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Linear Inequalities

Madhya Pradesh Board · Class 11 · Mathematics

NCERT Solutions for Linear Inequalities — Madhya Pradesh Board Class 11 Mathematics.

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48 Questions Solved · 2 Sections

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EXERCISE 5.1

1Solve 24x<10024x < 100, whenShow solution

24x<10024x<100

Divide both sides by 2424:

x<10024=256 x<\frac{100}{24}=\frac{25}{6}.

So the solutions depend on the set of numbers used in the subparts below.

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1(i)xx is a natural number.Show solution

For a natural number xx, we need x<256≈4.16x<\frac{25}{6}\approx 4.16.

So the natural numbers that satisfy the inequality are 1, 2, 3, 4.

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1(ii)xx is an integer.Show solution

For an integer xx, we need x<256≈4.16x<\frac{25}{6}\approx 4.16.

Hence all integers less than or equal to 4 satisfy it:

..., -2, -1, 0, 1, 2, 3, 4.

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2Solve −12x>30-12x > 30, whenShow solution

−12x>30-12x>30

Divide both sides by −12-12. Since we divide by a negative number, the inequality sign reverses:

x<30−12=−52 x<\frac{30}{-12}=-\frac{5}{2}.

So the solution is x<−52x<-\frac{5}{2}.

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2(i)xx is a natural number.Show solution

We get x<−52x< -\frac{5}{2}.

No natural number is less than a negative number, so there is no natural number solution.

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2(ii)xx is an integer.Show solution

We have x<−52x< -\frac{5}{2}.

So the integers satisfying this are all integers less than −2.5-2.5, namely ..., -5, -4, -3.

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3Solve 5x−3<75x - 3 < 7, whenShow solution

5x−3<75x-3<7

Add 33 to both sides:

5x<105x<10

Divide by 55:

x<2x<2.

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3(i)xx is an integer.Show solution

From x<2x<2, the integer solutions are all integers less than 22:

..., -4, -3, -2, -1, 0, 1.

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3(ii)xx is a real number.Show solution

From 5x−3<75x-3<7:

5x<105x<10

x<2x<2.

Since xx is a real number, the solution set is all real numbers less than 22, i.e. x<2x<2.

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4Solve 3x+8>23x + 8 > 2, whenShow solution

3x+8>23x+8>2

Subtract 88 from both sides:

3x>−63x>-6

Divide by 33:

x>−2x>-2.

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4(i)xx is an integer.Show solution

From x>−2x>-2, the integer solutions are ..., -1, 0, 1.

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4(ii)xx is a real number.Show solution

From 3x+8>23x+8>2:

3x>−63x>-6

x>−2x>-2.

So for real numbers, the solution is **x>−2x>-2**.

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54x+3<5x+74x + 3 < 5x + 7Show solution

4x+3<5x+74x+3<5x+7

Subtract 4x4x from both sides:

3<x+73<x+7

Subtract 77:

−4<x-4<x

So, **x>−4x>-4**.

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63x−7>5x−13x - 7 > 5x - 1Show solution

3x−7>5x−13x-7>5x-1

Subtract 3x3x:

−7>2x−1-7>2x-1

Add 11:

−6>2x-6>2x

Divide by 22:

−3>x-3>x, i.e. **x<−3x<-3**.

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73(x−1)≤2(x−3)3(x - 1) \leq 2(x - 3)Show solution
Expanding:

3x−3≤2x−63x-3\le 2x-6

Subtract 2x2x:

x−3≤−6x-3\le -6

Add 33:

x≤−3x\le -3. So the solution is **x≤−3x\le -3**.

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83(2−x)≥2(1−x)3(2 - x) \geq 2(1 - x)Show solution
Expand:

6−3x≥2−2x6-3x\ge 2-2x

Subtract 22:

4−3x≥−2x4-3x\ge -2x

Add 3x3x:

4≥x4\ge x,
so **x≤4x\le 4**.

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9x+x2+x3<11x + \frac{x}{2} + \frac{x}{3} < 11Show solution

x+x2+x3<11x+\frac{x}{2}+\frac{x}{3}<11

Take LCM 66:

6x+3x+2x6<11\frac{6x+3x+2x}{6}<11

11x6<11\frac{11x}{6}<11

Multiply by 66:

11x<6611x<66

Divide by 1111:

x<6\boxed{x<6}.

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10x3>x2+1\frac{x}{3} > \frac{x}{2} + 1Show solution

x3>x2+1\frac{x}{3} > \frac{x}{2} + 1

Multiply by 66:

2x>3x+62x>3x+6

Subtract 3x3x:

−x>6-x>6

Multiply by −1-1 and reverse the sign:

x<−6x<-6.

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113(x−2)5≤5(2−x)3\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}Show solution

3(x−2)5≤5(2−x)3\frac{3(x-2)}{5}\le \frac{5(2-x)}{3}

Multiply by 1515:

9(x−2)≤25(2−x)9(x-2)\le 25(2-x)

9x−18≤50−25x9x-18\le 50-25x

Add 25x25x and 1818:

34x≤6834x\le 68

Divide by 3434:

x≤2\boxed{x\le 2}.

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1212(3x5+4)≥13(x−6)\frac{1}{2}\left(\frac{3x}{5} + 4\right) \geq \frac{1}{3}(x - 6)Show solution

12(3x5+4)≥13(x−6)\frac12\left(\frac{3x}{5}+4\right)\ge \frac13(x-6)

Expand:

3x10+2≥x3−2\frac{3x}{10}+2\ge \frac{x}{3}-2

Multiply by 3030:

9x+60≥10x−609x+60\ge 10x-60

Subtract 9x9x:

60≥x−6060\ge x-60

Add 6060:

120≥x120\ge x

So, **x≤120x\le 120**.

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132(2x+3)−10<6(x−2)2(2x + 3) - 10 < 6(x - 2)Show solution

2(2x+3)−10<6(x−2)2(2x+3)-10<6(x-2)

Expand:

4x+6−10<6x−124x+6-10<6x-12

4x−4<6x−124x-4<6x-12

Add 1212:

4x+8<6x4x+8<6x

Subtract 4x4x:

8<2x8<2x

Divide by 22:

x>4\boxed{x>4}.

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1437−(3x+5)≥9x−8(x−3)37 - (3x + 5) \geq 9x - 8(x - 3)Show solution

37−(3x+5)≥9x−8(x−3)37-(3x+5)\ge 9x-8(x-3)

Simplify both sides:

32−3x≥9x−8x+2432-3x\ge 9x-8x+24

32−3x≥x+2432-3x\ge x+24

Subtract 2424:

8−3x≥x8-3x\ge x

Add 3x3x:

8≥4x8\ge 4x

Divide by 44:

x≤2\boxed{x\le 2}.

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15x4<(5x−2)3−(7x−3)5\frac{x}{4} < \frac{(5x - 2)}{3} - \frac{(7x - 3)}{5}Show solution

x4<5x−23−7x−35\frac{x}{4}<\frac{5x-2}{3}-\frac{7x-3}{5}

Take LCM 6060:

15x<20(5x−2)−12(7x−3)15x<20(5x-2)-12(7x-3)

15x<100x−40−84x+3615x<100x-40-84x+36

15x<16x−415x<16x-4

Subtract 15x15x:

0<x−40<x-4

So, **x>4x>4**.

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16(2x−1)3≥(3x−2)4−(2−x)5\frac{(2x - 1)}{3} \geq \frac{(3x - 2)}{4} - \frac{(2 - x)}{5}Show solution

2x−13≥3x−24−2−x5\frac{2x-1}{3}\ge \frac{3x-2}{4}-\frac{2-x}{5}

Take LCM 6060:

20(2x−1)≥15(3x−2)−12(2−x)20(2x-1)\ge 15(3x-2)-12(2-x)

40x−20≥45x−30−24+12x40x-20\ge 45x-30-24+12x

40x−20≥57x−5440x-20\ge 57x-54

Add 5454:

40x+34≥57x40x+34\ge 57x

Subtract 40x40x:

34≥17x34\ge 17x

Divide by 1717:

2≥x2\ge x, i.e. **x≤2x\le 2**.

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173x−2<2x+13x - 2 < 2x + 1
185x−3≥3x−55x - 3 \geq 3x - 5
193(1−x)<2(x+4)3(1 - x) < 2(x + 4)
20x2≥(5x−2)3−(7x−3)5\frac{x}{2} \geq \frac{(5x - 2)}{3} - \frac{(7x - 3)}{5}
21Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
22To receive Grade 'A' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade 'A' in the course.
23Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
24Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
25The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
26A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?

Miscellaneous Exercise on Chapter 5

12≤3x−4≤52 \leq 3x - 4 \leq 5
26≤−3(2x−4)<126 \leq -3(2x - 4) < 12
3−3≤4−7x2≤18-3 \leq 4 - \frac{7x}{2} \leq 18
4−15<3(x−2)5≤0-15 < \frac{3(x - 2)}{5} \leq 0
5−12<4−3x−5≤2-12 < 4 - \frac{3x}{-5} \leq 2
67≤(3x+11)2≤117 \leq \frac{(3x + 11)}{2} \leq 11.
75x+1>−245x + 1 > -24, 5x−1<245x - 1 < 24
82(x−1)<x+52(x - 1) < x + 5, 3(x+2)>2−x3(x + 2) > 2 - x
93x−7>2(x−6)3x - 7 > 2(x - 6), 6−x>11−2x6 - x > 11 - 2x
105(2x−7)−3(2x+3)≤05(2x - 7) - 3(2x + 3) \leq 0, 2x+19≤6x+472x + 19 \leq 6x + 47.
11A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by
12A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?
13How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
14IQ of a person is given by the formula

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Frequently Asked Questions

What are the important topics in Linear Inequalities for Madhya Pradesh Board Class 11 Mathematics?
Linear Inequalities covers several key topics that are frequently asked in Madhya Pradesh Board Class 11 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Linear Inequalities — Madhya Pradesh Board Class 11 Mathematics?
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