ML Aggarwal Solution Class 10 Chapter 4 Linear Inequations Exercise 4

Exercise 4

Question 1

Solve the inequation 3x -11 < 3 where x ∈ {1, 2, 3,……, 10}. Also represent its solution on a number line

Sol :

3x – 11 < 3

3x < 3 + 11

3x < 14 x < $\frac{14}{3}$

But x ∈ 6 {1, 2, 3, ……., 10}

Solution set is (1, 2, 3, 4}

Ans. Solution set on number line

Question 2

Solve 2(x – 3)< 1, x ∈ {1, 2, 3, …. 10}

Sol :

2(x – 3) < 1

$x-3<\frac{1}{2}$

$x<\frac{1}{2}+3$

$x<3 \frac{1}{2}$

But x ∈ {1, 2, 3 …..10}

Solution set = {1, 2, 3}

Question 3

Solve : 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.

Sol :

5 – 4x > 2 – 3x

– 4x + 3x > 2 – 5

=> – x > – 3

=> x < 3

x ∈ w,

solution set {0, 1, 2}

Solution set on Number Line :

Question 4

List the solution set of 30 – 4 (2.x – 1) < 30, given that x is a positive integer.

Sol :

30 – 4 (2x – 1) < 30

30 – 8x + 4 < 30

– 8x < 30 – 30 – 4

$-8 x<-4 x>\frac{-4}{-8}$

$x>\frac{1}{2}$

x is a positive integer

x = {1, 2, 3, 4…..} 

Question 5

Solve : 2 (x – 2) < 3x – 2, x ∈ { – 3, – 2, – 1, 0, 1, 2, 3} .

Sol :

2(x – 2) < 3x – 2

=> 2x – 4 < 3x – 2

=> 2x – 3x < – 2 + 4

=> – x < 2

=> x > – 2

Solution set = { – 1, 0, 1, 2, 3}

Question 6

If x is a negative integer, find the solution set of $\frac{2}{3}+\frac{1}{3}(x+1)>0$

Sol :

$\frac{2}{3}+\frac{1}{3} x+\frac{1}{3}>0$

$\frac{1}{3} x+1>0$

$\frac{1}{3} x>-1$

$x>-1 \times \frac{3}{1} \Rightarrow x>-3$

x is a negative integer

Solution set = {- 2, – 1}

Question 7

Solve : $\frac{2 x-3}{4} \geq \frac{1}{2}$∈ {0, 1, 2,…,8}

Sol :

$\frac{2 x-3}{4} \geq \frac{1}{2}$

$2 x-3 \geq \frac{4}{2}$

$2 x-3 \geq \frac{4}{2}$

2x-3≥2

2x≥2+3

∵x∈{0,1,2....8}

∴Solution set={3,4,5,6,7,8}

Question 8

Solve x – 3 (2 + x) > 2 (3x – 1), x ∈ { – 3, – 2, – 1, 0, 1, 2, 3}. Also represent its solution on the number line.

Sol :

x – 3 (2 + x) > 2 (3x – 1)

x – 6 – 3x > 6x – 2

x – 3x – 6x > – 2 + 6

– 8x > 4

$x<\frac{-4}{8}$

$x<-\frac{1}{2}$

x ∈ { – 3, – 2, – 1, 0, 1, 2}

.’. Solution set = { – 3, – 2, – 1}

Solution set on Number Line :

Question 9

Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.

Sol :

x – 3 < 2x – 1

x – 2x < – 1 + 3

– x < 2 x > – 2

But x ∈ {1, 2, 3, 4, 5, 6, 7, 9}

Solution set = {1, 2, 3, 4, 5, 6, 7, 9} Ans.

Question 10

Given A = {x : x ∈ I, – 4 ≤ x ≤ 4}, solve 2x – 3 < 3 where x has the domain A Graph the solution set on the number line.

Sol :

2x – 3 < 3 

2x < 3 + 3 

2x < 6 

x < 3

But x has the domain A = {x : x ∈ I – 4 ≤ x ≤ 4}

Solution set = { – 4, – 3, – 2, – 1, 0, 1, 2}

Solution set on Number line :

Question 11

List the solution set of the inequation

$\frac{1}{2}+8 x>5 x-\frac{3}{2}, x \in Z$

Sol :

$\frac{1}{2}+8 x>5 x-\frac{3}{2}$

$8 x-5 x>-\frac{3}{2}-\frac{1}{2}$

3x>-2

$x>-\frac{2}{3}$

∵x∈Z ,

∴Solution set={0,1,2,3,4.....}

Question 12

List the solution set of $\frac{11-2x}{5} \geq \frac{9-3 x}{8}+\frac{3}{4}$x∈N

Sol :

$\frac{11-2 x}{5} \geq \frac{9-3 x}{8}+\frac{3}{4}$

88 – 16x ≥ 45 – 15x + 30

(L.C.M. of 8, 5, 4 = 40}

– 16x + 15x ≥ 45 + 30 – 88

– x ≥ – 13

x ≤ 13

x ≤ N.

Solution set = {1, 2, 3, 4, 5, .. , 13} 

Question 13

Find the values of x, which satisfy the inequation : 

$-2 \leq \frac{1}{2}-\frac{2 x}{3} \leq 1 \frac{5}{6}, x \in N$

Graph the solution set on the number line. (2001)

Sol :

$-2 \leq \frac{1}{2}-\frac{2 x}{3} \leq 1 \frac{5}{6}, x \in N$

$-2-\frac{1}{2} \leq \frac{1}{2}-\frac{2 x}{3}-\frac{1}{2} \leq \frac{11}{6}-\frac{1}{2}$

[By subtracting $\frac{1}{2}$ on both sides of inequality]

⇒$-\frac{5}{2} \leq \frac{2 x}{3} \leq \frac{8}{6}$

⇒-15≤-4x≤8

⇒15≥4x≥-8

⇒$\frac{15}{4} \geq x \geq-2$

⇒$3 \frac{3}{4} \geq x \geq-2$

But x∈N , hence only possible solution for x={1,2,3}

Question 14

If x ∈ W, find the solution set of

$\frac{3}{5} x-\frac{2 x-1}{3}>1$

Also graph the solution set on the number line, if possible.

Sol :

$\frac{3}{5} x-\frac{2 x-1}{3}>1$

9x – (10x – 5) > 15 (L.C.M. of 5, 3 = 15)

9x – 10x + 5 > 15

– x > 15 – 5

– x > 10

x < – 10

But x ∈ W

Solution set = Φ

Hence it can’t be represented on number line.

Question 15

Solve:

(i) $\frac{x}{2}+5 \leq \frac{x}{3}+6$ where x is a positive odd integer.

(ii) $\frac{2 x+3}{3} \geq \frac{3 x-1}{4}$ where x is positive even integer.

Sol :

(i)

$\frac{x}{2}+5 \leq \frac{x}{3}+6$

$\frac{x}{2}-\frac{x}{3} \leq 6-5$

$\frac{3 x-2 x}{6} \leq 1$

$\frac{x}{6} \leq 1$

x≤6

∵x is a positive odd integer

∴x={1,3,5}

(ii)

$\frac{2 x+3}{3} \geq \frac{3 x-1}{4}$

⇒$\frac{2 x}{3}+\frac{3}{3} \geq \frac{3 x}{4}-\frac{1}{4}$

⇒$\frac{2 x}{3}-\frac{3 x}{4} \geq \frac{-1}{4}-1$

⇒$\frac{8 x-9 x}{12} \geq-\frac{5}{4}$

⇒$\frac{-x}{12} \geq \frac{-5}{4}$

⇒$\frac{x}{12} \leq \frac{5}{4}$

⇒$x \leq \frac{5}{4} \times 12$

⇒x≤15

∵x is positive even integer

∴x={2,4,6,8,10,12,14}

Question 16

Given that x ∈ I, solve the inequation and graph the solution on the number line :

$3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2$ (2004)

Sol :

$3 \geq \frac{x-4}{2}+\frac{x}{3}$ and $3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2$

(i) 

$3 \geq \frac{3 x-12+2 x}{6}$

⇒$3 \geq \frac{3 x-12+2 x}{6}$

⇒$3 \geq \frac{5 x-12}{6}$

⇒18≥5 x-12 

⇒5 x-12≤18

⇒5x≤18+12

⇒5x≤30

⇒x≤6

(ii)

⇒$\frac{x-4}{2}+\frac{x}{3} \geq 2$

⇒$\frac{3 x-12+2 x}{6} \geq 2$

⇒$\frac{5 x-12}{6} \geq 2$

⇒5x-12≥12

⇒$5 x \geq 12+12, x \geq \frac{24}{5}$

⇒$x \geq 4 \frac{4}{5}$

∴x={5,6}

Number line 

Question 17

Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1 < x + 4.

Sol :

⇒-3 < 2x – 1 < x + 4

⇒– 3 < 2x – 1 and 2x – 1 < x + 4

⇒ – 2x < – 1 + 3 and 2x – x < 4 + 1

⇒ – 2x < 2 and x < 5

⇒ – x < 1

⇒ x > – 1

⇒– 1 < x < 5

x ∈ {1, 2, 3, 4, 5, 6, 7, 9}

Solution set = {1, 2, 3, 4} Ans.

Question 18

Solve : 1 ≥ 15 – 7x > 2x – 27, x ∈ N

Sol :

⇒1≥15–7x>2x–27

⇒1≥15–7x and 15–7x>2x–27

⇒7x≥15-1 and -7x-2x>-27-15

⇒7x≥14 and -9x>-42

⇒$x \geq 2$ and $-x>-\frac{42}{9}$

⇒2≤x and $-x>-\frac{14}{3}$ and $x<\frac{14}{3}$

$2 \leq x<\frac{14}{3}$

But x∈N

∴Solution set={2,3,4}

Question 19

If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.

Sol :

2 + 4x < 2x – 5 ≤ 3x

2 + 4x < 2x – 5 and 2x – 5 ≤ 3x 

4x – 2x < – 5 – 2 ,and 2x – 3x ≤ 5

2x<-7 and -x≤5

$x<-\frac{7}{2}$ and x≥-5 and -5≤x

∴$-5 \leq x<-\frac{7}{2}$

∴x∈Z

∴Solution set={-5,-4}

Solution set on Number line

Question 20

Solve the inequation $=12+1 \frac{5}{6} x$≤ 5 + 3x, x ∈ R. Represent the solution on a number line. (1999)

Sol :

$12+1 \frac{5}{6} x \leq 5+3 x$

$12+\frac{11}{6} x \leq 5+3 x$

72+11x≤-42 

$-x \leq-\frac{42}{7}$

-x≤-6

x≥6

∴x∈R

∴Solution set={x : x∈R , x≥6 }

Solution set on Number line

Question 21

Solve : $\frac{4 x-10}{3} \leq \frac{5 x-7}{2} \times \in R$ and represent the solution set on the number line

Sol :

⇒$\frac{4 x-10}{3} \leq \frac{5 x-7}{2}$

⇒8x–20≤15x–21

(L.C.M. of 3, 2 = 6)

⇒8x-15x≤-21+20

⇒-7x≤-1

⇒$-x \leq-\frac{1}{7} \Rightarrow x>\frac{1}{7}$

∵x∈R

∴Solution set=$\left\{x: x \in R, x>\frac{1}{7}\right\}$

Solution set on the number line

Question 22

Solve $\frac{3 x}{5}-\frac{2 x-1}{3}>1$ , x ∈ R and represent the solution set on the number line.

Sol :

⇒$\frac{3 x}{5}-\frac{2 x-1}{3}>1$

⇒9x – (10x – 5) > 15

⇒9x – 10x + 5 > 15

⇒– x > 15 – 5

⇒– x > 10

⇒x < – 10

⇒x ∈ R.

∴ Solution set = {x : x ∈R, x < – 10}

Solution set on the number line

Question 23

Solve the inequation – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line. (2000)

Sol :

⇒– 3 ≤ 3 – 2x < 9

⇒– 3 ≤ 3 – 2x and 3 – 2x < 9

⇒2x ≤ 3 + 3 and – 2x < 9 – 3

⇒2x ≤ 6 and – 2x < 6

⇒ x ≤ 3 and – x < 3 

⇒ x ≤ – 3 and – 3 < x

⇒– 3 < x ≤ 3.

Solution set= {x : x ∈ R, – 3 < x ≤ 3)

Solution on number line

Question 24

Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on number line. (2003)

Sol :

⇒2 ≤ 2x – 3 ≤ 5 .

⇒2 ≤ 2x – 3 and 2x – 3 ≤ 5

⇒2 + 3 ≤ 2x and 2x ≤ 5 + 3

⇒5 ≤ 2x and 2x ≤ 8.

⇒$\frac{5}{2} \leq x$ and $x \leq 4$

∴$\frac{5}{2} \leq x \leq 4$

∴Solution set $=\left\{x: x \in R, \frac{5}{2} \leq x \leq 4\right\}$

Solution set on number line

Question 25

Given that x ∈ R, solve the following inequation and graph the solution on the number line: – 1 ≤ 3 + 4x < 23. (2006)

Sol :

We have

⇒– 1 ≤ 3 + 4x < 23

⇒– 1 – 3 ≤ 4x < 23 – 3

⇒– 4 ≤ 4x < 20 

⇒ – 1 ≤ x < 5, x ∈ R

Solution Set = { – 1 ≤ x < 5; x ∈ R}

The graph of the solution set is shown below

Question 26

Solve tlie following inequation and graph the solution on the number line. (2007)

$-2 \frac{2}{3} \leq x+\frac{1}{3}<3+\frac{1}{3} x \in R$

Sol :

Given 

$-2 \frac{2}{3} \leq x+\frac{1}{3}<3+\frac{1}{3} x \in R$

$-\frac{8}{3} \leq x+\frac{1}{3}<\frac{10}{3}$

Multiplying by 3, L.C.M. of fractions, we get

-8≤3x+1<10

-1-1≤3x+1-1<10-1  [add -1]

-9≤3x<3

-3≤x<3  [dividing by 3]

Hence , the solution set is {x : x∈R, -3≤x<3}

The graph of the solution set is shown by the thick portion of the number line. The solid circle at -3 indicates that the number -3 is included among the solutions whereas the open circle at 3 indicates that 3 is not included among the solutions.

Question 27

Solve the following inequation and represent the solution set on the number line :

$-3<-\frac{1}{2}-\frac{2 x}{3} \leq \frac{5}{6}, x \in R$

Sol :

(i)

⇒$-3<-\frac{1}{2}-\frac{2 x}{3} \leq \frac{5}{6}, x \in R$

⇒$-3<-\frac{1}{2}-\frac{2 x}{3}$

⇒$-3<-\left(\frac{1}{2}+\frac{2 x}{3}\right)$

⇒$-\frac{2 x}{3}>\frac{-5}{2}$

⇒$\frac{2 x}{3}<\frac{5}{2}$

⇒$x<\frac{5}{2} \times \frac{3}{2}$

⇒$x<\frac{15}{4}$...(i)

(ii)

⇒$-\frac{1}{2}-\frac{2 x}{3} \leq \frac{5}{6}$

⇒$-\frac{2 x}{3} \leq \frac{5}{6}+\frac{1}{2}$

⇒$\frac{-2 x}{3} \leq \frac{5+3}{6}$

⇒$\frac{-2}{3} x \leq \frac{8}{6}$

⇒$\frac{2}{3} x \geq \frac{-8}{6}$

⇒$x \geq \frac{-8}{6} \times \frac{3}{2}$

⇒x≥-2

⇒-2≤x ..(ii)

From (i) and (ii)

$-2 \leq x \leq \frac{15}{4}$

∴Solution $=\left\{x: x \in \mathrm{R},-2 \leq x<\frac{15}{4}\right\}$

Now solution on number line

Question 28

Solve $\frac{2 x+1}{2}+2(3-x) \geq 7$,x∈R Also graph the solution set on the number line

Sol :

⇒$\frac{2 x+1}{2}+2(3-x) \geq 7, x \in R$

⇒$\frac{2 x+1}{2}+6-2 x \geq 7$

⇒$\frac{2 x+1}{2}-2 x \geq 7-6$

⇒$\frac{2 x+1-4 x}{2} \geq 1$

⇒2x+1-4x≥2

⇒-2x≥2-1

⇒-2x≥1

⇒$-x \geq \frac{1}{2}$

⇒$x \leq-\frac{1}{2}$

∴Solution set $\left\{x: x \in R, x \leq-\frac{1}{2}\right\}$

Solution on number line

Question 29

Solving the following inequation, write the solution set and represent it on the number line. $-3(x-7) \geq 15-7 x>\frac{x+1}{3}, n \in R$

Sol :

⇒$-3(x-7)\geq15-7 x>\frac{x+1}{3}, n \in R$

⇒-3(x-7) ≥15-7 x

⇒-3 x+21≥15-7 x

⇒-3 x+7 x≥15-21

⇒4x≥-6

⇒$x \geq \frac{-6}{4}$

⇒$x \geq \frac{-3}{2}$

⇒$\frac{-3}{2} \leq x$

and $15-7 x>\frac{x+1}{3}$

⇒45-21 x>x+1

⇒45-1>x+21 x

⇒44>22x

⇒2>x

⇒x=2

∴$\frac{-3}{2} \leq x<2, x \in \mathrm{R}$

Question 30

Solve the inequation :

$-2 \frac{1}{2}+2 x \leq \frac{4 x}{3} \leq \frac{4}{3}+2 x,  x \in W$

Graph the solution set on the number line.

Sol :

⇒$-2 \frac{1}{2}+2 x \leq \frac{4 x}{3} \leq \frac{4}{3}+2 x, x \in W$

⇒$-\frac{5}{2}+2 x \leq \frac{4 x}{3} \leq \frac{4}{3}+2 x$

⇒$-\frac{5}{2}+2 x \leq \frac{4 x}{3}$ and $\frac{4 x}{3} \leq \frac{4}{3}+2 x$

⇒$2 x-\frac{4 x}{3} \leq \frac{5}{2}$ and $\frac{4 x}{3}-2 x \leq \frac{4}{3}$

⇒12x-8x 15 and 4x-6x 4

⇒4x≤15 and -2x≤4

⇒$x \leq \frac{15}{4}$ and $-x \leq 4$

⇒$x \leq \frac{15}{4}$ and $x \geq-4$

⇒$x \leq \frac{15}{4}$ and $-4 \leq x$

∴$-2 \leq x \leq \frac{15}{4}$ ∴x=0,1,2,3

Solution set {x : x∈W, x≤3}

Solution set on number line

Question 31

Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line. (2011)

Sol :

⇒2x – 5 ≤ 5x + 4 < 11 

⇒2x – 5 ≤ 5x + 4

⇒2x – 5 – 4 ≤ 5x and 5x + 4 < 11

⇒2x – 9 ≤ 5x and 5x < 11 – 4

and 5x < 7

⇒2x – 5x ≤ 9 and x < 75

⇒3x > – 9 and x< 1.4

⇒x > – 3

Question 32

If x ∈ I, A is the solution set of 2 (x – 1) < 3 x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩B.

Sol :

⇒2 (x – 1) < 3 x – 1

⇒2x – 2 < 3x – 1

⇒2x – 3x < – 1 + 2 

⇒ – x < 1 x > – 1

Solution set A = {0, 1, 2, 3, ..,.}

⇒4x – 3 ≤ 8 + x

⇒4x – x ≤ 8 + 3

⇒3x ≤ 11

⇒$x \leq \frac{11}{3}$

Solution set B = {3, 2, 1, 0, – 1…}

A ∩ B = {0, 1, 2, 3}

Question 33

If P is the solution set of – 3x + 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find

(i) P ∩ Q

(ii) Q – P.

Sol :

(i) 

⇒– 3 x + 4 < 2 x – 3

⇒– 3x – 2x < – 3 – 4

⇒– 5x < – 7

⇒$-x<-\frac{7}{5}$

⇒$x>\frac{7}{5}$

(ii)

4x-5>12

4x<12+5

4x<17

$x<\frac{17}{4}$

∵x∈W

∴Solution set Q={4,3,2,1,0}

(i) P⋂Q={2,3,4}

(ii) Q-P={1,0}

Question 34

A = {x : 11x – 5 > 7x + 3, x ∈R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}

Find the range of set A ∩ B and represent it on a number line

Sol :

A = {x : 11x – 5 > 7x + 3, x ∈R}

B = {x : 18x – 9 ≥ 15 + 12x, x ∈R}

Now, A = 11x – 5 > 7x + 3

⇒11x – 7x > 3 + 5

⇒4x > 8

⇒x > 2, x ∈ R

B=18x-9≥15+12x

⇒18x-12x≥15+9

⇒6x≥24

⇒x≥4

∴A⋂B=x≥4,x∈R

Hence Range of A⋂B={x : x≥4,x∈R} and its graph will be

Question 35

Given: P {x : 5 < 2x – 1 ≤ 11, x∈R)

Q{x : – 1 ≤ 3 + 4x < 23, x∈I) where

R = (real numbers), I = (integers)

Represent P and Q on number line. Write down the elements of P ∩ Q. (1996)

Sol :

⇒P={x : 5<2x–1≤11}

⇒5<2x–1≤11

⇒5<2x-1 and 2x-1≤11

⇒-2x<-5-1 and 2x≤11+1

⇒-2x<-6 and 2x≤12

⇒-x<-3 and x≤6

⇒x>3 or 3<x

∴Solution set=3<x≤6={4,5,6}

Solution set on number line

⇒Q={-1≤3+4x<23}

⇒-1≤3+4x<23

⇒-1<3+4x and 3+4x<23

⇒-4x<3+1 and 4x<23-3

⇒-4x<4 and 4x<20

⇒-x<1 and x<5

⇒x>-1

⇒-1<x

∴-1<x<5

∴Solution set={0,1,2,3,4}

Solution set on number line

P∩Q={4}

Question 36

If x ∈ I, find the smallest value of x which satisfies the inequation 

$2 x+\frac{5}{2}>\frac{5 x}{3}+2$

Sol :

⇒$2 x+\frac{5}{2}>\frac{5 x}{3}+2$

⇒$2 x-\frac{5 x}{3}>2-\frac{5}{2}$

⇒12x – 10x > 12 – 15

⇒2x > – 3

⇒$x>-\frac{3}{2}$

Smallest value of x = – 1

Question 37

Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when

(i) x ∈ I

(ii) x ∈ W

(iii) x ∈ N.

Sol :

20 – 5 x < 5 (x + 8)

⇒ 20 – 5x < 5x + 40

⇒ – 5x – 5x < 40 – 20

⇒ – 10x < 20

⇒ – x < 2

⇒ x > – 2

(i) When x ∈ I, then smallest value = – 1.

(ii) When x ∈ W, then smallest value = 0.

(iii) When x ∈ N, then smallest value = 1.

Question 38

Solve the following inequation and represent the solution set on the number line :

$4 x-19<\frac{3 x}{5}-2 \leq-\frac{2}{5}+x, x \in R$

Sol :

$4 x-19<\frac{3 x}{5}-2 \leq-\frac{2}{5}+x, x \in R$

Hence, solution set is {x : -4 < x < 5, x ∈ R}

The solution set is represented on the number line as below.

$\Rightarrow 4 x-19<\frac{3 x}{5}-2$ and $\frac{3 x}{5}-2 \leq \frac{-2}{5}+x,  x \in \mathrm{R}$

$\Rightarrow 4 x-\frac{3 x}{5}<17$ and $-2+\frac{2}{5} \leq x-\frac{3 x}{5},  x \in \mathrm{R}$

$\Rightarrow \frac{17 x}{5}<17$ and $\frac{-8}{5} \leq \frac{2 x}{5},  x \in \mathrm{R}$

⇒x<5 and -4≤x , x∈R

⇒-4≤x<5 , x∈R

Hence , solution set is {x : -4≤x<5, x∈R}

The solution set is represented on the number line as below

Question 39

Solve the given inequation and graph the solution on the number line :

2y–3<y+1≤4y+7; y∈R.

Sol :

2y–3<y+1≤4y+7; y∈R.

(a) 2y–3<y+1

⇒ 2y–y<1+3

⇒ y<4

⇒ 4>y ….(i)

(b) y+1≤4y+7

⇒ y-4y≤7-1

⇒ -3y≤6

⇒ 3y≥6

⇒ $y \geq \frac{6}{-3}$

⇒ y≥-2..(ii)

From (i) and (ii)

4>y≥-2 or -2≤y<4

Now representing it on a number given below

Question 40

Solve the inequation and represent the solution set on the number line.

$-3+x \leq \frac{8 x}{3}+2 \leq \frac{14}{3}+2 x$, Where x∈I

Sol :

$-3+x \leq \frac{8 x}{3}+2 \leq \frac{14}{3}+2 x$, Where x∈I

(i) $-3+x \leq \frac{8 x}{3}+2$

⇒$-3-2 \leq \frac{8 x}{3}-x$

⇒$-5 \leq \frac{5 x}{3}$

⇒$-1 \leq \frac{x}{3}$

⇒-3≤x...(i)

and $\frac{8 x}{3}=2 \leq \frac{14}{3}+2 x$

⇒$\frac{8 x}{3}-2 x \leq \frac{14}{3}-2$

⇒$\frac{2 x}{3} \leq \frac{8}{3}$

⇒x≤4..(ii)

From (i) and (ii)

⇒$-5 \leq \frac{5 x}{3}$ and $\frac{2 x}{3} \leq \frac{8}{3}$

⇒x≥-3 and x≤4

∴-3≤x≤4

Solution set={-3,-2,-1,0,1,2,3,4}

Solution set on number line

Question 41

Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.

Sol :

Let the greatest integer = x

According to the condition,

2x + 7 > 3x

⇒ 2x – 3x > – 7

⇒ – x > – 7

⇒ x < 7

Value of x which is greatest = 6 Ans.

Question 42

One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.

Sol :
Let the length of the shortest pole = x metre

Length of pole which is burried in mud $=\frac{x}{3}$

Length of pole which is in the water $=\frac{x}{6}$ 

According to this problem,

⇒$x-\left[\frac{x}{3}+\frac{x}{6}\right] \geq 3$

⇒$x-\left(\frac{2 x+x}{6}\right) \geq 3$

⇒$x-\frac{x}{2} \geq 3$

⇒$\frac{x}{2} \geq 3$

⇒x≥6

∴Length of pole (shortest in length) =6 metres