Linear Equations

Solve the following equation and check your answer:

5(z + 4) = 35

Answer

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.

Solve the following equation and check your answer:

$2\Big(y -\dfrac{5}{2}\Big) = 0.3$

Answer

We have:

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

∴ y = 2.65

Check:

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

= 2(2.65 - 2.5)

= 2(0.15) = 0.3

RHS = 0.3

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{5}{8}x - 6 = 9$

Answer

We have:

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

∴ x = 24

Check:

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

= $\dfrac{5}{8}$ × 24 - 6

= 15 - 6

= 9

RHS = 9

Hence, LHS = RHS.

Solve the following equation and check your answer:

$3x - \dfrac{5}{3} = x - 3$

Answer

We have:

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

∴ x = $-\dfrac{2}{3}$

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

0.6x - 1.9 = 0.2x + 0.5

Answer

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

⇒ $x = \dfrac{2.4}{0.4}$

∴ 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.

Solve the following equation and check your answer:

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

Answer

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 = $\dfrac{-7}{-2}$

∴ y = $3\dfrac{1}{2}$

Check:

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

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

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

= 17 - 8

= 9

RHS = 9

Hence, LHS = RHS.

Solve the following equation and check your answer:

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

Answer

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 = $\dfrac{4}{8}$

∴ x = $\dfrac{1}{2}$

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

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

Answer

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 = $\dfrac{-4}{2}$

⇒ x = $\dfrac{-2}{1}$

∴ 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.

Solve the following equation and check your answer:

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

Answer

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 = $\dfrac{-11}{11}$

⇒ x = $\dfrac{-1}{1}$

∴ 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.

Solve the following equation and check your answer:

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

Answer

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 = $\dfrac{28}{4}$

⇒ p = $\dfrac{7}{1}$

∴ 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.

Solve the following equation and check your answer:

$\dfrac{2x + 3}{3 +x} = \dfrac{3}{2}$

Answer

We have:

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

∴ x = 3

Check:

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

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

= $\dfrac{9}{6}$

= $\dfrac{3}{2}$

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{8 - 3x}{5x + 31} = \dfrac{2}{3}$

Answer

We have:

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

∴ x = -2

Check:

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

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

= $\dfrac{14}{21}$

= $\dfrac{2}{3}$

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{x}{3}+\dfrac{x}{4} = 14$

Answer

We have:

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

∴ x = 24

Check:

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

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

= 8 + 6 = 14

RHS = 14

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{2x}{3} + 4x = 42$

Answer

We have:

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

∴ x = 9

Check:

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

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

= $6 + 36$

= $42$

RHS = $42$

Hence, LHS = RHS.

Solve the following equation and check your answer:

x - 24% of x = 38

Answer

We have:

$\phantom{=} x - 24$

∴ x = 50

Check:

LHS = x - 24% of x

= 50 - 24% of 50

= 50 - 12

= 38

RHS = 38

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{x + 5}{2} + \dfrac{x}{3} = 20$

Answer

We have:

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

∴ x = 21

Check:

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

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

= 13 + 7

= 20

RHS = 20

Hence, LHS = RHS.

Solve the following equation and check your answer:

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

Answer

We have:

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

∴ x = 6

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{3y - 2}{7} - \dfrac{5y - 8}{4} = \dfrac{1}{14}$

Answer

We have:

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

∴ y = 2

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

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

Answer

We have:

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

∴ x = 11

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

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

Answer

We have:

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

∴ x = -9

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{2}{3}(3x - 2) = \dfrac{4}{5}(2x - 3)-\dfrac{4}{3}$

Answer

We have:

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

∴ x = -6

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

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

Answer

We have:

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

∴ x = $2\dfrac{7}{8}$

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

$x-\Big(2x - \dfrac{3x - 4}{7}\Big) = \dfrac{4x - 27}{3} - 3$

Answer

We have:

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

∴ x = 6

Check:

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

Hence, LHS = RHS.

Solve the following equation and check your answer:

$\dfrac{3}{4}(7x - 1)-\Big(2x -\dfrac{1-x}{2}\Big) = x + \dfrac{3}{2}$

Answer

We have:

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

∴ x = 1

Check:

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

Hence, LHS = RHS.

Three-sevenths of a number is 12. Find the number.

Answer

Let the required number be x.

Then, three-sevenths of this number = 12

$\therefore \dfrac{3}{7}x = 12 \\[1em] 3x = 12 \times 7 \\[1em] 3x = 84 \\[1em] x = \dfrac{84}{3} \\[1em] = 28$

Hence, the required number is 28.

A number increased by 9 gives 43. Find the number.

Answer

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.

A number diminished by 11 gives 57. Find the number.

Answer

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.

Thrice a number increased by 6 equals 39. Find the number.

Answer

Let the required number be x.

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

∴ 3x + 6 = 39

⇒ 3x = 39 - 6

⇒ 3x = 33

⇒ x = $\dfrac{33}{3}$

⇒ x = 11

Hence, the required number is 11.

Three-fourths of a number exceeds its one-third by 15. Find the number.

Answer

Let the required number be x.

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

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

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

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

Hence, the required number is 36.

A number when divided by 4 is reduced by 21. Find the number.

Answer

Let the required number be x.

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

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

Hence, the required number is 28.

A number is as much greater than 36 as is less than 86. Find the number.

Answer

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 = $\dfrac{122}{2}$

⇒ x = 61

Hence, the required number is 61.

A number exceeds its four-sevenths by 18. Find the number.

Answer

Let the required number be x.

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

According to the given condition we have:

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

Hence, the required number is 42.

A number exceeds 20% of itself by 40. Find the number.

Answer

Let the required number be x.

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

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

$\therefore x - \dfrac{1}{5}x = 40 \\[1em] \Rightarrow \dfrac{5x - x}{5} = 40 \\[1em] \Rightarrow \dfrac{4x}{5} = 40 \\[1em] \Rightarrow 4x = 40 \times 5 \\[1em] \Rightarrow 4x = 200 \\[1em] \Rightarrow x = \dfrac{200}{4} \\[1em] \Rightarrow x = 50$

Hence, the required number is 50.

If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.

Answer

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.

One fourth of a number is increased by 7 and the result is multiplied by 3. Thus, we obtain 36. Find the number.

Answer

Let the required number be x.

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

According to the given condition we have:

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

Hence, the required number is 20.

The sum of two consecutive odd numbers is 56. Find the numbers.

Answer

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 = $\dfrac{54}{2}$

⇒ x = 27

The numbers are 27 and 27 + 2 = 29.

Hence, the required numbers are 27 and 29.

The sum of three consecutive even numbers is 48. Find the numbers.

Answer

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 = $\dfrac{42}{3}$

⇒ x = 14

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

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

One of the two numbers exceeds the other by 9. Four times the smaller added to five times the larger gives 108. Find the numbers.

Answer

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 = $\dfrac{63}{9}$

⇒ x = 7

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

Hence, the required numbers are 16 and 7.

In a class of 40 students, the number of girls is three-fifths of the number of boys. Find the number of boys in the class.

Answer

Let the number of boys be x.

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

Total students = 40.

Boys + Girls = Total students

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

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

The length of a rectangular park is three times its breadth. If the perimeter of the park is 192 metres, find the dimensions of the park.

Answer

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 = $\dfrac{192}{8}$

⇒ 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.

Two equal sides of a triangle are each 5 metres less than twice the third side. If the perimeter of the triangle is 55 metres, find the lengths of its sides.

Answer

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 = $\dfrac{65}{5}$

⇒ 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.

Two supplementary angles differ by 44°. Find the angles.

Answer

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 = $\dfrac{136}{2}$

⇒ x = 68

⇒ (x + 44)° = (68 + 44)° = 112°.

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

The total cost of 3 tables and 2 chairs is ₹ 8745. If a table costs ₹ 40 more than a chair, find the price of each.

Answer

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 = ₹ $\dfrac{8625}{5}$

⇒ 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.

The denominator of a fraction is 3 more than the numerator. If 2 is added to the numerator and 5 is added to the denominator, the fraction becomes $\dfrac{1}{2}$. Find the fraction.

Answer

Let the numerator of the required fraction be x.

Then, its denominator = (x + 3).

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

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

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

Numerator = x = 4

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

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

A man is twice as old as his son. 20 years ago, the age of the man was 12 times the age of the son. Find their present ages.

Answer

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 = $\dfrac{-220}{-10}$

⇒ 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.

A man is 28 years older than his son. After 10 years, he will be thrice as old as his son. Find their present ages.

Answer

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 = $\dfrac{-8}{-2}$

⇒ 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.

Sunita is 24 years older than her daughter Kavita. 6 years ago, Sunita was thrice as old as Kavita. Find their present ages.

Answer

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 = $\dfrac{36}{2}$

⇒ 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.

Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8.

Answer

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:

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

First part = x = 72.

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

Hence, the two parts are 72 and 112.

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

Answer

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 = $\dfrac{-400}{-5}$

⇒ 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.

There are some 50 paisa and some 25 paisa coins in a bag. If the total number of coins is 30 and their total value is ₹ 11, find the number of coins of each kind.

Answer

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 = $\dfrac{350}{25}$

⇒ 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.

A labourer is engaged for 20 days on the condition that he will receive ₹ 280 for each day he works and will be fined ₹ 60 for each day he is absent. If he receives ₹ 2540 in all, for how many days did he remain absent?

Answer

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

⇒ 5600 - 280x - 60x = 2540 $\quad$ [Removing brackets]

⇒ 5600 - 340x = 2540

⇒ 5600 - 2540 = 340x $\quad$ [Transposing -340x to RHS and +2540 to LHS]

⇒ 3060 = 340x

⇒ x = $\dfrac{3060}{340}$

⇒ x = 9

Hence, the labourer remained absent for 9 days.

If 7x - 12 = 13 - 5x - 5, then the value of x is

1. $\dfrac{3}{4}$

2. $\dfrac{5}{3}$

3. $\dfrac{6}{5}$

4. $\dfrac{7}{4}$

Answer

Given:

7x - 12 = 13 - 5x - 5

⇒ 7x - 12 = 8 - 5x

⇒ 7x + 5x = 8 + 12

⇒ 12x = 20

⇒ x = $\dfrac{20}{12}$

⇒ x = $\dfrac{5}{3}$

Hence, option 2 is the correct option.

If 5(n - 3) = 4(n - 2), then the value of n is

1. 5
2. 6
3. 7
4. 8

Answer

Given:

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

⇒ 5n - 15 = 4n - 8

⇒ 5n - 4n = 15 - 8

⇒ n = 7

Hence, option 3 is the correct option.

If 7(5 - 6z) = 11(5 - 4z), then the value of z is

1. 10

2. $\dfrac{12}{5}$

3. 12

4. $\dfrac{18}{7}$

Answer

Given:

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

⇒ 35 - 42z = 55 - 44z

⇒ 44z - 42z = 55 - 35 $\quad$ [Transposing +35 to RHS and -44z to LHS]

⇒ 2z = 20

⇒ z = $\dfrac{20}{2}$

⇒ z = 10

Hence, option 1 is the correct option.

If $2x + \dfrac{3}{5} = \dfrac{2}{5}x + 3$, then the value of x is

1. 1

2. $\dfrac{3}{2}$

3. 2

4. $\dfrac{5}{2}$

Answer

We have:

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

Hence, option 2 is the correct option.

If $a - 3 = \dfrac{a}{2} + \dfrac{a}{3}$, then the value of $a$ is

1. 12
2. 15
3. 18
4. 24

Answer

We have:

$\phantom{=} a - 3 = \dfrac{a}{2} + \dfrac{a}{3} \\[1em] \Rightarrow a - 3 = \dfrac{3a + 2a}{6} \\[1em] \Rightarrow 6(a - 3) = 5a \\[1em] \Rightarrow 6a - 18 = 5a \\[1em] \Rightarrow 6a - 5a = 18 \quad \text{[Transposing -18 to RHS and +5a to LHS]} \\[1em] \Rightarrow a = 18$

Hence, option 3 is the correct option.

If $\dfrac{k-2}{k+3} = \dfrac{4}{9}$, then the value of k is

1. 5
2. 6
3. 8
4. 9

Answer

We have:

$\phantom{=} \dfrac{k - 2}{k + 3} = \dfrac{4}{9} \\[1em] \Rightarrow 9(k - 2) = 4(k + 3) \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 9k - 18 = 4k + 12 \\[1em] \Rightarrow 9k - 4k = 12 + 18 \quad \text{[Transposing -18 to RHS and +4k to LHS]} \\[1em] \Rightarrow 5k = 30 \\[1em] \Rightarrow k = \dfrac{30}{5} \\[1em] \Rightarrow k = 6$

Hence, option 2 is the correct option.

If $\dfrac{x+5}{6} - \dfrac{x+1}{9} = \dfrac{x+2}{4}$, then the value of x is

1. $\dfrac{5}{4}$

2. $\dfrac{6}{5}$

3. $\dfrac{7}{6}$

4. $\dfrac{8}{7}$

Answer

We have:

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

Hence, option 4 is the correct option.

A number is 6 more than another. If their sum is 32, then the smaller number is

1. 11
2. 12
3. 13
4. 14

Answer

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 = $\dfrac{26}{2}$

⇒ x = 13

Hence, option 3 is the correct option.

A sum of ₹ 158 is to be divided between A, B and C, so that A gets ₹ 8 more than B and C gets ₹ 10 more than A. Then, A gets

1. ₹ 44
2. ₹ 48
3. ₹ 52
4. ₹ 60

Answer

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 = $\dfrac{132}{3}$

⇒ x = 44

A gets ₹ (x + 8) = ₹ (44 + 8) = ₹ 52.

Hence, option 3 is the correct option.

Three times a number exceeds its one-third by 64. The number is

1. 24
2. 36
3. 42
4. 48

Answer

Let the number be x.

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

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

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

Hence, option 1 is the correct option.

Two supplementary angles differ by 42°. One of these angles is

1. 57°
2. 69°
3. 75°
4. 81°

Answer

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.

The sum of the digits of a two digit number is 9. On adding 27 to the number , its digits are reversed. The number is

1. 36
2. 45
3. 54
4. 72

Answer

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

⇒ 117 - 9 = 9x + 9x $\quad$ [Transposing -9x to RHS and +9 to LHS]

⇒ 108 = 18x

⇒ x = $\dfrac{108}{18}$

⇒ 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.

One fourth of a number exceeds one sixth of the number by $\dfrac{1}{2}$. The number is

1. 4
2. 6
3. 8
4. 12

Answer

Let the number be x.

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

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

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

$\therefore \dfrac{x}{4} - \dfrac{x}{6} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{3x - 2x}{12} = \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{1}{2} \times 12 \\[1em] \Rightarrow x = \dfrac{1}{1} \times 6 \\[1em] \Rightarrow x = 6$

Hence, option 2 is the correct option.

What number must be added to 54 so that the resulting number is 7 times the number added?

1. 6
2. 7
3. 8
4. 9

Answer

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 = $\dfrac{54}{6}$

⇒ x = 9

Hence, option 4 is the correct option.

The ages of A and B are in the ratio 2 : 5. After 6 years, their ages will be in the ratio 1 : 2. The present age of B is

1. 20 years
2. 24 years
3. 30 years
4. 36 years

Answer

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.

$\therefore \dfrac{2x + 6}{5x + 6} = \dfrac{1}{2} \\[1em] \Rightarrow 2(2x + 6) = 1(5x + 6) \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 4x + 12 = 5x + 6 \\[1em] \Rightarrow 4x - 5x = 6 - 12 \quad \text{[Transposing +12 to RHS and +5x to LHS]} \\[1em] \Rightarrow -x = -6 \\[1em] \Rightarrow x = 6$

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.

Fill in the blanks :

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

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

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

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

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

Answer

(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 = $\dfrac{15}{3}$

⇒ 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

Write true (T) or false (F) :

(i) The smaller of the two numbers whose sum is 26 and the difference is 6 is 10.

(ii) The solutions of an equation in variable z are the values of z for which L.H.S. = R.H.S.

(iii) Sum of two consecutive even natural numbers is 24. This statement can be represented by the equation x + (2x + 2) = 24.

(iv) 3x + 4 > 12 is a linear equation in one variable.

(v) x = 4 is a solution of the equation $\dfrac{5x}{2} - 1 = 4x - 7$.

Answer

(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 = $\dfrac{32}{2}$

⇒ 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:

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

Substitute x = 4 into both sides:

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

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

Sanjay runs a computer repair service company. For an HP laptop, he charges ₹ 320 for diagnosis and ₹ 90 per hour for repairs. For a Dell laptop, he charges ₹ 480 for diagnosis and ₹ 70 per hour for repairs.

(1) Sanjay repairs an HP laptop of Mr Lal. He charges ₹ 950 for the work. How long did the repair work take place?

1. 6 hours
2. 7 hours
3. 8 hours
4. 9 hours

(2) Sanjay repairs a Dell laptop of Mr Gupta. The total charges were ₹ 830. How long did the repair work take place?

1. 5 hours
2. 6 hours
3. 7 hours
4. 8 hours

(3) He repairs an HP laptop and a Dell laptop of Mr Subhash. He found that the repair work for both the laptops took equal time and also the service charges of the two laptops were the same. How long did the repair work take place for each laptop ?

1. 6 hours
2. 7 hours
3. 8 hours
4. 9 hours

(4) What were the total service charges that Mr. Subhash had to pay for both the laptops ?

1. ₹ 1040
2. ₹ 1260
3. ₹ 2040
4. ₹ 2520

Answer

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 = $\dfrac{160}{20}$

⇒ 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.

The ABS school organised Science competition for seventh class students. In this exam, there were 40 questions, all MCQ types. Three marks were awarded for each correct answer and one mark was deducted for each question that was answered incorrectly or left unanswered.

(1) Aryan scored 76 marks. How many questions did he answer correctly ?

1. 23
2. 25
3. 7
4. 29

(2) Ramya scored 84 marks. If she attempted all the questions, how many of her answers were incorrect ?

1. 9
2. 10
3. 11
4. 12

(3) Hema's score was double the number of questions she attempted correctly. How many of her answers were correct ?

1. 10
2. 15
3. 20
4. 25

(4) Bilok's score was the highest in the class. His score was 12 marks less than the maximum marks. How many of his answers were correct ?

1. 35
2. 36
3. 37
4. 38

Answer

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 = $\dfrac{116}{4}$

⇒ 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 = $\dfrac{124}{4}$

⇒ 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 = $\dfrac{40}{2}$

⇒ 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 = $\dfrac{148}{4}$

⇒ c = 37

∴ Bilok answered 37 questions correctly.

Hence, option 3 is the correct option.

Assertion: If 9 is the value of variable x in the equation $\dfrac{2x - 7}{2} = y$, then the value of y is 28.

Reason: The value which when substituted for the variable in an equation, makes LHS = RHS is called a solution of the given equation.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

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

Explanation

Assertion:

Given equation:

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

Substitute x = 9 into the equation:

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

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.

Assertion: Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the sign of the term.

Reason: Transposition is a process in which we drop a term from one side of an equation and put it on the other side with sign changed.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
3. Assertion (A) is true but Reason (R) is false.
4. Assertion (A) is false but Reason (R) is true.

Answer

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.