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 solution of (6x+5y=2) and (4x-5y=18)?
B. (x=2,\ y=-2)
दोनों समीकरण जोड़ने पर (10x=20), इसलिए (x=2)। फिर (6(2)+5y=2) से (y=-2)। / Adding both equations gives (10x=20), so (x=2). Then (6(2)+5y=2) gives (y=-2).
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For (3x+ay=7) and (6x+8y=20) to have no solution, what should be the value of (a)?
D. (a=4)
गुणांक समानुपाती करने के लिए (3:6=a:8) होना चाहिए। इससे (a=4), और स्थिरांक (7:20) समान अनुपात में नहीं हैं। / To make coefficients proportional, (3:6=a:8) must hold. This gives (a=4), and constants (7:20) are not in the same ratio.
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What is the value of (x+y) from (0.2x+0.5y=3.1) and (0.4x-0.1y=1.3)?
A. \(x+y=\frac{97}{11}\)
दशमलव हटाने पर (2x+5y=31) और (4x-y=13) मिलते हैं। हल \(x=\frac{48}{11},\ y=\frac{49}{11}\), इसलिए \(x+y=\frac{97}{11}\)। / Removing decimals gives (2x+5y=31) and (4x-y=13). The solution is \(x=\frac{48}{11},\ y=\frac{49}{11}\), so \(x+y=\frac{97}{11}\).
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On solving (8x-3y=31) and (2x+3y=29), what is (y)?
C. \(y=\frac{17}{3}\)
दोनों समीकरण जोड़ने पर (10x=60), इसलिए (x=6)। दूसरे समीकरण से \(y=\frac{17}{3}\)। / Adding both equations gives (10x=60), so (x=6). From the second equation, \(y=\frac{17}{3}\).
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What is the value of (x) from \(\frac{x}{5}+\frac{y}{2}=7\) and \(\frac{x}{3}-\frac{y}{4}=1\)?
B. \(x=\frac{135}{13}\)
हर हटाने पर (2x+5y=70) और (4x-3y=12) मिलते हैं। विलोपन से \(x=\frac{135}{13}\) मिलता है। / After clearing denominators, (2x+5y=70) and (4x-3y=12) are obtained. Elimination gives \(x=\frac{135}{13}\).
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D. (375) वर्ग सेमी(375) square cm
(l+b=40) और (l-b=10) से (l=25,\ b=15)। इसलिए क्षेत्रफल \(25\times15=375\) वर्ग सेमी है। / From (l+b=40) and (l-b=10), (l=25,\ b=15). Therefore the area is \(25\times15=375\) square cm.
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If (px+2y=18) and (3x-y=5) have solution (x=4,\ y=7), what is the value of (p)?
A. (p=1)
(x=4,\ y=7) को (px+2y=18) में रखें। (4p+14=18), इसलिए (p=1)। / Put (x=4,\ y=7) in (px+2y=18). (4p+14=18), so (p=1).
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What will be the value of (y) from (5x+3y=14) and (2x-3y=7)?
C. \(y=-\frac{1}{3}\)
दोनों समीकरण जोड़ने पर (7x=21), इसलिए (x=3)। पहले समीकरण से \(y=-\frac{1}{3}\)। / Adding both equations gives (7x=21), so (x=3). From the first equation, \(y=-\frac{1}{3}\).
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If (x+y=15) and (2x-3y=10), what is the value of (3x+y)?
B. (37)
(x=15-y) रखने पर (30-5y=10), इसलिए (y=4) और (x=11)। अतः (3x+y=37)। / Using (x=15-y) gives (30-5y=10), so (y=4) and (x=11). Hence (3x+y=37).
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What is the value of (x-y) from \(\frac{x+2y}{3}=8\) and \(\frac{2x-y}{5}=3\)?
D. \(\frac{21}{5}\)
दिए समीकरण (x+2y=24) और (2x-y=15) बनते हैं। हल \(x=\frac{54}{5},\ y=\frac{33}{5}\), इसलिए \(x-y=\frac{21}{5}\)। / The equations become (x+2y=24) and (2x-y=15). The solution is \(x=\frac{54}{5},\ y=\frac{33}{5}\), so \(x-y=\frac{21}{5}\).
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