Framing Algebraic Expressions

(i) $\bm{8 + x = y}$

(ii) $\bm{x - 5 = y}$

(iii) $\bm{15 \times m = 3n}$

(iv) $\bm{8 \times y = 3x}$

(v) $\bm{\frac{30}{b} = p}$

(vi) $\bm{z - 3x = y}$

(vii) $\bm{12 \times x = 5z}$

(viii) $\bm{45 - 3z = y}$

(ix) $\bm{\frac{8x}{y} = 2z}$

(x) $\bm{7y - 5x = 8z}$

By formula,

Perimeter of rectangle = 2 × (length + breadth)

Hence, $\bm{P = 2(l + b)}$.

By formula,

Perimeter of square = 4 × side

Hence, $\bm{P = 4 \times \text{side}}$.

By formula,

Area of square ($A$) = side × side = (side)²

Hence, $\bm{A = (\text{side})^2}$.

By formula,

Surface area of cube ($S$) = 6 × (edge)²

Hence, $\bm{S = 6(\text{edge})^2}$.

The sum of $x$ and $y$ is $(x + y)$, and subtracting $m$ from it gives :

Hence, $\bm{x + y - m}$.

The product of $x$ and $y$ is $xy$, and dividing it by $m$ gives :

Hence, $\bm{\frac{xy}{m}}$.

Subtracting $5m$ from $3n$ gives $(3n - 5m)$, and then adding $9p$ to it gives :

Hence, $\bm{3n - 5m + 9p}$.

The product of $12, x, y$ and $z$ is $12xyz$, and the product of $5, m$ and $n$ is $5mn$. Subtracting the second product from the first gives :

Hence, $\bm{12xyz - 5mn}$.

The sum of $p$ and $2r - s$ is $(p + 2r - s)$, and the sum of $a$ and $3n + 4x$ is $(a + 3n + 4x)$.

Subtracting the second sum from the first gives $(p + 2r - s) - (a + 3n + 4x)$.

Hence, $\bm{(p + 2r - s) - (a + 3n + 4x)}$.

Basic wage for $t$ hours = ₹ B

Since he works a total of $T$ hours, overtime = $(T - t)$ hours

Wages for overtime = ₹ $R(T - t)$

∴ Total wages = Basic wage + Overtime wages

⇒ $W = B + R(T - t)$

Hence, required formula is $\bm{W = B + R(T - t)}$.

Substituting $a = 3$ and $b = -1$ in $B = 2a^2 - b^2$, we get :

$$\begin{aligned}\Rightarrow B = 2(3)^{2} - (-1)^{2} \\[1em] \Rightarrow B = 2 \times 9 - 1 \\[1em] \Rightarrow B = 18 - 1 \\[1em] \Rightarrow B = 17.\end{aligned}$$

Hence, $\bm{B}$ = 17.

Substituting $x = 39.45$ and $t = 40$ in $W = xt$, we get :

$$\begin{aligned}\Rightarrow W = 39.45 \times 40 \\[1em] \Rightarrow W = 1578.\end{aligned}$$

Hence, his wages = ₹ 1578.

Substituting $C = 40°$ in $F = \frac{9}{5} \times C + 32°$, we get :

$$\begin{aligned}\Rightarrow F = \frac{9}{5} \times 40° + 32° \\[1em] \Rightarrow F = 9 \times 8° + 32° \\[1em] \Rightarrow F = 72° + 32° \\[1em] \Rightarrow F = 104°.\end{aligned}$$

Hence, $\bm{F}$ = 104°.

Average of four quantities = $\frac{\text{Sum of the quantities}}{\text{Number of quantities}}$

∴ Required formula is $A = \frac{p + q + r + s}{4}$

Substituting $A = 6, p = 3, q = 5$ and $r = 7$, we get :

$$\begin{aligned}\Rightarrow 6 = \frac{3 + 5 + 7 + s}{4} \\[1em] \Rightarrow 6 \times 4 = 15 + s \\[1em] \Rightarrow 24 = 15 + s \\[1em] \Rightarrow s = 24 - 15 \\[1em] \Rightarrow s = 9.\end{aligned}$$

Hence, $\bm{A = \frac{p + q + r + s}{4}}$ and $\bm{s = 9}$.

Substituting $a = 7$ and $b = 3$ in $T = 2a - b$, we get :

$$\begin{aligned}\Rightarrow T = 2(7) - 3 \\[1em] \Rightarrow T = 14 - 3 \\[1em] \Rightarrow T = 11.\end{aligned}$$

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

(i) $x + y$ = 3 + 6 = 9

(ii) $y - x$ = 6 − 3 = 3

(iii) $\frac{y}{x} = \frac{6}{3}$ = 2

(iv) $c \div b$ = 32 ÷ 8 = 4

(v) $z \div x$ = 18 ÷ 3 = 6

(vi) $y \times d$ = 6 × 0 = 0

(vii) $d \div x$ = 0 ÷ 3 = 0

(viii) $ab + y$ = (2 × 8) + 6 = 16 + 6 = 22

(ix) $a + b + x$ = 2 + 8 + 3 = 13

(x) $b + z - d$ = 8 + 18 − 0 = 26

(xi) $a - b + y$ = 2 − 8 + 6 = 0

(xii) $z - a - b$ = 18 − 2 − 8 = 8

(xiii) $d - a + x$ = 0 − 2 + 3 = 1

(xiv) $xy - bd$ = (3 × 6) − (8 × 0) = 18 − 0 = 18

(xv) $xz + cd$ = (3 × 18) + (32 × 0) = 54 + 0 = 54

Substituting $p = 1, q = 5$ and $r = 2$, we get :

$$\begin{aligned}\Rightarrow p + 2q + 3r \\[1em] \Rightarrow 1 + 2(5) + 3(2) \\[1em] \Rightarrow 1 + 10 + 6 \\[1em] \Rightarrow 17.\end{aligned}$$

Hence, $\bm{p + 2q + 3r = 17}$.

Substituting $a = 5, b = 10$ and $c = 20$, we get :

$$\begin{aligned}\Rightarrow 2a + 4b + 5c \\[1em] \Rightarrow 2(5) + 4(10) + 5(20) \\[1em] \Rightarrow 10 + 40 + 100 \\[1em] \Rightarrow 150.\end{aligned}$$

Hence, $\bm{2a + 4b + 5c = 150}$.

Substituting $a = 8$ and $b = 10$, we get :

$$\begin{aligned}\Rightarrow 3a - 2b \\[1em] \Rightarrow 3(8) - 2(10) \\[1em] \Rightarrow 24 - 20 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, $\bm{3a - 2b = 4}$.

Substituting $x = 3, y = 5$ and $z = 4$, we get :

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

Hence, $\bm{5x + 3y - 6z = 6}$.

Substituting $p = 10, q = 8, r = 6$ and $s = 2$, we get :

$$\begin{aligned}\Rightarrow 2p - 3q + 4r - 8s \\[1em] \Rightarrow 2(10) - 3(8) + 4(6) - 8(2) \\[1em] \Rightarrow 20 - 24 + 24 - 16 \\[1em] \Rightarrow 4.\end{aligned}$$

Hence, $\bm{2p - 3q + 4r - 8s = 4}$.

Substituting $m = 20, n = 10, p = 2$ and $q = 9$, we get :

$$\begin{aligned}\Rightarrow 6m - 2n - 5p - 3q \\[1em] \Rightarrow 6(20) - 2(10) - 5(2) - 3(9) \\[1em] \Rightarrow 120 - 20 - 10 - 27 \\[1em] \Rightarrow 63.\end{aligned}$$

Hence, $\bm{6m - 2n - 5p - 3q = 63}$.

Substituting $p = 5, q = 3$ and $r = \frac{1}{2}$, we get :

$$\begin{aligned}\Rightarrow 4pq \times 2r \\[1em] \Rightarrow 8pqr \\[1em] \Rightarrow 8 \times 5 \times 3 \times \frac{1}{2} \\[1em] \Rightarrow 60.\end{aligned}$$

Hence, $\bm{4pq \times 2r = 60}$.

Substituting $x = 8, y = 4$ and $z = 16$, we get :

$$\begin{aligned}\Rightarrow \frac{yx}{z} \\[1em] \Rightarrow \frac{4 \times 8}{16} \\[1em] \Rightarrow \frac{32}{16} \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, $\bm{\frac{yx}{z} = 2}$.

Substituting $a = 5, b = 7$ and $c = 2$, we get :

$$\begin{aligned}\Rightarrow \frac{a + b - c}{2a} \\[1em] \Rightarrow \frac{5 + 7 - 2}{2 \times 5} \\[1em] \Rightarrow \frac{10}{10} \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\bm{\frac{a + b - c}{2a} = 1}$.

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$$\begin{aligned}\Rightarrow 3a + 2b - 6c + 4d \\[1em] \Rightarrow 3(3) + 2(0) - 6(2) + 4(1) \\[1em] \Rightarrow 9 + 0 - 12 + 4 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\bm{3a + 2b - 6c + 4d = 1}$.

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$$\begin{aligned}\Rightarrow 6a - 3b - 4c - 2d \\[1em] \Rightarrow 6(3) - 3(0) - 4(2) - 2(1) \\[1em] \Rightarrow 18 - 0 - 8 - 2 \\[1em] \Rightarrow 8.\end{aligned}$$

Hence, $\bm{6a - 3b - 4c - 2d = 8}$.

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$$\begin{aligned}\Rightarrow ab - bc + cd - da \\[1em] \Rightarrow (3)(0) - (0)(2) + (2)(1) - (1)(3) \\[1em] \Rightarrow 0 - 0 + 2 - 3 \\[1em] \Rightarrow -1.\end{aligned}$$

Hence, $\bm{ab - bc + cd - da = -1}$.

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$$\begin{aligned}\Rightarrow abc - bcd + cda \\[1em] \Rightarrow (3)(0)(2) - (0)(2)(1) + (2)(1)(3) \\[1em] \Rightarrow 0 - 0 + 6 \\[1em] \Rightarrow 6.\end{aligned}$$

Hence, $\bm{abc - bcd + cda = 6}$.

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$$\begin{aligned}\Rightarrow a^{2} + 2b^{2} - 3c^{2} \\[1em] \Rightarrow (3)^{2} + 2(0)^{2} - 3(2)^{2} \\[1em] \Rightarrow 9 + 0 - 12 \\[1em] \Rightarrow -3.\end{aligned}$$

Hence, $\bm{a^2 + 2b^2 - 3c^2 = -3}$.

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$$\begin{aligned}\Rightarrow a^{2} + b^{2} - c^{2} + d^{2} \\[1em] \Rightarrow (3)^{2} + (0)^{2} - (2)^{2} + (1)^{2} \\[1em] \Rightarrow 9 + 0 - 4 + 1 \\[1em] \Rightarrow 6.\end{aligned}$$

Hence, $\bm{a^2 + b^2 - c^2 + d^2 = 6}$.

Substituting $a = 3, b = 0, c = 2$ and $d = 1$, we get :

$$\begin{aligned}\Rightarrow 2a^{2} - 3b^{2} + 4c^{2} - 5d^{2} \\[1em] \Rightarrow 2(3)^{2} - 3(0)^{2} + 4(2)^{2} - 5(1)^{2} \\[1em] \Rightarrow 2(9) - 0 + 4(4) - 5(1) \\[1em] \Rightarrow 18 + 16 - 5 \\[1em] \Rightarrow 29.\end{aligned}$$

Hence, $\bm{2a^2 - 3b^2 + 4c^2 - 5d^2 = 29}$.

Substituting $x$ = 2, we get :

$$\begin{aligned}\Rightarrow 5x^{2} - 3x + 2 \\[1em] \Rightarrow 5(2)^{2} - 3(2) + 2 \\[1em] \Rightarrow 5(4) - 6 + 2 \\[1em] \Rightarrow 20 - 6 + 2 \\[1em] \Rightarrow 16.\end{aligned}$$

Hence, $\bm{5x^2 - 3x + 2 = 16}$.

Substituting $x$ = −1, we get :

$$\begin{aligned}\Rightarrow 3x^{3} - 4x^{2} + 5x - 6 \\[1em] \Rightarrow 3(-1)^{3} - 4(-1)^{2} + 5(-1) - 6 \\[1em] \Rightarrow 3(-1) - 4(1) - 5 - 6 \\[1em] \Rightarrow -3 - 4 - 5 - 6 \\[1em] \Rightarrow -18.\end{aligned}$$

Hence, $\bm{3x^3 - 4x^2 + 5x - 6 = -18}$.

Substituting $x$ = 1, we get :

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

Hence, the value of $\bm{x^3 - 8x^2 + 12x - 5}$ is zero, when $\bm{x = 1}$.

(i) Substituting $x$ = 1 in $x$ + 5,

⇒ $x$ + 5 = 1 + 5 = 6

Since the value obtained is 6, the statement is true.

(ii) Substituting $x$ = 0 in 2$x$ − 3,

⇒ 2$x$ − 3 = 2(0) − 3 = −3

Since the value obtained is −3 (not 1), the statement is false.

(iii) Substituting $x$ = 1 in $\frac{2x - 4}{x + 1}$,

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

Since the value obtained is −1, the statement is true.

(i) Substituting $x$ = 2, we get :

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

Hence, $\bm{\frac{x}{2x^{2}} = \frac{1}{4}}$.

(ii) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$$\begin{aligned}\Rightarrow \frac{xz}{yz} = \frac{2 \times 4}{5 \times 4} \\[1em] \Rightarrow \frac{8}{20} = \frac{2}{5}.\end{aligned}$$

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

(iii) Substituting $z$ = 4 and $x$ = 2, we get :

$\Rightarrow z^{x} = 4^{2} = 16.$

Hence, $\bm{z^x = 16}$.

(iv) Substituting $y$ = 5 and $x$ = 2, we get :

$\Rightarrow y^{x} = 5^{2} = 25.$

Hence, y^x = 25.

(v) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$$\begin{aligned}\Rightarrow \frac{x^{2} , y^{2} , z^{2}}{xz} = x , y^{2} , z \\[1em] \Rightarrow 2 \times (5)^{2} \times 4 \\[1em] \Rightarrow 2 \times 25 \times 4 = 200.\end{aligned}$$

Hence, $\bm{\frac{x^{2} , y^{2} , z^{2}}{xz} = 200}$.

(vi) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$$\begin{aligned}\Rightarrow \frac{5x^{4} , y^{2} , z^{2}}{2x^{2}} = \frac{5}{2} , x^{2} , y^{2} , z^{2} \\[1em] \Rightarrow \frac{5}{2} \times (2)^{2} \times (5)^{2} \times (4)^{2} \\[1em] \Rightarrow \frac{5}{2} \times 4 \times 25 \times 16 \\[1em] \Rightarrow 4000.\end{aligned}$$

Hence, $\bm{\frac{5x^{4} , y^{2} , z^{2}}{2x^{2}} = 4000}$.

(vii) Substituting $x$ = 2, $y$ = 5 and $z$ = 4, we get :

$$\begin{aligned}\Rightarrow xy \div y^{2}z = \frac{xy}{y^{2}z} = \frac{x}{yz} \\[1em] \Rightarrow \frac{2}{5 \times 4} = \frac{2}{20} = \frac{1}{10}.\end{aligned}$$

Hence, $\bm{xy \div y^2z = \frac{1}{10}}$.

(viii) Substituting $x$ = 2 and $y$ = 5, we get :

$$\begin{aligned}\Rightarrow \frac{x^{2} , y^{x}}{x} = x , y^{x} \\[1em] \Rightarrow 2 \times (5)^{2} \\[1em] \Rightarrow 2 \times 25 = 50.\end{aligned}$$

Hence, $\bm{\frac{x^{2} , y^{x}}{x} = 50}$.

Substituting $a$ = 3, we get :

$$\begin{aligned}\Rightarrow a^{2} = (3)^{2} = 9 \\[1em] \Rightarrow 2^{a} = 2^{3} = 8.\end{aligned}$$

Hence, $\bm{a^2 = 9 \text{ and } 2^a = 8}$.

Substituting $m$ = 2, we get :

$$\begin{aligned}\Rightarrow 4m^{3} = 4(2)^{3} = 4 \times 8 = 32 \\[1em] \Rightarrow 3m^{4} = 3(2)^{4} = 3 \times 16 = 48.\end{aligned}$$

∴ Difference = $3m^4 - 4m^3$ = 48 − 32 = 16

Hence, the required difference is 16.

Solving,

$$\begin{aligned}\Rightarrow x + 2 = 6\\[1em] \text{ Transposing 2 to R.H.S.,}\\[1em] \Rightarrow x = 6 - 2\\[1em] \Rightarrow x = 4\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow x + 6 = 2\\[1em] \text{ Transposing 6 to R.H.S.,}\\[1em] \Rightarrow x = 2 - 6\\[1em] \Rightarrow x = -4\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow y + 8 = 5\\[1em] \text{ Transposing 8 to R.H.S.,}\\[1em] \Rightarrow y = 5 - 8\\[1em] \Rightarrow y = -3\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow x + 4 = -3\\[1em] \text{ Transposing 4 to R.H.S.,}\\[1em] \Rightarrow x = -3 - 4\\[1em] \Rightarrow x = -7\end{aligned}$$

Hence, $\bm{x} = -7$.

Solving,

$$\begin{aligned}\Rightarrow y + 2 = -8\\[1em] \text{ Transposing 2 to R.H.S.,}\\[1em] \Rightarrow y = -8 - 2\\[1em] \Rightarrow y = -10\end{aligned}$$

Hence, $\bm{y} = -10$.

Solving,

$$\begin{aligned}\Rightarrow b + 2.5 = 4.2\\[1em] \text{ Transposing 2.5 to R.H.S.,}\\[1em] \Rightarrow b = 4.2 - 2.5\\[1em] \Rightarrow b = 1.7\end{aligned}$$

Hence, $\bm{b} = 1.7$.

Solving,

$$\begin{aligned}\Rightarrow p + 4.6 = 8.5\\[1em] \text{ Transposing 4.6 to R.H.S.,}\\[1em] \Rightarrow p = 8.5 - 4.6\\[1em] \Rightarrow p = 3.9\end{aligned}$$

Hence, $\bm{p} = 3.9$.

Solving,

$$\begin{aligned}\Rightarrow y + 3.2 = -6.5\\[1em] \text{ Transposing 3.2 to R.H.S.,}\\[1em] \Rightarrow y = -6.5 - 3.2\\[1em] \Rightarrow y = -9.7\end{aligned}$$

Hence, $\bm{y} = -9.7$.

Solving,

$$\begin{aligned}\Rightarrow a + 8.9 = -12.6\\[1em] \text{ Transposing 8.9 to R.H.S.,}\\[1em] \Rightarrow a = -12.6 - 8.9\\[1em] \Rightarrow a = -21.5\end{aligned}$$

Hence, $\bm{a} = -21.5$.

Solving,

$$\begin{aligned}\Rightarrow x + 2\frac{1}{3} = 5\\[1em] \Rightarrow x + \frac{7}{3} = 5\\[1em] \text{ Transposing } \frac{7}{3} \text{ to R.H.S.,}\\[1em] \Rightarrow x = 5 - \frac{7}{3}\\[1em] \Rightarrow x = \frac{15 - 7}{3}\\[1em] \Rightarrow x = \frac{8}{3} = 2\frac{2}{3}\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow z + 2 = 4\frac{1}{5}\\[1em] \Rightarrow z + 2 = \frac{21}{5}\\[1em] \text{ Transposing 2 to R.H.S.,}\\[1em] \Rightarrow z = \frac{21}{5} - 2\\[1em] \Rightarrow z = \frac{21 - 10}{5}\\[1em] \Rightarrow z = \frac{11}{5} = 2\frac{1}{5}\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow m + 3\frac{1}{2} = 4\frac{1}{4}\\[1em] \Rightarrow m + \frac{7}{2} = \frac{17}{4}\\[1em] \text{ Transposing } \frac{7}{2} \text{ to R.H.S.,}\\[1em] \Rightarrow m = \frac{17}{4} - \frac{7}{2}\\[1em] \Rightarrow m = \frac{17 - 14}{4}\\[1em] \Rightarrow m = \frac{3}{4}\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow x + 2 = 1\frac{1}{4}\\[1em] \Rightarrow x + 2 = \frac{5}{4}\\[1em] \text{ Transposing 2 to R.H.S.,}\\[1em] \Rightarrow x = \frac{5}{4} - 2\\[1em] \Rightarrow x = \frac{5 - 8}{4}\\[1em] \Rightarrow x = -\frac{3}{4}\end{aligned}$$

Hence, $\bm{x} = -\frac{3}{4}$.

Solving,

$$\begin{aligned}\Rightarrow y + 5\frac{1}{3} = 4\\[1em] \Rightarrow y + \frac{16}{3} = 4\\[1em] \text{ Transposing } \frac{16}{3} \text{ to R.H.S.,}\\[1em] \Rightarrow y = 4 - \frac{16}{3}\\[1em] \Rightarrow y = \frac{12 - 16}{3}\\[1em] \Rightarrow y = -\frac{4}{3} = -1\frac{1}{3}\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow a + 3\frac{1}{5} = 1\frac{1}{2}\\[1em] \Rightarrow a + \frac{16}{5} = \frac{3}{2}\\[1em] \text{ Transposing } \frac{16}{5} \text{ to R.H.S.,}\\[1em] \Rightarrow a = \frac{3}{2} - \frac{16}{5}\\[1em] \Rightarrow a = \frac{15 - 32}{10}\\[1em] \Rightarrow a = -\frac{17}{10} = -1\frac{7}{10}\end{aligned}$$

Hence, $\bm{a} = -1\frac{7}{10}$.

Solving,

$$\begin{aligned}\Rightarrow x - 3 = 2\\[1em] \text{ Transposing −3 to R.H.S.,}\\[1em] \Rightarrow x = 2 + 3\\[1em] \Rightarrow x = 5\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow m - 2 = -5\\[1em] \text{ Transposing −2 to R.H.S.,}\\[1em] \Rightarrow m = -5 + 2\\[1em] \Rightarrow m = -3\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow b - 5 = 7\\[1em] \text{ Transposing −5 to R.H.S.,}\\[1em] \Rightarrow b = 7 + 5\\[1em] \Rightarrow b = 12\end{aligned}$$

Hence, $\bm{b} = 12$.

Solving,

$$\begin{aligned}\Rightarrow a - 2.5 = -4\\[1em] \text{ Transposing −2.5 to R.H.S.,}\\[1em] \Rightarrow a = -4 + 2.5\\[1em] \Rightarrow a = -1.5\end{aligned}$$

Hence, $\bm{a} = -1.5$.

Solving,

$$\begin{aligned}\Rightarrow y - 3\frac{1}{2} = 6\\[1em] \Rightarrow y - \frac{7}{2} = 6\\[1em] \text{ Transposing } -\frac{7}{2} \text{ to R.H.S.,}\\[1em] \Rightarrow y = 6 + \frac{7}{2}\\[1em] \Rightarrow y = \frac{12 + 7}{2}\\[1em] \Rightarrow y = \frac{19}{2} = 9\frac{1}{2}\end{aligned}$$

Hence, $\bm{y} = 9\frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow z - 2\frac{1}{3} = -6\\[1em] \Rightarrow z - \frac{7}{3} = -6\\[1em] \text{ Transposing } -\frac{7}{3} \text{ to R.H.S.,}\\[1em] \Rightarrow z = -6 + \frac{7}{3}\\[1em] \Rightarrow z = \frac{-18 + 7}{3}\\[1em] \Rightarrow z = -\frac{11}{3} = -3\frac{2}{3}\end{aligned}$$

Hence, $\bm{z} = -3\frac{2}{3}$.

Solving,

$$\begin{aligned}\Rightarrow p - 5.4 = 2.7\\[1em] \text{ Transposing −5.4 to R.H.S.,}\\[1em] \Rightarrow p = 2.7 + 5.4\\[1em] \Rightarrow p = 8.1\end{aligned}$$

Hence, $\bm{p} = 8.1$.

Solving,

$$\begin{aligned}\Rightarrow x - 1.5 = -4.9\\[1em] \text{ Transposing −1.5 to R.H.S.,}\\[1em] \Rightarrow x = -4.9 + 1.5\\[1em] \Rightarrow x = -3.4\end{aligned}$$

Hence, $\bm{x} = -3.4$.

Solving,

$$\begin{aligned}\Rightarrow n - 4 = -4\frac{1}{5}\\[1em] \Rightarrow n - 4 = -\frac{21}{5}\\[1em] \text{ Transposing −4 to R.H.S.,}\\[1em] \Rightarrow n = -\frac{21}{5} + 4\\[1em] \Rightarrow n = \frac{-21 + 20}{5}\\[1em] \Rightarrow n = -\frac{1}{5}\end{aligned}$$

Hence, $\bm{n} = -\frac{1}{5} = -0.2$.

Solving,

$$\begin{aligned}\Rightarrow 3x = 12\\[1em] \text{ Dividing both sides by 3,}\\[1em] \Rightarrow \frac{3x}{3} = \frac{12}{3}\\[1em] \Rightarrow x = 4\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 2y = 9\\[1em] \text{ Dividing both sides by 2,}\\[1em] \Rightarrow \frac{2y}{2} = \frac{9}{2}\\[1em] \Rightarrow y = \frac{9}{2} = 4\frac{1}{2}\end{aligned}$$

Hence, $\bm{y} = 4\frac{1}{2} = 4.5$.

Solving,

$$\begin{aligned}\Rightarrow 5z = 8.5\\[1em] \text{ Dividing both sides by 5,}\\[1em] \Rightarrow \frac{5z}{5} = \frac{8.5}{5}\\[1em] \Rightarrow z = 1.7\end{aligned}$$

Hence, $\bm{z} = 1.7$.

Solving,

$$\begin{aligned}\Rightarrow 2.5m = 7.5\\[1em] \text{ Dividing both sides by 2.5,}\\[1em] \Rightarrow \frac{2.5m}{2.5} = \frac{7.5}{2.5}\\[1em] \Rightarrow m = 3\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 3.2p = 16\\[1em] \text{ Dividing both sides by 3.2,}\\[1em] \Rightarrow \frac{3.2p}{3.2} = \frac{16}{3.2}\\[1em] \Rightarrow p = 5\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 2a = 4.6\\[1em] \text{ Dividing both sides by 2,}\\[1em] \Rightarrow \frac{2a}{2} = \frac{4.6}{2}\\[1em] \Rightarrow a = 2.3\end{aligned}$$

Hence, $\bm{a} = 2.3$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{2} = 5\\[1em] \text{ Multiplying both sides by 2,}\\[1em] \Rightarrow \frac{x}{2} \times 2 = 5 \times 2\\[1em] \Rightarrow x = 10\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{y}{3} = -2\\[1em] \text{ Multiplying both sides by 3,}\\[1em] \Rightarrow \frac{y}{3} \times 3 = -2 \times 3\\[1em] \Rightarrow y = -6\end{aligned}$$

Hence, $\bm{y} = -6$.

Solving,

$$\begin{aligned}\Rightarrow \frac{a}{5} = -15\\[1em] \text{ Multiplying both sides by 5,}\\[1em] \Rightarrow \frac{a}{5} \times 5 = -15 \times 5\\[1em] \Rightarrow a = -75\end{aligned}$$

Hence, $\bm{a} = -75$.

Solving,

$$\begin{aligned}\Rightarrow \frac{z}{4} = 3\frac{1}{4}\\[1em] \Rightarrow \frac{z}{4} = \frac{13}{4}\\[1em] \text{ Multiplying both sides by 4,}\\[1em] \Rightarrow \frac{z}{4} \times 4 = \frac{13}{4} \times 4\\[1em] \Rightarrow z = 13\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{m}{6} = 2\frac{1}{2}\\[1em] \Rightarrow \frac{m}{6} = \frac{5}{2}\\[1em] \text{ Multiplying both sides by 6,}\\[1em] \Rightarrow \frac{m}{6} \times 6 = \frac{5}{2} \times 6\\[1em] \Rightarrow m = 5 \times 3 = 15\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow \frac{n}{7} = -2.8\\[1em] \text{ Multiplying both sides by 7,}\\[1em] \Rightarrow \frac{n}{7} \times 7 = -2.8 \times 7\\[1em] \Rightarrow n = -19.6\end{aligned}$$

Hence, $\bm{n} = -19.6$.

Solving,

$$\begin{aligned}\Rightarrow -2x = 8\\[1em] \text{ Dividing both sides by −2,}\\[1em] \Rightarrow \frac{-2x}{-2} = \frac{8}{-2}\\[1em] \Rightarrow x = -4\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow -3.5y = 14\\[1em] \text{ Dividing both sides by −3.5,}\\[1em] \Rightarrow \frac{-3.5y}{-3.5} = \frac{14}{-3.5}\\[1em] \Rightarrow y = -4\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow -5z = 4\\[1em] \text{ Dividing both sides by −5,}\\[1em] \Rightarrow \frac{-5z}{-5} = \frac{4}{-5}\\[1em] \Rightarrow z = -\frac{4}{5} = -0.8\end{aligned}$$

Hence, $\bm{z} = -\frac{4}{5} = -0.8$.

Solving,

$$\begin{aligned}\Rightarrow -5 = a + 3\\[1em] \text{ Transposing 3 to L.H.S.,}\\[1em] \Rightarrow -5 - 3 = a\\[1em] \Rightarrow -8 = a\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 2 = p + 5\\[1em] \text{ Transposing 5 to L.H.S.,}\\[1em] \Rightarrow 2 - 5 = p\\[1em] \Rightarrow -3 = p\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 4.5 = m - 2.7\\[1em] \text{ Transposing −2.7 to L.H.S.,}\\[1em] \Rightarrow 4.5 + 2.7 = m\\[1em] \Rightarrow 7.2 = m\end{aligned}$$

Hence, $\bm{m} = 7.2$.

Solving,

$$\begin{aligned}\Rightarrow 3\frac{2}{5} = x - 2\frac{1}{3}\\[1em] \Rightarrow \frac{17}{5} = x - \frac{7}{3}\\[1em] \text{ Transposing } -\frac{7}{3} \text{ to L.H.S.,}\\[1em] \Rightarrow \frac{17}{5} + \frac{7}{3} = x\\[1em] \Rightarrow \frac{51 + 35}{15} = x\\[1em] \Rightarrow x = \frac{86}{15} = 5\frac{11}{15}\end{aligned}$$

Hence, $\bm{x} = 5\frac{11}{15}$.

Solving,

$$\begin{aligned}\Rightarrow 5 = m + 3\frac{4}{7}\\[1em] \Rightarrow 5 = m + \frac{25}{7}\\[1em] \text{ Transposing } \frac{25}{7} \text{ to L.H.S.,}\\[1em] \Rightarrow 5 - \frac{25}{7} = m\\[1em] \Rightarrow \frac{35 - 25}{7} = m\\[1em] \Rightarrow m = \frac{10}{7} = 1\frac{3}{7}\end{aligned}$$

Hence, $\bm{m} = 1\frac{3}{7}$.

Solving,

$$\begin{aligned}\Rightarrow -2\frac{1}{5} = y - 4\\[1em] \Rightarrow -\frac{11}{5} = y - 4\\[1em] \text{ Transposing −4 to L.H.S.,}\\[1em] \Rightarrow -\frac{11}{5} + 4 = y\\[1em] \Rightarrow \frac{-11 + 20}{5} = y\\[1em] \Rightarrow y = \frac{9}{5} = 1\frac{4}{5}\end{aligned}$$

Hence, $\bm{y} = 1\frac{4}{5}$.

Solving,

$$\begin{aligned}\Rightarrow 2x + 5 = 17 \\[1em] \Rightarrow 2x = 17 - 5 \\[1em] \Rightarrow 2x = 12 \\[1em] \Rightarrow x = \frac{12}{2} \\[1em] \Rightarrow x = 6.\end{aligned}$$

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

Solving,

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

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

Solving,

$$\begin{aligned}\Rightarrow 5p + 4 = 29 \\[1em] \Rightarrow 5p = 29 - 4 \\[1em] \Rightarrow 5p = 25 \\[1em] \Rightarrow p = \frac{25}{5} \\[1em] \Rightarrow p = 5.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 4a - 3 = -27 \\[1em] \Rightarrow 4a = -27 + 3 \\[1em] \Rightarrow 4a = -24 \\[1em] \Rightarrow a = \frac{-24}{4} \\[1em] \Rightarrow a = -6.\end{aligned}$$

Hence, $\bm{a} = -6$.

Solving,

$$\begin{aligned}\Rightarrow 2z + 3 = -19 \\[1em] \Rightarrow 2z = -19 - 3 \\[1em] \Rightarrow 2z = -22 \\[1em] \Rightarrow z = \frac{-22}{2} \\[1em] \Rightarrow z = -11.\end{aligned}$$

Hence, $\bm{z} = -11$.

Solving,

$$\begin{aligned}\Rightarrow 7m - 1 = 20 \\[1em] \Rightarrow 7m = 20 + 1 \\[1em] \Rightarrow 7m = 21 \\[1em] \Rightarrow m = \frac{21}{7} \\[1em] \Rightarrow m = 3.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 2.4x - 3 = 4.2 \\[1em] \Rightarrow 2.4x = 4.2 + 3 \\[1em] \Rightarrow 2.4x = 7.2 \\[1em] \Rightarrow x = \frac{7.2}{2.4} \\[1em] \Rightarrow x = 3.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 4m + 9.4 = 5 \\[1em] \Rightarrow 4m = 5 - 9.4 \\[1em] \Rightarrow 4m = -4.4 \\[1em] \Rightarrow m = \frac{-4.4}{4} \\[1em] \Rightarrow m = -1.1.\end{aligned}$$

Hence, $\bm{m} = -1.1$.

Solving,

$$\begin{aligned}\Rightarrow 6y + 4 = -4.4 \\[1em] \Rightarrow 6y = -4.4 - 4 \\[1em] \Rightarrow 6y = -8.4 \\[1em] \Rightarrow y = \frac{-8.4}{6} \\[1em] \Rightarrow y = -1.4.\end{aligned}$$

Hence, $\bm{y} = -1.4$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{3} - 5 = 2 \\[1em] \Rightarrow \frac{x}{3} = 2 + 5 \\[1em] \Rightarrow \frac{x}{3} = 7 \\[1em] \Rightarrow x = 7 \times 3 \\[1em] \Rightarrow x = 21.\end{aligned}$$

Hence, $\bm{x} = 21$.

Solving,

$$\begin{aligned}\Rightarrow \frac{y}{2} - 3 = 8 \\[1em] \Rightarrow \frac{y}{2} = 8 + 3 \\[1em] \Rightarrow \frac{y}{2} = 11 \\[1em] \Rightarrow y = 11 \times 2 \\[1em] \Rightarrow y = 22.\end{aligned}$$

Hence, $\bm{y} = 22$.

Solving,

$$\begin{aligned}\Rightarrow \frac{z}{7} + 1 = 2\frac{1}{2} \\[1em] \Rightarrow \frac{z}{7} + 1 = \frac{5}{2} \\[1em] \Rightarrow \frac{z}{7} = \frac{5}{2} - 1 \\[1em] \Rightarrow \frac{z}{7} = \frac{5 - 2}{2} \\[1em] \Rightarrow \frac{z}{7} = \frac{3}{2} \\[1em] \Rightarrow z = \frac{3}{2} \times 7 \\[1em] \Rightarrow z = \frac{21}{2} = 10\frac{1}{2}.\end{aligned}$$

Hence, $\bm{z} = 10\frac{1}{2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{a}{2.4} - 5 = 2.4 \\[1em] \Rightarrow \frac{a}{2.4} = 2.4 + 5 \\[1em] \Rightarrow \frac{a}{2.4} = 7.4 \\[1em] \Rightarrow a = 7.4 \times 2.4 \\[1em] \Rightarrow a = 17.76.\end{aligned}$$

Hence, $\bm{a} = 17.76$.

Solving,

$$\begin{aligned}\Rightarrow \frac{b}{1.6} + 3 = -2.5 \\[1em] \Rightarrow \frac{b}{1.6} = -2.5 - 3 \\[1em] \Rightarrow \frac{b}{1.6} = -5.5 \\[1em] \Rightarrow b = -5.5 \times 1.6 \\[1em] \Rightarrow b = -8.8.\end{aligned}$$

Hence, $\bm{b} = -8.8$.

Solving,

$$\begin{aligned}\Rightarrow \frac{m}{4} - 4.6 = -3.1 \\[1em] \Rightarrow \frac{m}{4} = -3.1 + 4.6 \\[1em] \Rightarrow \frac{m}{4} = 1.5 \\[1em] \Rightarrow m = 1.5 \times 4 \\[1em] \Rightarrow m = 6.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow -8m - 2 = -10 \\[1em] \Rightarrow -8m = -10 + 2 \\[1em] \Rightarrow -8m = -8 \\[1em] \Rightarrow m = \frac{-8}{-8} \\[1em] \Rightarrow m = 1.\end{aligned}$$

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

Solving,

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

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

Solving,

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

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

Solving,

$$\begin{aligned}\Rightarrow 6x + 2 = 2x + 10 \\[1em] \Rightarrow 6x - 2x = 10 - 2 \\[1em] \Rightarrow 4x = 8 \\[1em] \Rightarrow x = \frac{8}{4} \\[1em] \Rightarrow x = 2.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 18 - (2a - 12) = 8a \\[1em] \Rightarrow 18 - 2a + 12 = 8a \\[1em] \Rightarrow 30 - 2a = 8a \\[1em] \Rightarrow 30 = 8a + 2a \\[1em] \Rightarrow 30 = 10a \\[1em] \Rightarrow a = \frac{30}{10} \\[1em] \Rightarrow a = 3.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 3x + 5 + 2x + 6 + x = 4x + 21 \\[1em] \Rightarrow 6x + 11 = 4x + 21 \\[1em] \Rightarrow 6x - 4x = 21 - 11 \\[1em] \Rightarrow 2x = 10 \\[1em] \Rightarrow x = \frac{10}{2} \\[1em] \Rightarrow x = 5.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 3.5x - 9 - 3 = x + 1 \\[1em] \Rightarrow 3.5x - 12 = x + 1 \\[1em] \Rightarrow 3.5x - x = 1 + 12 \\[1em] \Rightarrow 2.5x = 13 \\[1em] \Rightarrow x = \frac{13}{2.5} \\[1em] \Rightarrow x = \frac{26}{5} = 5\frac{1}{5}.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 8x + 6 + 2x - 4 = 4x + 8 \\[1em] \Rightarrow 10x + 2 = 4x + 8 \\[1em] \Rightarrow 10x - 4x = 8 - 2 \\[1em] \Rightarrow 6x = 6 \\[1em] \Rightarrow x = \frac{6}{6} \\[1em] \Rightarrow x = 1.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow -m + (3m - 6m) = -8 - 14 \\[1em] \Rightarrow -m + (-3m) = -22 \\[1em] \Rightarrow -m - 3m = -22 \\[1em] \Rightarrow -4m = -22 \\[1em] \Rightarrow m = \frac{-22}{-4} \\[1em] \Rightarrow m = \frac{11}{2} = 5\frac{1}{2}.\end{aligned}$$

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

Solving,

$$\begin{aligned}\Rightarrow 5x - 14 = x - (24 + 4x) \\[1em] \Rightarrow 5x - 14 = x - 24 - 4x \\[1em] \Rightarrow 5x - 14 = -3x - 24 \\[1em] \Rightarrow 5x + 3x = -24 + 14 \\[1em] \Rightarrow 8x = -10 \\[1em] \Rightarrow x = \frac{-10}{8} \\[1em] \Rightarrow x = -\frac{5}{4} = -1\frac{1}{4}.\end{aligned}$$

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

Let the number be $x$.

Given,

The number increased by 17 becomes 54.

$$\begin{aligned}\Rightarrow x + 17 = 54 \\[1em] \Rightarrow x = 54 - 17 \\[1em] \Rightarrow x = 37\end{aligned}$$

Hence, the number is 37.

Let the number be $x$.

Given,

The number decreased by 8 equals 26.

$$\begin{aligned}\Rightarrow x - 8 = 26 \\[1em] \Rightarrow x = 26 + 8 \\[1em] \Rightarrow x = 34\end{aligned}$$

Hence, the number is 34.

Let the number be $x$.

Given,

One-fourth of the number when added to two-seventh of it, gives 135.

$$\begin{aligned}\Rightarrow \frac{1}{4}x + \frac{2}{7}x = 135 \\[1em] \Rightarrow \frac{7x + 8x}{28} = 135 \\[1em] \Rightarrow \frac{15x}{28} = 135 \\[1em] \Rightarrow 15x = 135 \times 28 \\[1em] \Rightarrow x = \frac{135 \times 28}{15} \\[1em] \Rightarrow x = 9 \times 28 = 252\end{aligned}$$

Hence, the number is 252.

Let the number be $x$.

Given,

Two-fifth of the number when subtracted from three-fourth of it, gives 56.

$$\begin{aligned}\Rightarrow \frac{3}{4}x - \frac{2}{5}x = 56 \\[1em] \Rightarrow \frac{15x - 8x}{20} = 56 \\[1em] \Rightarrow \frac{7x}{20} = 56 \\[1em] \Rightarrow 7x = 56 \times 20 \\[1em] \Rightarrow x = \frac{56 \times 20}{7} \\[1em] \Rightarrow x = 8 \times 20 = 160\end{aligned}$$

Hence, the number is 160.

Let the original number be $x$.

Given,

The number is increased by 12 and the new number obtained is multiplied by 5. The resulting number is 95.

$$\begin{aligned}\Rightarrow (x + 12) \times 5 = 95 \\[1em] \Rightarrow x + 12 = \frac{95}{5} \\[1em] \Rightarrow x + 12 = 19 \\[1em] \Rightarrow x = 19 - 12 \\[1em] \Rightarrow x = 7\end{aligned}$$

Hence, the original number is 7.

Let the original number be $x$.

Given,

The number is increased by 26 and the new number obtained is divided by 3. The resulting number is 18.

$$\begin{aligned}\Rightarrow \frac{x + 26}{3} = 18 \\[1em] \Rightarrow x + 26 = 18 \times 3 \\[1em] \Rightarrow x + 26 = 54 \\[1em] \Rightarrow x = 54 - 26 \\[1em] \Rightarrow x = 28\end{aligned}$$

Hence, the original number is 28.

Let the age of the son be $x$ years, then the age of the man = ($x$ + 27) years.

Given,

The age of a man is 27 years more than the age of his son. The sum of their ages is 47 years.

$$\begin{aligned}\Rightarrow x + (x + 27) = 47 \\[1em] \Rightarrow 2x + 27 = 47 \\[1em] \Rightarrow 2x = 47 - 27 \\[1em] \Rightarrow 2x = 20 \\[1em] \Rightarrow x = 10\end{aligned}$$

∴ Age of the man = $x$ + 27 = 10 + 27 = 37 years

Hence, the son is 10 years old and his father is 37 years old.

Let the age of Gopal be $x$ years, then the age of his father = ($x$ + 26) years.

Given,

The difference between the ages of Gopal and his father is 26 years. The sum of their ages is 56 years.

$$\begin{aligned}\Rightarrow x + (x + 26) = 56 \\[1em] \Rightarrow 2x + 26 = 56 \\[1em] \Rightarrow 2x = 56 - 26 \\[1em] \Rightarrow 2x = 30 \\[1em] \Rightarrow x = 15\end{aligned}$$

∴ Age of his father = $x$ + 26 = 15 + 26 = 41 years

Hence, Gopal is 15 years old and his father is 41 years old.

Let the two consecutive natural numbers be $x$ and $x + 1$.

Given,

When two consecutive natural numbers are added, the sum is 31.

$$\begin{aligned}\Rightarrow x + (x + 1) = 31 \\[1em] \Rightarrow 2x + 1 = 31 \\[1em] \Rightarrow 2x = 31 - 1 \\[1em] \Rightarrow 2x = 30 \\[1em] \Rightarrow x = 15\end{aligned}$$

∴ The numbers are 15 and 15 + 1 = 16.

Hence, the numbers are 15 and 16.

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

Given,

When three consecutive natural numbers are added, the sum is 66.

$$\begin{aligned}\Rightarrow x + (x + 1) + (x + 2) = 66 \\[1em] \Rightarrow 3x + 3 = 66 \\[1em] \Rightarrow 3x = 66 - 3 \\[1em] \Rightarrow 3x = 63 \\[1em] \Rightarrow x = 21\end{aligned}$$

∴ The numbers are 21, 21 + 1 = 22 and 21 + 2 = 23.

Hence, the numbers are 21, 22 and 23.

Let the number be $x$.

Given,

One-fourth of the number added to one-sixth of it is 15.

$$\begin{aligned}\Rightarrow \frac{1}{4}x + \frac{1}{6}x = 15 \\[1em] \Rightarrow \frac{3x + 2x}{12} = 15 \\[1em] \Rightarrow \frac{5x}{12} = 15 \\[1em] \Rightarrow 5x = 15 \times 12 \\[1em] \Rightarrow x = \frac{15 \times 12}{5} \\[1em] \Rightarrow x = 3 \times 12 = 36\end{aligned}$$

Hence, the number is 36.

Let the whole number be $x$.

Given,

The whole number is increased by 7 and the number so obtained is multiplied by 5, giving the result 45.

$$\begin{aligned}\Rightarrow (x + 7) \times 5 = 45 \\[1em] \Rightarrow x + 7 = \frac{45}{5} \\[1em] \Rightarrow x + 7 = 9 \\[1em] \Rightarrow x = 9 - 7 \\[1em] \Rightarrow x = 2\end{aligned}$$

Hence, the whole number is 2.

Given,

The age of a man and the age of his daughter differ by 23 years.

Let the age of the daughter be $x$ years, then the age of the man = ($x$ + 23) years.

Given,

The sum of their ages is 41 years.

$$\begin{aligned}\Rightarrow x + (x + 23) = 41 \\[1em] \Rightarrow 2x + 23 = 41 \\[1em] \Rightarrow 2x = 41 - 23 \\[1em] \Rightarrow 2x = 18 \\[1em] \Rightarrow x = 9\end{aligned}$$

∴ Age of the man = $x$ + 23 = 9 + 23 = 32 years

Hence, the age of the man is 32 years.

Let the smaller number be $x$.

(i) Since, the difference between the two numbers is 15,

Larger number = smaller number + 15 = $x$ + 15

Hence, the expression for the larger number is ($x$ + 15).

(ii) Given, the sum of the two numbers is 71.

$$\begin{aligned}\Rightarrow x + (x + 15) = 71 \\[1em] \Rightarrow 2x + 15 = 71 \\[1em] \Rightarrow 2x = 71 - 15 \\[1em] \Rightarrow 2x = 56 \\[1em] \Rightarrow x = 28\end{aligned}$$

∴ Larger number = $x$ + 15 = 28 + 15 = 43

Hence, the larger number is 43.

Let the larger number be $x$.

Given,

The difference between two numbers is 23.

(i) Since the difference between the two numbers is 23,

Smaller number = larger number − 23 = $x$ − 23

Hence, the expression for the smaller number is ($x$ − 23).

(ii) Given, the sum of the two numbers is 91.

$$\begin{aligned}\Rightarrow x + (x - 23) = 91 \\[1em] \Rightarrow 2x - 23 = 91 \\[1em] \Rightarrow 2x = 91 + 23 \\[1em] \Rightarrow 2x = 114 \\[1em] \Rightarrow x = 57\end{aligned}$$

∴ Smaller number = $x$ − 23 = 57 − 23 = 34

Hence, the smaller number is 34.

Given,

Sum of three consecutive integers is 78.

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

$$\begin{aligned}\Rightarrow x + (x + 1) + (x + 2) = 78 \\[1em] \Rightarrow 3x + 3 = 78 \\[1em] \Rightarrow 3x = 78 - 3 \\[1em] \Rightarrow 3x = 75 \\[1em] \Rightarrow x = 25\end{aligned}$$

∴ The integers are 25, 25 + 1 = 26 and 25 + 2 = 27.

Hence, the three consecutive integers are 25, 26 and 27.

Let the middle number be $x$, then the three consecutive numbers are $x - 1, x$ and $x + 1$.

Given,

The sum of three consecutive numbers is 54.

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

∴ The numbers are 18 − 1 = 17, 18 and 18 + 1 = 19.

Hence, the three numbers are 17, 18 and 19.

A regular pentagon has 5 equal sides.

∴ Each side = $\frac{\text{Perimeter}}{5} = \frac{x}{5}$ m

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow x - [y - {z - (x - \overline{y - z})}] \\[1em] \Rightarrow x - [y - {z - (x - y + z)}] \\[1em] \Rightarrow x - [y - {z - x + y - z}] \\[1em] \Rightarrow x - [y - {y - x}] \\[1em] \Rightarrow x - [y - y + x] \\[1em] \Rightarrow x - x = 0.\end{aligned}$$

Hence, option 1 is the correct option.

Substituting $a$ = 2 and $b$ = 1,

$$\begin{aligned}\Rightarrow a^{b} - b^{a} = 2^{1} - 1^{2} \\[1em] \Rightarrow 2 - 1 = 1.\end{aligned}$$

Hence, option 4 is the correct option.

Substituting $a$ = 3, $b$ = −3 and $c$ = 1,

$$\begin{aligned}\Rightarrow 2a - b + 3c = 2(3) - (-3) + 3(1) \\[1em] \Rightarrow 6 + 3 + 3 = 12.\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow x^{2}y \div xy^{2} \\[1em] \Rightarrow \frac{x^{2}y}{xy^{2}} \\[1em] \Rightarrow \frac{x}{y} \\[1em] \Rightarrow \frac{3}{2}.\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 3x - 2 = 7 \\[1em] \Rightarrow 3x = 7 + 2 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3.\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{x}{3} + 7 = 11 \\[1em] \Rightarrow \frac{x}{3} = 11 - 7 \\[1em] \Rightarrow \frac{x}{3} = 4 \\[1em] \Rightarrow x = 4 \times 3 = 12.\end{aligned}$$

Hence, option 3 is the correct option.

Let the number be $x$.

The number multiplied by 3 = $x \times 3$

This result divided by 2 = $\frac{x \times 3}{2}$

∴ The required equation is $\frac{x \times 3}{2} = 12$.

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 9.5 = x - 4.7 \\[1em] \Rightarrow 9.5 + 4.7 = x \\[1em] \Rightarrow x = 14.2.\end{aligned}$$

Hence, option 2 is the correct option.

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

$$\begin{aligned}\Rightarrow x + (x + 1) + (x + 2) = 21 \\[1em] \Rightarrow 3x + 3 = 21 \\[1em] \Rightarrow 3x = 18 \\[1em] \Rightarrow x = 6.\end{aligned}$$

∴ The integers are 6, 7 and 8.

Hence, option 1 is the correct option.

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

$$\begin{aligned}\Rightarrow x + (x + 2) = 14 \\[1em] \Rightarrow 2x + 2 = 14 \\[1em] \Rightarrow 2x = 12 \\[1em] \Rightarrow x = 6.\end{aligned}$$

∴ The numbers are 6 and 6 + 2 = 8.

Hence, option 2 is the correct option.

The number $x$ decreased by 12 = ($x$ − 12)

This new number multiplied by 2 = ($x$ − 12) × 2

∴ The required equation is ($x$ − 12) × 2 = 30.

Hence, option 3 is the correct option.

In $2^x − 32 = 0$, the variable $x$ appears in the exponent, so it is not a linear equation. So, statement 1 is true.

A linear equation has its variable raised to the power 1 only and cannot have the variable in the exponent. So, statement 2 is true.

Hence, option 1 is the correct option.

Two consecutive multiples of 4 differ by 4. Taking the smaller multiple as $x$, the next is $x + 4$.

$$\begin{aligned}\Rightarrow x + (x + 4) = 100 \\[1em] \Rightarrow 2x + 4 = 100 \\[1em] \Rightarrow 2x = 96 \\[1em] \Rightarrow x = 48.\end{aligned}$$

So the smaller number is 48, which means Assertion (A) is true.

The equation used, $(x) + (x + 4) = 100$, is exactly the correct setup that leads to the answer, so Reason (R) is true.

Hence, option 3 is the correct option.

Substituting $x$ = 2 in the given equation 5 − $x = \frac{x}{2}$ + 2 :

L.H.S. = 5 − 2 = 3

R.H.S. = $\frac{2}{2}$ + 2 = 1 + 2 = 3

Since L.H.S. = R.H.S., $x$ = 2 is a solution. So, Assertion (A) is true.

Reason (R) correctly states the method of checking a solution by substitution. So, Reason (R) is true.

Hence, option 3 is the correct option.