ML Aggarwal Solution Class 10 Chapter 4 Linear Inequations Test

 Test

Question 1

Solve the inequation : 5x – 2 ≤ 3(3 – x) where x ∈ { – 2, – 1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.

Sol :

⇒5x–2<3(3–x)

⇒5x–2≤9–3x

⇒5x+3x≤9+2

⇒8x≤11
⇒$x \leq \frac{11}{8}$
∵x∈{-2,-1,0,1,2,3,4}
∴Solution set on number line




Question 2

Solve the inequations :

6x – 5 < 3x + 4, x ∈ I.

Sol :

6x – 5 < 3x + 4

6x – 3x < 4 + 5

⇒ 3x <9

⇒ x < 3

x ∈ I

Solution Set = { -1, -2, 2, 1, 0….. }


Question 3

Find the solution set of the inequation

x + 5 < 2 x + 3 ; x ∈ R

Graph the solution set on the number line.

Sol :

x + 5 ≤ 2x + 3

x – 2x ≤ 3 – 5

⇒ -x ≤ -2

⇒ x ≥ 2

∵x∈{2,3,4,5...}
∴Solution set on number line




Question 4

If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.

Sol :

-1 < 3 – 2x ≤ 7

-1 < 3 – 2x and 3 – 2x ≤ 7

⇒ 2x < 3 + 1 and – 2x ≤ 7 – 3

⇒ 2x < 4 and -2x ≤ 4

⇒ x < 2 and -x ≤ 2

and x ≥ -2 or -2 ≤ x

x ∈ R

Solution set -2 ≤ x < 2

Solution set on number line





Question 5

Solve the inequation :

$\frac{5 x+1}{7}-4\left(\frac{x}{7}+\frac{2}{5}\right) \leq 1 \frac{3}{5}+\frac{3 x-1}{7}, x \in R$

Sol :

$\frac{5 x+1}{7}-4\left(\frac{x}{7}+\frac{2}{5}\right) \leq 1 \frac{3}{5}+\frac{3 x-1}{7}$

$\frac{5 x+1}{7}-4\left(\frac{x}{7}+\frac{2}{5}\right) \leq \frac{8}{5}+\frac{3 x-1}{7}$

Multiplying by L.C.M of 7 and 5 i.e. 35

25x+5-4(5x+14)≤56+15x-5

25x+5-20x-56≤56+15x-5

25x-20x-15x≤56-5-5+56

10x≤102

$-x \leq \frac{102}{10}$

$-x \leq \frac{51}{5}$

$x \geq-\frac{51}{5}$

∵x∈R

∴Solution set$=\left\{x: x \in R, x \geq-\frac{51}{5}\right\}$


Question 6

Find the range of values of a, which satisfy 7 ≤ – 4x + 2 < 12, x ∈ R. Graph these values of a on the real number line.

Sol :

7 < – 4x + 2 < 12

7<– 4x + 2 and – 4x + 2 < 12

4x≤2-7 and -4x<12-2

4x≤-5 and -4x<10

$x \leq \frac{-5}{4}$ and $-x<\frac{10}{4}$

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

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

∵x∈R

∴Solution set$-\frac{5}{2}<x \leq \frac{-5}{4}$

$\left\{x: x \in R,-\frac{5}{2}<x \leq \frac{-5}{4}\right\}$

Solution set on number line




Question 7

If x∈R, solve $2 x-3 \geq x+\frac{1-x}{3}>\frac{2}{5} x$

Sol :

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

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

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

6x-9≥3x+1-x and 15x+5-5x>6x

6x-3x+x≥1+9 and 15x-6x-5x>-5

4x≥10 and 4x>-5

$x \geq \frac{10}{4}$ and $x>-\frac{5}{4}$

$x \geq \frac{5}{2}$

∴$x \geq \frac{5}{2}$

∵x∈R

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

Solution set on number line




Question 8

Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.
Sol :
Let the positive integer = x
According to the problem,
5a – 6 < 4x
⇒ 5a – 4x < 6
⇒ x < 6
Solution set = {x : x < 6}
= { 1, 2, 3, 4, 5, 6}

Question 9

Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.
Sol :
Let first least natural number = x
then second number = x + 1
and third number = x + 2

According to the condition 
$\frac{1}{3}(x+2)-\frac{1}{5}(x) \geq 3$

5x+10-3x≥45 (Multiplying by 15 the LCM of 3 and 5)
2x≥45-10
2x≥35
$x \geq \frac{35}{2}$
$x \geq 17 \frac{1}{2}$

∵x is a natural least number
∴x=18
∴First least number=18

Second number=18+1=19
and third number=18+2=20

Hence least numbers are 18, 19 , 20

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