Concept-wise Practice

error-check MCQ Questions for Class 10

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

Practice Questions

13 questions tagged with error-check.

समीकरणों (5x+6y=37) और (5x-2y=13) को हल करने पर (xy) का मान क्या है?

Solving (5x+6y=37) and (5x-2y=13), what is the value of (xy)?

Explanation opens after your attempt
Correct Answer

A. (9)

Step 1

Concept

This question needs careful substitution after elimination; careless cancellation gives a wrong value. Check each obtained value in both equations before marking.

Step 2

Why this answer is correct

The correct answer is A. (9). This question needs careful substitution after elimination; careless cancellation gives a wrong value. Check each obtained value in both equations before marking.

Step 3

Exam Tip

घटाने पर (8y=24), इसलिए (y=3) और \(x=\frac{19}{5}\) नहीं बल्कि दूसरे में रखने से \(x=\frac{19}{5}\) नहीं आता; सही हल (x=5,y=2) नहीं है, इसलिए सावधानी चाहिए।

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विलोपन विधि से (4x+y=18) और (4x-y=10) का हल क्या है?

What is the solution of (4x+y=18) and (4x-y=10) by elimination?

Explanation opens after your attempt
Correct Answer

C. ( (3,6) )

Step 1

Concept

Adding gives (8x=28), so \(x=\frac{7}{2}\), which is not among the options. Check calculations carefully.

Step 2

Why this answer is correct

The correct answer is C. ( (3,6) ). Adding gives (8x=28), so \(x=\frac{7}{2}\), which is not among the options. Check calculations carefully.

Step 3

Exam Tip

जोड़ने पर (8x=28)? नहीं, सही जोड़ (8x=28) देता है इसलिए \(x=\frac{7}{2}\), यह विकल्पों में नहीं है। सावधानी से जाँचें।

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समीकरण (2x-3y=1) और (x+y=7) का प्रतिच्छेद बिंदु क्या है?

What is the intersection point of (2x-3y=1) and (x+y=7)?

Explanation opens after your attempt
Correct Answer

B. बिंदु (\left\(4,3\right\))Point (\left\(4,3\right\))

Step 1

Concept

Substituting (\left\(4,3\right\)) gives (2\left\(4\right\)-3\left\(3\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{22}{5},\frac{13}{5}\right\)).

Step 2

Why this answer is correct

The correct answer is B. बिंदु (\left\(4,3\right\)) / Point (\left\(4,3\right\)). Substituting (\left\(4,3\right\)) gives (2\left\(4\right\)-3\left\(3\right\)=-1), so it is not correct. The correct solution is (\left\(\frac{22}{5},\frac{13}{5}\right\)).

Step 3

Exam Tip

(\left\(4,3\right\)) रखने पर (2\left\(4\right\)-3\left\(3\right\)=-1), इसलिए यह नहीं है। सही हल (\left\(\frac{22}{5},\frac{13}{5}\right\)) है।

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समीकरण (2x+3y=18) और (x-y=1) का ग्राफीय हल क्या है?

What is the graphical solution of (2x+3y=18) and (x-y=1)?

Explanation opens after your attempt
Correct Answer

B. ( (4,3) )

Step 1

Concept

Checking ( (4,3) ) gives (2(4)+3(3)=17), not (18). Correct calculation gives (x=3) and (y=2).

Step 2

Why this answer is correct

The correct answer is B. ( (4,3) ). Checking ( (4,3) ) gives (2(4)+3(3)=17), not (18). Correct calculation gives (x=3) and (y=2).

Step 3

Exam Tip

( (4,3) ) पर (2(4)+3(3)=17) नहीं है, इसलिए पहले जाँचें। सही गणना से (x=3) और (y=2) मिलता है।

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यदि \(x^2-9x+20=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\) का मान क्या है?

If \(\alpha,\beta\) are the roots of \(x^2-9x+20=0\), what is \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\)?

Explanation opens after your attempt
Correct Answer

B. \(\frac{17}{3}\)

Step 1

Concept

The roots are (4) and (5). Direct substitution gives \(\frac{6}{2}+\frac{7}{3}=\frac{16}{3}\), so option (A) should be correct.

Step 2

Why this answer is correct

The correct answer is B. \(\frac{17}{3}\). The roots are (4) and (5). Direct substitution gives \(\frac{6}{2}+\frac{7}{3}=\frac{16}{3}\), so option (A) should be correct.

Step 3

Exam Tip

जड़ें (4) और (5) हैं। सीधे रखने पर \(\frac{6}{2}+\frac{7}{3}=\frac{16}{3}\) मिलता है, इसलिए विकल्प (A) सही होना चाहिए।

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(9x-2-6(a-1)x+a-2-4a-5=0) की जड़ें वास्तविक हों, तो (a) पर सही शर्त क्या है?

For (9x-2-6(a-1)x+a-2-4a-5=0) to have real roots, what is the correct condition on (a)?

Explanation opens after your attempt
Correct Answer

A. \(a\ge-\frac{7}{2}\)

Step 1

Concept

For real roots, \(D\ge0\) is required. Here (D=36(a-1)2-36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).

Step 2

Why this answer is correct

The correct answer is A. \(a\ge-\frac{7}{2}\). For real roots, \(D\ge0\) is required. Here (D=36(a-1)2-36\(a^2-4a-5\)=72a+216), so the exact condition is \(a\ge-3\), not \(a\ge-\frac{7}{2}\).

Step 3

Exam Tip

वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=36(a-1)2-36\(a^2-4a-5\)=72a+216), इसलिए \(a\ge-\frac{7}{2}\) नहीं बल्कि \(a\ge-3\) होगा।

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यदि \(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2-4\alpha+\beta^2-4\beta\) का मान क्या है?

If \(\alpha,\beta\) are roots of \(x^2-5x+6=0\), what is \(\alpha^2-4\alpha+\beta^2-4\beta\)?

Explanation opens after your attempt
Correct Answer

A. (5)

Step 1

Concept

(\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta=13). Thus the value is (13-4(5)=-7), so none of the options is correct.

Step 2

Why this answer is correct

The correct answer is A. (5). (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta=13). Thus the value is (13-4(5)=-7), so none of the options is correct.

Step 3

Exam Tip

(\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta=13) है। इसलिए मान (13-4(5)=-7) होगा, अतः कोई विकल्प सही नहीं है।

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यदि \(x^2-10x+m=0\) की दोनों जड़ें अभाज्य संख्याएँ हैं, तो (m) के संभव मानों का योग क्या है?

If both roots of \(x^2-10x+m=0\) are prime numbers, what is the sum of possible values of (m)?

Explanation opens after your attempt
Correct Answer

A. (42)

Step 1

Concept

The prime pairs with sum (10) are ((3,7)) and ((5,5)). Hence (m=21) or (m=25), and their sum is (46), so option (B) should be correct.

Step 2

Why this answer is correct

The correct answer is A. (42). The prime pairs with sum (10) are ((3,7)) and ((5,5)). Hence (m=21) or (m=25), and their sum is (46), so option (B) should be correct.

Step 3

Exam Tip

योग (10) वाली अभाज्य जोड़ियाँ ((3,7)) और ((5,5)) हैं। इसलिए (m=21) या (m=25), और योग (46) है, अतः विकल्प (B) सही होना चाहिए।

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यदि (x-2-(m-2)x+m-6=0) की एक जड़ (3) है, तो दूसरी जड़ क्या होगी?

If one root of (x-2-(m-2)x+m-6=0) is (3), what is the other root?

Explanation opens after your attempt
Correct Answer

A. (1)

Step 1

Concept

Putting (x=3) gives (9-3(m-2)+m-6=0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.

Step 2

Why this answer is correct

The correct answer is A. (1). Putting (x=3) gives (9-3(m-2)+m-6=0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.

Step 3

Exam Tip

(x=3) रखने पर (9-3(m-2)+m-6=0), इसलिए \(m=\frac{9}{2}\)। गुणनफल \(m-6=-\frac{3}{2}\) है, अतः दूसरी जड़ \(-\frac{1}{2}\) होगी, इसलिए कोई विकल्प सही नहीं है।

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यदि \(x^2-3x-1=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2+3\beta+\alpha\beta\) का मान क्या है जब \(\alpha\) बड़ी जड़ है?

If \(\alpha,\beta\) are the roots of \(x^2-3x-1=0\), what is \(\alpha^2+3\beta+\alpha\beta\) when \(\alpha\) is the larger root?

Explanation opens after your attempt
Correct Answer

A. (8)

Step 1

Concept

Since \(\alpha^2=3\alpha+1\), the expression becomes (3\alpha+1+3\beta-1=3\(\alpha+\beta\)=9), so no option is correct. The correct value is (9).

Step 2

Why this answer is correct

The correct answer is A. (8). Since \(\alpha^2=3\alpha+1\), the expression becomes (3\alpha+1+3\beta-1=3\(\alpha+\beta\)=9), so no option is correct. The correct value is (9).

Step 3

Exam Tip

क्योंकि \(\alpha^2=3\alpha+1\), व्यंजक (3\alpha+1+3\beta-1=3\(\alpha+\beta\)=9) बनता है, इसलिए कोई विकल्प सही नहीं है। सही मान (9) होगा।

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यदि \(x^2-8x+15=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\) का मान क्या है?

If \(\alpha,\beta\) are the roots of \(x^2-8x+15=0\), what is \(\frac{\alpha+2}{\alpha-2}+\frac{\beta+2}{\beta-2}\)?

Explanation opens after your attempt
Correct Answer

A. (8)

Step 1

Concept

The roots are (3) and (5). Substitution gives \(\frac{5}{1}+\frac{7}{3}=\frac{22}{3}\), so none of the options is correct; the correct value should be \(\frac{22}{3}\).

Step 2

Why this answer is correct

The correct answer is A. (8). The roots are (3) and (5). Substitution gives \(\frac{5}{1}+\frac{7}{3}=\frac{22}{3}\), so none of the options is correct; the correct value should be \(\frac{22}{3}\).

Step 3

Exam Tip

जड़ें (3) और (5) हैं। सीधे रखने पर \(\frac{5}{1}+\frac{7}{3}=\frac{22}{3}\) आता है, इसलिए विकल्पों में कोई सही नहीं; सही प्रश्न के लिए उत्तर \(\frac{22}{3}\) होना चाहिए।

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यदि \(x=1+\sqrt{2}\), तो \(x^3-3x\) का मान क्या है?

If \(x=1+\sqrt{2}\), what is the value of \(x^3-3x\)?

Explanation opens after your attempt
Correct Answer

A. \(2\sqrt{2}\)

Step 1

Concept

\(x^2=3+2\sqrt{2}\) and \(x^3=7+5\sqrt{2}\), so \(x^3-3x=4+2\sqrt{2}\). The correct value is not in the options so calculate carefully.

Step 2

Why this answer is correct

The correct answer is A. \(2\sqrt{2}\). \(x^2=3+2\sqrt{2}\) and \(x^3=7+5\sqrt{2}\), so \(x^3-3x=4+2\sqrt{2}\). The correct value is not in the options so calculate carefully.

Step 3

Exam Tip

\(x^2=3+2\sqrt{2}\) और \(x^3=7+5\sqrt{2}\), इसलिए \(x^3-3x=4+2\sqrt{2}\) होता है। सही मान विकल्पों में नहीं है इसलिए गणना सावधानी से करें।

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यदि \(x=2+\sqrt{7}\), तो \(x+\frac{1}{x}\) का मान क्या है?

If \(x=2+\sqrt{7}\), what is the value of \(x+\frac{1}{x}\)?

Explanation opens after your attempt
Correct Answer

A. \(2\sqrt{7}\)

Step 1

Concept

\(\frac{1}{2+\sqrt{7}}\) equals \(\frac{2-\sqrt{7}}{-3}=\frac{\sqrt{7}-2}{3}\). So \(x+\frac{1}{x}=2+\sqrt{7}+\frac{\sqrt{7}-2}{3}=\frac{4+4\sqrt{7}}{3}\).

Step 2

Why this answer is correct

The correct answer is A. \(2\sqrt{7}\). \(\frac{1}{2+\sqrt{7}}\) equals \(\frac{2-\sqrt{7}}{-3}=\frac{\sqrt{7}-2}{3}\). So \(x+\frac{1}{x}=2+\sqrt{7}+\frac{\sqrt{7}-2}{3}=\frac{4+4\sqrt{7}}{3}\).

Step 3

Exam Tip

\(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}\) नहीं बल्कि हर (4-7=-3) से मान \(\frac{2-\sqrt{7}}{3}\) है इसलिए सीधा विकल्प नहीं बनेगा। सही सरलीकरण जांचे बिना उत्तर न चुनें।

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