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 171-180 of 584 questions.
What is the correct conclusion about (6x+1>6x-5)?
A. सभी वास्तविक (x)All real (x)
दोनों तरफ से (6x) घटाने पर (1>-5) मिलता है जो हमेशा सत्य है। इसलिए सभी वास्तविक (x) हल हैं। / Subtracting (6x) from both sides gives (1>-5), which is always true. Therefore all real (x) are solutions.
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What is the correct conclusion for (4x+9<4x+2)?
A. कोई हल नहींNo solution
दोनों तरफ से (4x) घटाने पर (9<2) मिलता है जो असत्य है। अतः कोई हल नहीं है। / Subtracting (4x) from both sides gives (9<2), which is false. Hence there is no solution.
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If (a>0) and (ax+b<c), which form for (x) is correct?
A. \(x<\frac{c-b}{a}\)
पहले (b) घटाकर (ax<c-b) मिलता है। (a>0) होने से भाग देने पर चिन्ह नहीं बदलता। / First subtract (b) to get (ax<c-b). Since (a>0), the sign does not change on division.
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Four times a number is less than (28). What is the correct inequality form?
A. (4x<28)
किसी संख्या को (x) मानें तो उसका (4) गुना (4x) है। कम है का चिन्ह (<) होता है। / Let the number be (x), so four times it is (4x). The phrase less than uses the sign (<).
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When (7) is added to a number, the result is not more than (20). What is the correct solution?
A. \(x\leq 13\)
कथन \(x+7\leq 20\) देता है। अतः \(x\leq 13\) सही है। / The statement gives \(x+7\leq 20\). Therefore \(x\leq 13\) is correct.
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If three times a number decreased by (5) is at least (16), what is the solution?
A. \(x\geq 7\)
कथन \(3x-5\geq 16\) देता है। इससे \(3x\geq 21\) और \(x\geq 7\) मिलता है। / The statement gives \(3x-5\geq 16\). This gives \(3x\geq 21\) and \(x\geq 7\).
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A. (5)
शर्त \(50-8x\geq 10\) है जिससे \(x\leq 5\) मिलता है। अधिकतम पूर्ण संख्या (5) है। / The condition is \(50-8x\geq 10\), giving \(x\leq 5\). The maximum whole number is (5).
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A. (28)
\(x+12\geq 40\) से \(x\geq 28\) मिलता है। न्यूनतम मान सीमा बिंदु (28) है। / From \(x+12\geq 40\), we get \(x\geq 28\). The minimum value is the boundary point (28).
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What is the solution of (2(x+1)-3(x-2)<10)?
A. (x>-2)
सरल करने पर (2x+2-3x+6<10) यानी (-x+8<10) मिलता है। इसलिए (-x<2) और (x>-2) है। / Simplifying gives (2x+2-3x+6<10), that is (-x+8<10). So (-x<2) and (x>-2).
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Solve the inequality (4-(x-3)\geq 2x+1).
A. \(x\leq 2\)
\(4-x+3\geq 2x+1\) से \(7-x\geq 2x+1\) मिलता है। इसलिए \(6\geq 3x\) और \(x\leq 2\) है। / From \(4-x+3\geq 2x+1\), we get \(7-x\geq 2x+1\). Thus \(6\geq 3x\) and \(x\leq 2\).
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