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Mathematics Quiz - Class 10 Mathematics Hard Level 58 Quiz

Question 1/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (3x+ky=12) और (9x+15y=36) के अनंत हल होने के लिए (k) का मान क्या होगा?

What is the value of (k) for the equations (3x+ky=12) and (9x+15y=36) to have infinitely many solutions?

Explanation opens after your attempt

Correct Answer

C. (5)

Step 1

Concept

For infinitely many solutions, (3/9=k/15=12/36) must hold. Therefore, (k=5) is correct.

Step 2

Why this answer is correct

The correct answer is C. (5). For infinitely many solutions, (3/9=k/15=12/36) must hold. Therefore, (k=5) is correct.

Step 3

Exam Tip

अनंत हल के लिए (3/9=k/15=12/36) होना चाहिए। इसलिए (k=5) सही है।

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Question 2/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (ax+4y=10) और (6x+8y=25) का कोई हल न होने के लिए (a) का मान क्या होगा?

What is the value of (a) for the equations (ax+4y=10) and (6x+8y=25) to have no solution?

Explanation opens after your attempt

Correct Answer

B. (3)

Step 1

Concept

For no solution, (a/6=4/8) and (10/25) must be different. This gives (a=3).

Step 2

Why this answer is correct

The correct answer is B. (3). For no solution, (a/6=4/8) and (10/25) must be different. This gives (a=3).

Step 3

Exam Tip

कोई हल नहीं के लिए (a/6=4/8) और (10/25) अलग होना चाहिए। इससे (a=3) मिलता है।

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Question 3/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

यदि (2x+5y=17) और (4x+10y=m) असंगत युग्म हों, तो (m) के लिए सही शर्त क्या है?

If (2x+5y=17) and (4x+10y=m) form an inconsistent pair, what is the correct condition for (m)?

Explanation opens after your attempt

Correct Answer

B. \(m \ne 34\)

Step 1

Concept

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, \(m \ne 34\).

Step 2

Why this answer is correct

The correct answer is B. \(m \ne 34\). The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, \(m \ne 34\).

Step 3

Exam Tip

पहले दो अनुपात बराबर हैं, इसलिए असंगत होने के लिए स्थिर पद का अनुपात अलग होना चाहिए। अतः \(m \ne 34\) होगा।

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Question 4/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (5x-3y=11) और (10x-6y=22) के लिए सही निष्कर्ष कौन-सा है?

Which conclusion is correct for the equations (5x-3y=11) and (10x-6y=22)?

Explanation opens after your attempt

Correct Answer

C. अनंत हलInfinitely many solutions

Step 1

Concept

The second equation is (2) times the first. Therefore, both lines are coincident and have infinitely many solutions.

Step 2

Why this answer is correct

The correct answer is C. अनंत हल / Infinitely many solutions. The second equation is (2) times the first. Therefore, both lines are coincident and have infinitely many solutions.

Step 3

Exam Tip

दूसरा समीकरण पहले का (2) गुना है। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।

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Question 5/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (7x+4y=19) और (14x+8y=41) के लिए हलों की संख्या क्या होगी?

How many solutions will the equations (7x+4y=19) and (14x+8y=41) have?

Explanation opens after your attempt

Correct Answer

C. कोई हल नहींNo solution

Step 1

Concept

Here (7/14=4/8) but (19/41) is different. Therefore, the lines are parallel and distinct.

Step 2

Why this answer is correct

The correct answer is C. कोई हल नहीं / No solution. Here (7/14=4/8) but (19/41) is different. Therefore, the lines are parallel and distinct.

Step 3

Exam Tip

यहां (7/14=4/8) लेकिन (19/41) अलग है। इसलिए रेखाएं समानांतर और अलग हैं।

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Question 6/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (4x+7y=23) और (9x+15y=49) के लिए सही हल-स्थिति क्या है?

What is the correct solution status for the equations (4x+7y=23) and (9x+15y=49)?

Explanation opens after your attempt

Correct Answer

B. एक अद्वितीय हलOne unique solution

Step 1

Concept

Here \(4/9 \ne 7/15\), so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.

Step 2

Why this answer is correct

The correct answer is B. एक अद्वितीय हल / One unique solution. Here \(4/9 \ne 7/15\), so the lines intersect at one point. If the first two ratios are different, one unique solution is obtained.

Step 3

Exam Tip

यहां \(4/9 \ne 7/15\), इसलिए रेखाएं एक बिंदु पर कटती हैं। पहले दो अनुपात अलग हों तो एक अद्वितीय हल मिलता है।

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Question 7/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

यदि \(a_1/a_2=b_1/b_2=c_1/c_2\), तो युग्म की सही प्रकृति क्या है?

If \(a_1/a_2=b_1/b_2=c_1/c_2\), what is the correct nature of the pair?

Explanation opens after your attempt

Correct Answer

C. संगत और आश्रितConsistent and dependent

Step 1

Concept

If all three ratios are equal, both equations form the same line. Therefore, the pair is consistent and dependent.

Step 2

Why this answer is correct

The correct answer is C. संगत और आश्रित / Consistent and dependent. If all three ratios are equal, both equations form the same line. Therefore, the pair is consistent and dependent.

Step 3

Exam Tip

तीनों अनुपात बराबर हों तो दोनों समीकरण एक ही रेखा बनाते हैं। इसलिए युग्म संगत और आश्रित होता है।

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Question 8/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

यदि \(a_1/a_2=b_1/b_2 \ne c_1/c_2\), तो ग्राफ में कौन-सी स्थिति बनेगी?

If \(a_1/a_2=b_1/b_2 \ne c_1/c_2\), which situation will occur in the graph?

Explanation opens after your attempt

Correct Answer

C. अलग समानांतर रेखाएंDistinct parallel lines

Step 1

Concept

If the first two ratios are equal and the third differs, the lines are parallel and distinct. Therefore, there is no common solution.

Step 2

Why this answer is correct

The correct answer is C. अलग समानांतर रेखाएं / Distinct parallel lines. If the first two ratios are equal and the third differs, the lines are parallel and distinct. Therefore, there is no common solution.

Step 3

Exam Tip

पहले दो अनुपात बराबर और तीसरा अलग हो तो रेखाएं समानांतर और अलग होती हैं। इसलिए कोई सामान्य हल नहीं होता।

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Question 9/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (px+6y=18) और (10x+15y=45) के अनंत हल होने के लिए (p) क्या होगा?

What will (p) be for the equations (px+6y=18) and (10x+15y=45) to have infinitely many solutions?

Explanation opens after your attempt

Correct Answer

B. (3)

Step 1

Concept

For infinitely many solutions, (p/10=6/15=18/45) must hold. Since the common ratio is (2/5), (p=4) would be required.

Step 2

Why this answer is correct

The correct answer is B. (3). For infinitely many solutions, (p/10=6/15=18/45) must hold. Since the common ratio is (2/5), (p=4) would be required.

Step 3

Exam Tip

अनंत हल के लिए (p/10=6/15=18/45) होना चाहिए। इसलिए (p=4) नहीं, (p=4) नहीं बल्कि (p=4) अनुपात बिगाड़ता है और सही (p=4) नहीं है।

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Question 10/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (7x+qy=29) और (14x+10y=58) के अनंत हल होने के लिए (q) का मान क्या है?

What is the value of (q) for the equations (7x+qy=29) and (14x+10y=58) to have infinitely many solutions?

Explanation opens after your attempt

Correct Answer

C. (5)

Step 1

Concept

The second equation is (2) times the first, so (q/10=1/2) must hold. Hence, (q=5).

Step 2

Why this answer is correct

The correct answer is C. (5). The second equation is (2) times the first, so (q/10=1/2) must hold. Hence, (q=5).

Step 3

Exam Tip

दूसरा समीकरण पहले का (2) गुना है, इसलिए (q/10=1/2) होना चाहिए। अतः (q=5)।

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Question 11/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (6x+ky=24) और (3x+5y=17) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?

Which condition is correct for the equations (6x+ky=24) and (3x+5y=17) to have a unique solution?

Explanation opens after your attempt

Correct Answer

B. \(k \ne 10\)

Step 1

Concept

For a unique solution, \(6/3 \ne k/5\) must hold. Therefore, \(k \ne 10\) is the correct condition.

Step 2

Why this answer is correct

The correct answer is B. \(k \ne 10\). For a unique solution, \(6/3 \ne k/5\) must hold. Therefore, \(k \ne 10\) is the correct condition.

Step 3

Exam Tip

अद्वितीय हल के लिए \(6/3 \ne k/5\) होना चाहिए। इसलिए \(k \ne 10\) सही शर्त है।

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Question 12/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (5x+ay=35) और (15x+18y=105) के अनंत हल होने के लिए (a) क्या होगा?

What will (a) be for the equations (5x+ay=35) and (15x+18y=105) to have infinitely many solutions?

Explanation opens after your attempt

Correct Answer

C. (6)

Step 1

Concept

For infinitely many solutions, (5/15=a/18=35/105) must hold. This gives (a=6).

Step 2

Why this answer is correct

The correct answer is C. (6). For infinitely many solutions, (5/15=a/18=35/105) must hold. This gives (a=6).

Step 3

Exam Tip

अनंत हल के लिए (5/15=a/18=35/105) होना चाहिए। इससे (a=6) मिलता है।

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Question 13/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (bx+8y=32) और (12x+24y=71) का कोई हल न होने के लिए (b) का मान क्या होगा?

What is the value of (b) for the equations (bx+8y=32) and (12x+24y=71) to have no solution?

Explanation opens after your attempt

Correct Answer

B. (4)

Step 1

Concept

For no solution, (b/12=8/24) and (32/71) must be different. Hence, (b=4).

Step 2

Why this answer is correct

The correct answer is B. (4). For no solution, (b/12=8/24) and (32/71) must be different. Hence, (b=4).

Step 3

Exam Tip

कोई हल नहीं के लिए (b/12=8/24) और (32/71) अलग होना चाहिए। इसलिए (b=4)।

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Question 14/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (14x+21y=49) और (2x+3y=8) के ग्राफ के बारे में सही कथन कौन-सा है?

Which statement is correct about the graph of the equations (14x+21y=49) and (2x+3y=8)?

Explanation opens after your attempt

Correct Answer

C. रेखाएं अलग समानांतर हैंLines are distinct parallel

Step 1

Concept

Here (14/2=21/3) but (49/8) is different. Therefore, the graph will show distinct parallel lines.

Step 2

Why this answer is correct

The correct answer is C. रेखाएं अलग समानांतर हैं / Lines are distinct parallel. Here (14/2=21/3) but (49/8) is different. Therefore, the graph will show distinct parallel lines.

Step 3

Exam Tip

यहां (14/2=21/3) लेकिन (49/8) अलग है। इसलिए ग्राफ में अलग समानांतर रेखाएं बनेंगी।

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Question 15/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (15x+10y=40) और (3x+2y=8) के ग्राफ में क्या दिखेगा?

What will be shown in the graph of the equations (15x+10y=40) and (3x+2y=8)?

Explanation opens after your attempt

Correct Answer

A. एक ही रेखाSame line

Step 1

Concept

The first equation is (5) times the second. Therefore, both equations show the same line.

Step 2

Why this answer is correct

The correct answer is A. एक ही रेखा / Same line. The first equation is (5) times the second. Therefore, both equations show the same line.

Step 3

Exam Tip

पहला समीकरण दूसरे का (5) गुना है। इसलिए दोनों समीकरण एक ही रेखा दिखाते हैं।

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Question 16/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (2x+11y=27) और (5x+19y=46) के ग्राफ का सही वर्णन क्या है?

What is the correct description of the graph of the equations (2x+11y=27) and (5x+19y=46)?

Explanation opens after your attempt

Correct Answer

C. एक बिंदु पर कटती रेखाएंLines intersecting at one point

Step 1

Concept

Here \(2/5 \ne 11/19\), so the lines will intersect at one point. This gives one unique solution.

Step 2

Why this answer is correct

The correct answer is C. एक बिंदु पर कटती रेखाएं / Lines intersecting at one point. Here \(2/5 \ne 11/19\), so the lines will intersect at one point. This gives one unique solution.

Step 3

Exam Tip

यहां \(2/5 \ne 11/19\), इसलिए रेखाएं एक बिंदु पर कटेंगी। इससे एक अद्वितीय हल मिलेगा।

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Question 17/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (4x+9y-31=0) और (12x+27y-93=0) के लिए सही अनुपात संबंध कौन-सा है?

Which ratio relation is correct for the equations (4x+9y-31=0) and (12x+27y-93=0)?

Explanation opens after your attempt

Correct Answer

C. (412=9 / 27=(-31) / (-93))

Step 1

Concept

Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.

Step 2

Why this answer is correct

The correct answer is C. (4 / 12=9 / 27=(-31) / (-93)). Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.

Step 3

Exam Tip

यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।

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Question 18/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (8x-3y+22=0) और (16x-6y+47=0) के लिए सही अनुपात संबंध क्या है?

What is the correct ratio relation for the equations (8x-3y+22=0) and (16x-6y+47=0)?

Explanation opens after your attempt

Correct Answer

A. (816=(-3) / (-6) \ne 22 / 47)

Step 1

Concept

The first two ratios are equal and the third is different. Therefore, there will be no solution.

Step 2

Why this answer is correct

The correct answer is A. (8 / 16=(-3) / (-6) \ne 22 / 47). The first two ratios are equal and the third is different. Therefore, there will be no solution.

Step 3

Exam Tip

पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा।

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Question 19/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (9x+5y-28=0) और (18x+11y-56=0) के लिए सही निष्कर्ष क्या है?

What is the correct conclusion for the equations (9x+5y-28=0) and (18x+11y-56=0)?

Explanation opens after your attempt

Correct Answer

C. एक अद्वितीय हलOne unique solution

Step 1

Concept

Here \(9/18 \ne 5/11\), so the lines intersect at one point. In this case, one unique solution is obtained.

Step 2

Why this answer is correct

The correct answer is C. एक अद्वितीय हल / One unique solution. Here \(9/18 \ne 5/11\), so the lines intersect at one point. In this case, one unique solution is obtained.

Step 3

Exam Tip

यहां \(9/18 \ne 5/11\), इसलिए रेखाएं एक बिंदु पर कटती हैं। ऐसी स्थिति में एक अद्वितीय हल मिलता है।

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Question 20/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (3x+by=24) और (12x+20y=96) के अनंत हल होने के लिए (b) का मान क्या है?

What is the value of (b) for the equations (3x+by=24) and (12x+20y=96) to have infinitely many solutions?

Explanation opens after your attempt

Correct Answer

C. (5)

Step 1

Concept

For infinitely many solutions, (3/12=b/20=24/96) must hold. Therefore, (b=5).

Step 2

Why this answer is correct

The correct answer is C. (5). For infinitely many solutions, (3/12=b/20=24/96) must hold. Therefore, (b=5).

Step 3

Exam Tip

अनंत हल के लिए (3/12=b/20=24/96) होना चाहिए। इसलिए (b=5)।

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Question 21/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (11x+py=33) और (4x+2y=15) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?

Which condition is correct for the equations (11x+py=33) and (4x+2y=15) to have a unique solution?

Explanation opens after your attempt

Correct Answer

B. (p \ne 112)

Step 1

Concept

For a unique solution, \(11/4 \ne p/2\) must hold. Therefore, \(p \ne 11/2\) is the correct condition.

Step 2

Why this answer is correct

The correct answer is B. \(p \ne 11 / 2\). For a unique solution, \(11/4 \ne p/2\) must hold. Therefore, \(p \ne 11/2\) is the correct condition.

Step 3

Exam Tip

अद्वितीय हल के लिए \(11/4 \ne p/2\) होना चाहिए। इसलिए \(p \ne 11/2\) सही शर्त है।

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Question 22/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (6x+7y=41) और (18x+21y=123) को देखकर सबसे उचित निष्कर्ष क्या है?

What is the most suitable conclusion by observing the equations (6x+7y=41) and (18x+21y=123)?

Explanation opens after your attempt

Correct Answer

C. अनंत हल हैंThere are infinitely many solutions

Step 1

Concept

The second equation is (3) times the first. Therefore, both lines are coincident.

Step 2

Why this answer is correct

The correct answer is C. अनंत हल हैं / There are infinitely many solutions. The second equation is (3) times the first. Therefore, both lines are coincident.

Step 3

Exam Tip

दूसरा समीकरण पहले का (3) गुना है। इसलिए दोनों रेखाएं संपाती हैं।

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Question 23/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (12x+16y=36) और (3x+4y=11) को देखकर कौन-सा निष्कर्ष सही है?

Which conclusion is correct by observing the equations (12x+16y=36) and (3x+4y=11)?

Explanation opens after your attempt

Correct Answer

B. पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग हैFirst two ratios are equal but the constant ratio is different

Step 1

Concept

Here (12/3=16/4) but (36/11) is different. Therefore, there will be no solution.

Step 2

Why this answer is correct

The correct answer is B. पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है / First two ratios are equal but the constant ratio is different. Here (12/3=16/4) but (36/11) is different. Therefore, there will be no solution.

Step 3

Exam Tip

यहां (12/3=16/4) लेकिन (36/11) अलग है। इसलिए कोई हल नहीं होगा।

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Question 24/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (11x+6y=35) और (5x+3y=17) में (a) और (b) के अनुपातों की तुलना से क्या पता चलता है?

What is found by comparing the ratios of (a) and (b) in the equations (11x+6y=35) and (5x+3y=17)?

Explanation opens after your attempt

Correct Answer

C. (115 \ne 6 / 3) इसलिए एक अद्वितीय हल / 3) so one unique solution

Step 1

Concept

Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.

Step 2

Why this answer is correct

The correct answer is C. \(11 / 5 \ne 6 / 3\) इसलिए एक अद्वितीय हल / 3) so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.

Step 3

Exam Tip

यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं।

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Question 25/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (16x+24y=88) और (2x+3y=11) का युग्म किस प्रकार का है?

What type of pair is formed by the equations (16x+24y=88) and (2x+3y=11)?

Explanation opens after your attempt

Correct Answer

A. संगत और आश्रितConsistent and dependent

Step 1

Concept

The first equation is (8) times the second. Therefore, both are the same line and the pair is dependent.

Step 2

Why this answer is correct

The correct answer is A. संगत और आश्रित / Consistent and dependent. The first equation is (8) times the second. Therefore, both are the same line and the pair is dependent.

Step 3

Exam Tip

पहला समीकरण दूसरे का (8) गुना है। इसलिए दोनों एक ही रेखा हैं और युग्म आश्रित है।

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Question 26/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (18x+27y=63) और (2x+3y=8) का युग्म किस प्रकार का है?

What type of pair is formed by the equations (18x+27y=63) and (2x+3y=8)?

Explanation opens after your attempt

Correct Answer

C. असंगतInconsistent

Step 1

Concept

Here (18/2=27/3) but (63/8) is different. Therefore, this is an inconsistent pair.

Step 2

Why this answer is correct

The correct answer is C. असंगत / Inconsistent. Here (18/2=27/3) but (63/8) is different. Therefore, this is an inconsistent pair.

Step 3

Exam Tip

यहां (18/2=27/3) लेकिन (63/8) अलग है। इसलिए यह असंगत युग्म है।

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Question 27/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (13x+4y=39) और (6x+2y=17) का युग्म किस प्रकार का है?

What type of pair is formed by the equations (13x+4y=39) and (6x+2y=17)?

Explanation opens after your attempt

Correct Answer

C. संगत और स्वतंत्रConsistent and independent

Step 1

Concept

Here \(13/6 \ne 4/2\), so the lines intersect. Hence, the pair is consistent and independent.

Step 2

Why this answer is correct

The correct answer is C. संगत और स्वतंत्र / Consistent and independent. Here \(13/6 \ne 4/2\), so the lines intersect. Hence, the pair is consistent and independent.

Step 3

Exam Tip

यहां \(13/6 \ne 4/2\), इसलिए रेखाएं कटती हैं। अतः युग्म संगत और स्वतंत्र है।

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Question 28/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (5x+6y=32) और (15x+18y=96) में तीनों अनुपातों का संबंध क्या है?

What is the relation among all three ratios in the equations (5x+6y=32) and (15x+18y=96)?

Explanation opens after your attempt

Correct Answer

A. तीनों बराबर हैंAll three are equal

Step 1

Concept

Here (5/15=6/18=32/96). Therefore, both equations form the same line.

Step 2

Why this answer is correct

The correct answer is A. तीनों बराबर हैं / All three are equal. Here (5/15=6/18=32/96). Therefore, both equations form the same line.

Step 3

Exam Tip

यहां (5/15=6/18=32/96)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।

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Question 29/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (10x+15y=50) और (2x+3y=13) के लिए कौन-सा संबंध सही है?

Which relation is correct for the equations (10x+15y=50) and (2x+3y=13)?

Explanation opens after your attempt

Correct Answer

C. (102=15 / 3 \ne 50 / 13)

Step 1

Concept

The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.

Step 2

Why this answer is correct

The correct answer is C. \(10 / 2=15 / 3 \ne 50 / 13\). The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.

Step 3

Exam Tip

पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।

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Question 30/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (7x+8y=26) और (14x+15y=51) के लिए कौन-सा कथन सही है?

Which statement is correct for the equations (7x+8y=26) and (14x+15y=51)?

Explanation opens after your attempt

Correct Answer

B. (714 \ne 8 / 15), इसलिए एक अद्वितीय हल / 15), so one unique solution

Step 1

Concept

Here \(7/14 \ne 8/15\), so the lines intersect. Therefore, there is one unique solution.

Step 2

Why this answer is correct

The correct answer is B. \(7 / 14 \ne 8 / 15\), इसलिए एक अद्वितीय हल / 15), so one unique solution. Here \(7/14 \ne 8/15\), so the lines intersect. Therefore, there is one unique solution.

Step 3

Exam Tip

यहां \(7/14 \ne 8/15\), इसलिए रेखाएं कटती हैं। इस कारण एक अद्वितीय हल है।

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Question 31/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (4x+5y=29) और (8x+10y=s) के संगत और आश्रित होने के लिए (s) क्या होगा?

What should (s) be for the equations (4x+5y=29) and (8x+10y=s) to be consistent and dependent?

Explanation opens after your attempt

Correct Answer

C. (58)

Step 1

Concept

To be consistent and dependent, the second equation must be (2) times the first. Hence, (s=58).

Step 2

Why this answer is correct

The correct answer is C. (58). To be consistent and dependent, the second equation must be (2) times the first. Hence, (s=58).

Step 3

Exam Tip

संगत और आश्रित होने के लिए दूसरा समीकरण पहले का (2) गुना होना चाहिए। अतः (s=58)।

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Question 32/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (6x+7y=31) और (12x+14y=r) के असंगत होने के लिए कौन-सी शर्त सही है?

Which condition is correct for the equations (6x+7y=31) and (12x+14y=r) to be inconsistent?

Explanation opens after your attempt

Correct Answer

B. \(r \ne 62\)

Step 1

Concept

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, \(r \ne 62\).

Step 2

Why this answer is correct

The correct answer is B. \(r \ne 62\). The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, \(r \ne 62\).

Step 3

Exam Tip

पहले दो अनुपात बराबर हैं, इसलिए असंगत होने के लिए स्थिर पद का अनुपात अलग होना चाहिए। अतः \(r \ne 62\)।

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Question 33/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (5x+3y=14) और (10x+ay=33) का कोई हल न होने के लिए (a) क्या होगा?

What will (a) be for the equations (5x+3y=14) and (10x+ay=33) to have no solution?

Explanation opens after your attempt

Correct Answer

C. (6)

Step 1

Concept

For no solution, (5/10=3/a) and (14/33) must be different. This gives (a=6).

Step 2

Why this answer is correct

The correct answer is C. (6). For no solution, (5/10=3/a) and (14/33) must be different. This gives (a=6).

Step 3

Exam Tip

कोई हल नहीं के लिए (5/10=3/a) और (14/33) अलग होना चाहिए। इससे (a=6) मिलता है।

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Question 34/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (8x+ky=72) और (2x+7y=18) के अनंत हल होने के लिए (k) क्या होगा?

What will (k) be for the equations (8x+ky=72) and (2x+7y=18) to have infinitely many solutions?

Explanation opens after your attempt

Correct Answer

C. (28)

Step 1

Concept

The first equation must be (4) times the second. Therefore, (k=28).

Step 2

Why this answer is correct

The correct answer is C. (28). The first equation must be (4) times the second. Therefore, (k=28).

Step 3

Exam Tip

पहला समीकरण दूसरे का (4) गुना होना चाहिए। इसलिए (k=28) है।

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Question 35/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (12x+5y=41) और (24x+10y=82) को देखकर कौन-सा कथन सही है?

Which statement is correct by observing the equations (12x+5y=41) and (24x+10y=82)?

Explanation opens after your attempt

Correct Answer

B. रेखाएं एक ही हैंLines are the same

Step 1

Concept

The second equation is (2) times the first. Therefore, both lines are the same.

Step 2

Why this answer is correct

The correct answer is B. रेखाएं एक ही हैं / Lines are the same. The second equation is (2) times the first. Therefore, both lines are the same.

Step 3

Exam Tip

दूसरा समीकरण पहले का (2) गुना है। इसलिए दोनों रेखाएं एक ही हैं।

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Question 36/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (14x-10y=36) और (7x-5y=19) को देखकर सही निष्कर्ष क्या है?

What is the correct conclusion by observing the equations (14x-10y=36) and (7x-5y=19)?

Explanation opens after your attempt

Correct Answer

A. कोई हल नहींNo solution

Step 1

Concept

Here (14/7=(-10)/(-5)) but (36/19) is different. Therefore, the lines are parallel and distinct.

Step 2

Why this answer is correct

The correct answer is A. कोई हल नहीं / No solution. Here (14/7=(-10)/(-5)) but (36/19) is different. Therefore, the lines are parallel and distinct.

Step 3

Exam Tip

यहां (14/7=(-10)/(-5)) लेकिन (36/19) अलग है। इसलिए रेखाएं समानांतर और अलग हैं।

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Question 37/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (17x+6y=52) और (8x+3y=25) के लिए सही हल-स्थिति क्या है?

What is the correct solution status for the equations (17x+6y=52) and (8x+3y=25)?

Explanation opens after your attempt

Correct Answer

C. एक अद्वितीय हलOne unique solution

Step 1

Concept

Here \(17/8 \ne 6/3\), so the lines will intersect. Hence, there is one unique solution.

Step 2

Why this answer is correct

The correct answer is C. एक अद्वितीय हल / One unique solution. Here \(17/8 \ne 6/3\), so the lines will intersect. Hence, there is one unique solution.

Step 3

Exam Tip

यहां \(17/8 \ne 6/3\), इसलिए रेखाएं कटेंगी। अतः एक अद्वितीय हल है।

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Question 38/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (4x+9y=26) और (12x+27y=s) के असंगत होने के लिए (s) के लिए सही शर्त क्या है?

What is the correct condition on (s) for the equations (4x+9y=26) and (12x+27y=s) to be inconsistent?

Explanation opens after your attempt

Correct Answer

B. \(s \ne 78\)

Step 1

Concept

The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, \(s \ne 78\) is correct.

Step 2

Why this answer is correct

The correct answer is B. \(s \ne 78\). The first two ratios are equal, so the constant ratio must be different for inconsistency. Hence, \(s \ne 78\) is correct.

Step 3

Exam Tip

पहले दो अनुपात बराबर हैं, इसलिए असंगत होने के लिए स्थिर पद का अनुपात अलग होना चाहिए। अतः \(s \ne 78\) सही है।

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Question 39/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (mx+5y=15) और (18x+15y=45) के अनंत हल होने के लिए (m) का मान क्या होगा?

What will be the value of (m) for the equations (mx+5y=15) and (18x+15y=45) to have infinitely many solutions?

Explanation opens after your attempt

Correct Answer

C. (6)

Step 1

Concept

For infinitely many solutions, (m/18=5/15=15/45) must hold. Therefore, (m=6) is correct.

Step 2

Why this answer is correct

The correct answer is C. (6). For infinitely many solutions, (m/18=5/15=15/45) must hold. Therefore, (m=6) is correct.

Step 3

Exam Tip

अनंत हल के लिए (m/18=5/15=15/45) होना चाहिए। इसलिए (m=6) सही है।

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Question 40/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (7x+ay=20) और (14x+10y=47) का कोई हल न होने के लिए (a) क्या होगा?

What will (a) be for the equations (7x+ay=20) and (14x+10y=47) to have no solution?

Explanation opens after your attempt

Correct Answer

A. (5)

Step 1

Concept

For no solution, (7/14=a/10) and (20/47) must be different. Therefore, (a=5).

Step 2

Why this answer is correct

The correct answer is A. (5). For no solution, (7/14=a/10) and (20/47) must be different. Therefore, (a=5).

Step 3

Exam Tip

कोई हल नहीं के लिए (7/14=a/10) और (20/47) अलग होना चाहिए। इसलिए (a=5) है।

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Question 41/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (2x+3y=8) और (5x+ky=19) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?

Which condition is correct for the equations (2x+3y=8) and (5x+ky=19) to have a unique solution?

Explanation opens after your attempt

Correct Answer

D. (k \ne 152)

Step 1

Concept

For a unique solution, \(2/5 \ne 3/k\) must hold. Therefore, \(k \ne 15/2\) is the correct condition.

Step 2

Why this answer is correct

The correct answer is D. \(k \ne 15 / 2\). For a unique solution, \(2/5 \ne 3/k\) must hold. Therefore, \(k \ne 15/2\) is the correct condition.

Step 3

Exam Tip

अद्वितीय हल के लिए \(2/5 \ne 3/k\) होना चाहिए। इसलिए \(k \ne 15/2\) सही शर्त है।

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Question 42/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (6x-4y+11=0) और (15x-10y+28=0) के लिए सही हल-स्थिति क्या है?

What is the correct solution status for the equations (6x-4y+11=0) and (15x-10y+28=0)?

Explanation opens after your attempt

Correct Answer

C. कोई हल नहींNo solution

Step 1

Concept

Here (6/15=(-4)/(-10)) but (11/28) is different. Therefore, the lines are parallel and distinct.

Step 2

Why this answer is correct

The correct answer is C. कोई हल नहीं / No solution. Here (6/15=(-4)/(-10)) but (11/28) is different. Therefore, the lines are parallel and distinct.

Step 3

Exam Tip

यहां (6/15=(-4)/(-10)) लेकिन (11/28) अलग है। इसलिए रेखाएं समानांतर और अलग हैं।

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Question 43/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (3x+8y=27) और (9x+24y=81) का ग्राफ कैसा होगा?

What will be the graph of the equations (3x+8y=27) and (9x+24y=81)?

Explanation opens after your attempt

Correct Answer

B. एक ही रेखाSame line

Step 1

Concept

The second equation is (3) times the first. Therefore, both equations show the same line in the graph.

Step 2

Why this answer is correct

The correct answer is B. एक ही रेखा / Same line. The second equation is (3) times the first. Therefore, both equations show the same line in the graph.

Step 3

Exam Tip

दूसरा समीकरण पहले का (3) गुना है। इसलिए दोनों समीकरण ग्राफ में एक ही रेखा दिखाते हैं।

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Question 44/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (11x+7y=30) और (4x+3y=12) का युग्म किस प्रकार का है?

What type of pair is formed by the equations (11x+7y=30) and (4x+3y=12)?

Explanation opens after your attempt

Correct Answer

C. संगत और स्वतंत्रConsistent and independent

Step 1

Concept

Here \(11/4 \ne 7/3\), so the lines intersect at one point. Hence, the pair is consistent and independent.

Step 2

Why this answer is correct

The correct answer is C. संगत और स्वतंत्र / Consistent and independent. Here \(11/4 \ne 7/3\), so the lines intersect at one point. Hence, the pair is consistent and independent.

Step 3

Exam Tip

यहां \(11/4 \ne 7/3\), इसलिए रेखाएं एक बिंदु पर कटती हैं। अतः युग्म संगत और स्वतंत्र है।

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Question 45/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

दो वस्तुओं की कीमतों के लिए (6x+5y=240) और (12x+10y=490) समीकरण बने। यह प्रणाली कैसी है?

For prices of two items, the equations (6x+5y=240) and (12x+10y=490) are formed. What type of system is this?

Explanation opens after your attempt

Correct Answer

C. असंगतInconsistent

Step 1

Concept

The first two ratios are equal but (240/490) is different. Therefore, the system is inconsistent.

Step 2

Why this answer is correct

The correct answer is C. असंगत / Inconsistent. The first two ratios are equal but (240/490) is different. Therefore, the system is inconsistent.

Step 3

Exam Tip

पहले दो अनुपात बराबर हैं लेकिन (240/490) अलग है। इसलिए प्रणाली असंगत है।

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Question 46/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (5x+6y=31) और (15x+18y=r) के संगत और आश्रित होने के लिए (r) क्या होगा?

What should (r) be for the equations (5x+6y=31) and (15x+18y=r) to be consistent and dependent?

Explanation opens after your attempt

Correct Answer

D. (93)

Step 1

Concept

To be consistent and dependent, the second equation must be (3) times the first. Hence, (r=93).

Step 2

Why this answer is correct

The correct answer is D. (93). To be consistent and dependent, the second equation must be (3) times the first. Hence, (r=93).

Step 3

Exam Tip

संगत और आश्रित होने के लिए दूसरा समीकरण पहले का (3) गुना होना चाहिए। अतः (r=93)।

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Question 47/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (8x+12y=40) और (2x+3y=10) में तीनों अनुपातों का संबंध क्या है?

What is the relation among all three ratios in the equations (8x+12y=40) and (2x+3y=10)?

Explanation opens after your attempt

Correct Answer

A. तीनों बराबर हैंAll three are equal

Step 1

Concept

Here (8/2=12/3=40/10). Therefore, both equations form the same line.

Step 2

Why this answer is correct

The correct answer is A. तीनों बराबर हैं / All three are equal. Here (8/2=12/3=40/10). Therefore, both equations form the same line.

Step 3

Exam Tip

यहां (8/2=12/3=40/10)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।

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Question 48/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (16x-12y=44) और (4x-3y=12) को देखकर सही निष्कर्ष क्या है?

What is the correct conclusion by observing the equations (16x-12y=44) and (4x-3y=12)?

Explanation opens after your attempt

Correct Answer

B. कोई हल नहींNo solution

Step 1

Concept

Here (16/4=(-12)/(-3)) but (44/12) is different. Therefore, the lines are parallel and distinct.

Step 2

Why this answer is correct

The correct answer is B. कोई हल नहीं / No solution. Here (16/4=(-12)/(-3)) but (44/12) is different. Therefore, the lines are parallel and distinct.

Step 3

Exam Tip

यहां (16/4=(-12)/(-3)) लेकिन (44/12) अलग है। इसलिए रेखाएं समानांतर और अलग हैं।

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Question 49/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (9x+2y=25) और (18x+5y=54) के लिए कौन-सा कथन सही है?

Which statement is correct for the equations (9x+2y=25) and (18x+5y=54)?

Explanation opens after your attempt

Correct Answer

B. (918 \ne 2 / 5), इसलिए एक अद्वितीय हल / 5), so one unique solution

Step 1

Concept

Here \(9/18 \ne 2/5\), so the lines intersect. Therefore, there is one unique solution.

Step 2

Why this answer is correct

The correct answer is B. \(9 / 18 \ne 2 / 5\), इसलिए एक अद्वितीय हल / 5), so one unique solution. Here \(9/18 \ne 2/5\), so the lines intersect. Therefore, there is one unique solution.

Step 3

Exam Tip

यहां \(9/18 \ne 2/5\), इसलिए रेखाएं कटती हैं। इस कारण एक अद्वितीय हल है।

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Question 50/50 Hard Mathematics Pair of Linear Equations in Two Variables Conditions for solvability Class 10 Level 58

समीकरणों (cx+10y=50) और (9x+15y=80) का कोई हल न होने के लिए (c) का मान क्या होगा?

What is the value of (c) for the equations (cx+10y=50) and (9x+15y=80) to have no solution?

Explanation opens after your attempt

Correct Answer

B. (6)

Step 1

Concept

For no solution, (c/9=10/15) and (50/80) must be different. Therefore, (c=6).

Step 2

Why this answer is correct

The correct answer is B. (6). For no solution, (c/9=10/15) and (50/80) must be different. Therefore, (c=6).

Step 3

Exam Tip

कोई हल नहीं के लिए (c/9=10/15) और (50/80) अलग होना चाहिए। इसलिए (c=6) है।

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Class 10 Mathematics Hard Quiz

समीकरणों (3x+ky=12) और (9x+15y=36) के अनंत हल होने के लिए (k) का मान क्या होगा?

Concept

Why this answer is correct

Exam Tip

समीकरणों (ax+4y=10) और (6x+8y=25) का कोई हल न होने के लिए (a) का मान क्या होगा?

Concept

Why this answer is correct

Exam Tip

यदि (2x+5y=17) और (4x+10y=m) असंगत युग्म हों, तो (m) के लिए सही शर्त क्या है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (5x-3y=11) और (10x-6y=22) के लिए सही निष्कर्ष कौन-सा है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (7x+4y=19) और (14x+8y=41) के लिए हलों की संख्या क्या होगी?

Concept

Why this answer is correct

Exam Tip

समीकरणों (4x+7y=23) और (9x+15y=49) के लिए सही हल-स्थिति क्या है?

Concept

Why this answer is correct

Exam Tip

यदि \(a_1/a_2=b_1/b_2=c_1/c_2\), तो युग्म की सही प्रकृति क्या है?

Concept

Why this answer is correct

Exam Tip

यदि \(a_1/a_2=b_1/b_2 \ne c_1/c_2\), तो ग्राफ में कौन-सी स्थिति बनेगी?

Concept

Why this answer is correct

Exam Tip

समीकरणों (px+6y=18) और (10x+15y=45) के अनंत हल होने के लिए (p) क्या होगा?

Concept

Why this answer is correct

Exam Tip

समीकरणों (7x+qy=29) और (14x+10y=58) के अनंत हल होने के लिए (q) का मान क्या है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (6x+ky=24) और (3x+5y=17) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (5x+ay=35) और (15x+18y=105) के अनंत हल होने के लिए (a) क्या होगा?

Concept

Why this answer is correct

Exam Tip

समीकरणों (bx+8y=32) और (12x+24y=71) का कोई हल न होने के लिए (b) का मान क्या होगा?

Concept

Why this answer is correct

Exam Tip

समीकरणों (14x+21y=49) और (2x+3y=8) के ग्राफ के बारे में सही कथन कौन-सा है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (15x+10y=40) और (3x+2y=8) के ग्राफ में क्या दिखेगा?

Concept

Why this answer is correct

Exam Tip

समीकरणों (2x+11y=27) और (5x+19y=46) के ग्राफ का सही वर्णन क्या है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (4x+9y-31=0) और (12x+27y-93=0) के लिए सही अनुपात संबंध कौन-सा है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (8x-3y+22=0) और (16x-6y+47=0) के लिए सही अनुपात संबंध क्या है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (9x+5y-28=0) और (18x+11y-56=0) के लिए सही निष्कर्ष क्या है?

Concept

Why this answer is correct

Exam Tip

समीकरणों (3x+by=24) और (12x+20y=96) के अनंत हल होने के लिए (b) का मान क्या है?

Concept

Why this answer is correct