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 571-580 of 584 questions.
Solve the inequality \(\frac{5x+2}{6}<\frac{x-3}{2}+4\).
C. \(x<\frac{13}{2}\)
हर हटाने पर (5x+2<3x+15) मिलता है। इससे (2x<13), अतः \(x<\frac{13}{2}\) है। / Clearing denominators gives (5x+2<3x+15). Thus (2x<13), so \(x<\frac{13}{2}\).
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What is the solution of (8-3(2-x)\le 5x-10)?
A. \(x\ge 6\)
सरलीकरण पर \(12\le 2x\) मिलता है। इसलिए \(x\ge 6\) सही हल है। / Simplification gives \(12\le 2x\). Therefore \(x\ge 6\) is the correct solution.
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If (4x+1>2x+9) or (x-5<-12), what is the combined solution?
D. (x<-7) या (x>4)(x<-7) or (x>4)
पहली असमानता (x>4) देती है और दूसरी (x<-7) देती है। या वाली स्थिति में दोनों हलों का संघ लिया जाता है। / The first inequality gives (x>4), and the second gives (x<-7). For an OR condition, take the union of both solution sets.
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Solve the inequality \(\frac{7-2x}{3}\le \frac{x+5}{6}\).
B. \(x\ge \frac{9}{5}\)
हर हटाने पर \(14-4x\le x+5\) मिलता है। इसलिए \(9\le 5x\) और \(x\ge \frac{9}{5}\) है। / Clearing denominators gives \(14-4x\le x+5\). Therefore \(9\le 5x\), so \(x\ge \frac{9}{5}\).
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What is the solution of the compound inequality \(5\le 3x-1<2x+11\)?
C. \(2\le x<12\)
पहले भाग से \(x\ge 2\) और दूसरे भाग से (x<12) मिलता है। दोनों मिलकर \(2\le x<12\) देते हैं। / The first part gives \(x\ge 2\), and the second gives (x<12). Together they give \(2\le x<12\).
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Solve the inequality (-2(3x-4)+7\ge x-6).
A. \(x\le 3\)
सरलीकरण पर \(21\ge 7x\) मिलता है। इसलिए \(x\le 3\) सही है। / Simplification gives \(21\ge 7x\). Therefore \(x\le 3\) is correct.
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What is the solution of \(\frac{3x+8}{4}-\frac{x-6}{2}>5\)?
B. (x>0)
हर हटाने पर (x+20>20) मिलता है। इसलिए (x>0) हल है। / Clearing denominators gives (x+20>20). Therefore (x>0) is the solution.
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What is the solution set of (9x-4\le 3(3x-1)-2)?
D. \(\varnothing\)
सरलीकरण पर \(-4\le -5\) मिलता है, जो असत्य है। इसलिए हल समुच्चय रिक्त है। / Simplification gives \(-4\le -5\), which is false. Therefore the solution set is empty.
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Solve the inequality \(\frac{x}{2}-\frac{x-3}{6}\le 4\).
C. \(x\le \frac{21}{2}\)
हर हटाने पर \(2x+3\le 24\) मिलता है। इसलिए \(x\le \frac{21}{2}\) सही है। / Clearing denominators gives \(2x+3\le 24\). Therefore \(x\le \frac{21}{2}\) is correct.
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What is the solution of \(\frac{2x+1}{7}\ge \frac{x-5}{3}\)?
B. \(x\le 38\)
हर हटाने पर \(6x+3\ge 7x-35\) मिलता है। इसलिए \(x\le 38\) हल है। / Clearing denominators gives \(6x+3\ge 7x-35\). Therefore \(x\le 38\) is the solution.
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