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assumption MCQ Questions for Class 10

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3 questions tagged with assumption.

Question 1/3 Expert Mathematics Real Numbers Proof of irrationality of √2, √3, √5 Class 10 Level 17

\(\sqrt{2}\) के प्रमाण में (p) और (q) दोनों सम निकलने पर कौन-सा प्रारंभिक कथन झूठा सिद्ध होता है?

In the proof for \(\sqrt{2}\), when both (p) and (q) turn out even, which initial statement is proved false?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt{2}\) परिमेय है\(\sqrt{2}\) is rational

Step 1

Concept

We initially assumed that \(\sqrt{2}\) is rational.

Step 2

Why this answer is correct

That assumption led to a common factor in a lowest-form fraction.

Step 3

Exam Tip

Therefore the initial rational assumption is proved false. चरण 1: हमने शुरुआत में \(\sqrt{2}\) को परिमेय माना था। चरण 2: उसी मान्यता से सरलतम भिन्न में साझा गुणनखंड आ गया। चरण 3: इसलिए प्रारंभिक परिमेय मान्यता झूठी सिद्ध होती है।

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Question 2/3 Medium Mathematics Real Numbers Proof of irrationality of √2, √3, √5 Class 10 Level 16

\(\sqrt{3}\) को परिमेय मानने पर \(\sqrt{3}=\frac{p}{q}\) लिखा गया। यदि (p) और (q) में साझा गुणनखंड (3) मिल जाए, तो कौन सी मान्यता गलत सिद्ध होती है?

After assuming \(\sqrt{3}\) rational, \(\sqrt{3}=\frac{p}{q}\) is written. If common factor (3) is found in (p) and (q), which assumption is proved false?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt{3}\) परिमेय है\(\sqrt{3}\) is rational

Step 1

Concept

After assuming rationality, (p) and (q) were taken coprime.

Step 2

Why this answer is correct

Finding common factor (3) makes this assumption impossible.

Step 3

Exam Tip

So the rational assumption is false and \(\sqrt{3}\) is irrational. चरण 1: परिमेय मानकर (p) और (q) को सहअभाज्य माना गया था। चरण 2: साझा गुणनखंड (3) मिलना इस मान्यता को असंभव बनाता है। चरण 3: इसलिए मूल परिमेय मान्यता गलत और \(\sqrt{3}\) अपरिमेय है।

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Question 3/3 Easy Mathematics Real Numbers Proof of irrationality of √2, √3, √5 Class 10 Level 16

\(\sqrt{5}\) को अपरिमेय सिद्ध करने की शुरुआत में क्या माना जाता है?

What is assumed at the beginning to prove that \(\sqrt{5}\) is irrational?

Explanation opens after your attempt
Correct Answer

A. \(\sqrt{5}\) परिमेय है और \(\sqrt{5}=\frac{p}{q}\), जहां (p) और (q) सहअभाज्य हैं\(\sqrt{5}\) is rational and \(\sqrt{5}=\frac{p}{q}\), where (p) and (q) are coprime

Step 1

Concept

In the contradiction method, we assume the opposite.

Step 2

Why this answer is correct

So \(\sqrt{5}\) is assumed rational and written as \(\frac{p}{q}\).

Step 3

Exam Tip

Do not forget to mention that (p) and (q) are coprime. चरण 1: विरोधाभास विधि में उलटी बात मानते हैं। चरण 2: इसलिए \(\sqrt{5}\) को परिमेय मानकर \(\frac{p}{q}\) के रूप में लिखा जाता है। चरण 3: (p) और (q) को सहअभाज्य लिखना न भूलें।

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