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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What is the value of (y) from (9x+2y=10) and (3x-2y=14)?
B. (y=-4)
दोनों समीकरण जोड़ने पर (12x=24), इसलिए (x=2)। पहले समीकरण से (y=-4)। / Adding both equations gives (12x=24), so (x=2). The first equation gives (y=-4).
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If (x+y=31) and (4x-3y=19), what is the value of (2x-y)?
C. (17)
(x=31-y) रखने पर (124-7y=19), इसलिए (y=15) और (x=16)। अतः (2x-y=17)। / Using (x=31-y) gives (124-7y=19), so (y=15) and (x=16). Hence (2x-y=17).
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What is the value of (x-y) from \(\frac{x+4y}{5}=10\) and \(\frac{3x-y}{4}=7\)?
B. \(x-y=\frac{40}{13}\)
दिए समीकरण (x+4y=50) और (3x-y=28) बनते हैं। हल से \(x-y=\frac{40}{13}\)। / The equations become (x+4y=50) and (3x-y=28). Solving gives \(x-y=\frac{40}{13}\).
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What is the number of solutions of (4x+7y=31) and (8x+14y=62)?
D. अनंत हलInfinitely many solutions
दूसरा समीकरण पहले का (2) गुना है। इसलिए दोनों एक ही रेखा हैं और अनंत हल हैं। / The second equation is (2) times the first. Therefore both are the same line and have infinitely many solutions.
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Which statement is correct about (4x+7y=31) and (8x+14y=65)?
A. कोई हल नहीं हैThere is no solution
पहले समीकरण का (2) गुना (8x+14y=62) है, लेकिन दूसरा (8x+14y=65) है। इसलिए कोई हल नहीं। / Twice the first equation is (8x+14y=62), but the second is (8x+14y=65). Therefore there is no solution.
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What is the value of (y) from (6x+9y=117) and (8x-3y=37)?
C. \(y=\frac{119}{15}\)
दूसरे समीकरण को (3) से गुणा कर पहले में जोड़ें। हल करने पर \(y=\frac{119}{15}\) मिलता है। / Multiply the second equation by (3) and add it to the first. Solving gives \(y=\frac{119}{15}\).
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If (6a+5b=460) and (4a+7b=444), what is the value of (b)?
C. \(b=\frac{412}{11}\)
पहले समीकरण को (2) और दूसरे को (3) से गुणा कर घटाएं। इससे \(b=\frac{412}{11}\) मिलता है। / Multiply the first equation by (2) and the second by (3), then subtract. This gives \(b=\frac{412}{11}\).
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What is the value of (x+2y) from (9x-5y=42) and (3x+5y=30)?
C. \(x+2y=\frac{54}{5}\)
दोनों समीकरण जोड़ने पर (12x=72), इसलिए (x=6)। फिर \(y=\frac{12}{5}\), अतः \(x+2y=\frac{54}{5}\)। / Adding both equations gives (12x=72), so (x=6). Then \(y=\frac{12}{5}\), hence \(x+2y=\frac{54}{5}\).
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If (x=5y-8) and (4x+3y=61), what is the value of (y)?
C. \(y=\frac{93}{23}\)
(x=5y-8) को दूसरे समीकरण में रखें। (20y-32+3y=61), इसलिए \(y=\frac{93}{23}\)। / Substitute (x=5y-8) in the second equation. (20y-32+3y=61), so \(y=\frac{93}{23}\).
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For (9x+15y=45) and (kx+5y=18) to have no solution, what is the value of (k)?
B. (k=3)
पहला समीकरण (3x+5y=15) बनता है। (k=3) पर दूसरा (3x+5y=18) होगा, इसलिए कोई हल नहीं। / The first equation becomes (3x+5y=15). At (k=3), the second becomes (3x+5y=18), so there is no solution.
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