Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If \(\frac{x-a}{2}\le 3\) has solution \(x\le10\), what is (a)?
B. (4)
धनात्मक (2) से गुणा करने पर \(x-a\le6\), इसलिए \(x\le a+6\)। (a+6=10) से (a=4)। / Multiplying by positive (2) gives \(x-a\le6\), so \(x\le a+6\). From (a+6=10), (a=4).
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Solve the inequality \(-4<\frac{1-3x}{2}\le5\).
B. \(-3<x\le3\)
धनात्मक (2) से गुणा कर \(-8<1-3x\le10\) मिलता है। \(-9<-3x\le9\) से चिन्ह बदलकर \(-3\le x<3\) मिलता है। / Multiplying by positive (2) gives \(-8<1-3x\le10\). From \(-9<-3x\le9\), reversing signs gives \(-3\le x<3\).
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How many integers (x) satisfy \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\)?
D. अनंतinfinitely many
पहले भाग से \(x>-\frac{2}{3}\) और दूसरे से \(x\le10\) मिलता है। पूर्णांक हल (0) से (10) तक हैं, कुल (11)। / The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).
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For integer (x), how many solutions are there for \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\)?
C. (11)
दोनों भाग हल करने पर \(x>-\frac{2}{3}\) और \(x\le10\) मिलता है। अतः पूर्णांक \(x=0,1,\ldots,10\) हैं, कुल (11)। / Solving both parts gives \(x>-\frac{2}{3}\) and \(x\le10\). Thus integers are \(x=0,1,\ldots,10\), totaling (11).
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If \(x\in\mathbb{R}\), which is the solution of \(6-2x\ge x+9\)?
B. \(x\le-1\)
सरलीकरण से \(-3x\ge3\) मिलता है। ऋणात्मक संख्या से भाग देने पर चिन्ह बदलता है, इसलिए \(x\le-1\)। / Simplification gives \(-3x\ge3\). Dividing by a negative number reverses the sign, so \(x\le-1\).
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What is the solution of (9x-4(2x+1)>3x+8)?
A. (x<-6)
बाईं ओर (9x-8x-4=x-4) है। (x-4>3x+8) से (-12>2x), इसलिए (x<-6)। / The left side is (9x-8x-4=x-4). From (x-4>3x+8), (-12>2x), so (x<-6).
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If \(\frac{4x-1}{3}-\frac{x+2}{9}\ge x+1\), what is the solution for (x)?
A. \(x\ge7\)
धनात्मक (9) से गुणा करने पर (3(4x-1)-(x+2)\ge9x+9) मिलता है। इससे \(2x\ge14\), अतः \(x\ge7\)। / Multiplying by positive (9) gives (3(4x-1)-(x+2)\ge9x+9). Thus \(2x\ge14\), so \(x\ge7\).
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What is the solution of \(\frac{7-2x}{5}\le\frac{3x+1}{10}\)?
A. \(x\ge\frac{13}{7}\)
धनात्मक (10) से गुणा करने पर \(14-4x\le3x+1\) मिलता है। इससे \(13\le7x\), अतः \(x\ge\frac{13}{7}\)। / Multiplying by positive (10) gives \(14-4x\le3x+1\). Thus \(13\le7x\), so \(x\ge\frac{13}{7}\).
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If \(x\in\mathbb{Z}\) and \(1\le\frac{5x-2}{3}<8\), how many values of (x) are there?
B. (5)
इससे \(3\le5x-2<24\), यानी \(5\le5x<26\) मिलता है। इसलिए \(1\le x<\frac{26}{5}\), अतः (x=1,2,3,4,5)। / This gives \(3\le5x-2<24\), i.e. \(5\le5x<26\). Hence \(1\le x<\frac{26}{5}\), so (x=1,2,3,4,5).
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What is the solution of (2(1-3x)\le 5-4(2x+1))?
A. \(x\le-\frac{1}{2}\)
सरलीकरण से \(2-6x\le1-8x\) मिलता है। इससे \(2x\le-1\), इसलिए \(x\le-\frac{1}{2}\)। / Simplification gives \(2-6x\le1-8x\). Thus \(2x\le-1\), so \(x\le-\frac{1}{2}\).
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