Simple Linear Equations

⇒ $x$ + 5 = 10

Transposing 5 to R.H.S.,

⇒ $x$ = 10 − 5

⇒ $x$ = 5

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

⇒ 2 + $y$ = 7

Transposing 2 to R.H.S.,

⇒ $y$ = 7 − 2

⇒ $y$ = 5

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

⇒ $a$ − 2 = 6

Transposing −2 to R.H.S.,

⇒ $a$ = 6 + 2

⇒ $a$ = 8

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

⇒ $x$ − 5 = 8

Transposing −5 to R.H.S.,

⇒ $x$ = 8 + 5

⇒ $x$ = 13

Hence, $\bm{x}$ = 13.

⇒ 5 − $d$ = 12

Transposing 5 to R.H.S.,

⇒ −$d$ = 12 − 5

⇒ −$d$ = 7

Multiplying both sides by −1,

⇒ (−$d$) × (−1) = 7 × (−1)

⇒ $d$ = −7

Hence, $\bm{d}$ = −7.

⇒ $3p$ = 12

Dividing both sides by 3,

⇒ $\frac{3p}{3} = \frac{12}{3}$

⇒ $p$ = 4

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

⇒ 14 = $7m$

Dividing both sides by 7,

⇒ $\frac{14}{7} = \frac{7m}{7}$

⇒ 2 = $m$

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

⇒ $2x$ = 0

Dividing both sides by 2,

⇒ $\frac{2x}{2} = \frac{0}{2}$

⇒ $x$ = 0

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

⇒ $\frac{x}{9} = 2$

Multiplying both sides by 9,

⇒ $\frac{x}{9} \times 9$ = 2 × 9

⇒ $x$ = 18

Hence, $\bm{x}$ = 18.

⇒ $\frac{y}{-12} = -4$

Multiplying both sides by −12,

⇒ $\frac{y}{-12} \times (-12)$ = (−4) × (−12)

⇒ $y$ = 48

Hence, $\bm{y}$ = 48.

⇒ $8x$ − 2 = 38

Transposing −2 to R.H.S.,

⇒ $8x$ = 38 + 2

⇒ $8x$ = 40

Dividing both sides by 8,

⇒ $\frac{8x}{8} = \frac{40}{8}$

⇒ $x$ = 5

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

⇒ $2x$ + 5 = 5

Transposing 5 to R.H.S.,

⇒ $2x$ = 5 − 5

⇒ $2x$ = 0

Dividing both sides by 2,

⇒ $\frac{2x}{2} = \frac{0}{2}$

⇒ $x$ = 0

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

⇒ $5x$ − 1 = 74

Transposing −1 to R.H.S.,

⇒ $5x$ = 74 + 1

⇒ $5x$ = 75

Dividing both sides by 5,

⇒ $\frac{5x}{5} = \frac{75}{5}$

⇒ $x$ = 15

Hence, $\bm{x}$ = 15.

⇒ 14 = 27 − $x$

Transposing −$x$ to L.H.S. and 14 to R.H.S.,

⇒ $x$ = 27 − 14

⇒ $x$ = 13

Hence, $\bm{x}$ = 13.

⇒ 10 + $6a$ = 40

Transposing 10 to R.H.S.,

⇒ $6a$ = 40 − 10

⇒ $6a$ = 30

Dividing both sides by 6,

⇒ $\frac{6a}{6} = \frac{30}{6}$

⇒ $a$ = 5

Hence, $\bm{a}$ = 5.

⇒ $c - \frac{1}{2} = \frac{1}{3}$

Transposing $-\frac{1}{2}$ to R.H.S.,

⇒ $c = \frac{1}{3} + \frac{1}{2}$

⇒ $c = \frac{2 + 3}{6}$

⇒ $c = \frac{5}{6}$

⇒ $c$ = $\frac{5}{6}$

Hence, $\bm{c}$ = $\frac{5}{6}$.

⇒ $\frac{a}{15} - 2 = 0$

Transposing −2 to R.H.S.,

⇒ $\frac{a}{15} = 2$

Multiplying both sides by 15,

⇒ $\frac{a}{15} \times 15$ = 2 × 15

⇒ $a$ = 30

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

⇒ 12 = $c$ − 2

Transposing −2 to L.H.S.,

⇒ 12 + 2 = $c$

⇒ 14 = $c$

Hence, $\bm{c}$ = 14.

⇒ 4 = $x$ − 2.5

Transposing −2.5 to L.H.S.,

⇒ 4 + 2.5 = $x$

⇒ 6.5 = $x$

Hence, $\bm{x}$ = 6.5.

⇒ $y + 5 = 8\frac{1}{4}$

⇒ $y + 5 = \frac{33}{4}$

Transposing 5 to R.H.S.,

⇒ $y = \frac{33}{4} - 5$

⇒ $y = \frac{33 - 20}{4}$

⇒ $y = \frac{13}{4} = 3\frac{1}{4}$

Hence, $\bm{y}$ = $3\frac{1}{4}$.

⇒ $x + \frac{1}{4} = -\frac{3}{8}$

Transposing $\frac{1}{4}$ to R.H.S.,

⇒ $x = -\frac{3}{8} - \frac{1}{4}$

⇒ $x = \frac{-3 - 2}{8}$

⇒ $x = -\frac{5}{8}$

Hence, $\bm{x}$ = $-\frac{5}{8}$.

⇒ $p$ + 0.02 = 0.08

Transposing 0.02 to R.H.S.,

⇒ $p$ = 0.08 − 0.02

⇒ $p$ = 0.06

Hence, $\bm{p}$ = 0.06.

⇒ $p - 12 = 2\frac{2}{3}$

⇒ $p - 12 = \frac{8}{3}$

Transposing −12 to R.H.S.,

⇒ $p = \frac{8}{3} + 12$

⇒ $p = \frac{8 + 36}{3}$

⇒ $p = \frac{44}{3} = 14\frac{2}{3}$

Hence, $\bm{p}$ = $14\frac{2}{3}$.

⇒ −$3x$ = 15

Dividing both sides by −3,

⇒ $\frac{-3x}{-3} = \frac{15}{-3}$

⇒ $x$ = −5

Hence, $\bm{x}$ = −5.

⇒ $1.3b$ = 39

Dividing both sides by 1.3,

⇒ $b = \frac{39}{1.3} = \frac{390}{13}$

⇒ $b$ = 30

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

⇒ $\frac{5}{8}n = 20$

Multiplying both sides by $\frac{8}{5}$,

⇒ $\frac{5}{8}n \times \frac{8}{5} = 20 \times \frac{8}{5}$

⇒ $n$ = 4 × 8 = 32

Hence, $\bm{n}$ = 32.

⇒ $\frac{3}{16}m = 21$

Multiplying both sides by $\frac{16}{3}$,

⇒ $\frac{3}{16}m \times \frac{16}{3} = 21 \times \frac{16}{3}$

⇒ $m$ = 7 × 16 = 112

Hence, $\bm{m}$ = 112.

⇒ $2a$ − 3 = 5

Transposing −3 to R.H.S.,

⇒ $2a$ = 5 + 3

⇒ $2a$ = 8

Dividing both sides by 2,

⇒ $\frac{2a}{2} = \frac{8}{2}$

⇒ $a$ = 4

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

⇒ $3p$ − 1 = 8

Transposing −1 to R.H.S.,

⇒ $3p$ = 8 + 1

⇒ $3p$ = 9

Dividing both sides by 3,

⇒ $\frac{3p}{3} = \frac{9}{3}$

⇒ $p$ = 3

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

⇒ $9y$ − 7 = 20

Transposing −7 to R.H.S.,

⇒ $9y$ = 20 + 7

⇒ $9y$ = 27

Dividing both sides by 9,

⇒ $\frac{9y}{9} = \frac{27}{9}$

⇒ $y$ = 3

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

⇒ $2b$ − 14 = 8

Transposing −14 to R.H.S.,

⇒ $2b$ = 8 + 14

⇒ $2b$ = 22

Dividing both sides by 2,

⇒ $\frac{2b}{2} = \frac{22}{2}$

⇒ $b$ = 11

Hence, $\bm{b}$ = 11.

⇒ $\frac{7}{10}x + 6 = 41$

Transposing 6 to R.H.S.,

⇒ $\frac{7}{10}x = 41 - 6$

⇒ $\frac{7}{10}x = 35$

Multiplying both sides by $\frac{10}{7}$,

⇒ $\frac{7}{10}x \times \frac{10}{7} = 35 \times \frac{10}{7}$

⇒ $x$ = 5 × 10 = 50

Hence, $\bm{x}$ = 50.

⇒ $\frac{5}{12}m - 12 = 48$

Transposing −12 to R.H.S.,

⇒ $\frac{5}{12}m = 48 + 12$

⇒ $\frac{5}{12}m = 60$

Multiplying both sides by $\frac{12}{5}$,

⇒ $\frac{5}{12}m \times \frac{12}{5} = 60 \times \frac{12}{5}$

⇒ $m$ = 12 × 12 = 144

Hence, $\bm{m}$ = 144.

Solving,

$$\begin{aligned}\Rightarrow 8y - 4y = 20 \\[1em] \Rightarrow 4y = 20 \\[1em] \Rightarrow y = \frac{20}{4} \\[1em] \Rightarrow y = 5.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 9b - 4b + 3b = 16 \\[1em] \Rightarrow (9 - 4 + 3)b = 16 \\[1em] \Rightarrow 8b = 16 \\[1em] \Rightarrow b = \frac{16}{8} \\[1em] \Rightarrow b = 2.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 5y + 8 = 8y - 18 \\[1em] \Rightarrow 5y - 8y = -18 - 8 \\[1em] \Rightarrow -3y = -26 \\[1em] \Rightarrow y = \frac{-26}{-3} \\[1em] \Rightarrow y = \frac{26}{3} \\[1em] \Rightarrow y = 8\frac{2}{3}.\end{aligned}$$

Hence, $\bm{y}$ = $8\frac{2}{3}$.

Solving,

$$\begin{aligned}\Rightarrow 6 = 7 + 2p - 5 \\[1em] \Rightarrow 6 = 2 + 2p \\[1em] \Rightarrow 6 - 2 = 2p \\[1em] \Rightarrow 4 = 2p \\[1em] \Rightarrow \frac{4}{2} = \frac{2p}{2} \\[1em] \Rightarrow p = 2.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 8 - 7x = 13x + 8 \\[1em] \Rightarrow -7x - 13x = 8 - 8 \\[1em] \Rightarrow -20x = 0 \\[1em] \Rightarrow x = 0.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 4x - 5x + 2x = 28 + 3x \\[1em] \Rightarrow (4 - 5 + 2)x = 28 + 3x \\[1em] \Rightarrow x = 28 + 3x \\[1em] \Rightarrow x - 3x = 28 \\[1em] \Rightarrow -2x = 28 \\[1em] \Rightarrow x = \frac{28}{-2} \\[1em] \Rightarrow x = -14.\end{aligned}$$

Hence, $\bm{x}$ = −14.

Solving,

$$\begin{aligned}\Rightarrow 9 + m = 6m + 8 - m \\[1em] \Rightarrow 9 + m = 5m + 8 \\[1em] \Rightarrow m - 5m = 8 - 9 \\[1em] \Rightarrow -4m = -1 \\[1em] \Rightarrow m = \frac{-1}{-4} \\[1em] \Rightarrow m = \frac{1}{4}.\end{aligned}$$

Hence, $\bm{m}$ = $\frac{1}{4}$.

Solving,

$$\begin{aligned}\Rightarrow 24 = y + 2y + 3 + 4y \\[1em] \Rightarrow 24 = 7y + 3 \\[1em] \Rightarrow 24 - 3 = 7y \\[1em] \Rightarrow 21 = 7y \\[1em] \Rightarrow \frac{21}{7} = \frac{7y}{7} \\[1em] \Rightarrow y = 3.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 19x + 13 - 12x + 3 = 23 \\[1em] \Rightarrow (19 - 12)x + (13 + 3) = 23 \\[1em] \Rightarrow 7x + 16 = 23 \\[1em] \Rightarrow 7x = 23 - 16 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = \frac{7}{7} \\[1em] \Rightarrow x = 1.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 6b + 40 = -100 - b \\[1em] \Rightarrow 6b + b = -100 - 40 \\[1em] \Rightarrow 7b = -140 \\[1em] \Rightarrow b = \frac{-140}{7} \\[1em] \Rightarrow b = -20.\end{aligned}$$

Hence, $\bm{b}$ = −20.

Solving,

$$\begin{aligned}\Rightarrow 6 - 5m - 1 + 3m = 0 \\[1em] \Rightarrow (6 - 1) + (-5 + 3)m = 0 \\[1em] \Rightarrow 5 - 2m = 0 \\[1em] \Rightarrow -2m = -5 \\[1em] \Rightarrow m = \frac{-5}{-2} \\[1em] \Rightarrow m = \frac{5}{2} \\[1em] \Rightarrow m = 2\frac{1}{2}.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 0.4x - 1.2 = 0.3x + 0.6 \\[1em] \Rightarrow 0.4x - 0.3x = 0.6 + 1.2 \\[1em] \Rightarrow 0.1x = 1.8 \\[1em] \Rightarrow x = \frac{1.8}{0.1} \\[1em] \Rightarrow x = \frac{18}{1} \\[1em] \Rightarrow x = 18.\end{aligned}$$

Hence, $\bm{x}$ = 18.

Solving,

$$\begin{aligned}\Rightarrow 6(x + 4) = 36 \\[1em] \Rightarrow (x + 4) = \frac{36}{6} \\[1em] \Rightarrow x + 4 = 6 \\[1em] \Rightarrow x = 6 - 4 \\[1em] \Rightarrow x = 2.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 9(a + 5) + 2 = 11 \\[1em] \Rightarrow 9(a + 5) = 11 - 2 \\[1em] \Rightarrow 9(a + 5) = 9 \\[1em] \Rightarrow (a + 5) = \frac{9}{9} \\[1em] \Rightarrow a + 5 = 1 \\[1em] \Rightarrow a = 1 - 5 \\[1em] \Rightarrow a = -4.\end{aligned}$$

Hence, $\bm{a}$ = −4.

Solving,

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

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

Solving,

$$\begin{aligned}\Rightarrow -3(a - 6) = 24 \\[1em] \Rightarrow (a - 6) = \frac{24}{-3} \\[1em] \Rightarrow a - 6 = -8 \\[1em] \Rightarrow a = -8 + 6 \\[1em] \Rightarrow a = -2.\end{aligned}$$

Hence, $\bm{a}$ = −2.

Solving,

$$\begin{aligned}\Rightarrow 7(x - 2) = 2(2x - 4) \\[1em] \Rightarrow 7x - 14 = 4x - 8 \\[1em] \Rightarrow 7x - 4x = -8 + 14 \\[1em] \Rightarrow 3x = 6 \\[1em] \Rightarrow x = \frac{6}{3} \\[1em] \Rightarrow x = 2.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow (x - 4)(2x + 3) = 2x^{2} \\[1em] \Rightarrow 2x^{2} + 3x - 8x - 12 = 2x^{2} \\[1em] \Rightarrow 2x^{2} - 5x - 12 = 2x^{2} \\[1em] \Rightarrow -5x - 12 = 2x^{2} - 2x^{2} \\[1em] \Rightarrow -5x - 12 = 0 \\[1em] \Rightarrow -5x = 12 \\[1em] \Rightarrow x = \frac{12}{-5} \\[1em] \Rightarrow x = -\frac{12}{5} \\[1em] \Rightarrow x = -2\frac{2}{5}.\end{aligned}$$

Hence, $\bm{x}$ = $-2\frac{2}{5}$.

Solving,

$$\begin{aligned}\Rightarrow 21 - 3(b - 7) = b + 20 \\[1em] \Rightarrow 21 - 3b + 21 = b + 20 \\[1em] \Rightarrow 42 - 3b = b + 20 \\[1em] \Rightarrow -3b - b = 20 - 42 \\[1em] \Rightarrow -4b = -22 \\[1em] \Rightarrow b = \frac{-22}{-4} \\[1em] \Rightarrow b = \frac{11}{2} \\[1em] \Rightarrow b = 5\frac{1}{2}.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow x(x + 5) = x^{2} + x + 32 \\[1em] \Rightarrow x^{2} + 5x = x^{2} + x + 32 \\[1em] \Rightarrow x^{2} - x^{2} + 5x - x = 32 \\[1em] \Rightarrow 4x = 32 \\[1em] \Rightarrow x = \frac{32}{4} \\[1em] \Rightarrow x = 8.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{2} + x = 9 \\[1em] \Rightarrow \frac{x + 2x}{2} = 9 \\[1em] \Rightarrow \frac{3x}{2} = 9 \\[1em] \Rightarrow 3x = 18 \\[1em] \Rightarrow x = 6.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{5} + 2x = 33 \\[1em] \Rightarrow \frac{x + 10x}{5} = 33 \\[1em] \Rightarrow \frac{11x}{5} = 33 \\[1em] \Rightarrow 11x = 165 \\[1em] \Rightarrow x = 15.\end{aligned}$$

Hence, $\bm{x}$ = 15.

Solving,

$$\begin{aligned}\Rightarrow \frac{3x}{4} + 4x = 38 \\[1em] \Rightarrow \frac{3x + 16x}{4} = 38 \\[1em] \Rightarrow \frac{19x}{4} = 38 \\[1em] \Rightarrow 19x = 152 \\[1em] \Rightarrow x = 8.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{2} + \frac{x}{5} = 14 \\[1em] \Rightarrow \frac{5x + 2x}{10} = 14 \\[1em] \Rightarrow \frac{7x}{10} = 14 \\[1em] \Rightarrow 7x = 140 \\[1em] \Rightarrow x = 20.\end{aligned}$$

Hence, $\bm{x}$ = 20.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{3} - \frac{x}{4} = 2 \\[1em] \Rightarrow \frac{4x - 3x}{12} = 2 \\[1em] \Rightarrow \frac{x}{12} = 2 \\[1em] \Rightarrow x = 24.\end{aligned}$$

Hence, $\bm{x}$ = 24.

Solving,

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

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

Solving,

$$\begin{aligned}\Rightarrow \frac{4x}{3} - \frac{7x}{3} = 1 \\[1em] \Rightarrow \frac{4x - 7x}{3} = 1 \\[1em] \Rightarrow \frac{-3x}{3} = 1 \\[1em] \Rightarrow -x = 1 \\[1em] \Rightarrow x = -1.\end{aligned}$$

Hence, $\bm{x}$ = −1.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{2}m + \frac{3}{4}m - m = 2.5 \\[1em] \Rightarrow \frac{2m + 3m - 4m}{4} = 2.5 \\[1em] \Rightarrow \frac{m}{4} = 2.5 \\[1em] \Rightarrow m = 10.\end{aligned}$$

Hence, $\bm{m}$ = 10.

Solving,

$$\begin{aligned}\Rightarrow \frac{2x}{3} + \frac{x}{2} - \frac{3x}{4} = 1 \\[1em] \Rightarrow \frac{8x + 6x - 9x}{12} = 1 \\[1em] \Rightarrow \frac{5x}{12} = 1 \\[1em] \Rightarrow 5x = 12 \\[1em] \Rightarrow x = \frac{12}{5} \\[1em] \Rightarrow x = 2\frac{2}{5}.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{3a}{4} + \frac{a}{6} = 66 \\[1em] \Rightarrow \frac{9a + 2a}{12} = 66 \\[1em] \Rightarrow \frac{11a}{12} = 66 \\[1em] \Rightarrow 11a = 792 \\[1em] \Rightarrow a = 72.\end{aligned}$$

Hence, $\bm{a}$ = 72.

Solving,

$$\begin{aligned}\Rightarrow \frac{2p}{3} - \frac{p}{5} = 35 \\[1em] \Rightarrow \frac{10p - 3p}{15} = 35 \\[1em] \Rightarrow \frac{7p}{15} = 35 \\[1em] \Rightarrow 7p = 525 \\[1em] \Rightarrow p = 75.\end{aligned}$$

Hence, $\bm{p}$ = 75.

Solving,

$$\begin{aligned}\Rightarrow 0.6a + 0.2a = 0.4a + 8 \\[1em] \Rightarrow 0.8a = 0.4a + 8 \\[1em] \Rightarrow 0.8a - 0.4a = 8 \\[1em] \Rightarrow 0.4a = 8 \\[1em] \Rightarrow a = \frac{8}{0.4} \\[1em] \Rightarrow a = \frac{80}{4} \\[1em] \Rightarrow a = 20.\end{aligned}$$

Hence, $\bm{a}$ = 20.

Solving,

$$\begin{aligned}\Rightarrow p + 1.4p = 48 \\[1em] \Rightarrow 2.4p = 48 \\[1em] \Rightarrow p = \frac{48}{2.4} \\[1em] \Rightarrow p = \frac{480}{24} \\[1em] \Rightarrow p = 20.\end{aligned}$$

Hence, $\bm{p}$ = 20.

Solving,

$\Rightarrow 10$% of $x$ = 20

$\Rightarrow \frac{10}{100} \times x = 20$

$\Rightarrow x = \frac{20 \times 100}{10}$ = 200.

Hence, $\bm{x}$ = 200.

Solving,

$\Rightarrow y + 20$% of $y$ = 18

$\Rightarrow y + \frac{20}{100} \times y = 18$

$\Rightarrow y + \frac{y}{5} = 18$

$\Rightarrow \frac{5y + y}{5} = 18$

$\Rightarrow \frac{6y}{5} = 18$

$\Rightarrow 6y = 90$

$\Rightarrow y = \frac{90}{6}$ = 15.

Hence, $\bm{y}$ = 15.

Solving,

$\Rightarrow x - 30$% of $x$ = 35

$\Rightarrow x - \frac{30}{100} \times x = 35$

$\Rightarrow x - \frac{3x}{10} = 35$

$\Rightarrow \frac{10x - 3x}{10} = 35$

$\Rightarrow \frac{7x}{10} = 35$

$\Rightarrow 7x = 350$

$\Rightarrow x = \frac{350}{7}$ = 50.

Hence, $\bm{x}$ = 50.

Solving,

$$\begin{aligned}\Rightarrow \frac{x + 4}{2} + \frac{x}{3} = 7 \\[1em] \Rightarrow \frac{3(x + 4) + 2x}{6} = 7 \\[1em] \Rightarrow \frac{3x + 12 + 2x}{6} = 7 \\[1em] \Rightarrow \frac{5x + 12}{6} = 7 \\[1em] \Rightarrow 5x + 12 = 42 \\[1em] \Rightarrow 5x = 30 \\[1em] \Rightarrow x = 6.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{y + 2}{3} + \frac{y + 5}{4} = 6 \\[1em] \Rightarrow \frac{4(y + 2) + 3(y + 5)}{12} = 6 \\[1em] \Rightarrow \frac{4y + 8 + 3y + 15}{12} = 6 \\[1em] \Rightarrow \frac{7y + 23}{12} = 6 \\[1em] \Rightarrow 7y + 23 = 72 \\[1em] \Rightarrow 7y = 49 \\[1em] \Rightarrow y = 7.\end{aligned}$$

Hence, $\bm{y}$ = 7.

Solving,

$$\begin{aligned}\Rightarrow \frac{3a - 2}{7} - \frac{a - 2}{4} = 2 \\[1em] \Rightarrow \frac{4(3a - 2) - 7(a - 2)}{28} = 2 \\[1em] \Rightarrow \frac{12a - 8 - 7a + 14}{28} = 2 \\[1em] \Rightarrow \frac{5a + 6}{28} = 2 \\[1em] \Rightarrow 5a + 6 = 56 \\[1em] \Rightarrow 5a = 50 \\[1em] \Rightarrow a = 10.\end{aligned}$$

Hence, $\bm{a}$ = 10.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{2}(x + 5) - \frac{1}{3}(x - 2) = 4 \\[1em] \Rightarrow \frac{3(x + 5) - 2(x - 2)}{6} = 4 \\[1em] \Rightarrow \frac{3x + 15 - 2x + 4}{6} = 4 \\[1em] \Rightarrow \frac{x + 19}{6} = 4 \\[1em] \Rightarrow x + 19 = 24 \\[1em] \Rightarrow x = 5.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{x - 1}{2} - \frac{x - 2}{3} - \frac{x - 3}{4} = 0 \\[1em] \Rightarrow \frac{6(x - 1) - 4(x - 2) - 3(x - 3)}{12} = 0 \\[1em] \Rightarrow \frac{6x - 6 - 4x + 8 - 3x + 9}{12} = 0 \\[1em] \Rightarrow \frac{-x + 11}{12} = 0 \\[1em] \Rightarrow -x + 11 = 0 \\[1em] \Rightarrow x = 11.\end{aligned}$$

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

Given,

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

L.C.M. of 3, 5 and 7 = 105.

Multiplying both sides by 105,

$$\begin{aligned}\Rightarrow \left(\frac{x + 1}{3} + \frac{x + 4}{5}\right) \times 105 = \left(\frac{x - 4}{7}\right) \times 105 \\[1em] \Rightarrow 35(x + 1) + 21(x + 4) = 15(x - 4) \\[1em] \Rightarrow 35x + 35 + 21x + 84 = 15x - 60 \\[1em] \Rightarrow 56x + 119 = 15x - 60 \\[1em] \Rightarrow 56x - 15x = -60 - 119 \\[1em] \Rightarrow 41x = -179 \\[1em] \Rightarrow x = -\frac{179}{41} \\[1em] \Rightarrow x = -4\frac{15}{41}.\end{aligned}$$

Hence, $\bm{x}$ = $-4\frac{15}{41}$.

Solving,

$$\begin{aligned}\Rightarrow 15 - 2(5 - 3x) = 4(x - 3) + 13 \\[1em] \Rightarrow 15 - 10 + 6x = 4x - 12 + 13 \\[1em] \Rightarrow 5 + 6x = 4x + 1 \\[1em] \Rightarrow 6x - 4x = 1 - 5 \\[1em] \Rightarrow 2x = -4 \\[1em] \Rightarrow \frac{2x}{2} = \frac{-4}{2} \\[1em] \Rightarrow x = -2.\end{aligned}$$

Hence, $\bm{x}$ = −2.

Solving,

$$\begin{aligned}\Rightarrow \frac{2x + 1}{3x - 2} = 1\frac{1}{4} \\[1em] \Rightarrow \frac{2x + 1}{3x - 2} = \frac{5}{4} \\[1em] \Rightarrow 4(2x + 1) = 5(3x - 2) \\[1em] \Rightarrow 8x + 4 = 15x - 10 \\[1em] \Rightarrow 4 + 10 = 15x - 8x \\[1em] \Rightarrow 14 = 7x \\[1em] \Rightarrow x = 2.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 21 - 3(x - 7) = x + 20 \\[1em] \Rightarrow 21 - 3x + 21 = x + 20 \\[1em] \Rightarrow 42 - 3x = x + 20 \\[1em] \Rightarrow -3x - x = 20 - 42 \\[1em] \Rightarrow -4x = -22 \\[1em] \Rightarrow x = \frac{-22}{-4} \\[1em] \Rightarrow x = \frac{11}{2} \\[1em] \Rightarrow x = 5\frac{1}{2}.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{2(x - 3)}{7} - \frac{x - 4}{3} = 0 \\[1em] \Rightarrow \frac{6(x - 3) - 7(x - 4)}{21} = 0 \\[1em] \Rightarrow \frac{6x - 18 - 7x + 28}{21} = 0 \\[1em] \Rightarrow \frac{-x + 10}{21} = 0 \\[1em] \Rightarrow -x + 10 = 0 \\[1em] \Rightarrow x = 10.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{2x - 3}{3} - (x - 5) = \frac{x}{3} \\[1em] \Rightarrow (2x - 3) - 3(x - 5) = x \\[1em] \Rightarrow 2x - 3 - 3x + 15 = x \\[1em] \Rightarrow -x + 12 = x \\[1em] \Rightarrow 12 = 2x \\[1em] \Rightarrow x = 6.\end{aligned}$$

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

Given,

⇒ $\frac{x - 4}{7} = \frac{x + 3}{7} + \frac{x + 4}{5}$

L.C.M. of 7 and 5 = 35.

Multiplying both sides by 35,

$$\begin{aligned}\Rightarrow \left(\frac{x - 4}{7}\right) \times 35 = \left(\frac{x + 3}{7} + \frac{x + 4}{5}\right) \times 35 \\[1em] \Rightarrow 5(x - 4) = 5(x + 3) + 7(x + 4) \\[1em] \Rightarrow 5x - 20 = 5x + 15 + 7x + 28 \\[1em] \Rightarrow 5x - 20 = 12x + 43 \\[1em] \Rightarrow 5x - 12x = 43 + 20 \\[1em] \Rightarrow -7x = 63 \\[1em] \Rightarrow x = -9.\end{aligned}$$

Hence, $\bm{x}$ = −9.

Given,

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

L.C.M. of 5, 3 and 2 = 30.

Multiplying both sides by 30,

$$\begin{aligned}\Rightarrow \left(\frac{x - 1}{5} - \frac{x}{3}\right) \times 30 = \left(1 - \frac{x - 2}{2}\right) \times 30 \\[1em] \Rightarrow 6(x - 1) - 10x = 30 - 15(x - 2) \\[1em] \Rightarrow 6x - 6 - 10x = 30 - 15x + 30 \\[1em] \Rightarrow -4x - 6 = 60 - 15x \\[1em] \Rightarrow -4x + 15x = 60 + 6 \\[1em] \Rightarrow 11x = 66 \\[1em] \Rightarrow x = 6.\end{aligned}$$

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

Solving,

$\Rightarrow 2x + 20$% of $x$ = 12.1

$\Rightarrow 2x + \frac{20}{100} \times x = 12.1$

$\Rightarrow 2x + \frac{x}{5} = 12.1$

$\Rightarrow \frac{10x + x}{5} = 12.1$

$\Rightarrow \frac{11x}{5} = 12.1$

$\Rightarrow 11x = 60.5$

$\Rightarrow x = \frac{60.5}{11} = \frac{605}{110}$

$\Rightarrow x = \frac{11}{2}$ = $5\frac{1}{2}$.

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

Given,

One-fifth of a number is 5.

Let the number be $x$.

$$\begin{aligned}\therefore \frac{1}{5} \times x = 5 \\[1em] \Rightarrow \frac{x}{5} = 5 \\[1em] \Rightarrow x = 25\end{aligned}$$

Hence, the number is 25.

Given,

Six times a number is 72.

Let the number be $x$.

$$\begin{aligned}\therefore 6x = 72 \\[1em] \Rightarrow x = \frac{72}{6} \\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, the number is 12.

Given,

If 15 is added to a number, the result is 69.

Let the number be $x$.

$$\begin{aligned}\therefore x + 15 = 69 \\[1em] \Rightarrow x = 69 - 15 \\[1em] \Rightarrow x = 54\end{aligned}$$

Hence, the number is 54.

Given,

The sum of twice a number and 4 is 80.

Let the number be $x$.

$$\begin{aligned}\therefore 2x + 4 = 80 \\[1em] \Rightarrow 2x = 80 - 4 \\[1em] \Rightarrow 2x = 76 \\[1em] \Rightarrow x = \frac{76}{2} = 38\end{aligned}$$

Hence, the number is 38.

Given,

The difference between a number and one-fourth of itself is 24.

Let the number be $x$.

$$\begin{aligned}\therefore x - \frac{1}{4}x = 24 \\[1em] \Rightarrow \frac{4x - x}{4} = 24 \\[1em] \Rightarrow \frac{3x}{4} = 24 \\[1em] \Rightarrow 3x = 96 \\[1em] \Rightarrow x = 32\end{aligned}$$

Hence, the number is 32.

Given,

Find a number whose one-third part exceeds its one-fifth part by 20.

Let the number be $x$.

$$\begin{aligned}\therefore \frac{x}{3} - \frac{x}{5} = 20 \\[1em] \Rightarrow \frac{5x - 3x}{15} = 20 \\[1em] \Rightarrow \frac{2x}{15} = 20 \\[1em] \Rightarrow 2x = 300 \\[1em] \Rightarrow x = 150\end{aligned}$$

Hence, the number is 150.

Given,

A number is as much greater than 35 as is less than 53.

Let the number be $x$.

$$\begin{aligned}\therefore x - 35 = 53 - x \\[1em] \Rightarrow x + x = 53 + 35 \\[1em] \Rightarrow 2x = 88 \\[1em] \Rightarrow x = \frac{88}{2} = 44\end{aligned}$$

Hence, the number is 44.

Given,

The sum of two numbers is 18. If one is twice the other.

Let the smaller number be $x$, then the other number = $2x$.

$$\begin{aligned}\therefore x + 2x = 18 \\[1em] \Rightarrow 3x = 18 \\[1em] \Rightarrow x = 6\end{aligned}$$

∴ Other number = $2x$ = 2 × 6 = 12

Hence, the numbers are 6 and 12.

Given,

A number is 15 more than the other. The sum of the two numbers is 195.

Let the smaller number be $x$, then the greater number = $x$ + 15.

$$\begin{aligned}\therefore x + (x + 15) = 195 \\[1em] \Rightarrow 2x + 15 = 195 \\[1em] \Rightarrow 2x = 180 \\[1em] \Rightarrow x = 90\end{aligned}$$

∴ Greater number = $x$ + 15 = 90 + 15 = 105

Hence, the numbers are 90 and 105.

Given,

The sum of three consecutive even numbers is 54.

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

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) = 54 \\[1em] \Rightarrow 3x + 6 = 54 \\[1em] \Rightarrow 3x = 48 \\[1em] \Rightarrow x = 16\end{aligned}$$

∴ The numbers are 16, 16 + 2 = 18 and 16 + 4 = 20.

Hence, the numbers are 16, 18 and 20.

Given,

The sum of three consecutive odd numbers is 63.

Let the three consecutive odd numbers be $x, x + 2$ and $x + 4$.

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) = 63 \\[1em] \Rightarrow 3x + 6 = 63 \\[1em] \Rightarrow 3x = 57 \\[1em] \Rightarrow x = 19\end{aligned}$$

∴ The numbers are 19, 19 + 2 = 21 and 19 + 4 = 23.

Hence, the numbers are 19, 21 and 23.

Given,

A man has ₹ $x$ from which he spends ₹ 6. If twice of the money left with him is ₹ 86.

Total money = ₹ $x$

He spends = ₹ 6

Money left with the man = ₹ ($x$ − 6)

$$\begin{aligned}\therefore 2(x - 6) = 86 \\[1em] \Rightarrow x - 6 = 43 \\[1em] \Rightarrow x = 43 + 6 \\[1em] \Rightarrow x = 49\end{aligned}$$

Hence, $\bm{x}$ = 49.

Given,

A man is four times as old as his son. After 20 years, he will be twice as old as his son at that time.

Let the present age of the son be $x$ years, then the present age of the man = $4x$ years.

After 20 years:

Age of son = ($x$ + 20) years

Age of man = ($4x$ + 20) years

$$\begin{aligned}\therefore 4x + 20 = 2(x + 20) \\[1em] \Rightarrow 4x + 20 = 2x + 40 \\[1em] \Rightarrow 4x - 2x = 40 - 20 \\[1em] \Rightarrow 2x = 20 \\[1em] \Rightarrow x = 10\end{aligned}$$

∴ Present age of man = $4x$ = 4 × 10 = 40 years

Hence, the present age of the son is 10 years and that of the man is 40 years.

Given,

If 5 is subtracted from three times a number, the result is 16.

Let the number be $x$.

$$\begin{aligned}\therefore 3x - 5 = 16 \\[1em] \Rightarrow 3x = 16 + 5 \\[1em] \Rightarrow 3x = 21 \\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, the number is 7.

Given,

Find three consecutive natural numbers such that the sum of the first and the second is 15 more than the third.

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

$$\begin{aligned}\therefore x + (x + 1) = (x + 2) + 15 \\[1em] \Rightarrow 2x + 1 = x + 17 \\[1em] \Rightarrow 2x - x = 17 - 1 \\[1em] \Rightarrow x = 16\end{aligned}$$

∴ The numbers are 16, 16 + 1 = 17 and 16 + 2 = 18.

Hence, the numbers are 16, 17 and 18.

Given,

The difference between two numbers is 7. Six times the smaller plus the larger is 77.

Let the smaller number be $x$, then the larger number = $x$ + 7.

$$\begin{aligned}\therefore 6x + (x + 7) = 77 \\[1em] \Rightarrow 7x + 7 = 77 \\[1em] \Rightarrow 7x = 70 \\[1em] \Rightarrow x = 10\end{aligned}$$

∴ Larger number = $x$ + 7 = 10 + 7 = 17

Hence, the numbers are 10 and 17.

Given,

The length of a rectangular plot exceeds its breadth by 5 metres. If the perimeter of the plot is 142 metres.

Let the breadth of the plot be $x$ metres, then the length = ($x$ + 5) metres.

Perimeter of rectangle = 2(length + breadth)

Perimeter = 142 meters

$$\begin{aligned}\therefore 2[(x + 5) + x] = 142 \\[1em] \Rightarrow 2(2x + 5) = 142 \\[1em] \Rightarrow 2x + 5 = 71 \\[1em] \Rightarrow 2x = 66 \\[1em] \Rightarrow x = 33\end{aligned}$$

∴ Length = $x$ + 5 = 33 + 5 = 38 m

Hence, the length of the plot is 38 m and its breadth is 33 m.

Given,

The numerator of a fraction is four less than its denominator. If 1 is added to both the numerator and denominator, the fraction becomes $\frac{1}{2}$.

Let the denominator of the fraction be $x$, then the numerator = $x$ − 4.

∴ Fraction = $\frac{x - 4}{x}$

On adding 1 to both the numerator and the denominator,

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

∴ Numerator = $x$ − 4 = 7 − 4 = 3 and Denominator = 7

Hence, the fraction is $\frac{3}{7}$.

Given,

A man is thrice as old as his son. After 12 years, he will be twice as old as his son at that time.

Let the present age of the son be $x$ years, then the present age of the man = $3x$ years.

After 12 years:

Age of son = ($x$ + 12) years

Age of man = ($3x$ + 12) years

$$\begin{aligned}\therefore 3x + 12 = 2(x + 12) \\[1em] \Rightarrow 3x + 12 = 2x + 24 \\[1em] \Rightarrow 3x - 2x = 24 - 12 \\[1em] \Rightarrow x = 12\end{aligned}$$

∴ Present age of man = $3x$ = 3 × 12 = 36 years

Hence, the present age of the son is 12 years and that of the man is 36 years.

Given,

A sum of ₹ 500 is in the form of notes of denominations of ₹ 5 and ₹ 10. If the total number of notes is 90.

Let the number of ₹ 5 notes be $x$, then the number of ₹ 10 notes = 90 − $x$.

Value of ₹ 5 notes = ₹ $5x$

Value of ₹ 10 notes = ₹ 10(90 − $x$)

$$\begin{aligned}\therefore 5x + 10(90 - x) = 500 \\[1em] \Rightarrow 5x + 900 - 10x = 500 \\[1em] \Rightarrow -5x = 500 - 900 \\[1em] \Rightarrow -5x = -400 \\[1em] \Rightarrow x = 80\end{aligned}$$

∴ Number of ₹ 10 notes = 90 − 80 = 10

Hence, there are 80 notes of ₹ 5 each and 10 notes of ₹ 10 each.

Given,

⇒ $2x - 2 = \frac{1}{3}x - \frac{11}{4}$

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

Multiplying both sides by 12,

$$\begin{aligned}\Rightarrow (2x - 2) \times 12 = \left(\frac{1}{3}x - \frac{11}{4}\right) \times 12 \\[1em] \Rightarrow 2x \times 12 - 2 \times 12 = \frac{1}{3}x \times 12 - \frac{11}{4} \times 12 \\[1em] \Rightarrow 24x - 24 = 4x - 33 \\[1em] \Rightarrow 24x - 4x = -33 + 24 \\[1em] \Rightarrow 20x = -9 \\[1em] \Rightarrow x = -\frac{9}{20}.\end{aligned}$$

Hence, $\bm{x}$ = $-\frac{9}{20}$.

Given,

⇒ $7(x - 4) - 2(x - 5) - 3(x - 2) = 3(2 - x) - 8$

$$\begin{aligned}\Rightarrow 7x - 28 - 2x + 10 - 3x + 6 = 6 - 3x - 8 \\[1em] \Rightarrow 2x - 12 = -3x - 2 \\[1em] \Rightarrow 2x + 3x = -2 + 12 \\[1em] \Rightarrow 5x = 10 \\[1em] \Rightarrow x = 2.\end{aligned}$$

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

Solving,

$\Rightarrow x - 40$% of $(x − 10) = 64$

$\Rightarrow x - \frac{40}{100}(x - 10) = 64$

$\Rightarrow x - \frac{2}{5}(x - 10) = 64$

$\Rightarrow \frac{5x - 2(x - 10)}{5} = 64$

$\Rightarrow \frac{5x - 2x + 20}{5} = 64$

$\Rightarrow 3x + 20 = 320$

$\Rightarrow 3x = 300$

$\Rightarrow x = \frac{300}{3}$ = 100.

Hence, $\bm{x}$ = 100.

Given,

⇒ $\frac{2-3x}{3} - \frac{3-5x}{2} = \frac{2}{3}$

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

Multiplying both sides by 6,

$$\begin{aligned}\Rightarrow \left(\frac{2-3x}{3} - \frac{3-5x}{2}\right) \times 6 = \frac{2}{3} \times 6 \\[1em] \Rightarrow 2(2 - 3x) - 3(3 - 5x) = 4 \\[1em] \Rightarrow 4 - 6x - 9 + 15x = 4 \\[1em] \Rightarrow 9x - 5 = 4 \\[1em] \Rightarrow 9x = 9 \\[1em] \Rightarrow x = 1.\end{aligned}$$

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

Let the number be $x$.

Given,

Five times a certain number diminished by 25 gives 25.

$$\begin{aligned}\therefore 5x - 25 = 25 \\[1em] \Rightarrow 5x = 25 + 25 \\[1em] \Rightarrow 5x = 50 \\[1em] \Rightarrow x = 10\end{aligned}$$

Hence, the number is 10.

Let the number be $x$.

Given,

Three-fifths of a number is greater than one-third of the same number by 4.

$$\begin{aligned}\therefore \frac{3}{5}x - \frac{1}{3}x = 4 \\[1em] \Rightarrow \frac{9x - 5x}{15} = 4 \\[1em] \Rightarrow \frac{4x}{15} = 4 \\[1em] \Rightarrow 4x = 60 \\[1em] \Rightarrow x = 15\end{aligned}$$

Hence, the number is 15.

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

Given,

The sum of three consecutive even natural numbers is 36.

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) = 36 \\[1em] \Rightarrow 3x + 6 = 36 \\[1em] \Rightarrow 3x = 30 \\[1em] \Rightarrow x = 10\end{aligned}$$

∴ The numbers are 10, 10 + 2 = 12 and 10 + 4 = 14.

Hence, the numbers are 10, 12 and 14.

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

Given,

The sum of three consecutive odd integers is 51.

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) = 51 \\[1em] \Rightarrow 3x + 6 = 51 \\[1em] \Rightarrow 3x = 45 \\[1em] \Rightarrow x = 15\end{aligned}$$

∴ The integers are 15, 15 + 2 = 17 and 15 + 4 = 19.

Hence, the integers are 15, 17 and 19.

Let the present age of the son be $x$ years, then the present age of the man = ($x$ + 24) years.

Two years hence:

Age of son = ($x$ + 2) years

Age of man = ($x$ + 24 + 2) = ($x$ + 26) years

Given,

The difference between the ages of a man and his son is 24 years. Two years hence, the age of the man will be three times the age of his son at that time.

$$\begin{aligned}\therefore x + 26 = 3(x + 2) \\[1em] \Rightarrow x + 26 = 3x + 6 \\[1em] \Rightarrow 26 - 6 = 3x - x \\[1em] \Rightarrow 20 = 2x \\[1em] \Rightarrow x = 10\end{aligned}$$

∴ Present age of man = $x$ + 24 = 10 + 24 = 34 years

Hence, the present age of the son is 10 years and that of the man is 34 years.

Let the greater angle be $x$°, then the smaller angle = ($x$ − 20)°.

Since the angles are supplementary, their sum = 180°.

Given,

The difference between two supplementary angles is 20°.

$$\begin{aligned}\therefore x + (x - 20) = 180 \\[1em] \Rightarrow 2x - 20 = 180 \\[1em] \Rightarrow 2x = 200 \\[1em] \Rightarrow x = 100\end{aligned}$$

∴ Smaller angle = $x$ − 20 = 100 − 20 = 80°

Hence, the angles are 100° and 80°.

Let the number of articles bought at ₹ 16 each be $x$, then the number of articles bought at ₹ 5 each = 54 − $x$.

Given,

Some articles are bought at ₹ 16 each and some other articles are bought at ₹ 5 each. If the total number of articles bought is 54 and their total cost is ₹ 600.

$$\begin{aligned}\therefore 16x + 5(54 - x) = 600 \\[1em] \Rightarrow 16x + 270 - 5x = 600 \\[1em] \Rightarrow 11x = 600 - 270 \\[1em] \Rightarrow 11x = 330 \\[1em] \Rightarrow x = 30\end{aligned}$$

∴ Number of articles bought at ₹ 5 each = 54 − 30 = 24

Hence, the number of articles bought at ₹ 16 each is 30 and at ₹ 5 each is 24.

Let the third side of the isosceles triangle be $x$ cm.

Then, each of the equal sides = ($2x$ − 2) cm.

Perimeter of triangle = sum of its three sides

Given,

Perimeter = 24 cm

$$\begin{aligned}\therefore (2x - 2) + (2x - 2) + x = 24 \\[1em] \Rightarrow 5x - 4 = 24 \\[1em] \Rightarrow 5x = 28 \\[1em] \Rightarrow x = \frac{28}{5} \text{ cm} \\[1em] \Rightarrow 2x - 2 = 2 \times \frac{28}{5} - 2 \\[1em] \Rightarrow 2x - 2 = \frac{56}{5} - 2 \\[1em] \Rightarrow 2x - 2 = \frac{56 - 10}{5} \\[1em] \Rightarrow 2x - 2 = \frac{46}{5} \text{ cm }\end{aligned}$$

Hence, the sides of the triangle are $\frac{46}{5} \text{ cm, } \frac{46}{5} \text{ cm and } \frac{28}{5}$ cm.

Let one part be $x$, then the other part = 600 − $x$.

Given,

Divide 600 into two parts such that one-fourths of one part exceeds one-sixths of the other part by 20.

$$\begin{aligned}\therefore \frac{1}{4}x - \frac{1}{6}(600 - x) = 20 \\[1em] \Rightarrow \frac{3x - 2(600 - x)}{12} = 20 \\[1em] \Rightarrow 3x - 1200 + 2x = 240 \\[1em] \Rightarrow 5x = 1440 \\[1em] \Rightarrow x = 288\end{aligned}$$

∴ Other part = 600 − 288 = 312

Hence, the two parts are 288 and 312.

Given,

⇒ $\frac{3}{4} (2x - 5) - \frac{5}{6} (7 - 5x) = \frac{7x}{3}$

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

Multiplying both sides by 12,

$$\begin{aligned}\Rightarrow \frac{3}{4}(2x - 5) \times 12 - \frac{5}{6}(7 - 5x) \times 12 = \frac{7x}{3} \times 12 \\[1em] \Rightarrow 9(2x - 5) - 10(7 - 5x) = 28x \\[1em] \Rightarrow 18x - 45 - 70 + 50x = 28x \\[1em] \Rightarrow 68x - 115 = 28x \\[1em] \Rightarrow 68x - 28x = 115 \\[1em] \Rightarrow 40x = 115 \\[1em] \Rightarrow x = \frac{115}{40} \\[1em] \Rightarrow x = \frac{23}{8}.\end{aligned}$$

Hence, $\bm{x}$ = $\frac{23}{8}$.

Given,

⇒ $\frac{3}{4} (7x - 1) - \left(2x - \frac{1-x}{2}\right) = x + \frac{3}{2}$

Multiplying both sides by 4,

$$\begin{aligned}\Rightarrow \frac{3}{4}(7x - 1)\times 4 - \left(2x - \frac{1 - x}{2}\right)\times 4 = \Big(x + \frac{3}{2}\Big) \times 4 \\[1em] \Rightarrow 3(7x - 1) - \left(8x - 2(1 - x)\right) = 4x + 6 \\[1em] \Rightarrow 21x - 3 - (8x - 2 + 2x) = 4x + 6 \\[1em] \Rightarrow 21x - 3 - (10x - 2) = 4x + 6 \\[1em] \Rightarrow 21x - 3 - 10x + 2 = 4x + 6 \\[1em] \Rightarrow 11x - 1 = 4x + 6 \\[1em] \Rightarrow 11x - 4x = 6 + 1 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = 1.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 20x \div 3 = 40 \\[1em] \Rightarrow \frac{20x}{3} = 40 \\[1em] \Rightarrow x = \frac{40 \times 3}{20} \\[1em] \Rightarrow x = 2 \times 3 = 6.\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{2x}{3} - 5 = 3 \\[1em] \Rightarrow \frac{2x}{3} = 8 \\[1em] \Rightarrow 2x = 24 \\[1em] \Rightarrow x = 12.\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 2 - 3(5 - x) = 8 \\[1em] \Rightarrow 2 - 15 + 3x = 8 \\[1em] \Rightarrow 3x - 13 = 8 \\[1em] \Rightarrow 3x = 21 \\[1em] \Rightarrow x = 7.\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 5 + \frac{x - 1}{2} = x \\[1em] \Rightarrow 10 + x - 1 = 2x \\[1em] \Rightarrow 9 + x = 2x \\[1em] \Rightarrow x = 9.\end{aligned}$$

Hence, option 4 is the correct option.

Let the number be $x$.

Solving,

$$\begin{aligned}\Rightarrow 3x - 6 = 9 \\[1em] \Rightarrow 3x = 15 \\[1em] \Rightarrow x = 5.\end{aligned}$$

∴ The number is 5.

Hence, option 1 is the correct option.

Let the runs scored by Piyush be $x$, then the runs scored by Rohit = $2x$.

Their runs together = 100 − 4 = 96

Solving,

$$\begin{aligned}\Rightarrow x + 2x = 96 \\[1em] \Rightarrow 3x = 96 \\[1em] \Rightarrow x = 32.\end{aligned}$$

∴ The runs scored by Piyush are 32.

Hence, option 2 is the correct option.

Solving,

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

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{3} + 2x = 14 \\[1em] \Rightarrow \frac{x + 6x}{3} = 14 \\[1em] \Rightarrow \frac{7x}{3} = 14 \\[1em] \Rightarrow 7x = 42 \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, option 4 is the correct option.

Let the number be $x$.

Solving,

$$\begin{aligned}\Rightarrow 5x - 8 = 12 \\[1em] \Rightarrow 5x = 20 \\[1em] \Rightarrow x = 4.\end{aligned}$$

∴ The number is 4.

Hence, option 1 is the correct option.

Let the age of Ruhi's brother be $x$ years, then Ruhi's age = ($x$ + 7) years.

Solving,

$$\begin{aligned}\Rightarrow x + (x + 7) = 33 \\[1em] \Rightarrow 2x + 7 = 33 \\[1em] \Rightarrow 2x = 26 \\[1em] \Rightarrow x = 13.\end{aligned}$$

∴ Age of Ruhi = $x$ + 7 = 13 + 7 = 20 years

Hence, option 3 is the correct option.

Let the age of B be $x$ years, then A's age = ($3x$ + 5) years.

Solving,

$$\begin{aligned}\Rightarrow x + (3x + 5) = 57 \\[1em] \Rightarrow 4x + 5 = 57 \\[1em] \Rightarrow 4x = 52 \\[1em] \Rightarrow x = 13.\end{aligned}$$

∴ The age of B is 13 years.

Hence, option 3 is the correct option.

Let the lowest score be $x$.

Solving,

$$\begin{aligned}\Rightarrow 2x + 7 = 87 \\[1em] \Rightarrow 2x = 80 \\[1em] \Rightarrow x = 40.\end{aligned}$$

∴ The lowest score is 40.

Hence, option 1 is the correct option.

Let the number be $x$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{4} = 3 + 7 \\[1em] \Rightarrow \frac{x}{4} = 10 \\[1em] \Rightarrow x = 40.\end{aligned}$$

∴ The number is 40.

Hence, option 3 is the correct option.

Let the number be $x$.

Solving,

$$\begin{aligned}\Rightarrow \frac{3}{4}x + 3 = 21 \\[1em] \Rightarrow \frac{3}{4}x = 18 \\[1em] \Rightarrow 3x = 72 \\[1em] \Rightarrow x = 24.\end{aligned}$$

∴ The number is 24.

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 3(x - 1) = x - 11 \\[1em] \Rightarrow 3x - 3 = x - 11 \\[1em] \Rightarrow 3x - x = -11 + 3 \\[1em] \Rightarrow 2x = -8 \\[1em] \Rightarrow x = -4.\end{aligned}$$

Hence, option 2 is the correct option.

Dividing both the sides of an equation by the same non-zero number does not change the equation. Hence, Statement 1 is true.

If $n$ is a root of $ax + b$ = 0, then $an + b$ = 0, which gives $n = -\frac{b}{a}, a \neq 0$. Hence, Statement 2 is true.

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{0.5x + 2}{1.4x + 4} = \frac{5}{7} \\[1em] \Rightarrow 7(0.5x + 2) = 5(1.4x + 4) \\[1em] \Rightarrow 3.5x + 14 = 7x + 20 \\[1em] \Rightarrow 14 - 20 = 7x - 3.5x \\[1em] \Rightarrow -6 = 3.5x \\[1em] \Rightarrow x = \frac{-6}{3.5} = -\frac{12}{7} = -1\frac{5}{7}.\end{aligned}$$

So, Assertion (A) is true.

$ax + b = 0$, where $a \neq 0$ and $a, b$ ∈ R, is indeed a linear equation in one variable. So, Reason (R) is true.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 4(2x - 5)(x - 3) = (4x - 3)(2x + 1) \\[1em] \Rightarrow 4(2x^{2} - 6x - 5x + 15) = 8x^{2} + 4x - 6x - 3 \\[1em] \Rightarrow 8x^{2} - 44x + 60 = 8x^{2} - 2x - 3.\end{aligned}$$

On simplifying, the terms $8x^2$ cancel out and we get $−44x + 60 = −2x − 3$, in which the greatest exponent of $x$ is one.

So, the given equation is a linear equation and Assertion (A) is false.

An equation in which the greatest exponent of the variable after simplification is one, is called a linear equation in one unknown. So, Reason (R) is true.

Hence, option 2 is the correct option.

Let the number be $x$.

Solving,

$$\begin{aligned}\Rightarrow 73 - 3x = 7 \\[1em] \Rightarrow 73 - 7 = 3x \\[1em] \Rightarrow 66 = 3x \\[1em] \Rightarrow x = 22.\end{aligned}$$

Since the number is 22 and not 68, Assertion (A) is false.

The method described for solving a word problem involving one unknown is correct. So, Reason (R) is true.

Hence, option 2 is the correct option.

On transposing −3 to R.H.S. it becomes +3, and on transposing $3x$ to L.H.S. it becomes −$3x$.

Solving,

$$\begin{aligned}\Rightarrow 7x - 3 = 3x + 9 \\[1em] \Rightarrow 7x - 3x = 9 + 3.\end{aligned}$$

So, Assertion (A) is true.

Transposition is indeed the process of taking a term of an equation from one side to the other side with its sign changed. So, Reason (R) is true.

Hence, option 3 is the correct option.