Chapter 5: Linear Inequalities

Sol. We are given that $24 x<100$
Dividing both sides by 24 , which is positive
$$
x<\frac{100}{24} \quad \text { i.e., } \quad x<\frac{25}{6}=4.1 \text { nearly. }
$$
(i) For $x \in \mathrm{N}$, the values satisfying the inequality are $1, 2, 3, 4$.
$\therefore \quad$ The solution set $=\{1,2,3,4\}$.
(ii) For $x \in \mathrm{Z}$, the values satisfying the inequality are $\ldots,-3,-2,-1,0,1,2,3,4$.
$\therefore \quad$ The solution set $=\{\ldots,-3,-2,-1,0,1,2,3,4\}$.

Sol. We are given that $-12 x>30$
Dividing both sides by -12 , which is negative (so that the sign of inequality is reversed.)
$$
x<\frac{30}{-12}, \quad \text { i.e., } x<-\frac{5}{2}=-2.5
$$
(i) When $x \in \mathrm{~N}=\{1,2,3, \ldots\}$, no value of $x$ makes the statement true. Therefore the given inequality has no solution.
$\therefore \quad$ The solution set $=\phi$.
(ii) For $x \in \mathrm{Z}$, the values satisfying the inequality are
$$
\ldots,-5,-4,-3 .
$$
$\therefore \quad$ The solution set $=\{\ldots,-5,-4,-3\}$.

Sol. We are given that $5 x-3<7$
We add 3 to both sides to get
$$
\begin{aligned}
5 x & <10 \\
x & <2 .
\end{aligned}
$$

Dividing both sides by 5 ,
(i) For $x \in \mathrm{Z}$, the values satisfying the inequality are $\ldots,-2,-1,0,1$.
∴ $\quad$ The solution set $=\{\ldots,-2,-1,0,1\}$.
(ii) For $x \in \mathrm{R}$, all real numbers less than 2 satisfy the inequality.
∴ The solution set $=(-\infty, 2)$.

Sol. We are given that $3 x+8>2$
We add −8 to both sides to get
$$
\begin{aligned}
3 x & >-6 \\
x & >-2 .
\end{aligned}
$$

Dividing both sides by 3 gives us
(i) For $x \in \mathrm{Z}$, the values satisfying the inequality are $-1,0,1,2,3, \ldots$.
∴ $\quad$ The solution set $=\{-1,0,1,2,3, \ldots\}$.
(ii) For $x \in \mathrm{R}$, all real numbers greater than $-2$ satisfy the inequality.
∴ $\quad$ The solution set $=(-2, \infty)$.
Solve the inequalities in Exercises 5 to 16 for real $\boldsymbol{x}$.

Sol.
$$
\begin{aligned}
& & 4 x+3 & <5 x+7 \\
& \Rightarrow & 4 x-5 x & <7-3 \\
\Rightarrow & & x & >-4
\end{aligned} \Rightarrow-x<4
$$
∴ $\quad$ The solution set $=(-4, \infty)$.

Sol.
$$
\begin{array}{rlrl}
& & 3 x-7 & >5 x-1 \\
\Rightarrow & & 3 x-5 x & >-1+7 \Rightarrow-2 x>6 \\
\Rightarrow & & x<-3
\end{array}
$$
$\therefore \quad$ The solution set $=(-\infty,-3)$.

Sol. Starting with $3(x-1) \leq 2(x-3)$
$$
\begin{array}{lcl}
\Rightarrow & 3 x-3 & \leq 2 x-6 \\
\Rightarrow & x & y-3
\end{array} \quad \Rightarrow \quad 3 x-2 x \leq-6+3
$$
∴ $\quad$ The solution set $=(-\infty,-3]$.

Sol. Starting with $\quad 3(2-x) \geq 2(1-x)$
$\Rightarrow \quad 6-3 x \geq 2-2 x \quad \Rightarrow \quad-3 x+2 x \geq 2-6$
$\Rightarrow \quad-x \geq-4 \quad \Rightarrow \quad x \leq 4$
$\therefore \quad$ The solution set $=(-\infty, 4]$.

Sol. Starting with $x+\frac{x}{2}+\frac{x}{3}<11$
$$
\Rightarrow \frac{6 x+3 x+2 x}{6}<11
$$

Multiplying both sides by 6, we find
$$
\begin{aligned}
& 6 x+3 x+2 x<66 \\
& \Rightarrow \quad 11 x<66 \quad \Rightarrow \quad x<6
\end{aligned}
$$
∴ $\quad$ The solution set $=(-\infty, 6)$.

Sol. Starting with $\frac{x}{3}>\frac{x}{2}+1 \Rightarrow \frac{x}{3}>\frac{x+2}{2}$
After cross-multiplying, we obtain
$$
\begin{aligned}
& & 2 x & >3 x+6 \\
\Rightarrow & & 2 x-3 x & >6 \Rightarrow-x>6 \\
\Rightarrow & & x & <-6
\end{aligned}
$$
∴ $\quad$ The solution set $=(-\infty,-6)$.

Sol. Starting with $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$
After cross-multiplying, we obtain
$$
\begin{aligned}
& 9(x-2) \leq 25(2-x) & & \\
\Rightarrow & 9 x-18 \leq 50-25 x & & \Rightarrow 9 x+25 x \leq 50+18 \\
\Rightarrow & 34 x \leq 68 & & \Rightarrow \\
\Rightarrow & x \leq 2 & & x \leq \frac{68}{34}
\end{aligned}
$$
$\therefore \quad$ The solution set $=(-\infty, 2]$.

Sol. Starting with $\quad \frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$





$$
\Rightarrow \frac{1}{2}\left(\frac{3 x+20}{5}\right) \geq \frac{x-6}{3} \Rightarrow \frac{3 x+20}{10} \geq \frac{x-6}{3}
$$

After cross-multiplying, we obtain
$$
\begin{array}{rlrl}
& 3(3 x+20) \geq 10(x-6) & & \\
\Rightarrow & 9 x+60 \geq 10 x-60 & \Rightarrow & 9 x-10 x \geq-60-60 \\
\Rightarrow \quad-x \geq-120 & & \Rightarrow x \leq 120
\end{array}
$$
∴ $\quad$ The solution set $=(-\infty$, 120].

Sol. Starting with $2(2 x+3)-10<6(x-2)$
$$
\begin{array}{lclc}
\Rightarrow & 4 x+6-10<6 x-12 & \Rightarrow & 4 x-4<6 x-12 \\
\Rightarrow & 4 x-6 x<4-12 & & \Rightarrow \\
\Rightarrow & x>\frac{-8}{-2} & & \Rightarrow x>4
\end{array}
$$
∴ $\quad$ The solution set $=(4, \infty)$.

Sol. Starting with $\quad 37-(3 x+5) \geq 9 x-8(x-3)$
$$
\begin{array}{lrlrl}
\Rightarrow & 37-3 x-5 & \geq 9 x-8 x+24 & \\
\Rightarrow & 32-3 x & \geq x+24 & \Rightarrow & -3 x-x \geq 24-32 \\
\Rightarrow & -4 x & \geq-8 & \Rightarrow & x \leq \frac{-8}{-4} \\
\Rightarrow & & x \leq 2 & &
\end{array}
$$
$\therefore \quad$ The solution set $=(-\infty, 2]$.

Sol. Starting with $\quad \frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{x}{4}<\frac{5(5 x-2)-3(7 x-3)}{15} \\
& \Rightarrow \quad \frac{x}{4}<\frac{25 x-10-21 x+9}{15} \Rightarrow \frac{x}{4}<\frac{4 x-1}{15}
\end{aligned}
$$

After cross-multiplying, we obtain
$$
\begin{aligned}
15 x & <4(4 x-1) \\
\Rightarrow \quad 15 x & <16 x-4
\end{aligned} \quad \Rightarrow \quad 15 x-16 x<-4
$$





$$
\Rightarrow-x<-4 \quad \Rightarrow \quad x>4
$$
∴ The solution set $=(4, \infty)$.

Sol. Starting with $\quad \frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$
$$
\begin{array}{ll}
\Rightarrow & \frac{2 x-1}{3} \geq \frac{5(3 x-2)-4(2-x)}{20} \\
\Rightarrow & \frac{2 x-1}{3} \geq \frac{15 x-10-8+4 x}{20} \Rightarrow \frac{2 x-1}{3} \geq \frac{19 x-18}{20}
\end{array}
$$

After cross-multiplying, we obtain
$$
\begin{array}{rlrlrl}
& & 20(2 x-1) & \geq 3(19 x-18) & \\
\Rightarrow & 40 x-20 & \geq 57 x-54 \\
\Rightarrow & & \Rightarrow & \Rightarrow & 40 x-57 x \geq 20-54 \\
& \Rightarrow & & & \\
& & x & \leq 2 & &
\end{array}
$$
∴ $\quad$ The solution set $=(-\infty, 2]$.
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.

Sol. Starting with $\quad 3 x-2<2 x+1$
$$
\Rightarrow 3 x-2 x<1+2 \quad \Rightarrow \quad x<3
$$
$\therefore \quad$ The solution set $=(-\infty, 3)$.
The solution set is shown on the number line in the figure below.

Sol. Starting with $\quad 5 x-3 \geq 3 x-5$
$$
\begin{aligned}
& \Rightarrow \quad 5 x-3 x \geq 3-5 \\
& \Rightarrow \quad 2 x \geq-2
\end{aligned} \quad \Rightarrow \quad x \geq-1
$$
∴ $\quad$ The solution set $=[-1, \infty)$.
The solution set is shown on the number line in the figure below.

Sol. Starting with $\quad 3(1-x)<2(x+4)$
$$
\begin{array}{ll}
\Rightarrow \quad 3-3 x<2 x+8 & \Rightarrow \quad-3 x-2 x<8-3 \\
\Rightarrow \quad-5 x<5 & \Rightarrow \quad x>\frac{5}{-5} \\
\Rightarrow \quad x>-1 &
\end{array}
$$
∴ The solution set $=(-1, \infty)$.
The solution set is shown on the number line in the figure below.

Sol. Starting with $\quad \frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{x}{2} \geq \frac{5(5 x-2)-3(7 x-3)}{15} \\
& \Rightarrow \quad \frac{x}{2} \geq \frac{25 x-10-21 x+9}{15} \\
& \Rightarrow \quad \frac{x}{2} \geq \frac{4 x-1}{15}
\end{aligned}
$$

After cross-multiplying, we obtain
$$
\begin{array}{rlrl}
& 15 x & \geq 2(4 x-1) \\
\Rightarrow & 15 x & \geq 8 x-2 \\
\Rightarrow & 7 x & & \Rightarrow-2
\end{array} \Rightarrow 15 x-8 x \geq-2
$$
$\therefore \quad$ The solution set $=\left[-\frac{2}{7}, \infty\right)$.
The solution set is shown on the number line in the figure below.

Sol. Let $x$ denote the marks Ravi scores in the third test.
The average marks formula gives $=\frac{70+75+x}{3} \geq 60[\because$ At least $\Rightarrow \geq]$
$$
\begin{array}{rrrl}
\Rightarrow & 145+x & \geq 180 \\
\Rightarrow & x & \Rightarrow 35
\end{array} \quad \Rightarrow \quad x \geq 180-145
$$

Therefore, Ravi needs to score at least 35 marks in the third test to achieve an average of 60 or more.
Note: “At least 35 marks” means
$x \geq 35$, i.e., marks ≥ 35.

Sol. Let $x$ denote the marks Sunita scores in the fifth examination.
The average marks formula gives $=\frac{87+92+94+95+x}{5} \geq 90$
$$
\Rightarrow 368+x \geq 450 \quad \Rightarrow \quad x \geq 450-368
$$
$\Rightarrow \quad x \geq 82$
Therefore, Sunita must score at least 82 marks in the fifth examination to secure Grade A.

Sol. Let $x$ be the smaller consecutive odd positive integer; then the next one is $x+2$. [ ∵ Odd $+1=$ Even] Based on the given conditions,
$$
x<10, \quad x+2<10
$$
and $\quad x+(x+2)>11$
$\Rightarrow \quad x<10, \quad x<8$






and $\quad 2 x>9$
$\Rightarrow \quad x<8$, the smallest of the lesser than.
[ $\because x<8$ automatically $\Rightarrow x<10$ ]
and
$$
x>\frac{9}{2}
$$

Combining (i) and (ii), we get
$$
\frac{9}{2}(=4.5)$$

Since $x$ must be an odd positive integer,
$\therefore \quad x$ can take values 5 and 7 .
Therefore, the required pairs are $(x, x+2)=(5,7)$, (7, 9).

Sol. Let $x$ be the smaller consecutive even positive integer; then the next one is $x+2$. [ ∵ Even $+1=$ odd] Based on the given conditions,
$$
x>5, \quad x+2>5
$$
and $\quad x+(x+2)<23$
$\Rightarrow \quad x>5, x>3$
and $\quad 2 x<21$
$\Rightarrow \quad x>5$, largest of greater than
( $\because x>5$ automatically $\Rightarrow x>3$ ),
and
$$
x<\frac{21}{2}
$$

Combining (i) and (ii), we get
$$
5$$

Since $x$ must be an even positive integer,
$\therefore \quad x$ can take the values 6,8 and 10 .
Therefore, the required pairs are $(x, x+2)=(6,8)$, $(8,10),(10,12)$.

Sol. Suppose the shortest side has length $x \mathrm{~cm}$, then the length of the longest side is $3 x \mathrm{~cm}$ and the length of the third side is $(3 x-2) \mathrm{cm}$.
The perimeter (sum of all three sides) of the triangle
$$
\begin{aligned}
& =x+3 x+(3 x-2) \\
& =(7 x-2) \mathrm{cm}
\end{aligned}
$$

Based on the given condition,
$$
\begin{array}{rlrl}
& \text { Perimeter } & \geq 61 \mathrm{~cm} & (\because \text { At least } \Rightarrow \geq) \\
\Rightarrow & 7 x-2 & \geq 61 \\
\Rightarrow & x & \geq 9
\end{array} \quad \begin{aligned}
& \Rightarrow \Rightarrow 7 x \geq 63 \\
&
\end{aligned}
$$
∴ The shortest side must be at least $9 \mathrm{~cm}$ long.

Sol. Suppose the shortest piece has length $x \mathrm{~cm}$, then length of second is $(x+3) \mathrm{cm}$ and the length of third is $2 x \mathrm{~cm}$.
Since the total available length is 91 cm, we require
$$
\begin{aligned}
& & x+(x+3)+2 x & \leq 91 \\
\Rightarrow & & 4 x & \leq 88
\end{aligned} \quad \Rightarrow \quad x \leq 22
$$

Also, the third piece must be at least 5 cm longer than the second, so III − II $\geq 5$
$$
\begin{aligned}
& \therefore \quad 2 x-(x+3) \geq 5 \\
& \Rightarrow \quad 2 x-x-3 \geq 5
\end{aligned} \quad \begin{aligned}
& \quad \Rightarrow \quad x \geq 8
\end{aligned}
$$

Combining inequalities (i) and (ii), we get
$$
8 \leq x \leq 22
$$
∴ The shortest board must be between 8 cm and 22 cm (both values included).
Note: In the first inequality in the solution of above question, we have taken:
Sum of lengths of three pieces $=x+(x+3)+2 x \leq 91$ and not only $=91$ because even after cutting all the three pieces according to the requirement of the question, a part of the length of the board out of 91 cm may remain unused (left)

Sol. Starting with $2 \leq 3 x-4 \leq 5$
Adding 4 to all parts, we get $2+4 \leq 3 x-4+4 \leq 5+4$
$$
6 \leq 3 x \leq 9
$$

Dividing all parts by 3, we obtain
$$
\begin{aligned}
& 2 \leq x \leq 3 \\
\Rightarrow \quad x & \in[2,3] .
\end{aligned}
$$

Sol. Starting with $\quad 6 \leq-3(2 x-4)<12$
$$
\Rightarrow \quad 6 \leq-6 x+12<12
$$

Subtracting 12 from all parts, we get $6-12 \leq-6 x+12-12$ < 12-12
$$
-6 \leq-6 x<0
$$

Since −6 is negative, dividing all parts by −6 reverses the inequalities, giving
$$
\begin{aligned}
& 1 \geq x>0 \\
\Rightarrow & 0\end{aligned}
$$

Sol. Starting with $-3 \leq 4-\frac{7 x}{2} \leq 18$
Subtracting 4 from all parts, we get





$$
-7 \leq-\frac{7 x}{2} \leq 14
$$

Since $-\frac{2}{7}$ is negative, multiplying all parts by it reverses the inequalities, giving
$$
\begin{aligned}
& -\frac{2}{7}(-7) \geq-\frac{2}{7}\left(-\frac{7 x}{2}\right) \geq-\frac{2}{7} \\
& \Rightarrow \quad 2 \geq x \geq-4 \\
& \Rightarrow \quad-4 \leq x \leq 2 \\
& \Rightarrow \quad x \in[-4,2]
\end{aligned}
$$

Sol. Starting with $\quad-15<\frac{3(x-2)}{5} \leq 0$
Multiplying all parts by 5, we obtain $-75<3(x-2) \leq 0$
Dividing all parts by 3,
$$
-25$$

Adding 2 to all parts, we get
$$
\begin{array}{rlrl}
& & -23 & \Rightarrow & x & \in(-23,2] .
\end{array}
$$

Sol. Starting with $\quad-12<4-\frac{3 x}{-5} \leq 2$
$$
\Rightarrow \quad-12<4+\frac{3 x}{5} \leq 2
$$

Subtracting 4 from all parts, we get
$$
-16<\frac{3 x}{5} \leq-2
$$

Multiplying all parts by 5, we obtain $-80<3 x \leq-10$
Dividing all parts by 3,
$$
-\frac{80}{3}$$





$$
\Rightarrow \quad x \in\left(-\frac{80}{3},-\frac{10}{3}\right] .
$$

Sol.
$$
7 \leq \frac{(3 x+11)}{2} \leq 11
$$

Multiplying all parts by 2, we obtain
$$
14 \leq 3 x+11 \leq 22
$$

Subtracting 11 from all parts, we get
$$
3 \leq 3 x \leq 11
$$

Dividing all parts by 3, we obtain
$$
\begin{aligned}
& 1 \leq x \leq \frac{11}{3} \\
\Rightarrow & x \in\left[1, \frac{11}{3}\right] .
\end{aligned}
$$

Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.

Sol.
$$
\begin{aligned}
& 5 x+1>-24 \\
& \text { and } 5 x-1<24 \\
& \text { from }(i), \quad 5 x>-24-1 \\
& \Rightarrow \\
& \text { from }(i i), 5 x-1<24 \\
& \Rightarrow \\
& \Rightarrow \quad 5 x>-25 \quad \Rightarrow \quad x>-5 \\
& \quad \begin{aligned}
5 x & <24+1 \Rightarrow 5 x<25 \\
x & <5
\end{aligned} \Rightarrow \quad
\end{aligned}
$$

Combining results (iii) and (iv), we find $-5$$
\Rightarrow \quad x \in(-5,5)
$$

The solution set is marked as a bold segment on the number line in the figure below.

Sol. Starting with $2(x-1)and $3(x+2)>2-x$

Using (i), $\quad 2 x-2




$$
\Rightarrow \quad 2 x-x<5+2 \quad \Rightarrow \quad x<7
$$
$$
\begin{array}{lcl}
\text { Using (ii), } & & 3 x+6>2-x \\
\Rightarrow & & 3 x+x>2-6 \\
\Rightarrow & & x>-1
\end{array} \quad \Rightarrow 4 x>-4
$$

Combining results (iii) and (iv), we find $-1$$
\Rightarrow \quad x \in(-1,7)
$$

The solution set is marked as a bold segment on the number line in the figure below.

Sol.
$$
\begin{gathered}
3 x-7>2(x-6) \\
\text { and } 6-x>11-2 x
\end{gathered}
$$

Using (i), $\quad 3 x-7>2 x-12$
$$
\Rightarrow \quad 3 x-2 x>-12+7 \quad \Rightarrow \quad x>-5
$$

Using (ii), $\quad-x+2 x>11-6$
$$
\Rightarrow \quad x>5
$$

Combining results (iii) and (iv), we find $x>5$
$$
\Rightarrow \quad \begin{aligned}
(\because x>5 & \Rightarrow x>-5 \text { also }) \\
\Rightarrow \quad 5\end{aligned}
$$

The solution set is shown as a bold ray on the number line in the figure below.

Sol. Starting with $5(2 x-7)-3(2 x+3) \leq 0$
$$
\text { and } 2 x+19 \leq 6 x+47
$$

Using (i), $\quad 10 x-35-6 x-9 \leq 0$
$\Rightarrow \quad 4 x-44 \leq 0 \quad \Rightarrow \quad 4 x \leq 44$
$$
\Rightarrow \quad x \leq 11
$$

Using (ii), $\quad 2 x-6 x \Rightarrow 47-19$
$$
\Rightarrow \quad-4 x \leq 28 \quad \Rightarrow \quad x \geq-7
$$

Combining results (iii) and (iv), we find $-7 \leq x \leq 11$
$$
\Rightarrow \quad x \in[-7,11]
$$






The solution set is marked as a bold segment on the number line in the figure below.

Sol. We are given that $68<\mathrm{F}<77$
(Between $68^{\circ} \mathrm{F}$ and $77^{\circ} \mathrm{F} \Rightarrow \neq 68$ and $\neq 77$ )
Substituting $\mathrm{F}=\frac{9}{5} \mathrm{C}+32$ into the inequality, we get
$$
68<\frac{9}{5} \mathrm{C}+32<77
$$

Subtracting 32 from all parts, we get
$$
36<\frac{9}{5} \mathrm{C}<45
$$

Multiplying all parts by 5, we get $180<9 \mathrm{C}<225$
Dividing by 9,
$$
\Rightarrow \quad 20<\mathrm{C}<25
$$

Therefore, the required temperature range in Celsius is between $20^{\circ} \mathrm{C}$ and $25^{\circ} \mathrm{C}$.

Sol. We are told that we start with 640 litres of $8 \%$ boric acid solution.
Let $x$ represent the number of litres of $2 \%$ boric acid solution.
The total mixture then becomes $(640+x)$ litres.
Based on the given conditions,
$$
2 \% \text { of } x+8 \% \text { of }(640)>4 \% \text { of }(x+640)
$$
and $2 \%$ of $x+8 \%$ of $(640)<6 \%$ of $(x+640)$






i.e.
$$
\frac{2 x}{100}+\frac{8(640)}{100}>\frac{4 x}{100}+\frac{4(640)}{100}
$$
and
$$
\frac{2 x}{100}+\frac{8(640)}{100}<\frac{6 x}{100}+\frac{6(640)}{100}
$$

Multiplying every term by 100,
and
$$
2 x+5120>4 x+2560
$$
$$
2 x+5120<6 x+3840
$$
or $2560>2 x \quad$ and $1280<4 x$
or $1280>x$ and $320$$
\therefore \quad 320$$
∴ Amount $x$ of $2 \%$ boric acid solution to be added is more than 320 litres but less than 1280 litres.

Sol. Let $x$ litres of water need to be added. Since pure water contains $0 \%$ acid, i.e., no acid, we have
$$
0 \% \text { of } x+45 \% \text { of } 1125>25 \% \text { of }(1125+x)
$$
and $0 \%$ of $x+45 \%$ of $1125<30 \%$ of $(1125+x)$

From $(i), \quad \frac{45}{100}(1125)>\frac{25}{100}(1125+x)$
Multiplying every term by 100, we get
$$
45 \times 1125>25(1125+x)
$$

Dividing by 25, we obtain
$$
45 \times 45>1125+x
$$
$$
\begin{aligned}
\Rightarrow & & 2025 & >1125+x \\
\Rightarrow & & x & <900
\end{aligned}
$$

Using (ii), $\quad \frac{45}{100}(1125)<\frac{30}{100}(1125+x)$
Multiplying every term by 100, we get
$$
45 \times 1125<30 \times(1125+x)
$$

Dividing by 15,3 \times 1125<2(1125+x)$
$$
\begin{aligned}
& \Rightarrow 3375<2250+2 x \\
& \Rightarrow 1125<2 x
\end{aligned}
$$






Dividing by 2, we get $\frac{1125}{2}$\Rightarrow \quad 562.5
From (iii) and (iv), $562.5∴ The water to be added must be more than 562.5 litres but less than 900 litres.

Sol. Given $80 \leq \mathrm{IQ} \leq 140$

Let $x$ represent the mental age for a 12-year-old child.
$$
\begin{array}{ll}
\therefore \quad \mathrm{IQ} & =\frac{\mathrm{MA}}{\mathrm{CA}} \times 100 \\
& =\frac{x}{12} \times 100
\end{array}
$$
(given)

Substituting this expression for IQ into inequality (i),
$$
80 \leq \frac{x}{12} \times 100 \leq 140
$$

Multiplying all parts by 12,
$$
960 \leq 100 x \leq 1680
$$

Dividing by $100,9.6 \leq x \leq 16.8$
∴ The mental age of 12-year-old children in this group ranges from 9.6 to 16.8 years.
Remark: The use of $\leq$ means both endpoints 9.6 and 16.8 are included in the range.