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 421-430 of 584 questions.
What is the solution of \(\frac{9x+4}{5}\ge 2x-3\)?
A. \(x\le 19\)
(5) से गुणा करने पर \(9x+4\ge 10x-15\) मिलता है। अतः \(x\le 19\)। / Multiplying by (5) gives \(9x+4\ge 10x-15\). Hence \(x\le 19\).
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Solve the inequality \(3-\frac{2x+5}{7}\le \frac{1-x}{2}\).
C. \(x\ge -9\)
(14) से गुणा करने पर (42-2(2x+5)\le 7(1-x)) मिलता है। इससे \(32-4x\le 7-7x\), अतः \(x\le -\frac{25}{3}\) नहीं बल्कि \(3x\le -25\) है। / Multiplying by (14) gives (42-2(2x+5)\le 7(1-x)). This gives \(32-4x\le 7-7x\), so \(3x\le -25\).
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Choose the solution set of (8-3(2-x)\ge 4x-1).
A. \(x\le 3\)
बायाँ पक्ष (2+3x) है। \(2+3x\ge 4x-1\) से \(x\le 3\) मिलता है। / The left side is (2+3x). From \(2+3x\ge 4x-1\), \(x\le 3\).
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If \(\frac{x}{3}-\frac{x-5}{6}>2\), what is the correct condition for (x)?
A. (x>7)
बायाँ पक्ष \(\frac{x+5}{6}\) बनता है। \(\frac{x+5}{6}>2\) से (x>7) मिलता है। / The left side becomes \(\frac{x+5}{6}\). From \(\frac{x+5}{6}>2\), (x>7).
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What is the solution of \(-\frac{4x-1}{3}\le 5-x\)?
A. \(x\ge -14\)
(3) से गुणा करने पर \(-4x+1\le 15-3x\) मिलता है। इससे \(x\ge -14\)। / Multiplying by (3) gives \(-4x+1\le 15-3x\). This gives \(x\ge -14\).
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Which interval is the solution of (2(1-3x)<4-7x)?
A. (x<2)
(2-6x<4-7x) में (7x) जोड़ने पर (2+x<4) मिलता है। इसलिए (x<2)। / Adding (7x) to (2-6x<4-7x) gives (2+x<4). Hence (x<2).
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Solve the inequality \(5-\frac{x-3}{2}\ge \frac{3x+1}{4}\).
A. \(x\le \frac{21}{5}\)
(4) से गुणा करने पर (20-2(x-3)\ge 3x+1) मिलता है। इससे \(25\ge 5x\), इसलिए \(x\le 5\)। / Multiplying by (4) gives (20-2(x-3)\ge 3x+1). Thus \(25\ge 5x\), so \(x\le 5\).
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If (6x+2<3(2x+1)), what is the correct conclusion?
A. सभी वास्तविक संख्याएँAll real numbers
दाएँ पक्ष (6x+3) है। (6x+2<6x+3) हमेशा सत्य है, इसलिए सभी वास्तविक संख्याएँ हल हैं। / The right side is (6x+3). Since (6x+2<6x+3) is always true, all real numbers are solutions.
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Solve the inequality \(7x-4\le 2x+16\).
A. \(x\le 4\)
\(5x\le 20\) से \(x\le 4\) मिलता है। सरल रैखिक असमानता में पहले (x) वाले पद एक ओर लाएँ। / From \(5x\le 20\), we get \(x\le 4\). In a simple linear inequality, first collect (x)-terms on one side.
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What is the solution of (12-5x\ge 3(4-2x)+x)?
A. सभी वास्तविक संख्याएँAll real numbers
दाएँ पक्ष (12-5x) बनता है। समानता \(12-5x\ge 12-5x\) हमेशा सत्य है। / The right side becomes (12-5x). The statement \(12-5x\ge 12-5x\) is always true.
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