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 501-510 of 584 questions.
Which is the solution of \( \frac{x+4}{2}-\frac{x-3}{5}<4 \)?
A. \(x<\frac{8}{3}\)
(10) से गुणा करने पर (5x+20-2x+6<40) मिलता है। इससे (3x<14) नहीं बल्कि (3x<14) के कारण \(x<\frac{14}{3}\) होगा। / Multiplying by (10) gives (5x+20-2x+6<40). Therefore (3x<14), so \(x<\frac{14}{3}\).
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If ( 2(x-1)+3(x+2)\geq 4x+11 ), which statement is correct for (x)?
A. \(x\geq 7\)
बायाँ पक्ष (5x+4) है। \(5x+4\geq 4x+11\) से \(x\geq 7\) मिलता है। / The left side is (5x+4). From \(5x+4\geq 4x+11\), we get \(x\geq 7\).
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The solution of \( 0.4x-1.2\leq 0.1x+2.1 \) is:
A. \(x\leq 11\)
दोनों ओर (10) से गुणा करें तो \(4x-12\leq x+21\) मिलता है। इससे \(3x\leq 33\) और \(x\leq 11\) है। / Multiplying both sides by (10) gives \(4x-12\leq x+21\). Thus \(3x\leq 33\) and \(x\leq 11\).
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If \( \frac{2x+1}{3}-\frac{x-2}{4}\geq \frac{x+5}{6} \), what is the solution?
A. \(x\geq -4\)
(12) से गुणा करने पर \(8x+4-3x+6\geq 2x+10\) मिलता है। इससे \(3x\geq 0\) और \(x\geq 0\) है। / Multiplying by (12) gives \(8x+4-3x+6\geq 2x+10\). Hence \(3x\geq 0\) and \(x\geq 0\).
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Solve the inequality ( 9-4(x+1)<2x-7 ).
A. (x>2)
सरलीकरण से (5-4x<2x-7) और (12<6x) मिलता है। इसलिए (x>2) है। / Simplification gives (5-4x<2x-7) and (12<6x). Therefore (x>2).
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If \( \frac{3x-1}{2}\leq \frac{x+7}{3}<x+5 \), the correct solution for (x) is:
A. \(x\leq \frac{17}{7}\) और (x>-4)\(x\leq \frac{17}{7}\) and (x>-4)
पहले भाग से \(x\leq \frac{17}{7}\) और दूसरे से (x>-4) मिलता है। अतः संयुक्त हल (\(-4,\frac{17}{7}]\) है। / The first part gives \(x\leq \frac{17}{7}\) and the second gives (x>-4). Hence the combined solution is (\(-4,\frac{17}{7}]\).
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Which is the solution of ( -\frac{1}{2}(x-6)\leq \frac{3x+2}{4} )?
A. \(x\geq 2\)
(4) से गुणा करने पर \(-2x+12\leq 3x+2\) मिलता है। इससे \(10\leq 5x\) और \(x\geq 2\) है। / Multiplying by (4) gives \(-2x+12\leq 3x+2\). Thus \(10\leq 5x\) and \(x\geq 2\).
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If \(x\in \mathbb{N}\) and ( 2x-5<11 ), how many values of (x) are possible?
A. (7)
हल (x<8) है और प्राकृतिक मान (1) से (7) तक हैं। इसलिए कुल (7) मान हैं। / The solution is (x<8), and natural values are from (1) to (7). Therefore there are (7) values.
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The solution of \( 6-\frac{x}{3}\geq \frac{2x-1}{2} \) is:
A. \(x\leq \frac{39}{8}\)
(6) से गुणा करने पर \(36-2x\geq 6x-3\) मिलता है। इससे \(39\geq 8x\) और \(x\leq \frac{39}{8}\) है। / Multiplying by (6) gives \(36-2x\geq 6x-3\). Thus \(39\geq 8x\) and \(x\leq \frac{39}{8}\).
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If the solution of ( 3x+a<12 ) is (x<2), what is the value of (a)?
C. (6)
दिए गए हल से \(\frac{12-a}{3}=2\) होगा। इसलिए (12-a=6) और (a=6) है। / From the given solution \(\frac{12-a}{3}=2\). Therefore (12-a=6) and (a=6).
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