11\)

Step 1

Concept

Write \(110880=10080\times11\).

Step 2

Why this answer is correct

Since \(10080=2^5\times3^2\times5\times7\), the full form is \(2^5\times3^2\times5\times7\times11\).

Step 3

Exam Tip

Do not leave a composite factor such as 10080 in the final answer. चरण 1: \(110880=10080\times11\) लिखें। चरण 2: \(10080=2^5\times3^2\times5\times7\), इसलिए पूरा रूप \(2^5\times3^2\times5\times7\times11\) है। चरण 3: अंतिम उत्तर में 10080 जैसे संयुक्त गुणनखंड न छोड़ें।

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Question 9/35 Hard Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 20412 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 20412?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3^6\times7\)

Step 1

Concept

Write \(20412=2916\times7\).

Step 2

Why this answer is correct

\(2916=4\times729=2^2\times3^6\).

Step 3

Exam Tip

Therefore, \(20412=2^2\times3^6\times7\). चरण 1: \(20412=2916\times7\) लिखें। चरण 2: \(2916=4\times729=2^2\times3^6\)। चरण 3: इसलिए \(20412=2^2\times3^6\times7\)।

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Question 10/35 Hard Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 83160 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 83160?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^3\times5\times7\times11\)

Step 1

Concept

Write \(83160=7560\times11\).

Step 2

Why this answer is correct

Since \(7560=2^3\times3^3\times5\times7\), the full factorisation is \(2^3\times3^3\times5\times7\times11\).

Step 3

Exam Tip

Do not leave a composite factor like 7560 in the final answer. चरण 1: \(83160=7560\times11\) लिखें। चरण 2: \(7560=2^3\times3^3\times5\times7\), इसलिए पूरा गुणनखंडन \(2^3\times3^3\times5\times7\times11\) है। चरण 3: अंतिम उत्तर में 7560 जैसा संयुक्त गुणनखंड न छोड़ें।

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Question 11/35 Hard Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 14700 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 14700?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3\times5^2\times7^2\)

Step 1

Concept

Write \(14700=147\times100\).

Step 2

Why this answer is correct

\(147=3\times7^2\) and \(100=2^2\times5^2\), so \(14700=2^2\times3\times5^2\times7^2\).

Step 3

Exam Tip

147 and 100 are composite, so do not keep them in the final form. चरण 1: \(14700=147\times100\) लिखें। चरण 2: \(147=3\times7^2\) और \(100=2^2\times5^2\), इसलिए \(14700=2^2\times3\times5^2\times7^2\)। चरण 3: 147 और 100 संयुक्त हैं, इसलिए उन्हें अंतिम रूप में न रखें।

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Question 12/35 Hard Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 55440 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 55440?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^2\times5\times7\times11\)

Step 1

Concept

Write \(55440=5040\times11\).

Step 2

Why this answer is correct

Since \(5040=2^4\times3^2\times5\times7\), the complete factorisation is \(2^4\times3^2\times5\times7\times11\).

Step 3

Exam Tip

Do not leave a composite factor like 5040 in the final answer. चरण 1: \(55440=5040\times11\) लिखें। चरण 2: \(5040=2^4\times3^2\times5\times7\), इसलिए पूरा गुणनखंडन \(2^4\times3^2\times5\times7\times11\) है। चरण 3: अंतिम उत्तर में 5040 जैसे संयुक्त गुणनखंड को न छोड़ें।

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Question 13/35 Hard Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 4

संख्या 27720 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 27720?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^2\times5\times7\times11\)

Step 1

Concept

Write \(27720=2520\times11\).

Step 2

Why this answer is correct

Since \(2520=2^3\times3^2\times5\times7\), the full factorisation is \(2^3\times3^2\times5\times7\times11\).

Step 3

Exam Tip

Do not leave a composite factor like 2520 in the final answer. चरण 1: \(27720=2520\times11\) लिखें। चरण 2: \(2520=2^3\times3^2\times5\times7\), इसलिए पूरा गुणनखंडन \(2^3\times3^2\times5\times7\times11\) है। चरण 3: अंतिम उत्तर में 2520 जैसा संयुक्त गुणनखंड नहीं छोड़ना चाहिए।

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Question 14/35 Medium Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 2772 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 2772?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3^2\times7\times11\)

Step 1

Concept

Write \(2772=252\times11\).

Step 2

Why this answer is correct

\(252=2^2\times3^2\times7\), so \(2772=2^2\times3^2\times7\times11\).

Step 3

Exam Tip

252 is composite, so it should not remain in the final form. चरण 1: \(2772=252\times11\) लिखें। चरण 2: \(252=2^2\times3^2\times7\), इसलिए \(2772=2^2\times3^2\times7\times11\)। चरण 3: 252 संयुक्त है, इसलिए अंतिम रूप में नहीं रखना चाहिए।

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Question 15/35 Medium Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 1890 का अभाज्य गुणनखंडन क्या है?

What is the prime factorisation of 1890?

Explanation opens after your attempt
Correct Answer

A. \(2\times3^3\times5\times7\)

Step 1

Concept

Write \(1890=189\times10\).

Step 2

Why this answer is correct

\(189=3^3\times7\) and \(10=2\times5\).

Step 3

Exam Tip

Therefore, \(1890=2\times3^3\times5\times7\). चरण 1: \(1890=189\times10\) लिखें। चरण 2: \(189=3^3\times7\) और \(10=2\times5\)। चरण 3: इसलिए \(1890=2\times3^3\times5\times7\)।

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Question 16/35 Medium Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 924 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 924?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3\times7\times11\)

Step 1

Concept

Write \(924=84\times11\).

Step 2

Why this answer is correct

\(84=2^2\times3\times7\), so \(924=2^2\times3\times7\times11\).

Step 3

Exam Tip

84 is composite, so do not leave it in the final prime form. चरण 1: \(924=84\times11\) लिखें। चरण 2: \(84=2^2\times3\times7\), इसलिए \(924=2^2\times3\times7\times11\)। चरण 3: 84 संयुक्त है, इसलिए उसे अंतिम अभाज्य रूप में न छोड़ें।

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Question 17/35 Medium Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 1848 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 1848?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3\times7\times11\)

Step 1

Concept

Write \(1848=168\times11\).

Step 2

Why this answer is correct

\(168=2^3\times3\times7\), so \(1848=2^3\times3\times7\times11\).

Step 3

Exam Tip

168 is composite, so do not keep it in the final form. चरण 1: \(1848=168\times11\) लिखें। चरण 2: \(168=2^3\times3\times7\), इसलिए \(1848=2^3\times3\times7\times11\)। चरण 3: 168 संयुक्त है, इसलिए उसे अंतिम रूप में न रखें।

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Question 18/35 Medium Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 1260 का अभाज्य गुणनखंडन क्या है?

What is the prime factorisation of 1260?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3^2\times5\times7\)

Step 1

Concept

Write \(1260=126\times10\).

Step 2

Why this answer is correct

\(126=2\times3^2\times7\) and \(10=2\times5\).

Step 3

Exam Tip

Therefore, \(1260=2^2\times3^2\times5\times7\). चरण 1: \(1260=126\times10\) लिखें। चरण 2: \(126=2\times3^2\times7\) और \(10=2\times5\)। चरण 3: इसलिए \(1260=2^2\times3^2\times5\times7\)।

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Question 19/35 Easy Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 208 का अभाज्य गुणनखंडन क्या है?

What is the prime factorisation of 208?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times13\)

Step 1

Concept

Write \(208=16\times13\).

Step 2

Why this answer is correct

\(16=2^4\) and 13 is prime, so \(208=2^4\times13\).

Step 3

Exam Tip

Since 16 is composite, write it as \(2^4\) in the final form. चरण 1: \(208=16\times13\) लिखें। चरण 2: \(16=2^4\) और 13 अभाज्य है, इसलिए \(208=2^4\times13\)। चरण 3: 16 संयुक्त है, इसलिए अंतिम रूप में \(2^4\) लिखें।

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Question 20/35 Easy Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 6

संख्या 182 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 182?

Explanation opens after your attempt
Correct Answer

A. \(2\times7\times13\)

Step 1

Concept

Write \(182=2\times91\).

Step 2

Why this answer is correct

\(91=7\times13\), so \(182=2\times7\times13\).

Step 3

Exam Tip

14 is composite, so \(14\times13\) is not the final prime factorisation. चरण 1: \(182=2\times91\) लिखें। चरण 2: \(91=7\times13\), इसलिए \(182=2\times7\times13\)। चरण 3: 14 संयुक्त है, इसलिए \(14\times13\) अंतिम अभाज्य गुणनखंडन नहीं है।

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Question 21/35 Easy Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 176 का अभाज्य गुणनखंडन क्या है?

What is the prime factorisation of 176?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times11\)

Step 1

Concept

Write \(176=16\times11\).

Step 2

Why this answer is correct

\(16=2^4\) and 11 is prime, so \(176=2^4\times11\).

Step 3

Exam Tip

Since 16 is composite, write it as \(2^4\) in the final form. चरण 1: \(176=16\times11\) लिखें। चरण 2: \(16=2^4\) और 11 अभाज्य है, इसलिए \(176=2^4\times11\)। चरण 3: 16 संयुक्त है, इसलिए अंतिम रूप में \(2^4\) लिखें।

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Question 22/35 Easy Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 5

संख्या 154 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 154?

Explanation opens after your attempt
Correct Answer

A. \(2\times7\times11\)

Step 1

Concept

Write \(154=2\times77\).

Step 2

Why this answer is correct

\(77=7\times11\), so \(154=2\times7\times11\).

Step 3

Exam Tip

14 is composite, so \(14\times11\) is not the final prime factorisation. चरण 1: \(154=2\times77\) लिखें। चरण 2: \(77=7\times11\), इसलिए \(154=2\times7\times11\)। चरण 3: 14 संयुक्त है, इसलिए \(14\times11\) अंतिम अभाज्य गुणनखंडन नहीं है।

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Question 23/35 Easy Mathematics Chapter 1: Real Numbers 2: Fundamental Theorem of Arithmetic Class 10 Level 4

संख्या 132 का अभाज्य गुणनखंडन क्या है?

What is the prime factorisation of 132?

Explanation opens after your attempt
Correct Answer

A. \(2^2\times3\times11\)

Step 1

Concept

Write \(132=12\times11\).

Step 2

Why this answer is correct

\(12=2^2\times3\) and 11 is prime, so \(132=2^2\times3\times11\).

Step 3

Exam Tip

Do not leave 12 in the final form because it is composite. चरण 1: \(132=12\times11\) लिखें। चरण 2: \(12=2^2\times3\) और 11 अभाज्य है, इसलिए \(132=2^2\times3\times11\)। चरण 3: 12 को अंतिम रूप में न छोड़ें क्योंकि वह संयुक्त है।

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Question 24/35 Expert Mathematics Chapter 1: Real Numbers 1: Euclid’s Division Lemma Class 10 Level 3

किसी संख्या को 94 से भाग देने पर शेषफल 93 है। उस संख्या में 283 जोड़ने पर 94 से भाग देने पर शेषफल क्या होगा?

A number leaves remainder 93 when divided by 94. What is the remainder when 283 is added to the number and the result is divided by 94?

Explanation opens after your attempt
Correct Answer

A. 0

Step 1

Concept

The original remainder is 93.

Step 2

Why this answer is correct

283 leaves remainder 1 when divided by 94.

Step 3

Exam Tip

(93+1=94), so the final remainder is 0. चरण 1: मूल शेषफल 93 है। चरण 2: 283 को 94 से भाग देने पर शेषफल 1 है। चरण 3: (93+1=94), इसलिए अंतिम शेषफल 0 होगा।

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Question 25/35 Expert Mathematics Chapter 1: Real Numbers 1: Euclid’s Division Lemma Class 10 Level 3

यदि 9876 को 412 से भाग दिया जाए, तो शेषफल क्या होगा?

If 9876 is divided by 412, what is the remainder?

Explanation opens after your attempt
Correct Answer

A. 0

Step 1

Concept

\(412\times23=9476\) and \(412\times24=9888\).

Step 2

Why this answer is correct

The nearest lower multiple below 9876 is 9476, so the remainder is (400).

Step 3

Exam Tip

Since the remainder is less than 412, it is valid. चरण 1: \(412\times23=9476\) और \(412\times24=9888\) है। चरण 2: 9876 से छोटा निकट गुणज 9476 है, इसलिए शेषफल (400) होना चाहिए। चरण 3: शेषफल 412 से छोटा है, इसलिए वैध है।

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Question 26/35 Expert Mathematics Chapter 1: Real Numbers 1: Euclid’s Division Lemma Class 10 Level 2

किसी संख्या को 74 से भाग देने पर शेषफल 73 है। उस संख्या में 223 जोड़ने पर 74 से भाग देने पर शेषफल क्या होगा?

A number leaves remainder 73 when divided by 74. What is the remainder when 223 is added to the number and the result is divided by 74?

Explanation opens after your attempt
Correct Answer

A. 0

Step 1

Concept

The original remainder is 73.

Step 2

Why this answer is correct

223 leaves remainder 1 when divided by 74.

Step 3

Exam Tip

(73+1=74), so the final remainder is 0. चरण 1: मूल शेषफल 73 है। चरण 2: 223 को 74 से भाग देने पर शेषफल 1 है। चरण 3: (73+1=74), इसलिए अंतिम शेषफल 0 होगा।

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Question 27/35 Expert Mathematics Chapter 1: Real Numbers 1: Euclid’s Division Lemma Class 10 Level 2

यदि 7341 को 307 से भाग दिया जाए, तो शेषफल क्या होगा?

If 7341 is divided by 307, what is the remainder?

Explanation opens after your attempt
Correct Answer

A. 261

Step 1

Concept

\(307\times23=7061\).

Step 2

Why this answer is correct

(7341-7061=280), so the remainder is 280.

Step 3

Exam Tip

Choose a multiple that does not exceed the number and leaves a difference smaller than the divisor. चरण 1: \(307\times23=7061\) है। चरण 2: (7341-7061=280), इसलिए शेषफल 280 है। चरण 3: ऐसा गुणज चुनें जो दी गई संख्या से बड़ा न हो और अंतर भाजक से छोटा हो।

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Question 28/35 Expert Mathematics Chapter 1: Real Numbers 1: Euclid’s Division Lemma Class 10 Level 1

किसी संख्या को 58 से भाग देने पर शेषफल 57 है। उस संख्या में 175 जोड़ने पर 58 से भाग देने पर शेषफल क्या होगा?

A number leaves remainder 57 when divided by 58. What is the remainder when 175 is added to the number and the result is divided by 58?

Explanation opens after your attempt
Correct Answer

A. 0

Step 1

Concept

The original remainder is 57.

Step 2

Why this answer is correct

175 leaves remainder 1 when divided by 58.

Step 3

Exam Tip

(57+1=58), so the final remainder is 0. चरण 1: मूल शेषफल 57 है। चरण 2: 175 को 58 से भाग देने पर शेषफल 1 है। चरण 3: (57+1=58), इसलिए अंतिम शेषफल 0 होगा।

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Question 29/35 Expert Mathematics Chapter 1: Real Numbers 1: Euclid’s Division Lemma Class 10 Level 1

यदि 5289 को 211 से भाग दिया जाए, तो शेषफल क्या होगा?

If 5289 is divided by 211, what is the remainder?

Explanation opens after your attempt
Correct Answer

C. 14

Step 1

Concept

\(211\times25=5275\).

Step 2

Why this answer is correct

(5289-5275=14), so the remainder is 14.

Step 3

Exam Tip

While dividing, choose a multiple that does not exceed the given number. चरण 1: \(211\times25=5275\) है। चरण 2: (5289-5275=14), इसलिए शेषफल 14 है। चरण 3: भाग करते समय ऐसा गुणज चुनें जो दी गई संख्या से बड़ा न हो।

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Question 30/35 Hard Mathematics Chapter 1: Real Numbers 1: Euclid’s Division Lemma Class 10 Level 3

किसी संख्या को 42 से भाग देने पर शेषफल 41 है। उस संख्या में 85 जोड़ने पर 42 से भाग देने पर शेषफल क्या होगा?

A number leaves remainder 41 when divided by 42. What is the remainder when 85 is added to the number and the result is divided by 42?

Explanation opens after your attempt
Correct Answer

A. 0

Step 1

Concept

The original remainder is 41.

Step 2

Why this answer is correct

(85) leaves remainder 1 on division by 42, so total remainder (41+1=42), which becomes 0.

Step 3

Exam Tip

When adding a large number, first find its smaller remainder. चरण 1: मूल शेषफल 41 है। चरण 2: 85 को 42 से भाग देने पर शेषफल 1 है, इसलिए कुल शेषफल (41+1=42), जो 0 बनता है। चरण 3: बड़ी संख्या जोड़ने पर पहले उसका छोटा शेषफल निकालें।

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