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 584 questions.
If \(4x+7\le 2x+15\) or (5-3x<2), what is the solution set?
C. सभी वास्तविक (x)all real (x)
पहली से \(x\le4\) और दूसरी से (x>1) मिलता है। इनके संघ से सभी वास्तविक संख्याएँ शामिल हो जाती हैं। / The first gives \(x\le4\), and the second gives (x>1). Their union covers all real numbers.
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What is the solution of \(-3\le \frac{2x-5}{4}<1\)?
A. \(-\frac{7}{2}\le x<\frac{9}{2}\)
धनात्मक (4) से गुणा करने पर \(-12\le2x-5<4\) मिलता है। इससे \(-\frac{7}{2}\le x<\frac{9}{2}\)। / Multiplying by positive (4) gives \(-12\le2x-5<4\). Hence \(-\frac{7}{2}\le x<\frac{9}{2}\).
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For marks (s), the rule is \(40\le \frac{3s+10}{2}\le85\). What interval can (s) lie in?
B. \( \frac{70}{3}\le s\le\frac{160}{3}\)
धनात्मक (2) से गुणा कर \(80\le3s+10\le170\) मिलता है। इससे \(\frac{70}{3}\le s\le\frac{160}{3}\)। / Multiplying by positive (2) gives \(80\le3s+10\le170\). Thus \(\frac{70}{3}\le s\le\frac{160}{3}\).
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If (x) is real and \(\frac{x+2}{3}\ge\frac{2x-1}{5}\), which is the simplified solution?
A. \(x\le13\)
धनात्मक (15) से गुणा करने पर \(5x+10\ge6x-3\) मिलता है। अतः \(x\le13\)। / Multiplying by positive (15) gives \(5x+10\ge6x-3\). Therefore \(x\le13\).
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What is the solution of \(8-\frac{3x-2}{4}>2+\frac{x+6}{8}\)?
A. \(x<\frac{39}{7}\)
धनात्मक (8) से गुणा करने पर (64-2(3x-2)>16+x+6) मिलता है। इससे (68-6x>22+x), यानी \(x<\frac{46}{7}\)। / Multiplying by positive (8) gives (64-2(3x-2)>16+x+6). This gives (68-6x>22+x), so \(x<\frac{46}{7}\).
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Choose the correct solution of \(8-\frac{3x-2}{4}>2+\frac{x+6}{8}\).
B. \(x<\frac{46}{7}\)
हर हटाने पर (68-6x>22+x) बनता है। इससे (46>7x), अतः \(x<\frac{46}{7}\)। / Clearing denominators gives (68-6x>22+x). Hence (46>7x), so \(x<\frac{46}{7}\).
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What is the solution of (5(2-x)\ge 3(4-x)+2)?
A. \(x\le -2\)
सरलीकरण से \(10-5x\ge14-3x\) मिलता है। इससे \(-2x\ge4\), इसलिए \(x\le-2\)। / Simplification gives \(10-5x\ge14-3x\). Thus \(-2x\ge4\), so \(x\le-2\).
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For which (x) is (0.2x+1.5>0.7-0.3x) true?
A. (x>-1.6)
दशमलव हटाने के लिए (10) से गुणा करें: (2x+15>7-3x)। इससे (5x>-8), अतः (x>-1.6)। / Multiply by (10) to remove decimals: (2x+15>7-3x). Then (5x>-8), so (x>-1.6).
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What is the solution of \(3-\frac{x}{2}\le\frac{2x+9}{6}<5\)?
A. \(x\ge\frac{3}{5}\) और \(x<\frac{21}{2}\)\(x\ge\frac{3}{5}\) and \(x<\frac{21}{2}\)
बाईं असमानता से \(18-3x\le2x+9\), अतः \(x\ge\frac{9}{5}\) मिलता है; दाईं से \(x<\frac{21}{2}\)। सही प्रतिच्छेद \(x\ge\frac{9}{5}\) और \(x<\frac{21}{2}\) है। / The left inequality gives \(18-3x\le2x+9\), hence \(x\ge\frac{9}{5}\); the right gives \(x<\frac{21}{2}\). The correct intersection is \(x\ge\frac{9}{5}\) and \(x<\frac{21}{2}\).
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If the inequality \(kx-6\le 2x+3\) has solution \(x\ge -3\), what is the value of (k)?
A. (k=-1)
((k-2)x\le9) है। \(x\ge-3\) पाने के लिए (k-2<0) और \(\frac{9}{k-2}=-3\), इसलिए (k=-1)। / We have ((k-2)x\le9). To get \(x\ge-3\), need (k-2<0) and \(\frac{9}{k-2}=-3\), so (k=-1).
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