ML Aggarwal Solutions Linear Inequations Chapter-Test ICSE Class 10 Ch-4

ML Aggarwal Solutions Linear Inequations Chapter-Test ICSE Class 10 Ch-4. Understanding APC Maths is a popular publication among ICSE Student. Therefore Step by Step Solved Questions of Chapter Test Inequality is given.

ML Aggarwal Solutions Linear Inequations Chapter-Test ICSE Class 10 Ch-4

These Solutions are given as council Latest prescibe guideline for board exam. Visit official Website CISCE  for detail information about ICSE Board Class-10.

ML Aggarwal Solutions Linear Inequations Chapter-Test ICSE Class 10 Ch-4

 

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 10th
Chapter-4 Linear Inequations
writer ML Aggarwal
Book Name Understanding
Topics Solutions of Chapter Test Questions
Edition 2022-2023

Chapter-Test

ML Aggarwal Solutions Linear Inequations ICSE Class 10 Ch-4

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.

Answer- 1

5x – 2 < 3(3 – x)
⇒ 5x – 2 ≤ 9 – 3x
⇒ 5x + 3x ≤ 9 + 2

or  8x ≤ 11

⇒ x ≤ 11/8

∵ x ∈ {-2, – 1, 0, 1, 2, 3, 4}

∴ So Solution set = {- 2, – 1, 0, 1}

and Solution set on number line

Solve the inequation chapter test ans 1

Question- 2 Solve the inequations : 6x – 5 < 3x + 4, x ∈ I.

Answer- 2

6x – 5 < 3x + 4
6x – 3x < 4 + 5
⇒ 3x <9
⇒ x < 3 as x ∈ I
Therefore 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.

Answer -3

x + 5 ≤ 2x + 3
x – 2x ≤ 3 – 5
⇒ -x ≤ -2
⇒ x ≥ 2

∵ x ∈ R ,  ∴ Solution set = {2, 3, 4, 5, …….}

Hence Solution set on number line

quation chapter test ans 3

Question -4 If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.

Answer -4

-1 < 3 – 2x ≤ 7
-1 < 3 – 2x and 3 – 2x ≤ 7
⇒ 2x < 3 + 1 and – 2x ≤ 7 – 3
or  2x < 4 and -2x ≤ 4
⇒ x < 2 and -x ≤ 2
and x ≥ -2 or -2 ≤ x
x ∈ R
Solution set -2 ≤ x < 2
and Solution set on number line
Solve the inequation chapter test ans 4

Question- 5 Solve the inequation :

Solve the inequation chapter test ans 5

Answer -5

(5x + 1)/7 – 4(x/7 + 2/5) ≤ 1.3/5 + (3x – 1)/7

(5x + 1)/7 – 4(x/7 + 2/5) ≤ 8/5 + (3x – 1)/7

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

25x + 5 – 4 (5x + 14) ≤ 56 + 15x – 5

⇒ 25 + 5 – 20x – 56 ≤ 56 + 15x – 5

⇒ 25x – 20x – 15x ≤ 56 – 5 – 5 + 56

or   – 10x ≤ 102

⇒ – x ≤ 102/10

or    – x ≤ 51/5

⇒ x ≥ – 51/5

∵ x ∈ R

∴ Therefore Solution set = {x : x ∈ R, x ≥ – 51/5}

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.

Answer- 6

7 < – 4x + 2 < 12

⇒ 7 < – 4x + 2 and – 4x + 2 < 12

⇒ 4x ≤ 2 – 7 and – 4x < 12 – 2

or    4x ≤ – 5 and – 4x < 10

⇒ x ≤ -5/4 and – x < 10/4

=   x ≤ -5/4 and – x < 5/2

or x > – (5/2)

∵ x ∈ R

∴ Solution set – 5/2 < x ≤ -5/4

= {x : x ∈ R, – 5/2 < x ≤ – 5/4}

So Solution set on the number line

Solve the inequation chapter test ans 6

Question -7 If x ∈ R, solve

Solve the inequation chapter test ans 7

Answer- 7

Solve the inequation chapter test ans 7

⇒ 2x – 3 ≥ x + (1 – x)/3 and x + (1 – x)/3 > 2/5x

⇒ 2x – 3 ≥ (3x + 1 – x)/3 and (3x + 1 – x)/3 > 2/5x

on more solving

⇒ 6x – 9 ≥ 3x + 1 – x and 15x + 5 – 5x > 6x

⇒ 6x – 3x + x ≥ 1 + 9 and 15x – 6x – 5x > – 5

on further simplification

⇒ 4x ≥ 10 and 4x > – 5

⇒ x ≥ 10/4 and x > – 5/4

So ,  x ≥ 5/2 , ∴ x ≥ 5/2 ,  because ∵ x ∈ R

∴ Solution set = {x : x ∈ R, x ≥ 5/2}

Hence Solution set on number line

Solve the inequation chapter test ans 7.

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.

Answer -8

Let the positive integer = x
According to the problem,
5a – 6 < 4x
⇒ 5a – 4x < 6
⇒ x < 6
Therefore 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.

Answer- 9

Let first least natural number = x
then second number = x + 1
and third number = x + 2

According to the condition 1/3(x + 2) – 1/5 (x) ≥ 3

5x + 10 – 3x ≥ 45

(Multiplying by 15 the L.C.M. of 2 and 5)

2x ≥ 45 – 10

⇒ 2x ≥ 35

x ≥ 35/2

⇒ x ≥ 17.1/2

∵ x is a natural least number and if ∴ x = 18

∴ find least natural number = 18

Second number = 18 + 1 = 19

And third numbers = 18 + 2 = 20

Hence, least natural numbers are 18, 19, 20

— : End of ML Aggarwal Solutions Linear Inequations Chapter-Test ICSE Class 10  :–

Return to :- ML Aggarwal Solutions for ICSE Class-10

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