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 231-240 of 584 questions.
What is the solution of (2(4x-3)<5x+15)?
A. (x<7)
(8x-6<5x+15) से (3x<21) मिलता है। इसलिए (x<7) है। / From (8x-6<5x+15), we get (3x<21). Therefore (x<7).
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Solve the inequality (6-(x+2)\leq 2x-5).
B. \(x\geq 3\)
सरल करने पर \(4-x\leq 2x-5\) मिलता है। इसलिए \(9\leq 3x\) और \(x\geq 3\) है। / Simplifying gives \(4-x\leq 2x-5\). Thus \(9\leq 3x\) and \(x\geq 3\).
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What is the solution of \(\frac{2x+1}{3}\leq \frac{x+9}{6}\)?
A. \(x\leq 7\)
(6) से गुणा करने पर \(4x+2\leq x+9\) मिलता है। इसलिए \(3x\leq 7\) और \(x\leq \frac{7}{3}\) है। / Multiplying by (6) gives \(4x+2\leq x+9\). So \(3x\leq 7\) and \(x\leq \frac{7}{3}\).
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Which is the correct solution of \(\frac{2x+1}{3}\leq \frac{x+9}{6}\)?
A. \(x\leq \frac{7}{3}\)
(6) से गुणा करने पर \(4x+2\leq x+9\) मिलता है। इसलिए \(3x\leq 7\) और \(x\leq \frac{7}{3}\) है। / Multiplying by (6) gives \(4x+2\leq x+9\). Therefore \(3x\leq 7\) and \(x\leq \frac{7}{3}\).
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If the value of (2x+3) is greater than (15), what is the correct condition for (x)?
A. (x>6)
(2x+3>15) से (2x>12) मिलता है। इसलिए (x>6) है। / From (2x+3>15), we get (2x>12). Therefore (x>6).
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What is the solution of \(3\leq \frac{x+2}{5}\leq 6\)?
A. \(13\leq x\leq 28\)
(5) से गुणा करने पर \(15\leq x+2\leq 30\) मिलता है। फिर (2) घटाने पर \(13\leq x\leq 28\) मिलता है। / Multiplying by (5) gives \(15\leq x+2\leq 30\). Subtracting (2) gives \(13\leq x\leq 28\).
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Which interval satisfies both (2x-3>7) and (x<8)?
A. ((5,8))
पहली असमानता से (x>5) मिलता है और दूसरी (x<8) है। दोनों को मिलाकर (5<x<8) मिलता है। / The first inequality gives (x>5), and the second is (x<8). Combining both gives (5<x<8).
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What is the solution of (3(5-x)+2<4x-11)?
C. (x>4)
सरल करने पर (17-3x<4x-11) मिलता है। इसलिए (28<7x) और (x>4) है। / Simplifying gives (17-3x<4x-11). Therefore (28<7x) and (x>4).
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Which is the correct solution of \(\frac{x-4}{5}-\frac{x+1}{10}\geq 2\)?
B. \(x\geq 29\)
(10) से गुणा करने पर \(2x-8-x-1\geq 20\) मिलता है। इसलिए \(x-9\geq 20\) और \(x\geq 29\) है। / Multiplying by (10) gives \(2x-8-x-1\geq 20\). So \(x-9\geq 20\) and \(x\geq 29\).
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What is the solution of \(-2\leq \frac{3x+1}{4}<7\)?
C. \(-3\leq x<9\)
(4) से गुणा करने पर \(-8\leq 3x+1<28\) मिलता है। फिर (1) घटाकर और (3) से भाग देकर \(-3\leq x<9\) मिलता है। / Multiplying by (4) gives \(-8\leq 3x+1<28\). Subtracting (1) and dividing by (3) gives \(-3\leq x<9\).
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