ML Aggarwal Class 10 Solutions Linear Inequations Chapter Test
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test.
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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.
Solution:
5x – 2 < 3(3 – x)
=> 5x – 2 ≤ 9 – 3x
=> 5x + 3x ≤ 9 + 2
Question 2.
Solve the inequations :
6x – 5 < 3x + 4, x ∈ I.
Solution:
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.
Solution:
x + 5 ≤ 2x + 3
x – 2 x ≤ 3 – 5
=> – x ≤ – 2
=> x ≥ 2
Question 4.
If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.
Solution:
– 1 < 3 – 2x ≤ 7
– 1 < 3 – 2x and 3 – 2x ≤ 7
2 x < 3 + 1 and – 2x ≤ 7 – 3
2 x < 4 and – 2 x ≤ 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 :
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } ,x\in R\)
Solution:
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } \)
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le \frac { 8 }{ 5 } +\frac { 3x-1 }{ 7 } \)
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.
Solution:
7 < – 4x + 2 < 12
7 < – 4x + 2 and – 4x + 2 < 12
Question 7.
If x∈R, solve \(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
Solution:
\(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
\(2x-3\ge x+\frac { 1-x }{ 3 } \) and \(x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
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.
Solution:
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} Ans.
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 atleast 3.
Solution:
Let first least natural number = x
then second number = x + 1
and third number = x + 2
Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Linear Inequations Chapter Test are helpful to complete your math homework.
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