Linear Equations

Solving,

$$\begin{aligned}\Rightarrow 4x - 9 = 2x + 7 \\[1em] \Rightarrow 4x - 2x = 7 + 9 \\[1em] \Rightarrow 2x = 16 \\[1em] \Rightarrow x = \frac{16}{2} \\[1em] \Rightarrow x = 8\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 5y + 18 = 11 - 2y \\[1em] \Rightarrow 5y + 2y = 11 - 18 \\[1em] \Rightarrow 7y = - 7 \\[1em] \Rightarrow y = - \frac{7}{7} \\[1em] \Rightarrow y = - 1\end{aligned}$$

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

Solving,

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

Hence, $\bm{x = \frac{11}{2}}$.

Solving,

$$\begin{aligned}\Rightarrow 3(y - 7) - 2(3y - 4) = (2 - 5y) \\[1em] \Rightarrow 3y - 21 - 6y + 8 = 2 - 5y \\[1em] \Rightarrow - 3y - 13 = 2 - 5y \\[1em] \Rightarrow 5y - 3y = 2 + 13 \\[1em] \Rightarrow 2y = 15 \\[1em] \Rightarrow y = \frac{15}{2}\end{aligned}$$

Hence, $\bm{y = \frac{15}{2}}$.

Solving,

$$\begin{aligned}\Rightarrow 3(t - 5) - 16t = 12 - 2(t - 3) \\[1em] \Rightarrow 3t - 15 - 16t = 12 - 2t + 6 \\[1em] \Rightarrow - 13t - 15 = 18 - 2t \\[1em] \Rightarrow - 15 - 18 = 13t - 2t \\[1em] \Rightarrow 11t = - 33 \\[1em] \Rightarrow t = - \frac{33}{11} \\[1em] \Rightarrow t = - 3\end{aligned}$$

Hence, $\bm{t = -3}$.

Given,

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

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

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow 12 \times \Big(\frac{3x}{4} - \frac{(x - 4)}{3}\Big) = 12 \times \frac{5}{3} \\[1em] \Rightarrow \frac{12 \times 3x}{4} - \frac{12(x - 4)}{3} = \frac{12 \times 5}{3} \\[1em] \Rightarrow 9x - 4(x - 4) = 20 \\[1em] \Rightarrow 9x - 4x + 16 = 20 \\[1em] \Rightarrow 5x = 20 - 16 \\[1em] \Rightarrow 5x = 4 \\[1em] \Rightarrow x = \frac{4}{5}\end{aligned}$$

Hence, $\bm{x = \frac{4}{5}}$.

Given,

$\frac{(4x + 1)}{3} + \frac{(2x - 1)}{2} - \frac{(3x - 7)}{5} = 6$

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

Multiplying both sides by 30, we get:

$$\begin{aligned}\Rightarrow 30 \times \Big(\frac{(4x + 1)}{3} + \frac{(2x - 1)}{2} - \frac{(3x - 7)}{5}\Big) = 30 \times 6 \\[1em] \Rightarrow \frac{30(4x + 1)}{3} + \frac{30(2x - 1)}{2} - \frac{30(3x - 7)}{5} = 30 \times 6 \\[1em] \Rightarrow 10(4x + 1) + 15(2x - 1) - 6(3x - 7) = 180 \\[1em] \Rightarrow 40x + 10 + 30x - 15 - 18x + 42 = 180 \\[1em] \Rightarrow 52x + 37 = 180 \\[1em] \Rightarrow 52x = 180 - 37 \\[1em] \Rightarrow 52x = 143 \\[1em] \Rightarrow x = \frac{143}{52} \\[1em] \Rightarrow x = \frac{11}{4}\end{aligned}$$

Hence, $\bm{x = \frac{11}{4}}$.

Given,

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

Calculating L.C.M. of 6, 9 and 4:

$$\begin{array}{l|rrr} 2 & 6, & 9, & 4 \\ \hline 2 & 3, & 9, & 2 \\ \hline 3 & 3, & 9, & 1 \\ \hline 3 & 1, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$$

∴ L.C.M. of 6, 9 and 4 = 2 × 2 × 3 × 3 = 36

Multiplying both sides by 36, we get:

$$\begin{aligned}\Rightarrow 36 \times \Big(\frac{(z + 5)}{6} - \frac{(z + 1)}{9}\Big) = 36 \times \frac{(z + 3)}{4} \\[1em] \Rightarrow \frac{36(z + 5)}{6} - \frac{36(z + 1)}{9} = \frac{36(z + 3)}{4} \\[1em] \Rightarrow 6(z + 5) - 4(z + 1) = 9(z + 3) \\[1em] \Rightarrow 6z + 30 - 4z - 4 = 9z + 27 \\[1em] \Rightarrow 2z + 26 = 9z + 27 \\[1em] \Rightarrow 26 - 27 = 9z - 2z \\[1em] \Rightarrow 7z = - 1 \\[1em] \Rightarrow z = - \frac{1}{7}\end{aligned}$$

Hence, $\bm{z = -\frac{1}{7}}$.

Given,

$\frac{2 - 9z}{17 - 4z} = \frac{4}{5}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 5(2 - 9z) = 4(17 - 4z) \\[1em] \Rightarrow 10 - 45z = 68 - 16z \\[1em] \Rightarrow 10 - 68 = 45z - 16z \\[1em] \Rightarrow 29z = - 58 \\[1em] \Rightarrow z = - \frac{58}{29} \\[1em] \Rightarrow z = - 2\end{aligned}$$

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

Given,

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

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow (2x - 3)(3x + 4) = (2x + 3)(3x - 1) \\[1em] \Rightarrow 6x^{2} + 8x - 9x - 12 = 6x^{2} - 2x + 9x - 3 \\[1em] \Rightarrow 6x^{2} - x - 12 = 6x^{2} + 7x - 3 \\[1em] \Rightarrow - x - 12 = 7x - 3 \\[1em] \Rightarrow 3 - 12 = 7x + x \\[1em] \Rightarrow 8x = - 9 \\[1em] \Rightarrow x = - \frac{9}{8}\end{aligned}$$

Hence, $\bm{x = -\frac{9}{8}}$.

Given,

$\frac{2 - 9x}{16 + 5x} = 0$

As the value of the fraction is zero, its numerator must be zero.

$$\begin{aligned}\Rightarrow 2 - 9x = 0 \\[1em] \Rightarrow - 9x = - 2 \\[1em] \Rightarrow x = \frac{-2}{-9} \\[1em] \Rightarrow x = \frac{2}{9}\end{aligned}$$

Hence, $\bm{x = \frac{2}{9}}$.

Given,

$\frac{0.5z + 4}{1.2z + 6} = \frac{5}{3}$

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 3(0.5z + 4) = 5(1.2z + 6) \\[1em] \Rightarrow 1.5z + 12 = 6z + 30 \\[1em] \Rightarrow 12 - 30 = 6z - 1.5z \\[1em] \Rightarrow 4.5z = - 18 \\[1em] \Rightarrow z = - \frac{18}{4.5} \\[1em] \Rightarrow z = - 4\end{aligned}$$

Hence, $\bm{z = -4}$.

Given,

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

Taking L.C.M. of the denominators on L.H.S., we get:

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

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow 2x(14x - 1) = 7(4x^{2} - 1) \\[1em] \Rightarrow 28x^{2} - 2x = 28x^{2} - 7 \\[1em] \Rightarrow - 2x = - 7 \\[1em] \Rightarrow x = \frac{7}{2}\end{aligned}$$

Hence, $\bm{x = \frac{7}{2}}$.

Given,

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

Taking L.C.M. of the denominators on both the sides, we get:

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

By cross multiplication, we get:

$$\begin{aligned}\Rightarrow (x - 5)(x - 1) = (x + 5)(x - 2) \\[1em] \Rightarrow x^{2} - x - 5x + 5 = x^{2} - 2x + 5x - 10 \\[1em] \Rightarrow x^{2} - 6x + 5 = x^{2} + 3x - 10 \\[1em] \Rightarrow 5 + 10 = 3x + 6x \\[1em] \Rightarrow 9x = 15 \\[1em] \Rightarrow x = \frac{15}{9} \\[1em] \Rightarrow x = \frac{5}{3}\end{aligned}$$

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

Solving,

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

Hence, $\bm{z = -3}$.

Solving,

$$\begin{aligned}\Rightarrow y(2y + 3) - 2y(y - 5) = 26 \\[1em] \Rightarrow 2y^{2} + 3y - 2y^{2} + 10y = 26 \\[1em] \Rightarrow 13y = 26 \\[1em] \Rightarrow y = \frac{26}{13} \\[1em] \Rightarrow y = 2\end{aligned}$$

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

Given,

$\frac{x + 6}{4} - \frac{5x - 4}{8} + \frac{x - 3}{5} = 0$

L.C.M. of 4, 8 and 5 = 40

Multiplying both sides by 40, we get:

$$\begin{aligned}\Rightarrow 40 \times \Big(\frac{x + 6}{4} - \frac{5x - 4}{8} + \frac{x - 3}{5}\Big) = 40 \times 0 \\[1em] \Rightarrow \frac{40(x + 6)}{4} - \frac{40(5x - 4)}{8} + \frac{40(x - 3)}{5} = 40 \times 0 \\[1em] \Rightarrow 10(x + 6) - 5(5x - 4) + 8(x - 3) = 0 \\[1em] \Rightarrow 10x + 60 - 25x + 20 + 8x - 24 = 0 \\[1em] \Rightarrow - 7x + 56 = 0 \\[1em] \Rightarrow 7x = 56 \\[1em] \Rightarrow x = \frac{56}{7} \\[1em] \Rightarrow x = 8\end{aligned}$$

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

Given,

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

Simplifying the terms inside the brackets and the R.H.S., we get:

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

L.C.M. of 4 and 2 = 4

Multiplying both sides by 4, we get:

$$\begin{aligned}\Rightarrow 4 \times \Big[\frac{3}{4}(7x - 1) - \left(\frac{5x - 1}{2}\right)\Big] = 4 \times \frac{2x + 3}{2} \\[1em] \Rightarrow 3(7x - 1) - 2(5x - 1) = 2(2x + 3) \\[1em] \Rightarrow 21x - 3 - 10x + 2 = 4x + 6 \\[1em] \Rightarrow 11x - 1 = 4x + 6 \\[1em] \Rightarrow 11x - 4x = 6 + 1 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = \frac{7}{7} \\[1em] \Rightarrow x = 1\end{aligned}$$

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

Let the number be $x$.

Given,

17 less than four times a number is 11.

$$\begin{aligned}\therefore 4x - 17 = 11 \\[1em] \Rightarrow 4x = 11 + 17 \\[1em] \Rightarrow 4x = 28 \\[1em] \Rightarrow x = \frac{28}{4} \\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, the required number = 7.

Let the number be $x$.

Given,

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

$$\begin{aligned}\therefore 4x + 10 = 5x - 5 \\[1em] \Rightarrow 10 + 5 = 5x - 4x \\[1em] \Rightarrow x = 15\end{aligned}$$

Hence, the required number = 15.

Let the original number be $x$.

Given,

$\frac{2}{3}$ of a number is 20 less than the original number.

$\therefore \frac{2}{3}x = x - 20$

Multiplying both sides by 3, we get:

$$\begin{aligned}\Rightarrow \frac{3 \times 2x}{3} = 3 \times \Big(x - 20\Big) \\[1em] \Rightarrow 2x = 3x - 60 \\[1em] \Rightarrow 60 = 3x - 2x \\[1em] \Rightarrow x = 60\end{aligned}$$

Hence, the original number = 60.

Let the number be $x$.

Given,

A number is 25 more than its $\frac{5}{6}$th part.

$\therefore x = \frac{5}{6}x + 25$

Multiplying both sides by 6, we get:

$$\begin{aligned}\Rightarrow 6 \times x = 6 \times \Big(\frac{5}{6}x + 25\Big) \\[1em] \Rightarrow 6x = \frac{6 \times 5x}{6} + 6 \times 25 \\[1em] \Rightarrow 6x = 5x + 150 \\[1em] \Rightarrow 6x - 5x = 150 \\[1em] \Rightarrow x = 150\end{aligned}$$

Hence, the required number = 150.

Let the number be $x$.

Given,

A number is as much greater than 21 as is less than 71.

$$\begin{aligned}\therefore x - 21 = 71 - x \\[1em] \Rightarrow x + x = 71 + 21 \\[1em] \Rightarrow 2x = 92 \\[1em] \Rightarrow x = \frac{92}{2} \\[1em] \Rightarrow x = 46\end{aligned}$$

Hence, the required number = 46.

Let the number be $x$.

Given,

6 more than one-fourth of a number is two-fifths of the number.

$\therefore \frac{x}{4} + 6 = \frac{2x}{5}$

L.C.M. of 4 and 5 = 20

Multiplying both sides by 20, we get:

$$\begin{aligned}\Rightarrow 20 \times \Big(\frac{x}{4} + 6\Big) = 20 \times \frac{2x}{5} \\[1em] \Rightarrow \frac{20x}{4} + 20 \times 6 = \frac{20 \times 2x}{5} \\[1em] \Rightarrow 5x + 120 = 8x \\[1em] \Rightarrow 120 = 8x - 5x \\[1em] \Rightarrow 3x = 120 \\[1em] \Rightarrow x = \frac{120}{3} \\[1em] \Rightarrow x = 40\end{aligned}$$

Hence, the required number = 40.

Let the number be $x$.

Given,

One-third of a number exceeds one-fourth of the number by 15.

$\therefore \frac{x}{3} - \frac{x}{4} = 15$

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

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow 12 \times \Big(\frac{x}{3} - \frac{x}{4}\Big) = 12 \times 15 \\[1em] \Rightarrow \frac{12x}{3} - \frac{12x}{4} = 12 \times 15 \\[1em] \Rightarrow 4x - 3x = 180 \\[1em] \Rightarrow x = 180\end{aligned}$$

Hence, the required number = 180.

Let the number be $x$.

Given,

One-fifth of a number decreased by 5 is 16.

$\therefore \frac{x}{5} - 5 = 16$

Multiplying both sides by 5, we get:

$$\begin{aligned}\Rightarrow 5 \times \Big(\frac{x}{5} - 5\Big) = 5 \times 16 \\[1em] \Rightarrow \frac{5x}{5} - 5 \times 5 = 5 \times 16 \\[1em] \Rightarrow x - 25 = 80 \\[1em] \Rightarrow x = 80 + 25 \\[1em] \Rightarrow x = 105\end{aligned}$$

Hence, the required number = 105.

Let the number be $x$.

Given,

A number when divided by 6 is diminished by 40.

$\therefore \frac{x}{6} = x - 40$

Multiplying both sides by 6, we get:

$$\begin{aligned}\Rightarrow \frac{6x}{6} = 6 \times \Big(x - 40\Big) \\[1em] \Rightarrow x = 6x - 240 \\[1em] \Rightarrow 240 = 6x - x \\[1em] \Rightarrow 5x = 240 \\[1em] \Rightarrow x = \frac{240}{5} \\[1em] \Rightarrow x = 48\end{aligned}$$

Hence, the required number = 48.

Let the number be $x$.

Given,

Four-fifths of a number exceeds two-thirds of the number by 10.

$\therefore \frac{4x}{5} - \frac{2x}{3} = 10$

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

Multiplying both sides by 15, we get:

$$\begin{aligned}\Rightarrow 15 \times \Big(\frac{4x}{5} - \frac{2x}{3}\Big) = 15 \times 10 \\[1em] \Rightarrow \frac{15 \times 4x}{5} - \frac{15 \times 2x}{3} = 15 \times 10 \\[1em] \Rightarrow 12x - 10x = 150 \\[1em] \Rightarrow 2x = 150 \\[1em] \Rightarrow x = \frac{150}{2} \\[1em] \Rightarrow x = 75\end{aligned}$$

Hence, the required number = 75.

Given,

Two numbers are in the ratio 3 : 4.

Let the two numbers be $3x$ and $4x$.

Given,

The sum of the two numbers is 84.

$$\begin{aligned}\therefore 3x + 4x = 84 \\[1em] \Rightarrow 7x = 84 \\[1em] \Rightarrow x = \frac{84}{7} \\[1em] \Rightarrow x = 12\end{aligned}$$

So, the numbers are 3 × 12 = 36 and 4 × 12 = 48.

Hence, the required numbers are 36 and 48.

Given,

Three numbers are in the ratio 4 : 5 : 6.

Let the three numbers be $4x, 5x$ and $6x$.

Given,

The sum of the three numbers is 135.

$$\begin{aligned}\therefore 4x + 5x + 6x = 135 \\[1em] \Rightarrow 15x = 135 \\[1em] \Rightarrow x = \frac{135}{15} \\[1em] \Rightarrow x = 9\end{aligned}$$

So, the numbers are 4 × 9 = 36, 5 × 9 = 45 and 6 × 9 = 54.

Hence, the required numbers are 36, 45 and 54.

Given,

Two numbers are in the ratio 3 : 5.

Let the two numbers be $3x$ and $5x$.

Given,

If each number is increased by 10, the ratio between the new numbers so formed is 5 : 7.

$$\begin{aligned}\therefore \frac{3x + 10}{5x + 10} = \frac{5}{7} \\[1em] \Rightarrow 7(3x + 10) = 5(5x + 10) \\[1em] \Rightarrow 21x + 70 = 25x + 50 \\[1em] \Rightarrow 70 - 50 = 25x - 21x \\[1em] \Rightarrow 4x = 20 \\[1em] \Rightarrow x = \frac{20}{4} \\[1em] \Rightarrow x = 5\end{aligned}$$

So, the original numbers are 3 × 5 = 15 and 5 × 5 = 25.

Hence, the original numbers are 15 and 25.

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

Given,

The sum of three consecutive odd numbers is 75.

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) = 75 \\[1em] \Rightarrow 3x + 6 = 75 \\[1em] \Rightarrow 3x = 75 - 6 \\[1em] \Rightarrow 3x = 69 \\[1em] \Rightarrow x = \frac{69}{3} \\[1em] \Rightarrow x = 23\end{aligned}$$

So, the numbers are 23, 23 + 2 = 25 and 23 + 4 = 27.

Hence, the required numbers are 23, 25 and 27.

Given,

25 is divided into two parts.

Let the first part be $x$ and the second part be $(25 - x)$.

Given,

7 times the first part added to 5 times the second part makes 139.

$$\begin{aligned}\therefore 7x + 5(25 - x) = 139 \\[1em] \Rightarrow 7x + 125 - 5x = 139 \\[1em] \Rightarrow 2x = 139 - 125 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \frac{14}{2} \\[1em] \Rightarrow x = 7\end{aligned}$$

So, the second part = 25 - 7 = 18

Hence, the two parts are 7 and 18.

Given,

180 is divided into two parts.

Let the second part be $x$ and the first part be $(180 - x)$.

Given,

The first part is 12 less than twice the second part.

$$\begin{aligned}\therefore 180 - x = 2x - 12 \\[1em] \Rightarrow 180 + 12 = 2x + x \\[1em] \Rightarrow 3x = 192 \\[1em] \Rightarrow x = \frac{192}{3} \\[1em] \Rightarrow x = 64\end{aligned}$$

So, the first part = 180 - 64 = 116

Hence, the two parts are 116 and 64.

Given,

The denominator of a fraction is 4 more than its numerator.

Let the numerator of the fraction be $x$ and its denominator be $(x + 4)$.

Given,

On subtracting 1 from each of the numerator and the denominator, the fraction becomes $\frac{1}{2}$.

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

So, the denominator = 5 + 4 = 9

Hence, the original fraction = $\mathbf{\frac{5}{9}}$.

Given,

The denominator of a fraction is 1 more than double the numerator.

Let the numerator of the fraction be $x$ and its denominator be $(2x + 1)$.

Given,

On adding 2 to the numerator and subtracting 3 from the denominator, we obtain 1.

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

So, the denominator = 2 × 4 + 1 = 9

Hence, the original fraction = $\mathbf{\frac{4}{9}}$.

Given,

The sum of the digits of a two-digit number is 5.

Let the digit at unit's place be $x$ and the digit at ten's place be $(5 - x)$.

So, the original number = $10(5 - x) + x = 50 - 9x$

On reversing the digits, the new number = $10x + (5 - x) = 9x + 5$

Given,

On adding 27 to the number, its digits are reversed.

$$\begin{aligned}\therefore (50 - 9x) + 27 = 9x + 5 \\[1em] \Rightarrow 77 - 9x = 9x + 5 \\[1em] \Rightarrow 77 - 5 = 9x + 9x \\[1em] \Rightarrow 18x = 72 \\[1em] \Rightarrow x = \frac{72}{18} \\[1em] \Rightarrow x = 4\end{aligned}$$

So, the digit at ten's place = 5 - 4 = 1

and the original number = 10 × 1 + 4 = 14

Hence, the original number = 14.

Let the number to be added be $x$.

Given,

On adding the same number to each one of the numbers 15, 23, 29 and 44, the numbers obtained are in proportion.

$$\begin{aligned}\therefore \frac{15 + x}{23 + x} = \frac{29 + x}{44 + x} \\[1em] \Rightarrow (15 + x)(44 + x) = (23 + x)(29 + x) \\[1em] \Rightarrow 660 + 15x + 44x + x^{2} = 667 + 23x + 29x + x^{2} \\[1em] \Rightarrow 660 + 59x + x^{2} = 667 + 52x + x^{2} \\[1em] \Rightarrow 59x - 52x = 667 - 660 \\[1em] \Rightarrow 7x = 7 \\[1em] \Rightarrow x = \frac{7}{7} \\[1em] \Rightarrow x = 1\end{aligned}$$

Hence, the required number to be added = 1.

Given,

The sum of two numbers is 110.

Let the larger number be $x$ and the smaller number be $(110 - x)$.

Given,

One-fifth of the larger number is 8 more than one-ninth of the smaller number.

$\therefore \frac{x}{5} = \frac{110 - x}{9} + 8$

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

Multiplying both sides by 45, we get:

$$\begin{aligned}\Rightarrow 45 \times \frac{x}{5} = 45 \times \Big(\frac{110 - x}{9} + 8\Big) \\[1em] \Rightarrow \frac{45x}{5} = \frac{45(110 - x)}{9} + 45 \times 8 \\[1em] \Rightarrow 9x = 5(110 - x) + 360 \\[1em] \Rightarrow 9x = 550 - 5x + 360 \\[1em] \Rightarrow 9x + 5x = 910 \\[1em] \Rightarrow 14x = 910 \\[1em] \Rightarrow x = \frac{910}{14} \\[1em] \Rightarrow x = 65\end{aligned}$$

So, the smaller number = 110 - 65 = 45

Hence, the required numbers are 65 and 45.

Let the number be $x$.

Given,

When the number is subtracted from the numerator of the fraction $\frac{12}{13}$ and six times that number is added to the denominator, the new fraction is $\frac{1}{11}$.

$$\begin{aligned}\therefore \frac{12 - x}{13 + 6x} = \frac{1}{11} \\[1em] \Rightarrow 11(12 - x) = 13 + 6x \\[1em] \Rightarrow 132 - 11x = 13 + 6x \\[1em] \Rightarrow 132 - 13 = 6x + 11x \\[1em] \Rightarrow 17x = 119 \\[1em] \Rightarrow x = \frac{119}{17} \\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, the required number = 7.

Given,

The two perpendicular sides of the right angled triangle are in the ratio 5 : 12.

Let the two perpendicular sides of the right angled triangle be $5x$ cm and $12x$ cm.

By Pythagoras Theorem,

$$\begin{aligned}\Rightarrow \text{Hypotenuse} = \sqrt{(5x)^{2} + (12x)^{2}} \\[1em] \Rightarrow \text{Hypotenuse} = \sqrt{25x^{2} + 144x^{2}} \\[1em] \Rightarrow \text{Hypotenuse} = \sqrt{169x^{2}} \\[1em] \Rightarrow \text{Hypotenuse} = 13x \text{ cm}\end{aligned}$$

Given,

The perimeter of the right angled triangle is 120 cm.

Perimeter of the triangle = Sum of all the sides

$$\begin{aligned}\therefore 5x + 12x + 13x = 120 \\[1em] \Rightarrow 30x = 120 \\[1em] \Rightarrow x = \frac{120}{30} \\[1em] \Rightarrow x = 4\end{aligned}$$

So, the sides are 5 × 4 = 20 cm, 12 × 4 = 48 cm and 13 × 4 = 52 cm.

Hence, the lengths of the sides of the triangle are 20 cm, 48 cm and 52 cm.

Given,

The sum of the digits of a two-digit number is 9.

Let the digit at unit's place be $x$ and the digit at ten's place be $(9 - x)$.

So, the original number = $10(9 - x) + x = 90 - 9x$

On reversing the digits, the new number = $10x + (9 - x) = 9x + 9$

Given,

If 9 is added to the number formed by reversing the digits, then the result is thrice the original number.

$$\begin{aligned}\therefore (9x + 9) + 9 = 3(90 - 9x) \\[1em] \Rightarrow 9x + 18 = 270 - 27x \\[1em] \Rightarrow 9x + 27x = 270 - 18 \\[1em] \Rightarrow 36x = 252 \\[1em] \Rightarrow x = \frac{252}{36} \\[1em] \Rightarrow x = 7\end{aligned}$$

So, the digit at ten's place = 9 - 7 = 2

and the original number = 10 × 2 + 7 = 27

Hence, the original number = 27.

Given,

The length of a rectangular plot of land exceeds its breadth by 23 m.

Let the breadth of the plot be $x$ m and its length be $(x + 23)$ m.

Given,

If the length is decreased by 15 m and the breadth is increased by 7 m, the area is reduced by 360 m².

We know that,

Area of a rectangle = length × breadth

$$\begin{aligned}\therefore x(x + 23) - (x + 23 - 15)(x + 7) = 360 \\[1em] \Rightarrow x(x + 23) - (x + 8)(x + 7) = 360 \\[1em] \Rightarrow x^{2} + 23x - (x^{2} + 7x + 8x + 56) = 360 \\[1em] \Rightarrow x^{2} + 23x - x^{2} - 15x - 56 = 360 \\[1em] \Rightarrow 8x - 56 = 360 \\[1em] \Rightarrow 8x = 360 + 56 \\[1em] \Rightarrow 8x = 416 \\[1em] \Rightarrow x = \frac{416}{8} \\[1em] \Rightarrow x = 52\end{aligned}$$

So, the breadth of the plot = 52 m

and the length of the plot = 52 + 23 = 75 m

Hence, the length and breadth of the plot are 75 m and 52 m respectively.

Given,

The length of a rectangular park is twice its breadth.

Let the breadth of the park be $x$ m and its length be $2x$ m.

Given,

The perimeter of the park is 186 m.

We know that,

Perimeter of a rectangle = 2(length + breadth)

$$\begin{aligned}\therefore 2(2x + x) = 186 \\[1em] \Rightarrow 2 \times 3x = 186 \\[1em] \Rightarrow 6x = 186 \\[1em] \Rightarrow x = \frac{186}{6} \\[1em] \Rightarrow x = 31\end{aligned}$$

So, the breadth of the park = 31 m

and the length of the park = 2 × 31 = 62 m

Hence, the length and breadth of the park are 62 m and 31 m respectively.

Given,

The length of a rectangle is 7 cm more than its breadth.

Let the breadth of the rectangle be $x$ cm and its length be $(x + 7)$ cm.

Given,

The perimeter of the rectangle is 90 cm.

We know that,

Perimeter of a rectangle = 2(length + breadth)

$$\begin{aligned}\therefore 2[(x + 7) + x] = 90 \\[1em] \Rightarrow 2(2x + 7) = 90 \\[1em] \Rightarrow 4x + 14 = 90 \\[1em] \Rightarrow 4x = 90 - 14 \\[1em] \Rightarrow 4x = 76 \\[1em] \Rightarrow x = \frac{76}{4} \\[1em] \Rightarrow x = 19\end{aligned}$$

So, the breadth of the rectangle = 19 cm

and the length of the rectangle = 19 + 7 = 26 cm

Hence, the length and breadth of the rectangle are 26 cm and 19 cm respectively.

Given,

The length of a rectangle is 7 cm less than twice its breadth.

Let the breadth of the original rectangle be $x$ cm and its length be $(2x - 7)$ cm.

Given,

If the length is decreased by 2 cm and breadth increased by 3 cm, the perimeter of the resulting rectangle is 66 cm.

We know that,

Perimeter of a rectangle = 2(length + breadth)

$$\begin{aligned}\therefore 2[(2x - 7 - 2) + (x + 3)] = 66 \\[1em] \Rightarrow 2[(2x - 9) + (x + 3)] = 66 \\[1em] \Rightarrow 2(3x - 6) = 66 \\[1em] \Rightarrow 3x - 6 = \frac{66}{2} \\[1em] \Rightarrow 3x - 6 = 33 \\[1em] \Rightarrow 3x = 33 + 6 \\[1em] \Rightarrow 3x = 39 \\[1em] \Rightarrow x = \frac{39}{3} \\[1em] \Rightarrow x = 13\end{aligned}$$

So, the breadth of the original rectangle = 13 cm

and the length of the original rectangle = 2 × 13 - 7 = 19 cm

Hence, the length and breadth of the original rectangle are 19 cm and 13 cm respectively.

Given,

A man is five times as old as his son.

Let the present age of the son be $x$ years and the present age of the man be $5x$ years.

Given,

In two years time, the man will be four times as old as his son.

$$\begin{aligned}\therefore 5x + 2 = 4(x + 2) \\[1em] \Rightarrow 5x + 2 = 4x + 8 \\[1em] \Rightarrow 5x - 4x = 8 - 2 \\[1em] \Rightarrow x = 6\end{aligned}$$

So, the present age of the man = 5 × 6 = 30 years

Hence, the present ages of the man and his son are 30 years and 6 years respectively.

Given,

A man is twice as old as his son.

Let the present age of the son be $x$ years and the present age of the man be $2x$ years.

Given,

Twelve years ago, the man was thrice as old as his son.

$$\begin{aligned}\therefore 2x - 12 = 3(x - 12) \\[1em] \Rightarrow 2x - 12 = 3x - 36 \\[1em] \Rightarrow 36 - 12 = 3x - 2x \\[1em] \Rightarrow x = 24\end{aligned}$$

So, the present age of the man = 2 × 24 = 48 years

Hence, the present ages of the man and his son are 48 years and 24 years respectively.

Given,

The ratio of the ages of Seema and Rekha is 5 : 3.

Let the ages of Seema and Rekha be $5x$ years and $3x$ years respectively.

Given,

Seema is 10 years elder than Rekha.

$$\begin{aligned}\therefore 5x - 3x = 10 \\[1em] \Rightarrow 2x = 10 \\[1em] \Rightarrow x = \frac{10}{2} \\[1em] \Rightarrow x = 5\end{aligned}$$

So, Seema's age = 5 × 5 = 25 years

and Rekha's age = 3 × 5 = 15 years

Hence, the ages of Seema and Rekha are 25 years and 15 years respectively.

Given,

The sum of the present ages of Parvathi and her son is 55 years.

Let Parvathi's present age be $x$ years and her son's present age be $(55 - x)$ years.

Given,

5 years ago, the age of Parvathi was 4 times the age of her son.

$$\begin{aligned}\therefore x - 5 = 4(55 - x - 5) \\[1em] \Rightarrow x - 5 = 4(50 - x) \\[1em] \Rightarrow x - 5 = 200 - 4x \\[1em] \Rightarrow x + 4x = 200 + 5 \\[1em] \Rightarrow 5x = 205 \\[1em] \Rightarrow x = \frac{205}{5} \\[1em] \Rightarrow x = 41\end{aligned}$$

Hence, Parvathi's present age = 41 years.

Let after $x$ years the father become twice as old as his son.

Given,

The man is 56 years old and his son is 24 years old.

$$\begin{aligned}\therefore 56 + x = 2(24 + x) \\[1em] \Rightarrow 56 + x = 48 + 2x \\[1em] \Rightarrow 56 - 48 = 2x - x \\[1em] \Rightarrow x = 8\end{aligned}$$

Hence, after 8 years the father will be twice as old as his son.

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

Given,

9 years hence, the girl will be 3 times as old as she was 9 years ago.

$$\begin{aligned}\therefore x + 9 = 3(x - 9) \\[1em] \Rightarrow x + 9 = 3x - 27 \\[1em] \Rightarrow 9 + 27 = 3x - x \\[1em] \Rightarrow 2x = 36 \\[1em] \Rightarrow x = \frac{36}{2} \\[1em] \Rightarrow x = 18\end{aligned}$$

Hence, the present age of the girl = 18 years.

Given,

The man made a trip of 480 km.

Let the part of the trip covered at 60 km/hr be $x$ km and the part of the trip covered at 45 km/hr be $(480 - x)$ km.

Given,

The trip was made in 9 hours.

We know that,

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

$\therefore \frac{480 - x}{45} + \frac{x}{60} = 9$

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

Multiplying both sides by 180, we get:

$$\begin{aligned}\Rightarrow 180 \times \Big(\frac{480 - x}{45} + \frac{x}{60}\Big) = 180 \times 9 \\[1em] \Rightarrow \frac{180(480 - x)}{45} + \frac{180x}{60} = 180 \times 9 \\[1em] \Rightarrow 4(480 - x) + 3x = 1620 \\[1em] \Rightarrow 1920 - 4x + 3x = 1620 \\[1em] \Rightarrow 1920 - x = 1620 \\[1em] \Rightarrow 1920 - 1620 = x \\[1em] \Rightarrow x = 300\end{aligned}$$

Hence, the part of the trip covered at 60 km/hr = 300 km.

Let the distance between the two towns be $x$ km.

Given,

The motorist travelled from town A to town B at 54 km/hr and returned at 60 km/hr, and the total time taken is $9\frac{1}{2}$ hours.

We know that,

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

$$\begin{aligned}\therefore \frac{x}{54} + \frac{x}{60} = 9\frac{1}{2} \\[1em] \Rightarrow \frac{x}{54} + \frac{x}{60} = \frac{19}{2}\end{aligned}$$

L.C.M. of 54, 60 and 2 = 540

Multiplying both sides by 540, we get:

$$\begin{aligned}\Rightarrow 540 \times \Big(\frac{x}{54} + \frac{x}{60}\Big) = 540 \times \frac{19}{2} \\[1em] \Rightarrow \frac{540x}{54} + \frac{540x}{60} = \frac{540 \times 19}{2} \\[1em] \Rightarrow 10x + 9x = 270 \times 19 \\[1em] \Rightarrow 19x = 5130 \\[1em] \Rightarrow x = \frac{5130}{19} \\[1em] \Rightarrow x = 270\end{aligned}$$

Hence, the distance between the two towns = 270 km.

Given,

The speed of one of the motor-cyclists is 7 km/hr faster than that of the other.

Let the speed of the slower motor-cyclist be $x$ km/hr and the speed of the faster motor-cyclist be $(x + 7)$ km/hr.

Let the slower motor-cyclist start from station A and the faster motor-cyclist start from station B.

Let C and D be their positions after 2 hours.

We know that,

Distance = Speed × Time

AC = $2x$ km, BD = $2(x + 7)$ km and CD = 34 km

Given,

The distance between the two stations is 300 km.

$$\begin{aligned}\therefore \text{AC} + \text{BD} + \text{CD} = 300 \\[1em] \Rightarrow 2x + 2(x + 7) + 34 = 300 \\[1em] \Rightarrow 2x + 2x + 14 + 34 = 300 \\[1em] \Rightarrow 4x + 48 = 300 \\[1em] \Rightarrow 4x = 300 - 48 \\[1em] \Rightarrow 4x = 252 \\[1em] \Rightarrow x = \frac{252}{4} \\[1em] \Rightarrow x = 63\end{aligned}$$

So, the speed of the faster motor-cyclist = 63 + 7 = 70 km/hr

Hence, the speeds of the two motor-cyclists are 63 km/hr and 70 km/hr.

Given,

The rate of stream is 5 kmph.

Let the speed of the boat in still water be $x$ kmph.

So, speed of the boat upstream = $(x - 5)$ kmph

and speed of the boat downstream = $(x + 5)$ kmph

Given,

The boat travels 30 km upstream in the same period of time as it takes to travel 50 km downstream.

We know that,

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

$$\begin{aligned}\therefore \frac{30}{x - 5} = \frac{50}{x + 5} \\[1em] \Rightarrow 30(x + 5) = 50(x - 5) \\[1em] \Rightarrow 30x + 150 = 50x - 250 \\[1em] \Rightarrow 150 + 250 = 50x - 30x \\[1em] \Rightarrow 20x = 400 \\[1em] \Rightarrow x = \frac{400}{20} \\[1em] \Rightarrow x = 20\end{aligned}$$

Hence, the speed of the boat in still water = 20 kmph.

Given,

The length of each of the equal sides of an isosceles triangle is 4 cm longer than the base.

Let the length of the base of the isosceles triangle be $x$ cm and the length of each of the equal sides be $(x + 4)$ cm.

Given,

The perimeter of the triangle is 62 cm.

Perimeter of the triangle = Sum of all the sides

$$\begin{aligned}\therefore (x + 4) + (x + 4) + x = 62 \\[1em] \Rightarrow 3x + 8 = 62 \\[1em] \Rightarrow 3x = 62 - 8 \\[1em] \Rightarrow 3x = 54 \\[1em] \Rightarrow x = \frac{54}{3} \\[1em] \Rightarrow x = 18\end{aligned}$$

So, the base = 18 cm

and each of the equal sides = 18 + 4 = 22 cm

Hence, the lengths of the sides of the triangle are 22 cm, 22 cm and 18 cm.

Let the total number of candidates appeared be $x$.

Given,

One-fifth of the whole plus 16 secured first division, one-fourth plus 15 secured second division, one-fourth minus 25 secured third division and the remaining 60 candidates failed.

$\therefore \left(\frac{x}{5} + 16\right) + \left(\frac{x}{4} + 15\right) + \left(\frac{x}{4} - 25\right) + 60 = x$

L.C.M. of 5 and 4 = 20

Multiplying both sides by 20, we get:

$$\begin{aligned}\Rightarrow 20 \times \Big[\left(\frac{x}{5} + 16\right) + \left(\frac{x}{4} + 15\right) + \left(\frac{x}{4} - 25\right) + 60\Big] = 20 \times x \\[1em] \Rightarrow \frac{20x}{5} + 20 \times 16 + \frac{20x}{4} + 20 \times 15 + \frac{20x}{4} - 20 \times 25 + 20 \times 60 = 20x \\[1em] \Rightarrow 4x + 320 + 5x + 300 + 5x - 500 + 1200 = 20x \\[1em] \Rightarrow 14x + 1320 = 20x \\[1em] \Rightarrow 1320 = 20x - 14x \\[1em] \Rightarrow 6x = 1320 \\[1em] \Rightarrow x = \frac{1320}{6} \\[1em] \Rightarrow x = 220\end{aligned}$$

Hence, the total number of candidates appeared = 220.

Given,

Raman has three times as much money as Kamal.

Let the money Kamal had originally be ₹ $x$ and the money Raman had originally be ₹ $3x$.

Given,

If Raman gives ₹ 750 to Kamal, then Kamal will have twice as much as left with Raman.

$$\begin{aligned}\therefore x + 750 = 2(3x - 750) \\[1em] \Rightarrow x + 750 = 6x - 1500 \\[1em] \Rightarrow 750 + 1500 = 6x - x \\[1em] \Rightarrow 5x = 2250 \\[1em] \Rightarrow x = \frac{2250}{5} \\[1em] \Rightarrow x = 450\end{aligned}$$

So, the money Raman had originally = 3 × 450 = ₹ 1,350

Hence, originally Kamal had ₹ 450 and Raman had ₹ 1,350.

Given,

The angles of a triangle are in the ratio 2 : 3 : 4.

Let the angles of the triangle be $(2x)^{\circ}, (3x)^{\circ}$ and $(4x)^{\circ}$.

We know that,

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

$$\begin{aligned}\therefore (2x)^{\circ} + (3x)^{\circ} + (4x)^{\circ} = 180^{\circ} \\[1em] \Rightarrow (9x)^{\circ} = 180^{\circ} \\[1em] \Rightarrow x^{\circ} = \frac{180^{\circ}}{9} \\[1em] \Rightarrow x^{\circ} = 20^{\circ}\end{aligned}$$

So, the angles are $2 \times 20 = 40^{\circ}, 3 \times 20 = 60^{\circ}$ and $4 \times 20 = 80^{\circ}$.

Hence, the angles of the triangle are 40°, 60° and 80°.

Given,

A certain number of men can finish a piece of work in 50 days.

Let the number of men originally there be $x$.

⇒ 1 man completes the work in $50x$ days.

Given,

If there are 7 more men, the work can be completed 10 days earlier.

⇒ $(x + 7)$ men complete the work in (50 - 10) days = 40 days.

⇒ 1 man completes the work in $40(x + 7)$ days.

As the work is the same in both the cases,

$$\begin{aligned}\therefore 50x = 40(x + 7) \\[1em] \Rightarrow 50x = 40x + 280 \\[1em] \Rightarrow 50x - 40x = 280 \\[1em] \Rightarrow 10x = 280 \\[1em] \Rightarrow x = \frac{280}{10} \\[1em] \Rightarrow x = 28\end{aligned}$$

Hence, originally there were 28 men.

Given,

600 is divided into two parts.

Let the first part be $x$ and the second part be $(600 - x)$.

Given,

40% of the first part exceeds 60% of the second part by 120.

$$\begin{aligned}\therefore \frac{40}{100} \times x - \frac{60}{100} \times (600 - x) = 120 \\[1em] \Rightarrow \frac{2x}{5} - \frac{3(600 - x)}{5} = 120\end{aligned}$$

Multiplying both sides by 5, we get:

$$\begin{aligned}\Rightarrow 5 \times \Big(\frac{2x}{5} - \frac{3(600 - x)}{5}\Big) = 5 \times 120 \\[1em] \Rightarrow 2x - 3(600 - x) = 5 \times 120 \\[1em] \Rightarrow 2x - 1800 + 3x = 600 \\[1em] \Rightarrow 5x = 600 + 1800 \\[1em] \Rightarrow 5x = 2400 \\[1em] \Rightarrow x = \frac{2400}{5} \\[1em] \Rightarrow x = 480\end{aligned}$$

So, the second part = 600 - 480 = 120

Hence, the two parts are 480 and 120.

Let the number of days for which the workman remained absent be $x$ and the number of days for which he worked be $(30 - x)$.

Given,

The workman is paid ₹ 150 for each day he works and is fined ₹ 50 for each day he is absent. In a month of 30 days he earned ₹ 2,100.

$$\begin{aligned}\therefore 150(30 - x) - 50x = 2100 \\[1em] \Rightarrow 4500 - 150x - 50x = 2100 \\[1em] \Rightarrow 4500 - 200x = 2100 \\[1em] \Rightarrow 4500 - 2100 = 200x \\[1em] \Rightarrow 200x = 2400 \\[1em] \Rightarrow x = \frac{2400}{200} \\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, the workman remained absent for 12 days.

Given,

$\frac{x}{2} - \frac{3x}{4} + \frac{5x}{6} = 21$

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

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow 12 \times \Big(\frac{x}{2} - \frac{3x}{4} + \frac{5x}{6}\Big) = 12 \times 21 \\[1em] \Rightarrow \frac{12x}{2} - \frac{12 \times 3x}{4} + \frac{12 \times 5x}{6} = 12 \times 21 \\[1em] \Rightarrow 6x - 9x + 10x = 252 \\[1em] \Rightarrow 7x = 252 \\[1em] \Rightarrow x = \frac{252}{7} \\[1em] \Rightarrow x = 36\end{aligned}$$

Hence, option 3 is the correct option.

Given,

$\frac{x + 5}{2} + \frac{x - 5}{3} = \frac{25}{6}$

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

Multiplying both sides by 6, we get:

$$\begin{aligned}\Rightarrow 6 \times \Big(\frac{x + 5}{2} + \frac{x - 5}{3}\Big) = 6 \times \frac{25}{6} \\[1em] \Rightarrow \frac{6(x + 5)}{2} + \frac{6(x - 5)}{3} = \frac{6 \times 25}{6} \\[1em] \Rightarrow 3(x + 5) + 2(x - 5) = 25 \\[1em] \Rightarrow 3x + 15 + 2x - 10 = 25 \\[1em] \Rightarrow 5x + 5 = 25 \\[1em] \Rightarrow 5x = 25 - 5 \\[1em] \Rightarrow 5x = 20 \\[1em] \Rightarrow x = \frac{20}{5} \\[1em] \Rightarrow x = 4\end{aligned}$$

Hence, option 3 is the correct option.

Let the number be $x$.

Given,

$\frac{4}{5}$ of a number exceeds its $\frac{2}{3}$ by 8.

$\therefore \frac{4x}{5} - \frac{2x}{3} = 8$

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

Multiplying both sides by 15, we get:

$$\begin{aligned}\Rightarrow 15 \times \Big(\frac{4x}{5} - \frac{2x}{3}\Big) = 15 \times 8 \\[1em] \Rightarrow \frac{15 \times 4x}{5} - \frac{15 \times 2x}{3} = 15 \times 8 \\[1em] \Rightarrow 12x - 10x = 120 \\[1em] \Rightarrow 2x = 120 \\[1em] \Rightarrow x = \frac{120}{2} \\[1em] \Rightarrow x = 60\end{aligned}$$

Hence, option 2 is the correct option.

Let the number be $x$.

Given,

The fifth part of the number increased by 5 is equal to its fourth part diminished by 5.

$\therefore \frac{x}{5} + 5 = \frac{x}{4} - 5$

L.C.M. of 5 and 4 = 20

Multiplying both sides by 20, we get:

$$\begin{aligned}\Rightarrow 20 \times \Big(\frac{x}{5} + 5\Big) = 20 \times \Big(\frac{x}{4} - 5\Big) \\[1em] \Rightarrow \frac{20x}{5} + 20 \times 5 = \frac{20x}{4} - 20 \times 5 \\[1em] \Rightarrow 4x + 100 = 5x - 100 \\[1em] \Rightarrow 100 + 100 = 5x - 4x \\[1em] \Rightarrow x = 200\end{aligned}$$

Hence, option 3 is the correct option.

Let the original number be $x$.

Given,

The student found $\frac{7}{18}$ of the number instead of $\frac{7}{8}$ of it, and his answer was 770 less than the correct one.

$\therefore \frac{7x}{8} - \frac{7x}{18} = 770$

L.C.M. of 8 and 18 = 72

Multiplying both sides by 72, we get:

$$\begin{aligned}\Rightarrow 72 \times \Big(\frac{7x}{8} - \frac{7x}{18}\Big) = 72 \times 770 \\[1em] \Rightarrow \frac{72 \times 7x}{8} - \frac{72 \times 7x}{18} = 72 \times 770 \\[1em] \Rightarrow 63x - 28x = 55440 \\[1em] \Rightarrow 35x = 55440 \\[1em] \Rightarrow x = \frac{55440}{35} \\[1em] \Rightarrow x = 1584\end{aligned}$$

As 1584 is not one of the numbers given in options 1, 2 and 3, the correct option is "None of these".

Hence, option 4 is the correct option.

Let the capacity of the drum be $x$ litres.

Given,

The drum of kerosene is $\frac{3}{4}$ full. When 80 litres of kerosene is drawn from it, it remains $\frac{7}{12}$ full.

$\therefore \frac{3x}{4} - \frac{7x}{12} = 80$

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

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow 12 \times \Big(\frac{3x}{4} - \frac{7x}{12}\Big) = 12 \times 80 \\[1em] \Rightarrow \frac{12 \times 3x}{4} - \frac{12 \times 7x}{12} = 12 \times 80 \\[1em] \Rightarrow 9x - 7x = 960 \\[1em] \Rightarrow 2x = 960 \\[1em] \Rightarrow x = \frac{960}{2} \\[1em] \Rightarrow x = 480\end{aligned}$$

So, the capacity of the drum = 480 litres

Hence, option 4 is the correct option.

Let the seven consecutive integers be $x, x + 1, x + 2, x + 3, x + 4, x + 5$ and $x + 6$.

Given,

The sum of seven consecutive integers is 175.

$$\begin{aligned}\therefore x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) = 175 \\[1em] \Rightarrow 7x + 21 = 175 \\[1em] \Rightarrow 7x = 175 - 21 \\[1em] \Rightarrow 7x = 154 \\[1em] \Rightarrow x = \frac{154}{7} \\[1em] \Rightarrow x = 22\end{aligned}$$

So, the smallest number = 22 and the largest number = 22 + 6 = 28

Twice the largest number = 2 × 28 = 56

Thrice the smallest number = 3 × 22 = 66

⇒ Required difference = 56 - 66 = -10

Hence, option 3 is the correct option.

Let the five consecutive odd numbers be $x, x + 2, x + 4, x + 6$ and $x + 8$.

Given,

The sum of five consecutive odd numbers is 575.

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 575 \\[1em] \Rightarrow 5x + 20 = 575 \\[1em] \Rightarrow 5x = 575 - 20 \\[1em] \Rightarrow 5x = 555 \\[1em] \Rightarrow x = \frac{555}{5} \\[1em] \Rightarrow x = 111\end{aligned}$$

So, the five consecutive odd numbers are 111, 113, 115, 117 and 119.

The next set of five consecutive odd numbers = 121, 123, 125, 127 and 129.

⇒ Required sum = 121 + 123 + 125 + 127 + 129 = 625

Hence, option 3 is the correct option.

Let the four consecutive even integers be $x, x + 2, x + 4$ and $x + 6$.

Given,

The sum of four consecutive even integers is 1284.

$$\begin{aligned}\therefore x + (x + 2) + (x + 4) + (x + 6) = 1284 \\[1em] \Rightarrow 4x + 12 = 1284 \\[1em] \Rightarrow 4x = 1284 - 12 \\[1em] \Rightarrow 4x = 1272 \\[1em] \Rightarrow x = \frac{1272}{4} \\[1em] \Rightarrow x = 318\end{aligned}$$

So, the greatest integer = 318 + 6 = 324

Hence, option 3 is the correct option.

Given,

The ratio of the ages of a father and his son is 17 : 7.

Let the present ages of the father and his son be $17x$ years and $7x$ years respectively.

Given,

Six years ago, the ratio of their ages was 3 : 1.

$$\begin{aligned}\therefore \frac{17x - 6}{7x - 6} = \frac{3}{1} \\[1em] \Rightarrow 17x - 6 = 3(7x - 6) \\[1em] \Rightarrow 17x - 6 = 21x - 18 \\[1em] \Rightarrow 18 - 6 = 21x - 17x \\[1em] \Rightarrow 4x = 12 \\[1em] \Rightarrow x = \frac{12}{4} \\[1em] \Rightarrow x = 3\end{aligned}$$

So, the father's present age = 17 × 3 = 51 years

Hence, option 2 is the correct option.

Given,

The age of a mother today is thrice that of her daughter.

Let the present age of the daughter be $x$ years and the present age of the mother be $3x$ years.

Given,

After 12 years, the age of the mother will be twice that of her daughter.

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

Hence, option 1 is the correct option.

Given,

The present ages of three persons are in the proportion 4 : 7 : 9.

Let the present ages of the three persons be $4x$ years, $7x$ years and $9x$ years.

Given,

Eight years ago, the sum of their ages was 56 years.

$$\begin{aligned}\therefore (4x - 8) + (7x - 8) + (9x - 8) = 56 \\[1em] \Rightarrow 20x - 24 = 56 \\[1em] \Rightarrow 20x = 56 + 24 \\[1em] \Rightarrow 20x = 80 \\[1em] \Rightarrow x = \frac{80}{20} \\[1em] \Rightarrow x = 4\end{aligned}$$

So, the present age of the eldest person = 9 × 4 = 36 years

Hence, option 2 is the correct option.

Let the number of benches be $x$.

Given,

If 6 students are made to sit on a bench, then to accommodate all of them, one more bench is needed.

⇒ Number of students = $6(x + 1)$

Given,

If 7 students are made to sit on a bench, then after accommodating all of them, space for 5 students is left.

⇒ Number of students = $7x - 5$

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

So, the total number of students = 6(11 + 1) = 6 × 12 = 72

Hence, option 3 is the correct option.

Given,

180 oranges are distributed among 70 boys and girls.

Let the number of boys be $x$ and the number of girls be $(70 - x)$.

Given,

Each boy gets 2 oranges and each girl gets 3 oranges.

$$\begin{aligned}\therefore 2x + 3(70 - x) = 180 \\[1em] \Rightarrow 2x + 210 - 3x = 180 \\[1em] \Rightarrow 210 - x = 180 \\[1em] \Rightarrow 210 - 180 = x \\[1em] \Rightarrow x = 30\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Rahul receives twice as much as Sumit, who receives one-fifth as much as Gaurav.

Let the money received by Sumit be ₹ $x$, the money received by Rahul be ₹ $2x$ and the money received by Gaurav be ₹ $5x$.

Given,

A sum of ₹ 1000 is divided among the three boys.

$$\begin{aligned}\therefore 2x + x + 5x = 1000 \\[1em] \Rightarrow 8x = 1000 \\[1em] \Rightarrow x = \frac{1000}{8} \\[1em] \Rightarrow x = 125\end{aligned}$$

So, the money received by Gaurav = 5 × 125 = ₹ 625

Hence, option 4 is the correct option.

Let the total number of sweets be $x$.

Number of children present on that day = 112 - 32 = 80

Given,

When the sweets were distributed equally among the 80 children present, each child got 6 sweets more than he would have got if all the 112 children were present.

$\therefore \frac{x}{80} - \frac{x}{112} = 6$

L.C.M. of 80 and 112 = 560

Multiplying both sides by 560, we get:

$$\begin{aligned}\Rightarrow 560 \times \Big(\frac{x}{80} - \frac{x}{112}\Big) = 560 \times 6 \\[1em] \Rightarrow \frac{560x}{80} - \frac{560x}{112} = 560 \times 6 \\[1em] \Rightarrow 7x - 5x = 3360 \\[1em] \Rightarrow 2x = 3360 \\[1em] \Rightarrow x = \frac{3360}{2} \\[1em] \Rightarrow x = 1680\end{aligned}$$

So, the number of sweets each child was originally supposed to get = $\frac{1680}{112} = 15$

Hence, option 1 is the correct option.

Given,

The employee travelled 90 km in one week.

Let the distance travelled by taxi be $x$ km and the distance travelled by his own car be $(90 - x)$ km.

Given,

He claims ₹ 7 for each km when he travels by taxi and ₹ 6 for each km if he drives his own car, and he claimed ₹ 675 in that week.

$$\begin{aligned}\therefore 7x + 6(90 - x) = 675 \\[1em] \Rightarrow 7x + 540 - 6x = 675 \\[1em] \Rightarrow x + 540 = 675 \\[1em] \Rightarrow x = 675 - 540 \\[1em] \Rightarrow x = 135\end{aligned}$$

Hence, option 1 is the correct option.

Given,

The student attempted all the 200 questions.

Let the number of questions answered correctly be $x$ and the number of questions answered wrongly be $(200 - x)$.

Given,

The student scores 4 marks for every correct answer, loses 1 mark for every wrong answer and scored in all 200 marks.

$$\begin{aligned}\therefore 4x - (200 - x) = 200 \\[1em] \Rightarrow 4x - 200 + x = 200 \\[1em] \Rightarrow 5x = 200 + 200 \\[1em] \Rightarrow 5x = 400 \\[1em] \Rightarrow x = \frac{400}{5} \\[1em] \Rightarrow x = 80\end{aligned}$$

Hence, option 3 is the correct option.

Given,

The person aimed at the target one hundred times.

Let the number of times he hits the target be $x$ and the number of times he misses the target be $(100 - x)$.

Given,

He gets one rupee each time he hits the target, loses one rupee each time he misses it and gets ₹ 30 in all.

$$\begin{aligned}\therefore x - (100 - x) = 30 \\[1em] \Rightarrow x - 100 + x = 30 \\[1em] \Rightarrow 2x = 30 + 100 \\[1em] \Rightarrow 2x = 130 \\[1em] \Rightarrow x = \frac{130}{2} \\[1em] \Rightarrow x = 65\end{aligned}$$

So, the number of times he misses the target = 100 - 65 = 35

Hence, option 2 is the correct option.

Given,

There are a total of 324 coins of 20 paise and 25 paise.

Let the number of 25 paise coins be $x$ and the number of 20 paise coins be $(324 - x)$.

Given,

The coins make a sum of ₹ 71.

Total sum = ₹ 71 = 71 × 100 paise = 7100 paise

$$\begin{aligned}\therefore 25x + 20(324 - x) = 7100 \\[1em] \Rightarrow 25x + 6480 - 20x = 7100 \\[1em] \Rightarrow 5x = 7100 - 6480 \\[1em] \Rightarrow 5x = 620 \\[1em] \Rightarrow x = \frac{620}{5} \\[1em] \Rightarrow x = 124\end{aligned}$$

Hence, option 2 is the correct option.

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

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

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

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

Number of deer drinking water from the pond = 9

As the sum of the deer grazing, playing and drinking water is equal to the total number of deer in the herd,

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

Hence, option (d) is the correct option.

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

From part 1, we have

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

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

Multiplying both sides by 8, we get:

$$\begin{aligned}\Rightarrow 8 \times \Big(\frac{x}{2} + \frac{3x}{8}\Big) = 8 \times \Big(x - 9\Big) \\[1em] \Rightarrow \frac{8x}{2} + \frac{8 \times 3x}{8} = 8x - 8 \times 9 \\[1em] \Rightarrow 4x + 3x = 8x - 72 \\[1em] \Rightarrow 7x = 8x - 72 \\[1em] \Rightarrow 72 = 8x - 7x \\[1em] \Rightarrow x = 72\end{aligned}$$

Hence, option (b) is the correct option.

3. From part 2, the total number of deer in the herd = 72.

Number of deer grazing in the field = $\frac{72}{2} = 36$

As there is one attendant for every four grazing deer,

⇒ Number of attendants = $\frac{36}{4} = 9$

Hence, option (b) is the correct option.

4. From part 2, the total number of deer in the herd = 72.

Number of deer grazing in the field = $\frac{72}{2} = 36$

Number of deer playing nearby = $\frac{3 \times 72}{8} = 27$

Number of deer drinking water from the pond = 9

⇒ Required ratio = 36 : 27 : 9

Calculating H.C.F. of 36, 27 and 9:

36 = 2² × 3²

27 = 3³

9 = 3²

∴ H.C.F. of 36, 27 and 9 = 3² = 9

Dividing each term of the ratio by 9, we get:

⇒ Required ratio = 4 : 3 : 1

Hence, option (b) is the correct option.

Given,

$\frac{m}{3} + 1 = \frac{7}{15}$

On transposing 1 from L.H.S. to R.H.S., its sign changes from + to -.

$\Rightarrow \frac{m}{3} = \frac{7}{15} - 1$

So, Assertion (A) is true.

Transposition is the process in which any term of an equation may be taken to the other side with sign changed, so Reason (R) is true.

As Assertion (A) is obtained by transposing 1 to R.H.S., Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Given, the two numbers differ by 40.

If one number is $x$, then the other number must differ from $x$ by 40.

So, the other number = $x - 40$ or $x + 40$ and not $(40 - x)$.

So, Assertion (A) is false.

Solving an equation means finding all its solutions (roots), so Reason (R) is true.

Hence, option 4 is the correct option.

Let the smaller number be $x$.

As the numbers differ by 24, the bigger number = $x + 24$

We know that,

If one number is a multiple of the other, then their L.C.M. is the bigger number.

$$\begin{aligned}\Rightarrow x + 24 = 48 \\[1em] \Rightarrow x = 48 - 24 \\[1em] \Rightarrow x = 24\end{aligned}$$

So, the two numbers are 24 and 48.

⇒ Sum of the numbers = 24 + 48 = 72

Hence, option 2 is the correct option.

Let us check each of the given numbers.

Option 1 : For the number 24, the digits are 2 and 4.

Difference between the digits = 4 - 2 = 2 and twice the difference = 2 × 2 = 4, which is the ones digit.

⇒ $\frac{24 \times 24}{\frac{24}{2}} = \frac{576}{12} = 48$

⇒ $\frac{48 + 36}{2} = \frac{84}{2} = 42$

As 42 has the same digits as 24 but in reverse order, 24 satisfies all the given conditions.

Option 2 : For the number 32, the digits are 3 and 2.

⇒ $\frac{32 \times 32}{\frac{32}{2}} = \frac{1024}{16} = 64$

⇒ $\frac{64 + 36}{2} = \frac{100}{2} = 50$

As 50 is not 23, the number 32 does not satisfy the given conditions.

Option 3 : For the number 42, the ones digit is 2 but twice the difference between the digits = 2 × (4 - 2) = 4.

So, the number 42 does not satisfy the given conditions.

Option 4 : For the number 46, the ones digit is 6 but twice the difference between the digits = 2 × (6 - 4) = 4.

So, the number 46 does not satisfy the given conditions.

Hence, option 1 is the correct option.

Let the number be $x$.

Given,

5 less than $\frac{1}{3}$ of a number is same as $\frac{5}{4}$ less than four times of the same number.

$\therefore \frac{x}{3} - 5 = 4x - \frac{5}{4}$

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

Multiplying both sides by 12, we get:

$$\begin{aligned}\Rightarrow 12 \times \Big(\frac{x}{3} - 5\Big) = 12 \times \Big(4x - \frac{5}{4}\Big) \\[1em] \Rightarrow \frac{12x}{3} - 12 \times 5 = 12 \times 4x - \frac{12 \times 5}{4} \\[1em] \Rightarrow 4x - 60 = 48x - 15 \\[1em] \Rightarrow 15 - 60 = 48x - 4x \\[1em] \Rightarrow 44x = - 45 \\[1em] \Rightarrow x = - \frac{45}{44}\end{aligned}$$

6 more than $\frac{11}{15}$ of the number,

$$\begin{aligned}\Rightarrow \frac{11}{15} \times \left(- \frac{45}{44}\right) + 6 \\[1em] \Rightarrow - \frac{3}{4} + 6 \\[1em] \Rightarrow \frac{-3 + 24}{4} \\[1em] \Rightarrow \frac{21}{4}\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 0.16(5x - 2) = 0.4x + 7 \\[1em] \Rightarrow 0.8x - 0.32 = 0.4x + 7 \\[1em] \Rightarrow 0.8x - 0.4x = 7 + 0.32 \\[1em] \Rightarrow 0.4x = 7.32 \\[1em] \Rightarrow x = \frac{7.32}{0.4} \\[1em] \Rightarrow x = 18.3\end{aligned}$$

Hence, option 4 is the correct option.

Let the number Rakhi thought of be $x$.

Given,

Rakhi doubled the number and added 20 to it. On dividing the resulting number by 25, she got 4.

$$\begin{aligned}\therefore \frac{2x + 20}{25} = 4 \\[1em] \Rightarrow 2x + 20 = 25 \times 4 \\[1em] \Rightarrow 2x + 20 = 100 \\[1em] \Rightarrow 2x = 100 - 20 \\[1em] \Rightarrow 2x = 80 \\[1em] \Rightarrow x = \frac{80}{2} \\[1em] \Rightarrow x = 40\end{aligned}$$

Hence, option 2 is the correct option.

Given,

In a zoo, total number of monkeys and peacocks is 82.

Let the number of monkeys be $x$ and the number of peacocks be $(82 - x)$.

Given,

The total number of legs is 200.

We know that,

A monkey has 4 legs and a peacock has 2 legs.

$$\begin{aligned}\therefore 4x + 2(82 - x) = 200 \\[1em] \Rightarrow 4x + 164 - 2x = 200 \\[1em] \Rightarrow 2x = 200 - 164 \\[1em] \Rightarrow 2x = 36 \\[1em] \Rightarrow x = \frac{36}{2} \\[1em] \Rightarrow x = 18\end{aligned}$$

So, the number of monkeys = 18

and the number of peacocks = 82 - 18 = 64

⇒ Required difference = 64 - 18 = 46

Hence, option 3 is the correct option.

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

Given,

Tuffy is now the same age as that Tomy was four years ago, and Tomy is now 12 years old.

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

Verification :

Four years ago, Tuffy's age = 8 - 4 = 4 years

and Tomy's age = 8 years

⇒ $\frac{1}{2} \times 8 = 4$ years, which is Tuffy's age four years ago.

So, Tuffy was half his mother's age four years ago.

Hence, option 1 is the correct option.

Given,

The number of 100-rupee notes was twice the number of 50-rupee notes.

Let the number of 50-rupee notes be $x$ and the number of 100-rupee notes be $2x$.

Given,

The total money she had was ₹ 750.

$$\begin{aligned}\therefore 50x + 100 \times 2x = 750 \\[1em] \Rightarrow 50x + 200x = 750 \\[1em] \Rightarrow 250x = 750 \\[1em] \Rightarrow x = \frac{750}{250} \\[1em] \Rightarrow x = 3\end{aligned}$$

So, the number of 50-rupee notes = 3

and the number of 100-rupee notes = 2 × 3 = 6

⇒ Total number of notes = 3 + 6 = 9

Hence, option 4 is the correct option.