Linear Equations and Inequalities in One Variable

Solving,

$$\begin{aligned}\Rightarrow 5x - 3 = 3x - 5 \\[1em] \Rightarrow 5x - 3x = - 5 + 3 \\[1em] \Rightarrow 2x = - 2 \\[1em] \Rightarrow x = - \frac{2}{2} \\[1em] \Rightarrow x = - 1\end{aligned}$$

Hence, $\bm{x = -1}$.

Solving,

$$\begin{aligned}\Rightarrow 3x - 7 = 3(5 - x) \\[1em] \Rightarrow 3x - 7 = 15 - 3x \\[1em] \Rightarrow 3x + 3x = 15 + 7 \\[1em] \Rightarrow 6x = 22 \\[1em] \Rightarrow x = \frac{22}{6} \\[1em] \Rightarrow x = \frac{11}{3}\end{aligned}$$

Hence, $\bm{x = \frac{11}{3}}$.

Solving,

$$\begin{aligned}\Rightarrow 4(2x + 1) = 3(x - 1) + 7 \\[1em] \Rightarrow 8x + 4 = 3x - 3 + 7 \\[1em] \Rightarrow 8x + 4 = 3x + 4 \\[1em] \Rightarrow 8x - 3x = 4 - 4 \\[1em] \Rightarrow 5x = 0 \\[1em] \Rightarrow x = \frac{0}{5} \\[1em] \Rightarrow x = 0\end{aligned}$$

Hence, $\bm{x = 0}$.

Solving,

$$\begin{aligned}\Rightarrow 3(2p - 1) = 5 - (3p - 2) \\[1em] \Rightarrow 6p - 3 = 5 - 3p + 2 \\[1em] \Rightarrow 6p - 3 = 7 - 3p \\[1em] \Rightarrow 6p + 3p = 7 + 3 \\[1em] \Rightarrow 9p = 10 \\[1em] \Rightarrow p = \frac{10}{9} \\[1em] \Rightarrow p = 1\frac{1}{9}\end{aligned}$$

Hence, $\bm{p = 1\frac{1}{9}}$.

Solving,

$$\begin{aligned}\Rightarrow 5y - 2[y - 3(y - 5)] = 6 \\[1em] \Rightarrow 5y - 2[y - 3y + 15] = 6 \\[1em] \Rightarrow 5y - 2[- 2y + 15] = 6 \\[1em] \Rightarrow 5y + 4y - 30 = 6 \\[1em] \Rightarrow 9y = 6 + 30 \\[1em] \Rightarrow 9y = 36 \\[1em] \Rightarrow y = \frac{36}{9} \\[1em] \Rightarrow y = 4\end{aligned}$$

Hence, $\bm{y = 4}$.

Solving,

$$\begin{aligned}\Rightarrow 0.3(6 - x) = 0.4(x + 8) \\[1em] \Rightarrow 1.8 - 0.3x = 0.4x + 3.2 \\[1em] \Rightarrow 1.8 - 3.2 = 0.4x + 0.3x \\[1em] \Rightarrow - 1.4 = 0.7x \\[1em] \Rightarrow x = - \frac{1.4}{0.7} \\[1em] \Rightarrow x = - 2\end{aligned}$$

Hence, $\bm{x = -2}$.

Given,

$\frac{x - 1}{3} = \frac{x + 2}{6} + 3$

L.C.M. of 3 and 6 = 6

Multiplying both sides by 6, we get :

$$\begin{aligned}\Rightarrow \frac{(x - 1)}{3} \times 6 = \Big[\frac{(x + 2)}{6} + 3\Big] \times 6 \\[1em] \Rightarrow 2(x - 1) = \frac{(x + 2)}{6} \times 6 + 3 \times 6 \\[1em] \Rightarrow 2(x - 1) = (x + 2) + 18 \\[1em] \Rightarrow 2x - 2 = x + 20 \\[1em] \Rightarrow 2x - x = 20 + 2 \\[1em] \Rightarrow x = 22\end{aligned}$$

Hence, $\bm{x = 22}$.

Given,

$\frac{x + 7}{3} = 1 + \frac{3x - 2}{5}$

L.C.M. of 3 and 5 = 15

Multiplying both sides by 15, we get:

$$\begin{aligned}\Rightarrow \frac{(x + 7)}{3} \times 15 = \Big[1 + \frac{(3x - 2)}{5}\Big] \times 15 \\[1em] \Rightarrow 5(x + 7) = 1 \times 15 + \frac{(3x - 2)}{5} \times 15 \\[1em] \Rightarrow 5(x + 7) = 15 + 3(3x - 2) \\[1em] \Rightarrow 5x + 35 = 15 + 9x - 6 \\[1em] \Rightarrow 5x + 35 = 9x + 9 \\[1em] \Rightarrow 35 - 9 = 9x - 5x \\[1em] \Rightarrow 26 = 4x \\[1em] \Rightarrow x = \frac{26}{4} \\[1em] \Rightarrow x = \frac{13}{2} = 6\frac{1}{2}\end{aligned}$$

Hence, $\bm{x = 6\frac{1}{2}}$.

Given,

$\frac{y + 1}{3} - \frac{y - 1}{2} = \frac{1 + 2y}{3}$

L.C.M. of 3 and 2 = 6

Multiplying both sides by 6, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{(y + 1)}{3} - \frac{(y - 1)}{2}\Big] \times 6 = \frac{(1 + 2y)}{3} \times 6 \\[1em] \Rightarrow \frac{(y + 1)}{3} \times 6 - \frac{(y - 1)}{2} \times 6 = 2(1 + 2y) \\[1em] \Rightarrow 2(y + 1) - 3(y - 1) = 2(1 + 2y) \\[1em] \Rightarrow 2y + 2 - 3y + 3 = 2 + 4y \\[1em] \Rightarrow - y + 5 = 2 + 4y \\[1em] \Rightarrow 5 - 2 = 4y + y \\[1em] \Rightarrow 3 = 5y \\[1em] \Rightarrow y = \frac{3}{5}\end{aligned}$$

Hence, $\bm{y = \frac{3}{5}}$.

Given,

$\frac{p}{3} + \frac{p}{4} = 55 - \frac{p + 40}{5}$

L.C.M. of 3, 4 and 5 = 60

Multiplying both sides by 60, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{p}{3} + \frac{p}{4}\Big] \times 60 = \Big[55 - \frac{(p + 40)}{5}\Big] \times 60 \\[1em] \Rightarrow \frac{p}{3} \times 60 + \frac{p}{4} \times 60 = 55 \times 60 - \frac{(p + 40)}{5} \times 60 \\[1em] \Rightarrow 20p + 15p = 3300 - 12(p + 40) \\[1em] \Rightarrow 35p = 3300 - 12p - 480 \\[1em] \Rightarrow 35p = 2820 - 12p \\[1em] \Rightarrow 35p + 12p = 2820 \\[1em] \Rightarrow 47p = 2820 \\[1em] \Rightarrow p = \frac{2820}{47} \\[1em] \Rightarrow p = 60\end{aligned}$$

Hence, $\bm{p = 60}$.

Given,

$n - \frac{n - 1}{2} = 1 - \frac{n - 2}{3}$

L.C.M. of 2 and 3 = 6

Multiplying both sides by 6, we get:

$$\begin{aligned}\Rightarrow \Big[n - \frac{(n - 1)}{2}\Big] \times 6 = \Big[1 - \frac{(n - 2)}{3}\Big] \times 6 \\[1em] \Rightarrow n \times 6 - \frac{(n - 1)}{2} \times 6 = 1 \times 6 - \frac{(n - 2)}{3} \times 6 \\[1em] \Rightarrow 6n - 3(n - 1) = 6 - 2(n - 2) \\[1em] \Rightarrow 6n - 3n + 3 = 6 - 2n + 4 \\[1em] \Rightarrow 3n + 3 = 10 - 2n \\[1em] \Rightarrow 3n + 2n = 10 - 3 \\[1em] \Rightarrow 5n = 7 \\[1em] \Rightarrow n = \frac{7}{5} = 1\frac{2}{5}\end{aligned}$$

Hence, $\bm{n = 1\frac{2}{5}}$.

Given,

$\frac{3t - 2}{3} + \frac{2t + 3}{2} = t + \frac{7}{6}$

L.C.M. of 3, 2 and 6 = 6

Multiplying both sides by 6, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{(3t - 2)}{3} + \frac{(2t + 3)}{2}\Big] \times 6 = \Big[t + \frac{7}{6}\Big] \times 6 \\[1em] \Rightarrow \frac{(3t - 2)}{3} \times 6 + \frac{(2t + 3)}{2} \times 6 = t \times 6 + \frac{7}{6} \times 6 \\[1em] \Rightarrow 2(3t - 2) + 3(2t + 3) = 6t + 7 \\[1em] \Rightarrow 6t - 4 + 6t + 9 = 6t + 7 \\[1em] \Rightarrow 12t + 5 = 6t + 7 \\[1em] \Rightarrow 12t - 6t = 7 - 5 \\[1em] \Rightarrow 6t = 2 \\[1em] \Rightarrow t = \frac{2}{6} \\[1em] \Rightarrow t = \frac{1}{3}\end{aligned}$$

Hence, $\bm{t = \frac{1}{3}}$.

Solving,

$$\begin{aligned}\Rightarrow 4(3x + 2) - 5(6x - 1) = 2(x - 8) - 6(7x - 4) \\[1em] \Rightarrow 12x + 8 - 30x + 5 = 2x - 16 - 42x + 24 \\[1em] \Rightarrow - 18x + 13 = - 40x + 8 \\[1em] \Rightarrow - 18x + 40x = 8 - 13 \\[1em] \Rightarrow 22x = - 5 \\[1em] \Rightarrow x = - \frac{5}{22}\end{aligned}$$

Hence, $\bm{x = -\frac{5}{22}}$.

Solving,

$$\begin{aligned}\Rightarrow 3(5x + 7) + 5(2x - 11) = 3(8x - 5) - 15 \\[1em] \Rightarrow 15x + 21 + 10x - 55 = 24x - 15 - 15 \\[1em] \Rightarrow 25x - 34 = 24x - 30 \\[1em] \Rightarrow 25x - 24x = - 30 + 34 \\[1em] \Rightarrow x = 4\end{aligned}$$

Hence, $\bm{x = 4}$.

Given,

$\frac{3 - 2x}{2x + 5} = -\frac{3}{11}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 11(3 - 2x) = - 3(2x + 5) \\[1em] \Rightarrow 33 - 22x = - 6x - 15 \\[1em] \Rightarrow 33 + 15 = - 6x + 22x \\[1em] \Rightarrow 48 = 16x \\[1em] \Rightarrow x = \frac{48}{16} \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, $\bm{x = 3}$.

Given,

$\frac{5p + 2}{8 - 2p} = \frac{7}{6}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 6(5p + 2) = 7(8 - 2p) \\[1em] \Rightarrow 30p + 12 = 56 - 14p \\[1em] \Rightarrow 30p + 14p = 56 - 12 \\[1em] \Rightarrow 44p = 44 \\[1em] \Rightarrow p = \frac{44}{44} \\[1em] \Rightarrow p = 1\end{aligned}$$

Hence, $\bm{p = 1}$.

Given,

$\frac{5}{x} = \frac{7}{x - 4}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 5(x - 4) = 7x \\[1em] \Rightarrow 5x - 20 = 7x \\[1em] \Rightarrow - 20 = 7x - 5x \\[1em] \Rightarrow 2x = - 20 \\[1em] \Rightarrow x = - \frac{20}{2} \\[1em] \Rightarrow x = - 10\end{aligned}$$

Hence, $\bm{x = -10}$.

Given,

$\frac{4}{2x + 3} = \frac{5}{x + 4}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 4(x + 4) = 5(2x + 3) \\[1em] \Rightarrow 4x + 16 = 10x + 15 \\[1em] \Rightarrow 16 - 15 = 10x - 4x \\[1em] \Rightarrow 1 = 6x \\[1em] \Rightarrow x = \frac{1}{6}\end{aligned}$$

Hence, $\bm{x = \frac{1}{6}}$.

$\frac{2x + 5}{2} - \frac{5x}{x - 1} = x$

L.C.M. of 2 and $(x - 1)$ = $2(x - 1)$

Multiplying both sides by $2(x - 1)$, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{(2x + 5)}{2} - \frac{5x}{(x - 1)}\Big] \times 2(x - 1) = x \times 2(x - 1) \\[1em] \Rightarrow \frac{(2x + 5)}{2} \times 2(x - 1) - \frac{5x}{(x - 1)} \times 2(x - 1) = 2x(x - 1) \\[1em] \Rightarrow (x - 1)(2x + 5) - 10x = 2x(x - 1) \\[1em] \Rightarrow 2x^{2} + 5x - 2x - 5 - 10x = 2x^{2} - 2x \\[1em] \Rightarrow 2x^{2} - 7x - 5 = 2x^{2} - 2x \\[1em] \Rightarrow - 7x + 2x = 5 \\[1em] \Rightarrow - 5x = 5 \\[1em] \Rightarrow x = - \frac{5}{5} \\[1em] \Rightarrow x = - 1\end{aligned}$$

Hence, $\bm{x = -1}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{5}\left(\frac{1}{3x} - 5\right) = \frac{1}{3}\left(3 - \frac{1}{x}\right) \\[1em] \Rightarrow \frac{1}{15x} - 1 = 1 - \frac{1}{3x} \\[1em] \Rightarrow \frac{1}{15x} + \frac{1}{3x} = 1 + 1 \\[1em] \Rightarrow \frac{1 + 5}{15x} = 2 \\[1em] \Rightarrow \frac{6}{15x} = 2\end{aligned}$$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 6 = 30x \\[1em] \Rightarrow x = \frac{6}{30} \\[1em] \Rightarrow x = \frac{1}{5}\end{aligned}$$

Hence, $\bm{x = \frac{1}{5}}$.

Given,

$\frac{2x - 3}{2x - 1} = \frac{3x - 1}{3x + 1}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow (2x - 3)(3x + 1) = (3x - 1)(2x - 1) \\[1em] \Rightarrow 6x^{2} + 2x - 9x - 3 = 6x^{2} - 3x - 2x + 1 \\[1em] \Rightarrow 6x^{2} - 7x - 3 = 6x^{2} - 5x + 1 \\[1em] \Rightarrow - 7x + 5x = 1 + 3 \\[1em] \Rightarrow - 2x = 4 \\[1em] \Rightarrow x = - \frac{4}{2} \\[1em] \Rightarrow x = - 2\end{aligned}$$

Hence, $\bm{x = -2}$.

Given,

$\frac{2y + 3}{3y + 2} = \frac{4y + 5}{6y + 7}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow (2y + 3)(6y + 7) = (4y + 5)(3y + 2) \\[1em] \Rightarrow 12y^{2} + 14y + 18y + 21 = 12y^{2} + 8y + 15y + 10 \\[1em] \Rightarrow 12y^{2} + 32y + 21 = 12y^{2} + 23y + 10 \\[1em] \Rightarrow 32y - 23y = 10 - 21 \\[1em] \Rightarrow 9y = - 11 \\[1em] \Rightarrow y = - \frac{11}{9}\end{aligned}$$

Hence, $\bm{y = -\frac{11}{9}}$.

Given,

$x = p + 1$

Substituting the value of $x$ in the given equation, we get:

$$\begin{aligned}\Rightarrow \frac{1}{2}[5(p + 1) - 30] - \frac{1}{3}(1 + 7p) = \frac{1}{4} \\[1em] \Rightarrow \frac{1}{2}(5p + 5 - 30) - \frac{1}{3}(1 + 7p) = \frac{1}{4} \\[1em] \Rightarrow \frac{1}{2}(5p - 25) - \frac{1}{3}(1 + 7p) = \frac{1}{4}\end{aligned}$$

L.C.M. of 2, 3 and 4 = 12

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{1}{2}(5p - 25) - \frac{1}{3}(1 + 7p)\Big] \times 12 = \frac{1}{4} \times 12 \\[1em] \Rightarrow \frac{1}{2}(5p - 25) \times 12 - \frac{1}{3}(1 + 7p) \times 12 = 3 \\[1em] \Rightarrow 6(5p - 25) - 4(1 + 7p) = 3 \\[1em] \Rightarrow 30p - 150 - 4 - 28p = 3 \\[1em] \Rightarrow 2p - 154 = 3 \\[1em] \Rightarrow 2p = 3 + 154 \\[1em] \Rightarrow 2p = 157 \\[1em] \Rightarrow p = \frac{157}{2} = 78\frac{1}{2}\end{aligned}$$

Hence, $\bm{p = 78\frac{1}{2}}$.

Given,

$\frac{x + 3}{3} - \frac{x - 2}{2} = 1$

L.C.M. of 3 and 2 = 6

Multiplying both sides by 6, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{(x + 3)}{3} - \frac{(x - 2)}{2}\Big] \times 6 = 1 \times 6 \\[1em] \Rightarrow \frac{(x + 3)}{3} \times 6 - \frac{(x - 2)}{2} \times 6 = 6 \\[1em] \Rightarrow 2(x + 3) - 3(x - 2) = 6 \\[1em] \Rightarrow 2x + 6 - 3x + 6 = 6 \\[1em] \Rightarrow - x + 12 = 6 \\[1em] \Rightarrow - x = 6 - 12 \\[1em] \Rightarrow - x = - 6 \\[1em] \Rightarrow x = 6\end{aligned}$$

Substituting $x = 6$ in $\frac{1}{x} + p = 1$, we get:

$$\begin{aligned}\Rightarrow \frac{1}{6} + p = 1 \\[1em] \Rightarrow p = 1 - \frac{1}{6} \\[1em] \Rightarrow p = \frac{6 - 1}{6} \\[1em] \Rightarrow p = \frac{5}{6}\end{aligned}$$

Hence, $\bm{x = 6}$ and $\bm{p = \frac{5}{6}}$.

Let the number be $x$.

Three more than twice the number = $2x + 3$

Four less than the number = $x - 4$

Given,

Three more than twice a number is equal to four less than the number.

$$\begin{aligned}\therefore 2x + 3 = x - 4 \\[1em] \Rightarrow 2x - x = - 4 - 3 \\[1em] \Rightarrow x = - 7\end{aligned}$$

Hence, the required number = -7.

Let the four consecutive integers be $x, x + 1, x + 2$ and $x + 3$.

According to the given condition,

$$\begin{aligned}\Rightarrow x + (x + 1) + (x + 2) + (x + 3) = 46 \\[1em] \Rightarrow 4x + 6 = 46 \\[1em] \Rightarrow 4x = 46 - 6 \\[1em] \Rightarrow 4x = 40 \\[1em] \Rightarrow x = \frac{40}{4} \\[1em] \Rightarrow x = 10\end{aligned}$$

So, the integers are 10, 10 + 1 = 11, 10 + 2 = 12 and 10 + 3 = 13.

Hence, the required integers are 10, 11, 12 and 13.

Let the number thought of by Manjula be $x$.

On subtracting $\frac{7}{3}$ from it, we get $x - \frac{7}{3}$

On multiplying the result by 6, we get $6\left(x - \frac{7}{3}\right)$

2 less than twice the original number = $2x - 2$

According to the given condition,

$$\begin{aligned}\Rightarrow 6\left(x - \frac{7}{3}\right) = 2x - 2 \\[1em] \Rightarrow 6x - 14 = 2x - 2 \\[1em] \Rightarrow 6x - 2x = - 2 + 14 \\[1em] \Rightarrow 4x = 12 \\[1em] \Rightarrow x = \frac{12}{4} \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, the required number = 3.

Let the smaller number be $x$.

Then, the other number = $7x$

On adding 15 to both the numbers, the new numbers are $x + 15$ and $7x + 15$.

According to the given condition,

⇒ $7x + 15 = \frac{5}{2}(x + 15)$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 2(7x + 15) = 5(x + 15) \\[1em] \Rightarrow 14x + 30 = 5x + 75 \\[1em] \Rightarrow 14x - 5x = 75 - 30 \\[1em] \Rightarrow 9x = 45 \\[1em] \Rightarrow x = \frac{45}{9} \\[1em] \Rightarrow x = 5\end{aligned}$$

So, the other number = 7 × 5 = 35

Hence, the required numbers are 35 and 5.

Let the three consecutive even integers be $x, x + 2$ and $x + 4$.

The sum of three consecutive even integers = 0.

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) = 0 \\[1em] \Rightarrow 3x + 6 = 0 \\[1em] \Rightarrow 3x = - 6 \\[1em] \Rightarrow x = - \frac{6}{3} \\[1em] \Rightarrow x = - 2\end{aligned}$$

So, the integers are -2, -2 + 2 = 0 and -2 + 4 = 2.

Hence, the required integers are -2, 0 and 2.

Let the smaller odd integer be $x$.

Then, the greater odd integer = $x + 2$

Given,

Two-fifth of the smaller exceeds two-ninth of the greater by 4.

⇒ $\frac{2}{5}x - \frac{2}{9}(x + 2) = 4$

L.C.M. of 5 and 9 = 45

Multiplying both sides by 45, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{2}{5}x - \frac{2}{9}(x + 2)\Big] \times 45 = 4 \times 45 \\[1em] \Rightarrow \frac{2}{5}x \times 45 - \frac{2}{9}(x + 2) \times 45 = 180 \\[1em] \Rightarrow 18x - 10(x + 2) = 180 \\[1em] \Rightarrow 18x - 10x - 20 = 180 \\[1em] \Rightarrow 8x = 180 + 20 \\[1em] \Rightarrow 8x = 200 \\[1em] \Rightarrow x = \frac{200}{8} \\[1em] \Rightarrow x = 25\end{aligned}$$

So, the greater odd integer = 25 + 2 = 27

Hence, the required integers are 25 and 27.

Let the numerator of the fraction be $x$.

Then, its denominator = $2x + 1$

So, the original fraction = $\frac{x}{2x + 1}$

On increasing the numerator and the denominator by 5, we get $\frac{x + 5}{2x + 1 + 5} = \frac{x + 5}{2x + 6}$

According to the given condition,

⇒ $\frac{x + 5}{2x + 6} = \frac{3}{5}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 5(x + 5) = 3(2x + 6) \\[1em] \Rightarrow 5x + 25 = 6x + 18 \\[1em] \Rightarrow 25 - 18 = 6x - 5x \\[1em] \Rightarrow x = 7\end{aligned}$$

So, the denominator = 2 × 7 + 1 = 15

Hence, the original fraction = $\mathbf{\frac{7}{15}}$.

Let the two positive numbers be $2x$ and $5x$.

According to the given condition,

$$\begin{aligned}\Rightarrow 5x - 2x = 15 \\[1em] \Rightarrow 3x = 15 \\[1em] \Rightarrow x = \frac{15}{3} \\[1em] \Rightarrow x = 5\end{aligned}$$

So, the numbers are 2 × 5 = 10 and 5 × 5 = 25.

Hence, the required numbers are 10 and 25.

Let the number to be added be $x$.

On adding $x$ to each of the given numbers, the numbers become $12 + x, 22 + x, 42 + x$ and $72 + x$.

As the resulting numbers are in proportion,

⇒ $\frac{12 + x}{22 + x} = \frac{42 + x}{72 + x}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow (12 + x)(72 + x) = (22 + x)(42 + x) \\[1em] \Rightarrow 864 + 12x + 72x + x^{2} = 924 + 22x + 42x + x^{2} \\[1em] \Rightarrow 864 + 84x + x^{2} = 924 + 64x + x^{2} \\[1em] \Rightarrow 84x - 64x = 924 - 864 \\[1em] \Rightarrow 20x = 60 \\[1em] \Rightarrow x = \frac{60}{20} \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, the required number to be added = 3.

Let the digit at unit's place be $x$ and the digit at ten's place be $y$.

So, the original number = $10y + x$

On interchanging the digits, the new number = $10x + y$

According to the given condition,

$$\begin{aligned}\Rightarrow (10y + x) + (10x + y) = 143 \\[1em] \Rightarrow 11x + 11y = 143 \\[1em] \Rightarrow 11(x + y) = 143 \\[1em] \Rightarrow x + y = \frac{143}{11} \\[1em] \Rightarrow x + y = 13 \quad \text{...(1)}\end{aligned}$$

Also given, the digits differ by 3.

Case 1 : Let $y - x = 3$ ...(2)

Adding equations (1) and (2), we get

$$\begin{aligned}\Rightarrow 2y = 16 \\[1em] \Rightarrow y = 8\end{aligned}$$

Substituting $y = 8$ in equation (1), we get $x = 13 - 8 = 5$

So, the number = 10 × 8 + 5 = 85

Case 2 : Let $x - y = 3$ ...(3)

Adding equations (1) and (3), we get

$$\begin{aligned}\Rightarrow 2x = 16 \\[1em] \Rightarrow x = 8\end{aligned}$$

Substituting $x = 8$ in equation (1), we get $y = 13 - 8 = 5$

So, the number = 10 × 5 + 8 = 58

Hence, the original number can be 58 or 85.

Let the digit at unit's place be $x$ and the digit at ten's place be $y$.

So, the original number = $10y + x$

On interchanging the digits, the new number = $10x + y$

Given, sum of the digits = 11

⇒ $x + y = 11 \quad \text{...(1)}$

According to the given condition,

$$\begin{aligned}\Rightarrow (10x + y) - (10y + x) = 63 \\[1em] \Rightarrow 9x - 9y = 63 \\[1em] \Rightarrow 9(x - y) = 63 \\[1em] \Rightarrow x - y = \frac{63}{9} \\[1em] \Rightarrow x - y = 7 \quad \text{...(2)}\end{aligned}$$

Adding equations (1) and (2), we get

$$\begin{aligned}\Rightarrow 2x = 18 \\[1em] \Rightarrow x = 9\end{aligned}$$

Substituting $x = 9$ in equation (1), we get $y = 11 - 9 = 2$

So, the number = 10 × 2 + 9 = 29

Hence, the required two digit number = 29.

Let the present age of Raju be $x$ years.

Then, the present age of Ritu = $4x$ years

After 4 years, Raju's age = $(x + 4)$ years and Ritu's age = $(4x + 4)$ years

According to the given condition,

$$\begin{aligned}\Rightarrow 4x + 4 = 2(x + 4) \\[1em] \Rightarrow 4x + 4 = 2x + 8 \\[1em] \Rightarrow 4x - 2x = 8 - 4 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \frac{4}{2} \\[1em] \Rightarrow x = 2\end{aligned}$$

So, Ritu's present age = 4 × 2 = 8 years

Hence, the present ages of Ritu and Raju are 8 years and 2 years respectively.

Let the present age of the son be $x$ years.

Then, the present age of the father = $7x$ years

Two years ago, son's age = $(x - 2)$ years and father's age = $(7x - 2)$ years

According to the given condition,

$$\begin{aligned}\Rightarrow 7x - 2 = 13(x - 2) \\[1em] \Rightarrow 7x - 2 = 13x - 26 \\[1em] \Rightarrow 26 - 2 = 13x - 7x \\[1em] \Rightarrow 6x = 24 \\[1em] \Rightarrow x = \frac{24}{6} \\[1em] \Rightarrow x = 4\end{aligned}$$

So, father's present age = 7 × 4 = 28 years

Hence, the present ages of the father and the son are 28 years and 4 years respectively.

Let the present ages of Sona and Sonali be $5x$ years and $3x$ years respectively.

Five years hence, Sona's age = $(5x + 5)$ years and Sonali's age = $(3x + 5)$ years

According to the given condition,

⇒ $\frac{5x + 5}{3x + 5} = \frac{10}{7}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 7(5x + 5) = 10(3x + 5) \\[1em] \Rightarrow 35x + 35 = 30x + 50 \\[1em] \Rightarrow 35x - 30x = 50 - 35 \\[1em] \Rightarrow 5x = 15 \\[1em] \Rightarrow x = \frac{15}{5} \\[1em] \Rightarrow x = 3\end{aligned}$$

So, Sona's present age = 5 × 3 = 15 years

and Sonali's present age = 3 × 3 = 9 years

Hence, the present ages of Sona and Sonali are 15 years and 9 years respectively.

Let the number of days for which the employee remained absent be $x$.

Then, the number of days for which he worked = $30 - x$

Amount received for the days he worked = ₹ $800(30 - x)$

Amount fined for the days he remained absent = ₹ $80x$

According to the given condition,

$$\begin{aligned}\Rightarrow 800(30 - x) - 80x = 15200 \\[1em] \Rightarrow 24000 - 800x - 80x = 15200 \\[1em] \Rightarrow 24000 - 880x = 15200 \\[1em] \Rightarrow 24000 - 15200 = 880x \\[1em] \Rightarrow 880x = 8800 \\[1em] \Rightarrow x = \frac{8800}{880} \\[1em] \Rightarrow x = 10\end{aligned}$$

Hence, the employee remained absent for 10 days.

Let the number of ₹ 5 coins be $x$.

Then, the number of ₹ 2 coins = $3x$

As the total number of coins is 160,

number of ₹ 1 coins = $160 - x - 3x = 160 - 4x$

Value of ₹ 5 coins = ₹ $5x$

Value of ₹ 2 coins = ₹ $2 \times 3x$ = ₹ $6x$

Value of ₹ 1 coins = ₹ $1 \times (160 - 4x)$ = ₹ $(160 - 4x)$

According to the given condition,

$$\begin{aligned}\Rightarrow 5x + 6x + (160 - 4x) = 300 \\[1em] \Rightarrow 7x + 160 = 300 \\[1em] \Rightarrow 7x = 300 - 160 \\[1em] \Rightarrow 7x = 140 \\[1em] \Rightarrow x = \frac{140}{7} \\[1em] \Rightarrow x = 20\end{aligned}$$

So, the number of ₹ 5 coins = 20,

the number of ₹ 2 coins = 3 × 20 = 60

and the number of ₹ 1 coins = 160 - 4 × 20 = 160 - 80 = 80

Hence, there are 80 coins of ₹ 1, 60 coins of ₹ 2 and 20 coins of ₹ 5.

Let the number of passengers with ₹ 5 tickets be $x$.

Then, the number of passengers with ₹ 7.50 tickets = $40 - x$

Amount received from ₹ 5 tickets = ₹ $5x$

Amount received from ₹ 7.50 tickets = ₹ $7.50(40 - x)$

According to the given condition,

$$\begin{aligned}\Rightarrow 5x + 7.50(40 - x) = 230 \\[1em] \Rightarrow 5x + 300 - 7.50x = 230 \\[1em] \Rightarrow 300 - 2.50x = 230 \\[1em] \Rightarrow 300 - 230 = 2.50x \\[1em] \Rightarrow 2.50x = 70 \\[1em] \Rightarrow x = \frac{70}{2.50} \\[1em] \Rightarrow x = 28\end{aligned}$$

Hence, the number of passengers with ₹ 5 tickets = 28.

Let the number of students in the group be $x$.

As each student pays ₹ 50, the total fare of the boat = ₹ $50x$

If there had been 3 more students, the number of students = $x + 3$

and the amount paid by each student = ₹ (50 - 10) = ₹ 40

So, the total fare of the boat = ₹ $40(x + 3)$

As the fare of the boat remains the same,

$$\begin{aligned}\Rightarrow 40(x + 3) = 50x \\[1em] \Rightarrow 40x + 120 = 50x \\[1em] \Rightarrow 120 = 50x - 40x \\[1em] \Rightarrow 10x = 120 \\[1em] \Rightarrow x = \frac{120}{10} \\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, there were 12 students in the group.

Let the number of deer in the herd be $x$.

Number of deer grazing in the field = $\frac{x}{2}$

Remaining deer = $x - \frac{x}{2} = \frac{x}{2}$

Number of deer playing nearby = $\frac{3}{4} \times \frac{x}{2} = \frac{3x}{8}$

According to the given condition,

⇒ $x - \frac{x}{2} - \frac{3x}{8} = 9$

L.C.M. of 2 and 8 = 8

Multiplying both sides by 8, we get:

$$\begin{aligned}\Rightarrow \Big[x - \frac{x}{2} - \frac{3x}{8}\Big] \times 8 = 9 \times 8 \\[1em] \Rightarrow x \times 8 - \frac{x}{2} \times 8 - \frac{3x}{8} \times 8 = 72 \\[1em] \Rightarrow 8x - 4x - 3x = 72 \\[1em] \Rightarrow x = 72\end{aligned}$$

Hence, the number of deer in the herd = 72.

Let the number of flowers Sakshi had in the beginning be $x$.

At the first temple, she offers $\frac{x}{2}$ flowers, so flowers left = $x - \frac{x}{2} = \frac{x}{2}$

At the second temple, she offers half of $\frac{x}{2}$, so flowers left = $\frac{1}{2} \times \frac{x}{2} = \frac{x}{4}$

At the third temple, she offers half of $\frac{x}{4}$, so flowers left = $\frac{1}{2} \times \frac{x}{4} = \frac{x}{8}$

According to the given condition,

⇒ $\frac{x}{8} = 6$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow x = 6 \times 8 \\[1em] \Rightarrow x = 48\end{aligned}$$

Hence, Sakshi had 48 flowers in the beginning.

Let the smaller angle be $x$°.

As the angles are supplementary, their sum = 180°

So, the greater angle = $(180 - x)$°

According to the given condition,

$$\begin{aligned}\Rightarrow (180 - x)^{\circ} - x^{\circ} = 50^{\circ} \\[1em] \Rightarrow 180 - 2x = 50^{\circ} \\[1em] \Rightarrow 180^{\circ} - 50^{\circ} = (2x)^{\circ} \\[1em] \Rightarrow (2x)^{\circ} = 130^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{130}{2} \\[1em] \Rightarrow x^{\circ} = 65^{\circ}\end{aligned}$$

So, the greater angle = 180° - 65° = 115°

Hence, the two angles are 65° and 115°.

Let the angles of the triangle be $(5x)^{\circ}, (6x)^{\circ}$ and $(7x)^{\circ}$.

We know that,

The sum of the angles of a triangle is 180°.

$$\begin{aligned}\Rightarrow (5x)^{\circ} + (6x)^{\circ} + (7x)^{\circ} = 180^{\circ} \\[1em] \Rightarrow (18x)^{\circ} = 180^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{180}{18} \\[1em] \Rightarrow x^{\circ} = 10^{\circ}\end{aligned}$$

So, the angles are 5 × 10 = 50°, 6 × 10 = 60° and 7 × 10 = 70°.

Hence, the angles of the triangle are 50°, 60° and 70°.

As the triangle is isosceles, its two equal sides are equal in length.

$$\begin{aligned}\Rightarrow 3x - 1 = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 + 1 \\[1em] \Rightarrow x = 3\end{aligned}$$

So, the equal sides = 3 × 3 - 1 = 8 units and 2 × 3 + 2 = 8 units

and the third side = 2 × 3 = 6 units

Perimeter of the triangle = Sum of all the sides

⇒ Perimeter = 8 + 8 + 6 = 22 units

Hence, $\bm{x = 3}$ and the perimeter of the triangle = 22 units.

Let the perimeter of the given triangle be $x$ cm.

As each of the three sides is increased by 4 cm, the perimeter increases by (3 × 4) cm = 12 cm.

So, the perimeter of the new triangle = $(x + 12)$ cm

According to the given condition,

⇒ $\frac{x + 12}{x} = \frac{7}{5}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 5(x + 12) = 7x \\[1em] \Rightarrow 5x + 60 = 7x \\[1em] \Rightarrow 60 = 7x - 5x \\[1em] \Rightarrow 2x = 60 \\[1em] \Rightarrow x = \frac{60}{2} \\[1em] \Rightarrow x = 30\end{aligned}$$

Hence, the perimeter of the given triangle = 30 cm.

Let the breadth of the original rectangle be $x$ cm.

Then, its length = $(2x - 5)$ cm

New length = $(2x - 5 - 3)$ cm = $(2x - 8)$ cm

New breadth = $(x + 2)$ cm

We know that,

Perimeter of a rectangle = 2(length + breadth)

According to the given condition,

$$\begin{aligned}\Rightarrow 2[(2x - 8) + (x + 2)] = 72 \\[1em] \Rightarrow 2(3x - 6) = 72 \\[1em] \Rightarrow 3x - 6 = \frac{72}{2} \\[1em] \Rightarrow 3x - 6 = 36 \\[1em] \Rightarrow 3x = 36 + 6 \\[1em] \Rightarrow 3x = 42 \\[1em] \Rightarrow x = \frac{42}{3} \\[1em] \Rightarrow x = 14\end{aligned}$$

So, breadth of the original rectangle = 14 cm

and length of the original rectangle = 2 × 14 - 5 = 23 cm

Area of the original rectangle = length × breadth

⇒ Area = 23 × 14 = 322 cm²

Hence, the area of the original rectangle = 322 cm².

Length of the given rectangle = 10 cm and breadth = 8 cm

When each side is increased by $x$ cm,

new length = $(10 + x)$ cm and new breadth = $(8 + x)$ cm

We know that,

Perimeter of a rectangle = 2(length + breadth)

As the perimeter of the new rectangle is double the perimeter of the given rectangle,

⇒ $2[(x + 10) + (x + 8)] = 2 \times 2(10 + 8)$

Solving the above equation, we get

$$\begin{aligned}\Rightarrow 2(2x + 18) = 2 \times 36 \\[1em] \Rightarrow 2x + 18 = 36 \\[1em] \Rightarrow 2x = 36 - 18 \\[1em] \Rightarrow 2x = 18 \\[1em] \Rightarrow x = \frac{18}{2} \\[1em] \Rightarrow x = 9\end{aligned}$$

So, new length = 10 + 9 = 19 cm and new breadth = 8 + 9 = 17 cm

Area of the new rectangle = length × breadth

⇒ Area = 19 × 17 = 323 cm²

Hence, the equation is $\bm{2[(x + 10) + (x + 8)] = 2 \times 2(10 + 8)}$ and the area of the new rectangle = 323 cm².

Let the speed of the steamer in still water be $x$ km/h.

Given, speed of the stream = 5 km/h

So, speed of the steamer downstream = $(x + 5)$ km/h

and speed of the steamer upstream = $(x - 5)$ km/h

We know that,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken to travel 90 km downstream = $\frac{90}{x + 5}$ hours

Time taken to travel 60 km upstream = $\frac{60}{x - 5}$ hours

According to the given condition,

⇒ $\frac{90}{x + 5} = \frac{60}{x - 5}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 90(x - 5) = 60(x + 5) \\[1em] \Rightarrow 90x - 450 = 60x + 300 \\[1em] \Rightarrow 90x - 60x = 300 + 450 \\[1em] \Rightarrow 30x = 750 \\[1em] \Rightarrow x = \frac{750}{30} \\[1em] \Rightarrow x = 25\end{aligned}$$

Hence, the speed of the steamer in still water = 25 km/h.

Let the speed of the streamer in still water be $x$ km/h.

Given, speed of the stream = 1 km/h

So, speed of the streamer downstream = $(x + 1)$ km/h

and speed of the streamer upstream = $(x - 1)$ km/h

We know that,

Distance = Speed × Time

Distance covered downstream in 5 hours = $5(x + 1)$ km

Distance covered upstream in 6 hours = $6(x - 1)$ km

As the distance between the two ports is the same,

$$\begin{aligned}\Rightarrow 5(x + 1) = 6(x - 1) \\[1em] \Rightarrow 5x + 5 = 6x - 6 \\[1em] \Rightarrow 5 + 6 = 6x - 5x \\[1em] \Rightarrow x = 11\end{aligned}$$

So, the distance between the two ports = 5(11 + 1) = 5 × 12 = 60 km

Hence, the speed of the streamer in still water = 11 km/h and the distance between the two ports = 60 km.

Let the speed of the faster car be $x$ km/h.

Then, the speed of the other car = $(x - 8)$ km/h

Let the faster car start from place A and the slower car start from place B.

Let P and Q be their positions after 4 hours.

We know that,

Distance = Speed × Time

AP = $4x$ km, BQ = $4(x - 8)$ km and PQ = 62 km

According to the given condition,

$$\begin{aligned}\Rightarrow \text{AP} + \text{PQ} + \text{QB} = 350 \\[1em] \Rightarrow 4x + 62 + 4(x - 8) = 350 \\[1em] \Rightarrow 4x + 62 + 4x - 32 = 350 \\[1em] \Rightarrow 8x + 30 = 350 \\[1em] \Rightarrow 8x = 350 - 30 \\[1em] \Rightarrow 8x = 320 \\[1em] \Rightarrow x = \frac{320}{8} \\[1em] \Rightarrow x = 40\end{aligned}$$

So, the speed of the other car = 40 - 8 = 32 km/h

Hence, the speeds of the two cars are 40 km/h and 32 km/h.

$x \gt - 2$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 2 are - 1, 0, 1 and 3.

Hence, solution set = {-1, 0, 1, 3}.

$x \lt - 2$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are less than - 2 are - 7, - 5 and - 3.

Hence, solution set = {-7, -5, -3}.

$x \gt 2$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The only element of the replacement set which is greater than 2 is 3.

Hence, solution set = {3}.

$- 5 \lt x \le 5$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 5 and less than or equal to 5 are - 3, - 1, 0, 1 and 3.

Hence, solution set = {-3, -1, 0, 1, 3}.

$- 8 \lt x \lt 1$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 8 and less than 1 are - 7, - 5, - 3, - 1 and 0.

Hence, solution set = {-7, -5, -3, -1, 0}.

$0 \le x \le 4$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than or equal to 0 and less than or equal to 4 are 0, 1 and 3.

Hence, solution set = {0, 1, 3}.

$x \le 4$

∵ $x$ is a natural number.

Hence, solution set = {1, 2, 3, 4}.

The solution set is shown by thick dots on the number line.

$x \lt 5$

∵ $x$ is a whole number.

Hence, solution set = {0, 1, 2, 3, 4}.

The solution set is shown by thick dots on the number line.

$- 3 \le x \lt 3$

∵ $x$ is an integer.

Hence, solution set = {-3, -2, -1, 0, 1, 2}.

The solution set is shown by thick dots on the number line.

$- 4 \le x \lt 4$

∵ The replacement set = {-6, -4, -2, 0, 2, 4, 6}.

The elements of the replacement set which are greater than or equal to - 4 and less than 4 are - 4, - 2, 0 and 2.

Hence, solution set = {-4, -2, 0, 2}.

The solution set is shown by thick dots on the number line.

$x \lt 4$

∵ The replacement set = {1, 2, 3, ..., 10}.

The elements of the replacement set which are less than 4 are 1, 2 and 3.

Hence, solution set = {1, 2, 3}.

$x \lt 4$

∵ The replacement set = {-1, 0, 1, 2, 5, 8}.

The elements of the replacement set which are less than 4 are - 1, 0, 1 and 2.

Hence, solution set = {-1, 0, 1, 2}.

$x \lt 4$

∵ The replacement set = {-5, 10}.

The only element of the replacement set which is less than 4 is - 5.

Hence, solution set = {-5}.

$x \lt 4$

∵ The replacement set = {5, 6, 7, 8, 9, 10}.

No element of the replacement set is less than 4.

Hence, solution set = φ.

Solving,

$$\begin{aligned}\Rightarrow 2x - 3 \gt 7 \\[1em] \Rightarrow 2x \gt 7 + 3 \\[1em] \Rightarrow 2x \gt 10 \\[1em] \Rightarrow x \gt \frac{10}{2} \\[1em] \Rightarrow x \gt 5\end{aligned}$$

∵ The replacement set = {-6, -3, 0, 3, 6, 9, 12}.

Hence, truth set = {6, 9, 12}.

Solving,

$$\begin{aligned}\Rightarrow 3x + 8 \le 2 \\[1em] \Rightarrow 3x \le 2 - 8 \\[1em] \Rightarrow 3x \le - 6 \\[1em] \Rightarrow x \le - \frac{6}{3} \\[1em] \Rightarrow x \le - 2\end{aligned}$$

∵ The replacement set = {-6, -3, 0, 3, 6, 9, 12}.

Hence, truth set = {-6, -3}.

Solving,

$$\begin{aligned}\Rightarrow - 3 \lt 1 - 2x \\[1em] \Rightarrow 2x \lt 1 + 3 \\[1em] \Rightarrow 2x \lt 4 \\[1em] \Rightarrow x \lt \frac{4}{2} \\[1em] \Rightarrow x \lt 2\end{aligned}$$

∵ The replacement set = {-6, -3, 0, 3, 6, 9, 12}.

Hence, truth set = {-6, -3, 0}.

Solving,

$$\begin{aligned}\Rightarrow 4x + 1 \lt 17 \\[1em] \Rightarrow 4x \lt 17 - 1 \\[1em] \Rightarrow 4x \lt 16 \\[1em] \Rightarrow x \lt \frac{16}{4} \\[1em] \Rightarrow x \lt 4\end{aligned}$$

∵ $x$ is a natural number.

Hence, solution set = {1, 2, 3}.

Solving,

$$\begin{aligned}\Rightarrow 4x + 1 \le 17 \\[1em] \Rightarrow 4x \le 17 - 1 \\[1em] \Rightarrow 4x \le 16 \\[1em] \Rightarrow x \le \frac{16}{4} \\[1em] \Rightarrow x \le 4\end{aligned}$$

∵ $x$ is a whole number.

Hence, solution set = {0, 1, 2, 3, 4}.

Solving,

$$\begin{aligned}\Rightarrow 4 \gt 3x - 11 \\[1em] \Rightarrow 4 + 11 \gt 3x \\[1em] \Rightarrow 15 \gt 3x \\[1em] \Rightarrow x \lt \frac{15}{3} \\[1em] \Rightarrow x \lt 5\end{aligned}$$

∵ $x$ is a natural number.

Hence, solution set = {1, 2, 3, 4}.

Solving,

$$\begin{aligned}\Rightarrow - 17 \le 9x - 8 \\[1em] \Rightarrow - 17 + 8 \le 9x \\[1em] \Rightarrow - 9 \le 9x \\[1em] \Rightarrow x \ge - \frac{9}{9} \\[1em] \Rightarrow x \ge - 1\end{aligned}$$

∵ $x$ is an integer.

Hence, solution set = {-1, 0, 1, 2, .........}.

$\frac{2y - 1}{5} \le 2$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 2y - 1 \le 2 \times 5 \\[1em] \Rightarrow 2y - 1 \le 10 \\[1em] \Rightarrow 2y \le 10 + 1 \\[1em] \Rightarrow 2y \le 11 \\[1em] \Rightarrow y \le \frac{11}{2} \\[1em] \Rightarrow y \le 5.5\end{aligned}$$

∵ $y$ is a natural number.

Hence, solution set = {1, 2, 3, 4, 5}.

Solving,

$$\begin{aligned}\Rightarrow \frac{2y + 1}{3} + 1 \le 3 \\[1em] \Rightarrow \frac{2y + 1}{3} \le 3 - 1 \\[1em] \Rightarrow \frac{2y + 1}{3} \le 2\end{aligned}$$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 2y + 1 \le 2 \times 3 \\[1em] \Rightarrow 2y + 1 \le 6 \\[1em] \Rightarrow 2y \le 6 - 1 \\[1em] \Rightarrow 2y \le 5 \\[1em] \Rightarrow y \le \frac{5}{2} \\[1em] \Rightarrow y \le 2.5\end{aligned}$$

∵ $y$ is a whole number.

Hence, solution set = {0, 1, 2}.

Solving,

$$\begin{aligned}\Rightarrow \frac{2}{3}p + 5 \lt 9 \\[1em] \Rightarrow \frac{2}{3}p \lt 9 - 5 \\[1em] \Rightarrow \frac{2}{3}p \lt 4 \\[1em] \Rightarrow p \lt \frac{4 \times 3}{2} \\[1em] \Rightarrow p \lt \frac{12}{2} \\[1em] \Rightarrow p \lt 6\end{aligned}$$

∵ $p$ is a whole number.

Hence, solution set = {0, 1, 2, 3, 4, 5}.

Solving,

$$\begin{aligned}\Rightarrow - 2(p + 3) \gt 5 \\[1em] \Rightarrow - 2p - 6 \gt 5 \\[1em] \Rightarrow - 2p \gt 5 + 6 \\[1em] \Rightarrow - 2p \gt 11 \\[1em] \Rightarrow 2p \lt - 11 \\[1em] \Rightarrow p \lt - \frac{11}{2} \\[1em] \Rightarrow p \lt - 5.5\end{aligned}$$

∵ $p$ is an integer.

Hence, solution set = {........., -8, -7, -6}.

Solving,

$$\begin{aligned}\Rightarrow 2x - 3 \lt x + 2 \\[1em] \Rightarrow 2x - x \lt 2 + 3 \\[1em] \Rightarrow x \lt 5\end{aligned}$$

∵ $x$ is a natural number.

Hence, solution set = {1, 2, 3, 4}.

Solving,

$$\begin{aligned}\Rightarrow 3 - x \le 5 - 3x \\[1em] \Rightarrow 3x - x \le 5 - 3 \\[1em] \Rightarrow 2x \le 2 \\[1em] \Rightarrow x \le \frac{2}{2} \\[1em] \Rightarrow x \le 1\end{aligned}$$

∵ $x$ is a whole number.

Hence, solution set = {0, 1}.

Solving,

$$\begin{aligned}\Rightarrow 3(x - 2) \lt 2(x - 1) \\[1em] \Rightarrow 3x - 6 \lt 2x - 2 \\[1em] \Rightarrow 3x - 2x \lt - 2 + 6 \\[1em] \Rightarrow x \lt 4\end{aligned}$$

∵ $x$ is a whole number.

Hence, solution set = {0, 1, 2, 3}.

$\frac{3}{2} - \frac{x}{2} \gt -1$

Multiplying both sides by 2, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{3}{2} - \frac{x}{2}\Big] \times 2 \gt -1 \times 2 \\[1em] \Rightarrow \frac{3}{2} \times 2 - \frac{x}{2} \times 2 \gt - 2 \\[1em] \Rightarrow 3 - x \gt - 2 \\[1em] \Rightarrow - x \gt - 2 - 3 \\[1em] \Rightarrow - x \gt - 5 \\[1em] \Rightarrow x \lt 5\end{aligned}$$

∵ $x$ is a natural number.

Hence, solution set = {1, 2, 3, 4}.

$\frac{3x - 1}{2} \lt 2$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 3x - 1 \lt 2 \times 2 \\[1em] \Rightarrow 3x - 1 \lt 4 \\[1em] \Rightarrow 3x \lt 4 + 1 \\[1em] \Rightarrow 3x \lt 5 \\[1em] \Rightarrow x \lt \frac{5}{3} \\[1em] \Rightarrow x \lt 1.66.....\end{aligned}$$

∵ The replacement set = {-3, -2, -1, 0, 1, 2, 3}.

Hence, solution set = {-3, -2, -1, 0, 1}.

The solution set is shown by thick dots on the number line.

$\frac{x}{3} + \frac{1}{4} \lt \frac{x}{6} + \frac{1}{2}$

L.C.M. of 3, 4, 6 and 2 = 12

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{x}{3} + \frac{1}{4}\Big] \times 12 \lt \Big[\frac{x}{6} + \frac{1}{2}\Big] \times 12 \\[1em] \Rightarrow \frac{x}{3} \times 12 + \frac{1}{4} \times 12 \lt \frac{x}{6} \times 12 + \frac{1}{2} \times 12 \\[1em] \Rightarrow 4x + 3 \lt 2x + 6 \\[1em] \Rightarrow 4x - 2x \lt 6 - 3 \\[1em] \Rightarrow 2x \lt 3 \\[1em] \Rightarrow x \lt \frac{3}{2} \\[1em] \Rightarrow x \lt 1.5\end{aligned}$$

∵ $x$ is a whole number.

Hence, solution set = {0, 1}.

The solution set is shown by thick dots on the number line.

$- 4 \le 4x \lt 14$

Solving L.H.S., we get:

$$\begin{aligned}\Rightarrow - 4 \le 4x \\[1em] \Rightarrow x \ge - \frac{4}{4} \\[1em] \Rightarrow x \ge - 1 \quad \text{...(1)}\end{aligned}$$

Solving R.H.S., we get:

$$\begin{aligned}\Rightarrow 4x \lt 14 \\[1em] \Rightarrow x \lt \frac{14}{4} \\[1em] \Rightarrow x \lt 3.5 \quad \text{...(2)}\end{aligned}$$

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

⇒ $- 1 \le x \lt 3.5$

∵ $x$ is a natural number.

Hence, solution set = {1, 2, 3}.

The solution set is shown by thick dots on the number line.

$-1 \lt \frac{x}{2} + 1 \le 3$

Solving L.H.S., we get:

$$\begin{aligned}\Rightarrow - 1 \lt \frac{x}{2} + 1 \\[1em] \Rightarrow - 1 - 1 \lt \frac{x}{2} \\[1em] \Rightarrow - 2 \lt \frac{x}{2} \\[1em] \Rightarrow - 4 \lt x \quad \text{...(1)}\end{aligned}$$

Solving R.H.S., we get:

$$\begin{aligned}\Rightarrow \frac{x}{2} + 1 \le 3 \\[1em] \Rightarrow \frac{x}{2} \le 3 - 1 \\[1em] \Rightarrow \frac{x}{2} \le 2 \\[1em] \Rightarrow x \le 4 \quad \text{...(2)}\end{aligned}$$

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

⇒ $- 4 \lt x \le 4$

∵ $x$ is an integer.

Hence, solution set = {-3, -2, -1, 0, 1, 2, 3, 4}.

The solution set is shown by thick dots on the number line.

(i) An equation of the type $ax + b = 0$ where $a \ne 0$ is called a linear equation in variable $x$.

(ii) Any value of the variable which satisfies the equation is called a solution of the equation.

(iii) The process of finding all the solutions of an equation is called solving the equation.

(iv) We can add the same number to both sides of an equation.

(v) We can divide both sides of an equation by the same non-zero number.

(vi) $3x \le 10$

$$\begin{aligned}\Rightarrow x \le \frac{10}{3} \\[1em] \Rightarrow x \le 3.33.....\end{aligned}$$

∵ $x$ is a natural number.

The solution set of the inequality $3x \le 10, x \in N$ is {1, 2, 3}.

(i) An equation is a statement that two expressions are equal.

This statement is True.

(ii) We cannot subtract the same number from both sides of an equation.

We can subtract the same number from both sides of an equation.

So, this statement is False.

(iii) $3x + 2 = 4(x + 7) + 9$ is a linear equation in variable $x$

On simplifying, we get

$$\begin{aligned}\Rightarrow 3x + 2 = 4x + 28 + 9 \\[1em] \Rightarrow 3x + 2 = 4x + 37 \\[1em] \Rightarrow - x - 35 = 0\end{aligned}$$

As it is of the type $ax + b = 0$, where $a \ne 0$, it is a linear equation in variable $x$.

So, this statement is True.

(iv) $x = 1$ is the solution of equation $4(x + 5) = 24$

Substituting $x = 1$ in L.H.S., we get

⇒ L.H.S. = 4(1 + 5) = 4 × 6 = 24 = R.H.S.

So, this statement is True.

We know that,

A linear equation in one variable is of the type $ax + b = 0$, where $a \ne 0$ and it contains only one variable.

The equation $x + 2y = 7$ contains two variables $x$ and $y$.

So, it is not a linear equation in one variable.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{2}{3}x + 1 = \frac{15}{9} \\[1em] \Rightarrow \frac{2}{3}x = \frac{15}{9} - 1 \\[1em] \Rightarrow \frac{2}{3}x = \frac{15 - 9}{9} \\[1em] \Rightarrow \frac{2}{3}x = \frac{6}{9} \\[1em] \Rightarrow \frac{2}{3}x = \frac{2}{3} \\[1em] \Rightarrow x = 1\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 4z + 3 = 6 + 2z \\[1em] \Rightarrow 4z - 2z = 6 - 3 \\[1em] \Rightarrow 2z = 3 \\[1em] \Rightarrow z = \frac{3}{2}\end{aligned}$$

Hence, option 2 is the correct option.

$\frac{3x}{5} + 1 = \frac{4x}{15} + 7$

L.C.M. of 5 and 15 = 15

Multiplying both sides by 15, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{3x}{5} + 1\Big] \times 15 = \Big[\frac{4x}{15} + 7\Big] \times 15 \\[1em] \Rightarrow \frac{3x}{5} \times 15 + 1 \times 15 = \frac{4x}{15} \times 15 + 7 \times 15 \\[1em] \Rightarrow 9x + 15 = 4x + 105 \\[1em] \Rightarrow 9x - 4x = 105 - 15 \\[1em] \Rightarrow 5x = 90 \\[1em] \Rightarrow x = \frac{90}{5} \\[1em] \Rightarrow x = 18\end{aligned}$$

Hence, option 4 is the correct option.

$\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$

L.C.M. of 2, 5, 3 and 4 = 60

Multiplying both sides by 60, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{x}{2} - \frac{1}{5}\Big] \times 60 = \Big[\frac{x}{3} + \frac{1}{4}\Big] \times 60 \\[1em] \Rightarrow \frac{x}{2} \times 60 - \frac{1}{5} \times 60 = \frac{x}{3} \times 60 + \frac{1}{4} \times 60 \\[1em] \Rightarrow 30x - 12 = 20x + 15 \\[1em] \Rightarrow 30x - 20x = 15 + 12 \\[1em] \Rightarrow 10x = 27 \\[1em] \Rightarrow x = \frac{27}{10} \\[1em] \Rightarrow x = 2.7\end{aligned}$$

Hence, option 1 is the correct option.

$\frac{8x - 3}{3x} = 2$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 8x - 3 = 2 \times 3x \\[1em] \Rightarrow 8x - 3 = 6x \\[1em] \Rightarrow 8x - 6x = 3 \\[1em] \Rightarrow 2x = 3 \\[1em] \Rightarrow x = \frac{3}{2}\end{aligned}$$

Hence, option 3 is the correct option.

Let the number be $x$.

According to the given condition,

$$\begin{aligned}\Rightarrow \left(x - \frac{1}{2}\right) \times \frac{1}{2} = \frac{1}{8} \\[1em] \Rightarrow x - \frac{1}{2} = \frac{1}{8} \times 2 \\[1em] \Rightarrow x - \frac{1}{2} = \frac{1}{4} \\[1em] \Rightarrow x = \frac{1}{4} + \frac{1}{2} \\[1em] \Rightarrow x = \frac{1 + 2}{4} \\[1em] \Rightarrow x = \frac{3}{4}\end{aligned}$$

Hence, option 2 is the correct option.

Let Ravi's present age be $x$ years.

Fifteen years from now, Ravi's age = $(x + 15)$ years

According to the given condition,

$$\begin{aligned}\Rightarrow x + 15 = 4x \\[1em] \Rightarrow 15 = 4x - x \\[1em] \Rightarrow 3x = 15 \\[1em] \Rightarrow x = \frac{15}{3} \\[1em] \Rightarrow x = 5\end{aligned}$$

Hence, option 2 is the correct option.

Let the three consecutive integers be $x, x + 1$ and $x + 2$.

Given,

Sum of three consecutive integers is 51.

$$\begin{aligned}\Rightarrow x + (x + 1) + (x + 2) = 51 \\[1em] \Rightarrow 3x + 3 = 51 \\[1em] \Rightarrow 3x = 51 - 3 \\[1em] \Rightarrow 3x = 48 \\[1em] \Rightarrow x = \frac{48}{3} \\[1em] \Rightarrow x = 16\end{aligned}$$

So, the largest integer = 16 + 2 = 18

Hence, option 3 is the correct option.

Let the length of the rectangle be $x$ cm.

Given, width = $2\frac{3}{4}$ cm = $\frac{11}{4}$ cm

We know that,

Perimeter of a rectangle = 2(length + width)

According to the given condition,

$$\begin{aligned}\Rightarrow 2\left(x + \frac{11}{4}\right) = 13 \\[1em] \Rightarrow x + \frac{11}{4} = \frac{13}{2} \\[1em] \Rightarrow x = \frac{13}{2} - \frac{11}{4} \\[1em] \Rightarrow x = \frac{26 - 11}{4} \\[1em] \Rightarrow x = \frac{15}{4} = 3\frac{3}{4}\end{aligned}$$

Hence, option 2 is the correct option.

Let the required number be $x$.

Twice the rational number $\frac{-7}{3} = 2 \times \frac{-7}{3} = \frac{-14}{3}$

According to the given condition,

$$\begin{aligned}\Rightarrow x + \left(\frac{-14}{3}\right) = \frac{3}{7} \\[1em] \Rightarrow x = \frac{3}{7} + \frac{14}{3} \\[1em] \Rightarrow x = \frac{9 + 98}{21} \\[1em] \Rightarrow x = \frac{107}{21}\end{aligned}$$

Hence, option 4 is the correct option.

Let the digit at unit's place be $x$ and the digit at ten's place be $y$.

So, the original number = $10y + x$

On reversing the digits, the new number = $10x + y$

Given, sum of the digits = 8

⇒ $x + y = 8 \quad \text{...(1)}$

According to the given condition,

$$\begin{aligned}\Rightarrow (10x + y) - (10y + x) = 18 \\[1em] \Rightarrow 9x - 9y = 18 \\[1em] \Rightarrow x - y = \frac{18}{9} \\[1em] \Rightarrow x - y = 2 \quad \text{...(2)}\end{aligned}$$

Adding equations (1) and (2), we get

$$\begin{aligned}\Rightarrow 2x = 10 \\[1em] \Rightarrow x = 5\end{aligned}$$

Substituting $x = 5$ in equation (1), we get $y = 8 - 5 = 3$

So, the original number = 10 × 3 + 5 = 35

Hence, option 1 is the correct option.

Let Shriya's present age be $x$ years.

Then, Arjun's present age = $2x$ years

Five years ago, Shriya's age = $(x - 5)$ years and Arjun's age = $(2x - 5)$ years

According to the given condition,

$$\begin{aligned}\Rightarrow 2x - 5 = 3(x - 5) \\[1em] \Rightarrow 2x - 5 = 3x - 15 \\[1em] \Rightarrow 15 - 5 = 3x - 2x \\[1em] \Rightarrow x = 10\end{aligned}$$

So, Arjun's present age = 2 × 10 = 20 years

Hence, option 3 is the correct option.

$- 3 \lt x \lt 3$

∵ The replacement set = {-5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 3 and less than 3 are - 1, 0 and 1.

So, solution set = {-1, 0, 1}

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 4y - 3 = 7y + 2 \\[1em] \Rightarrow - 3 - 2 = 7y - 4y \\[1em] \Rightarrow 3y = - 5 \\[1em] \Rightarrow y = - \frac{5}{3}\end{aligned}$$

As $y = -\frac{5}{3}$ and not -1, Statement I is false.

We can add or subtract the same expression to both sides of a linear equation, so Statement II is true.

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 2(x - 3) = 10 \\[1em] \Rightarrow 2x - 6 = 10 \\[1em] \Rightarrow 2x = 10 + 6 \\[1em] \Rightarrow 2x = 16 \\[1em] \Rightarrow x = \frac{16}{2} \\[1em] \Rightarrow x = 8\end{aligned}$$

As $x = 8$, Statement I is true.

We can multiply or divide both sides of a linear equation by any non-zero expression, so Statement II is true.

Hence, option 3 is the correct option.

Given, cost price of the table = ₹ 1,000

Profit required = atleast 20% of the cost price

⇒ Profit = $\frac{20}{100} \times 1,000$ = ₹ 200

So, selling price = 1,000 + 200 = ₹ 1,200

As the shopkeeper must sell the table for atleast ₹ 1,200 and not for greater than ₹ 1,100, Statement I is false.

Profit % is always calculated on the cost price, so Statement II is true.

Hence, option 2 is the correct option.

$\frac{3}{5}x + 2 = \frac{2}{3}x$

L.C.M. of 5 and 3 = 15

Multiplying both sides by 15, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{3}{5}x + 2\Big] \times 15 = \frac{2}{3}x \times 15 \\[1em] \Rightarrow \frac{3}{5}x \times 15 + 2 \times 15 = 10x \\[1em] \Rightarrow 9x + 30 = 10x \\[1em] \Rightarrow 30 = 10x - 9x \\[1em] \Rightarrow x = 30\end{aligned}$$

Hence, $\bm{x = 30}$.

$\frac{x}{2} - \frac{3x}{4} + \frac{x}{6} = \frac{-2}{3}$

L.C.M. of 2, 4, 6 and 3 = 12

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{x}{2} - \frac{3x}{4} + \frac{x}{6}\Big] \times 12 = \frac{-2}{3} \times 12 \\[1em] \Rightarrow \frac{x}{2} \times 12 - \frac{3x}{4} \times 12 + \frac{x}{6} \times 12 = - 8 \\[1em] \Rightarrow 6x - 9x + 2x = - 8 \\[1em] \Rightarrow - x = - 8 \\[1em] \Rightarrow x = 8\end{aligned}$$

Hence, $\bm{x = 8}$.

Solving,

$$\begin{aligned}\Rightarrow 5x - 3(4x - 3) = 2(4 - 5x) + 10 \\[1em] \Rightarrow 5x - 12x + 9 = 8 - 10x + 10 \\[1em] \Rightarrow - 7x + 9 = 18 - 10x \\[1em] \Rightarrow - 7x + 10x = 18 - 9 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = \frac{9}{3} \\[1em] \Rightarrow x = 3\end{aligned}$$

Hence, $\bm{x = 3}$.

$\frac{3x + 2}{5x + 4} = \frac{3}{4}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 4(3x + 2) = 3(5x + 4) \\[1em] \Rightarrow 12x + 8 = 15x + 12 \\[1em] \Rightarrow 8 - 12 = 15x - 12x \\[1em] \Rightarrow 3x = - 4 \\[1em] \Rightarrow x = - \frac{4}{3}\end{aligned}$$

Hence, $\bm{x = -\frac{4}{3}}$.

$\frac{2x - 3}{5x - 4} = \frac{2x + 1}{5x + 20}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow (2x - 3)(5x + 20) = (2x + 1)(5x - 4) \\[1em] \Rightarrow 10x^{2} + 40x - 15x - 60 = 10x^{2} - 8x + 5x - 4 \\[1em] \Rightarrow 10x^{2} + 25x - 60 = 10x^{2} - 3x - 4 \\[1em] \Rightarrow 25x + 3x = - 4 + 60 \\[1em] \Rightarrow 28x = 56 \\[1em] \Rightarrow x = \frac{56}{28} \\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, $\bm{x = 2}$.

$\frac{2x + 5}{2} - \frac{5x}{x - 1} = x$

L.C.M. of 2 and $(x - 1)$ = $2(x - 1)$

Multiplying both sides by $2(x - 1)$, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{(2x + 5)}{2} - \frac{5x}{(x - 1)}\Big] \times 2(x - 1) = x \times 2(x - 1) \\[1em] \Rightarrow \frac{(2x + 5)}{2} \times 2(x - 1) - \frac{5x}{(x - 1)} \times 2(x - 1) = 2x(x - 1) \\[1em] \Rightarrow (x - 1)(2x + 5) - 10x = 2x(x - 1) \\[1em] \Rightarrow 2x^{2} + 5x - 2x - 5 - 10x = 2x^{2} - 2x \\[1em] \Rightarrow 2x^{2} - 7x - 5 = 2x^{2} - 2x \\[1em] \Rightarrow - 7x + 2x = 5 \\[1em] \Rightarrow - 5x = 5 \\[1em] \Rightarrow x = - \frac{5}{5} \\[1em] \Rightarrow x = - 1\end{aligned}$$

Hence, $\bm{x = -1}$.

Let the three consecutive multiples of 11 be $11x, 11x + 11$ and $11x + 22$.

According to the given condition,

$$\begin{aligned}\Rightarrow 11x + (11x + 11) + (11x + 22) = 363 \\[1em] \Rightarrow 33x + 33 = 363 \\[1em] \Rightarrow 33x = 363 - 33 \\[1em] \Rightarrow 33x = 330 \\[1em] \Rightarrow x = \frac{330}{33} \\[1em] \Rightarrow x = 10\end{aligned}$$

So, the multiples are 11 × 10 = 110, 110 + 11 = 121 and 110 + 22 = 132.

Hence, the required multiples are 110, 121 and 132.

Let the smaller number be $x$.

As one number exceeds the other by 15, the greater number = $x + 15$

According to the given condition,

$$\begin{aligned}\Rightarrow x + (x + 15) = 95 \\[1em] \Rightarrow 2x + 15 = 95 \\[1em] \Rightarrow 2x = 95 - 15 \\[1em] \Rightarrow 2x = 80 \\[1em] \Rightarrow x = \frac{80}{2} \\[1em] \Rightarrow x = 40\end{aligned}$$

So, the greater number = 40 + 15 = 55

Hence, the required numbers are 55 and 40.

Let the first number be $x$.

Then, its succeeding number = $x + 1$

According to the given condition,

⇒ $\frac{1}{2}x = \frac{1}{3}(x + 1)$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 3x = 2(x + 1) \\[1em] \Rightarrow 3x = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 \\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, the first number = 2.

Let the denominator of the rational number be $x$.

Then, its numerator = $x - 8$

So, the rational number = $\frac{x - 8}{x}$

On increasing the numerator by 2 and decreasing the denominator by 1, we get $\frac{x - 8 + 2}{x - 1} = \frac{x - 6}{x - 1}$

According to the given condition,

⇒ $\frac{x - 6}{x - 1} = \frac{1}{2}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 2(x - 6) = x - 1 \\[1em] \Rightarrow 2x - 12 = x - 1 \\[1em] \Rightarrow 2x - x = - 1 + 12 \\[1em] \Rightarrow x = 11\end{aligned}$$

So, the numerator = 11 - 8 = 3

Hence, the required rational number = $\mathbf{\frac{3}{11}}$.

Let the present ages of Rohit and Mayank be $11x$ years and $8x$ years respectively.

After 8 years, Rohit's age = $(11x + 8)$ years and Mayank's age = $(8x + 8)$ years

According to the given condition,

$$\begin{aligned}\Rightarrow (11x + 8) + (8x + 8) = 54 \\[1em] \Rightarrow 19x + 16 = 54 \\[1em] \Rightarrow 19x = 54 - 16 \\[1em] \Rightarrow 19x = 38 \\[1em] \Rightarrow x = \frac{38}{19} \\[1em] \Rightarrow x = 2\end{aligned}$$

So, Rohit's present age = 11 × 2 = 22 years

and Mayank's present age = 8 × 2 = 16 years

Hence, the present ages of Rohit and Mayank are 22 years and 16 years respectively.

Let the sum of the present ages of the two sons be $x$ years.

Then, the present age of the father = $3x$ years

Five years later, each son will be 5 years older, so the sum of the ages of the two sons = $(x + 10)$ years

and the age of the father = $(3x + 5)$ years

According to the given condition,

$$\begin{aligned}\Rightarrow 3x + 5 = 2(x + 10) \\[1em] \Rightarrow 3x + 5 = 2x + 20 \\[1em] \Rightarrow 3x - 2x = 20 - 5 \\[1em] \Rightarrow x = 15\end{aligned}$$

So, the present age of the father = 3 × 15 = 45 years

Hence, the present age of the father = 45 years.

Let the digit at unit's place be $x$ and the digit at ten's place be $y$.

So, the original number = $10y + x$

On interchanging the digits, the new number = $10x + y$

According to the given condition,

$$\begin{aligned}\Rightarrow (10y + x) + (10x + y) = 121 \\[1em] \Rightarrow 11x + 11y = 121 \\[1em] \Rightarrow 11(x + y) = 121 \\[1em] \Rightarrow x + y = \frac{121}{11} \\[1em] \Rightarrow x + y = 11 \quad \text{...(1)}\end{aligned}$$

Also given, the digits differ by 7.

Case 1 : Let $x - y = 7$ ...(2)

Adding equations (1) and (2), we get

$$\begin{aligned}\Rightarrow 2x = 18 \\[1em] \Rightarrow x = 9\end{aligned}$$

Substituting $x = 9$ in equation (1), we get $y = 11 - 9 = 2$

So, the number = 10 × 2 + 9 = 29

Case 2 : Let $y - x = 7$ ...(3)

Adding equations (1) and (3), we get

$$\begin{aligned}\Rightarrow 2y = 18 \\[1em] \Rightarrow y = 9\end{aligned}$$

Substituting $y = 9$ in equation (1), we get $x = 11 - 9 = 2$

So, the number = 10 × 9 + 2 = 92

Hence, the original number is 29 or 92.

Let the digit at unit's place be $x$.

As the ten's digit exceeds the unit's digit by 5, the digit at ten's place = $x + 5$

So, the original number = $10(x + 5) + x$

On reversing the digits, the new number = $10x + (x + 5)$

According to the given condition,

$$\begin{aligned}\Rightarrow [10(x + 5) + x] + [10x + (x + 5)] = 99 \\[1em] \Rightarrow 10x + 50 + x + 10x + x + 5 = 99 \\[1em] \Rightarrow 22x + 55 = 99 \\[1em] \Rightarrow 22x = 99 - 55 \\[1em] \Rightarrow 22x = 44 \\[1em] \Rightarrow x = \frac{44}{22} \\[1em] \Rightarrow x = 2\end{aligned}$$

So, the digit at ten's place = 2 + 5 = 7

and the original number = 10 × 7 + 2 = 72

Hence, the original number = 72.

Let the number of ₹ 500 notes be $x$.

As she got 250 currency notes in all, the number of ₹ 200 notes = $250 - x$

Value of ₹ 500 notes = ₹ $500x$

Value of ₹ 200 notes = ₹ $200(250 - x)$

According to the given condition,

$$\begin{aligned}\Rightarrow 500x + 200(250 - x) = 80000 \\[1em] \Rightarrow 500x + 50000 - 200x = 80000 \\[1em] \Rightarrow 300x = 80000 - 50000 \\[1em] \Rightarrow 300x = 30000 \\[1em] \Rightarrow x = \frac{30000}{300} \\[1em] \Rightarrow x = 100\end{aligned}$$

So, the number of ₹ 200 notes = 250 - 100 = 150

Hence, Sonia got 100 notes of ₹ 500 and 150 notes of ₹ 200.

Let the distance covered at the rate of 60 km/h be $x$ km.

Then, the distance covered at the rate of 45 km/h = $(240 - x)$ km

We know that,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken to cover $(240 - x)$ km at 45 km/h = $\frac{240 - x}{45}$ hours

Time taken to cover $x$ km at 60 km/h = $\frac{x}{60}$ hours

Total time = $4\frac{1}{4}$ hours = $\frac{17}{4}$ hours

According to the given condition,

⇒ $\frac{240 - x}{45} + \frac{x}{60} = \frac{17}{4}$

L.C.M. of 45, 60 and 4 = 180

Multiplying both sides by 180, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{(240 - x)}{45} + \frac{x}{60}\Big] \times 180 = \frac{17}{4} \times 180 \\[1em] \Rightarrow \frac{(240 - x)}{45} \times 180 + \frac{x}{60} \times 180 = 765 \\[1em] \Rightarrow 4(240 - x) + 3x = 765 \\[1em] \Rightarrow 960 - 4x + 3x = 765 \\[1em] \Rightarrow 960 - x = 765 \\[1em] \Rightarrow x = 960 - 765 \\[1em] \Rightarrow x = 195\end{aligned}$$

Hence, the distance covered at the rate of 60 km/h = 195 km.

$- 5 \le x \lt 5$

∵ $x$ is an even integer.

The even integers which are greater than or equal to - 5 and less than 5 are - 4, - 2, 0, 2 and 4.

Hence, solution set = {-4, -2, 0, 2, 4}.

The solution set is shown by thick dots on the number line.

$-\frac{1}{4} \le \frac{1}{2} - \frac{x}{3} \lt 2$

Subtracting $\frac{1}{2}$ from each term, we get

$$\begin{aligned}\Rightarrow -\frac{1}{4} - \frac{1}{2} \le - \frac{x}{3} \lt 2 - \frac{1}{2} \\[1em] \Rightarrow \frac{-1 - 2}{4} \le - \frac{x}{3} \lt \frac{4 - 1}{2} \\[1em] \Rightarrow -\frac{3}{4} \le - \frac{x}{3} \lt \frac{3}{2}\end{aligned}$$

Multiplying each term by -3 and reversing the symbol of inequality, we get

$$\begin{aligned}\Rightarrow \frac{9}{4} \ge x \gt - \frac{9}{2} \\[1em] \Rightarrow - \frac{9}{2} \lt x \le \frac{9}{4} \\[1em] \Rightarrow - 4.5 \lt x \le 2.25\end{aligned}$$

∵ $x$ is an integer.

Hence, solution set = {-4, -3, -2, -1, 0, 1, 2}.

The solution set is shown by thick dots on the number line.

Let the distance covered on foot be $x$ km.

Then, the distance covered on bicycle = $(24 - x)$ km

We know that,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken to cover $x$ km on foot = $\frac{x}{4.5}$ hours = $\frac{2x}{9}$ hours

Time taken to cover $(24 - x)$ km on bicycle = $\frac{24 - x}{10}$ hours

Total time = $3\frac{1}{2}$ hours = $\frac{7}{2}$ hours

According to the given condition,

⇒ $\frac{2x}{9} + \frac{24 - x}{10} = \frac{7}{2}$

L.C.M. of 9, 10 and 2 = 90

Multiplying both sides by 90, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{2x}{9} + \frac{(24 - x)}{10}\Big] \times 90 = \frac{7}{2} \times 90 \\[1em] \Rightarrow \frac{2x}{9} \times 90 + \frac{(24 - x)}{10} \times 90 = 315 \\[1em] \Rightarrow 20x + 9(24 - x) = 315 \\[1em] \Rightarrow 20x + 216 - 9x = 315 \\[1em] \Rightarrow 11x = 315 - 216 \\[1em] \Rightarrow 11x = 99 \\[1em] \Rightarrow x = \frac{99}{11} \\[1em] \Rightarrow x = 9\end{aligned}$$

Hence, the distance covered on foot = 9 km.

Let the quantity of pure alcohol to be mixed be $x$ mL.

Quantity of alcohol in 400 mL of 15% alcohol solution = $\frac{15}{100} \times 400 = 60$ mL

On mixing $x$ mL of pure alcohol,

quantity of alcohol in the new solution = $(60 + x)$ mL

and quantity of the new solution = $(400 + x)$ mL

As the new solution is a 32% alcohol solution,

⇒ $\frac{60 + x}{400 + x} = \frac{32}{100}$

By cross multiplication, we get

$$\begin{aligned}\Rightarrow 100(60 + x) = 32(400 + x) \\[1em] \Rightarrow 6000 + 100x = 12800 + 32x \\[1em] \Rightarrow 100x - 32x = 12800 - 6000 \\[1em] \Rightarrow 68x = 6800 \\[1em] \Rightarrow x = \frac{6800}{68} \\[1em] \Rightarrow x = 100\end{aligned}$$

Hence, 100 mL of pure alcohol should be mixed.

Let the distance between P and Q be $x$ km.

We know that,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken by Rahul = $\frac{x}{10} + \frac{x}{9}$ hours

Time taken by Anuj = $\frac{x}{12} + \frac{x}{12} = \frac{x}{6}$ hours

Given, Anuj took 10 minutes less than Rahul.

10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours

According to the given condition,

⇒ $\frac{x}{10} + \frac{x}{9} - \frac{x}{6} = \frac{1}{6}$

L.C.M. of 10, 9 and 6 = 90

Multiplying both sides by 90, we get:

$$\begin{aligned}\Rightarrow \Big[\frac{x}{10} + \frac{x}{9} - \frac{x}{6}\Big] \times 90 = \frac{1}{6} \times 90 \\[1em] \Rightarrow \frac{x}{10} \times 90 + \frac{x}{9} \times 90 - \frac{x}{6} \times 90 = 15 \\[1em] \Rightarrow 9x + 10x - 15x = 15 \\[1em] \Rightarrow 4x = 15 \\[1em] \Rightarrow x = \frac{15}{4} \\[1em] \Rightarrow x = 3.75\end{aligned}$$

Hence, the distance between P and Q = 3.75 km.

$\frac{x - b - c}{a} + \frac{x - c - a}{b} + \frac{x - a - b}{c} = 3$

Subtracting 3 from both the sides, we get

$$\begin{aligned}\Rightarrow \frac{x - b - c}{a} + \frac{x - c - a}{b} + \frac{x - a - b}{c} - 3 = 3 - 3 \\[1em] \Rightarrow \left(\frac{x - b - c}{a} - 1\right) + \left(\frac{x - c - a}{b} - 1\right) + \left(\frac{x - a - b}{c} - 1\right) = 3 - 3 \\[1em] \Rightarrow \frac{x - b - c - a}{a} + \frac{x - c - a - b}{b} + \frac{x - a - b - c}{c} = 0 \\[1em] \Rightarrow (x - a - b - c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = 0\end{aligned}$$

As $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ne 0$, we get

$$\begin{aligned}\Rightarrow x - a - b - c = 0 \\[1em] \Rightarrow x = a + b + c\end{aligned}$$

Hence, $\bm{x = a + b + c}$.