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 581-584 of 584 questions.
Solve the inequality (3(1-x)-2(4-3x)>5x-10).
C. \(x<\frac{5}{2}\)
सरलीकरण पर (5>2x) मिलता है। इसलिए \(x<\frac{5}{2}\) सही है। / Simplification gives (5>2x). Therefore \(x<\frac{5}{2}\) is correct.
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What is the solution of \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{6}\ge 4\)?
A. \(x\ge \frac{17}{3}\)
हर हटाने पर \(6x-10\ge 24\) मिलता है। अतः \(x\ge \frac{17}{3}\) हल है। / Clearing denominators gives \(6x-10\ge 24\). Hence \(x\ge \frac{17}{3}\) is the solution.
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Solve the compound inequality \(-5<2-\frac{x}{3}\le 4\).
D. \(-6\le x<21\)
दोनों भाग हल करने पर (x<21) और \(x\ge -6\) मिलता है। इसलिए \(-6\le x<21\) सही है। / Solving both parts gives (x<21) and \(x\ge -6\). Therefore \(-6\le x<21\) is correct.
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What is the solution of \(2-\frac{3x-1}{4}\ge \frac{x+5}{2}\)?
C. \(x\le -\frac{1}{5}\)
हर हटाने पर \(9-3x\ge 2x+10\) मिलता है। इसलिए \(-5x\ge 1\) और \(x\le -\frac{1}{5}\) है। / Clearing denominators gives \(9-3x\ge 2x+10\). Therefore \(-5x\ge 1\), so \(x\le -\frac{1}{5}\).
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