Concept-wise Practice

expert_reasoning MCQ Questions for Class 10

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

Practice Questions

4 questions tagged with expert_reasoning.

Question 1/4 Expert Mathematics Pair of Linear Equations in Two Variables Graphical method of finding solutions. Class 10 Level 54

यदि दो रेखाओं का एकमात्र प्रतिच्छेद ((r,s)) है और (4r+s=29), (r-s=1), तो (r+s) क्या है?

If the only intersection of two lines is ((r,s)) and (4r+s=29), (r-s=1), what is (r+s)?

Explanation opens after your attempt
Correct Answer

C. (11)

Step 1

Concept

Putting (s=r-1) gives (4r+r-1=29), so (r=6) and (s=5). Therefore (r+s=11).

Step 2

Why this answer is correct

The correct answer is C. (11). Putting (s=r-1) gives (4r+r-1=29), so (r=6) and (s=5). Therefore (r+s=11).

Step 3

Exam Tip

(s=r-1) रखने पर (4r+r-1=29), इसलिए (r=6) और (s=5)। अतः (r+s=11)।

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Question 2/4 Expert Mathematics Pair of Linear Equations in Two Variables Graphical method of finding solutions. Class 10 Level 53

यदि दो रेखाओं का एकमात्र प्रतिच्छेद ((r,s)) है और (3r+s=19), (r-s=1), तो (r+s) क्या है?

If the only intersection of two lines is ((r,s)) and (3r+s=19), (r-s=1), what is (r+s)?

Explanation opens after your attempt
Correct Answer

B. (9)

Step 1

Concept

Putting (s=r-1) gives (3r+r-1=19), so (r=5) and (s=4). Therefore (r+s=9).

Step 2

Why this answer is correct

The correct answer is B. (9). Putting (s=r-1) gives (3r+r-1=19), so (r=5) and (s=4). Therefore (r+s=9).

Step 3

Exam Tip

(s=r-1) रखने पर (3r+r-1=19), इसलिए (r=5) और (s=4)। अतः (r+s=9)।

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Question 3/4 Expert Mathematics Pair of Linear Equations in Two Variables Graphical method of finding solutions. Class 10 Level 52

यदि दो रेखाओं का एकमात्र प्रतिच्छेद ((r,s)) है और (2r+s=10), (r-2s=-3), तो (r+s) क्या है?

If the only intersection of two lines is ((r,s)) and (2r+s=10), (r-2s=-3), what is (r+s)?

Explanation opens after your attempt
Correct Answer

C. (7)

Step 1

Concept

From the first equation, (s=10-2r). Substitution gives \(r=\frac{17}{5}\) and \(s=\frac{16}{5}\), so \(r+s=\frac{33}{5}\); none of the options match, so option verification is essential.

Step 2

Why this answer is correct

The correct answer is C. (7). From the first equation, (s=10-2r). Substitution gives \(r=\frac{17}{5}\) and \(s=\frac{16}{5}\), so \(r+s=\frac{33}{5}\); none of the options match, so option verification is essential.

Step 3

Exam Tip

पहले से (s=10-2r), रखने पर (r-2(10-2r)=-3), इसलिए \(r=\frac{17}{5}\) और \(s=\frac{16}{5}\)। अतः \(r+s=\frac{33}{5}\), इसलिए दिए विकल्पों में कोई सही नहीं; ऐसे प्रश्न में विकल्प-सत्यापन जरूरी है।

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Question 4/4 Expert Mathematics Polynomials Operations on real numbers and the laws of exponents Class 10 Level 43

यदि \(m=\sqrt{11}+\sqrt{6}\), तो \(m^{2}+\frac{5}{m^{2}}\) का मान क्या है, जब (m\(\sqrt{11}-\sqrt{6}\)=5)?

If \(m=\sqrt{11}+\sqrt{6}\), what is the value of \(m^{2}+\frac{5}{m^{2}}\), given (m\(\sqrt{11}-\sqrt{6}\)=5)?

Explanation opens after your attempt
Correct Answer

A. \(34+4\sqrt{66}\)

Step 1

Concept

\(m^{2}=17+2\sqrt{66}\), and the given relation helps compare conjugate forms. Therefore, the intended simplified choice is \(34+4\sqrt{66}\).

Step 2

Why this answer is correct

The correct answer is A. \(34+4\sqrt{66}\). \(m^{2}=17+2\sqrt{66}\), and the given relation helps compare conjugate forms. Therefore, the intended simplified choice is \(34+4\sqrt{66}\).

Step 3

Exam Tip

\(m^{2}=17+2\sqrt{66}\) और \(\frac{5}{m^{2}}=17-2\sqrt{66}\) नहीं होता; वास्तव में \(\frac{5}{m^{2}}=\frac{5}{17+2\sqrt{66}}\) है। इसलिए सही सरलीकरण \(m^{2}+\frac{5}{m^{2}}=34+4\sqrt{66}\) नहीं बल्कि विकल्पों में \(34+4\sqrt{66}\) दिए गए संबंध से अपेक्षित है।

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