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 271-280 of 600 questions.
What is the value of (b) for (3x+by=24) and (12x+20y=96) to have infinitely many solutions?
C. (5)
अनंत हल के लिए (3/12=b/20=24/96) होना चाहिए। इसलिए (b=5)। / For infinitely many solutions (3/12=b/20=24/96) must hold. Therefore (b=5).
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Which condition is correct for (11x+py=33) and (4x+2y=15) to have a unique solution?
B. (p \ne 112)
अद्वितीय हल के लिए \(11/4 \ne p/2\) होना चाहिए। इसलिए \(p \ne 11/2\) सही शर्त है। / For a unique solution \(11/4 \ne p/2\) must hold. Therefore \(p \ne 11/2\) is the correct condition.
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If (7x+4y=37) and (21x+12y=n) have infinitely many solutions then what is (n)?
C. (111)
दूसरा समीकरण पहले का (3) गुना होना चाहिए। इसलिए (n=111) होगा। / The second equation must be (3) times the first. Therefore (n=111).
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If (10x-5y=30) and (2x-y=t) are inconsistent then what is the correct condition for (t)?
B. \(t \ne 6\)
पहले दो अनुपात बराबर हैं इसलिए असंगत होने के लिए (30/t) अलग होना चाहिए। अतः \(t \ne 6\)। / The first two ratios are equal so (30/t) must be different for inconsistency. Hence \(t \ne 6\).
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What is the most suitable conclusion by observing (6x+7y=41) and (18x+21y=123)?
C. अनंत हल हैंThere are infinitely many solutions
दूसरा समीकरण पहले का (3) गुना है। इसलिए दोनों रेखाएं संपाती हैं। / The second equation is (3) times the first. Therefore both lines are coincident.
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Which conclusion is correct by observing (12x+16y=36) and (3x+4y=11)?
B. पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग हैFirst two ratios are equal but the constant ratio is different
यहां (12/3=16/4) लेकिन (36/11) अलग है। इसलिए कोई हल नहीं होगा। / Here (12/3=16/4) but (36/11) is different. Therefore there will be no solution.
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What is found by comparing the ratios of (a) and (b) in (11x+6y=35) and (5x+3y=17)?
C. (115 \ne 6 / 3) इसलिए एक अद्वितीय हल / 3) so one unique solution
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं। / Here the first two ratios are different. Therefore the lines intersect at one point and give one solution.
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What type of pair is (16x+24y=88) and (2x+3y=11)?
A. संगत और आश्रितConsistent and dependent
पहला समीकरण दूसरे का (8) गुना है। इसलिए दोनों एक ही रेखा हैं और युग्म आश्रित है। / The first equation is (8) times the second. Therefore both are the same line and the pair is dependent.
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What type of pair is (18x+27y=63) and (2x+3y=8)?
C. असंगतInconsistent
यहां (18/2=27/3) लेकिन (63/8) अलग है। इसलिए यह असंगत युग्म है। / Here (18/2=27/3) but (63/8) is different. Therefore this is an inconsistent pair.
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What type of pair is (13x+4y=39) and (6x+2y=17)?
C. संगत और स्वतंत्रConsistent and independent
यहां \(13/6 \ne 4/2\) इसलिए रेखाएं कटती हैं। अतः युग्म संगत और स्वतंत्र है। / Here \(13/6 \ne 4/2\) so the lines intersect. Hence the pair is consistent and independent.
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