ICSE Class 10 Mathematics · Selina Concise Mathematics · 56 solved questions
If x ∈ W, then solution set of inequation -x > -7, is :
Given,
-x > -7
x < 7
Since x ∈ W and x < 7:
The solution set is: {0, 1, 2, 3, 4, 5, 6}
Hence, Option 2 is the correct option.
The value of x, for 4(2x - 5) < 2x + 28, x ∈ R, is :
Given,
⇒ 4(2x - 5) < 2x + 28
⇒ 8x - 20 < 2x + 28
⇒ 8x - 2x < 28 + 20
⇒ 6x < 48
⇒ x < $\frac{48}{6}$
⇒ x < 8.
Hence, Option 2 is the correct option.
The solution set for the inequation -2x + 7 ≤ 3, x ∈ R is :
Given,
⇒ -2x + 7 ≤ 3
⇒ 2x ≥ 7 - 3
⇒ 2x ≥ 4
⇒ x ≥ $\frac{4}{2}$
⇒ x ≥ 2.
Since, x ∈ R.
Solution set = {x : x ∈ R, x ≥ 2}
Hence, Option 4 is the correct option.
For 7 - 3x < x - 5, the solution set is :
Given,
⇒ 7 - 3x < x - 5
⇒ x + 3x > 7 + 5
⇒ 4x > 12
⇒ x > $\frac{12}{4}$
⇒ x > 3.
Hence, Option 1 is the correct option.
x(8 - x) > 0 and x ∈ N gives :
For x(8 - x) > 0
Either,
⇒ x > 0 and (8 - x) > 0
⇒ x > 0 and x < 8 ...........(1)
or,
⇒ x < 0 and (8 - x) < 0
⇒ x < 0 and x > 8 .............(2)
But both conditions of equation (2) are not possible simultaneously.
From equation (1),
Solution set = {0 < x < 8}.
Hence, Option 3 is the correct option.
State, true or false :
(i) x < -y ⇒ -x > y
(ii) -5x ≥ 15 ⇒ x ≥ -3
(iii) 2x ≤ -7 ⇒ $\frac{2x}{-4} \ge \frac{-7}{-4}$
(iv) 7 > 5 ⇒ $\frac{1}{7} \lt \frac{1}{5}$
(i) Given,
x < -y
∴ -x > y [Using rule 5]
Hence, the statement is True.
(ii) Given,
-5x ≥ 15
Dividing both sides of the above inequation by -5,
⇒ x ≤ -3 [Using rule 4]
Hence, the statement is False.
(iii) Given,
2x ≤ -7
Dividing both sides of the above inequation by -4,
⇒ $\frac{2x}{-4} \ge \frac{-7}{-4}$ [Using rule 4]
Hence, the statement is True.
(iv) Given,
7 > 5
Taking reciprocals,
$\frac{1}{7} \lt \frac{1}{5}$ [Using rule 6]
Hence, the statement is True.
State, whether the following statements are true or false.
(i) If a < b, then a - c < b - c
(ii) If a > b, then a + c > b + c
(iii) If a < b, then ac > bc
(iv) If a > b, then $\frac{a}{c} \lt \frac{b}{c}$
(v) If a - c > b - d; then a + d > b + c
(vi) If a < b, and c > 0, then a - c > b - c
where a, b, c, and d are real numbers c ≠ 0.
(i) Given,
a < b
Subtracting both sides by c,
a - c < b - c.
Hence, the statement is True.
(ii) Given,
a > b
Adding both sides by c,
a + c > b + c.
Hence, the statement is True.
(iii) Given,
a < b
If c is a positive number,
Multiplying both sides by c we get,
ac < bc
If c is a negative number,
Multiplying both sides by c we get,
ac > bc [Using rule 4]
Hence, the statement is False.
(iv) Given,
a > b
If c is a positive number,
Dividing both sides by c we get,
$\frac{a}{c} \gt \frac{b}{c}$
If c is a negative number,
Dividing both sides by c we get,
$\frac{a}{c} \lt \frac{b}{c}$ [Using rule 4]
Hence, the statement is False.
(v) Given,
a - c > b - d
Adding both sides by (c + d) we get,
⇒ a - c + (c + d) > b - d + (c + d)
⇒ a - c + c + d > b + c - d + d
⇒ a + d > b + c
Hence, the statement is True.
(vi) Given,
a < b and c > 0
Subtracting both sides by c we get,
a - c < b - c [As c is a positive number.]
Hence, the statement is False.
Solve the inequation :
3 - 2x ≥ x - 12 given that x ∈ N.
Given,
⇒ 3 - 2x ≥ x - 12
⇒ x + 2x ≤ 3 + 12
⇒ 3x ≤ 15
Dividing both sides by 3 we get,
⇒ x ≤ 5
Since, x ∈ N
∴ Solution set = {1, 2, 3, 4, 5}.
If 25 - 4x ≤ 16, find :
(i) the smallest value of x when x is a real number,
(ii) the smallest value of x when x is an integer.
Given,
⇒ 25 - 4x ≤ 16
⇒ -4x ≤ 16 - 25
⇒ -4x ≤ -9
Multiplying both sides by -1 we get,
⇒ 4x ≥ 9 (As on multiplying by negative no. the sign reverses.)
Dividing both sides by 4 we get,
⇒ x ≥ $\frac{9}{4}$
⇒ x ≥ 2.25
(i) Given,
x ≥ 2.25
Hence, smallest value of x when x is a real number is 2.25
(ii) Given,
x ≥ 2.25
Hence, smallest value of x when x is an integer is 3.
If the replacement set is the set of real numbers, solve :
(i) -4x ≥ -16
(ii) 8 - 3x ≤ 20
(i) -4x ≥ -16
$$\begin{aligned}\Rightarrow \frac{-4x}{-4} \le \frac{-16}{-4} \\[1em] \Rightarrow x \le 4\end{aligned}$$
∴ Solution set = {x : x ∈ R and x ≤ 4}
(ii) 8 - 3x ≤ 20
⇒ -3x ≤ 20 - 8
⇒ -3x ≤ 12
Dividing both sides by -3 we get.
⇒ x ≥ -4 (As on dividing by negative no. the sign reverses.)
∴ Solution set = {x : x ∈ R and x ≥ -4}.
Find the smallest value of x for which 5 - 2x < $5\frac{1}{2} - \frac{5}{3}x$, where x is an integer.
Given,
$$\begin{aligned}\Rightarrow 5 - 2x \lt 5\frac{1}{2} - \frac{5}{3}x \\[1em] \Rightarrow 5 - 2x \lt \frac{11}{2} - \frac{5}{3}x \\[1em] \Rightarrow 5 - \frac{11}{2} \lt -\frac{5}{3}x + 2x \\[1em] \Rightarrow \frac{10 - 11}{2} \lt \frac{-5x + 6x}{3} \\[1em] \Rightarrow -\frac{1}{2} \lt \frac{x}{3} \\[1em] \Rightarrow \frac{x}{3} \times 3 \gt -\frac{1}{2} \times 3 \\[1em] \Rightarrow x \gt -\frac{3}{2} \\[1em] \Rightarrow x \gt -1.5\end{aligned}$$
Since, x is an integer.
Hence, smallest value of x = -1.
Find the largest value of x for which
2(x - 1) ≤ 9 - x and x ∈ W.
Given,
⇒ 2(x - 1) ≤ 9 - x
⇒ 2x - 2 ≤ 9 - x
⇒ 2x + x ≤ 9 + 2
⇒ 3x ≤ 11
⇒ x ≤ $\frac{11}{3}$
⇒ x ≤ 3.67
Since, x ∈ W
Hence, largest value of x for which 2(x - 1) ≤ 9 - x is 3.
For the following real number line, the solution set is :
From real number line, we get :
x ∈ R, x > -2 and x ≤ 4.
∴ Solution set = {x : x ∈ R and -2 < x ≤ 4}.
Hence, Option 3 is the correct option.
The solution set for the following number line is :
From number line :
x ∈ Integers (Z) and x ≥ -3.
Solution set = {x : x ∈ Z and -3 ≤ x}.
Hence, Option 2 is the correct option.
The following number line represents :
From the number line, we get :
Solution set = {{(x < 10) ∪ (x > 10)}}
Hence, Option 2 is the correct option.
The solution set for the following number line is :
From number line, we get :
x ∈ R, x ≤ -2 or x ≥ 3.
∴ Solution set = {x : x ∈ R, x ≤ -2 or x ≥ 3}
Hence, Option 4 is the correct option.
The number line for the solution of inequation x > 5 and x < 10 (x ∈ R) is :
For x > 5, x < 10 and x ∈ R.
Solution set = {x : x ∈ R and 5 < x < 10}.
Hence, Option 2 is the correct option.
For each graph given alongside, write an inequation taking x as the variable :
(i) From graph we get,
x ≤ -1 and x ∈ R.
(ii) From graph we get,
x ≥ 2 and x ∈ R.
(iii) From graph we get,
-4 ≤ x < 3 and x ∈ R.
(iv) From graph we get,
-1 < x ≤ 5 and x ∈ R.
For the following inequations, graph the solution set on the real number line :
-4 ≤ 3x - 1 < 8
-4 ≤ 3x - 1 < 8
Solving L.H.S. of the equation,
⇒ -4 ≤ 3x - 1
⇒ 3x ≥ -4 + 1
⇒ 3x ≥ -3
⇒ x ≥ -1 .........(i)
Solving R.H.S. of the equation,
⇒ 3x - 1 < 8
⇒ 3x < 8 + 1
⇒ 3x < 9
⇒ x < 3 .........(ii)
From (i) and (ii) we get,
-1 ≤ x < 3.
∴ Solution set = {-1 ≤ x < 3 : x ∈ R }
Solution on the number line is :
For the following inequations, graph the solution set on the real number line :
-1 < 3 - 2x ≤ 7
Given,
-1 < 3 - 2x ≤ 7
Solving L.H.S. of the equation,
⇒ -1 < 3 - 2x
⇒ -1 - 3 < -2x
⇒ -4 < -2x
Dividing both sides by -2 we get,
⇒ 2 > x (As sign reverses on dividing by negative no.)
⇒ x < 2 .........(i)
Solving R.H.S. of the equation,
⇒ 3 - 2x ≤ 7
⇒ 3 - 7 ≤ 2x
⇒ -4 ≤ 2x
Dividing both sides by 2 we get,
⇒ -2 ≤ x
⇒ x ≥ -2 .........(ii)
From (i) and (ii) we get,
-2 ≤ x < 2
∴ Solution set = {x : x ∈ R and -2 ≤ x < 2}
Solution on the number line is :
For the following inequations, graph the solution set on the real number line :
x - 1 < 3 - x ≤ 5
x - 1 < 3 - x ≤ 5
Solving L.H.S. of the equation,
⇒ x - 1 < 3 - x
⇒ x + x < 3 + 1
⇒ 2x < 4
⇒ x < 2 .........(i)
Solving R.H.S. of the equation,
⇒ 3 - x ≤ 5
⇒ x ≥ 3 - 5
⇒ x ≥ -2 .........(ii)
From (i) and (ii) we get,
-2 ≤ x < 2.
∴ Solution set = {-2 ≤ x < 2 : x ∈ R}
Solution on the number line is :
List the elements of the solution set of inequation -3 < x - 2 ≤ 9 - 2x; x ∈ N.
Given,
⇒ -3 < x - 2 ≤ 9 - 2x
Solving L.H.S. of the equation,
⇒ -3 < x - 2
⇒ x > -3 + 2
⇒ x > -1 .......(i)
Solving R.H.S. of the equation,
⇒ x - 2 ≤ 9 - 2x
⇒ x + 2x ≤ 9 + 2
⇒ 3x ≤ 11
⇒ x ≤ $\frac{11}{3}$ .........(ii)
From (i) and (ii) we get,
⇒ -1 < x ≤ $\frac{11}{3}$.
Since, x ∈ N
∴ Solution set = {1, 2, 3}.
Find the range of values of x which satisfies
$-2\frac{2}{3} \le x + \frac{1}{3} \lt 3\frac{1}{3}$, x ∈ R.
Graph these values on number line.
Given,
$$\begin{aligned}\Rightarrow -2\frac{2}{3} \le x + \frac{1}{3} \lt 3\frac{1}{3} \\[1em] \Rightarrow -\frac{8}{3} \le x + \frac{1}{3} \lt \frac{10}{3} \\[1em]\end{aligned}$$
Solving L.H.S. of the equation,
$$\begin{aligned}\Rightarrow -\frac{8}{3} \le x + \frac{1}{3} \\[1em] \Rightarrow x \ge -\frac{8}{3} - \frac{1}{3} \\[1em] \Rightarrow x \ge -\frac{9}{3} \\[1em] \Rightarrow x \ge -3 ........(i)\end{aligned}$$
Solving R.H.S. of the equation,
$$\begin{aligned}\Rightarrow x + \frac{1}{3} \lt \frac{10}{3} \\[1em] \Rightarrow x \lt \frac{10}{3} - \frac{1}{3} \\[1em] \Rightarrow x \lt \frac{9}{3} \\[1em] \Rightarrow x \lt 3 ..........(ii)\end{aligned}$$
From (i) and (ii) we get,
-3 ≤ x < 3.
Solution set = {x : x ∈ R and -3 ≤ x < 3}.
Solution on the number line is :
Find the values of x, which satisfy the inequation :
$-2 \le \frac{1}{2} - \frac{2x}{3} \le 1\frac{5}{6}$, x ∈ N.
Graph the solution on the number line.
Given,
$$\begin{aligned}\Rightarrow -2 \le \frac{1}{2} - \frac{2x}{3} \le 1\frac{5}{6} \\[1em] \Rightarrow -2 \le \frac{3 - 4x}{6} \le \frac{11}{6} \\[1em]\end{aligned}$$
Solving L.H.S. of the equation,
$$\begin{aligned}\Rightarrow -2 \le \frac{3 - 4x}{6} \\[1em] \Rightarrow -12 \le 3 - 4x \\[1em] \Rightarrow 4x \le 3 + 12 \\[1em] \Rightarrow x \le \frac{15}{4} ........(i)\end{aligned}$$
Solving R.H.S. of the equation,
$$\begin{aligned}\Rightarrow \frac{1}{2} - \frac{2x}{3} \le \frac{11}{6} \\[1em] \Rightarrow \frac{2x}{3} \ge \frac{1}{2} - \frac{11}{6} \\[1em] \Rightarrow \frac{2x}{3} \ge \frac{3 - 11}{6} \\[1em] \Rightarrow \frac{2x}{3} \ge -\frac{8}{6} \\[1em] \Rightarrow x \ge -\frac{8}{6} \times \frac{3}{2} \\[1em] \Rightarrow x \ge -2 .........(ii)\end{aligned}$$
From (i) and (ii) we get,
$-2 \le x \le \frac{15}{4}$
Since, x ∈ N,
∴ Solution set = {1, 2, 3}.
Solution on the number line is :
Given x ∈ {real numbers}, find the range of values of x for which -5 ≤ 2x - 3 < x + 2 and represent it on a real number line.
Given,
-5 ≤ 2x - 3 < x + 2
Solving L.H.S. of the equation,
⇒ -5 ≤ 2x - 3
⇒ 2x ≥ -5 + 3
⇒ 2x ≥ -2
Dividing both sides by 2 we get,
⇒ x ≥ -1 ........(i)
Solving R.H.S. of the equation,
⇒ 2x - 3 < x + 2
⇒ 2x - x < 2 + 3
⇒ x < 5 ........(ii)
From (i) and (ii) we get,
-1 ≤ x < 5.
∴ Solution set = {x : x ∈ R and -1 ≤ x < 5}.
Solution on the number line is :
If 5x - 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and b.
Given,
⇒ 5x - 3 ≤ 5 + 3x ≤ 4x + 2
Solving L.H.S. of the equation,
⇒ 5x - 3 ≤ 5 + 3x
⇒ 5x - 3x ≤ 5 + 3
⇒ 2x ≤ 8
⇒ x ≤ 4 ........(i)
Solving R.H.S. of the equation,
⇒ 5 + 3x ≤ 4x + 2
⇒ 3x - 4x ≤ 2 - 5
⇒ -x ≤ -3
⇒ x ≥ 3 .......(ii)
From (i) and (ii) we get,
3 ≤ x ≤ 4
Comparing above equation with a ≤ x ≤ b we get,
a = 3 and b = 4.
Hence, a = 3 and b = 4.
Solve the following inequation and graph the solution set on the number line :
2x - 3 < x + 2 ≤ 3x + 5; x ∈ R.
Given,
2x - 3 < x + 2 ≤ 3x + 5
Solving L.H.S. of the equation,
⇒ 2x - 3 < x + 2
⇒ 2x - x < 2 + 3
⇒ x < 5 ...........(i)
Solving R.H.S. of the equation,
⇒ x + 2 ≤ 3x + 5
⇒ x - 3x ≤ 5 - 2
⇒ -2x ≤ 3
⇒ 2x ≥ -3
Dividing both sides by 2 we get,
⇒ x ≥ -1.5 .........(ii)
From (i) and (ii) we get,
-1.5 ≤ x < 5
∴ Solution set = {x : x ∈ R and -1.5 ≤ x < 5}.
Solution on the number line is :
Solve and graph the solution set of :
2x - 9 < 7 and 3x + 9 ≤ 25; x ∈ R.
Given,
2x - 9 < 7 and 3x + 9 ≤ 25
Solving, 2x - 9 < 7
⇒ 2x < 7 + 9
⇒ 2x < 16
⇒ x < 8 .........(i)
Solving, 3x + 9 ≤ 25
⇒ 3x ≤ 25 - 9
⇒ 3x ≤ 16
⇒ x ≤ $\frac{16}{3}$
⇒ x ≤ $5\frac{1}{3}$ ........(ii)
From (i) and (ii) we get,
⇒ x ≤ $5\frac{1}{3}$
∴ Solution set = {x : x ≤ $5\frac{1}{3}$ and x ∈ R}.
Solution on the number line is :
Solve and graph the solution set of :
2x - 9 ≤ 7 and 3x + 9 > 25; x ∈ I
Given,
2x - 9 ≤ 7 and 3x + 9 > 25
Solving, 2x - 9 ≤ 7
⇒ 2x ≤ 7 + 9
⇒ 2x ≤ 16
⇒ x ≤ 8 .........(i)
Solving, 3x + 9 > 25
⇒ 3x > 25 - 9
⇒ 3x > 16
⇒ x > $\frac{16}{3}$
⇒ x > $5\frac{1}{3}$ ........(ii)
From (i) and (ii) we get,
⇒ $5\frac{1}{3} \lt x \le 8$
Since, x ∈ I,
∴ Solution set = {6, 7, 8}.
Solution on the number line is :
Solve and graph the solution set of :
x + 5 ≥ 4(x - 1) and 3 - 2x < -7; x ∈ R.
Given,
x + 5 ≥ 4(x - 1) and 3 - 2x < -7
Solving, x + 5 ≥ 4(x - 1)
⇒ x + 5 ≥ 4x - 4
⇒ 4x - x ≤ 5 + 4
⇒ 3x ≤ 9
Dividing both sides by 3 we get,
⇒ x ≤ 3 .......(i)
Solving, 3 - 2x < -7
⇒ 2x > 3 + 7
⇒ 2x > 10
⇒ x > 5 .......(ii)
From (i) and (ii) we get,
x ≤ 3 and x > 5
There is no number possible which is less than or equal to 3 and greater than 5 hence, no solution.
Hence, solution set is an empty set.
Solve and graph the solution set of :
3x - 2 > 19 or 3 - 2x ≥ -7; x ∈ R
Given,
3x - 2 > 19 or 3 - 2x ≥ -7
Solving, 3x - 2 > 19
⇒ 3x > 19 + 2
⇒ 3x > 21
⇒ x > 7
Solving, 3 - 2x ≥ -7
⇒ 2x ≤ 3 + 7
⇒ 2x ≤ 10
⇒ x ≤ 5.
Hence, x > 7 or x ≤ 5.
Solution on the number line is :
The diagram represents two inequations A and B on a real number lines :
(i) Write down A and B in set builder notation.
(ii) Represent A ∩ B and A ∩ B' on two different number lines.
(i) A = {x : -2 ≤ x < 5 and x ∈ R}
B = {x : -4 ≤ x < 3 and x ∈ R}
(ii) A ∩ B = Numbers common to both A and B
= {x : -2 ≤ x < 3}
A ∩ B' = Numbers which belong to A but do not belong to B
= {x : 3 ≤ x < 5}
Given A = {x : -1 < x ≤ 5, x ∈ R} and B = {x : -4 ≤ x < 3, x ∈ R}
Represent on different number lines :
(i) A ∩ B
(ii) A' ∩ B
(iii) A - B
A = {x : -1 < x ≤ 5, x ∈ R} and B = {x : -4 ≤ x < 3, x ∈ R}
(i) A ∩ B = Numbers common to both A and B.
= {x : -1 < x < 3, x ∈ R}
Solution on the number line is :
(ii) A' ∩ B = Numbers which do not belong to A but belong to B
= {x : -4 ≤ x ≤ -1, x ∈ R}
Solution on the number line is :
(iii) A - B = Numbers which belong to A but do not belong to B
= {x : 3 ≤ x ≤ 5}
Find the range of values of x, which satisfy :
$-\frac{1}{3} \le \frac{x}{2} + 1\frac{2}{3} \lt 5\frac{1}{6}$
Graph, in each of the following cases, the values of x on different real number lines:
(i) x ∈ W
(ii) x ∈ Z
(iii) x ∈ R
Given,
$\Rightarrow -\frac{1}{3} \le \frac{x}{2} + 1\frac{2}{3} \lt 5\frac{1}{6}$
Solving L.H.S of the above equation,
$$\begin{aligned}\Rightarrow -\frac{1}{3} \le \frac{x}{2} + 1\frac{2}{3} \\[1em] \Rightarrow -\frac{1}{3} \le \frac{x}{2} + \frac{5}{3} \\[1em] \Rightarrow \frac{x}{2} \ge -\frac{1}{3} - \frac{5}{3} \\[1em] \Rightarrow \frac{x}{2} \ge -\frac{6}{3} \\[1em] \Rightarrow \frac{x}{2} \ge -2 \\[1em] \Rightarrow x \ge -2 \times 2 \\[1em] \Rightarrow x \ge -4 \space .......(i)\end{aligned}$$
Solving R.H.S of the above equation,
$$\begin{aligned}\Rightarrow \frac{x}{2} + 1\frac{2}{3} \lt 5\frac{1}{6} \\[1em] \Rightarrow \frac{x}{2} + \frac{5}{3} \lt \frac{31}{6} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{31}{6} - \frac{5}{3} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{31 - 10}{6} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{21}{6} \\[1em] \Rightarrow x \lt 2 \times \frac{21}{6} \\[1em] \Rightarrow x \lt \frac{42}{6} \\[1em] \Rightarrow x \lt 7 \space .......(ii)\end{aligned}$$
From (i) and (ii) we get,
⇒ -4 ≤ x < 7
(i) In this case x ∈ W
∴ Solution set = {0, 1, 2, 3, 4, 5, 6}
Solution on the number line is :
(ii) In this case x ∈ Z
∴ Solution set = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}
Solution on the number line is :
(iii) In this case x ∈ R
∴ Solution set = {x : -4 ≤ x < 7, x ∈ R}
Solution on the number line is :
Given : A = {x : -8 < 5x + 2 ≤ 17, x ∈ I}
B = {x : -2 ≤ 7 + 3x < 17, x ∈ R}
Where R = {real numbers} and I = {integers}
Represent A and B on two different numbers lines. Write down elements of A ∩ B.
Given,
A = {x : -8 < 5x + 2 ≤ 17, x ∈ I}
Solving L.H.S. of the equation,
⇒ -8 < 5x + 2
⇒ 5x > -8 - 2
⇒ 5x > -10
⇒ x > -2 ......(i)
Solving R.H.S. of the equation,
⇒ 5x + 2 ≤ 17
⇒ 5x ≤ 17 - 2
⇒ 5x ≤ 15
⇒ x ≤ 3 .......(ii)
From (i) and (ii) we get,
-2 < x ≤ 3
Since, x ∈ I
∴ Solution set = {-1, 0, 1, 2, 3}
Given,
B = {x : -2 ≤ 7 + 3x < 17, x ∈ R}
Solving L.H.S. of the equation,
⇒ -2 ≤ 7 + 3x
⇒ 3x ≥ -2 - 7
⇒ 3x ≥ -9
⇒ x ≥ -3 .......(iii)
Solving R.H.S. of the equation,
⇒ 7 + 3x < 17
⇒ 3x < 17 - 7
⇒ 3x < 10
⇒ x < $\frac{10}{3}$ ........(iv)
From (iii) and (iv) we get,
-3 ≤ x < $\frac{10}{3}$
A ∩ B = Elements common to both A and B,
Hence, A ∩ B ={-1, 0, 1, 2, 3}.
Solve the following inequation and represent the solution set on the number line 2x - 5 ≤ 5x + 4 < 11, where x ∈ I.
Given,
2x - 5 ≤ 5x + 4 < 11
Solving L.H.S. of the equation,
⇒ 2x - 5 ≤ 5x + 4
⇒ 5x - 2x ≥ -5 - 4
⇒ 3x ≥ -9
⇒ x ≥ -3 .......(i)
Solving R.H.S. of the equation,
⇒ 5x + 4 < 11
⇒ 5x < 11 - 4
⇒ 5x < 7
⇒ x < $\frac{7}{5}$ .......(ii)
From (i) and (ii) we get,
-3 ≤ x < $\frac{7}{5}$
Since, x ∈ I
∴ Solution set = {-3, -2, -1, 0, 1}.
Solution on the number line is :
The maximum value of x for the inequation 4x ≤ 12 + x is :
Given,
⇒ 4x ≤ 12 + x
⇒ 4x - x ≤ 12
⇒ 3x ≤ 12
⇒ x ≤ $\frac{12}{3}$
⇒ x ≤ 4
∴ Maximum value of x will be 4.
Hence, Option 2 is the correct option.
The minimum value of x for the inequation 5x - 4 ≥ 18 - 6x is :
Given,
⇒ 5x - 4 ≥ 18 - 6x
⇒ 5x + 6x ≥ 18 + 4
⇒ 11x ≥ 22
⇒ x ≥ $\frac{22}{11}$
⇒ x ≥ 2
∴ Minimum value of x will be 2.
Hence, Option 1 is the correct option.
If 1 ≤ -x < 5, x is an integer then the sum of smallest and greatest values of x is :
Given,
⇒ 1 ≤ -x < 5
⇒ -5 < x ≤ -1
Since, x is an integer
The possible integer values are: {-4, -3, -2, -1}
The sum of smallest and greatest values = -4 + (-1) = -5
Hence, Option 1 is the correct option.
The value of x for the inequation 3x + 15 < 5x + 13, x ∈ Z is :
Given,
⇒ 3x + 15 < 5x + 13
⇒ 5x + 13 > 3x + 15
⇒ 5x - 3x > 15 - 13
⇒ 2x > 2
⇒ x > 1
Hence, Option 1 is the correct option.
The real number lines for two inequations A and B are as given below, A ∩ B is :
A = {x : x ∈ R and -3 < x ≤ 1}
B = {x : x ∈ R and -4 ≤ x < 0}
A ∩ B = {x : x ∈ R and -3 < x < 0}
Hence, Option 1 is the correct option.
For the inequations A and B [as given above in part (d)], A ∪ B is :
![For the inequations A and B [as given above in part (d)], A ∪ B is : Linear Inequations, Concise Mathematics Solutions ICSE Class 10.](/assets/img/f9cf5f15d065d54389.png)
A = {x : x ∈ R and -3 < x ≤ 1}
B = {x : x ∈ R and -4 ≤ x < 0}
A ∪ B = {x : x ∈ R and -4 ≤ x ≤ 1}
Hence, Option 1 is the correct option.
$-\frac{3}{2} \le -\frac{2x}{3}$ where x ∈ R.
Assertion (A): The largest value of x is $\frac{9}{4}$.
Reason (R): When the signs of both the sides of an inequalities are changed, the sign of inequality reverses.
Both A and R are true and R is the correct reason for A.
Reason
According to the assertion:
$$\begin{aligned}\Rightarrow -\frac{3}{2} \le -\frac{2x}{3}\\[1em] \Rightarrow \frac{3}{2} \ge \frac{2x}{3}\\[1em] \Rightarrow x \le \frac{3 \times 3}{2 \times 2}\\[1em] \Rightarrow x \le \frac{9}{4}\end{aligned}$$
So, Assertion (A) is true.
According to the reason:
When you multiply or divide both sides of an inequality by a negative number, the direction of the inequality sign must be reversed to maintain the validity of the inequality
So, Reason (R) is true.
Hence, option 3 is correct.
Inequation 5 - 2x ≥ x - 10, where x ∈ N (Natural numbers)
Assertion (A): 5 - 2x ≥ x - 10 ⇒ -3x ≥ -15 ⇒ x ≥ 5
∴ Solution set = {5, 6, 7, 8, ..........}
Reason (R): 5 - 2x ≥ x - 10 ⇒ 5 + 10 ≥ 3x ⇒ x ≤ 5
∴ Solution set = {1, 2, 3, 4, 5}
A is false, R is true.
Reason
According to Assertion: 5 - 2x ≥ x - 10
⇒ 5 - 2x + 10 ≥ x
⇒ -2x + 15 ≥ x
⇒ 15 ≥ x + 2x
⇒ 15 ≥ 3x
⇒ x ≤ $\frac{15}{3}$
⇒ x ≤ 5
∴ Solution set = {1, 2, 3, 4, 5}
So, Assertion (A) is false.
According to Reason:
Solution set = {1, 2, 3, 4, 5}
So, Reason (R) is true.
Hence, A is false, R is true.
x ∈ W, x ≥ -3 and x < 5.
Statement (1) : There will be no solution for the given inequalities.
Statement (2) : The real number line for the given inequations is :
Statement 1 is false, and statement 2 is true.
Reason
x ≥ -3
Solution set of x = {-3, -2, -1, 0, 1, 2, ..........} .......... (1)
And, x < 5
Solution set of x = {.........., 1, 2, 3, 4} .......... (2)
From (1) and (2), we get
Solution set = {-3, -2, -1, 0, 1, 2, 3, 4}
So, statement 1 is false.
The real number line for the given inequations is :
So, statement 2 is true.
Hence, option 4 is correct.
5 + x ≤ 2x < x - 2, x ∈ R.
Statement (1) : There is no value of x ∈ R that satisfies the given inequation.
Statement (2) : 5 + x - x ≤ 2x - x < x - 2 - x ⇒ 5 ≤ x < -2
Both the statements are true.
Reason
Given,
Equation : 5 + x ≤ 2x < x - 2
⇒ 5 + x ≤ 2x
⇒ 5 ≤ 2x - x
⇒ 5 ≤ x .......... (1)
And, 2x < x - 2
⇒ 2x - x < -2
⇒ x < -2 .......... (2)
There can be no real number which is greater than or equal to 5 and less than -2 at the same time.
∴ Both the statements are true.
Hence, option 1 is correct.
Solve the inequation :
$12 + 1\frac{5}{6}x \le 5 + 3x$ and x ∈ R.
Given,
$$\begin{aligned}\Rightarrow 12 + 1\frac{5}{6}x \le 5 + 3x \\[1em] \Rightarrow 12 + \frac{11}{6}x \le 5 + 3x \\[1em] \Rightarrow 3x - \frac{11}{6}x \ge 12 - 5 \\[1em] \Rightarrow \frac{18x - 11x}{6} \ge 7 \\[1em] \Rightarrow \frac{7x}{6} \ge 7 \\[1em] \Rightarrow x \ge \frac{7 \times 6}{7} \\[1em] \Rightarrow x \ge 6.\end{aligned}$$
∴ Solution set = {x : x ∈ R and x ≥ 6}.
Given x ∈ {whole numbers}, find the solution set of :
-1 ≤ 3 + 4x < 23
Given,
-1 ≤ 3 + 4x < 23
Solving L.H.S. of the equation,
⇒ -1 ≤ 3 + 4x
⇒ 4x ≥ -1 - 3
⇒ 4x ≥ -4
⇒ x ≥ -1 ........(i)
Solving R.H.S. of the equation,
⇒ 3 + 4x < 23
⇒ 4x < 23 - 3
⇒ 4x < 20
⇒ x < 5 .........(ii)
From (i) and (ii) we get,
-1 ≤ x < 5
Since, x ∈ {whole numbers},
∴ Solution set = {0, 1, 2, 3, 4}.
Find the set of values of x, satisfying :
7x + 3 ≥ 3x - 5 and $\frac{x}{4} - 5 \le \frac{5}{4} - x$, where x ∈ N.
Solving,
⇒ 7x + 3 ≥ 3x - 5
⇒ 7x - 3x ≥ -5 - 3
⇒ 4x ≥ -8
⇒ x ≥ -2 .......(i)
Solving,
$$\begin{aligned}\Rightarrow \frac{x}{4} - 5 \le \frac{5}{4} - x \\[1em] \Rightarrow \frac{x}{4} + x \le \frac{5}{4} + 5 \\[1em] \frac{x + 4x}{4} \le \frac{25}{4} \\[1em] \frac{5x}{4} \le \frac{25}{4} \\[1em] x \le \frac{25}{4} \times \frac{4}{5} \\[1em] x \le 5 ........(ii)\end{aligned}$$
From (i) and (ii) we get,
-2 ≤ x ≤ 5
Since, x ∈ N
∴ Solution set = {1, 2, 3, 4, 5}.
Solve :
(i) $\frac{x}{2} + 5 \le \frac{x}{3} + 6$, where x is a positive odd integer
(ii) $\frac{2x + 3}{3} \ge \frac{3x - 1}{4}$, where x is a positive even integer
(i) Solving,
$$\begin{aligned}\Rightarrow \frac{x}{2} + 5 \le \frac{x}{3} + 6 \\[1em] \Rightarrow \frac{x}{2} - \frac{x}{3} \le 6 - 5 \\[1em] \Rightarrow \frac{3x - 2x}{6} \le 1 \\[1em] \Rightarrow \frac{x}{6} \le 1 \\[1em] \Rightarrow x \le 6.\end{aligned}$$
Since, x is a positive odd integer
∴ Solution set = {1, 3, 5}.
(ii) Solving,
$$\begin{aligned}\Rightarrow \frac{2x + 3}{3} \ge \frac{3x - 1}{4} \\[1em] \Rightarrow \frac{2x + 3}{3} - \frac{3x - 1}{4} \ge 0 \\[1em] \Rightarrow \frac{4(2x + 3) - 3(3x - 1)}{12} \ge 0 \\[1em] \Rightarrow 8x + 12 - 9x + 3 \ge 0 \\[1em] \Rightarrow -x + 15 \ge 0 \\[1em] \Rightarrow x \le 15.\end{aligned}$$
Since, x is a positive even integer
∴ Solution set = {2, 4, 6, 8, 10, 12, 14}.
Solve the inequation :
$-2\frac{1}{2} + 2x \le \frac{4x}{5} \le \frac{4}{3} + 2x$, x ∈ W.
Graph the solution set on the number line.
Given,
$-2\frac{1}{2} + 2x \le \frac{4x}{5} \le \frac{4}{3} + 2x$
Solving L.H.S. of the equation,
$$\begin{aligned}\Rightarrow -2\frac{1}{2} + 2x \le \frac{4x}{5} \\[1em] \Rightarrow -\frac{5}{2} + 2x \le \frac{4x}{5} \\[1em] \Rightarrow 2x - \frac{4x}{5} \le \frac{5}{2} \\[1em] \Rightarrow \frac{10x - 4x}{5} \le \frac{5}{2} \\[1em] \Rightarrow \frac{6x}{5} \le \frac{5}{2} \\[1em] \Rightarrow x \le \frac{5}{2} \times \frac{5}{6} \\[1em] \Rightarrow x \le \frac{25}{12} \\[1em] \Rightarrow x \le 2\frac{1}{12} ........(i)\end{aligned}$$
Solving R.H.S. of the equation,
$$\begin{aligned}\Rightarrow \frac{4x}{5} \le \frac{4}{3} + 2x \\[1em] \Rightarrow \frac{4x}{5} - 2x \le \frac{4}{3} \\[1em] \Rightarrow \frac{4x - 10x}{5} \le \frac{4}{3} \\[1em] \Rightarrow -\frac{6x}{5} \le \frac{4}{3} \\[1em] \Rightarrow \frac{6x}{5} \ge -\frac{4}{3} \\[1em] \Rightarrow x \ge -\frac{4}{3} \times \frac{5}{6} \\[1em] \Rightarrow x \ge -\frac{10}{9} \\[1em] \Rightarrow x \ge -1\frac{1}{9} .......(ii)\end{aligned}$$
From (i) and (ii) we get,
$-1\frac{1}{9} \le x \le 2\frac{1}{12}$
Since, x ∈ W
∴ Solution set = {0, 1, 2}.
Solution on the number line is :
Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is at most 20.
Let three consecutive positive integers be x, x + 1 and x + 2.
Given, sum of one-third of first, one-fourth of second and one-fifth of third is at most 20
$$\begin{aligned}\therefore \frac{1}{3}x + \frac{1}{4}(x + 1) + \frac{1}{5}(x + 2) \le 20 \\[1em] \Rightarrow \frac{x}{3} + \frac{x + 1}{4} + \frac{x + 2}{5} \le 20 \\[1em] \Rightarrow \frac{20x + 15(x + 1) + 12(x + 2)}{60} \le 20 \\[1em] \Rightarrow \frac{20x + 15x + 15 + 12x + 24}{60} \le 20 \\[1em] \Rightarrow 47x + 39 \le 1200 \\[1em] \Rightarrow 47x \le 1161 \\[1em] \Rightarrow x \le \frac{1161}{47} \\[1em] \Rightarrow x \le 24.702\end{aligned}$$
∴ x = 24, x + 1 = 25, x + 2 = 26.
Hence, three consecutive numbers are 24, 25 and 26.
Solve the following inequation and represent the solution set on the number line :
4x - 19 < $\frac{3x}{5} - 2 \le -\frac{2}{5} + x$, x ∈ R
Given,
4x - 19 < $\frac{3x}{5} - 2 \le -\frac{2}{5} + x$
Solving L.H.S. of the equation,
$$\begin{aligned}\Rightarrow 4x - 19 \lt \frac{3x}{5} - 2 \\[1em] \Rightarrow 4x - 19 \lt \frac{3x - 10}{5} \\[1em] \Rightarrow 5(4x - 19) \lt 3x - 10 \\[1em] \Rightarrow 20x - 95 \lt 3x - 10 \\[1em] \Rightarrow 20x - 3x \lt 95 - 10 \\[1em] \Rightarrow 17x \lt 85 \\[1em] \Rightarrow x \lt 5 ........(i)\end{aligned}$$
Solving R.H.S. of the equation,
$$\begin{aligned}\Rightarrow \frac{3x}{5} - 2 \le -\frac{2}{5} + x \\[1em] \Rightarrow x - \frac{3x}{5} \ge -2 + \frac{2}{5} \\[1em] \Rightarrow \frac{5x - 3x}{5} \ge \frac{-10 + 2}{5} \\[1em] \Rightarrow \frac{2x}{5} \ge \frac{-8}{5} \\[1em] \Rightarrow x \ge -\frac{8}{5} \times \frac{5}{2} \\[1em] \Rightarrow x \ge -4 ........(ii)\end{aligned}$$
From (i) and (ii) we get,
-4 ≤ x < 5
∴ Solution set = {x : -4 ≤ x < 5, x ∈ R}.
Solution on the number line is :
Find the greatest value of x ∈ Z, so that :
-1 ≤ 3 + 4x < 23
Given,
-1 ≤ 3 + 4x < 23
Solving L.H.S. of the inequation,
⇒ -1 ≤ 3 + 4x
⇒ -1 - 3 ≤ 4x
⇒ -4 ≤ 4x
⇒ $\frac{-4}{4}$ ≤ x
⇒ -1 ≤ x
⇒ x ≥ -1 ....(1)
Solving R.H.S. of the inequation,
⇒ 3 + 4x < 23
⇒ 4x < 23 - 3
⇒ 4x < 20
⇒ x < $\frac{20}{4}$
⇒ x < 5 ....(2)
From (1) and (2) we get,
-1 ≤ x < 5
Since x ∈ Z,
x = {-1, 0, 1, 2, 3, 4}
Hence, greatest value of x is 4.
If 7 ≥ -2x + 1 > -7 ; find the sum of greatest and smallest values of x ∈ I.
Given,
7 ≥ -2x + 1 > -7
Solving L.H.S. of the inequation,
⇒ 7 ≥ -2x + 1
⇒ 7 - 1 ≥ -2x
⇒ 6 ≥ -2x
⇒ 2x ≥ -6
⇒ x ≥ $\frac{-6}{2}$
⇒ x ≥ -3 ....(1)
Solving R.H.S. of the inequation,
⇒ -2x + 1 > -7
⇒ -2x > -7 - 1
⇒ -2x > -8
⇒ 2x < 8
⇒ x < $\frac{8}{2}$
⇒ x < 4 ....(2)
From (1) and (2) we get,
-3 ≤ x < 4
Since x ∈ I,
x = {-3, -2, -1, 0, 1, 2, 3}
Greatest value of x is 3 and smallest value of x is -3
The sum of greatest and smallest values of x = -3 + 3 = 0
Hence, sum of greatest and smallest values of x = 0.
Case study :
A teacher asked to Rohan to draw a triangle with following condition: The longest side of the triangle is 7 cm less than twice the shortest side and third side is 7 cm shorter than longest side. The perimeter of the triangle is atleast 84 cm.
Based on the above information, form a linear inequation and answer the following questions :
(i) What is the minimum length of the shortest side ?
(ii) What is the minimum length of the longest side ?
(iii) Identify the type of triangle that Rohan has drawn along with the length possible sides he got.
(iv) What is the least area of the triangle drawn ?
Let the shortest side of triangle be x cm.
Given,
Longest side of triangle = 2x - 7
Third side of triangle = (2x - 7) - 7 = 2x - 14
(i) Given,
Perimeter of triangle is atleast 84 cm.
⇒ x + (2x - 7) + (2x - 14) ≥ 84
⇒ 5x - 21 ≥ 84
⇒ 5x ≥ 84 + 21
⇒ 5x ≥ 105
⇒ x ≥ $\frac{105}{5}$
⇒ x ≥ 21.
Hence, minimum length of the shortest side of triangle = 21 cm.
(ii) Length of the longest side of triangle = 2x - 7
Minimum length of longest side will be when x = 21, substituting value we get :
⇒ 2(21) - 7
⇒ 42 - 7
⇒ 35 cm.
Hence, minimum length of the longest side of triangle = 35 cm.
(iii) Minimum length of third side :
Third side of triangle = 2x - 14
= 2(21) - 14
= 42 - 14
= 28 cm.
Three sides of triangle = 21 cm, 28 cm and 35 cm.
⇒ 212 + 282 = 441 + 784 = 1225
⇒ 352 = 1225.
Thus, we can say that 212 + 282 = 352.
Thus, it is a right angled triangle with hypotenuse 35 cm and other two sides are 21 cm and 28 cm.
Hence, the triangle is right-angled triangle.
(iv) By formula,
Area of right angled triangle = $\frac{1}{2}$ × product of sides containing right angle
Least area of the triangle = $\frac{1}{2}$ × 21 × 28
= 294 cm2.
Hence, the least area of the triangle = 294 cm2.
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