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
x118
∵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 :

5x+174(x7+25)135+3x17,xR

Sol :

5x+174(x7+25)135+3x17

5x+174(x7+25)85+3x17

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

x10210

x515

x515

∵x∈R

∴Solution set={x:xR,x515}


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

x54 and x<104

x54 and x<52

x>52

∵x∈R

∴Solution set52<x54

{x:xR,52<x54}

Solution set on number line




Question 7

If x∈R, solve 2x3x+1x3>25x

Sol :

2x3x+1x3>25x

2x3x+1x3 and x+1x3>25x

2x33x+13x and 3x+1x3>25x

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

x104 and x>54

x52

x52

∵x∈R

∴Solution set={x:xR,x52}

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 
13(x+2)15(x)3

5x+10-3x≥45 (Multiplying by 15 the LCM of 3 and 5)
2x≥45-10
2x≥35
x352
x1712

∵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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