Linear Equations

We have:

5(z + 4) = 35

⇒ z + 4 = 7 $\quad$[Dividing both sides by 5]

⇒ z = 7 - 4 $\quad$[Transposing +4 to RHS]

∴ z = 3

Check:

LHS = 5(z + 4)

= 5(3 + 4)

= 5 × 7

= 35

RHS = 35

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} 2\Big(y -\frac{5}{2}\Big) = 0.3 \\[1em] \Rightarrow 2y - 2 \times \frac{5}{2} = 0.3 \quad \text{[Removing brackets]} \\[1em] \Rightarrow 2y - 5 = 0.3 \\[1em] \Rightarrow 2y = 0.3 + 5 \quad\text{[Transposing -5 to RHS]} \\[1em] \Rightarrow 2y = 5.3 \\[1em] \Rightarrow y = \frac{5.3}{2}\end{aligned}$$

∴ y = 2.65

Check:

LHS = $2\left(y - \frac{5}{2}\right)$

= 2(2.65 - 2.5)

= 2(0.15) = 0.3

RHS = 0.3

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \Rightarrow \frac{5}{8}x - 6 = 9 \\[1em] \Rightarrow \frac{5}{8}x = 9 + 6 \quad\text{[Transposing -6 to RHS]} \\[1em] \Rightarrow \frac{5}{8}x = 15 \\[1em] \Rightarrow x = 15 \times \frac{8}{5} \\[1em] \Rightarrow x = 3 \times \frac{8}{1}\end{aligned}$$

∴ x = 24

Check:

LHS = $\frac{5}{8}x$ - 6

= $\frac{5}{8}$ × 24 - 6

= 15 - 6

= 9

RHS = 9

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} 3x - \frac{5}{3} = x - 3 \\[1em] \Rightarrow 3x - x = \frac{5}{3} - 3 \quad \text{[Transposing } -\frac{5}{3} \text{ to RHS and +x to LHS]} \\[1em] \Rightarrow 2x = \frac{5 - 9}{3} \\[1em] \Rightarrow 2x = -\frac{4}{3} \\[1em] \Rightarrow x = -\frac{4}{3 \times 2} \\[1em] \Rightarrow x = -\frac{2}{3 \times 1}\end{aligned}$$

∴ x = $-\frac{2}{3}$

Check:

$$\begin{aligned}\text{LHS} = 3x - \frac{5}{3} \\[1em] \phantom{\text{LHS}} = 3\left(-\frac{2}{3}\right) - \frac{5}{3} \\[1em] \phantom{\text{LHS}} = -2 - \frac{5}{3} \\[1em] \phantom{\text{LHS}} = -\frac{11}{3} \\[2em] \text{RHS} = x - 3 \\[1em] \phantom{\text{RHS}} = -\frac{2}{3} - 3 \\[1em] \phantom{\text{RHS}} = -\frac{11}{3}\end{aligned}$$

Hence, LHS = RHS.

We have:

0.6x - 1.9 = 0.2x + 0.5

⇒ 0.6x - 0.2x = 0.5 + 1.9 $\quad$[Transposing -1.9 to RHS and +0.2x to LHS]

⇒ 0.4x = 2.4

⇒ $x = \frac{2.4}{0.4}$

∴ x = 6

Check:

LHS = 0.6x - 1.9

= 0.6 × 6 - 1.9

= 3.6 - 1.9

= 1.7

RHS = 0.2x + 0.5

= 0.2 × 6 + 0.5

= 1.2 + 0.5

= 1.7

Hence, LHS = RHS.

We have:

2(3y - 2) - 4(2y - 5) = 9

⇒ 6y - 4 - 8y + 20 = 9 $\quad$[Removing brackets]

⇒ -2y + 16 = 9

⇒ -2y = 9 - 16 $\quad$[Transposing +16 to RHS]

⇒ -2y = -7

⇒ y = $\frac{-7}{-2}$

∴ y = $3\frac{1}{2}$

Check:

LHS = 2(3y - 2) - 4(2y - 5)

= $2\left(3 × \frac{7}{2} - 2\right)$ - $4\left(2 × \frac{7}{2} - 5\right)$

= $2\left(\frac{17}{2}\right)$ - 4(2)

= 17 - 8

= 9

RHS = 9

Hence, LHS = RHS.

We have:

5(3 - x) + 1 = 3(x + 4)

⇒ 15 - 5x + 1 = 3x + 12 $\quad$[Removing brackets]

⇒ 16 - 5x = 3x + 12

⇒ 16 - 12 = 3x + 5x $\quad$[Transposing -5x to RHS and +12 to LHS]

⇒ 4 = 8x

⇒ x = $\frac{4}{8}$

∴ x = $\frac{1}{2}$

Check:

$$\begin{aligned}\text{LHS} = 5(3 - x) + 1 \\[1em] \phantom{\text{LHS}} = 5\left(3 - \frac{1}{2}\right) + 1 \\[1em] \phantom{\text{LHS}} = 5\left(\frac{5}{2}\right) + 1 \\[1em] \phantom{\text{LHS}} = \frac{25}{2} + 1 \\[1em] \phantom{\text{LHS}} = \frac{27}{2} \\[2em] \text{RHS} = 3(x + 4) \\[1em] \phantom{\text{RHS}} = 3\left(\frac{1}{2} + 4\right) \\[1em] \phantom{\text{RHS}} = 3\left(\frac{9}{2}\right) \\[1em] \phantom{\text{RHS}} = \frac{27}{2}\end{aligned}$$

Hence, LHS = RHS.

We have:

7 - 2(5 - 3x) = 4(x - 3) + 5

⇒ 7 - 10 + 6x = 4x - 12 + 5 $\quad$[Removing brackets]

⇒ -3 + 6x = 4x - 7

⇒ 6x - 4x = -7 + 3 $\quad$[Transposing -3 to RHS and +4x to LHS]

⇒ 2x = -4

⇒ x = $\frac{-4}{2}$

⇒ x = $\frac{-2}{1}$

∴ x = -2

Check:

LHS = 7 - 2(5 - 3x)

= 7 - 2[5 - 3(-2)]

= 7 - 2(11)

= -15

RHS = 4(x - 3) + 5

= 4(-2 - 3) + 5

= -20 + 5

= -15

Hence, LHS = RHS.

We have:

6(3x + 2) - 5(6x - 1) = 3(x - 8) - 5(7x - 6) + 9x

⇒ 18x + 12 - 30x + 5 = 3x - 24 - 35x + 30 + 9x $\quad$[Removing brackets]

⇒ -12x + 17 = -23x + 6

⇒ -12x + 23x = 6 - 17 $\quad$[Transposing +17 to RHS and -23x to LHS]

⇒ 11x = -11

⇒ x = $\frac{-11}{11}$

⇒ x = $\frac{-1}{1}$

∴ x = -1

Check:

LHS = 6(3x + 2) - 5(6x - 1)

= 6(-1) - 5(-7)

= -6 + 35

= 29

RHS = 3(x - 8) - 5(7x - 6) + 9x

= 3(-9) - 5(-13) - 9

= -27 + 65 - 9

= 29

Hence, LHS = RHS.

We have:

p - (2p + 5) - 5(1 - 2p) = 2(3 + 4p) - 3(p - 4)

⇒ p - 2p - 5 - 5 + 10p = 6 + 8p - 3p + 12 $\quad$[Removing brackets]

⇒ 9p - 10 = 5p + 18

⇒ 9p - 5p = 18 + 10 $\quad$[Transposing -10 to RHS and +5p to LHS]

⇒ 4p = 28

⇒ p = $\frac{28}{4}$

⇒ p = $\frac{7}{1}$

∴ p = 7

Check:

LHS = p - (2p + 5) - 5(1 - 2p)

= 7 - 19 - 5(-13)

= -12 + 65

= 53

RHS = 2(3 + 4p) - 3(p - 4)

= 2(31) - 3(3)

= 62 - 9

= 53

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{2x + 3}{3 +x} = \frac{3}{2} \\[1em] \Rightarrow 2(2x + 3) = 3(3 + x) \\[1em] \text{[By cross multiplication]} \\[1em] \Rightarrow 4x + 6 = 9 + 3x \\[1em] \Rightarrow 4x - 3x = 9 - 6 \\[1em] \quad \text{[Transposing +6 to RHS and +3x to LHS]}\end{aligned}$$

∴ x = 3

Check:

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

= $\frac{2(3) + 3}{3 + 3}$

= $\frac{9}{6}$

= $\frac{3}{2}$

RHS = $\frac{3}{2}$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{8 - 3x}{5x + 31} = \frac{2}{3} \\[1em] \Rightarrow 3(8 - 3x) = 2(5x + 31) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 24 - 9x = 10x + 62 \\[1em] \Rightarrow 24 - 62 = 10x + 9x \\[1em] \quad \text{[Transposing -9x to RHS and +62 to LHS]} \\[1em] \Rightarrow -38 = 19x \\[1em] \Rightarrow x = \frac{-38}{19} \\[1em]\end{aligned}$$

∴ x = -2

Check:

LHS = $\frac{8 - 3x}{5x + 31}$

= $\frac{8 - 3(-2)}{5(-2) + 31}$

= $\frac{14}{21}$

= $\frac{2}{3}$

RHS = $\frac{2}{3}$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{x}{3}+\frac{x}{4} = 14 \\[1em] \Rightarrow \frac{4x + 3x}{12} = 14 \\[1em] \Rightarrow 4x + 3x = 14 \times 12 \\[1em] \Rightarrow 7x = 168 \\[1em] \Rightarrow x = \frac{168}{7}\end{aligned}$$

∴ x = 24

Check:

LHS = $\frac{x}{3} + \frac{x}{4}$

= $\frac{24}{3} + \frac{24}{4}$

= 8 + 6 = 14

RHS = 14

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{2x}{3} + 4x = 42 \\[1em] \Rightarrow \frac{2x + 12x}{3} = 42 \\[1em] \Rightarrow 2x + 12x = 42 \times 3 \\[1em] \Rightarrow 14x = 126 \\[1em] \Rightarrow x = \frac{126}{14}\end{aligned}$$

∴ x = 9

Check:

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

= $\frac{2(9)}{3} + 4(9)$

= $6 + 36$

= $42$

RHS = $42$

Hence, LHS = RHS.

We have:

x - 24% of x = 38

$$\begin{aligned}\Rightarrow x - \frac{24}{100}x = 38 \\[1em] \Rightarrow \frac{100x - 24x}{100} = 38 \\[1em] \Rightarrow \frac{76x}{100} = 38 \\[1em] \Rightarrow 76x = 38 \times 100 \\[1em] \Rightarrow 76x = 3800 \\[1em] \Rightarrow x = \frac{3800}{76}\end{aligned}$$

∴ x = 50

Check:

LHS = x - 24% of x

= 50 - 24% of 50

= 50 - 12

= 38

RHS = 38

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{x + 5}{2} + \frac{x}{3} = 20 \\[1em] \Rightarrow \frac{3(x + 5) + 2x}{6} = 20 \\[1em] \Rightarrow 3x + 15 + 2x = 20 \times 6 \\[1em] \Rightarrow 5x + 15 = 120 \\[1em] \Rightarrow 5x = 120 - 15 \quad \text{[Transposing +15 to RHS]} \\[1em] \Rightarrow 5x = 105 \\[1em] \Rightarrow x = \frac{105}{5} \\[1em]\end{aligned}$$

∴ x = 21

Check:

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

= $\frac{21 + 5}{2} + \frac{21}{3}$

= 13 + 7

= 20

RHS = 20

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{2x + 3}{3}-\frac{3x - 2}{4} = 1 \\[1em] \Rightarrow \frac{4(2x + 3) - 3(3x - 2)}{12} = 1 \\[1em] \Rightarrow 8x + 12 - 9x + 6 = 1 \times 12 \\[1em] \Rightarrow -x + 18 = 12 \\[1em] \Rightarrow -x = 12 - 18 \quad \text{[Transposing +18 to RHS]} \\[1em] \Rightarrow -x = -6 \\[1em] \Rightarrow x = 6 \\[1em]\end{aligned}$$

∴ x = 6

Check:

$$\begin{aligned}\text{LHS} = \frac{2x + 3}{3} - \frac{3x - 2}{4} \\[1em] \phantom{\text{LHS}} = \frac{15}{3} - \frac{16}{4} \\[1em] \phantom{\text{LHS}} = 5 - 4 \\[1em] \phantom{\text{LHS}} = 1 \\[2em] \text{RHS} = 1\end{aligned}$$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{3y - 2}{7} - \frac{5y - 8}{4} = \frac{1}{14} \\[1em] \Rightarrow \frac{4(3y - 2) - 7(5y - 8)}{28} = \frac{1}{14} \\[1em] \Rightarrow 12y - 8 - 35y + 56 = \frac{1}{14} \times 28 \\[1em] \Rightarrow -23y + 48 = \frac{1}{1} \times 2 \\[1em] \Rightarrow -23y + 48 = 2 \\[1em] \Rightarrow -23y = 2 - 48 \quad \text{[Transposing +48 to RHS]} \\[1em] \Rightarrow -23y = - 46 \\[1em] y = \frac{-46}{-23} \\[1em]\end{aligned}$$

∴ y = 2

Check:

$$\begin{aligned}\text{LHS} = \frac{3y - 2}{7} - \frac{5y - 8}{4} \\[1em] \phantom{\text{LHS}} = \frac{6 - 2}{7} - \frac{10 - 8}{4} \\[1em] \phantom{\text{LHS}} = \frac{4}{7} - \frac{1}{2} \\[1em] \phantom{\text{LHS}} = \frac{8 - 7}{14} \\[1em] \phantom{\text{LHS}} = \frac{1}{14} \\[2em] \text{RHS} = \frac{1}{14}\end{aligned}$$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{x-2}{3}+\frac{x-3}{4} = \frac{x-1}{2} \\[1em] \Rightarrow \frac{4(x - 2) + 3(x - 3)}{12} = \frac{x-1}{2} \\[1em] \Rightarrow 4x - 8 + 3x - 9 = \left(\frac{x-1}{2}\right) \times 12 \\[1em] \Rightarrow 7x - 17 = \left(\frac{x-1}{1}\right) \times 6 \\[1em] \Rightarrow 7x - 17 = 6(x-1) \\[1em] \Rightarrow 7x - 17 = 6x - 6 \\[1em] \Rightarrow 7x - 6x = -6 + 17 \\[1em] \quad \text{[Transposing -17 to RHS and +6x to LHS]}\end{aligned}$$

∴ x = 11

Check:

$$\begin{aligned}\text{LHS} = \frac{x-2}{3} + \frac{x-3}{4} \\[1em] \phantom{\text{LHS}} = \frac{11-2}{3} + \frac{11-3}{4} \\[1em] \phantom{\text{LHS}} = 3 + 2 \\[1em] \phantom{\text{LHS}} = 5 \\[2em] \text{RHS} = \frac{x-1}{2} \\[1em] \phantom{\text{RHS}} = \frac{11-1}{2} \\[1em] \phantom{\text{RHS}} = 5\end{aligned}$$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{x-4}{7}-\frac{x+4}{5} = \frac{x+3}{7} \\[1em] \Rightarrow \frac{5(x - 4) - 7(x + 4)}{35} = \frac{x+3}{7} \\[1em] \Rightarrow 5x - 20 - 7x - 28 = \left(\frac{x+3}{7}\right) \times 35 \\[1em] \Rightarrow -2x - 48 = \left(\frac{x+3}{1}\right) \times 5 \\[1em] \Rightarrow -2x - 48 = 5(x+3) \\[1em] \Rightarrow -2x - 48 = 5x + 15 \\[1em] \Rightarrow -2x - 5x = 15 + 48 \\[1em] \quad \text{[Transposing -48 to RHS and +5x to LHS]} \\[1em] \Rightarrow -7x = 63 \\[1em] \Rightarrow -x = \frac{63}{7} \\[1em] \Rightarrow -x = 9\end{aligned}$$

∴ x = -9

Check:

$$\begin{aligned}\text{LHS} = \frac{x-4}{7} - \frac{x+4}{5} \\[1em] \phantom{\text{LHS}} = \frac{-13}{7} - \frac{-5}{5} \\[1em] \phantom{\text{LHS}} = -\frac{13}{7} + 1 \\[1em] \phantom{\text{LHS}} = -\frac{6}{7} \\[2em] \text{RHS} = \frac{x+3}{7} \\[1em] \phantom{\text{RHS}} = \frac{-9+3}{7} \\[1em] \phantom{\text{RHS}} = -\frac{6}{7}\end{aligned}$$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{2}{3}(3x - 2) = \frac{4}{5}(2x - 3)-\frac{4}{3} \\[1em] \Rightarrow \frac{6x - 4}{3} = \frac{12(2x - 3) - 20}{15} \\[1em] \Rightarrow \frac{6x - 4}{3} = \frac{24x - 36 - 20}{15} \\[1em] \Rightarrow \frac{6x - 4}{3} = \frac{24x - 56}{15} \\[1em] \Rightarrow 15(6x - 4) = 3(24x - 56) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 90x - 60 = 72x - 168 \\[1em] \Rightarrow 90x - 72x = 60 - 168 \\[1em] \quad \text{[Transposing -60 to RHS and +72x to LHS]} \\[1em] \Rightarrow 18x = -108 \\[1em] \Rightarrow x = \frac{-108}{18} \\[1em]\end{aligned}$$

∴ x = -6

Check:

$$\begin{aligned}\text{LHS} = \frac{2}{3}(3x - 2) \\[1em] \phantom{\text{LHS}} = \frac{2}{3}(-20) \\[1em] \phantom{\text{LHS}} = -\frac{40}{3} \\[2em] \text{RHS} = \frac{4}{5}(2x - 3) - \frac{4}{3} \\[1em] \phantom{\text{RHS}} = \frac{4}{5}(-15) - \frac{4}{3} \\[1em] \phantom{\text{RHS}} = -12 - \frac{4}{3} \\[1em] \phantom{\text{RHS}} = -\frac{40}{3}\end{aligned}$$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{3}{4}(2x - 5) - \frac{5}{6}(7 - 5x) = \frac{7x}{3} \\[1em] \Rightarrow \frac{9(2x - 5) - 10(7 - 5x)}{12} = \frac{7x}{3} \\[1em] \Rightarrow \frac{18x - 45 - 70 + 50x}{12} = \frac{7x}{3} \\[1em] \Rightarrow \frac{68x - 115}{12} = \frac{7x}{3} \\[1em] \Rightarrow 3(68x - 115) = 12(7x) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 68x - 115 = 4(7x) \\[1em] \quad \text{[Dividing by 3 on both sides]} \\[1em] \Rightarrow 68x - 115 = 28x \\[1em] \Rightarrow 68x - 28x = 115 \\[1em] \quad \text{[Transposing -115 to RHS and +28x to LHS]} \\[1em] \Rightarrow 40x = 115 \\[1em] \Rightarrow x = \frac{115}{40} \\[1em] \Rightarrow x = \frac{23}{8}\end{aligned}$$

∴ x = $2\frac{7}{8}$

Check:

$$\begin{aligned}\text{LHS} = \frac{3}{4}(2x - 5) - \frac{5}{6}(7 - 5x) \\[1em] \phantom{\text{LHS}} = \frac{3}{4}\left(\frac{23}{4} - 5\right) - \frac{5}{6}\left(7 - \frac{115}{8}\right) \\[1em] \phantom{\text{LHS}} = \frac{3}{4}\left(\frac{3}{4}\right) - \frac{5}{6}\left(-\frac{59}{8}\right) \\[1em] \phantom{\text{LHS}} = \frac{9}{16} + \frac{295}{48} \\[1em] \phantom{\text{LHS}} = \frac{322}{48} \\[1em] \phantom{\text{LHS}} = \frac{161}{24} \\[2em] \text{RHS} = \frac{7x}{3} \\[1em] \phantom{\text{RHS}} = \frac{7}{3} \times \frac{23}{8} \\[1em] \phantom{\text{RHS}} = \frac{161}{24}\end{aligned}$$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} x-\Big(2x - \frac{3x - 4}{7}\Big) = \frac{4x - 27}{3} - 3 \\[1em] \Rightarrow x - 2x + \frac{3x - 4}{7} = \frac{4x - 27 - 9}{3} \\[1em] \Rightarrow -x + \frac{3x - 4}{7} = \frac{4x - 36}{3} \\[1em] \Rightarrow \frac{-7x + 3x - 4}{7} = \frac{4x - 36}{3} \\[1em] \Rightarrow \frac{-4x - 4}{7} = \frac{4x - 36}{3} \\[1em] \Rightarrow 3(-4x - 4) = 7(4x - 36) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow -12x - 12 = 28x - 252 \\[1em] \Rightarrow -12x - 28x = 12 - 252 \\[1em] \quad \text{[Transposing -12 to RHS and +28x to LHS]} \\[1em] \Rightarrow -40x = -240 \\[1em] \Rightarrow x = \frac{-240}{-40}\end{aligned}$$

∴ x = 6

Check:

$$\begin{aligned}\text{LHS} = x - \left(2x - \frac{3x - 4}{7}\right) \\[1em] \phantom{\text{LHS}} = 6 - \left(12 - 2\right) \\[1em] \phantom{\text{LHS}} = -4 \\[2em] \text{RHS} = \frac{4x - 27}{3} - 3 \\[1em] \phantom{\text{RHS}} = \frac{24 - 27}{3} - 3 \\[1em] \phantom{\text{RHS}} = -1 - 3 \\[1em] \phantom{\text{RHS}} = -4\end{aligned}$$

Hence, LHS = RHS.

We have:

$$\begin{aligned}\phantom{=} \frac{3}{4}(7x - 1)-\Big(2x -\frac{1-x}{2}\Big) = x + \frac{3}{2} \\[1em] \Rightarrow \frac{21x - 3}{4}-\left(\frac{4x -(1 - x)}{2}\right) = \frac{2x + 3}{2} \\[1em] \Rightarrow \frac{21x - 3}{4}-\left(\frac{4x - 1 + x}{2}\right) = \frac{2x + 3}{2} \\[1em] \Rightarrow \frac{21x - 3}{4}-\frac{5x - 1}{2} = \frac{2x + 3}{2} \\[1em] \Rightarrow \frac{21x - 3 - 10x + 2}{4} = \frac{2x + 3}{2} \\[1em] \Rightarrow \frac{11x - 1}{4} = \frac{2x + 3}{2} \\[1em] \Rightarrow 2(11x - 1) = 4(2x + 3) \\[1em] \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 22x - 2 = 8x + 12 \\[1em] \Rightarrow 22x - 8x = 12 + 2 \\[1em] \quad \text{[Transposing -2 to RHS and +8x to LHS]} \\[1em] \Rightarrow 14x = 14 \\[1em] \Rightarrow x = \frac{14}{14}\end{aligned}$$

∴ x = 1

Check:

$$\begin{aligned}\text{LHS} = \frac{3}{4}(7x - 1) - \left(2x - \frac{1-x}{2}\right) \\[1em] \phantom{\text{LHS}} = \frac{3}{4}(6) - \left(2 - 0\right) \\[1em] \phantom{\text{LHS}} = \frac{9}{2} - 2 \\[1em] \phantom{\text{LHS}} = \frac{5}{2} \\[2em] \text{RHS} = x + \frac{3}{2} \\[1em] \phantom{\text{RHS}} = 1 + \frac{3}{2} \\[1em] \phantom{\text{RHS}} = \frac{5}{2}\end{aligned}$$

Hence, LHS = RHS.

Let the required number be x.

Then, three-sevenths of this number = 12

$$\begin{aligned}\therefore \frac{3}{7}x = 12 \\[1em] 3x = 12 \times 7 \\[1em] 3x = 84 \\[1em] x = \frac{84}{3} \\[1em] = 28\end{aligned}$$

Hence, the required number is 28.

Let the required number be x.

Then, the number increased by 9 = 43

∴ x + 9 = 43

⇒ x = 43 - 9

⇒ x = 34

Hence, the required number is 34.

Let the required number be x.

Then, the number diminished by 11 = 57

∴ x - 11 = 57

⇒ x = 57 + 11

⇒ x = 68

Hence, the required number is 68.

Let the required number be x.

Then, thrice the number (3x) increased by 6 = 39

∴ 3x + 6 = 39

⇒ 3x = 39 - 6

⇒ 3x = 33

⇒ x = $\frac{33}{3}$

⇒ x = 11

Hence, the required number is 11.

Let the required number be x.

Three-fourths of the number = $\frac{3}{4}x$

One-third of the number = $\frac{1}{3}x$

According to the question, the difference between them is 15.

$$\begin{aligned}\therefore \frac{3}{4}x - \frac{1}{3}x = 15 \\[1em] \Rightarrow \frac{9x - 4x}{12} = 15 \\[1em] \Rightarrow \frac{5x}{12} = 15 \\[1em] \Rightarrow 5x = 15 \times 12 \\[1em] \Rightarrow 5x = 180 \\[1em] \Rightarrow x = \frac{180}{5} \\[1em] \Rightarrow x = 36\end{aligned}$$

Hence, the required number is 36.

Let the required number be x.

When the number is divided by 4, the result is equal to 21 less than the original number.

$$\begin{aligned}\therefore \frac{x}{4} = x - 21 \\[1em] \Rightarrow x = 4(x - 21) \\[1em] \Rightarrow x = 4x - 84 \\[1em] \Rightarrow 84 = 4x - x \quad \text{[Transposing +x to RHS]} \\[1em] \Rightarrow 84 = 3x \\[1em] \Rightarrow x = \frac{84}{3} \\[1em] \Rightarrow x = 28\end{aligned}$$

Hence, the required number is 28.

Let the required number be x.

"As much greater than 36" means: (x - 36)

"As is less than 86" means: (86 - x)

Since these two differences are equal:

∴ x - 36 = 86 - x

⇒ x + x = 86 + 36 $\quad$ [Transposing -36 to RHS and -x to LHS]

⇒ 2x = 122

⇒ x = $\frac{122}{2}$

⇒ x = 61

Hence, the required number is 61.

Let the required number be x.

Four-sevenths of the number = $\frac{4}{7}x$

According to the given condition we have:

$$\begin{aligned}\phantom{=} x - \frac{4}{7}x = 18 \\[1em] \Rightarrow \frac{7x - 4x}{7} = 18 \\[1em] \Rightarrow \frac{3x}{7} = 18 \\[1em] \Rightarrow 3x = 18 \times 7 \\[1em] \Rightarrow 3x = 126 \\[1em] \Rightarrow x = \frac{126}{3} \\[1em] \Rightarrow x = 42\end{aligned}$$

Hence, the required number is 42.

Let the required number be x.

20% of the number = $\frac{20}{100}x = \frac{1}{5}x$

According to the question x exceeds 20% of itself by 40:

$$\begin{aligned}\therefore x - \frac{1}{5}x = 40 \\[1em] \Rightarrow \frac{5x - x}{5} = 40 \\[1em] \Rightarrow \frac{4x}{5} = 40 \\[1em] \Rightarrow 4x = 40 \times 5 \\[1em] \Rightarrow 4x = 200 \\[1em] \Rightarrow x = \frac{200}{4} \\[1em] \Rightarrow x = 50\end{aligned}$$

Hence, the required number is 50.

Let the required number be x.

"10 added to four times the number" = 4x + 10

"5 less than five times the number" = 5x - 5

According to the given condition we have:

4x + 10 = 5x - 5

⇒ 4x - 5x = - 5 - 10 $\quad$ [Transposing +10 to RHS and +5x to LHS]

⇒ -x = -15

⇒ x = 15

Hence, the required number is 15.

Let the required number be x.

One-fourth of the number increased by 7 is $\Big(\frac{x}{4} + 7\Big)$.

According to the given condition we have:

$$\begin{aligned}\phantom{=} 3 \times \Big(\frac{x}{4} + 7\Big) = 36 \\[1em] \Rightarrow \frac{x}{4} + 7 = \frac{36}{3} \\[1em] \Rightarrow \frac{x}{4} + 7 = 12 \\[1em] \Rightarrow \frac{x}{4} = 12 - 7 \quad \text{[Transposing +7 to RHS]} \\[1em] \Rightarrow \frac{x}{4} = 5 \\[1em] \Rightarrow x = 5 \times 4 \\[1em] \Rightarrow x = 20\end{aligned}$$

Hence, the required number is 20.

Let the two consecutive odd numbers be x and x + 2.

The sum of two consecutive odd numbers is 56:

∴ x + (x + 2) = 56

⇒ 2x + 2 = 56

⇒ 2x = 56 - 2 $\quad$ [Transposing +2 to RHS]

⇒ 2x = 54

⇒ x = $\frac{54}{2}$

⇒ x = 27

The numbers are 27 and 27 + 2 = 29.

Hence, the required numbers are 27 and 29.

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

The sum of three consecutive even numbers is 48:

∴ x + (x + 2) + (x + 4) = 48

⇒ 3x + 6 = 48

⇒ 3x = 48 - 6 $\quad$ [Transposing +6 to RHS]

⇒ 3x = 42

⇒ x = $\frac{42}{3}$

⇒ x = 14

The numbers are 14, 14 + 2 = 16, and 14 + 4 = 18.

Hence, the required numbers are 14, 16 and 18.

Let the smaller number be x.

Then the larger number = x + 9.

Four times the smaller added to five times the larger gives 108:

∴ 4(x) + 5(x + 9) = 108

⇒ 4x + 5x + 45 = 108

⇒ 9x + 45 = 108

⇒ 9x = 108 - 45 $\quad$ [Transposing +45 to RHS]

⇒ 9x = 63

⇒ x = $\frac{63}{9}$

⇒ x = 7

Smaller number = 7, Larger number = 7 + 9 = 16.

Hence, the required numbers are 16 and 7.

Let the number of boys be x.

Then the number of girls = $\frac{3}{5}x$.

Total students = 40.

Boys + Girls = Total students

$$\begin{aligned}\therefore x + \frac{3}{5}x = 40 \\[1em] \Rightarrow \frac{5x + 3x}{5} = 40 \\[1em] \Rightarrow \frac{8x}{5} = 40 \\[1em] \Rightarrow 8x = 40 \times 5 \\[1em] \Rightarrow 8x = 200 \\[1em] \Rightarrow x = \frac{200}{8} \\[1em] \Rightarrow x = 25\end{aligned}$$

Hence, the number of boys in the class is 25.

Let the breadth of the rectangular park be x metres.

Then, the length of the park = 3x metres.

Perimeter of the park = 192 metres

We know the formula:

Perimeter of a rectangle = 2(Length + Breadth)

∴ 192 = 2(3x + x)

⇒ 192 = 6x + 2x

⇒ 192 = 8x

⇒ x = $\frac{192}{8}$

⇒ x = 24

∴ Breadth = x = 24 m

And

Length = 3x = 3 × 24 m = 72 m

Hence, the dimensions of the park are : Length = 72 m and Breadth = 24 m.

Let the length of the third side be x metres.

Then, each of the two equal sides = (2x - 5) metres.

Perimeter of the triangle = 55 metres.

We know the formula:

Perimeter of a triangle = Sum of all three sides

∴ 55 = (2x - 5) + (2x - 5) + x

⇒ 55 = 2x + 2x + x - 5 - 5

⇒ 55 = 5x - 10

⇒ 55 + 10 = 5x $\quad$ [Transposing -10 to LHS]

⇒ 65 = 5x

⇒ x = $\frac{65}{5}$

⇒ x = 13

∴ Third side = x = 13 m

Each equal side = (2x - 5) m = (2 x 13 - 5) m = (26 - 5) m= 21 m

Hence, the lengths of the sides of the triangle are 13 m, 21 m and 21 m.

Let the required angles be x° and (x + 44)°.

Since the sum of supplementary angles is 180°, we have:

x + (x + 44) = 180

⇒ 2x + 44 = 180

⇒ 2x = 180 - 44 $\quad$ [Transposing +44 to RHS]

⇒ 2x = 136

⇒ x = $\frac{136}{2}$

⇒ x = 68

⇒ (x + 44)° = (68 + 44)° = 112°.

Hence, the required angles are 68° and 112°.

Let the price of a chair be ₹ x.

Then, the price of a table = ₹ (x + 40).

Total cost = Cost of 3 tables + Cost of 2 chairs

Substituting the values in above, we get:

₹ 8745 = 3(x + 40) + 2x

⇒ ₹ 8745 = 3x + 120 + 2x

⇒ ₹ 8745 = 5x + 120

⇒ ₹ 8745 - 120 = 5x $\quad$ [Transposing +120 to LHS]

⇒ ₹ 8625 = 5x

⇒ x = ₹ $\frac{8625}{5}$

⇒ x = ₹ 1725

Price of a chair = ₹ x = ₹ 1725.

Price of a table = ₹ (x + 40) = ₹ (1725 + 40) = ₹ 1765.

Hence, the price of a chair is ₹ 1725 and price of a table is ₹ 1765.

Let the numerator of the required fraction be x.

Then, its denominator = (x + 3).

The fraction is $\frac{x}{x + 3}$.

According to the question, if 2 is added to the numerator and 5 is added to the denominator, the fraction becomes $\frac{1}{2}$.

$$\begin{aligned}\therefore \frac{x + 2}{(x + 3) + 5} = \frac{1}{2} \\[1em] \Rightarrow \frac{x + 2}{x + 8} = \frac{1}{2} \\[1em] \Rightarrow 2(x + 2) = 1(x + 8) \\[1em] \text{[By cross-multiplication]} \\[1em] \Rightarrow 2x + 4 = x + 8 \\[1em] \Rightarrow 2x - x = 8 - 4 \\[1em] \text{[Transposing +4 to RHS and +x to LHS]} \\[1em] \Rightarrow x = 4\end{aligned}$$

Numerator = x = 4

Denominator = (x + 3) = 4 + 3 = 7.

Hence, the required fraction is $\frac{4}{7}$.

Let the son's present age be x years.

Then, the man's present age = 2x years.

20 years ago:

Son's age = (x - 20) years.

Man's age = (2x - 20) years.

According to the question, 20 years ago, the age of the man was 12 times the age of the son:

∴ 2x - 20 = 12(x - 20)

⇒ 2x - 20 = 12x - 240

⇒ 2x - 12x = -240 + 20 $\quad$ [Transposing -20 to RHS and +12x to LHS]

⇒ -10x = -220

⇒ x = $\frac{-220}{-10}$

⇒ x = 22

Son's present age = x = 22 years.

Man's present age = 2x years = (2 × 22) years = 44 years.

Hence, the present age of the man is 44 years and his son's age is 22 years.

Let the son's present age be x years.

Then, the man's present age = (x + 28) years.

After 10 years:

Son's age = (x + 10) years.

Man's age = (x + 28 + 10) years = (x + 38) years.

According to the question, after 10 years, the man will be thrice as old as his son:

∴ x + 38 = 3(x + 10)

⇒ x + 38 = 3x + 30

⇒ x - 3x = 30 - 38 $\quad$ [Transposing +38 to RHS and +3x to LHS]

⇒ -2x = -8

⇒ x = $\frac{-8}{-2}$

⇒ x = 4

Son's present age = x years = 4 years.

Man's present age = (x + 28) years = (4 + 28) years = 32 years.

Hence, the man's present age is 32 years and his son's age is 4 years.

Let Kavita's present age be x years.

Then, Sunita's present age = (x + 24) years.

6 years ago:

Kavita's age = (x - 6) years.

Sunita's age = (x + 24 - 6) years = (x + 18) years.

According to the question, 6 years ago, Sunita was thrice as old as Kavita:

∴ x + 18 = 3(x - 6)

⇒ x + 18 = 3x - 18

⇒ 18 + 18 = 3x - x $\quad$ [Transposing +x to RHS and -18 to LHS]

⇒ 36 = 2x

⇒ x = $\frac{36}{2}$

⇒ x = 18

Kavita's present age = x years = 18 years.

Sunita's present age = (x + 24) years = (18 + 24) years = 42 years.

Hence, Sunita's present age is 42 years and Kavita's present age is 18 years.

Let the first part be x.

Then, the second part = (184 - x).

According to the question, one-third of the first part exceeds one-seventh of the second part by 8:

$$\begin{aligned}\therefore \frac{1}{3}x - \frac{1}{7}(184 - x) = 8 \\[1em] \Rightarrow \frac{7x - 3(184 - x)}{21} = 8 \\[1em] \Rightarrow 7x - 552 + 3x = 8 \times 21 \\[1em] \Rightarrow 10x - 552 = 168 \\[1em] \Rightarrow 10x = 168 + 552 \\[1em] \text{[Transposing -552 to RHS]} \\[1em] \Rightarrow 10x = 720 \\[1em] \Rightarrow x = \frac{720}{10} \\[1em] \Rightarrow x = 72\end{aligned}$$

First part = x = 72.

Second part = (184 - x) = (184 - 72) = 112.

Hence, the two parts are 72 and 112.

Let the number of ₹ 5 notes be x.

Then, the number of ₹ 10 notes = (90 - x).

Total value of ₹ 5 notes = 5x

Total value of ₹ 10 notes = 10 × (90 - x) = 900 - 10x

According to the question, the total sum is ₹ 500:

∴ 5x + (900 - 10x) = 500

⇒ 5x + 900 - 10x = 500 $\quad$ [Removing brackets]

⇒ 5x - 10x = 500 - 900 $\quad$ [Transposing +900 to RHS]

⇒ -5x = -400

⇒ x = $\frac{-400}{-5}$

⇒ x = 80

Number of ₹ 5 notes = x = 80.

Number of ₹ 10 notes = (90 - x) = (90 - 80) = 10.

Hence, number of ₹ 5 notes = 80 and number of ₹ 10 notes = 10.

First, convert the total value to paisa:

₹ 1 = 100 paisa

∴ ₹ 11 = (11 × 100) paisa = 1100 paisa.

Let the number of 50 paisa coins be x.

Then, the number of 25 paisa coins = (30 - x).

Value of 50 paisa coins = 50x

Value of 25 paisa coins = 25 × (30 - x) = 750 - 25x

According to the question, total value is ₹ 11:

∴ 50x + (750 - 25x) = 1100

⇒ 50x + 750 - 25x = 1100 $\quad$ [Removing brackets]

⇒ 25x = 1100 - 750 $\quad$ [Transposing +750 to RHS]

⇒ 25x = 350

⇒ x = $\frac{350}{25}$

⇒ x = 14

Number of 50 paisa coins = x = 14.

Number of 25 paisa coins = (30 - x) = (30 - 14) = 16.

Hence, the number of 50 paisa coins = 14 and the number of 25 paisa coins = 16.

Let the number of days the labourer was absent be x.

Then, the number of days he worked = (20 - x).

Amount earned for working = 280 x (20 - x) = 5600 - 280x

Amount fined for being absent = 60 × x = 60x

Total amount received = Earnings - Fines

∴ (5600 - 280x) - 60x = 2540

Remove brackets:

⇒ 5600 - 280x - 60x = 2540

⇒ 5600 - 340x = 2540

Transpose -340x to RHS and +2540 to LHS:

⇒ 5600 - 2540 = 340x

⇒ 3060 = 340x

⇒ x = $\frac{3060}{340}$

⇒ x = 9

Hence, the labourer remained absent for 9 days.

Given:

7x - 12 = 13 - 5x - 5

⇒ 7x - 12 = 8 - 5x

⇒ 7x + 5x = 8 + 12

⇒ 12x = 20

⇒ x = $\frac{20}{12}$

⇒ x = $\frac{5}{3}$

Hence, option 2 is the correct option.

Given:

5(n - 3) = 4(n - 2)

⇒ 5n - 15 = 4n - 8

⇒ 5n - 4n = 15 - 8

⇒ n = 7

Hence, option 3 is the correct option.

Given:

7(5 - 6z) = 11(5 - 4z)

⇒ 35 - 42z = 55 - 44z

Transpose +35 to RHS and -44z to LHS

⇒ 44z - 42z = 55 - 35

⇒ 2z = 20

⇒ z = $\frac{20}{2}$

⇒ z = 10

Hence, option 1 is the correct option.

We have:

$$\begin{aligned}\phantom{=} 2x + \frac{3}{5} = \frac{2}{5}x + 3 \\[1em] \Rightarrow 2x - \frac{2}{5}x = 3 - \frac{3}{5} \\[1em] \Rightarrow \frac{10x - 2x}{5} = \frac{15 - 3}{5} \\[1em] \Rightarrow \frac{8x}{5} = \frac{12}{5} \\[1em] \text{[Since the denominators are same, we cancel them]} \\[1em] \Rightarrow 8x = 12 \\[1em] \Rightarrow x = \frac{12}{8} \\[1em] \Rightarrow x = \frac{3}{2}\end{aligned}$$

Hence, option 2 is the correct option.

We have:

$$\begin{aligned}\phantom{=} a - 3 = \frac{a}{2} + \frac{a}{3} \\[1em] \Rightarrow a - 3 = \frac{3a + 2a}{6} \\[1em] \Rightarrow 6(a - 3) = 5a \\[1em] \Rightarrow 6a - 18 = 5a \\[1em] \text{[Transpose -18 to RHS and +5a to LHS]} \\[1em] \Rightarrow 6a - 5a = 18 \\[1em] \Rightarrow a = 18\end{aligned}$$

Hence, option 3 is the correct option.

We have:

$$\begin{aligned}\phantom{=} \frac{k - 2}{k + 3} = \frac{4}{9} \\[1em] \text{By cross multiplication:} \\[1em] \Rightarrow 9(k - 2) = 4(k + 3) \\[1em] \Rightarrow 9k - 18 = 4k + 12 \\[1em] \text{Transpose -18 to RHS and +4k to LHS} \\[1em] \Rightarrow 9k - 4k = 12 + 18 \\[1em] \Rightarrow 5k = 30 \\[1em] \Rightarrow k = \frac{30}{5} \\[1em] \Rightarrow k = 6\end{aligned}$$

Hence, option 2 is the correct option.

We have:

$$\begin{aligned}\phantom{=} \frac{x+5}{6} - \frac{x+1}{9} = \frac{x+2}{4} \\[1em] \Rightarrow \frac{3(x + 5) - 2(x + 1)}{18} = \frac{x + 2}{4} \\[1em] \Rightarrow \frac{(3x + 15) - (2x + 2)}{18} = \frac{x + 2}{4} \\[1em] \Rightarrow \frac{3x + 15 - 2x - 2}{18} = \frac{x + 2}{4} \\[1em] \Rightarrow \frac{x + 13}{18} = \frac{x + 2}{4} \\[1em] \text{By cross multiplication:} \\[1em] \Rightarrow 4(x + 13) = 18(x + 2) \\[1em] \Rightarrow 4x + 52 = 18x + 36 \\[1em] \text{Transpose +4x to RHS and +36 to LHS} \\[1em] \Rightarrow 52 - 36 = 18x - 4x \\[1em] \Rightarrow 16 = 14x \\[1em] \Rightarrow x = \frac{16}{14} \\[1em] \Rightarrow x = \frac{8}{7}\end{aligned}$$

Hence, option 4 is the correct option.

Let smaller number be x.

Larger number is (x + 6).

According to the question their sum is 32:

∴ x + (x + 6) = 32

⇒ 2x + 6 = 32

⇒ 2x = 32 - 6

⇒ 2x = 26

⇒ x = $\frac{26}{2}$

⇒ x = 13

Hence, option 3 is the correct option.

Let B gets ₹ x.

Then A gets ₹ (x + 8)

C gets ₹ (x + 8 + 10) = ₹ (x + 18).

Total amount = ₹ 158

∴ x + (x + 8) + (x + 18) = 158

⇒ x + x + 8 + x + 18 = 158 $\quad$ [Removing brackets]

⇒ 3x + 26 = 158

⇒ 3x = 158 - 26 $\quad$ [Transposing +26 to RHS]

⇒ 3x = 132

⇒ x = $\frac{132}{3}$

⇒ x = 44

A gets ₹ (x + 8) = ₹ (44 + 8) = ₹ 52.

Hence, option 3 is the correct option.

Let the number be x.

One-third of a number = $\frac{1}{3}x = \frac{x}{3}$

According to the question, three times a number exceeds its one-third by 64:

$$\begin{aligned}\therefore 3x - \frac{x}{3} = 64 \\[1em] \Rightarrow \frac{9x - x}{3} = 64 \\[1em] \Rightarrow 8x = 64 \times 3 \\[1em] \Rightarrow 8x = 192 \\[1em] \Rightarrow x = \frac{192}{8} \\[1em] \Rightarrow x = 24\end{aligned}$$

Hence, option 1 is the correct option.

Let the required angles be x° and (x + 42)°.

Since the sum of supplementary angles is 180°, we have:

x + (x + 42) = 180

⇒ x + x + 42 = 180 $\quad$ [Removing brackets]

⇒ 2x + 42 = 180

⇒ 2x = 180 - 42 $\quad$ [Transposing +42 to RHS]

⇒ 2x = 138

⇒ x = 69°

Hence, option 2 is the correct option.

Let the digit in the units place be x.

Since the sum of the digits is 9, the digit in the tens place must be (9 - x).

A two-digit number is written as: 10 x (Tens digit) + (Units digit)

Original Number = 10(9 - x) + x

= 90 - 10x + x

= 90 - 9x

When digits are reversed, x becomes the tens digit and (9 - x) becomes the units digit.

Reversed Number = 10(x) + (9 - x)

= 10x + 9 - x

= 9x + 9

The problem states that adding 27 to the original number results in the reversed number.

Original Number + 27 = Reversed Number

(90 - 9x) + 27 = 9x + 9

⇒ 117 - 9x = 9x + 9

Transpose -9x to RHS and +9 to LHS

⇒ 117 - 9 = 9x + 9x

⇒ 108 = 18x

⇒ x = $\frac{108}{18}$

⇒ x = 6

Units digit = x = 6

Tens digit = (9 - x) = (9 - 6) = 3

∴ The original number is 36.

Hence, option 1 is the correct option.

Let the number be x.

One fourth of a number = $\frac{1}{4}x = \frac{x}{4}$

One sixth of the number = $\frac{1}{6}x = \frac{x}{6}$

According to the question, one fourth of a number exceeds one sixth of the number by $\frac{1}{2}$:

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

Hence, option 2 is the correct option.

Let the number be x.

According to the question, the resulting number is 7 times the number added:

∴ 54 + x = 7x

⇒ 54 = 7x - x

⇒ 54 = 6x

⇒ x = $\frac{54}{6}$

⇒ x = 9

Hence, option 4 is the correct option.

The ages of A and B are in the ratio 2 : 5.

Let the present age of A be 2x years.

Let the present age of B be 5x years.

After 6 years:

Age of A = (2x + 6) years.

Age of B = (5x + 6) years.

The problem states that after 6 years, their ages will be in the ratio 1 : 2.

$$\begin{aligned}\therefore \frac{2x + 6}{5x + 6} = \frac{1}{2} \\[1em] \text{By cross multiplication:} \\[1em] \Rightarrow 2(2x + 6) = 1(5x + 6) \\[1em] \Rightarrow 4x + 12 = 5x + 6 \\[1em] \text{Transpose +12 to RHS and +5x to LHS} \\[1em] \Rightarrow 4x - 5x = 6 - 12 \\[1em] \Rightarrow -x = -6 \\[1em] \Rightarrow x = 6\end{aligned}$$

Present age of A = 2x = 2(6) = 12 years.

Present age of B = 5x = 5(6) = 30 years.

∴ The present age of B is 30 years.

Hence, option 3 is the correct option.

(i) An equation involving one variable with highest power 1, is called a linear equation in that variable.

(ii) The solution of an equation is also called root of the given equation.

(iii) The process in which we drop a term from one side of an equation and put it on the other side with sign changed is called transposition.

(iv) A linear equation can have at the most 1 solution(s).

(v) If thrice a number increased by 8 gives 23, then the number is 5.

(vi) If x - 7 = 4, then 3x - 15 is equal to 18.

Explanation

(i) A linear equation is defined by the fact that the variable has an exponent of 1. If the power were 2, it would be a quadratic equation.

(ii) In algebra, the specific value of the variable that makes the equation true (LHS = RHS) is formally called the "root" or the solution.

(iii) Transposition is the mathematical term for moving terms across the equal sign (=) while reversing their operation (plus becomes minus, multiply becomes divide).

(iv) A linear equation in one variable can have at the most one solution.

(v)

Let the number be x.

According to the condition:

3x + 8 = 23

⇒ 3x = 23 - 8 $\quad$ [Transposing +8 to RHS]

⇒ 3x = 15

⇒ x = $\frac{15}{3}$

⇒ x = 5

∴ The required number is 5.

(vi)

Given equation:

x - 7 = 4

First, find the value of x:

x - 7 = 4

⇒ x = 4 + 7 $\quad$ [Transposing -7 to RHS]

⇒ x = 11

Now, substitute x = 11 into the expression 3x - 15:

3x - 15 = 3(11) - 15

= 33 - 15

= 18

∴ 3x - 15 = 18

(i) True
Reason —

Let the numbers be x and y.

Sum = x + y = 26

Difference = x - y = 6.

Adding the equations:

x + y + x - y = 26 + 6

⇒ 2x = 32

⇒ x = $\frac{32}{2}$

⇒ x = 16.

Substituting the value of x in (x + y = 26) we get:

x + y = 26

⇒ 16 + y = 26

⇒ y = 26 - 16 $\quad$ [Transposing +16 to RHS]

⇒ y = 10.

The smaller number is 10.

(ii) True
Reason — The statement is the fundamental definition of a solution. A value is only a "root" or "solution" if it makes the Left-Hand Side (L.H.S.) equal to the Right-Hand Side (R.H.S.).

(iii) False
Reason —

Consecutive even numbers are represented as x and x + 2. The equation should be:

x + (x + 2) = 24.

The expression 2x + 2 represents twice the first number plus two, which is not the definition of the next consecutive even number.

(iv) False
Reason — The symbol > makes this an inequality, not an equation. A linear equation must have an equal sign (=).

(v) True
Reason —

Given equation:

$\frac{5x}{2} - 1 = 4x - 7$

Substitute x = 4 into both sides:

$$\begin{aligned}\therefore \frac{5x}{2} - 1 = 4x - 7 \\[1em] \Rightarrow \frac{5(4)}{2} - 1 = 4(4) - 7 \\[1em] \Rightarrow \frac{20}{2} - 1 = 16 - 7 \\[1em] \Rightarrow \frac{20 - 2}{2} = 9 \\[1em] \Rightarrow \frac{18}{2} = 9 \\[1em] \Rightarrow 9 = 9\end{aligned}$$

Since, L.H.S. = R.H.S., x = 4 is the correct solution.

Given:

Sanjay's Charge Rates:

HP: ₹ 320 (fixed) + ₹ 90 per hour

Dell: ₹ 480 (fixed) + ₹ 70 per hour

(1)

Let the repair time be x hours.

∴ 320 + 90x = 950

⇒ 90x = 950 - 320 $\quad$ [Transposing +320 to RHS]

⇒ 90x = 630

⇒ x = 7 hours.

Hence, option 2 is the correct option.

(2)

Let the repair time be y hours.

∴ 480 + 70y = 830

⇒ 70y = 830 - 480 $\quad$ [Transposing +480 to RHS]

⇒ 70y = 350

⇒ y = 5 hours.

Hence, option 1 is the correct option.

(3)

Let the time for each be t hours. Since the charges are equal:

HP Charge = Dell Charge

320 + 90t = 480 + 70t

⇒ 90t - 70t = 480 - 320 $\quad$ [Transposing +320 to RHS and +70t to LHS]

⇒ 20t = 160

⇒ t = $\frac{160}{20}$

⇒ t = 8 hours

Hence, option 3 is the correct option.

(4)

Let's calculate the cost for one laptop using t = 8:

HP: ₹ (320 + 90(8)) = ₹ (320 + 720) = ₹ 1040

Dell: ₹ (480 + 70(8)) = ₹ (480 + 560) = ₹ 1040

Total for both (HP + Dell) = ₹ 1040 + ₹ 1040 = ₹ 2080

The total service charges that Mr. Subhash had to pay for both the laptops is ₹ 2080. None of the options are correct.

Given:

Total Questions: 40

Correct Answer: +3 marks

Incorrect/Unanswered: -1 mark

Let c = correct answers.

Then (40 - c) = incorrect/unanswered.

(1)

Aryan scores 76 marks:

∴ 3(c) - 1(40 - c) = 76

⇒ 3c - 40 + c = 76

⇒ 4c - 40 = 76

⇒ 4c = 76 + 40 $\quad$ [Transposing -40 to RHS]

⇒ 4c = 116

⇒ c = $\frac{116}{4}$

⇒ c = 29

Aryan answered 29 questions correctly.

Hence, option 4 is the correct option.

(2)

Ramya scored 84 marks:

∴ 3(c) - 1(40 - c) = 84

⇒ 3c - 40 + c = 84

⇒ 4c - 40 = 84

⇒ 4c = 84 + 40 $\quad$ [Transposing -40 to RHS]

⇒ 4c = 124

⇒ c = $\frac{124}{4}$

⇒ c = 31

Correct answers = 31

Incorrect answers = (40 - c) = (40 - 31) = 9.

Hence, option 1 is the correct option.

(3)

Hema's score was double the number of questions she attempted correctly:

Score = 2c

∴ 3c - (40 - c) = 2c

⇒ 3c - 40 + c = 2c

⇒ 4c - 40 = 2c

⇒ 4c - 2c = 40 $\quad$ [Transposing +2c to LHS and -40 to RHS]

⇒ 2c = 40

⇒ c = $\frac{40}{2}$

⇒ c = 20

∴ Hema answered 20 questions correctly.

Hence, option 3 is the correct option.

(4)

Bilok's score was 12 marks less than the maximum marks

Maximum possible marks = 40 x 3 = 120.

Bilok's score = 120 - 12 = 108.

∴ 3c - (40 - c) = 108

⇒ 3c - 40 + c = 108

⇒ 4c - 40 = 108

⇒ 4c = 108 + 40 $\quad$ [Transposing -40 to RHS]

⇒ 4c = 148

⇒ c = $\frac{148}{4}$

⇒ c = 37

∴ Bilok answered 37 questions correctly.

Hence, option 3 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion:

Given equation:

$\frac{2x - 7}{2} = y$

Substitute x = 9 into the equation:

$$\begin{aligned}\phantom{=} \frac{2x - 7}{2} = y \\[1em] \Rightarrow \frac{2(9) - 7}{2} = y \\[1em] \Rightarrow \frac{18 - 7}{2} = y \\[1em] \Rightarrow \frac{11}{2} = y \\[1em] 5.5 = y\end{aligned}$$

Since y = 5.5 and not 28, the Assertion is False.

Reason:

The definition provided for a "solution" or "root" of an equation is mathematically perfect. Therefore, the Reason is True.

Hence, option 4 is the correct option.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

Assertion:

The statement describes the standard rule of algebra. If we move +5 to the other side, it becomes -5. This is a true statement.

Reason provides the formal definition of the process mentioned in the assertion. It is also true and correct explanation of assertion.

Hence, option 1 is the correct option.

Statement 1: x − 6 = 16

The expression x − 6 means 6 is subtracted from x, which represents the difference between x and 6. So, the equation can be read as "the difference between x and 6 is 16."

Hence, Statement 1 is True.

Statement 2: (2 + 3)t = 35

The given expression (2 + 3)t means (2 + 3) is multiplied by t, i.e., 5t = 35.

The statement "3 times a number plus 2 gives 35" would be written as 3t + 2 = 35, which is not the same as (2 + 3)t = 35.

Hence, Statement 2 is False.

Statement 3: p − 1 = 11

The expression p − 1 means 1 is subtracted from p. So, the equation can be read as "1 subtracted from p is 11."

Hence, Statement 3 is True.

Hence, option 2 is the correct option.

Given equation:

2(x − 4) = 16

Kartik's Method — Dividing both sides by 2:

$$\begin{aligned}\phantom{=} 2(x - 4) = 16 \\[1em] \Rightarrow \frac{2(x - 4)}{2} = \frac{16}{2} \\[1em] \Rightarrow x - 4 = 8\end{aligned}$$

So, Kartik's rewriting is correct.

Sanjay's Method — Removing the bracket by multiplication:

$$\begin{aligned}\phantom{=} 2(x - 4) = 16 \\[1em] \Rightarrow 2 \times x - 2 \times 4 = 16 \\[1em] \Rightarrow 2x - 8 = 16\end{aligned}$$

So, Sanjay's rewriting is correct.

Both methods are valid ways of simplifying the same equation and will lead to the same solution x = 12.

Hence, option 3 is the correct option.

Let the number of bananas in each small box be x.

Then, the number of bananas in 6 small boxes = 6x.

According to the question, a large box contains as many as 6 small boxes plus 2 loose bananas, and the total number of bananas in 1 large box is 50.

∴ 6x + 2 = 50

Hence, option 3 is the correct option.

Statement P:

Given equation:

3(x + 3) − 2(x − 1) = 5(x − 5)

⇒ 3x + 9 − 2x + 2 = 5x − 25 $\quad$ [Removing brackets]

⇒ x + 11 = 5x − 25

⇒ 11 + 25 = 5x − x $\quad$ [Transposing -25 to LHS and +x to RHS]

⇒ 36 = 4x

⇒ x = $\frac{36}{4}$

⇒ x = 9

Since the value of x is 9, Statement P is True.

Statement Q:

Given equation:

$\frac{y}{2} + \frac{3}{2} = \frac{2y}{5} + 1$

$$\begin{aligned}\phantom{=} \frac{y}{2} + \frac{3}{2} = \frac{2y}{5} + 1 \\[1em] \Rightarrow \frac{y + 3}{2} = \frac{2y + 5}{5} \\[1em] \Rightarrow 5(y + 3) = 2(2y + 5) \quad\text{[By cross multiplication]} \\[1em] \Rightarrow 5y + 15 = 4y + 10 \quad\text{[Removing brackets]} \\[1em] \Rightarrow 5y - 4y = 10 - 15 \quad\text{[Transposing +15 to RHS and +4y to LHS]} \\[1em] \Rightarrow y = -5\end{aligned}$$

Since the value of y is −5 and not 5, Statement Q is False.

Hence, option 3 is the correct option.

Let the number of people in the second room be x.

Then, the number of people in the first room = (x + 9).

And, the number of people in the third room = (x − 1).

According to the question, the total number of people in all three rooms is 41:

∴ (x + 9) + x + (x − 1) = 41

⇒ x + 9 + x + x − 1 = 41 $\quad$[Removing brackets]

⇒ 3x + 8 = 41

⇒ 3x = 41 − 8 $\quad$[Transposing +8 to RHS]

⇒ 3x = 33

⇒ x = $\frac{33}{3}$

⇒ x = 11

∴ The number of people in the second room is 11.

Hence, option 2 is the correct option.