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 401-410 of 584 questions.
Solve the inequality \(\frac{3x+5}{4}-\frac{x-1}{2}\ge 6\).
A. \(x\ge 15\)
बायाँ पक्ष \(\frac{x+7}{4}\) बनता है। \(\frac{x+7}{4}\ge 6\) से \(x\ge 17\) होना चाहिए। / The left side becomes \(\frac{x+7}{4}\). From \(\frac{x+7}{4}\ge 6\), \(x\ge 17\) should result.
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If \(\frac{4-x}{6}>\frac{x+2}{3}\), what is the correct interval for (x)?
A. (x<0)
(6) से गुणा करने पर (4-x>2x+4) मिलता है। इससे (-3x>0), इसलिए (x<0)। / Multiplying by (6) gives (4-x>2x+4). Thus (-3x>0), so (x<0).
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What is the solution of (1.5x+2.4<0.6x-3)?
A. (x<-6)
(0.9x<-5.4) से (x<-6) मिलता है। दशमलव वाले प्रश्नों में स्थान मान ध्यान से रखें। / From (0.9x<-5.4), we get (x<-6). In decimal problems, handle place values carefully.
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What is the lower bound for (x) in \(2x-\frac{3}{5}\ge \frac{x}{2}+\frac{9}{10}\)?
A. \(x\ge 1\)
(10) से गुणा करने पर \(20x-6\ge 5x+9\) मिलता है। इससे \(15x\ge 15\), इसलिए \(x\ge 1\)। / Multiplying by (10) gives \(20x-6\ge 5x+9\). Thus \(15x\ge 15\), so \(x\ge 1\).
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Solve the inequality (6-5(x-2)\le 3(2-x)).
A. \(x\ge 5\)
सरलीकरण से \(16-5x\le 6-3x\) मिलता है। इससे \(10\le 2x\), अतः \(x\ge 5\)। / Simplification gives \(16-5x\le 6-3x\). Thus \(10\le 2x\), hence \(x\ge 5\).
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If (3(x-4)+2(x+1)\le 5x-7), what will be the solution?
B. कोई हल नहींNo solution
बायाँ पक्ष (5x-10) है और असमानता \(5x-10\le 5x-7\) हमेशा सत्य है। इसलिए सही उत्तर सभी वास्तविक संख्याएँ होना चाहिए। / The left side is (5x-10), and \(5x-10\le 5x-7\) is always true. Hence the answer should be all real numbers.
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What is the correct conclusion for (4(2x+3)>8x+15)?
C. कोई हल नहींNo solution
सरलीकरण से (8x+12>8x+15) अर्थात (12>15) मिलता है। यह असत्य है, इसलिए कोई हल नहीं। / Simplification gives (8x+12>8x+15), i.e. (12>15). This is false, so there is no solution.
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Choose the solution of \(\frac{7-2x}{5}\le \frac{3x+1}{10}\).
A. \(x\ge \frac{13}{7}\)
(10) से गुणा करने पर \(14-4x\le 3x+1\) मिलता है। इससे \(13\le 7x\), अतः \(x\ge \frac{13}{7}\)। / Multiplying by (10) gives \(14-4x\le 3x+1\). Thus \(13\le 7x\), so \(x\ge \frac{13}{7}\).
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What is the solution of (-3(x+2)+7\ge 2(4-x)-5)?
A. \(x\le -2\)
सरलीकरण से \(1-3x\ge 3-2x\) मिलता है। इससे \(-2\ge x\), यानी \(x\le -2\)। / Simplification gives \(1-3x\ge 3-2x\). Thus \(-2\ge x\), i.e. \(x\le -2\).
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Solve the inequality \(\frac{x+6}{8}-\frac{x-2}{4}<1\).
A. (x>-2)
बायाँ पक्ष \(\frac{10-x}{8}\) बनता है। \(\frac{10-x}{8}<1\) से (x>2) मिलता है। / The left side becomes \(\frac{10-x}{8}\). From \(\frac{10-x}{8}<1\), we get (x>2).
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