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Chapter 4

Linear Inequations — Chapter Test

Class - 10 ML Aggarwal Understanding ICSE Mathematics



Chapter 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.

Answer

Given,

5x−2≤3(3−x)⇒5x−2≤9−3x⇒5x+3x≤9+2⇒8x≤11⇒x≤1185x - 2 \le 3(3 - x) \\[0.5em] \Rightarrow 5x - 2 \le 9 - 3x \\[0.5em] \Rightarrow 5x + 3x \le 9 + 2 \\[0.5em] \Rightarrow 8x \le 11 \\[0.5em] \Rightarrow x \le \dfrac{11}{8}

Since, x ∈ { -2, -1, 0, 1, 2, 3, 4}.

∴ Solution set ={-2, -1, 0, 1}.

The graph of the solution set is represented by thick black dots.

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. Linear Inequations, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Question 2

Solve the inequations : 6x - 5 < 3x + 4, x ∈ I

Answer

Given,

6x−5<3x+4⇒6x−3x<4+5⇒3x<9⇒x<36x - 5 \lt 3x + 4 \\[0.5em] \Rightarrow 6x - 3x \lt 4 + 5 \\[0.5em] \Rightarrow 3x \lt 9 \\[0.5em] \Rightarrow x \lt 3

x ∈ I

∴ Solution Set = {..., -2, -1, 0, 1, 2}.

Question 3

Find the solution set of the inequation x + 5 ≤ 2x + 3 ; x ∈ R. Graph the solution set on the number line.

Answer

Given,

x+5≤2x+3⇒x−2x≤3−5⇒−x≤−2⇒x≥2x + 5 \le 2x + 3 \\[0.5em] \Rightarrow x - 2x \le 3 - 5\\[0.5em] \Rightarrow -x \le -2\\[0.5em] \Rightarrow x \ge 2

∴ Solution set = {x : x ∈ R, x ≥ 2 }.

The graph of the solution set is represented by thick line starting from and including 2.

Find the solution set of the inequation x + 5 ≤ 2x + 3 ; x ∈ R. Graph the solution set on the number line. Linear Inequations, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Question 4

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

Answer

Given,

−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⇒x<2 and x≥−2-1 \lt 3 - 2x \le 7 \\[0.5em] \Rightarrow -1 \lt 3 - 2x \text{ and } 3 - 2x \le 7 \\[0.5em] \Rightarrow 2x \lt 3 + 1 \text{ and } -2x \le 7 - 3 \\[0.5em] \Rightarrow 2x \lt 4 \text{ and } -2x \le 4 \\[0.5em] \Rightarrow x \lt 2 \text{ and } -x \le 2 \\[0.5em] \Rightarrow x \lt 2 \text{ and } x \ge -2

∴ Solution set = {x : x ∈ R, -2 ≤ x < 2}.

The graph of this inequation is represented by thick black line starting from -2 (including -2) till (not including) 2.

If x ∈ R (real numbers) and -1 < 3 - 2x ≤ 7, find solution set and represent it on a number line. Linear Inequations, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Question 5

Solve the inequation :

5x+17−4(x7+25)≤135+3x−17,x∈R.\dfrac{5x+1}{7} - 4\Big(\dfrac{x}{7}+ \dfrac{2}{5}\Big) \le 1\dfrac{3}{5}+\dfrac{3x-1}{7}, x ∈ \bold{R}.

Answer

Given,

5x+17−4(x7+25)≤135+3x−17,x∈R.\dfrac{5x+1}{7} - 4\Big(\dfrac{x}{7}+ \dfrac{2}{5}\Big) \le 1\dfrac{3}{5}+\dfrac{3x-1}{7}, x ∈ \bold{R}.

Multiplying both sides by 35

⇒25x+5−4(5x+14)≤56+15x−5⇒25x+5−20x−56≤56+15x−5⇒5x−51≤51+15x⇒5x−15x≤51+51⇒−10x≤102⇒x≥−10210⇒x≥−515\Rightarrow 25x + 5-4(5x + 14) \le 56 + 15x -5 \\[0.5em] \Rightarrow 25x + 5 - 20x - 56 \le 56 + 15x -5 \\[0.5em] \Rightarrow 5x - 51 \le 51 + 15x \\[0.5em] \Rightarrow 5x - 15x \le 51 + 51 \\[0.5em] \Rightarrow -10x \le 102 \\[0.5em] \Rightarrow x \ge -\dfrac{102}{10} \\[0.5em] \Rightarrow x \ge -\dfrac{51}{5}

∴ Solution set = {x : x ∈ R, x ≥−515\ge -\dfrac{51}{5} }.

Question 6

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

Answer

Given,
7 ≤ –4x + 2 < 12

⇒7≤−4x+2 and −4x+2<12\Rightarrow 7 \le - 4x + 2 \text{ and } - 4x + 2 \lt 12 \\[0.5em]

Solving left side,

7≤−4x+2⇒4x≤2−7⇒4x≤−5⇒x≤−547 \le - 4x + 2 \\[0.5em] \Rightarrow 4x \le 2-7 \\[0.5em] \Rightarrow 4x \le -5 \\[0.5em] \Rightarrow x \le \dfrac{-5}{4}

Solving right side,

−4x+2<12⇒−4x<12−2⇒−4x<10⇒4x>−10⇒x>−52-4x + 2 \lt 12 \\[0.5em] \Rightarrow -4x \lt 12 - 2 \\[0.5em] \Rightarrow -4x \lt 10 \\[0.5em] \Rightarrow 4x \gt -10 \\[0.5em] \Rightarrow x \gt \dfrac{-5}{2}

∴ Solution set = {x : x ∈ R, −52-\dfrac{5}{2} < x ≤ −54-\dfrac{5}{4}}.

The graph of the inequation is represented by thick black line starting from −52-\dfrac{5}{2} (excluding −52-\dfrac{5}{2}) till −54-\dfrac{5}{4} (including −54-\dfrac{5}{4}).

Find the range of values of x, which satisfy 7 ≤ –4x + 2 < 12, x ∈ R. Graph these values of x on the real number line. Linear Inequations, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Question 7

If x ∈ R, solve 3 - 2x ≥ x + 1−x3\dfrac{1 - x}{3} > 25x\dfrac{2}{5}x. Also represent the solution on the number line.

Answer

To prove:

3 - 2x ≥ x + 1−x3\dfrac{1 - x}{3} > 2x5\dfrac{2x}{5}

Solving L.H.S. of the above inequation, we get :

⇒ 3 - 2x ≥ x + 1−x3\dfrac{1 - x}{3}

⇒ 3 - 2x ≥ 3x+1−x3\dfrac{3x + 1 - x}{3}

⇒ 3(3 - 2x) ≥ 2x + 1

⇒ 9 - 6x ≥ 2x + 1

⇒ 2x + 6x ≤ 9 - 1

⇒ 8x ≤ 8

⇒ x ≤ 88\dfrac{8}{8}

⇒ x ≤ 1 ............(1)

Solving R.H.S. of the above equation, we get :

⇒ x + 1−x3>2x5\dfrac{1 - x}{3} \gt \dfrac{2x}{5}

⇒ 3x+1−x3>2x5\dfrac{3x + 1 - x}{3} \gt \dfrac{2x}{5}

⇒ 5(2x + 1) > 3 × 2x

⇒ 10x + 5 > 6x

⇒ 10x - 6x > -5

⇒ 4x > -5

⇒ x > −54-\dfrac{5}{4} ...........(2)

From equation (1) and (2), we get :

Solution set = {x : −54-\dfrac{5}{4} < x ≤ 1, x ∈ R}

Representation of solution set on real number line is :

Solve the following inequation and write the solution and represent it on the real number line. ICSE 2024 Maths Specimen Solved Question Paper.

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

Let the positive integer = x
According to the problem,

5x−6<4x⇒5x−4x<6⇒x<65x - 6 \lt 4x \\[0.5em] \Rightarrow 5x - 4x \lt 6 \\[0.5em] \Rightarrow x \lt 6

∴ Solution set = { 1, 2, 3, 4, 5}.

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

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

Given,

13(x+2)−15(x)≥3⇒x3−x5≥3−23⇒5x−3x15≥9−23⇒2x15≥73⇒2x≥7×153⇒2x≥35⇒x≥352⇒x≥1712\dfrac{1}{3}(x+2) - \dfrac{1}{5}(x) \ge 3 \\[0.5em] \Rightarrow \dfrac{x}{3} - \dfrac{x}{5} \ge 3 - \dfrac{2}{3} \\[0.5em] \Rightarrow \dfrac{5x- 3x}{15} \ge \dfrac{9-2}{3} \\[0.5em] \Rightarrow \dfrac{2x}{15} \ge \dfrac{7}{3} \\[0.5em] \Rightarrow 2x \ge \dfrac{7 \times 15}{3} \\[0.5em] \Rightarrow 2x \ge 35 \\[0.5em] \Rightarrow x \ge \dfrac{35}{2} \\[0.5em] \Rightarrow x \ge 17\dfrac{1}{2}

Since the three consecutive numbers should be natural numbers
∴ x = 18
    x + 1 = 19
    x + 2 = 20

Hence, the three smallest consecutive natural numbers are 18, 19, 20

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