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 481-490 of 584 questions.
If \(\frac{x}{2}-\frac{x-3}{4}>\frac{2x+5}{8}\), what is the solution?
A. (x<1)
धनात्मक (8) से गुणा करने पर (4x-2(x-3)>2x+5) मिलता है। इससे (2x+6>2x+5), जो सदैव सत्य है। / Multiplying by positive (8) gives (4x-2(x-3)>2x+5). This reduces to (2x+6>2x+5), always true.
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What is the correct solution set of \(\frac{x}{2}-\frac{x-3}{4}>\frac{2x+5}{8}\)?
C. सभी वास्तविक (x)all real (x)
सरलीकरण के बाद (1>0) जैसा सत्य कथन मिलता है। इसलिए हर वास्तविक (x) इस असमानता को संतुष्ट करता है। / After simplification, a true statement like (1>0) remains. Therefore every real (x) satisfies the inequality.
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For which (x) is \(-1<\frac{x-4}{2}\le 6\)?
A. \(2<x\le16\)
धनात्मक (2) से गुणा करने पर \(-2<x-4\le12\) मिलता है। (4) जोड़ने पर \(2<x\le16\)। / Multiplying by positive (2) gives \(-2<x-4\le12\). Adding (4) gives \(2<x\le16\).
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If \(x\in\mathbb{R}\), what is the combined solution of (5x-2<3x+10) and \(x+4\ge2\)?
A. \(-2\le x<6\)
पहली असमानता (x<6) देती है और दूसरी \(x\ge-2\) देती है। दोनों का प्रतिच्छेद \(-2\le x<6\) है। / The first inequality gives (x<6), and the second gives \(x\ge-2\). Their intersection is \(-2\le x<6\).
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What is the solution of \(4-\frac{2x+1}{3}\ge\frac{1-x}{2}\)?
A. \(x\le19\)
धनात्मक (6) से गुणा करने पर (24-2(2x+1)\ge3(1-x)) मिलता है। इससे \(22-4x\ge3-3x\), इसलिए \(x\le19\)। / Multiplying by positive (6) gives (24-2(2x+1)\ge3(1-x)). Thus \(22-4x\ge3-3x\), so \(x\le19\).
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If (x) is an integer and \(-10<4x+2\le18\), what is the greatest value of (x)?
B. (4)
घटाने पर \(-12<4x\le16\), इसलिए \(-3<x\le4\) मिलता है। पूर्णांक हलों में सबसे बड़ा (4) है। / Subtracting gives \(-12<4x\le16\), so \(-3<x\le4\). The greatest integer solution is (4).
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For which (x) is (3(2x-1)-2(x+5)\le x+4) true?
A. \(x\le17\)
बाईं ओर (6x-3-2x-10=4x-13) है। \(4x-13\le x+4\) से \(3x\le17\), इसलिए \(x\le\frac{17}{3}\)। / The left side is (6x-3-2x-10=4x-13). From \(4x-13\le x+4\), \(3x\le17\), so \(x\le\frac{17}{3}\).
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Choose the correct solution of (3(2x-1)-2(x+5)\le x+4).
C. \(x\le\frac{17}{3}\)
सरलीकरण से \(4x-13\le x+4\) मिलता है। इसलिए \(3x\le17\) और अंतिम हल \(x\le\frac{17}{3}\) है। / Simplification gives \(4x-13\le x+4\). Hence \(3x\le17\), and the final solution is \(x\le\frac{17}{3}\).
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Which is the solution of the inequality \( \frac{3x-5}{4}-\frac{x+1}{3}\leq 2 \)?
A. \(x\leq \frac{31}{5}\)
हर पद को (12) से गुणा करने पर \(5x-31\leq 0\) मिलता है। परीक्षा में हरात्मक हटाते समय चिह्न ध्यान से रखें। / Multiplying every term by (12) gives \(5x-31\leq 0\). In exams keep the inequality sign unchanged when multiplying by a positive number.
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Which is the solution of the inequality \( \frac{3x-2}{5}-\frac{x+4}{2}\geq -1 \)?
B. \(x\geq 14\)
(10) से गुणा करने पर \(6x-4-5x-20\geq -10\) मिलता है। इसलिए \(x\geq 14\) सही है। / Multiplying by (10) gives \(6x-4-5x-20\geq -10\). Therefore \(x\geq 14\) is correct.
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