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

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