NCERT Exemplar MCQs

Hint

\(
\text { Solve the inequality, } 3 x-5<x+7 \text {, when } x 
\)

We have \(3 x-5<x+7\)
\(
\Rightarrow 3 \mathrm{x}<\mathrm{x}+12
\) ….(Adding 5 to both sides)
\(\Rightarrow 2 \mathrm{x}<12\) (Subtracting \(\mathrm{x}\) from both sides)
\(\Rightarrow x<6\)
(Dividing by 2 on both sides)
Solution set is \(\{1,2,3,4,5\}\)

Hint

We have \(3 x-5<x+7\)
\(
\Rightarrow 3 \mathrm{x}<\mathrm{x}+12
\) ….(Adding 5 to both sides)
\(\Rightarrow 2 \mathrm{x}<12\)
(Subtracting \(\mathrm{x}\) from both sides)
\(\Rightarrow x<6\)
(Dividing by 2 on both sides)
Solution set is \(\{0,1,2,3,4,5\}\)

Hint

We have \(3 x-5<x+7\)
\(\Rightarrow 3 x<x+12 \quad \ldots\). (Adding 5 to both sides)
\(\Rightarrow 2 \mathrm{x}<12 \ldots .\). (Subtracting \(\mathrm{x}\) from both sides)
\(\Rightarrow x<6\) (Dividing by 2 on both sides)
Solution set is \(\{\ldots .-3,-2,-1,0,1,2,3,4,5\}\)

Hint

We have \(\frac{x-2}{x+5}>2\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{x-2}{x+5}-2>0 \\
& \Rightarrow \quad \frac{-(x+12)}{x+5}>0 \\
& \Rightarrow \quad \frac{x+12}{x+5}<0 \\
& \Rightarrow \quad x+12>0 \text { and } x+5<0 \quad \text { [Since } \frac{a}{b}<0 \Rightarrow a \text { and } b \text { are of opposite signs] } \\
& \text { or } \\
& x+12<0 \text { and } x+5>0 \\
& \Rightarrow \quad x>-12 \text { and } x<-5 \\
& x<-12 \text { and } x>-5 \quad \text { (Not possible) } \\
& \text { Therefore, } \quad-12<x<-5, \quad \text { i.e. } x \in(-12,-5) \\
&
\end{aligned}
\)

Hint

We have \(|3-4 x| \geq 9\).
\(
\begin{array}{lll}
\Rightarrow & 3-4 x \leq-9 \text { or } 3-4 x \geq 9 & \text { (Since }|x| \geq a \Rightarrow x \leq-a \text { or } x \geq a) \\
\Rightarrow & -4 x \leq-12 \text { or }-4 x \geq 6 \\
\Rightarrow & \left.x \geq 3 \quad \text { or } \quad x \leq \frac{-3}{2} \quad \text { (Dividing both sides by }-4\right) \\
\Rightarrow \quad & x \in\left(-\infty, \frac{-3}{2}\right] \cup[3, \infty) &
\end{array}
\)

Hint

e have \(1 \leq|x-2| \leq 3\)
\(
\begin{array}{ll}
\Rightarrow & |x-2| \geq 1 \text { and }|x-2| \leq 3 \\
\Rightarrow & (x-2 \leq-1 \text { or } x-2 \geq 1) \text { and }(-3 \leq x-2 \leq 3) \\
\Rightarrow & (x \leq 1 \text { or } x \geq 3) \text { and }(-1 \leq x \leq 5) \\
\Rightarrow & x \in(-\infty, 1] \cup[3, \infty) \text { and } x \in[-1,5]
\end{array}
\)
Combining the solutions of two inequalities, we have
\(
x \in[-1,1] \cup[3,5]
\)

Hint

We have, profit \(=\) Revenue \(–\) Cost
\(
\begin{aligned}
& =(60 x+2000)-(20 x+4000) \\
& =40 x-2000
\end{aligned}
\)
To earn some profit, \(40 x-2000>0\)
\(
\Rightarrow \quad x>50
\)
Hence, the manufacturer must sell more than 50 items to realise some profit.

Hint

On LHS of the given inequality, we have two terms both containing modulus. By equating the expression within the modulus to zero, we get \(x=-1,0\) as critical points. These critical points divide the real line in three parts as \((-\infty,-1),[-1,0),[0, \infty)\).

Case I When \(-\infty<x<-1\)
\(
|x+1|+|x|>3 \Rightarrow-x-1-x>3 \Rightarrow x<-2
\)
Case II When \(-1 \leq x<0[latex],
[latex]
|x+1|+|x|>3 \Rightarrow x+1-x>3 \Rightarrow 1>3 \quad \text { (not possible) }
\)
Case III When \(0 \leq x<\infty[latex],
[latex]
|x+1|+|x|>3 \Rightarrow x+1+x>3 \Rightarrow x>1 .
\)
Combining the results of cases (I), (II) and (III), we get
\(
x \in(-\infty,-2) \cup(1, \infty)
\)

Hint

We have \(\frac{|x+3|+x}{x+2}>1\)
\(
\begin{array}{ll}
\Rightarrow & \frac{|x+3|+x}{x+2}-1>0 \\
\Rightarrow & \frac{|x+3|-2}{x+2}>0
\end{array}
\)
Now two cases arise:
Case I: When \(x+3 \geq 0\), i.e., \(x \geq-3\). Then
\(
\begin{array}{ll}
& \frac{|x+3|-2}{x+2}>0 \Rightarrow \frac{x+3-2}{x+2}>0 \\
\Rightarrow & \frac{x+1}{x+2}>0 \\
\Rightarrow & \{(x+1)>0 \text { and } x+2>0\} \text { or }\{x+1<0 \text { and } x+2<0\} \\
\Rightarrow & \{x>-1 \text { and } x>-2\} \text { or }\{x<-1 \text { and } x<-2\} \\
\Rightarrow & x>-1 \text { or } x<-2 \\
\Rightarrow & x \in(-1, \infty) \text { or } x \in(-\infty,-2) \\
\Rightarrow & x \in(-3,-2) \cup(-1, \infty) \quad[\text { Since } x \geq-3] \dots(1)
\end{array}
\)
Case II: When \(x+3<0\), i.e., \(x<-3\)
\(
\begin{array}{ll}
& \frac{|x+3|-2}{x+2}>0 \quad \Rightarrow \quad \frac{-x-3-2}{x+2}>0 \\
\Rightarrow \quad & \frac{-(x+5)}{x+2}>0 \quad \Rightarrow \quad \frac{x+5}{x+2}<0 \\
\Rightarrow \quad & (x+5<0 \text { and } x+2>0) \quad \text { or }(x+5>0 \text { and } x+2<0) \\
\Rightarrow \quad & (x<-5 \text { and } x>-2) \quad \text { or } \quad(x>-5 \text { and } x<-2) \\
\Rightarrow \quad & \text { it is not possible. } \\
\Rightarrow & x \in(-5,-2) \dots(2)
\end{array}
\)
Combining (I) and (II), the required solution is
\(
x \in(-5,-2) \cup(-1, \infty)
\)

Hint

From the first inequality, we have \(\frac{x}{2 x+1}-\frac{1}{4} \geq 0\)
\(
\begin{array}{ll}
\Rightarrow & \frac{2 x-1}{2 x+1} \geq 0 \\
\Rightarrow & (2 x-1 \geq 0 \text { and } 2 x+1>0) \text { or }(2 x-1 \leq 0 \text { and } 2 x+1<0)[\text { Since } 2 x+1 \neq 0) \\
\Rightarrow & \left(x \geq \frac{1}{2} \text { and } x>-\frac{1}{2}\right) \text { or }\left(x \leq \frac{1}{2} \text { and } x<-\frac{1}{2}\right) \\
\Rightarrow & x \geq \frac{1}{2} \text { or } x<-\frac{1}{2} \\
\Rightarrow & x \in\left(-\infty,-\frac{1}{2}\right) \cup\left[\frac{1}{2}, \infty\right) \dots(1)
\end{array}
\)
From the second inequality, we have \(\frac{6 x}{4 x-1}-\frac{1}{2}<0\)
\(
\begin{array}{llll}
\Rightarrow & \frac{8 x+1}{4 x-1}<0 & \\
\Rightarrow & (8 x+1<0 \text { and } 4 x-1>0) \quad & \text { or } & (8 x+1>0 \text { and } 4 x-1<0) \\
\Rightarrow & \left(x<-\frac{1}{8} \text { and } x>\frac{1}{4}\right) \quad \text { or } \quad\left(x>-\frac{1}{8} \text { and } x<\frac{1}{4}\right) \\
\Rightarrow & x \in\left(-\frac{1}{8}, \frac{1}{4}\right) \quad \text { (Since the first is not possible) } \dots(2)
\end{array}
\)

Note that the common solution of (1) and (2) is null set. Hence, the given system of
inequalities has no solution.

Hint

ave \(2 x+3 y \geq 3\) as linear inequality corresponding to the line \(2 x+3 y=3\).

(ii) Consider \(3 x+4 y=18\). We observe that the shaded region and the origin lie on the same side of this line and \((0,0)\) satisfies \(3 x+4 y \leq 18\). Therefore, \(3 x+4 y \leq 18\) is the linear inequality corresponding to the line \(3 x+4 y=18\).

(iii) Consider \(-7 x+4 y=14\). It is clear from the figure that the shaded region and the origin lie on the same side of this line and \((0,0)\) satisfies the inequality \(-7 x+4 y \leq 14\). Therefore, \(-7 x+4 y \leq 14\) is the inequality corresponding to the line \(-7 x+4 y=14\)

(iv) Consider \(x-6 y=3\). It may be noted that the shaded portion and origin lie on the same side of this line and \((0,0)\) satisfies \(x-6 y \leq 3\). Therefore, \(x-6 y \leq 3\) is the inequality corresponding to the line \(x-6 y=3\).

(v) Also the shaded region lies in the first quadrant only. Therefore, \(x \geq 0, y \geq 0\). Hence, in view of (i), (ii), (iii), (iv) and (v) above, the linear inequalities corresponding to the given solution set are :
\(
2 x+3 y \geq 3,3 x+4 y \leq 18-7 x+4 y \leq 14, x-6 y \leq 3, x \geq 0, y \geq 0 \text {. }
\)

Hint

\(
\text { (b) is the correct choice. Since } \frac{|x-2|}{x-2} \geq 0 \text {, for }|x-2| \geq 0 \text {, and } x-2 \neq 0 \text {. }
\)

Hint

(c) is the correct choice. If \(x \mathrm{~cm}\) is the breadth, then
\(
2(3 x+x) \geq 160 \Rightarrow x \geq 20
\)

Hint

(a) is the correct choice. Common solution of the inequalities is from \(-\infty\) to \(-4\) and 3 to \(\infty\).

Hint

(d) is the correct choice, since \(|x+3| \geq 10, \Rightarrow x+3 \leq-10[latex] or [latex]x+3 \geq 10[latex]
[latex]
\begin{array}{ll}
\Rightarrow & x \leq-13 \text { or } \quad x \geq 7 \\
\Rightarrow & x \in(-\infty,-13] \cup[7, \infty)
\end{array}
\)

Hint

True, because the sign of inequality is reversed when we multiply both sides of an inequality by a
negative quantity.

Hint

False, product of two numbers is positive if they have the same sign.

Hint

False, product of two numbers is negative if they have opposite signs.

Hint

\(
\text { True if }|x|<5 \Rightarrow-5<x<5 \Rightarrow x \in(-5,5) \text {. }
\)

Hint

\((\geq)\), because same number can be added to both sides of inequality without changing the sign of inequality.

Hint

\(
(\geq) \text {, after multiplying both sides by }-2, \text { the sign of inequality is reversed. }
\)

Hint

\(
(<) \text {, because if } \frac{a}{b}<0 \text { and } a>0 \text {, then } b<0 \text {. }
\)

Hint

(>), if both sides are divided by the same negative quantity, then the sign of inequality is reversed.

Hint

\(
(\leq, \leq),|x-1| \leq 2 \Rightarrow-2 \leq x-1 \leq 2 \Rightarrow-1 \leq x \leq 3
\)

Hint

\(
\begin{aligned}
& (<,>),|3 x-7|>2 \Rightarrow 3 x-7<-2 \text { or } 3 x-7>2 \\
& \Rightarrow x<\frac{5}{3} \quad \text { or } x>3
\end{aligned}
\)

Hint

\(
(<) \text {, as } p \text { is positive and } q \text { is negative, therefore, } p+q \text { is always smaller than } p \text {. }
\)

Hint

Given that
\(
\begin{aligned}
& \frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x>0 \\
& \Rightarrow \quad 4 \leq 3(x+1) \leq 6 \Rightarrow 4 \leq 3 x+3 \leq 6 \\
& \Rightarrow \quad 4-3 \leq 3 x \leq 6-3 \Rightarrow 1 \leq 3 x \leq 3 \\
& \Rightarrow \quad \frac{1}{3} \leq x \leq 1
\end{aligned}
\)
Hence, the solution is \((1 / 3) \leq x \leq 1\).

Hint

\(
\begin{aligned}
& \text { Let }|\mathrm{x}-2|=\mathrm{y} \\
& \therefore \frac{y-1}{y-2} \leq 0 \\
& \Rightarrow y-1 \leq 0 \text { and } y-2>0 \quad \text { or } y-1 \geq 0 \text { and } y-2<0 \\
& \Rightarrow y \leq 1 \text { and } y>2 \text { or } y \geq 1 \text { and } y<2 \\
& \Rightarrow 1 \leq y<2 \\
& \Rightarrow 1 \leq|x-2|<2 \\
& \Rightarrow-2<x-2 \leq-1 \text { or } 1 \leq x-2<2 \\
& \Rightarrow 0<x \leq 1 \text { or } 3 \leq x<4 \\
& \Rightarrow x \in(0,1] \cup[3,4)
\end{aligned}
\)

Hint

Given that
\(
\begin{aligned}
& \frac{1}{|x|-3} \leq \frac{1}{2} \\
& \Rightarrow \quad|x|-3 \geq 2 \\
& {\left[\because \frac{1}{x}<\frac{1}{y} \Rightarrow x>y\right]} \\
& \Rightarrow \quad|x| \geq 5 \\
& \text { So, } x \in(-\infty,-5)] \cup[5, \infty) \dots(i)\\
& \text { Here } \quad|x|-3 \neq 0 \\
& \Rightarrow \quad|x|-3<0 \text { (or) }|x|-3>0 \\
& \Rightarrow \quad|x|<3 \quad \text { (or) } \quad|x|>3 \\
& \Rightarrow-3<x<3 \text { (or) } x<-3 \text { or } x>3 \ldots \text { (ii) } \\
& \text { From (i) and (ii) we get } \\
& x \in(-\infty,-5] \cup(-3,3) \cup[5, \infty)
\end{aligned}
\)

Hint

\(
\mid x-1 \mid \leq 5
\)
There are two cases,
\(
x-1 \leq 5
\)
Adding 1 to L.H.S. and R.H.S
\(
\begin{aligned}
& \Rightarrow x \leq 6 \\
& \Rightarrow-(x-1) \leq 5 \\
& \Rightarrow-x+1 \leq 5
\end{aligned}
\)
Subtracting 1 from L.H.S. and R.H.S.,
\(
\begin{aligned}
& \Rightarrow-x \leq 4 \\
& \Rightarrow x \geq-4
\end{aligned}
\)
From cases 1 and 2, we have
\(
\Rightarrow-4 \leq x \leq 6 \ldots \ldots .[i]
\)
Also,
\(
\begin{aligned}
& |x| \geq 2 \\
& \Rightarrow x \geq 2 \text { and } \\
& \Rightarrow-x \geq 2 \\
& \Rightarrow x \leq-2 \\
& \Rightarrow x \in(\infty,-2] \cup[2, \infty) \dots(ii)
\end{aligned}
\)
Combining equation [i] and [ii], we get
\(
x \in[-4,-2] \cup[2,6]
\)

Hint

\(
5 \leq \frac{2-3 x}{4} \leq 9
\)
Multiplying each term by 4 , we get
\(
\Rightarrow-20 \leq 2-3 x \leq 36
\)
Adding \(-2\) each term, we get
\(
\Rightarrow-22 \leq-3 x \leq 34
\)
Dividing each term by 3 , we get
\(
\Rightarrow-\frac{22}{3} \leq-x \leq \frac{34}{3}
\)
We know that,
Multiplication by \(-1\) inverts the inequality.
So, multiplying each term by \(-1\), we get
\(
\Rightarrow-\frac{34}{3} \leq \mathrm{x} \leq \frac{22}{3}
\)

Hint

\(
\begin{aligned}
& 4 x+3 \geq 2 x+17 \\
& \Rightarrow 4 x-2 x \geq 17-3 \\
& \Rightarrow 2 x \geq 14 \\
& \Rightarrow x \geq 7 \dots(i)
\end{aligned}
\)
Also,
\(
\begin{aligned}
& 3 x-5<-2 \\
& \Rightarrow 3 x<3 \\
& \Rightarrow x<1 \dots(ii)
\end{aligned}
\)
Since, equations [i] and [ii] cannot be possible
Simultaneously
We conclude that \(x\) has no solution.

Hint

We know that
Profit \(=\) Revenue \(–\) cost
The requirement is, profit \(>0\)
According to the question,
Revenue, \(R(x)=43 x\)
Cost, \(C(x)=26,000+30 x\)
Where \(\mathrm{x}\) is number of cassettes
\(
\begin{aligned}
& \Rightarrow \text { Profit }=43 x-(26,000+30 x)>0 \\
& \Rightarrow 13 x-26,000>0 \\
& \Rightarrow 13 x>26000 \\
& \Rightarrow x>2000
\end{aligned}
\)
Therefore, the company should sell more than 2000 cassettes to realise profit.

Hint

Let the third \(\mathrm{pH}\) value be \(\mathrm{x}\).
Given that first \(\mathrm{pH}\) value \(=8.48\)
And second \(\mathrm{pH}\) value \(=8.35\)
\(\therefore\) Average value of \(\mathrm{pH}=\frac{8.48+8.35+x}{3}\)
But average value of \(\mathrm{pH}\) lies between \(8.2\) and \(8.5\)
\(
\begin{aligned}
& \therefore 8 \cdot 2<\frac{8.48+8 \cdot 35+x}{x}<8 \cdot 5 \\
& \Rightarrow 24.6<16.83+\mathrm{x}<25.5 \\
& \Rightarrow 24.6-16.83<\mathrm{x}<25.5-16.83 \\
& \Rightarrow 7.77<\mathrm{x}<8.67
\end{aligned}
\)
Hence, the third pH value lies between \(7.77\) and \(8.67\).

Hint

Let \(\mathrm{x}\) litres of \(3 \%\) solution be added to 460 litres of \(9 \%\).
\(\therefore\) Total amount of mixture \(=(460+\mathrm{x})\) litres
Given that the acid contents in the resulting mixture is more than \(5 \%\) but less than \(7 \%\) acid.
\(
\begin{aligned}
& \therefore 5 \% \text { of }(460+\mathrm{x})<\frac{9}{100}+\frac{3}{100} \times x<7 \% \text { of }(460+\mathrm{x}) \\
& \Rightarrow \frac{5}{100}(460+x)<\frac{4140+3 x}{100}<\frac{7}{100}(460+x) \\
& \Rightarrow 5(460+\mathrm{x})<4140+3 \mathrm{x}<3220+7 \mathrm{x} \\
& \Rightarrow 2300+5 \mathrm{x}<4140+3 \mathrm{x}<3220+7 \mathrm{x} \\
& \Rightarrow 2300+5 \mathrm{x}<4140+3 \mathrm{x} \text { and } 4140+3 \mathrm{x}<3220+7 \mathrm{x} \\
& \Rightarrow 5 \mathrm{x}-3 \mathrm{x}<4140-2300 \text { and } 3 \mathrm{x}-7 \mathrm{x}<3220-4140 \\
& \Rightarrow 2 \mathrm{x}<1840 \text { and }-4 \mathrm{x}<-920 \\
& \Rightarrow x<\frac{1840}{2} \text { and } 4 \mathrm{x}>920 \\
& \Rightarrow \mathrm{x}<920 \\
& \therefore x>\frac{920}{4} \text { and } \mathrm{x}>230
\end{aligned}
\)
Hence the required amount of acid solution is more than 230 litres and less than 920 litres.

Hint

Let temperature in Celsius be \(C\)
Let temperature in Fahrenheit be \(\mathrm{F}\)
Solution should be kept between \(40^{\circ} \mathrm{C}\) and \(45^{\circ} \mathrm{C}\)
\(
\Rightarrow 40<C<45
\)
Multiplying each term by \(\frac{9}{5}\), we get
\(
\Rightarrow 72<\frac{9}{5} C<81
\)
Adding 32 to each term, we get
\(
\begin{aligned}
& \Rightarrow 104<\frac{9}{5} C+32<113 \\
& \Rightarrow 104<F<113
\end{aligned}
\)
Hence, the range of temperature in Fahrenheit should be between \(104^{\circ} \mathrm{F}\) and \(113^{\circ} \mathrm{F}\).

Hint

Let the length of shortest side \(={ }^{\prime} x\) ‘ \(\mathrm{cm}\)
The longest side of a triangle is twice the shortest side
\(\Rightarrow\) Length of largest side \(=2 x\)
Also, the third side is \(2 \mathrm{~cm}\) longer than the shortest side
\(\Rightarrow\) Length of third side \(=(x+2) \mathrm{cm}\)
Perimeter of triangle \(=\) sum of three sides
\(
\begin{aligned}
& =x+2 x+x+2 \\
& =4 x+2 \mathrm{~cm}
\end{aligned}
\)
Now, we know that,
Perimeter is more than \(166 \mathrm{~cm}\)
\(
\begin{aligned}
& \Rightarrow 4 x+2 \geq 166 \\
& \Rightarrow 4 x \geq 164 \\
& \Rightarrow x \geq 41
\end{aligned}
\)
Hence, the minimum length of the shortest side should be \(=41 \mathrm{~cm}\).

Hint

\(
T=30+25(x-3), 3 \leq x \leq 15
\)
Where, \(\mathrm{T}=\) temperature and \(\mathrm{x}=\) depth inside the earth
The Temperature should be between \(155^{\circ} \mathrm{C}\) and \(205^{\circ} \mathrm{C}\)
So, we get,
\(
\begin{aligned}
& \Rightarrow 155<\mathrm{T}<205 \\
& \Rightarrow 155<30+25(\mathrm{x}-3)<205 \\
& \Rightarrow 155<30+25 \mathrm{x}-75<205 \\
& \Rightarrow 155<25 \mathrm{x}-45<205
\end{aligned}
\)
Adding 45 to each term, we get
\(
\Rightarrow 200<25 \mathrm{x}<250
\)
Dividing each term by 25 , we get
\(
\Rightarrow 8<\mathrm{x}<10
\)
Hence, temperature varies from \(155^{\circ} \mathrm{C}\) to \(205^{\circ} \mathrm{C}\) at a depth of \(8 \mathrm{~km}\) to \(10 \mathrm{~km}\).

Hint

We have,
\(
(2 x+1) /(7 x-1)>5,(x+7) /(x-8)>2
\)
Now, \(\frac{2 x+1}{7 x-1}-5>0[latex]
[latex]
\begin{aligned}
& \Rightarrow \quad \frac{(2 x+1)-5(7 x-1)}{7 x-1}>0 \Rightarrow \frac{2 x+1-35 x+5}{7 x-1}>0 \\
& \Rightarrow \quad \frac{-33 x+6}{7 x-1}>0 \Rightarrow \frac{11 x-2}{7 x-1}<0 \\
& \Rightarrow \quad 11 x-2<0 \text { and } 7 x-1>0 \text { or } 11 x-2>0 \text { and } 7 x-1<0 \\
& \Rightarrow \quad x<2 / 11 \text { and } x>1 / 7 \quad \text { or } x>2 / 11 \text { and } x<1 / 7 \\
& \Rightarrow \quad x \in(1 / 7,2 / 11) \quad \text { (i) }
\end{aligned}
\)
Also \(\frac{x+7}{x-8}>2\)
\(
\begin{aligned}
& \Rightarrow \frac{x+7}{x-8}-2>0 \Rightarrow \frac{x+7-2(x-8)}{x-8}>0 \\
& \Rightarrow \frac{x+7-2 x+16}{x-8}>0 \Rightarrow \frac{-x+23}{x-8}>0 \Rightarrow \frac{x-23}{x-8}<0 \\
& \Rightarrow \quad x-23<0 \text { and } x-8>0 \text { or } x-23>0 \text { and } x-8<0 \\
& \Rightarrow \quad x<23 \text { and } x>8 \text { or } x>23 \text { and } x<8 \\
& \Rightarrow \quad x \in(8,23) \quad \text { (ii) } \\
&
\end{aligned}
\)
From (i) and (ii), we can find that there is no common set of values of \(x\). So, the given system of equation has no solution.

Hint

Consider the line \(3 x+2 y=48\), we observe that the shaded and the origin are on the same side of the line \(3 x+2 y=48\) and \((0,0)\) satisfy the linear constraint \(3 x+2 y \leq 48\). So, we must have one inequation as \(3 x+2 y \leq 48\).
Now, condider the line \(x+y=20\). We find that the shaded region and the origin are on the same side of the line \(x+y=20\) and \((0,0)\) satisfy the constraints \(x+y \leq 20\). So, the second inequation is \(x+y \leq 20\).
We also notice that the shaded region is above \(\mathrm{X}\) – axis and is on the right side of \(y\)-axis, so we must have \(x \geq 0, y \geq 0\).
Thus, the linear inequations corresponding to the given solution set are \(3 x+2 y \leq 48, x+y \leq 20\) and \(x \geq 0, y \geq 0\).

Hint

(i) Consider the line \(x+y=8\). We observe that the shaded region and origin lie on the same side of this line and \((0,0)\) satisfies \(x+y \leq 8\)
Therefore, \(x+y \leq 8\) is the linear inequality corresponding to the line \(x+y=8\).

(ii) Consider \(x+y=4\). We observe that shaded region and origin are on the opposite side of this line and \((0,0)\) satisfies \(x+y \leq 4\)
Therefore, we must have \(x+y \geq 4\) as linear inequalities corresponding to the line \(x+y=4\)

(iii) Shaded portion lie below the line \(y=5\). So, \(y \leq 5\) is the linear inequality corresponding to \(y=5\).

(iv) Shaded portion lie on the left side of the line \(x=5\), So, \(x \leq 5\) is the linear inequality corresponding to \(x=5\)

(v) Also, the shaded region lies in the first quadrant only. Therefore, \(x \geq 0, y \geq 0\).
In view of (i), (ii), (iii), (iv) and (v) above the linear inequalities corresponding to the given solutions are: \(x+y \leq 8, x+y \geq 4, y \leq 5, x \leq 5, x \geq 0\) and \(y \geq 0\)

Hint

Given: \(x+2 y \leq 3 \quad\) Line: \(x+2 y=3\)

\(
\begin{array}{|l|l|l|}
\hline x & 3 & 1 \\
\hline y & 0 & 1 \\
\hline
\end{array}
\)
Now, \((0,0)\) doesn’t satisfies \(3 x+4 y \geq 12\),
Thus, the region is not towards the origin
region is to the right of the \(y\) axis, thus,
\(x \geq 0\)
& Region is above the line \(x=1\), thus,
\(y \geq 1\)
Thus, the graph can be plotted as, 

Therefore, the system has no common region as solution.

Hint

\(
3 x+2 y \geq 24
\)
Line: \(3 x+2 y=24\)

\(
\begin{array}{|l|l|l|}
\hline x & 0 & 8 \\
\hline y & 12 & 0 \\
\hline
\end{array}
\)

Also, \((0,0)\) doesn’t satisfy the \(3 x+2 y \geq 24\), hence region is away from the origin
\(
3 x+y \leq 15
\)
Line: \(3 x+y=15\)

\(
\begin{array}{|l|l|l|}
\hline \mathrm{x} & 0 & 5 \\
\hline \mathrm{y} & 15 & 0 \\
\hline
\end{array}
\)

Also, \((0,0)\) satisfies the \(3 x+y \leq 15\), hence region is towards the origin \(x \geq 4\) implies that region is right to the line \(x=4\), therefore graph is
It is clear from the graph the above system has no common region as solution.

Hint

Let’s plot the region of each inequality and then find the common region of all
\(
2 x+y \geq 8
\)
Line: \(2 x+y=8\)
\(
\begin{array}{|c|c|c|}
\hline x & 0 & 4 \\
\hline y & 8 & 0 \\
\hline
\end{array}
\)
Also, \((0,0)\) doesn’t satisy the \(2 x+y \geq 8\), hence region is away from the origin
\(
x+2 y \geq 10
\)
Line: \(x+2 y=10\)
\(
\begin{array}{|c|c|c|}
\hline \mathrm{X} & 0 & 10 \\
\hline \mathrm{Y} & 5 & 0 \\
\hline
\end{array}
\)
\(
\text { Also, }(0,0) \text { doesn’t satisy the } x+2 y \geq 10 \text {, hence region is away from the origin }
\)

Hint

\(
x<5
\)
Multiplication or dividing by negative number inverts the inequality sign.
\(
\Rightarrow-x>-5
\)

Hint

Given that \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{b}\) are real numbers and \(\mathrm{x}<\mathrm{y}, \mathrm{b}<0\), then \(\frac{x}{b}>\frac{y}{b}\).
Explanation:
Given that \(x<y, b<0\)
\(
\Rightarrow \frac{x}{b}>\frac{y}{b}, b<0
\)

Hint

\(
\begin{aligned}
& -3 x+17<-13 \\
& \Rightarrow-3 x<-30 \text { [Subtracting } 17 \text { both side] } \\
& \Rightarrow x>10 \text { [Division by negative number inverts the inequality sign] } \\
& \Rightarrow x \in(10, \infty)
\end{aligned}
\)

Hint

\(|x|<3\)
Hence, there are two cases,
\(x<3 \dots(1)\)
and
\(
\begin{aligned}
& -x<3 \\
& \Rightarrow x>-3 \dots(2)
\end{aligned}
\)
From [1] and [2], we get
\(
\Rightarrow-3<x<3
\)

Hint

\(
|x|>b
\)
Hence, there are two cases,
\(
\begin{aligned}
& x>b \\
& \Rightarrow x \in(b, \infty)……[1]
\end{aligned}
\)
and
\(
\begin{aligned}
& -x>b \\
& \Rightarrow x<-b \\
& \Rightarrow x \in(-\infty,-b)…..[2]
\end{aligned}
\)
From [1] and [2], we get
\(
\Rightarrow x \in(-\infty,-b) \cup(b, \infty)
\)

Hint

Given that \(|x-1|>5\)
\(
\begin{aligned}
& \Rightarrow(x-1)<-5 \text { or }(x-1)>5 \\
& \Rightarrow x<-5+1 \text { or } x>5+1 \\
& \Rightarrow x<-4 \text { or } x>6 \\
& \Rightarrow x \in[-\infty,-4) \cup(6, \infty)
\end{aligned}
\)

Hint

Given that \(|x+2| \leq 9\)
\(
\begin{aligned}
& \Rightarrow-9 \leq x+2 \leq 9 \\
& \Rightarrow-9-2 \leq x \leq 9-2 \quad \ldots \ldots[|x|<a \Rightarrow-a \leq x \leq a] \\
& \Rightarrow-11 \leq x \leq 7 \\
& \Rightarrow x \in[-11,7]
\end{aligned}
\)

Hint

The inequality representing the following graph is \(|\underline{\mathbf{x}}| \mathbf{\leq 5}\).

Explanation:

The given graph represents \(x>-5[latex] and [latex]x<5\)
Combining the two inequalities \(|x|<5\)

Hint

The gives graph represents all values of \(x\) greater than 5 including 5 on the real number line
\(
\text { So } x \in(5, \infty)
\)

Hint

The given graph has all real values of \(x\) greater than and equal to \(\frac{9}{2}\).
So, \(x \geq \frac{9}{2}\)
\(
\Rightarrow x \in\left[\frac{9}{2}, \infty\right)
\)

Hint

The given graph represents all the values of \(\mathrm{x}\) less than \(\frac{7}{2}\) on real number line. So, \(x \in\left(-\infty, \frac{7}{2}\right)\)

Hint

The given graph represents all real values of \(\mathrm{x}\) less than and equal to \(-2\)
So, \(x \in(-\infty,-2]\)

Hint

Given that \(x<y, b<0\)
\(
\Rightarrow \frac{x}{b}>\frac{y}{b}, b<0
\)

Hint

\(
\text { If } x, y>0 \text { then } x>0, y>0 \text { or } x<0, y<0
\)

Hint

\(x<-5\) and \(x<-2\)
\(
\Rightarrow x \in(-\infty,-5)
\)

Hint

\(
\text { If } x<-5 \text { and } x>2 \text {, then } x \text { have no value. }
\)

Hint

\(
\begin{aligned}
& \text { If }|x|>5 \text { then } x<-5 \text { or } x>5 \\
& \Rightarrow x \in(-\infty,-5) \cup(5, \infty)
\end{aligned}
\)

Hint

If \(|x| \leq 4\), then \(-4 \leq x \leq 4\)
\(
\Rightarrow x \in[-4,4]
\)