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 461-470 of 600 questions.
What is the value of (p) for (px-10y=30) and (3x-5y=15) to have infinitely many solutions?
C. (p=6)
अनंत हल के लिए पहला समीकरण दूसरे का (2) गुना होना चाहिए। इसलिए (p=6)। / For infinitely many solutions, the first equation must be (2) times the second. Hence (p=6).
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Which ordered pair satisfies (8x-5y=7) and (3x+5y=48)?
A. \(x=5,\ y=\frac{33}{5}\)
दोनों समीकरण जोड़ने पर (11x=55), इसलिए (x=5)। दूसरे समीकरण से \(y=\frac{33}{5}\)। / Adding both equations gives (11x=55), so (x=5). The second equation gives \(y=\frac{33}{5}\).
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B. (95)
दहाई अंक (x) और इकाई अंक (y) मानें। (x+y=14) और (9(x-y)=36) से (x=9,\ y=5)। / Let the tens digit be (x) and units digit be (y). From (x+y=14) and (9(x-y)=36), (x=9,\ y=5).
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C. (39) वर्ष(39) years
मान लें पिता (x) और पुत्र (y) है। (x+y=56) और (x-6=3(y-6)) से (x=39)। / Let the father be (x) and the son be (y). From (x+y=56) and (x-6=3(y-6)), (x=39).
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On solving (7x+4y=58) and (3x-4y=22), what is the value of (y)?
A. \(y=\frac{1}{2}\)
दोनों समीकरण जोड़ने पर (10x=80), इसलिए (x=8)। पहले समीकरण से \(y=\frac{1}{2}\)। / Adding both equations gives (10x=80), so (x=8). The first equation gives \(y=\frac{1}{2}\).
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If \(\frac{x+y}{4}=6\) and \(\frac{x-y}{5}=2\), what are the values of (x) and (y)?
C. (x=17,\ y=7)
दिए समीकरण (x+y=24) और (x-y=10) बनते हैं। जोड़ने पर (x=17) और (y=7)। / The given equations become (x+y=24) and (x-y=10). Adding gives (x=17) and (y=7).
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C. \(q=\frac{200}{7}\)
विलोपन से \(p=\frac{340}{7}\) मिलता है। इसे किसी एक समीकरण में रखने पर \(q=\frac{200}{7}\)। / Elimination gives \(p=\frac{340}{7}\). Substituting it in either equation gives \(q=\frac{200}{7}\).
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If (x=5,\ y=2) is a solution of (3x+my=29), what will be the value of (m)?
C. (m=7)
(x=5,\ y=2) रखने पर (15+2m=29) मिलता है। इसलिए (m=7)। / Substituting (x=5,\ y=2) gives (15+2m=29). Therefore (m=7).
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For (12x+18y=54) and (2x+3y=c) to have no solution, which value of (c) is correct?
C. (c=10)
पहला समीकरण (2x+3y=9) बनता है। (c=10) होने पर समान बायां पक्ष और अलग दायां पक्ष होगा। / The first equation becomes (2x+3y=9). When (c=10), the left side is the same but the right side is different.
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What is the value of (x) from \(\frac{x}{5}+\frac{y}{6}=7\) and (x-y=6)?
C. \(x=\frac{240}{11}\)
पहले समीकरण को (30) से गुणा कर (6x+5y=210) बनाएं। (x=y+6) रखने पर \(x=\frac{240}{11}\)। / Multiply the first equation by (30) to get (6x+5y=210). Using (x=y+6) gives \(x=\frac{240}{11}\).
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