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Class 11 Mathematics Medium Quiz

Level 50 • 50/50 questions • 35 seconds per question.

Level readiness 50/50 Questions
Time Left 29:10 35 sec/question
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Question 1 / 50 0 score
Answered 0/50 Correct 0 Time 29:10

असमानता \(3x+y\le 15\) के लिए सीमा रेखा के अवरोध कौन-से हैं?

What are the intercepts of the boundary line for \(3x+y\le 15\)?

Explanation opens after your attempt
Correct Answer

D. ((5,0)) और ((0,15))((5,0)) and ((0,15))

Step 1

Concept

The boundary line is (3x+y=15). At (y=0), (x=5), and at (x=0), (y=15).

Step 2

Why this answer is correct

The correct answer is D. ((5,0)) और ((0,15)) / ((5,0)) and ((0,15)). The boundary line is (3x+y=15). At (y=0), (x=5), and at (x=0), (y=15).

Step 3

Exam Tip

सीमा रेखा (3x+y=15) है। (y=0) पर (x=5) और (x=0) पर (y=15) मिलता है।

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रेखा (2x-3y=12) के सापेक्ष मूल बिंदु परीक्षण करने पर असमानता (2x-3y<12) के लिए कौन-सा निष्कर्ष सही है?

Using the origin test relative to the line (2x-3y=12), which conclusion is correct for (2x-3y<12)?

Explanation opens after your attempt
Correct Answer

B. मूल बिंदु वाली ओर शेड होगीside containing origin is shaded

Step 1

Concept

Substituting ((0,0)) gives (0<12), which is true. Hence the half-plane containing the origin is the solution.

Step 2

Why this answer is correct

The correct answer is B. मूल बिंदु वाली ओर शेड होगी / side containing origin is shaded. Substituting ((0,0)) gives (0<12), which is true. Hence the half-plane containing the origin is the solution.

Step 3

Exam Tip

((0,0)) रखने पर (0<12) सत्य है। इसलिए मूल बिंदु वाली ओर का अर्ध-समतल हल है।

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असमानता (x+4y>8) के आलेख में सीमा रेखा और शेडिंग के बारे में सही कथन कौन-सा है?

Which statement about the boundary line and shading is correct for (x+4y>8)?

Explanation opens after your attempt
Correct Answer

C. टूटी रेखा और मूल बिंदु के विपरीत ओरdashed line and opposite to origin

Step 1

Concept

The sign (>) makes the boundary dashed. Substituting ((0,0)) gives (0>8), false, so shade the opposite side.

Step 2

Why this answer is correct

The correct answer is C. टूटी रेखा और मूल बिंदु के विपरीत ओर / dashed line and opposite to origin. The sign (>) makes the boundary dashed. Substituting ((0,0)) gives (0>8), false, so shade the opposite side.

Step 3

Exam Tip

चिन्ह (>) होने से रेखा टूटी होगी। ((0,0)) रखने पर (0>8) असत्य है इसलिए विपरीत ओर शेड होगा।

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रेखा (x+2y=10) पर स्थित कौन-सा बिंदु असमानता \(x+2y\le 10\) के हल में शामिल होगा?

Which point on the line (x+2y=10) is included in the solution of \(x+2y\le 10\)?

Explanation opens after your attempt
Correct Answer

A. ((4,3))

Step 1

Concept

Substituting ((4,3)) gives (4+6=10). The boundary line is included with \(\le\).

Step 2

Why this answer is correct

The correct answer is A. ((4,3)). Substituting ((4,3)) gives (4+6=10). The boundary line is included with \(\le\).

Step 3

Exam Tip

((4,3)) रखने पर (4+6=10) मिलता है। \(\le\) में सीमा रेखा शामिल होती है।

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असमानता \(y\le -3x+9\) का हल क्षेत्र किस ओर होगा?

Where is the solution region of \(y\le -3x+9\)?

Explanation opens after your attempt
Correct Answer

D. रेखा के नीचे सहितbelow the line including it

Step 1

Concept

In \(y\le -3x+9\), (y) is less than or equal to the line value. Therefore the region below the line including the boundary is the solution.

Step 2

Why this answer is correct

The correct answer is D. रेखा के नीचे सहित / below the line including it. In \(y\le -3x+9\), (y) is less than or equal to the line value. Therefore the region below the line including the boundary is the solution.

Step 3

Exam Tip

\(y\le -3x+9\) में (y) रेखा के मान से कम या बराबर है। इसलिए रेखा के नीचे का भाग सीमा सहित हल है।

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असमानता \(2y+x\ge 14\) को (y) के रूप में लिखने पर कौन-सा रूप सही है?

Which form is correct when \(2y+x\ge 14\) is written in terms of (y)?

Explanation opens after your attempt
Correct Answer

B. \(y\ge \frac{14-x}{2}\)

Step 1

Concept

From \(2y+x\ge 14\), we get \(2y\ge 14-x\). Hence \(y\ge \frac{14-x}{2}\) is correct.

Step 2

Why this answer is correct

The correct answer is B. \(y\ge \frac{14-x}{2}\). From \(2y+x\ge 14\), we get \(2y\ge 14-x\). Hence \(y\ge \frac{14-x}{2}\) is correct.

Step 3

Exam Tip

\(2y+x\ge 14\) से \(2y\ge 14-x\) मिलता है। इसलिए \(y\ge \frac{14-x}{2}\) सही है।

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प्रथम चतुर्थांश में \(2x+y\le 12\) से बने क्षेत्र के शीर्ष कौन-से हैं?

What are the vertices of the region formed by \(2x+y\le 12\) in the first quadrant?

Explanation opens after your attempt
Correct Answer

C. ((0,0)), ((6,0)), ((0,12))

Step 1

Concept

The intercepts of (2x+y=12) are ((6,0)) and ((0,12)). With the axes, ((0,0)) is the third vertex.

Step 2

Why this answer is correct

The correct answer is C. ((0,0)), ((6,0)), ((0,12)). The intercepts of (2x+y=12) are ((6,0)) and ((0,12)). With the axes, ((0,0)) is the third vertex.

Step 3

Exam Tip

रेखा (2x+y=12) के अवरोध ((6,0)) और ((0,12)) हैं। अक्षों के साथ ((0,0)) तीसरा शीर्ष है।

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असमानताओं \(x\ge 2\), \(y\ge 1\), \(x+y\le 7\) से बने क्षेत्र का कौन-सा बिंदु हल में है?

Which point lies in the solution region of \(x\ge 2\), \(y\ge 1\), and \(x+y\le 7\)?

Explanation opens after your attempt
Correct Answer

A. ((3,2))

Step 1

Concept

The point ((3,2)) satisfies all conditions because \(3\ge 2\), \(2\ge 1\), and \(3+2\le 7\). Check each condition separately.

Step 2

Why this answer is correct

The correct answer is A. ((3,2)). The point ((3,2)) satisfies all conditions because \(3\ge 2\), \(2\ge 1\), and \(3+2\le 7\). Check each condition separately.

Step 3

Exam Tip

((3,2)) सभी शर्तों को संतुष्ट करता है क्योंकि \(3\ge 2\), \(2\ge 1\) और \(3+2\le 7\)। हर शर्त अलग से जांचें।

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असमानताओं \(x+y\le 6\) और \(x+y\ge 6\) का संयुक्त हल क्या है?

What is the common solution of \(x+y\le 6\) and \(x+y\ge 6\)?

Explanation opens after your attempt
Correct Answer

B. रेखा (x+y=6)line (x+y=6)

Step 1

Concept

Both conditions together give (x+y=6). Opposite inequalities with equality give the same line.

Step 2

Why this answer is correct

The correct answer is B. रेखा (x+y=6) / line (x+y=6). Both conditions together give (x+y=6). Opposite inequalities with equality give the same line.

Step 3

Exam Tip

दोनों शर्तें मिलकर (x+y=6) देती हैं। विपरीत दिशाओं की बराबरी वाली असमानताएं समान रेखा देती हैं।

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रेखाओं (x+2y=8) और (3x-y=3) का प्रतिच्छेद बिंदु कौन-सा है?

Which is the intersection point of (x+2y=8) and (3x-y=3)?

Explanation opens after your attempt
Correct Answer

D. ((2,3)) नहीं बल्कि ((2,3))not ((2,3)) but ((2,3))

Step 1

Concept

Solving (x+2y=8) and (3x-y=3) gives (x=2) and (y=3). Verify the intersection in both equations.

Step 2

Why this answer is correct

The correct answer is D. ((2,3)) नहीं बल्कि ((2,3)) / not ((2,3)) but ((2,3)). Solving (x+2y=8) and (3x-y=3) gives (x=2) and (y=3). Verify the intersection in both equations.

Step 3

Exam Tip

(x+2y=8) और (3x-y=3) हल करने पर (x=2) और (y=3) मिलता है। प्रतिच्छेद बिंदु को दोनों समीकरणों में जांचें।

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असमानता \(5x-2y\le 10\) को (y) के रूप में लिखने पर सही शेडिंग नियम कौन-सा है?

What is the correct shading rule when \(5x-2y\le 10\) is written in terms of (y)?

Explanation opens after your attempt
Correct Answer

C. \(y\ge \frac{5x-10}{2}\) के ऊपरabove \(y\ge \frac{5x-10}{2}\)

Step 1

Concept

From \(-2y\le 10-5x\), reversing the sign gives \(y\ge \frac{5x-10}{2}\). The inequality sign changes when dividing by a negative.

Step 2

Why this answer is correct

The correct answer is C. \(y\ge \frac{5x-10}{2}\) के ऊपर / above \(y\ge \frac{5x-10}{2}\). From \(-2y\le 10-5x\), reversing the sign gives \(y\ge \frac{5x-10}{2}\). The inequality sign changes when dividing by a negative.

Step 3

Exam Tip

\(-2y\le 10-5x\) से चिन्ह पलटकर \(y\ge \frac{5x-10}{2}\) मिलता है। ऋणात्मक से भाग देने पर असमानता का चिन्ह बदलता है।

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यदि सीमा रेखा (x=3) ठोस है और शेडिंग इसके दाईं ओर है तो असमानता कौन-सी होगी?

If the boundary line (x=3) is solid and shading is to its right, which inequality is represented?

Explanation opens after your attempt
Correct Answer

A. \(x\ge 3\)

Step 1

Concept

To the right, (x)-values are greater than (3), and a solid line includes equality. Therefore \(x\ge 3\) is correct.

Step 2

Why this answer is correct

The correct answer is A. \(x\ge 3\). To the right, (x)-values are greater than (3), and a solid line includes equality. Therefore \(x\ge 3\) is correct.

Step 3

Exam Tip

दाईं ओर (x) के मान (3) से बड़े होते हैं और ठोस रेखा बराबरी शामिल करती है। इसलिए \(x\ge 3\) सही है।

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यदि सीमा रेखा (y=-2) टूटी है और शेडिंग इसके ऊपर है तो असमानता कौन-सी होगी?

If the boundary line (y=-2) is dashed and shading is above it, which inequality is represented?

Explanation opens after your attempt
Correct Answer

B. (y>-2)

Step 1

Concept

In the region above, (y) is greater than (-2). A dashed line does not include equality.

Step 2

Why this answer is correct

The correct answer is B. (y>-2). In the region above, (y) is greater than (-2). A dashed line does not include equality.

Step 3

Exam Tip

ऊपर के क्षेत्र में (y) का मान (-2) से बड़ा होता है। टूटी रेखा बराबरी को शामिल नहीं करती।

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असमानताओं \(x\ge 0\), \(y\ge 0\), \(x+y\ge 9\) का हल क्षेत्र कैसा है?

What type of solution region is given by \(x\ge 0\), \(y\ge 0\), and \(x+y\ge 9\)?

Explanation opens after your attempt
Correct Answer

D. असीमित क्षेत्रunbounded region

Step 1

Concept

In the first quadrant, the region above (x+y=9) extends infinitely. Hence the solution region is unbounded.

Step 2

Why this answer is correct

The correct answer is D. असीमित क्षेत्र / unbounded region. In the first quadrant, the region above (x+y=9) extends infinitely. Hence the solution region is unbounded.

Step 3

Exam Tip

प्रथम चतुर्थांश में (x+y=9) के ऊपर का क्षेत्र अनंत तक फैलता है। इसलिए हल क्षेत्र असीमित है।

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असमानताओं \(x\ge 0\), \(y\ge 0\), \(x\le 4\), \(y\le 3\) से बने क्षेत्र का क्षेत्रफल क्या होगा?

What is the area of the region formed by \(x\ge 0\), \(y\ge 0\), \(x\le 4\), and \(y\le 3\)?

Explanation opens after your attempt
Correct Answer

C. (12) वर्ग इकाई(12) square units

Step 1

Concept

The region is a rectangle of width (4) and height (3). Hence the area is \(4\times 3=12\).

Step 2

Why this answer is correct

The correct answer is C. (12) वर्ग इकाई / (12) square units. The region is a rectangle of width (4) and height (3). Hence the area is \(4\times 3=12\).

Step 3

Exam Tip

क्षेत्र (4) चौड़ाई और (3) ऊंचाई वाला आयत है। इसलिए क्षेत्रफल \(4\times 3=12\) है।

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असमानताओं \(x\ge 0\), \(y\ge 0\), \(2x+3y\le 18\) से बने त्रिभुज का क्षेत्रफल क्या है?

What is the area of the triangle formed by \(x\ge 0\), \(y\ge 0\), and \(2x+3y\le 18\)?

Explanation opens after your attempt
Correct Answer

A. (27) वर्ग इकाई(27) square units

Step 1

Concept

The intercepts are ((9,0)) and ((0,6)). The area is \(\frac{1}{2}\times 9\times 6=27\).

Step 2

Why this answer is correct

The correct answer is A. (27) वर्ग इकाई / (27) square units. The intercepts are ((9,0)) and ((0,6)). The area is \(\frac{1}{2}\times 9\times 6=27\).

Step 3

Exam Tip

रेखा के अवरोध ((9,0)) और ((0,6)) हैं। त्रिभुज का क्षेत्रफल \(\frac{1}{2}\times 9\times 6=27\) है।

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रेखाओं (x=1), (y=2), और (x+y=8) से बने क्षेत्र का शीर्ष ((1,2)) किस दो सीमाओं से बनता है?

The vertex ((1,2)) of the region formed by (x=1), (y=2), and (x+y=8) is formed by which two boundaries?

Explanation opens after your attempt
Correct Answer

D. (x=1) और (y=2)(x=1) and (y=2)

Step 1

Concept

The point ((1,2)) is the direct intersection of (x=1) and (y=2). To identify vertices, intersect boundary lines pairwise.

Step 2

Why this answer is correct

The correct answer is D. (x=1) और (y=2) / (x=1) and (y=2). The point ((1,2)) is the direct intersection of (x=1) and (y=2). To identify vertices, intersect boundary lines pairwise.

Step 3

Exam Tip

((1,2)) सीधे (x=1) और (y=2) का प्रतिच्छेद है। शीर्ष पहचानने के लिए सीमाओं को जोड़ी में काटें।

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असमानता (x-4y<8) में मूल बिंदु परीक्षण से क्या पता चलता है?

What does the origin test show for (x-4y<8)?

Explanation opens after your attempt
Correct Answer

B. मूल बिंदु हल क्षेत्र में हैorigin is in the solution region

Step 1

Concept

Substituting ((0,0)) gives (0<8), which is true. Hence the side containing the origin is the solution region.

Step 2

Why this answer is correct

The correct answer is B. मूल बिंदु हल क्षेत्र में है / origin is in the solution region. Substituting ((0,0)) gives (0<8), which is true. Hence the side containing the origin is the solution region.

Step 3

Exam Tip

((0,0)) रखने पर (0<8) सत्य है। इसलिए मूल बिंदु वाला भाग हल क्षेत्र है।

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असमानता \(4x+2y\le 16\) को सरल करने पर कौन-सी सीमा रेखा मिलेगी?

Which boundary line is obtained after simplifying \(4x+2y\le 16\)?

Explanation opens after your attempt
Correct Answer

A. (2x+y=8)

Step 1

Concept

Dividing the inequality by (2) gives \(2x+y\le 8\). Its boundary line is (2x+y=8).

Step 2

Why this answer is correct

The correct answer is A. (2x+y=8). Dividing the inequality by (2) gives \(2x+y\le 8\). Its boundary line is (2x+y=8).

Step 3

Exam Tip

असमानता को (2) से भाग देने पर \(2x+y\le 8\) मिलता है। इसकी सीमा रेखा (2x+y=8) होगी।

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असमानताओं \(x+y\le 10\), \(x\ge 2\), \(y\ge 3\) से बने क्षेत्र का कौन-सा बिंदु शीर्ष है?

Which point is a vertex of the region formed by \(x+y\le 10\), \(x\ge 2\), and \(y\ge 3\)?

Explanation opens after your attempt
Correct Answer

C. ((2,3))

Step 1

Concept

The intersection of (x=2) and (y=3) is ((2,3)), and it satisfies \(x+y\le 10\). Therefore it is a vertex.

Step 2

Why this answer is correct

The correct answer is C. ((2,3)). The intersection of (x=2) and (y=3) is ((2,3)), and it satisfies \(x+y\le 10\). Therefore it is a vertex.

Step 3

Exam Tip

(x=2) और (y=3) का प्रतिच्छेद ((2,3)) है और यह \(x+y\le 10\) को संतुष्ट करता है। इसलिए यह एक शीर्ष है।

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रेखा (3x+2y=18) और (x=4) का प्रतिच्छेद बिंदु क्या है?

What is the intersection point of (3x+2y=18) and (x=4)?

Explanation opens after your attempt
Correct Answer

D. ((4,3))

Step 1

Concept

Putting (x=4) gives (12+2y=18), so (y=3). The intersection is ((4,3)).

Step 2

Why this answer is correct

The correct answer is D. ((4,3)). Putting (x=4) gives (12+2y=18), so (y=3). The intersection is ((4,3)).

Step 3

Exam Tip

(x=4) रखने पर (12+2y=18) इसलिए (y=3)। प्रतिच्छेद ((4,3)) है।

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रेखा (2x+5y=20) और (y=2) का प्रतिच्छेद बिंदु कौन-सा है?

Which is the intersection point of (2x+5y=20) and (y=2)?

Explanation opens after your attempt
Correct Answer

B. ((5,2))

Step 1

Concept

Putting (y=2) gives (2x+10=20), so (x=5). Hence the intersection is ((5,2)).

Step 2

Why this answer is correct

The correct answer is B. ((5,2)). Putting (y=2) gives (2x+10=20), so (x=5). Hence the intersection is ((5,2)).

Step 3

Exam Tip

(y=2) रखने पर (2x+10=20) इसलिए (x=5)। इसलिए प्रतिच्छेद ((5,2)) है।

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असमानताओं \(y\ge x+1\) और \(y\le 5\) के संयुक्त क्षेत्र में कौन-सा बिंदु आता है?

Which point lies in the common region of \(y\ge x+1\) and \(y\le 5\)?

Explanation opens after your attempt
Correct Answer

C. ((2,4))

Step 1

Concept

At ((2,4)), both \(4\ge 3\) and \(4\le 5\) are true. For a common region, check all inequalities.

Step 2

Why this answer is correct

The correct answer is C. ((2,4)). At ((2,4)), both \(4\ge 3\) and \(4\le 5\) are true. For a common region, check all inequalities.

Step 3

Exam Tip

((2,4)) पर \(4\ge 3\) और \(4\le 5\) दोनों सत्य हैं। संयुक्त क्षेत्र के लिए सभी असमानताएं जांचें।

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असमानता \(2x+y\ge 6\) में बिंदु ((1,4)) की स्थिति क्या है?

What is the position of ((1,4)) with respect to \(2x+y\ge 6\)?

Explanation opens after your attempt
Correct Answer

A. हल क्षेत्र में और सीमा परin solution region and on boundary

Step 1

Concept

Substituting ((1,4)) gives (2+4=6). Equality puts the point on the boundary and it is included because of \(\ge\).

Step 2

Why this answer is correct

The correct answer is A. हल क्षेत्र में और सीमा पर / in solution region and on boundary. Substituting ((1,4)) gives (2+4=6). Equality puts the point on the boundary and it is included because of \(\ge\).

Step 3

Exam Tip

((1,4)) रखने पर (2+4=6) मिलता है। बराबरी होने से बिंदु सीमा पर है और \(\ge\) के कारण शामिल है।

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असमानताओं \(x\ge 0\), \(y\ge 0\), \(x+2y\le 8\), \(x\le 6\) से बने क्षेत्र का कौन-सा शीर्ष संभव है?

Which vertex is possible for the region formed by \(x\ge 0\), \(y\ge 0\), \(x+2y\le 8\), and \(x\le 6\)?

Explanation opens after your attempt
Correct Answer

D. ((6,0))

Step 1

Concept

The intersection of (x=6) and (y=0) is ((6,0)), and it satisfies \(x+2y\le 8\). Hence it is a vertex of the region.

Step 2

Why this answer is correct

The correct answer is D. ((6,0)). The intersection of (x=6) and (y=0) is ((6,0)), and it satisfies \(x+2y\le 8\). Hence it is a vertex of the region.

Step 3

Exam Tip

(x=6) और (y=0) मिलने पर ((6,0)) आता है और यह \(x+2y\le 8\) को संतुष्ट करता है। इसलिए यह क्षेत्र का शीर्ष है।

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असमानताओं \(x\le 1\) और \(x\ge 4\) का संयुक्त हल क्या है?

What is the common solution of \(x\le 1\) and \(x\ge 4\)?

Explanation opens after your attempt
Correct Answer

A. रिक्त समुच्चयempty set

Step 1

Concept

No (x) can be both less than or equal to (1) and greater than or equal to (4). Therefore there is no common solution.

Step 2

Why this answer is correct

The correct answer is A. रिक्त समुच्चय / empty set. No (x) can be both less than or equal to (1) and greater than or equal to (4). Therefore there is no common solution.

Step 3

Exam Tip

कोई (x) एक साथ (1) से छोटा या बराबर और (4) से बड़ा या बराबर नहीं हो सकता। इसलिए कोई संयुक्त हल नहीं है।

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असमानताओं \(2\le y\le 6\) का आलेख किस प्रकार का क्षेत्र है?

What type of region is represented by \(2\le y\le 6\)?

Explanation opens after your attempt
Correct Answer

C. दो क्षैतिज रेखाओं के बीच पट्टीstrip between two horizontal lines

Step 1

Concept

The lines (y=2) and (y=6) are horizontal. The closed region between them is the common solution.

Step 2

Why this answer is correct

The correct answer is C. दो क्षैतिज रेखाओं के बीच पट्टी / strip between two horizontal lines. The lines (y=2) and (y=6) are horizontal. The closed region between them is the common solution.

Step 3

Exam Tip

(y=2) और (y=6) क्षैतिज रेखाएं हैं। इनके बीच का बंद क्षेत्र संयुक्त हल है।

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असमानताओं (1<x<5) और \(y\ge 0\) से बने क्षेत्र की (x=1) सीमा कैसी होगी?

For the region (1<x<5) and \(y\ge 0\), what type of boundary is (x=1)?

Explanation opens after your attempt
Correct Answer

B. टूटी रेखाdashed line

Step 1

Concept

Equality is not included in (1<x). Therefore the boundary (x=1) is shown by a dashed line.

Step 2

Why this answer is correct

The correct answer is B. टूटी रेखा / dashed line. Equality is not included in (1<x). Therefore the boundary (x=1) is shown by a dashed line.

Step 3

Exam Tip

(1<x) में बराबरी शामिल नहीं है। इसलिए (x=1) की सीमा टूटी रेखा से दिखाई जाएगी।

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रेखा (x-y=4) को (y) के रूप में लिखने पर सीमा रेखा कौन-सी होगी?

What is the boundary line in terms of (y) for (x-y=4)?

Explanation opens after your attempt
Correct Answer

D. (y=x-4)

Step 1

Concept

From (x-y=4), (-y=4-x), so (y=x-4). This is the slope-intercept form of the boundary line.

Step 2

Why this answer is correct

The correct answer is D. (y=x-4). From (x-y=4), (-y=4-x), so (y=x-4). This is the slope-intercept form of the boundary line.

Step 3

Exam Tip

(x-y=4) से (-y=4-x) और (y=x-4) मिलता है। यह ढाल-अवरोध रूप में सीमा रेखा है।

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असमानता \(y\ge -x+6\) की सीमा रेखा पर (x=2) होने पर (y) का मान क्या होगा?

On the boundary line of \(y\ge -x+6\), what is the value of (y) when (x=2)?

Explanation opens after your attempt
Correct Answer

C. (4)

Step 1

Concept

The boundary line is (y=-x+6). Substituting (x=2) gives (y=4).

Step 2

Why this answer is correct

The correct answer is C. (4). The boundary line is (y=-x+6). Substituting (x=2) gives (y=4).

Step 3

Exam Tip

सीमा रेखा (y=-x+6) है। (x=2) रखने पर (y=4) मिलता है।

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असमानता \(3x+y\le 9\) के आलेख में ((4,0)) किस स्थिति में है?

What is the position of ((4,0)) in the graph of \(3x+y\le 9\)?

Explanation opens after your attempt
Correct Answer

A. हल क्षेत्र के अंदरinside solution region

Step 1

Concept

Substituting ((4,0)) gives \(12\le 9\), which is false. Therefore it lies outside the solution region.

Step 2

Why this answer is correct

The correct answer is A. हल क्षेत्र के अंदर / inside solution region. Substituting ((4,0)) gives \(12\le 9\), which is false. Therefore it lies outside the solution region.

Step 3

Exam Tip

((4,0)) रखने पर \(12\le 9\) असत्य है। इसलिए यह हल क्षेत्र के बाहर है।

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असमानताओं \(x+2y\le 10\) और \(x+2y\ge 4\) का संयुक्त क्षेत्र क्या है?

What is the common region of \(x+2y\le 10\) and \(x+2y\ge 4\)?

Explanation opens after your attempt
Correct Answer

D. दो समानांतर रेखाओं के बीच की पट्टीstrip between two parallel lines

Step 1

Concept

The boundary lines (x+2y=10) and (x+2y=4) are parallel. The closed region between them is the common solution.

Step 2

Why this answer is correct

The correct answer is D. दो समानांतर रेखाओं के बीच की पट्टी / strip between two parallel lines. The boundary lines (x+2y=10) and (x+2y=4) are parallel. The closed region between them is the common solution.

Step 3

Exam Tip

सीमा रेखाएं (x+2y=10) और (x+2y=4) समानांतर हैं। इनके बीच का बंद क्षेत्र संयुक्त हल है।

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यदि किसी हल क्षेत्र में \(x\ge 0\), \(y\ge 0\), और \(x+y\le 0\) हैं तो हल क्षेत्र क्या होगा?

If a solution region has \(x\ge 0\), \(y\ge 0\), and \(x+y\le 0\), what is the solution region?

Explanation opens after your attempt
Correct Answer

B. केवल ((0,0))only ((0,0))

Step 1

Concept

In the first quadrant, both (x) and (y) are non-negative. The condition \(x+y\le 0\) is possible only when both are (0).

Step 2

Why this answer is correct

The correct answer is B. केवल ((0,0)) / only ((0,0)). In the first quadrant, both (x) and (y) are non-negative. The condition \(x+y\le 0\) is possible only when both are (0).

Step 3

Exam Tip

प्रथम चतुर्थांश में (x) और (y) दोनों गैर-ऋणात्मक हैं। \(x+y\le 0\) तभी संभव है जब दोनों (0) हों।

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असमानता \(y\le 2x-1\) में बिंदु ((2,3)) की स्थिति क्या है?

What is the position of ((2,3)) with respect to \(y\le 2x-1\)?

Explanation opens after your attempt
Correct Answer

C. सीमा रेखा पर और हल मेंon boundary and in solution

Step 1

Concept

Substituting ((2,3)) gives \(3=2\times 2-1\). Equality places the point on the boundary and it is included because of \(\le\).

Step 2

Why this answer is correct

The correct answer is C. सीमा रेखा पर और हल में / on boundary and in solution. Substituting ((2,3)) gives \(3=2\times 2-1\). Equality places the point on the boundary and it is included because of \(\le\).

Step 3

Exam Tip

((2,3)) रखने पर \(3=2\times 2-1\) मिलता है। बराबरी होने से बिंदु सीमा पर है और \(\le\) के कारण हल में है।

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असमानताओं \(x\ge 0\), \(y\ge 0\), \(x+y\le 8\), \(y\le 2\) से बने क्षेत्र में कौन-सा बिंदु नहीं है?

Which point is not in the region formed by \(x\ge 0\), \(y\ge 0\), \(x+y\le 8\), and \(y\le 2\)?

Explanation opens after your attempt
Correct Answer

C. ((3,3))

Step 1

Concept

For ((3,3)), the condition \(y\le 2\) is false. To lie in the common solution, all conditions must be true.

Step 2

Why this answer is correct

The correct answer is C. ((3,3)). For ((3,3)), the condition \(y\le 2\) is false. To lie in the common solution, all conditions must be true.

Step 3

Exam Tip

((3,3)) में \(y\le 2\) असत्य है। संयुक्त हल में आने के लिए सभी शर्तें सत्य होनी चाहिए।

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रेखा (6x+3y=18) का सरल सीमा रेखा रूप क्या है?

What is the simplified boundary-line form of (6x+3y=18)?

Explanation opens after your attempt
Correct Answer

B. (2x+y=6)

Step 1

Concept

Dividing the whole equation by (3) gives (2x+y=6). Simplify the equation before drawing the graph.

Step 2

Why this answer is correct

The correct answer is B. (2x+y=6). Dividing the whole equation by (3) gives (2x+y=6). Simplify the equation before drawing the graph.

Step 3

Exam Tip

पूरे समीकरण को (3) से भाग देने पर (2x+y=6) मिलता है। ग्राफ बनाने से पहले समीकरण सरल कर लेना चाहिए।

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असमानता (y>-4) के आलेख में कौन-सा बिंदु हल क्षेत्र में है?

Which point lies in the solution region of (y>-4)?

Explanation opens after your attempt
Correct Answer

D. ((3,-3))

Step 1

Concept

For ((3,-3)), (y=-3), and (-3>-4) is true. In a strict inequality, boundary points are not included.

Step 2

Why this answer is correct

The correct answer is D. ((3,-3)). For ((3,-3)), (y=-3), and (-3>-4) is true. In a strict inequality, boundary points are not included.

Step 3

Exam Tip

((3,-3)) में (y=-3) है और (-3>-4) सत्य है। सख्त असमानता में सीमा रेखा के बिंदु शामिल नहीं होते।

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असमानता (x< -2) का आलेख किस क्षेत्र को दर्शाता है?

Which region is represented by (x< -2)?

Explanation opens after your attempt
Correct Answer

A. रेखा (x=-2) के बाईं ओर और रेखा शामिल नहींleft of (x=-2) excluding the line

Step 1

Concept

In (x< -2), (x) is less than (-2). Hence shade the left side and draw a dashed boundary.

Step 2

Why this answer is correct

The correct answer is A. रेखा (x=-2) के बाईं ओर और रेखा शामिल नहीं / left of (x=-2) excluding the line. In (x< -2), (x) is less than (-2). Hence shade the left side and draw a dashed boundary.

Step 3

Exam Tip

(x< -2) में (x) का मान (-2) से कम है। इसलिए बाईं ओर शेड होगा और सीमा रेखा टूटी होगी।

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असमानताओं \(x\ge 1\), \(x\le 5\), \(y\ge 0\), \(y\le 4\) से बने क्षेत्र का आकार क्या है?

What is the shape of the region formed by \(x\ge 1\), \(x\le 5\), \(y\ge 0\), and \(y\le 4\)?

Explanation opens after your attempt
Correct Answer

C. आयतrectangle

Step 1

Concept

(x) lies between two vertical boundaries and (y) lies between two horizontal boundaries. Hence a closed rectangle is formed.

Step 2

Why this answer is correct

The correct answer is C. आयत / rectangle. (x) lies between two vertical boundaries and (y) lies between two horizontal boundaries. Hence a closed rectangle is formed.

Step 3

Exam Tip

(x) दो ऊर्ध्वाधर सीमाओं और (y) दो क्षैतिज सीमाओं के बीच है। इसलिए बंद आयत बनता है।

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असमानता \(2x-y\le 5\) के लिए कौन-सा बिंदु मूल बिंदु वाली ओर के बारे में सही परीक्षण देता है?

For \(2x-y\le 5\), which test correctly describes the side containing the origin?

Explanation opens after your attempt
Correct Answer

B. मूल बिंदु हल में है क्योंकि \(0\le 5\) सत्य हैorigin is in solution because \(0\le 5\) is true

Step 1

Concept

Substituting ((0,0)) gives \(0\le 5\), which is true. Therefore shade the side containing the origin.

Step 2

Why this answer is correct

The correct answer is B. मूल बिंदु हल में है क्योंकि \(0\le 5\) सत्य है / origin is in solution because \(0\le 5\) is true. Substituting ((0,0)) gives \(0\le 5\), which is true. Therefore shade the side containing the origin.

Step 3

Exam Tip

((0,0)) रखने पर \(0\le 5\) सत्य है। इसलिए मूल बिंदु वाली ओर शेड होगी।

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

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

Explanation opens after your attempt
Correct Answer

D. ((3,1))

Step 1

Concept

Using (y=x-2) from (x-y=2), we get (2x+x-2=7). Hence (x=3) and (y=1).

Step 2

Why this answer is correct

The correct answer is D. ((3,1)). Using (y=x-2) from (x-y=2), we get (2x+x-2=7). Hence (x=3) and (y=1).

Step 3

Exam Tip

(x-y=2) से (y=x-2) रखकर (2x+x-2=7) मिलता है। इसलिए (x=3) और (y=1) हैं।

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असमानताओं (x+y<4) और (x+y>4) का संयुक्त हल क्या होगा?

What is the common solution of (x+y<4) and (x+y>4)?

Explanation opens after your attempt
Correct Answer

C. रिक्त समुच्चयempty set

Step 1

Concept

For the same point, (x+y) cannot be both less than (4) and greater than (4). Hence the common solution is empty.

Step 2

Why this answer is correct

The correct answer is C. रिक्त समुच्चय / empty set. For the same point, (x+y) cannot be both less than (4) and greater than (4). Hence the common solution is empty.

Step 3

Exam Tip

एक ही बिंदु के लिए (x+y) एक साथ (4) से कम और (4) से अधिक नहीं हो सकता। इसलिए संयुक्त हल रिक्त है।

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असमानता \(3x+3y\ge 21\) को सरल करने पर कौन-सी असमानता मिलेगी?

Which inequality is obtained by simplifying \(3x+3y\ge 21\)?

Explanation opens after your attempt
Correct Answer

A. \(x+y\ge 7\)

Step 1

Concept

Dividing the whole inequality by (3) gives \(x+y\ge 7\). Dividing by a positive number does not change the sign.

Step 2

Why this answer is correct

The correct answer is A. \(x+y\ge 7\). Dividing the whole inequality by (3) gives \(x+y\ge 7\). Dividing by a positive number does not change the sign.

Step 3

Exam Tip

पूरी असमानता को (3) से भाग देने पर \(x+y\ge 7\) मिलता है। धनात्मक संख्या से भाग देने पर चिन्ह नहीं बदलता।

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असमानता \(-2x-y\le -6\) को सरल करने पर सही रूप कौन-सा है?

Which form is correct after simplifying \(-2x-y\le -6\)?

Explanation opens after your attempt
Correct Answer

D. \(2x+y\ge 6\)

Step 1

Concept

Multiplying the whole inequality by (-1) reverses the sign. Hence \(2x+y\ge 6\) is obtained.

Step 2

Why this answer is correct

The correct answer is D. \(2x+y\ge 6\). Multiplying the whole inequality by (-1) reverses the sign. Hence \(2x+y\ge 6\) is obtained.

Step 3

Exam Tip

पूरी असमानता को (-1) से गुणा करने पर चिन्ह पलटता है। इसलिए \(2x+y\ge 6\) मिलता है।

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सीमा रेखा (x+2y=4) पर कौन-सा बिंदु नहीं है?

Which point is not on the boundary line (x+2y=4)?

Explanation opens after your attempt
Correct Answer

B. ((1,2))

Step 1

Concept

Substituting ((1,2)) gives (1+4=5). Therefore it is not on the boundary line.

Step 2

Why this answer is correct

The correct answer is B. ((1,2)). Substituting ((1,2)) gives (1+4=5). Therefore it is not on the boundary line.

Step 3

Exam Tip

((1,2)) रखने पर (1+4=5) मिलता है। इसलिए यह सीमा रेखा पर नहीं है।

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असमानताओं \(x\ge 0\), \(y\ge 0\), \(x+3y\le 9\) से बने त्रिभुज का (x)-अवरोध कौन-सा है?

What is the (x)-intercept of the triangle formed by \(x\ge 0\), \(y\ge 0\), and \(x+3y\le 9\)?

Explanation opens after your attempt
Correct Answer

C. ((9,0))

Step 1

Concept

Putting (y=0) gives (x=9). Hence the vertex on the (x)-axis is ((9,0)).

Step 2

Why this answer is correct

The correct answer is C. ((9,0)). Putting (y=0) gives (x=9). Hence the vertex on the (x)-axis is ((9,0)).

Step 3

Exam Tip

(y=0) रखने पर (x=9) मिलता है। इसलिए (x)-अक्ष पर शीर्ष ((9,0)) है।

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असमानता \(x+5y\ge 15\) में कौन-सा बिंदु हल क्षेत्र में है?

Which point lies in the solution region of \(x+5y\ge 15\)?

Explanation opens after your attempt
Correct Answer

A. ((5,2))

Step 1

Concept

Substituting ((5,2)) gives (5+10=15), and equality is included in \(\ge\). Therefore it lies in the solution region.

Step 2

Why this answer is correct

The correct answer is A. ((5,2)). Substituting ((5,2)) gives (5+10=15), and equality is included in \(\ge\). Therefore it lies in the solution region.

Step 3

Exam Tip

((5,2)) रखने पर (5+10=15) मिलता है और \(\ge\) में बराबरी शामिल है। इसलिए यह हल क्षेत्र में है।

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यदि सीमा रेखा (x-y=0) है और शेडिंग उस ओर है जहां (y) का मान (x) से बड़ा है तो असमानता कौन-सी होगी?

If the boundary line is (x-y=0) and shading is on the side where (y) is greater than (x), which inequality is represented?

Explanation opens after your attempt
Correct Answer

D. \(y\ge x\)

Step 1

Concept

Where (y) is greater than or equal to (x), the inequality is \(y\ge x\). If the boundary (y=x) is included, equality is also used.

Step 2

Why this answer is correct

The correct answer is D. \(y\ge x\). Where (y) is greater than or equal to (x), the inequality is \(y\ge x\). If the boundary (y=x) is included, equality is also used.

Step 3

Exam Tip

जहां (y) का मान (x) से बड़ा या बराबर है वहां \(y\ge x\) होगा। सीमा रेखा (y=x) शामिल होने पर बराबरी भी आती है।

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असमानताओं \(x\ge 0\), \(y\ge 0\), \(x+2y\le 6\), \(2x+y\le 6\) के हल क्षेत्र में दोनों तिरछी रेखाओं का प्रतिच्छेद कौन-सा है?

In the solution region of \(x\ge 0\), \(y\ge 0\), \(x+2y\le 6\), and \(2x+y\le 6\), what is the intersection of the two oblique lines?

Explanation opens after your attempt
Correct Answer

B. ((2,2))

Step 1

Concept

Solving (x+2y=6) and (2x+y=6) gives (x=2) and (y=2). This is an important vertex of the common region.

Step 2

Why this answer is correct

The correct answer is B. ((2,2)). Solving (x+2y=6) and (2x+y=6) gives (x=2) and (y=2). This is an important vertex of the common region.

Step 3

Exam Tip

(x+2y=6) और (2x+y=6) हल करने पर (x=2) और (y=2) मिलता है। यह संयुक्त क्षेत्र का महत्वपूर्ण शीर्ष है।

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असमानताओं \(x\ge 1\), \(y\ge 0\), \(x+y\le 5\), \(x\le 4\) से बने क्षेत्र का क्षेत्रफल क्या है?

What is the area of the region formed by \(x\ge 1\), \(y\ge 0\), \(x+y\le 5\), and \(x\le 4\)?

Explanation opens after your attempt
Correct Answer

C. \(\frac{15}{2}\) वर्ग इकाई\(\frac{15}{2}\) square units

Step 1

Concept

The vertices are ((1,0)), ((4,0)), ((4,1)), and ((1,4)). With parallel sides (4) and (1) and distance (3), the area is (\frac{1}{2}(4+1)3=\frac{15}{2}).

Step 2

Why this answer is correct

The correct answer is C. \(\frac{15}{2}\) वर्ग इकाई / \(\frac{15}{2}\) square units. The vertices are ((1,0)), ((4,0)), ((4,1)), and ((1,4)). With parallel sides (4) and (1) and distance (3), the area is (\frac{1}{2}(4+1)3=\frac{15}{2}).

Step 3

Exam Tip

क्षेत्र के शीर्ष ((1,0)), ((4,0)), ((4,1)), ((1,4)) हैं। समांतर भुजाओं (4) और (1) तथा दूरी (3) से क्षेत्रफल (\frac{1}{2}(4+1)3=\frac{15}{2}) है।

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