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Mathematics algebraic solution of linear inequalities in one variable MCQ Questions for Class 11 General

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01 Practice Topic Graphical solution of linear inequalities in two variables 575 MCQs 02 Practice Topic Introduction to inequalities 591 MCQs 03 Practice Topic representation on number line 592 MCQs

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Question 561/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

यदि (7x-2(3x+5)\ge 4) और \(\frac{x-1}{2}>3\), तो संयुक्त हल क्या है?

If (7x-2(3x+5)\ge 4) and \(\frac{x-1}{2}>3\), what is the combined solution?

Explanation opens after your attempt
Correct Answer

A. \(x\ge 14\)

Explanation

Simple Explanation

पहली असमानता से \(x\ge 14\) और दूसरी से (x>7) मिलता है। प्रतिच्छेद \(x\ge 14\) है। / The first inequality gives \(x\ge 14\), and the second gives (x>7). Their intersection is \(x\ge 14\).

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Question 562/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

यदि \(2x-9\le x-4\) या (5x+3>18), तो संयुक्त हल समुच्चय क्या है?

If \(2x-9\le x-4\) or (5x+3>18), what is the combined solution set?

Explanation opens after your attempt
Correct Answer

D. \(x\in\mathbb{R}\)

Explanation

Simple Explanation

पहली शर्त \(x\le 5\) देती है और दूसरी (x>3) देती है। दोनों का संघ सभी वास्तविक संख्याएँ है। / The first condition gives \(x\le 5\), and the second gives (x>3). Their union is all real numbers.

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Question 563/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

असमानता (\frac{1}{3}(6x-9)-\frac{1}{2}(4x+2)\ge -5) का हल समुच्चय क्या है?

What is the solution set of (\frac{1}{3}(6x-9)-\frac{1}{2}(4x+2)\ge -5)?

Explanation opens after your attempt
Correct Answer

B. \(x\in\mathbb{R}\)

Explanation

Simple Explanation

बायाँ पक्ष (-4) बनता है और \(-4\ge -5\) सत्य है। इसलिए हर वास्तविक (x) हल है। / The left side becomes (-4), and \(-4\ge -5\) is true. Therefore every real (x) is a solution.

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Question 564/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

असमानता (\frac{3}{4}(x-8)<\frac{1}{2}(x+2)-5) को हल कीजिए।

Solve the inequality (\frac{3}{4}(x-8)<\frac{1}{2}(x+2)-5).

Explanation opens after your attempt
Correct Answer

C. (x<8)

Explanation

Simple Explanation

सरलीकरण पर \(\frac{x}{4}<2\) मिलता है। इसलिए (x<8) हल है। / Simplification gives \(\frac{x}{4}<2\). Therefore (x<8) is the solution.

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Question 565/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

असमानता (11-2(5-x)>3x+4) का हल क्या है?

What is the solution of (11-2(5-x)>3x+4)?

Explanation opens after your attempt
Correct Answer

A. (x<-3)

Explanation

Simple Explanation

सरलीकरण से (-x>3) मिलता है। इसलिए (x<-3) सही है। / Simplification gives (-x>3). Therefore (x<-3) is correct.

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Question 566/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

असमानता \(\frac{x+7}{3}-\frac{2x-1}{4}\ge 0\) को हल कीजिए।

Solve the inequality \(\frac{x+7}{3}-\frac{2x-1}{4}\ge 0\).

Explanation opens after your attempt
Correct Answer

D. \(x\le \frac{31}{2}\)

Explanation

Simple Explanation

हर हटाने पर \(31-2x\ge 0\) मिलता है। अतः \(x\le \frac{31}{2}\) सही है। / Clearing denominators gives \(31-2x\ge 0\). Hence \(x\le \frac{31}{2}\) is correct.

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Question 567/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

असमानता (6x-5<2(3x-4)) का हल समुच्चय क्या है?

What is the solution set of (6x-5<2(3x-4))?

Explanation opens after your attempt
Correct Answer

C. \(\varnothing\)

Explanation

Simple Explanation

सरलीकरण पर (-5<-8) मिलता है, जो असत्य है। इसलिए कोई वास्तविक हल नहीं है। / Simplification gives (-5<-8), which is false. Therefore there is no real solution.

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Question 568/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

असमानता (2(4-x)\le \frac{x+9}{3}) को हल कीजिए।

Solve the inequality (2(4-x)\le \frac{x+9}{3}).

Explanation opens after your attempt
Correct Answer

B. \(x\ge \frac{15}{7}\)

Explanation

Simple Explanation

हर हटाने पर \(24-6x\le x+9\) मिलता है। इसलिए \(15\le 7x\) और \(x\ge \frac{15}{7}\) आता है। / Clearing the denominator gives \(24-6x\le x+9\). Hence \(15\le 7x\), so \(x\ge \frac{15}{7}\).

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Question 569/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

युग्म असमानता \(-6\le \frac{4-x}{2}<3\) को हल कीजिए।

Solve the compound inequality \(-6\le \frac{4-x}{2}<3\).

Explanation opens after your attempt
Correct Answer

D. \(-2<x\le 16\)

Explanation

Simple Explanation

दोनों भाग हल करने पर \(x\le 16\) और (x>-2) मिलता है। इसलिए \(-2<x\le 16\) हल है। / Solving both parts gives \(x\le 16\) and (x>-2). Therefore \(-2<x\le 16\) is the solution.

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Question 570/584 Expert Mathematics Linear Inequalities algebraic solution of linear inequalities in one variable Class 11 Level 45

यदि \(\frac{x-2}{3}\ge \frac{2x+1}{5}\) और \(x+4\le 10\), तो संयुक्त हल क्या है?

If \(\frac{x-2}{3}\ge \frac{2x+1}{5}\) and \(x+4\le 10\), what is the combined solution?

Explanation opens after your attempt
Correct Answer

B. \(x\le -13\)

Explanation

Simple Explanation

पहली असमानता से \(x\le -13\) और दूसरी से \(x\le 6\) मिलता है। दोनों का प्रतिच्छेद \(x\le -13\) है। / The first inequality gives \(x\le -13\), and the second gives \(x\le 6\). Their intersection is \(x\le -13\).

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