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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

अभाज्य गुणनखंडन में \(64\times3969\) को अंतिम उत्तर क्यों नहीं माना जाएगा?

Why will \(64\times3969\) not be considered the final answer in prime factorisation?

Explanation opens after your attempt
Correct Answer

A. क्योंकि 64 और 3969 संयुक्त रूप हैंBecause 64 and 3969 are composite forms

Step 1

Concept

In final prime factorisation, every base should be prime.

Step 2

Why this answer is correct

\(64=2^6\) and \(3969=3^4\times7^2\).

Step 3

Exam Tip

Therefore, the final form is \(2^6\times3^4\times7^2\). चरण 1: अंतिम अभाज्य गुणनखंडन में हर आधार अभाज्य होना चाहिए। चरण 2: \(64=2^6\) और \(3969=3^4\times7^2\) है। चरण 3: इसलिए अंतिम रूप \(2^6\times3^4\times7^2\) होगा।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 2138400 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 2138400?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(2138400=32\times66825\).

Step 2

Why this answer is correct

\(66825=3^5\times5^2\times11\), so \(2138400=2^5\times3^5\times5^2\times11\).

Step 3

Exam Tip

Give 66825 its complete prime form. चरण 1: \(2138400=32\times66825\) लिखें। चरण 2: \(66825=3^5\times5^2\times11\), इसलिए \(2138400=2^5\times3^5\times5^2\times11\)। चरण 3: 66825 को पूरा अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 254016 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 254016?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^4\times7^2\)

Step 1

Concept

Write \(254016=64\times3969\).

Step 2

Why this answer is correct

\(64=2^6\) and \(3969=3^4\times7^2\), so \(254016=2^6\times3^4\times7^2\).

Step 3

Exam Tip

Convert 3969 into prime powers. चरण 1: \(254016=64\times3969\) लिखें। चरण 2: \(64=2^6\) और \(3969=3^4\times7^2\), इसलिए \(254016=2^6\times3^4\times7^2\)। चरण 3: 3969 को अभाज्य घातों में बदलें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 582120 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 582120?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^3\times5^2\times7^2\times11\)

Step 1

Concept

Write the number using prime-base groups.

Step 2

Why this answer is correct

\(16\times27\times25\times49\times11=2^4\times3^3\times5^2\times7^2\times11\).

Step 3

Exam Tip

Avoid decimal-based options and write prime bases. चरण 1: \(582120=52920\times11\) लिखें। चरण 2: \(52920=2^3\times3^3\times5\times7^2\) नहीं बल्कि यहां \(16\times27\times25\times49\times11\) से \(2^4\times3^3\times5^2\times7^2\times11\) मिलता है। चरण 3: दशमलव वाले विकल्प से बचें और अभाज्य आधार लिखें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(2^3\times3^3\times5^2\times7\times11\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?

If \(2^3\times3^3\times5^2\times7\times11\) is the prime factorisation of a number, what is the number?

Explanation opens after your attempt
Correct Answer

A. 415800

Step 1

Concept

Calculate \(2^3=8\), \(3^3=27\), and \(5^2=25\).

Step 2

Why this answer is correct

\(8\times27\times25\times7\times11=415800\).

Step 3

Exam Tip

When there are many factors, multiply in small groups. चरण 1: \(2^3=8\), \(3^3=27\) और \(5^2=25\) निकालें। चरण 2: \(8\times27\times25\times7\times11=415800\)। चरण 3: कई गुणनखंड हों तो छोटे समूह बनाकर गुणा करें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(2^4\times3^5\times5^2\times7\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?

If \(2^4\times3^5\times5^2\times7\) is the prime factorisation of a number, what is the number?

Explanation opens after your attempt
Correct Answer

A. 680400

Step 1

Concept

Take \(2^4=16\), \(3^5=243\), \(5^2=25\), and (7).

Step 2

Why this answer is correct

\(16\times243\times25\times7=680400\).

Step 3

Exam Tip

Solving powers first helps find the correct option quickly. चरण 1: \(2^4=16\), \(3^5=243\), \(5^2=25\) और (7) लें। चरण 2: \(16\times243\times25\times7=680400\)। चरण 3: घातों को पहले हल करने से सही विकल्प जल्दी मिलता है।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस विकल्प में अभाज्य गुणनखंडन अधूरा है?

Which option has incomplete prime factorisation?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times81\times121\)

Step 1

Concept

In an incomplete form, composite bases remain.

Step 2

Why this answer is correct

81 and 121 are composite bases.

Step 3

Exam Tip

\(2^6\times81\times121\) must be changed into \(2^6\times3^4\times11^2\). चरण 1: अधूरे रूप में संयुक्त आधार बच जाता है। चरण 2: 81 और 121 संयुक्त आधार हैं। चरण 3: \(2^6\times81\times121\) को \(2^6\times3^4\times11^2\) में बदलना होगा।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस विकल्प में केवल अंतिम अभाज्य गुणनखंडन दिया गया है?

Which option gives only the final prime factorisation?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^4\times5^2\times7\)

Step 1

Concept

In the final form, bases must be prime.

Step 2

Why this answer is correct

In the first option, bases 2, 3, 5, and 7 are prime.

Step 3

Exam Tip

32, 81, 25, 405, 35, and 160 are composite, so they are not final forms. चरण 1: अंतिम रूप में आधार अभाज्य होने चाहिए। चरण 2: पहले विकल्प में आधार 2, 3, 5 और 7 अभाज्य हैं। चरण 3: 32, 81, 25, 405, 35 और 160 संयुक्त हैं, इसलिए वे अंतिम रूप नहीं हैं।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस विकल्प में \(2^4\times3^3\times5^2\times7^2\) का सही मान है?

Which option gives the correct value of \(2^4\times3^3\times5^2\times7^2\)?

Explanation opens after your attempt
Correct Answer

A. 529200

Step 1

Concept

Calculate \(2^4=16\), \(3^3=27\), \(5^2=25\), and \(7^2=49\).

Step 2

Why this answer is correct

\(16\times27\times25\times49=529200\).

Step 3

Exam Tip

Do the multiplication step by step. चरण 1: \(2^4=16\), \(3^3=27\), \(5^2=25\) और \(7^2=49\) निकालें। चरण 2: \(16\times27\times25\times49=529200\)। चरण 3: गुणा को चरणों में करें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस विकल्प में \(2^6\times3^4\times11^2\) का सही मान है?

Which option gives the correct value of \(2^6\times3^4\times11^2\)?

Explanation opens after your attempt
Correct Answer

A. 627264

Step 1

Concept

Calculate \(2^6=64\), \(3^4=81\), and \(11^2=121\).

Step 2

Why this answer is correct

\(64\times81\times121=627264\).

Step 3

Exam Tip

Solve all three powers separately first. चरण 1: \(2^6=64\), \(3^4=81\) और \(11^2=121\) निकालें। चरण 2: \(64\times81\times121=627264\)। चरण 3: तीनों घातों को पहले अलग-अलग हल करें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(x=2^6\times3^4\times7^2\) और \(y=2^3\times3^5\times5\times7\), तो (xy) में 3 की घात क्या होगी?

If \(x=2^6\times3^4\times7^2\) and \(y=2^3\times3^5\times5\times7\), what will be the power of 3 in (xy)?

Explanation opens after your attempt
Correct Answer

A. 9

Step 1

Concept

Powers with the same base 3 are added in multiplication.

Step 2

Why this answer is correct

The power of 3 in (x) is 4 and in (y) is 5.

Step 3

Exam Tip

The total power will be (4+5=9). चरण 1: समान आधार 3 की घातें गुणा में जुड़ती हैं। चरण 2: (x) में 3 की घात 4 है और (y) में 3 की घात 5 है। चरण 3: कुल घात (4+5=9) होगी।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(a=2^5\times3^3\times5^2\times11\) और \(b=2^4\times3^5\times7\times11^2\), तो (ab) में 11 की घात क्या होगी?

If \(a=2^5\times3^3\times5^2\times11\) and \(b=2^4\times3^5\times7\times11^2\), what will be the power of 11 in (ab)?

Explanation opens after your attempt
Correct Answer

A. 3

Step 1

Concept

In multiplication, powers of the same prime base are added.

Step 2

Why this answer is correct

The power of 11 in (a) is 1 and in (b) is 2.

Step 3

Exam Tip

In (ab), the power of 11 will be (1+2=3). चरण 1: गुणा में समान अभाज्य आधार की घातें जुड़ती हैं। चरण 2: (a) में 11 की घात 1 है और (b) में 11 की घात 2 है। चरण 3: (ab) में 11 की घात (1+2=3) होगी।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि किसी संख्या का अभाज्य गुणनखंडन \(2^6\times3^5\times5^4\times7^3\times11^2\times13\) है, तो उसमें अलग-अलग अभाज्य गुणनखंड कितने हैं?

If a number has prime factorisation \(2^6\times3^5\times5^4\times7^3\times11^2\times13\), how many distinct prime factors does it have?

Explanation opens after your attempt
Correct Answer

A. 6

Step 1

Concept

While counting distinct primes, only bases are counted.

Step 2

Why this answer is correct

The bases are 2, 3, 5, 7, 11, and 13.

Step 3

Exam Tip

Therefore, the number of distinct prime factors is 6. चरण 1: अलग-अलग अभाज्य गिनते समय केवल आधार गिने जाते हैं। चरण 2: आधार 2, 3, 5, 7, 11 और 13 हैं। चरण 3: इसलिए अलग-अलग अभाज्य गुणनखंडों की संख्या 6 है।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि किसी संख्या का अभाज्य गुणनखंडन \(2^8\times3^6\times5^4\times7^3\times11^2\) है, तो दोहराव सहित अभाज्य गुणनखंडों की संख्या कितनी है?

If a number has prime factorisation \(2^8\times3^6\times5^4\times7^3\times11^2\), how many prime factors does it have with repetition?

Explanation opens after your attempt
Correct Answer

A. 23

Step 1

Concept

To count with repetition, add the exponents.

Step 2

Why this answer is correct

(8+6+4+3+2=23).

Step 3

Exam Tip

Counting only bases and counting with repetition are different. चरण 1: दोहराव सहित गिनती में घातों को जोड़ा जाता है। चरण 2: (8+6+4+3+2=23)। चरण 3: केवल आधार गिनना और दोहराव सहित गिनना अलग बातें हैं।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(n=2^9\times3^4\times5^5\times13^2\), तो (n) किस संख्या से अवश्य विभाज्य होगा?

If \(n=2^9\times3^4\times5^5\times13^2\), by which number must (n) be divisible?

Explanation opens after your attempt
Correct Answer

A. \(2^8\times3^4\times5^4\times13^2\)

Step 1

Concept

For divisibility, the exponents of the divisor must not exceed those in the number.

Step 2

Why this answer is correct

In the first option, all exponents are less than or equal to those in (n).

Step 3

Exam Tip

Therefore, (n) must be divisible by that number. चरण 1: विभाज्यता के लिए भाजक की घातें संख्या की घातों से अधिक नहीं होनी चाहिए। चरण 2: पहले विकल्प में सभी घातें (n) की घातों से कम या बराबर हैं। चरण 3: इसलिए (n) उस संख्या से अवश्य विभाज्य होगा।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(n=2^8\times3^6\times5^2\times7^4\), तो (n) किस संख्या से विभाज्य नहीं होगा?

If \(n=2^8\times3^6\times5^2\times7^4\), by which number will (n) not be divisible?

Explanation opens after your attempt
Correct Answer

A. \(5^3\)

Step 1

Concept

Every prime power of a divisor must be available in the number.

Step 2

Why this answer is correct

(n) has power 2 of 5, but \(5^3\) needs power 3.

Step 3

Exam Tip

Therefore, (n) is not divisible by \(5^3\). चरण 1: भाजक की हर अभाज्य घात संख्या में उपलब्ध होनी चाहिए। चरण 2: (n) में 5 की घात 2 है, पर \(5^3\) के लिए घात 3 चाहिए। चरण 3: इसलिए (n), \(5^3\) से विभाज्य नहीं होगा।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि संख्या \(2^{10}\times3^8\times5^5\times11^4\) है, तो इसे पूर्ण घन बनाने के लिए सबसे छोटी किस संख्या से भाग देना होगा?

If the number is \(2^{10}\times3^8\times5^5\times11^4\), by which smallest number should it be divided to make a perfect cube?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

In a perfect cube, exponents are multiples of 3.

Step 2

Why this answer is correct

Reducing 10 to 9, 8 to 6, 5 to 3, and 4 to 3 is the smallest way.

Step 3

Exam Tip

Therefore, the divisor is \(2\times3^2\times5^2\times11\). चरण 1: पूर्ण घन में घातें 3 के गुणज होती हैं। चरण 2: 10 को 9, 8 को 6, 5 को 3 और 4 को 3 तक घटाना सबसे छोटा तरीका है। चरण 3: इसलिए भाजक \(2\times3^2\times5^2\times11\) है।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि संख्या \(2^5\times3^4\times5^2\times7^7\) है, तो इसे पूर्ण घन बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?

If the number is \(2^5\times3^4\times5^2\times7^7\), by which smallest number should it be multiplied to make a perfect cube?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

In a perfect cube, every exponent must be a multiple of 3.

Step 2

Why this answer is correct

Powers 5, 4, 2, and 7 must become 6, 6, 3, and 9.

Step 3

Exam Tip

The smallest multiplier is \(2\times3^2\times5\times7^2\). चरण 1: पूर्ण घन में हर घात 3 का गुणज होनी चाहिए। चरण 2: 5 को 6, 4 को 6, 2 को 3 और 7 को 9 बनाना होगा। चरण 3: सबसे छोटा गुणक \(2\times3^2\times5\times7^2\) है।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि संख्या \(2^8\times3^5\times5^4\times7^3\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से भाग देना होगा?

If the number is \(2^8\times3^5\times5^4\times7^3\), by which smallest number should it be divided to make a perfect square?

Explanation opens after your attempt
Correct Answer

A. 21

Step 1

Concept

For a perfect square, exponents should be even.

Step 2

Why this answer is correct

The powers of 3 and 7 are odd.

Step 3

Exam Tip

Dividing by \(3\times7=21\) makes the powers 4 and 2. चरण 1: पूर्ण वर्ग के लिए घातें सम चाहिए। चरण 2: 3 की घात 5 और 7 की घात 3 विषम हैं। चरण 3: \(3\times7=21\) से भाग देने पर घातें 4 और 2 हो जाएंगी।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि संख्या \(2^7\times3^4\times5^2\times11^3\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?

If the number is \(2^7\times3^4\times5^2\times11^3\), by which smallest number should it be multiplied to make a perfect square?

Explanation opens after your attempt
Correct Answer

A. 22

Step 1

Concept

In a perfect square, all exponents are even.

Step 2

Why this answer is correct

The powers of 2 and 11 are odd.

Step 3

Exam Tip

Multiplying by \(2\times11=22\) makes both powers even. चरण 1: पूर्ण वर्ग में सभी घातें सम होती हैं। चरण 2: 2 की घात 7 और 11 की घात 3 विषम हैं। चरण 3: \(2\times11=22\) से गुणा करने पर दोनों घातें सम हो जाएंगी।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस संख्या का अभाज्य गुणनखंडन \(2^3\times3^6\times5\times13\) है?

Which number has prime factorisation \(2^3\times3^6\times5\times13\)?

Explanation opens after your attempt
Correct Answer

A. 379080

Step 1

Concept

Calculate \(2^3=8\) and \(3^6=729\).

Step 2

Why this answer is correct

\(8\times729\times5\times13=379080\).

Step 3

Exam Tip

Simplify the higher-power part first. चरण 1: \(2^3=8\) और \(3^6=729\) निकालें। चरण 2: \(8\times729\times5\times13=379080\)। चरण 3: बड़े घात वाले भाग को पहले सरल करें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस संख्या का अभाज्य गुणनखंडन \(3^4\times5^2\times7^2\) है?

Which number has prime factorisation \(3^4\times5^2\times7^2\)?

Explanation opens after your attempt
Correct Answer

A. 99225

Step 1

Concept

Calculate \(3^4=81\), \(5^2=25\), and \(7^2=49\).

Step 2

Why this answer is correct

\(81\times25\times49=99225\).

Step 3

Exam Tip

Do not skip square powers in a hurry. चरण 1: \(3^4=81\), \(5^2=25\) और \(7^2=49\) निकालें। चरण 2: \(81\times25\times49=99225\)। चरण 3: वर्ग घातों को जल्दबाजी में न छोड़ें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस संख्या का अभाज्य गुणनखंडन \(2^7\times3^2\times11^2\) है?

Which number has prime factorisation \(2^7\times3^2\times11^2\)?

Explanation opens after your attempt
Correct Answer

A. 139392

Step 1

Concept

Calculate \(2^7=128\), \(3^2=9\), and \(11^2=121\).

Step 2

Why this answer is correct

\(128\times9\times121=139392\).

Step 3

Exam Tip

Find the values of powers first, then multiply. चरण 1: \(2^7=128\), \(3^2=9\) और \(11^2=121\) निकालें। चरण 2: \(128\times9\times121=139392\)। चरण 3: पहले घातों का मान निकालें, फिर गुणा करें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस संख्या का अभाज्य गुणनखंडन \(2^4\times3^5\times7^2\) है?

Which number has prime factorisation \(2^4\times3^5\times7^2\)?

Explanation opens after your attempt
Correct Answer

A. 190512

Step 1

Concept

Calculate \(2^4=16\), \(3^5=243\), and \(7^2=49\).

Step 2

Why this answer is correct

\(16\times243\times49=190512\).

Step 3

Exam Tip

Finding all three powers separately is safer. चरण 1: \(2^4=16\), \(3^5=243\) और \(7^2=49\) निकालें। चरण 2: \(16\times243\times49=190512\)। चरण 3: तीनों घातों का मान अलग-अलग निकालना सुरक्षित रहता है।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

किस संख्या का अभाज्य गुणनखंडन \(2^6\times3^5\times7^2\) है?

Which number has prime factorisation \(2^6\times3^5\times7^2\)?

Explanation opens after your attempt
Correct Answer

A. 762048

Step 1

Concept

Calculate \(2^6=64\), \(3^5=243\), and \(7^2=49\).

Step 2

Why this answer is correct

\(64\times243\times49=762048\).

Step 3

Exam Tip

In such calculations, solve powers first. चरण 1: \(2^6=64\), \(3^5=243\) और \(7^2=49\) निकालें। चरण 2: \(64\times243\times49=762048\)। चरण 3: ऐसी गणना में घातों को पहले हल करें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(137200=2^4\times5^2\times7^p\), तो (p) का मान क्या है?

If \(137200=2^4\times5^2\times7^p\), what is the value of (p)?

Explanation opens after your attempt
Correct Answer

A. 3

Step 1

Concept

Write \(137200=16\times8575\).

Step 2

Why this answer is correct

\(8575=5^2\times7^3\), so the power of 7 is 3.

Step 3

Exam Tip

From the given form, (p=3). चरण 1: \(137200=16\times8575\) लिखें। चरण 2: \(8575=5^2\times7^3\), इसलिए 7 की घात 3 है। चरण 3: दिए गए रूप से (p=3) मिलेगा।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(680400=2^4\times3^b\times5^2\times7\), तो (b) का मान क्या है?

If \(680400=2^4\times3^b\times5^2\times7\), what is the value of (b)?

Explanation opens after your attempt
Correct Answer

A. 5

Step 1

Concept

\(680400=16\times42525\).

Step 2

Why this answer is correct

\(42525=3^5\times5^2\times7\), so the power of 3 is 5.

Step 3

Exam Tip

Comparing gives (b=5). चरण 1: \(680400=16\times42525\) है। चरण 2: \(42525=3^5\times5^2\times7\), इसलिए 3 की घात 5 है। चरण 3: तुलना करने पर (b=5) होगा।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(627264=2^a\times3^4\times11^2\), तो (a) का मान क्या है?

If \(627264=2^a\times3^4\times11^2\), what is the value of (a)?

Explanation opens after your attempt
Correct Answer

A. 6

Step 1

Concept

Write \(627264=64\times9801\).

Step 2

Why this answer is correct

\(64=2^6\) and \(9801=3^4\times11^2\), so the power of 2 is 6.

Step 3

Exam Tip

From the given form, (a=6). चरण 1: \(627264=64\times9801\) लिखें। चरण 2: \(64=2^6\) और \(9801=3^4\times11^2\), इसलिए 2 की घात 6 है। चरण 3: दिए गए रूप से (a=6) है।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

यदि \(529200=2^4\times3^3\times5^2\times7^q\), तो (q) का मान क्या है?

If \(529200=2^4\times3^3\times5^2\times7^q\), what is the value of (q)?

Explanation opens after your attempt
Correct Answer

A. 2

Step 1

Concept

\(529200=16\times33075\).

Step 2

Why this answer is correct

\(33075=3^3\times5^2\times7^2\), so the power of 7 is 2.

Step 3

Exam Tip

Comparing gives (q=2). चरण 1: \(529200=16\times33075\) है। चरण 2: \(33075=3^3\times5^2\times7^2\), इसलिए 7 की घात 2 है। चरण 3: तुलना करने पर (q=2) मिलता है।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 518400 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 518400?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(518400=256\times2025\).

Step 2

Why this answer is correct

\(256=2^8\) and \(2025=3^4\times5^2\), so \(518400=2^8\times3^4\times5^2\).

Step 3

Exam Tip

Give 2025 its final prime form. चरण 1: \(518400=256\times2025\) लिखें। चरण 2: \(256=2^8\) और \(2025=3^4\times5^2\), इसलिए \(518400=2^8\times3^4\times5^2\)। चरण 3: 2025 को अंतिम अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 99225 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 99225?

Explanation opens after your attempt
Correct Answer

A. \(3^4\times5^2\times7^2\)

Step 1

Concept

Write \(99225=81\times1225\).

Step 2

Why this answer is correct

\(81=3^4\) and \(1225=5^2\times7^2\), so \(99225=3^4\times5^2\times7^2\).

Step 3

Exam Tip

Convert 1225 into prime powers. चरण 1: \(99225=81\times1225\) लिखें। चरण 2: \(81=3^4\) और \(1225=5^2\times7^2\), इसलिए \(99225=3^4\times5^2\times7^2\)। चरण 3: 1225 को अभाज्य घातों में बदलें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 475200 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 475200?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^3\times5^2\times11\)

Step 1

Concept

Write \(475200=64\times7425\).

Step 2

Why this answer is correct

\(7425=3^3\times5^2\times11\), so \(475200=2^6\times3^3\times5^2\times11\).

Step 3

Exam Tip

Do not leave 7425 in the final form. चरण 1: \(475200=64\times7425\) लिखें। चरण 2: \(7425=3^3\times5^2\times11\), इसलिए \(475200=2^6\times3^3\times5^2\times11\)। चरण 3: 7425 को अंतिम रूप में न छोड़ें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 117000 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 117000?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^2\times5^3\times13\)

Step 1

Concept

Write \(117000=8\times14625\).

Step 2

Why this answer is correct

\(14625=3^2\times5^3\times13\), so \(117000=2^3\times3^2\times5^3\times13\).

Step 3

Exam Tip

Give 14625 its complete prime form. चरण 1: \(117000=8\times14625\) लिखें। चरण 2: \(14625=3^2\times5^3\times13\), इसलिए \(117000=2^3\times3^2\times5^3\times13\)। चरण 3: 14625 को पूरा अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 139392 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 139392?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(139392=128\times1089\).

Step 2

Why this answer is correct

\(128=2^7\) and \(1089=3^2\times11^2\), so \(139392=2^7\times3^2\times11^2\).

Step 3

Exam Tip

Convert 1089 into prime powers. चरण 1: \(139392=128\times1089\) लिखें। चरण 2: \(128=2^7\) और \(1089=3^2\times11^2\), इसलिए \(139392=2^7\times3^2\times11^2\)। चरण 3: 1089 को अभाज्य घातों में बदलें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 453600 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 453600?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times3^4\times5^2\times7\)

Step 1

Concept

Write \(453600=32\times14175\).

Step 2

Why this answer is correct

\(14175=3^4\times5^2\times7\), so \(453600=2^5\times3^4\times5^2\times7\).

Step 3

Exam Tip

Do not leave 14175 in the final form. चरण 1: \(453600=32\times14175\) लिखें। चरण 2: \(14175=3^4\times5^2\times7\), इसलिए \(453600=2^5\times3^4\times5^2\times7\)। चरण 3: 14175 को अंतिम रूप में न छोड़ें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 127575 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 127575?

Explanation opens after your attempt
Correct Answer

A. \(3^6\times5^2\times7\)

Step 1

Concept

Write \(127575=729\times175\).

Step 2

Why this answer is correct

\(729=3^6\) and \(175=5^2\times7\), so \(127575=3^6\times5^2\times7\).

Step 3

Exam Tip

Give prime form to both 729 and 175. चरण 1: \(127575=729\times175\) लिखें। चरण 2: \(729=3^6\) और \(175=5^2\times7\), इसलिए \(127575=3^6\times5^2\times7\)। चरण 3: 729 और 175 दोनों को अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 137200 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 137200?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times5^2\times7^3\)

Step 1

Concept

Write \(137200=16\times8575\).

Step 2

Why this answer is correct

\(16=2^4\) and \(8575=5^2\times7^3\), so the correct form is \(2^4\times5^2\times7^3\).

Step 3

Exam Tip

Give 8575 its final prime form. चरण 1: \(137200=16\times8575\) लिखें। चरण 2: \(16=2^4\) और \(8575=5^2\times7^3\), इसलिए सही रूप \(2^4\times5^2\times7^3\) है। चरण 3: 8575 को अंतिम अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 762048 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 762048?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^5\times7^2\)

Step 1

Concept

Write \(762048=64\times11907\).

Step 2

Why this answer is correct

\(64=2^6\) and \(11907=3^5\times7^2\), so \(762048=2^6\times3^5\times7^2\).

Step 3

Exam Tip

Write 11907 as prime powers. चरण 1: \(762048=64\times11907\) लिखें। चरण 2: \(64=2^6\) और \(11907=3^5\times7^2\), इसलिए \(762048=2^6\times3^5\times7^2\)। चरण 3: 11907 को अभाज्य घातों में लिखें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 415800 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 415800?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(415800=8\times51975\).

Step 2

Why this answer is correct

\(51975=3^3\times5^2\times7\times11\), so \(415800=2^3\times3^3\times5^2\times7\times11\).

Step 3

Exam Tip

Do not leave 51975 in the final form. चरण 1: \(415800=8\times51975\) लिखें। चरण 2: \(51975=3^3\times5^2\times7\times11\), इसलिए \(415800=2^3\times3^3\times5^2\times7\times11\)। चरण 3: 51975 को अंतिम रूप में न छोड़ें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 79200 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 79200?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(79200=32\times2475\).

Step 2

Why this answer is correct

\(2475=3^2\times5^2\times11\), so \(79200=2^5\times3^2\times5^2\times11\).

Step 3

Exam Tip

Give 2475 its complete prime form. चरण 1: \(79200=32\times2475\) लिखें। चरण 2: \(2475=3^2\times5^2\times11\), इसलिए \(79200=2^5\times3^2\times5^2\times11\)। चरण 3: 2475 को पूरा अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 174636 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 174636?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(174636=4\times43659\).

Step 2

Why this answer is correct

\(4=2^2\) and \(43659=3^4\times7^2\times11\), so the correct form is \(2^2\times3^4\times7^2\times11\).

Step 3

Exam Tip

Convert 43659 into prime powers. चरण 1: \(174636=4\times43659\) लिखें। चरण 2: \(4=2^2\) और \(43659=3^4\times7^2\times11\), इसलिए सही रूप \(2^2\times3^4\times7^2\times11\) है। चरण 3: 43659 को अभाज्य घातों में बदलें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 155520 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 155520?

Explanation opens after your attempt
Correct Answer

A. \(2^7\times3^5\times5\)

Step 1

Concept

Write \(155520=128\times1215\).

Step 2

Why this answer is correct

\(128=2^7\) and \(1215=3^5\times5\), so \(155520=2^7\times3^5\times5\).

Step 3

Exam Tip

Convert 1215 into prime powers. चरण 1: \(155520=128\times1215\) लिखें। चरण 2: \(128=2^7\) और \(1215=3^5\times5\), इसलिए \(155520=2^7\times3^5\times5\)। चरण 3: 1215 को अभाज्य घातों में बदलें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 529200 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 529200?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^3\times5^2\times7^2\)

Step 1

Concept

Write \(529200=16\times33075\).

Step 2

Why this answer is correct

\(33075=3^3\times5^2\times7^2\), so \(529200=2^4\times3^3\times5^2\times7^2\).

Step 3

Exam Tip

Do not leave 33075 in the final form. चरण 1: \(529200=16\times33075\) लिखें। चरण 2: \(33075=3^3\times5^2\times7^2\), इसलिए \(529200=2^4\times3^3\times5^2\times7^2\)। चरण 3: 33075 को अंतिम रूप में न छोड़ें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 177408 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 177408?

Explanation opens after your attempt
Correct Answer

A. \(2^8\times3^2\times7\times11\)

Step 1

Concept

Write \(177408=256\times693\).

Step 2

Why this answer is correct

\(256=2^8\) and \(693=3^2\times7\times11\), so the correct form is \(2^8\times3^2\times7\times11\).

Step 3

Exam Tip

Convert 693 into prime form. चरण 1: \(177408=256\times693\) लिखें। चरण 2: \(256=2^8\) और \(693=3^2\times7\times11\), इसलिए सही रूप \(2^8\times3^2\times7\times11\) है। चरण 3: 693 को अभाज्य रूप में बदलें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 289575 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 289575?

Explanation opens after your attempt
Correct Answer

A. \(3^4\times5^2\times11\times13\)

Step 1

Concept

Write \(289575=81\times3575\).

Step 2

Why this answer is correct

\(81=3^4\) and \(3575=5^2\times11\times13\), so \(289575=3^4\times5^2\times11\times13\).

Step 3

Exam Tip

Give 3575 its complete prime form too. चरण 1: \(289575=81\times3575\) लिखें। चरण 2: \(81=3^4\) और \(3575=5^2\times11\times13\), इसलिए \(289575=3^4\times5^2\times11\times13\)। चरण 3: 3575 को भी पूरा अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 196000 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 196000?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times5^3\times7^2\)

Step 1

Concept

Write \(196000=32\times6125\).

Step 2

Why this answer is correct

\(32=2^5\) and \(6125=5^3\times7^2\), so \(196000=2^5\times5^3\times7^2\).

Step 3

Exam Tip

Do not keep 6125 in the final answer. चरण 1: \(196000=32\times6125\) लिखें। चरण 2: \(32=2^5\) और \(6125=5^3\times7^2\), इसलिए \(196000=2^5\times5^3\times7^2\)। चरण 3: 6125 को अंतिम उत्तर में न रखें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 379080 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 379080?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^6\times5\times13\)

Step 1

Concept

Write \(379080=8\times47385\).

Step 2

Why this answer is correct

\(8=2^3\) and \(47385=3^6\times5\times13\), so the correct form is \(2^3\times3^6\times5\times13\).

Step 3

Exam Tip

Give 47385 its complete prime form. चरण 1: \(379080=8\times47385\) लिखें। चरण 2: \(8=2^3\) और \(47385=3^6\times5\times13\), इसलिए सही रूप \(2^3\times3^6\times5\times13\) है। चरण 3: 47385 को पूरी तरह अभाज्य रूप दें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 627264 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 627264?

Explanation opens after your attempt
Correct Answer

A. \(2^6\times3^4\times11^2\)

Step 1

Concept

Write \(627264=64\times9801\).

Step 2

Why this answer is correct

\(64=2^6\) and \(9801=3^4\times11^2\), so \(627264=2^6\times3^4\times11^2\).

Step 3

Exam Tip

Convert 9801 into prime powers. चरण 1: \(627264=64\times9801\) लिखें। चरण 2: \(64=2^6\) और \(9801=3^4\times11^2\), इसलिए \(627264=2^6\times3^4\times11^2\)। चरण 3: 9801 को अभाज्य घातों में बदलें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

संख्या 680400 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 680400?

Explanation opens after your attempt
Correct Answer

A. \(2^4\times3^5\times5^2\times7\)

Step 1

Concept

Write \(680400=16\times42525\).

Step 2

Why this answer is correct

\(16=2^4\) and \(42525=3^5\times5^2\times7\), so the correct form is \(2^4\times3^5\times5^2\times7\).

Step 3

Exam Tip

Do not leave 42525 in the final form. चरण 1: \(680400=16\times42525\) लिखें। चरण 2: \(16=2^4\) और \(42525=3^5\times5^2\times7\), इसलिए सही रूप \(2^4\times3^5\times5^2\times7\) है। चरण 3: 42525 को अंतिम रूप में न छोड़ें।

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Question Expert Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 8

अंतिम अभाज्य गुणनखंडन में संयुक्त आधार बच जाने पर उत्तर अधूरा क्यों माना जाता है?

Why is the answer considered incomplete if a composite base remains in final prime factorisation?

Explanation opens after your attempt
Correct Answer

A. क्योंकि अंतिम रूप में केवल अभाज्य आधार रहने चाहिएBecause only prime bases should remain in the final form

Step 1

Concept

The aim of prime factorisation is to write the number using prime bases only.

Step 2

Why this answer is correct

If a base like (45) or (121) remains, it is composite and must be broken further.

Step 3

Exam Tip

Before writing the final answer, check whether every base is prime. चरण 1: अभाज्य गुणनखंडन का उद्देश्य संख्या को केवल अभाज्य आधारों में लिखना है। चरण 2: यदि (45) या (121) जैसा आधार बचा है, तो वह संयुक्त है और आगे टूटेगा। चरण 3: अंतिम उत्तर लिखने से पहले हर आधार की अभाज्यता जांचें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

अभाज्य गुणनखंडन में \(8\times27225\) को अंतिम उत्तर क्यों नहीं माना जाएगा?

Why will \(8\times27225\) not be considered the final answer in prime factorisation?

Explanation opens after your attempt
Correct Answer

A. क्योंकि 8 और 27225 संयुक्त रूप हैंBecause 8 and 27225 are composite forms

Step 1

Concept

In final prime factorisation, every base should be prime.

Step 2

Why this answer is correct

\(8=2^3\) and \(27225=3^2\times5^2\times11^2\).

Step 3

Exam Tip

Therefore, the final form is \(2^3\times3^2\times5^2\times11^2\). चरण 1: अंतिम अभाज्य गुणनखंडन में हर आधार अभाज्य होना चाहिए। चरण 2: \(8=2^3\) और \(27225=3^2\times5^2\times11^2\) है। चरण 3: इसलिए अंतिम रूप \(2^3\times3^2\times5^2\times11^2\) होगा।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

संख्या 279936 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 279936?

Explanation opens after your attempt
Correct Answer

A. \(2^7\times3^7\)

Step 1

Concept

(279936) can be written as \(128\times2187\).

Step 2

Why this answer is correct

\(128=2^7\) and \(2187=3^7\), so \(279936=2^7\times3^7\).

Step 3

Exam Tip

Do not leave 128 and 2187 in the final form. चरण 1: \(279936=128\times2187\) लिखा जा सकता है। चरण 2: \(128=2^7\) और \(2187=3^7\), इसलिए \(279936=2^7\times3^7\)। चरण 3: 128 और 2187 को अंतिम रूप में न छोड़ें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

संख्या 217800 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 217800?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(217800=8\times27225\).

Step 2

Why this answer is correct

\(27225=3^2\times5^2\times11^2\), so \(217800=2^3\times3^2\times5^2\times11^2\).

Step 3

Exam Tip

Convert 27225 into prime powers. चरण 1: \(217800=8\times27225\) लिखें। चरण 2: \(27225=3^2\times5^2\times11^2\), इसलिए \(217800=2^3\times3^2\times5^2\times11^2\)। चरण 3: 27225 को अभाज्य घातों में बदलें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

संख्या 158760 का सही अभाज्य गुणनखंडन क्या है?

What is the correct prime factorisation of 158760?

Explanation opens after your attempt
Correct Answer

A. \(2^3\times3^4\times5\times7^2\)

Step 1

Concept

Write \(158760=8\times19845\).

Step 2

Why this answer is correct

\(19845=3^4\times5\times7^2\), so \(158760=2^3\times3^4\times5\times7^2\).

Step 3

Exam Tip

Break 19845 completely. चरण 1: \(158760=8\times19845\) लिखें। चरण 2: \(19845=3^4\times5\times7^2\), इसलिए \(158760=2^3\times3^4\times5\times7^2\)। चरण 3: 19845 को पूरी तरह तोड़ें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(2^2\times3^2\times5^2\times11^2\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?

If \(2^2\times3^2\times5^2\times11^2\) is the prime factorisation of a number, what is the number?

Explanation opens after your attempt
Correct Answer

A. 108900

Step 1

Concept

Calculate \(2^2=4\), \(3^2=9\), \(5^2=25\), and \(11^2=121\).

Step 2

Why this answer is correct

\(4\times9\times25\times121=108900\).

Step 3

Exam Tip

This is a square form, so observe the powers carefully. चरण 1: \(2^2=4\), \(3^2=9\), \(5^2=25\) और \(11^2=121\) निकालें। चरण 2: \(4\times9\times25\times121=108900\)। चरण 3: यह वर्ग रूप है, इसलिए घातों को ध्यान से देखें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(2^6\times3^3\times5^2\) किसी संख्या का अभाज्य गुणनखंडन है, तो संख्या क्या है?

If \(2^6\times3^3\times5^2\) is the prime factorisation of a number, what is the number?

Explanation opens after your attempt
Correct Answer

A. 43200

Step 1

Concept

\(2^6=64\), \(3^3=27\), and \(5^2=25\).

Step 2

Why this answer is correct

\(64\times27\times25=43200\).

Step 3

Exam Tip

Solving powers first helps find the correct option quickly. चरण 1: \(2^6=64\), \(3^3=27\) और \(5^2=25\) हैं। चरण 2: \(64\times27\times25=43200\)। चरण 3: घातों को पहले हल करने से सही विकल्प जल्दी मिलता है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस विकल्प में अभाज्य गुणनखंडन अधूरा है?

Which option has incomplete prime factorisation?

Explanation opens after your attempt
Correct Answer

A. \(2^5\times81\times49\)

Step 1

Concept

In an incomplete form, composite bases remain.

Step 2

Why this answer is correct

81 and 49 are composite bases.

Step 3

Exam Tip

\(2^5\times81\times49\) must be changed into \(2^5\times3^4\times7^2\). चरण 1: अधूरे रूप में संयुक्त आधार बच जाता है। चरण 2: 81 और 49 संयुक्त आधार हैं। चरण 3: \(2^5\times81\times49\) को \(2^5\times3^4\times7^2\) में बदलना होगा।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस विकल्प में केवल अंतिम अभाज्य गुणनखंडन दिया गया है?

Which option gives only the final prime factorisation?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

In the final form, every base must be prime.

Step 2

Why this answer is correct

In the first option, bases 2, 3, 5, and 7 are prime.

Step 3

Exam Tip

8, 135, 49, 27, 245, and 216 are composite, so they are not final forms. चरण 1: अंतिम रूप में हर आधार अभाज्य होना चाहिए। चरण 2: पहले विकल्प में आधार 2, 3, 5 और 7 अभाज्य हैं। चरण 3: 8, 135, 49, 27, 245 और 216 संयुक्त हैं, इसलिए वे अंतिम रूप नहीं हैं।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस विकल्प में \(2^3\times3^5\times7^2\) का सही मान है?

Which option gives the correct value of \(2^3\times3^5\times7^2\)?

Explanation opens after your attempt
Correct Answer

A. 95256

Step 1

Concept

Calculate \(2^3=8\), \(3^5=243\), and \(7^2=49\).

Step 2

Why this answer is correct

\(8\times243\times49=95256\).

Step 3

Exam Tip

Simplify higher powers first. चरण 1: \(2^3=8\), \(3^5=243\) और \(7^2=49\) निकालें। चरण 2: \(8\times243\times49=95256\)। चरण 3: बड़ी घातों को पहले सरल करें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस विकल्प में \(2^5\times3^4\times7^2\) का सही मान है?

Which option gives the correct value of \(2^5\times3^4\times7^2\)?

Explanation opens after your attempt
Correct Answer

A. 127008

Step 1

Concept

Calculate \(2^5=32\), \(3^4=81\), and \(7^2=49\).

Step 2

Why this answer is correct

\(32\times81\times49=127008\).

Step 3

Exam Tip

Solve all three powers separately first. चरण 1: \(2^5=32\), \(3^4=81\) और \(7^2=49\) निकालें। चरण 2: \(32\times81\times49=127008\)। चरण 3: तीनों घातों को पहले अलग-अलग हल करें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(x=2^8\times5^3\times7^2\times13\) और \(y=2^5\times3^2\times5^4\times13^3\), तो (xy) में 13 की घात क्या होगी?

If \(x=2^8\times5^3\times7^2\times13\) and \(y=2^5\times3^2\times5^4\times13^3\), what will be the power of 13 in (xy)?

Explanation opens after your attempt
Correct Answer

A. 4

Step 1

Concept

Powers with the same base 13 are added in multiplication.

Step 2

Why this answer is correct

The power of 13 in (x) is 1 and in (y) is 3.

Step 3

Exam Tip

The total power will be (1+3=4). चरण 1: समान आधार 13 की घातें गुणा में जुड़ेंगी। चरण 2: (x) में 13 की घात 1 है और (y) में 13 की घात 3 है। चरण 3: कुल घात (1+3=4) होगी।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(a=2^6\times3^4\times5^2\times11\) और \(b=2^3\times3^5\times7\times11^2\), तो (ab) में 3 की घात क्या होगी?

If \(a=2^6\times3^4\times5^2\times11\) and \(b=2^3\times3^5\times7\times11^2\), what will be the power of 3 in (ab)?

Explanation opens after your attempt
Correct Answer

A. 9

Step 1

Concept

In multiplication, powers of the same prime base are added.

Step 2

Why this answer is correct

The power of 3 in (a) is 4 and in (b) is 5.

Step 3

Exam Tip

In (ab), the power of 3 will be (4+5=9). चरण 1: गुणा में समान अभाज्य आधार की घातें जुड़ती हैं। चरण 2: (a) में 3 की घात 4 है और (b) में 3 की घात 5 है। चरण 3: (ab) में 3 की घात (4+5=9) होगी।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि किसी संख्या का अभाज्य गुणनखंडन \(2^6\times3^5\times7^4\times11^3\times13^2\) है, तो उसमें अलग-अलग अभाज्य गुणनखंड कितने हैं?

If a number has prime factorisation \(2^6\times3^5\times7^4\times11^3\times13^2\), how many distinct prime factors does it have?

Explanation opens after your attempt
Correct Answer

A. 5

Step 1

Concept

While counting distinct primes, only bases are counted.

Step 2

Why this answer is correct

The bases are 2, 3, 7, 11, and 13.

Step 3

Exam Tip

Therefore, the number of distinct prime factors is 5. चरण 1: अलग-अलग अभाज्य गिनते समय केवल आधार गिने जाते हैं। चरण 2: आधार 2, 3, 7, 11 और 13 हैं। चरण 3: इसलिए अलग-अलग अभाज्य गुणनखंडों की संख्या 5 है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि किसी संख्या का अभाज्य गुणनखंडन \(2^9\times3^7\times5^4\times7^3\times11^2\) है, तो दोहराव सहित अभाज्य गुणनखंडों की संख्या कितनी है?

If a number has prime factorisation \(2^9\times3^7\times5^4\times7^3\times11^2\), how many prime factors does it have with repetition?

Explanation opens after your attempt
Correct Answer

A. 25

Step 1

Concept

To count with repetition, add the exponents.

Step 2

Why this answer is correct

(9+7+4+3+2=25).

Step 3

Exam Tip

Keep the number of bases and the total count with repetition separate. चरण 1: दोहराव सहित गिनती में घातों को जोड़ा जाता है। चरण 2: (9+7+4+3+2=25)। चरण 3: आधारों की संख्या और दोहराव सहित कुल संख्या को अलग रखें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(n=2^{10}\times3^5\times7^4\times13^2\), तो (n) किस संख्या से अवश्य विभाज्य होगा?

If \(n=2^{10}\times3^5\times7^4\times13^2\), by which number must (n) be divisible?

Explanation opens after your attempt
Correct Answer

A. \(2^9\times3^4\times7^3\times13\)

Step 1

Concept

For divisibility, exponents in the divisor must not exceed those in the given number.

Step 2

Why this answer is correct

\(2^9\), \(3^4\), \(7^3\), and 13 are all available in (n).

Step 3

Exam Tip

Therefore, (n) must be divisible by the first option. चरण 1: विभाज्यता के लिए भाजक की घातें दी गई संख्या की घातों से अधिक नहीं होनी चाहिए। चरण 2: \(2^9\), \(3^4\), \(7^3\) और 13 सभी (n) में उपलब्ध हैं। चरण 3: इसलिए (n) पहले विकल्प से अवश्य विभाज्य होगा।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(n=2^8\times3^6\times5^4\times11^2\), तो (n) किस संख्या से विभाज्य नहीं होगा?

If \(n=2^8\times3^6\times5^4\times11^2\), by which number will (n) not be divisible?

Explanation opens after your attempt
Correct Answer

A. \(11^3\)

Step 1

Concept

Every prime power of a divisor must be available in the number.

Step 2

Why this answer is correct

(n) has power 2 of 11, but \(11^3\) needs power 3.

Step 3

Exam Tip

Therefore, (n) is not divisible by \(11^3\). चरण 1: भाजक की हर अभाज्य घात संख्या में उपलब्ध होनी चाहिए। चरण 2: (n) में 11 की घात 2 है, पर \(11^3\) के लिए घात 3 चाहिए। चरण 3: इसलिए (n), \(11^3\) से विभाज्य नहीं होगा।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि संख्या \(2^{10}\times3^8\times5^5\times7^4\) है, तो इसे पूर्ण घन बनाने के लिए सबसे छोटी किस संख्या से भाग देना होगा?

If the number is \(2^{10}\times3^8\times5^5\times7^4\), by which smallest number should it be divided to make a perfect cube?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

For a perfect cube, exponents should be multiples of 3.

Step 2

Why this answer is correct

Reducing 10 to 9, 8 to 6, 5 to 3, and 4 to 3 is the smallest way.

Step 3

Exam Tip

Therefore, the divisor is \(2\times3^2\times5^2\times7\). चरण 1: पूर्ण घन के लिए घातें 3 के गुणज चाहिए। चरण 2: 10 को 9, 8 को 6, 5 को 3 और 4 को 3 तक घटाना सबसे छोटा तरीका है। चरण 3: इसलिए भाजक \(2\times3^2\times5^2\times7\) है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि संख्या \(2^5\times3^7\times5^4\times13^2\) है, तो इसे पूर्ण घन बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?

If the number is \(2^5\times3^7\times5^4\times13^2\), by which smallest number should it be multiplied to make a perfect cube?

Explanation opens after your attempt
Correct Answer

A. \(2\times3^2\times5^2\times13\)

Step 1

Concept

In a perfect cube, every exponent must be a multiple of 3.

Step 2

Why this answer is correct

Powers 5, 7, 4, and 2 must become 6, 9, 6, and 3.

Step 3

Exam Tip

The smallest multiplier is \(2\times3^2\times5^2\times13\). चरण 1: पूर्ण घन में हर घात 3 का गुणज होनी चाहिए। चरण 2: 5 को 6, 7 को 9, 4 को 6 और 2 को 3 बनाना होगा। चरण 3: सबसे छोटा गुणक \(2\times3^2\times5^2\times13\) है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि संख्या \(2^9\times3^4\times5^7\times11^2\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से भाग देना होगा?

If the number is \(2^9\times3^4\times5^7\times11^2\), by which smallest number should it be divided to make a perfect square?

Explanation opens after your attempt
Correct Answer

A. \(2\times5\)

Step 1

Concept

For a perfect square, exponents should be even.

Step 2

Why this answer is correct

The powers of 2 and 5 are odd.

Step 3

Exam Tip

Dividing by \(2\times5\) makes the powers 8 and 6. चरण 1: पूर्ण वर्ग के लिए घातें सम चाहिए। चरण 2: 2 की घात 9 और 5 की घात 7 विषम हैं। चरण 3: \(2\times5\) से भाग देने पर घातें 8 और 6 हो जाएंगी।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि संख्या \(2^8\times3^5\times5^2\times7^3\) है, तो इसे पूर्ण वर्ग बनाने के लिए सबसे छोटी किस संख्या से गुणा करना होगा?

If the number is \(2^8\times3^5\times5^2\times7^3\), by which smallest number should it be multiplied to make a perfect square?

Explanation opens after your attempt
Correct Answer

A. 21

Step 1

Concept

In a perfect square, all exponents should be even.

Step 2

Why this answer is correct

The powers of 3 and 7 are odd.

Step 3

Exam Tip

Multiplying by \(3\times7=21\) makes both powers even. चरण 1: पूर्ण वर्ग में सभी घातें सम होनी चाहिए। चरण 2: 3 की घात 5 और 7 की घात 3 विषम हैं। चरण 3: \(3\times7=21\) से गुणा करने पर दोनों घातें सम हो जाएंगी।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस संख्या का अभाज्य गुणनखंडन \(2^5\times3^4\times7^2\) है?

Which number has prime factorisation \(2^5\times3^4\times7^2\)?

Explanation opens after your attempt
Correct Answer

A. 127008

Step 1

Concept

Calculate \(2^5=32\), \(3^4=81\), and \(7^2=49\).

Step 2

Why this answer is correct

\(32\times81\times49=127008\).

Step 3

Exam Tip

Solving powers first keeps multiplication clear. चरण 1: \(2^5=32\), \(3^4=81\) और \(7^2=49\) निकालें। चरण 2: \(32\times81\times49=127008\)। चरण 3: घातों को पहले हल करने से गुणा साफ रहता है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस संख्या का अभाज्य गुणनखंडन \(2^6\times3^3\times5^2\) है?

Which number has prime factorisation \(2^6\times3^3\times5^2\)?

Explanation opens after your attempt
Correct Answer

A. 43200

Step 1

Concept

Calculate \(2^6=64\), \(3^3=27\), and \(5^2=25\).

Step 2

Why this answer is correct

\(64\times27\times25=43200\).

Step 3

Exam Tip

Simplify all three powers separately. चरण 1: \(2^6=64\), \(3^3=27\) और \(5^2=25\) निकालें। चरण 2: \(64\times27\times25=43200\)। चरण 3: तीनों घातों को अलग-अलग सरल करें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस संख्या का अभाज्य गुणनखंडन \(2\times3\times5\times7\times11^2\) है?

Which number has prime factorisation \(2\times3\times5\times7\times11^2\)?

Explanation opens after your attempt
Correct Answer

A. 25410

Step 1

Concept

Calculate \(11^2=121\).

Step 2

Why this answer is correct

\(2\times3\times5\times7\times121=25410\).

Step 3

Exam Tip

First take the product of smaller factors as 210, then multiply by 121. चरण 1: \(11^2=121\) निकालें। चरण 2: \(2\times3\times5\times7\times121=25410\)। चरण 3: पहले छोटे गुणनखंडों का गुणनफल 210 लें, फिर 121 से गुणा करें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस संख्या का अभाज्य गुणनखंडन \(2^5\times3^4\times7\) है?

Which number has prime factorisation \(2^5\times3^4\times7\)?

Explanation opens after your attempt
Correct Answer

A. 18144

Step 1

Concept

Calculate \(2^5=32\) and \(3^4=81\).

Step 2

Why this answer is correct

\(32\times81\times7=18144\).

Step 3

Exam Tip

It is better to find higher powers first. चरण 1: \(2^5=32\) और \(3^4=81\) निकालें। चरण 2: \(32\times81\times7=18144\)। चरण 3: बड़ी घातों का मान पहले निकालना ठीक रहता है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

किस संख्या का अभाज्य गुणनखंडन \(2^4\times3^4\times11\) है?

Which number has prime factorisation \(2^4\times3^4\times11\)?

Explanation opens after your attempt
Correct Answer

A. 14256

Step 1

Concept

Calculate \(2^4=16\) and \(3^4=81\).

Step 2

Why this answer is correct

\(16\times81\times11=14256\).

Step 3

Exam Tip

Solve powers first, then multiply by 11. चरण 1: \(2^4=16\) और \(3^4=81\) निकालें। चरण 2: \(16\times81\times11=14256\)। चरण 3: घातों को पहले हल करें, फिर 11 से गुणा करें।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(68040=2^3\times3^p\times5\times7\), तो (p) का मान क्या है?

If \(68040=2^3\times3^p\times5\times7\), what is the value of (p)?

Explanation opens after your attempt
Correct Answer

A. 5

Step 1

Concept

Write \(68040=8\times8505\).

Step 2

Why this answer is correct

\(8505=3^5\times5\times7\), so the power of 3 is 5.

Step 3

Exam Tip

Comparing gives (p=5). चरण 1: \(68040=8\times8505\) लिखें। चरण 2: \(8505=3^5\times5\times7\), इसलिए 3 की घात 5 है। चरण 3: तुलना करने पर (p=5) मिलेगा।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(108900=2^2\times3^2\times5^m\times11^2\), तो (m) का मान क्या है?

If \(108900=2^2\times3^2\times5^m\times11^2\), what is the value of (m)?

Explanation opens after your attempt
Correct Answer

A. 2

Step 1

Concept

\(108900=900\times121\).

Step 2

Why this answer is correct

\(900=2^2\times3^2\times5^2\) and \(121=11^2\), so the power of 5 is 2.

Step 3

Exam Tip

Comparing gives (m=2). चरण 1: \(108900=900\times121\) है। चरण 2: \(900=2^2\times3^2\times5^2\) और \(121=11^2\), इसलिए 5 की घात 2 है। चरण 3: तुलना करने पर (m=2) होगा।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(95256=2^3\times3^b\times7^2\), तो (b) का मान क्या है?

If \(95256=2^3\times3^b\times7^2\), what is the value of (b)?

Explanation opens after your attempt
Correct Answer

A. 5

Step 1

Concept

Write \(95256=8\times11907\).

Step 2

Why this answer is correct

\(11907=3^5\times7^2\), so \(95256=2^3\times3^5\times7^2\).

Step 3

Exam Tip

From the given form, (b=5). चरण 1: \(95256=8\times11907\) लिखें। चरण 2: \(11907=3^5\times7^2\), इसलिए \(95256=2^3\times3^5\times7^2\)। चरण 3: दिए गए रूप से (b=5) है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

यदि \(127008=2^a\times3^4\times7^2\), तो (a) का मान क्या है?

If \(127008=2^a\times3^4\times7^2\), what is the value of (a)?

Explanation opens after your attempt
Correct Answer

A. 5

Step 1

Concept

\(127008=32\times3969\).

Step 2

Why this answer is correct

\(32=2^5\) and \(3969=3^4\times7^2\), so the power of 2 is 5.

Step 3

Exam Tip

Comparing gives (a=5). चरण 1: \(127008=32\times3969\) है। चरण 2: \(32=2^5\) और \(3969=3^4\times7^2\), इसलिए 2 की घात 5 है। चरण 3: तुलना करने पर (a=5) मिलता है।

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Question Hard Mathematics Chapter 1: Real Numbers 3: Prime factorisation Class 10 Level 9

संख्या 130680 का अभाज्य गुणनखंडन कौन सा है?

Which is the prime factorisation of 130680?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

Write \(130680=1080\times121\).

Step 2

Why this answer is correct

\(1080=2^3\times3^3\times5\) and \(121=11^2\), so \(130680=2^3\times3^3\times5\times11^2\).

Step 3

Exam Tip

Break 1080 and 121 completely. चरण 1: \(130680=1080\times121\) लिखें। चरण 2: \(1080=2^3\times3^3\times5\) और \(121=11^2\), इसलिए \(130680=2^3\times3^3\times5\times11^2\)। चरण 3: 1080 और 121 को पूरा तोड़ें।

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