Linear Equations in One Variable

$\frac{x + 0.5}{6x - 1} = \frac{1}{2}$

On cross multiplying, we get

⇒ (x + 0.5) $\times$ 2 = 1 $\times$ (6x - 1)

⇒ 2 $\times$ x + 2 $\times$ 0.5 = 1 $\times$ 6x - 1 $\times$ 1

⇒ 2x + 1 = 6x - 1

⇒ 6x - 2x = 1 + 1

⇒ 4x = 2

⇒ x = $\frac{2}{4}$

⇒ x = $\frac{1}{2}$

Hence, option 1 is the correct option.

3(x + 1) - 4x = 0

⇒ 3x + 3 - 4x = 0

⇒ 3 - x = 0

⇒ x = 3

Hence, option 3 is the correct option.

7x - 8 = 5x + 2

⇒ 7x - 5x = 2 + 8

⇒ 2x = 10

⇒ x = $\frac{10}{2}$

⇒ x = 5

Hence, option 3 is the correct option.

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

On cross multiplying, we get :

⇒ 5 $\times$ (x - 4) = 7 $\times$ x

⇒ 5 $\times$ x - 5 $\times$ 4 = 7x

⇒ 5x - 20 = 7x

⇒ 5x - 7x = 20

⇒ - 2x = 20

⇒ x = - $\frac{20}{2}$

⇒ x = - 10

Hence, option 1 is the correct option.

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

Since L.C.M. of denominators 4 and 5 = 20, multiply each term by 20 to get:

⇒ $\frac{20(x - 2)}{4} - \frac{20(x - 3)}{5} = 1 \times 20$

⇒ 5(x - 2) - 4(x - 3) = 20

⇒ 5x - 10 - 4x + 12 = 20

⇒ x + 2 = 20

⇒ x = 20 - 2

⇒ x = 18

Hence, option 4 is the correct option.

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

On cross multiplying, we get :

⇒ (3x + 2) = - 7 $\times$ (x - 6)

⇒ 3x + 2 = - 7x + 7 $\times$ 6

⇒ 3x + 2 = - 7x + 42

⇒ 3x + 7x = 42 - 2

⇒ 10x = 40

⇒ x = $\frac{40}{10}$

⇒ x = 4

Hence, the value of x is 4.

3a - 4 = 2(4 - a)

⇒ 3a - 4 = 8 - 2a

⇒ 3a + 2a = 8 + 4

⇒ 5a = 12

⇒ a = $\frac{12}{5}$

⇒ a = 2.4

Hence, the value of a is 2.4.

3(b - 4) = 2(4 - b)

⇒ 3b - 12 = 8 - 2b

⇒ 3b + 2b = 8 + 12

⇒ 5b = 20

⇒ b = $\frac{20}{5}$

⇒ b = 4

Hence, the value of b is 4.

$\frac{x + 2}{9} = \frac{x + 4}{11}$

On cross multiplying, we get :

⇒ (x + 2) $\times$ 11 = (x + 4) $\times$ 9

⇒ 11x + 22 = 9x + 36

⇒ 11x - 9x = 36 - 22

⇒ 2x = 14

⇒ x = $\frac{14}{2}$

⇒ x = 7

Hence, the value of x is 7.

$\frac{x - 8}{5} = \frac{x - 12}{9}$

On cross multiplying , we get :

⇒ (x - 8) $\times$ 9 = (x - 12) $\times$ 5

⇒ 9x - 72 = 5x - 60

⇒ 9x - 5x = - 60 + 72

⇒ 4x = 12

⇒ x = $\frac{12}{4}$

⇒ x = 3

Hence, the value of x is 3.

5(8x + 3) = 9(4x + 7)

⇒ 40x + 15 = 36x + 63

⇒ 40x - 36x = 63 - 15

⇒ 4x = 48

⇒ x = $\frac{48}{4}$

⇒ x = 12

Hence, the value of x is 12.

3(x + 1) = 12 + 4(x - 1)

⇒ 3x + 3 = 12 + 4x - 4

⇒ 3x + 3 = 8 + 4x

⇒ 3x - 4x = 8 - 3

⇒ - x = 5

⇒ x = -5

Hence, the value of x is - 5.

$$\begin{aligned}\frac{3x}{4} - \frac{1}{4}(x - 20) = \frac{x}{4} + 32\\[1em] ⇒ \frac{3x}{4} - \frac{x}{4} + \frac{20}{4} = \frac{x}{4} + 32\\[1em] ⇒ \frac{3x}{4} - \frac{x}{4} + 5 = \frac{x}{4} + 32\\[1em] ⇒ \frac{3x - x}{4} + 5 = \frac{x}{4} + 32\\[1em] ⇒ \frac{2x}{4} + 5 = \frac{x}{4} + 32\\[1em] ⇒ \frac{2x}{4} - \frac{x}{4} = 32 - 5\\[1em] ⇒ \frac{2x - x}{4} = 32 - 5\\[1em] ⇒ \frac{x}{4} = 27\\[1em] ⇒ x = 27 \times 4\\[1em] ⇒ x = 108\end{aligned}$$

Hence, the value of x is 108.

$$\begin{aligned}3a - \frac{1}{5} = \frac{a}{5} + 5\frac{2}{5}\\[1em] ⇒ 3a - \frac{1}{5} = \frac{a}{5} + \frac{27}{5}\\[1em] ⇒ 3a - \frac{a}{5} = \frac{27}{5} + \frac{1}{5}\\[1em] ⇒ \frac{15a}{5} - \frac{a}{5} = \frac{27}{5} + \frac{1}{5}\\[1em] ⇒ \frac{15a - a}{5} = \frac{27 + 1}{5} \\[1em] ⇒ \frac{14a}{5} = \frac{28}{5} \\[1em] ⇒ a = \frac{28 \times 5}{5 \times 14}\\[1em] ⇒ a = \frac{140}{70}\\[1em] ⇒ a = 2\end{aligned}$$

Hence, the value of a is 2.

$$\begin{aligned}\frac{x}{3} - 2\frac{1}{2} = \frac{4x}{9} - \frac{2x}{3}\\[1em] ⇒ \frac{x}{3} - \frac{5}{2} = \frac{4x}{9} - \frac{2x}{3}\\[1em]\end{aligned}$$

Since L.C.M. of denominators 3, 2 and 9 = 18, multiply each term by 18 to get:

⇒ $\frac{18 \times x}{3} - \frac{18 \times 5}{2} = \frac{18 \times 4x}{9} - \frac{18 \times 2x}{3}$

⇒ 6x - 9 $\times$ 5 = 2 $\times$ 4x - 6 $\times$ 2x

⇒ 6x - 45 = 8x - 12x

⇒ 6x + 4x = 45

⇒ 10x = 45

⇒ x = $\frac{45}{10}$

⇒ x = 4.5

Hence, the value of x is 4.5.

$$\begin{aligned}\frac{4(y + 2)}{5} = 7 + \frac{5y}{13}\\[1em] ⇒ \frac{4y + 8}{5} = 7 + \frac{5y}{13}\\[1em] ⇒ \frac{4y}{5} + \frac{8}{5} = 7 + \frac{5y}{13}\\[1em] ⇒ \frac{4y}{5} - \frac{5y}{13} = 7 - \frac{8}{5}\\[1em]\end{aligned}$$

Since L.C.M. of denominators 5 and 13 = 65, multiply each term by 65 to get:

$⇒ \frac{4y \times 65}{5} - \frac{5y \times 65}{13} = 7 \times 65 - \frac{8 \times 65}{5}$

⇒ 4y $\times$ 13 - 5y $\times$ 5 = 455 - 8 $\times$ 13

⇒ 52y - 25y = 455 - 104

⇒ 27y = 351

⇒ y = $\frac{351}{27}$

⇒ y = 13

Hence, the value of y is 13.

$\frac{a + 5}{6} - \frac{a + 1}{9} = \frac{a + 3}{4}$

Since L.C.M. of denominators 6, 9 and 4 = 36, multiply each term by 36 to get:

$⇒ \frac{36(a + 5)}{6} - \frac{36(a + 1)}{9} = \frac{36(a + 3)}{4}$

⇒ 6(a + 5) - 4(a + 1) = 9(a +3)

⇒ (6a + 30) - (4a + 4) = (9a + 27)

⇒ 6a + 30 - 4a - 4 = 9a + 27

⇒ 2a + 26 = 9a + 27

⇒ 2a - 9a = 27 - 26

⇒ - 7a = 1

⇒ a = -$\frac{1}{7}$

Hence, the value of a is $-\frac{1}{7}$.

$\frac{2x - 13}{5} - \frac{x - 3}{11} = \frac{x - 9}{5} + 1$

Since L.C.M. of denominators 5 and 11 = 55, multiply each term by 55 to get:

$$\begin{aligned}⇒ \frac{(2x - 13) \times 55}{5} - \frac{(x - 3) \times 55}{11} = \frac{(x - 9) \times 55}{5} + 1 \times 55\\[1em]\end{aligned}$$

⇒ 11 (2x - 13) - 5 (x - 3) = 11 (x - 9) + 55

⇒ (22x - 143) - (5x - 15) = (11x - 99) + 55

⇒ 22x - 143 - 5x + 15 = 11x - 99 + 55

⇒ 17x - 128 = 11x - 44

⇒ 17x - 11x = 128 - 44

⇒ 6x = 84

⇒ x = $\frac{84}{6}$

⇒ x = 14

Hence, the value of x is 14.

6(6x - 5) - 5(7x - 8) = 12(4 - x) + 1

⇒ 36x - 30 - 35x + 40 = 48 - 12x + 1

⇒ 1x + 10 = 49 - 12x

⇒ 1x + 12x = 49 - 10

⇒ 13x = 39

⇒ x = $\frac{39}{13}$

⇒ x = 3

Hence, the value of x is 3.

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

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

⇒ x² + 3x - 5x - 15 = x² + 4x - 7x - 28

⇒ x² - 2x - 15 = x² - 3x - 28

⇒ x² - x² - 2x + 3x = - 28 + 15

⇒ 1x = - 13

Hence, the value of x is - 13.

(x - 5)² - (x + 2)² = -2

Using the formula,

[∵ (x - y)² = x² + y² - 2xy]

And,

[∵ (x + y)² = x² + y² + 2xy]

⇒ x² + 5² - 2 $\times$ x $\times$ 5 - (x² + 2² + 2 $\times$ x $\times$ 2) = -2

⇒ x² + 25 - 10x - (x² + 4 + 4x) = -2

⇒ x² + 25 - 10x - x² - 4 - 4x = -2

⇒ 21 - 14x = -2

⇒ - 14x = -2 - 21

⇒ - 14x = - 23

⇒ x = $\frac{23}{14}$

⇒ x = $1\frac{9}{14}$

Hence, the value of x is $1\frac{9}{14}$.

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

⇒ x(x + 6) - 1(x + 6) - x(x - 3) + 2(x - 3) = 3

⇒ x² + 6x - 1x - 6 - x² + 3x + 2x - 6 = 3

⇒ x² - x² + 6x - 1x + 3x + 2x - 6 - 6 = 3

⇒ 10x - 12 = 3

⇒ 10x = 3 + 12

⇒ 10x = 15

⇒ x = $\frac{15}{10}$

⇒ x = $\frac{3}{2}$

⇒ x = $1\frac{1}{2}$

Hence, the value of x is $1\frac{1}{2}$.

$$\begin{aligned}\frac{3x}{x + 6} - \frac{x}{x + 5} = 2\\[1em]\end{aligned}$$

Since L.C.M. of denominators (x + 6) and (x + 5) = (x + 6)(x + 5), multiply each term with (x + 6)(x + 5) to get:

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

⇒ 3x(x + 5) - x(x + 6) = 2(x + 6)(x + 5)

⇒ (3x² + 15x) - (x² + 6x) = 2(x² + 6x + 5x + 30)

⇒ 3x² + 15x - x² - 6x = 2(x² + 11x + 30)

⇒ 2x² + 9x = 2x² + 22x + 60

⇒ 2x² - 2x² + 9x - 22x = 60

⇒ -13x = 60

⇒ x = - $\frac{60}{13}$

⇒ x = - $4\frac{8}{13}$

Hence, the value of x is $- 4\frac{8}{13}$.

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

Since L.C.M. of denominators (x - 1), (x - 2) and (x - 3) = (x - 1)(x - 2)(x - 3), multiply each term with (x - 1)(x - 2)(x - 3) to get :

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

⇒ (x - 2)(x - 3) + 2(x - 1)(x - 3) = 3(x - 1)(x - 2)

⇒ (x² - 2x - 3x + 6) + 2(x² - 1x - 3x + 3) = 3(x² - 1x - 2x + 2)

⇒ (x² - 5x + 6) + 2(x² - 4x + 3) = 3(x² - 3x + 2)

⇒ x² - 5x + 6 + 2x² - 8x + 6 = 3x² - 9x + 6

⇒ 3x² - 13x + 12 = 3x² - 9x + 6

⇒ 3x² - 3x² - 13x + 9x = 6 - 12

⇒ - 4x = - 6

⇒ x = $\frac{6}{4}$

⇒ x = $\frac{3}{2}$

⇒ x = $1\frac{1}{2}$

Hence, the value of x is $1\frac{1}{2}$.

$\frac{x - 1}{7x - 14} = \frac{x - 3}{7x - 26}$

By cross multiplying, we get :

⇒ (x - 1)(7x - 26) = (x - 3)( 7x - 14)

⇒ x(7x - 26) - 1(7x - 26) = x( 7x - 14) - 3( 7x - 14)

⇒ 7x² - 26x - 7x + 26 = 7x² - 14x - 21x + 42

⇒ 7x² - 33x + 26 = 7x² - 35x + 42

⇒ 7x² - 7x² - 33x + 35x = 42 - 26

⇒ 2x = 16

⇒ x = $\frac{16}{2}$

⇒ x = 8

Hence, the value of x is 8.

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

Since L.C.M. of denominators x and (x - 1) = x(x - 1) and L.C.M. of (x + 3) and (x + 4) = (x + 3)(x + 4),

$$\begin{aligned}⇒ \frac{1 \times x}{x(x - 1)} - \frac{1 \times (x - 1)}{x(x - 1)} = \frac{1 \times (x + 4)}{(x + 3) \times (x + 4)} - \frac{1 \times (x + 3)}{(x + 4) \times (x + 3)}\\[1em] ⇒ \frac{x}{x^2 - x} - \frac{x - 1}{x^2 - x} = \frac{x + 4}{x^2 + 3x + 4x + 12} - \frac{x + 3}{x^2 + 3x + 4x + 12}\\[1em] ⇒ \frac{x - (x - 1)}{x^2 - x} = \frac{x + 4}{x^2 + 7x + 12} - \frac{x + 3}{x^2 + 7x + 12}\\[1em] ⇒ \frac{x - x + 1}{x^2 - x} = \frac{x + 4}{x^2 + 7x + 12} - \frac{x + 3}{x^2 + 7x + 12}\\[1em] ⇒ \frac{x - x + 1}{x^2 - x} = \frac{(x + 4) - (x + 3)}{x^2 + 7x + 12}\\[1em] ⇒ \frac{1}{x^2 - x} = \frac{x + 4 - x - 3}{x^2 + 7x + 12}\\[1em] ⇒ \frac{1}{x^2 - x} = \frac{1}{x^2 + 7x + 12}\\[1em]\end{aligned}$$

By cross multiplying, we get :

$$\begin{aligned}⇒ x^2 + 7x + 12 = x^2 - x\\[1em] ⇒ x^2 - x^2 + 7x + x = - 12\\[1em] ⇒ 8x = - 12\\[1em] ⇒ x = - \frac{12}{8}\\[1em] ⇒ x = - \frac{3}{2}\\[1em] ⇒ x = - 1\frac{1}{2}\end{aligned}$$

Hence, the value of x is $- 1\frac{1}{2}$.

$\frac{2x}{3} - \frac{x - 1}{6} + \frac{7x - 1}{4} = 2\frac{1}{6}$

$⇒ \frac{2x}{3} - \frac{x - 1}{6} + \frac{7x - 1}{4} = \frac{13}{6}$

Since L.C.M. of denominators 3, 6 and 4 = 12, multiply each term with 12 to get:

$⇒ \frac{2x \times 12}{3} - \frac{(x - 1) \times 12}{6} + \frac{(7x - 1) \times 12}{4} = \frac{13 \times 12}{6}$

⇒ 2x $\times$ 4 - (x - 1) $\times$ 2 + (7x - 1) $\times$ 3 = 13 $\times$ 2

⇒ 8x - (2x - 2) + (21x - 3) = 26

⇒ 8x - 2x + 2 + 21x - 3 = 26

⇒ 27x - 1 = 26

⇒ 27x = 26 + 1

⇒ 27x = 27

⇒ x = $\frac{27}{27}$

⇒ x = 1

Now, when x = 1

$$\begin{aligned}\frac{1}{a} + 5x = 8\\[1em] ⇒ \frac{1}{a} + 5 \times 1 = 8\\[1em] ⇒ \frac{1}{a} + 5 = 8\\[1em] ⇒ \frac{1}{a} = 8 - 5\\[1em] ⇒ \frac{1}{a} = 3\\[1em] ⇒ a = \frac{1}{3}\end{aligned}$$

Hence, the value of x is 1 and a is $\frac{1}{3}$.

$$\begin{aligned}\frac{4 - 3x}{5} + \frac{7 - x}{3} + 4\frac{1}{3} = 0\\[1em] ⇒ \frac{4 - 3x}{5} + \frac{7 - x}{3} + \frac{13}{3} = 0\\[1em]\end{aligned}$$

Since L.C.M. of denominators 5 and 3 = 15, multiply each term with 15 to get:

$⇒ \frac{(4 - 3x) \times 15}{5} + \frac{(7 - x) \times 15}{3} + \frac{13 \times 15}{3} = 0 \times 15$

⇒ (4 - 3x) $\times$ 3 + (7 - x) $\times$ 5 + 13 $\times$ 5 = 0

⇒ (12 - 9x) + (35 - 5x) + 65 = 0

⇒ 12 - 9x + 35 - 5x + 65 = 0

⇒ 112 - 14x = 0

⇒ 14x = 112

⇒ x = $\frac{112}{14}$

⇒ x = 8

Now, when x = 8

3p - 2x + 1 = 0

⇒ 3p - 2 $\times$ 8 + 1 = 0

⇒ 3p - 16 + 1 = 0

⇒ 3p - 15 = 0

⇒ 3p = 15

⇒ p = $\frac{15}{3}$

⇒ p = 5

Hence, the value of x is 8 and p is 5.

$0.25 + \frac{1.95}{x} = 0.9$

⇒ 0.25 $\times$ x + 1.95 = 0.9 $\times$ x

⇒ 0.25x + 1.95 = 0.9x

⇒ 0.25x - 0.9x = - 1.95

⇒ -0.65x = - 1.95

⇒ x = $\frac{1.95}{0.65}$

⇒ x = 3

Hence, the value of x is 3.

$$\begin{aligned}5x - (4x + \frac{5x - 4}{7}) = \frac{4x - 14}{3}\\[1em] ⇒ 5x - 4x - \frac{5x - 4}{7} = \frac{4x - 14}{3}\\[1em] ⇒ x - \frac{5x - 4}{7} = \frac{4x - 14}{3}\\[1em] ⇒ \frac{x}{1} - \frac{5x - 4}{7} = \frac{4x - 14}{3}\\[1em]\end{aligned}$$

Since L.C.M. of denominators 3 and 7 = 21, multiply each term with 21 to get:

$⇒ x \times 21 - \frac{(5x - 4)\times 21}{7} = \frac{(4x - 14) \times 21}{3}$

⇒ 21x - (5x - 4) $\times$ 3 = (4x - 14) \times 7

⇒ 21x - (15x - 12) = (28x - 98)

⇒ 21x - 15x + 12 = 28x - 98

⇒ 6x + 12 = 28x - 98

⇒ 6x - 28x = - 98 - 12

⇒ - 22x = - 110

⇒ x = $\frac{110}{22}$

⇒ x = 5

Hence, the value of x is 5.

Let the 3 consecutive number be (x - 2), x and (x + 2).

(x - 2) + x + (x + 2) = 90

⇒ x - 2 + x + x + 2 = 90

⇒ 3x = 90

⇒ x = $\frac{90}{3}$

⇒ x = 30

Hence, option 2 is the correct option.

Let the breadth of the rectangle be a.

So, length of the rectangle = 8 + a

Perimeter of the rectangle = 40

As we know, perimeter = 2(l + b)

Putting the value,

⇒ 40 = 2((8 + a) + a)

⇒ 40 = 2(8 + a + a)

⇒ 40 = 2(8 + 2a)

⇒ 40 = 16 + 4a

⇒ 4a = 40 - 16

⇒ 4a = 24

⇒ a = $\frac{24}{4}$

⇒ a = 6

Length = a + 6 = 8 + 6 = 14 m

Hence, option 3 is the correct option.

Given,

Two whole numbers are in the ratio 5 : 3.

Let the 2 numbers be 5x and 3x.

Difference of two number be 18.

⇒ 5x - 3x = 18

⇒ 2x = 18

⇒ x = $\frac{18}{2}$

⇒ x = 9

So, the 2 numbers are 5 x 9 = 45 and 3 x 9 = 27.

Hence, option 3 is the correct option.

Let the number be x.

So,

3 + 2x = x - 4

⇒ 2x - x = - 4 - 3

⇒ x = - 7

Hence, option 2 is the correct option.

Let the two numbers be x and y.

So,

x + y = 70

x - y = 16

Adding two equations, we get

⇒ (x + y) + (x - y) = 70 + 16

⇒ x + y + x - y = 70 + 16

⇒ 2x = 86

⇒ x = $\frac{86}{2}$

⇒ x = 43

So the other number be y = 70 - x

⇒ y = 70 - 43

⇒ y = 27

Hence, option 1 is the correct option.

Let the number be x.

According to the question,

⇒ 4x - 15 = 9

⇒ 4x = 9 + 15

⇒ 4x = 24

⇒ x = $\frac{24}{4}$

⇒ x = 6

Hence, the number be 6.

Let Megha's age be x years.

So,

x + 3x = 60

⇒ 4x = 60

⇒ x = $\frac{60}{4}$

⇒ x = 15

Hence, the age of Megha is 15 years.

Let the number be x.

So,

⇒ 4x - 12 = 28

⇒ 4x = 28 + 12

⇒ 4x = 40

⇒ x = $\frac{40}{4}$

⇒ x = 10

Hence, the number be 10.

Let the number be x.

So,

⇒ 3x - 5 = - 20

⇒ 3x = - 20 + 5

⇒ 3x = - 15

⇒ x = - $\frac{15}{3}$

⇒ x = - 5

Hence, the number is - 5.

Let the Neetu's age be x years.

⇒ 3x + 15 = 4x

⇒ 4x - 3x = 15

⇒ 1x = 15

Hence, the age of Neetu is 15 years.

Let the number be x.

So,

⇒ x - 30 = 14 - 3x

⇒ x + 3x = 14 + 30

⇒ 4x = 44

⇒ x = $\frac{44}{4}$

⇒ x = 11

Hence, the number is 11.

Let the B's salary be ₹ x.

So, A's salary = ₹ 4x

Therefore,

A's salary + B's salary = 3,750

⇒ 4x + x = 3,750

⇒ 5x = 3,750

⇒ x = $\frac{3,750}{5}$

⇒ x = 750

So, A's salary = 4x

= 4 x 750

= 3,000

Hence, A' salary = ₹ 3,000 and B's salary = ₹ 750.

Let the first part be x.

So the other part is 178 - x.

⇒ x = 2(178 - x) - 8

⇒ x = 356 - 2x - 8

⇒ x = 348 - 2x

⇒ x + 2x = 348

⇒ 3x = 348

⇒ x = $\frac{348}{3}$

⇒ x = 116

Other number = 178 - x

= 178 - 116

= 62

Hence, the two numbers be 116 and 62.

Let the number be x.

So,

$$\begin{aligned}⇒ 6 + \frac{1}{4}x = \frac{2}{5}x\\[1em] ⇒ 6 = \frac{2}{5}x - \frac{1}{4}x\end{aligned}$$

Since L.C.M. of denominators 5 and 4 = 20, multiply each term with 20 to get:

$⇒ 6 \times 20 = \frac{2 \times 20}{5}x - \frac{1 \times 20}{4}x$

⇒ 120 = 2 $\times$ 4x - 1 $\times$ 5x

⇒ 120 = 8x - 5x

⇒ 120 = 3x

⇒ x = $\frac{120}{3}$

⇒ x = 40

Hence, the number is 40.

Let the breadth of the rectangle be b cm.

The length of a rectangle is twice its width.

l = 2b

Perimeter of the rectangle = 54cm

Perimeter = 2(l + b)

⇒ 2(2b + b) = 54 cm

⇒ 2(3b) = 54 cm

⇒ 6b = 54 cm

⇒ b = $\frac{54}{6}$ cm

⇒ b = 9 cm

l = 2b

= 2 x 9

= 18 cm

Hence, length of the rectangle is 18cm.

Let the width of the rectangle be w cm

Length of the rectangle is 5 cm less than twice its width.

l = 2w - 5 cm

New width = old width + 2

= w + 2 cm

New length = old length - 5

= (2w - 5) - 5 cm

= 2w - 5 - 5 cm

= 2w - 10 cm

New perimeter = 2(new length + new width)

⇒ 2((2w - 10) + (w + 2)) = 74 cm

⇒ 2(2w - 10 + w + 2) = 74 cm

⇒ 2(3w - 8) = 74 cm

⇒ 6w - 16 = 74 cm

⇒ 6w = 74 + 16 cm

⇒ 6w = 90 cm

⇒ w = $\frac{90}{6}$ cm

⇒ w = 15 cm

Length = 2w - 5 cm

= 2 x 15 - 5 cm

= 30 - 5 cm

= 25 cm

Hence, the length is 25 cm and breadth is 15 cm.

Let the three consecutive odd numbers be (x - 2), x, (x + 2).

So,

(x - 2) + x + (x + 2) = 57

⇒ x - 2 + x + x + 2 = 57

⇒ 3x = 57

⇒ x = $\frac{57}{3}$

⇒ x = 19

Other numbers = (x - 2), (x + 2)

= (19 - 2), (19 + 2)

= 17, 21

Hence, the three numbers are 17, 19 and 21.

Let the age of son be x years.

The man's age is three times that of his son, so it equals 3x.

In twelve years, he will be twice as old as his son will be,

Man's age + 12 = 2(son's age + 12)

⇒ 3x + 12 = 2(x + 12)

⇒ 3x + 12 = 2x + 24

⇒ 3x - 2x = 24 - 12

⇒ x = 12

Man's age = 3x

= 3 $\times$ 12

= 36

Hence, the man's age is 36 years and the son's age is 12 years.

The age of man = 42 years

The age of son = 12 years

Let the son's age be half of the man's age after x years.

Son's age + x = $\frac{1}{2}$ (Man's age + x)

⇒ 12 + x = $\frac{1}{2}$ (42 + x)

⇒ 2 $\times$ (12 + x) = 1 $\times$ (42 + x)

⇒ 24 + 2x = 42 + x

⇒ 2x - x = 42 - 24

⇒ x = 18

Hence, after 18 years, the son's age will be half of the man's age.

Total distance = 136 km

Total time = 8 hours

Let the part of the trip that was covered at 15 km/h be x km.

So, the remaining part, (136 - x)km, was covered at 18km/hr.

As we know, Time = $\frac{Distance}{Speed}$

And

Total time = time taken to cover x km at 15 km/h + time taken to cover (136−x) km at 18 km/h.

$$\begin{aligned}⇒ \frac{x}{15} + \frac{136 - x}{18} = 8\\[1em]\end{aligned}$$

Since L.C.M. of 15 and 18 = 90, multiply each term with 90 to get:

$⇒ \frac{x \times 90}{15} + \frac{(136 - x) \times 90}{18} = 8 \times 90$

⇒ x $\times$ 6 + (136 - x) $\times$ 5 = 720

⇒ 6x + 680 - 5x = 720

⇒ x + 680 = 720

⇒ x = 720 - 680

⇒ x = 40

Other part = 136 - x km

= 136 - 40 km

= 96 km

Hence, the man covered 96 km at the speed of 18 km/hr.

Let the two numbers be x and y.

So,

x - y = 3 .............(1)

And,

x² - y² = 69

Using the formula

[∵ (x² - y²) = (x - y)(x + y)]

⇒ (x - y)(x + y) = 69

Using eq (1), we get

⇒ 3(x + y) = 69

⇒ x + y = $\frac{69}{3}$

⇒ x + y = 23 .............(2)

Adding (1) and (2)

⇒ (x - y) + (x + y) = 3 + 23

⇒ x - y + x + y = 26

⇒ 2x = 26

⇒ x = $\frac{26}{2}$

⇒ x = 13

Using eq (2)

⇒ y = 23 - 13

⇒ y = 10

Hence, the numbers are 13 and 10.

Let the consecutive natural numbers be x and (x + 1).

So,

$$\begin{aligned}\frac{1}{4} \times x - \frac{1}{5} \times (x + 1) = 1 \\[1em] ⇒ \frac{x}{4} - \frac{x + 1}{5} = 1 \\[1em]\end{aligned}$$

Since L.C.M. of denominators 5 and 4 = 20, multiply each term with 20 to get:

$⇒ \frac{x \times 20}{4} - \frac{(x + 1) \times 20}{5} = 1\times 20$

⇒ x $\times$ 5 - (x + 1) $\times$ 4 = 20

⇒ 5x - (4x + 4) = 20

⇒ 5x - 4x - 4 = 20

⇒ x - 4 = 20

⇒ x = 20 + 4

⇒ x = 24

Other number = (x + 1)

= 24 + 1

= 25

Hence, the numbers are 24 and 25.

Let the three consecutive whole numbers be (x - 1), x, (x + 1).

According to the question, the sum of the quotients is 40.

$⇒ \frac{x - 1}{5} + \frac{x}{3} + \frac{x + 1}{4} = 40$

Since L.C.M. of denominators 5, 3 and 4 = 60, multiply each term with 60 to get:

$⇒ \frac{(x - 1) \times 60}{5} + \frac{x \times 60}{3} + \frac{(x + 1) \times 60}{4} = 40 \times 60$

⇒ (x - 1) $\times$ 12 + x $\times$ 20 + (x + 1) $\times$ 15 = 2400

⇒ (12x - 12) + 20x + (15x + 15) = 2400

⇒ 12x - 12 + 20x + 15x + 15 = 40

⇒ 47x + 3 = 2400

⇒ 47x = 2400 - 3

⇒ 47x = 2397

⇒ x = $\frac{2397}{47}$

⇒ x = 51

Other number = (x - 1), (x + 1)

= (51 - 1), (51 + 1)

= 50, 52

Hence, the numbers are 50, 51 and 52.

Let x be added to each number. The numbers will then be 5 + x, 11 + x, 15 + x and 31 + x.

According to question,

$\frac{5 + x}{11 + x} = \frac{15 + x}{31 + x}$

By cross multiplying, we get :

⇒ (5 + x)(31 + x) = (15 + x)(11 + x)

⇒ 155 + 5x + 31x + x² = 165 + 11x + 15x + x²

⇒ 155 + 36x + x² = 165 + 26x + x²

⇒ 36x - 26x + x² - x² = 165 - 155

⇒ 10x = 10

⇒ x = $\frac{10}{10}$

⇒ x = 1

Hence, 1 is the number that should be added to the given numbers.

Let the age of son be x years.

The present age of a man is twice that of his son.

Man's age = 2x

After eight years, their ages will be in the ratio 7 : 4.

$$\begin{aligned}⇒ \frac{\text{Man's age after 8 years}}{\text{Son's age after 8 years}} = \frac{7}{4}\\[1em] ⇒ \frac{\text{Man's present age + 8 years}}{\text{Son's present age + 8 years}} = \frac{7}{4}\\[1em] ⇒ \frac{2x + 8}{x + 8} = \frac{7}{4}\\[1em]\end{aligned}$$

By cross multiplying, we get

$$\begin{aligned}⇒ 4(2x + 8) = 7(x + 8)\\[1em] ⇒ 8x + 32 = 7x + 56\\[1em] ⇒ 8x - 7x = 56 - 32\\[1em] ⇒ x = 24\\[1em]\end{aligned}$$

Present son's age = 24 years

Present man's age = 2x years

= 2 $\times$ 24 years

= 48 years

Hence, son's age is 24 years and man's age is 48 years.

$\frac{x}{2} + \frac{x}{3} = 15$

Since L.C.M. of denominators 2 and 3 = 6, multiply each term with 6 to get:

$⇒ \frac{x \times 6}{2} + \frac{x \times 6}{3} = 15 \times 6$

⇒ x $\times$ 3 + x $\times$ 2 = 90

⇒ 3x + 2x = 90

⇒ 5x = 90

⇒ x = $\frac{90}{5}$

⇒ x = 18

Hence, option 1 is the correct option.

Let the three consecutive odd whole numbers be (x - 2), x and (x + 2).

According to question,

⇒ (x - 2) + x + (x + 2) = 87

⇒ x - 2 + x + x + 2 = 87

⇒ 3x = 87

⇒ x = $\frac{87}{3}$

⇒ x = 29

Other numbers are (x - 2), (x + 2)

= (29 - 2) , (29 + 2)

= 27, 31

The larger number is 31.

Hence, option 2 is the correct option.

$\frac{x + 3}{(x + 4) + 3} = \frac{4}{5}$

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

⇒ $\frac{x + 3}{x + 7} = \frac{4}{5}$

By cross multiplying, we get :

⇒ 5(x + 3) = 4(x + 7)

⇒ 5x + 15 = 4x + 28

⇒ 5x - 4x = 28 - 15

⇒ x = 13

Hence, option 4 is the correct option.

Let the four consecutive integers be (x - 1), x, (x + 1) and (x + 2).

⇒ (x - 1) + x + (x + 1) + (x + 2) = -46

⇒ x - 1 + x + x + 1 + x + 2 = -46

⇒ 4x + 2 = -46

⇒ 4x = -46 - 2

⇒ 4x = -48

⇒ x = - $\frac{48}{4}$

⇒ x = - 12

Other numbers are (x - 1), (x + 1), (x + 2)

= (-12 - 1), (-12 + 1), (-12 + 2)

= - 13, - 11, - 10

Hence, option 2 is the correct option.

Let the width of the rectangular plot be w.

Length of the rectangular plot = 3w

Perimeter = 560 m

As we know, perimeter of a rectangle = 2(l + w)

⇒ 2 (3w + w) = 560 m

⇒ 2 x 4w = 560 m

⇒ 8w = 560 m

⇒ w = $\frac{560}{8}$ m

⇒ w = 70 m

Length of rectangle = 3w

= 3 x 70 m

= 210 m

Hence, option 2 is the correct option.

Given, $\frac{ax + b}{cx + d} = \frac{p}{q}$

On cross-multiplication,

⇒ q(ax + b) = p(cx + d)

So, statement 1 is false.

Transposition is a fundamental algebraic technique used to isolate a variable by moving terms from one side of an equation to the other. When a term is transposed, its sign changes;

• A term added to one side gets subtracted on the other.

• A term subtracted from one side gets added on the other.

∴ ax + b = c becomes ax = c - b after transposition.

So, statement 2 is true.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

An equation in which the greatest exponent of the variable after simplification is one is called a linear equation.

This definition aligns with the standard definition of a linear equation. A linear equation in one variable is of the form ax + b = 0, where a ≠ 0 and the highest power of x is 1.

So, reason (R) is true.

Given,

⇒ (2x + 5)² - (2x - 5)² = 40

⇒ 4x² + 25 + 20x - (4x² + 25 - 20x) = 40

⇒ 4x² + 25 + 20x - 4x² - 25 + 20x = 40

⇒ 40x = 40

⇒ 40x - 40 = 0

The above equation is in the form ax + b = 0, where a = 40, b = -40 and the highest power of x is 1.

So, assertion (A) is true.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{x}{2} + \frac{x}{3} = 15 \\[1em] \Rightarrow \frac{3x + 2x}{6} = 15 \\[1em] \Rightarrow \frac{5x}{6} = 15 \\[1em] \Rightarrow x = \frac{6 \times 15}{5} \\[1em] \Rightarrow x = \frac{90}{5} \\[1em] \Rightarrow x = 18\end{aligned}$$

So, assertion (A) is true.

In the process of solving a linear equation, multiple methods can be applied, not just transposition.

So, reason (R) is false.

∴ A is true, but R is false.

Hence, option 3 is the correct option.

Given, $\frac{ax + b}{cx + d} = \frac{p}{q}$

On cross-multiplication,

⇒ q(ax + b) = p(cx + d)

So, reason (R) is true.

Given,

$$\begin{aligned}\Rightarrow \frac{x - 3}{x + 4} = \frac{x + 1}{x - 2} \\[1em] \Rightarrow (x - 3) \times (x - 2) = (x + 4) \times (x + 1) \\[1em] \Rightarrow x(x - 2) - 3(x - 2) = x(x + 1) + 4(x + 1) \\[1em] \Rightarrow x^2 - 2x - 3x + 6 = x^2 + x + 4x + 4 \\[1em] \Rightarrow x^2 - 5x + 6 = x^2 + 5x + 4 \\[1em] \Rightarrow x^2 - 5x + 6 - x^2 - 5x - 4 = 0 \\[1em] \Rightarrow -10x + 2 = 0 \\[1em] \Rightarrow 10x = 2 \\[1em] \Rightarrow x = \frac{2}{10} = \frac{1}{5}.\end{aligned}$$

So, assertion (A) is false.

∴ A is false, but R is true.

Hence, option 4 is the correct option.

In algebra, on dividing both sides of an equation by the same non-zero number, the equation remains the same.

So, reason (R) is true.

Given, 7x = -21

Dividing both sides by 7, we get

$\Rightarrow \frac{7x}{7} = -\frac{21}{7}$

⇒ x = -3

So, assertion (A) is true.

∴ Both A and R are correct, and R is the correct explanation for A.

Hence, option 1 is the correct option.

$$\begin{aligned}\frac{1}{3}x - 6 = \frac{5}{2}\\[1em] ⇒ \frac{1}{3}x = \frac{5}{2} + 6\\[1em] ⇒ \frac{1}{3}x = \frac{5}{2} + \frac{6}{1}\end{aligned}$$

Since L.C.M. of denominators 3, 2 and 1 = 6, multiply each term with 6 to get:

$⇒ \frac{1 \times 6}{3}x = \frac{5 \times 6}{2} + \frac{6 \times 6}{1}$

⇒ 1 $\times$ 2x = 5 $\times$ 3 + 6 $\times$ 6

⇒ 2x = 15 + 36

⇒ 2x = 51

⇒ x = $\frac{51}{2}$

⇒ x = $25\frac{1}{2}$

Hence, the value of x is $25\frac{1}{2}$.

$\frac{2x}{3} - \frac{3x}{8} = \frac{7}{12}$

Since L.C.M. of denominators 3 and 8 = 24, multiply each term with 24 to get:

$⇒ \frac{2x \times 24}{3} - \frac{3x \times 24}{8} = \frac{7 \times 24}{12}$

⇒ 2x $\times$ 8 - 3x $\times$ 3 = 7 $\times$ 2

⇒ 16x - 9x = 14

⇒ 7x = 14

⇒ x = $\frac{14}{7}$

⇒ x = 2

Hence, the value of x is 2.

(x + 2) (x + 3) + (x - 3) (x - 2) - 2x(x + 1) = 0

⇒ x (x + 3) + 2 (x + 3) + x (x - 2) - 3 (x - 2) - 2x(x + 1) = 0

⇒ x² + 3x + 2x + 6 + x² - 2x - 3x + 6 - 2x² - 2x = 0

⇒ (x² + x² - 2x²) + (3x + 2x - 2x - 3x - 2x) + (6 + 6) = 0

⇒ - 2x + 12 = 0

⇒ 2x = 12

⇒ x = $\frac{12}{2}$

⇒ x = 6

Hence, the value of x is 6.

$$\begin{aligned}\frac{1}{10} - \frac{7}{x} = 35\\[1em] ⇒ \frac{7}{x} = \frac{1}{10} - 35\\[1em] ⇒ \frac{7}{x} = \frac{1}{10} - \frac{35}{1}\end{aligned}$$

Since L.C.M. of denominators 10 and 1 = 10, multiply each term with 10 to get:

$$\begin{aligned}⇒ \frac{7 \times 10}{x} = \frac{1 \times 10}{10} - \frac{35 \times 10}{1}\\[1em] ⇒ \frac{70}{x} = 1 - 350\\[1em] ⇒ \frac{70}{x} = - 349\\[1em] ⇒ x = - \frac{70}{349}\end{aligned}$$

Hence, the value of x is $-\frac{70}{349}$.

13(x - 4) - 3(x - 9) - 5(x + 4) = 0

⇒ 13x - 52 - 3x + 27 - 5x - 20 = 0

⇒ (13x - 3x - 5x) + (- 52 + 27 - 20) = 0

⇒ 5x - 45 = 0

⇒ 5x = 45

⇒ x = $\frac{45}{5}$

⇒ x = 9

Hence, the value of x is 9.

$x + 7 -\frac{8x}{3} = \frac{17x}{6} - \frac{5x}{8}$

Since L.C.M. of denominators 6 and 8 = 24, multiply each term with 24 to get:

$⇒ x\times 24 + 7\times 24 -\frac{8x\times 24}{3} = \frac{17x \times 24}{6} - \frac{5x \times 24}{8}$

⇒ 24x + 168 - 8x $\times$ 8 = 17x $\times$ 4 - 5x $\times$ 3

⇒ 24x + 168 - 64x = 68x - 15x

⇒ 168 - 40x = 53x

⇒ - 53x - 40x = - 168

⇒ - 93x = - 168

⇒ x = $\frac{168}{93}$

⇒ x = $\frac{56}{31}$

⇒ x = $1\frac{25}{31}$

Hence, the value of x is $1\frac{25}{31}$

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

Since L.C.M. of denominators 3 and 4 = 12, multiply each term with 12 to get:

$⇒ \frac{12(3x - 2)}{4} - \frac{12(2x + 3)}{3} + \frac{x \times 12}{1} = \frac{2 \times 12}{3}$

⇒ 3(3x - 2) - 4(2x + 3) + x $\times$ 12 = 2 $\times$ 4

⇒ (9x - 6) - (8x + 12) + 12x = 8

⇒ 9x - 6 - 8x - 12 + 12x = 8

⇒ 13x - 18 = 8

⇒ 13x = 8 + 18

⇒ 13x = 26

⇒ x = $\frac{26}{13}$

⇒ x = 2

Hence, the value of x is 2.

$$\begin{aligned}\frac{x + 2}{6} - \Big(\frac{11 - x}{3} - \frac{1}{4}\Big) = \frac{3x - 4}{12}\\[1em] ⇒ \frac{x + 2}{6} - \frac{11 - x}{3} + \frac{1}{4} = \frac{3x - 4}{12}\end{aligned}$$

Since L.C.M. of denominators 6, 3 and 4 = 12, multiply each term with 12 to get:

$⇒ \frac{12(x + 2)}{6} - \frac{12(11 - x)}{3} + \frac{1 \times 12}{4} = \frac{12(3x - 4)}{12}$

⇒ 2(x + 2) - 4(11 - x) + 1 $\times$ 3 = 1(3x - 4)

⇒ (2x + 4) - (44 - 4x) + 3 = 3x - 4

⇒ 2x + 4 - 44 + 4x + 3 = 3x - 4

⇒ 6x - 37 = 3x - 4

⇒ 6x - 3x = 37 - 4

⇒ 3x = 33

⇒ x = $\frac{33}{3}$

⇒ x = 11

Hence, the value of x is 11.

$\frac{2}{5x} - \frac{5}{3x} = \frac{1}{15}$

Since L.C.M. of denominators 5x and 3x = 15x, multiply each term with 15x to get:

$⇒ \frac{2 \times 15x}{5x} - \frac{5 \times 15x}{3x} = \frac{1 \times 15x}{15}$

⇒ 2 $\times$ 3 - 5 $\times$ 5 = 1 $\times$ x

⇒ 6 - 25 = x

⇒ - 19 = x

Hence, the value of x is -19.

$$\begin{aligned}\frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1\\[1em] ⇒ \frac{x + 2}{3} - \frac{x + 1}{5} - \frac{x - 3}{4} = - 1\end{aligned}$$

Since L.C.M. of denominators 3, 5 and 4 = 60, multiply each term with 60 to get:

$⇒ \frac{60(x + 2)}{3} - \frac{60(x + 1)}{5} - \frac{60(x - 3)}{4} = - 1 \times 60$

⇒ 20(x + 2) - 12(x + 1) - 15(x - 3) = - 60

⇒ (20x + 40) - (12x + 12) - (15x - 45) = - 60

⇒ 20x + 40 - 12x - 12 - 15x + 45 = - 60

⇒ - 7x + 73 = - 60

⇒ - 7x = - 60 - 73

⇒ - 7x = - 133

⇒ x = $\frac{-133}{-7}$

⇒ x = 19

Hence, the value of x is 19.

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

Since L.C.M. of denominators 3 and 2 = 6, multiply each term with 6 to get:

$⇒ \frac{6(3x - 2)}{3} + \frac{6(2x + 3)}{2} = x \times 6 + \frac{7 \times 6}{6}$

⇒ 2(3x - 2) + 3(2x + 3) = 6x + 7 $\times$ 1

⇒ (6x - 4) + (6x + 9) = 6x + 7

⇒ 6x - 4 + 6x + 9 = 6x + 7

⇒ 12x + 5 = 6x + 7

⇒ 12x - 6x = 7 - 5

⇒ 6x = 2

⇒ x = $\frac{2}{6}$

⇒ x = $\frac{1}{3}$

Hence, the value of x is $\frac{1}{3}$.

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

Since L.C.M. of denominators 1, 2 and 3 = 6, multiply each term with 6 to get:

$⇒ \frac{x \times 6}{1} -\frac{6(x - 1)}{2} = 1 \times 6 -\frac{6(x - 2)}{3}$

⇒ x $\times$ 6 - 3(x - 1) = 6 - 2(x - 2)

⇒ 6x - (3x - 3) = 6 - (2x - 4)

⇒ 6x - 3x + 3 = 6 - 2x + 4

⇒ 3x + 3 = 10 - 2x

⇒ 3x + 2x = 10 - 3

⇒ 5x = 7

⇒ x = $\frac{7}{5}$

Hence, the value of x is $\frac{7}{5}$.

$$\begin{aligned}\frac{9x + 7}{2} - (x - \frac{x - 2}{7}) = 36\\[1em] ⇒ \frac{9x + 7}{2} - x + \frac{x - 2}{7} = 36\\[1em] ⇒ \frac{9x + 7}{2} - \frac{x}{1} + \frac{x - 2}{7} = 36\end{aligned}$$

Since L.C.M. of denominators 2 and 7 = 14, multiply each term with 14 to get:

$⇒ \frac{14(9x + 7)}{2} - \frac{x \times 14}{1} + \frac{14(x - 2)}{7} = 36 \times 14$

⇒ 7(9x + 7) - x $\times$ 14 + 2(x - 2) = 504

⇒ (63x + 49) - 14x + (2x - 4) = 504

⇒ 63x + 49 - 14x + 2x - 4 = 504

⇒ 51x + 45 = 504

⇒ 51x = 504 - 45

⇒ 51x = 459

⇒ x = $\frac{459}{51}$

⇒ x = 9

Hence, the value of x is 9.

$$\begin{aligned}\frac{6x + 1}{2} + 1 = \frac{7x - 3}{3}\\[1em] ⇒ \frac{6x + 1}{2} - \frac{7x - 3}{3} = - 1\end{aligned}$$

Since L.C.M. of denominators 2 and 3 = 6, multiply each term with 6 to get:

$⇒ \frac{6(6x + 1)}{2} - \frac{6(7x - 3)}{3} = - 1 \times 6$

⇒ 3(6x + 1) - 2(7x - 3) = - 6

⇒ (18x + 3) - (14x - 6) = - 6

⇒ 18x + 3 - 14x + 6 = - 6

⇒ 4x + 9 = - 6

⇒ 4x = - 6 - 9

⇒ 4x = - 15

⇒ x = - $\frac{15}{4}$

Hence, the value of x is $-\frac{15}{4}$.

Let my present age be x years.

According to question, after 12 years, I shall be 3 times as old as I was 4 years ago.

⇒ (present age + 12 years) = 3(present age - 4 years)

⇒ (x + 12) = 3(x - 4)

⇒ x + 12 = 3x - 12

⇒ 3x - x = 12 + 12

⇒ 2x = 24

⇒ x = $\frac{24}{2}$

⇒ x = 12

Hence, my present age is 12 years.

Given:

S.P. = ₹ 396

Profit = 10%

Let the C.P. of the article be ₹ x.

As we know,

Profit % = $\frac{\text{ Profit}}{\text{C.P.}} \times 100$

⇒ 10 = $\frac{\text{ Profit}}{x} \times 100$

⇒ Profit = $\frac{10 \times x}{100}$

⇒ Profit = $\frac{10x}{100}$

⇒ Profit = $\frac{x}{10}$

And,

Profit = S.P. - C.P.

⇒ $\frac{x}{10}$ = 396 - x

⇒ $\frac{x}{10} + \frac{x}{1}$ = 396

Since L.C.M. of 10 and 1 = 10, multiply each term with 10 to get:

⇒ $\frac{1x \times 10}{10} + \frac{x \times 10}{1}$ = 396 $\times$ 10

⇒ 1x + 10x = 3960

⇒ 11x = 3960

⇒ x = $\frac{3960}{11}$

⇒ x = 360

Hence, cost price of the article is ₹ 360.

The sum of two numbers is 4500.

Let the number be x. The other number is (4500 - x).

So, 10% of one number is 12.5% of the other number.

10% of x = 12.5% of (4500 - x)

$$\begin{aligned}⇒ \frac{10}{100} \times x = \frac{12.5}{100} \times (4500 - x)\\[1em] ⇒ \frac{x}{10} = \frac{125}{1000} \times (4500 - x)\\[1em] ⇒ \frac{x}{10} = \frac{(4500 - x)}{8} \\[1em]\end{aligned}$$

By cross multiplying, we get :

⇒ 8x = 10(4500 - x)

⇒ 8x = 45000 - 10x

⇒ 8x + 10x = 45000

⇒ 18x = 45000

⇒ x = $\frac{45000}{18}$

⇒ x = 2500

Other number = (4500 - x)

= (4500 - 2500)

= 2000

Hence, the two numbers are 2500 and 2000.

The ratio of two numbers is 8 : 7.

Let the numbers be 8x and 7x.

The sum of numbers is 405.

⇒ 8x + 7x = 405

⇒ 15x = 405

⇒ x = $\frac{405}{15}$

⇒ x = 27

Numbers = 8x = 27 x 8

= 216

And, 7x = 27 x 7

= 189

Hence, the numbers are 216 and 189.

Let the present age of A be 7x years and the present age of B be 5x years.

According to the question:

Ten years hence, the ratio of their ages will be 9 : 7.

$$\begin{aligned}⇒ \frac{\text{Present age of A + 10}}{\text{Present age of B + 10}} = \frac{9}{7}\\[1em] ⇒ \frac{7x + 10}{5x + 10} = \frac{9}{7}\\[1em]\end{aligned}$$

By cross multiplying, we get :

⇒ 7(7x + 10) = 9(5x + 10)

⇒ 49x + 70 = 45x + 90

⇒ 49x - 45x = 90 - 70

⇒ 4x = 20

⇒ x = $\frac{20}{4}$

⇒ x = 5

So, present age of A = 7x

= 7 x 5

= 35 years

And, present age of B = 5x

= 5 x 5

= 25 years

Hence, the present ages of A and B are 35 years and 25 years, respectively.

Let the number be x.

Double of the number is 45 greater than its half.

$$\begin{aligned}⇒ 2x = 45 + \frac{1}{2}x\\[1em] ⇒ 2x - \frac{1}{2}x = 45 \\[1em] ⇒ \frac{2x}{1} - \frac{x}{2} = 45\\[1em]\end{aligned}$$

Since L.C.M. of denominators 1 and 2 = 2, multiply each term with 2 to get:

$⇒ \frac{2x \times 2}{1} - \frac{x \times 2}{2} = 45\times 2$

⇒ 2x $\times$ 2 - x = 90

⇒ 4x - x = 90

⇒ 3x = 90

⇒ x = $\frac{90}{3}$

⇒ x = 30

Hence, the number is 30.

Let the two consecutive numbers be x, (x + 1).

According to the question, the difference between the squares of two consecutive numbers is 31.

⇒ (x + 1)² - x² = 31

Using the formula,

[∵ (x + y)² = x² + y² + 2xy]

⇒ x² + 1² + 2 $\times$ x $\times$ 1 - x² = 31

⇒ x² + 1 + 2x - x² = 31

⇒ 1 + 2x = 31

⇒ 2x = 31 - 1

⇒ 2x = 30

⇒ x = $\frac{30}{2}$

⇒ x = 15

Other number = (x + 1)

= 15 + 1

= 16

Hence, the two consecutive numbers are 15 and 16.

Let the number be x.

According to the question,

when 5 is subtracted from 5 times the number, the result is 4 more than twice the number

⇒ 5x - 5 = 2x + 4

⇒ 5x - 2x = 4 + 5

⇒ 3x = 9

⇒ x = $\frac{9}{3}$

⇒ x = 3

Hence, the number is 3.

Let the denominator of the fraction be x.

The numerator of a fraction is 5 less than its denominator

So, numerator = x - 5

According to the question,

when 3 is added to the numerator and denominator both, the fraction becomes $\frac{4}{5}$

$$\begin{aligned}⇒ \frac{\text{numerator} + 3}{\text{denominator} + 3} = \frac{4}{5}\\[1em] ⇒ \frac{(x - 5) + 3}{x + 3} = \frac{4}{5}\\[1em] ⇒ \frac{x - 5 + 3}{x + 3} = \frac{4}{5}\\[1em] ⇒ \frac{x - 2}{x + 3} = \frac{4}{5}\\[1em]\end{aligned}$$

By cross multiplying, we get :

⇒ 5(x - 2) = 4(x + 3)

⇒ 5x - 10 = 4x + 12

⇒ 5x - 4x = 10 + 12

⇒ x = 22

Denominator = 22

Numerator = (x - 5)

= 22 - 5

= 17

Hence, the fraction is $\frac{17}{22}$.

Let the number be x.

It is given that the difference between the numbers is 36. So, the other number is x−36.

According to the question, the quotient when one number is divided by the other is 4.

$⇒ \frac{x}{x - 36} = 4$

By cross multiplying, we get :

⇒ x = 4(x - 36)

⇒ x = 4x - 144

⇒ 4x - x = 144

⇒ 3x = 144

⇒ x = $\frac{144}{3}$

⇒ x = 48

Other number be (x - 36)

= (48 - 36)

= 12

Hence, the two numbers are 48 and 12.

Let the age of Manish be x.

According to the question, five years ago, Mohit was three times as old as Manish.

⇒ Mohit's age - 5 years = 3 x (Manish's age - 5 years)

⇒ Mohit's age - 5 = 3(x - 5)

⇒ Mohit's age - 5 = 3x - 15

⇒ Mohit's age = 3x - 15 + 5

⇒ Mohit's age = 3x - 10 ...............(1)

And,

Ten years later, Mohit will be twice as old as Manish.

⇒ Mohit's age + 10 years = 2 x (Manish's age + 10 years)

⇒ Mohit's age + 10 = 2(x + 10)

⇒ Mohit's age + 10 = 2x + 20

⇒ Mohit's age = 2x + 20 - 10

⇒ Mohit's age = 2x + 10

From equation (1),

⇒ 3x - 10 = 2x + 10

⇒ 3x - 2x = 10 + 10

⇒ x = 20

Mohit's age = 3x - 10

= 3 $\times$ 20 - 10

= 60 - 10

= 50

Hence, the present age of Mohit is 50 years, and the present age of Manish is 20 years.

Let the numerator of the fraction be x.

The denominator of a fraction is 3 more than its numerator.

So, denominator = x + 3

If the numerator is increased by 7 and the denominator is decreased by 2, the fraction is 2.

$$\begin{aligned}⇒ \frac{numerator + 7}{denominator - 2} = 2\\[1em] ⇒ \frac{x + 7}{(x + 3) - 2} = 2\\[1em] ⇒ \frac{x + 7}{x + 3 - 2} = 2\\[1em] ⇒ \frac{x + 7}{x + 1} = 2\\[1em]\end{aligned}$$

By cross multiplying, we get :

⇒ (x + 7) = 2(x + 1)

⇒ x + 7 = 2x + 2

⇒ 2x - x = 7 - 2

⇒ x = 5

Denominator = (x + 3)

= 5 + 3

= 8

The fraction is $\frac{5}{8}$.

The sum of the numerator and the denominator of the fraction = 5 + 8 = 13.

Hence, the sum of the numerator and the denominator of the fraction is 13.

Let the original number be 10x + y.

As it is given that, the digits of a two-digit number differ by 3.

⇒ x - y = 3

⇒ x = 3 + y ...............(1)

If the digits are interchanged, the new number is 10y + x

The sum of interchanged number and original number is 143.

⇒ (10y + x) + (10x + y) = 143

⇒ 10y + x + 10x + y = 143

⇒ 11y + 11x = 143

⇒ 11(y + x) = 143

⇒ y + x = $\frac{143}{11}$

⇒ y + x = 13

From equation (1),

⇒ y + (3 + y) = 13

⇒ y + 3 + y = 13

⇒ 2y + 3 = 13

⇒ 2y = 13 - 3

⇒ 2y = 10

⇒ y = $\frac{10}{2}$

⇒ y = 5

x = y + 3

x = 5 + 3

x = 8

Original number = 10x + y

= 10 $\times$ 8 + 5

= 80 + 5

= 85

Hence, the original number is 85.