Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 181-190 of 575 questions.
Which form is correct when \(2y+x\ge 14\) is written in terms of (y)?
B. \(y\ge \frac{14-x}{2}\)
\(2y+x\ge 14\) से \(2y\ge 14-x\) मिलता है। इसलिए \(y\ge \frac{14-x}{2}\) सही है। / From \(2y+x\ge 14\), we get \(2y\ge 14-x\). Hence \(y\ge \frac{14-x}{2}\) is correct.
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What are the vertices of the region formed by \(2x+y\le 12\) in the first quadrant?
C. ((0,0)), ((6,0)), ((0,12))
रेखा (2x+y=12) के अवरोध ((6,0)) और ((0,12)) हैं। अक्षों के साथ ((0,0)) तीसरा शीर्ष है। / The intercepts of (2x+y=12) are ((6,0)) and ((0,12)). With the axes, ((0,0)) is the third vertex.
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Which point lies in the solution region of \(x\ge 2\), \(y\ge 1\), and \(x+y\le 7\)?
A. ((3,2))
((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। / The point ((3,2)) satisfies all conditions because \(3\ge 2\), \(2\ge 1\), and \(3+2\le 7\). Check each condition separately.
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What is the common solution of \(x+y\le 6\) and \(x+y\ge 6\)?
B. रेखा (x+y=6)line (x+y=6)
दोनों शर्तें मिलकर (x+y=6) देती हैं। विपरीत दिशाओं की बराबरी वाली असमानताएं समान रेखा देती हैं। / Both conditions together give (x+y=6). Opposite inequalities with equality give the same line.
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Which is the intersection point of (x+2y=8) and (3x-y=3)?
D. ((2,3)) नहीं बल्कि ((2,3))not ((2,3)) but ((2,3))
(x+2y=8) और (3x-y=3) हल करने पर (x=2) और (y=3) मिलता है। प्रतिच्छेद बिंदु को दोनों समीकरणों में जांचें। / Solving (x+2y=8) and (3x-y=3) gives (x=2) and (y=3). Verify the intersection in both equations.
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What is the correct shading rule when \(5x-2y\le 10\) is written in terms of (y)?
C. \(y\ge \frac{5x-10}{2}\) के ऊपरabove \(y\ge \frac{5x-10}{2}\)
\(-2y\le 10-5x\) से चिन्ह पलटकर \(y\ge \frac{5x-10}{2}\) मिलता है। ऋणात्मक से भाग देने पर असमानता का चिन्ह बदलता है। / From \(-2y\le 10-5x\), reversing the sign gives \(y\ge \frac{5x-10}{2}\). The inequality sign changes when dividing by a negative.
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If the boundary line (x=3) is solid and shading is to its right, which inequality is represented?
A. \(x\ge 3\)
दाईं ओर (x) के मान (3) से बड़े होते हैं और ठोस रेखा बराबरी शामिल करती है। इसलिए \(x\ge 3\) सही है। / To the right, (x)-values are greater than (3), and a solid line includes equality. Therefore \(x\ge 3\) is correct.
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If the boundary line (y=-2) is dashed and shading is above it, which inequality is represented?
B. (y>-2)
ऊपर के क्षेत्र में (y) का मान (-2) से बड़ा होता है। टूटी रेखा बराबरी को शामिल नहीं करती। / In the region above, (y) is greater than (-2). A dashed line does not include equality.
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What type of solution region is given by \(x\ge 0\), \(y\ge 0\), and \(x+y\ge 9\)?
D. असीमित क्षेत्रunbounded region
प्रथम चतुर्थांश में (x+y=9) के ऊपर का क्षेत्र अनंत तक फैलता है। इसलिए हल क्षेत्र असीमित है। / In the first quadrant, the region above (x+y=9) extends infinitely. Hence the solution region is unbounded.
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What is the area of the region formed by \(x\ge 0\), \(y\ge 0\), \(x\le 4\), and \(y\le 3\)?
C. (12) वर्ग इकाई(12) square units
क्षेत्र (4) चौड़ाई और (3) ऊंचाई वाला आयत है। इसलिए क्षेत्रफल \(4\times 3=12\) है। / The region is a rectangle of width (4) and height (3). Hence the area is \(4\times 3=12\).
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