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.
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When will the pair (kx+5y=10) and (6x+15y=18) have a unique solution?
B. \(k\neq2\)
अद्वितीय हल के लिए \(\frac{k}{6}\neq\frac{5}{15}\) होना चाहिए। इसलिए \(k\neq2\) सही है। / For a unique solution, \(\frac{k}{6}\neq\frac{5}{15}\) must hold. Therefore, \(k\neq2\) is correct.
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What is the value of (m) for no solution in (4x+my=8) and (10x+15y=21)?
C. (m=6)
गुणांक अनुपात समान करने पर (m=6) मिलता है। स्थिर पद का अनुपात अलग है इसलिए कोई हल नहीं होगा। / Equating coefficient ratios gives (m=6). The constant ratio is different, so there is no solution.
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For infinitely many solutions of ((p-2)x+7y=14) and (9x+21y=42), what is (p)?
C. (p=5)
अनंत हलों के लिए तीनों अनुपात समान होने चाहिए। \(\frac{p-2}{9}=\frac{7}{21}\) से (p=5) मिलता है। / For infinitely many solutions, all three ratios must be equal. From \(\frac{p-2}{9}=\frac{7}{21}\), (p=5).
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Find the value of (a) for infinitely many solutions of (8x+ay=12) and (20x+10y=30).
B. (a=4)
यहाँ \(\frac{8}{20}=\frac{12}{30}\) है। अतः \(\frac{a}{10}=\frac{2}{5}\) से (a=4) होगा। / Here \(\frac{8}{20}=\frac{12}{30}\). Thus \(\frac{a}{10}=\frac{2}{5}\) gives (a=4).
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For (3x-2y=5) and (12x-8y=c) to have no solution, what condition is required on (c)?
B. \(c\neq20\)
गुणांक अनुपात \(\frac{1}{4}\) है। कोई हल नहीं के लिए \(\frac{5}{c}\neq\frac{1}{4}\) इसलिए \(c\neq20\)। / The coefficient ratio is \(\frac{1}{4}\). For no solution, \(\frac{5}{c}\neq\frac{1}{4}\), so \(c\neq20\).
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When will (qx+4y=9) and (15x+10y=18) have a unique solution?
A. \(q\neq6\)
अद्वितीय हल के लिए \(\frac{q}{15}\neq\frac{4}{10}\) होना चाहिए। इसलिए \(q\neq6\) सही शर्त है। / For a unique solution, \(\frac{q}{15}\neq\frac{4}{10}\) must hold. Hence \(q\neq6\) is correct.
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For infinitely many solutions of (5x+(r+1)y=11) and (15x+18y=33), what is (r)?
B. (r=5)
अनंत हलों के लिए \(\frac{5}{15}=\frac{r+1}{18}=\frac{11}{33}\) चाहिए। इससे (r=5) मिलता है। / For infinitely many solutions, \(\frac{5}{15}=\frac{r+1}{18}=\frac{11}{33}\) is needed. This gives (r=5).
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What is the value of (n) for no solution in (2x+ny=7) and (6x+15y=19)?
C. (n=5)
गुणांक अनुपात समान करने पर \(\frac{2}{6}=\frac{n}{15}\) से (n=5) मिलता है। स्थिर अनुपात अलग है इसलिए युग्म असंगत है। / Equating coefficient ratios, \(\frac{2}{6}=\frac{n}{15}\) gives (n=5). The constant ratio is different, so the pair is inconsistent.
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When will (11x+3y=4) and (22x+sy=10) have a unique solution?
B. \(s\neq6\)
अद्वितीय हल के लिए \(\frac{11}{22}\neq\frac{3}{s}\) होना चाहिए। अतः \(s\neq6\) सही है। / For a unique solution, \(\frac{11}{22}\neq\frac{3}{s}\) must hold. Hence \(s\neq6\) is correct.
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What is the value of (t) for infinitely many solutions of (6x-ty=18) and (14x-21y=42)?
C. (t=9)
अनंत हलों के लिए \(\frac{6}{14}=\frac{-t}{-21}=\frac{18}{42}\) होना चाहिए। इससे (t=9) मिलता है। / For infinitely many solutions, \(\frac{6}{14}=\frac{-t}{-21}=\frac{18}{42}\) must hold. This gives (t=9).
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