NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (Ex 4.2) Exercise 4.2

Q.1

$\begin{array}{l}\mathrm{Give}\mathrm{first}\mathrm{the}\mathrm{step}\mathrm{you}\mathrm{will}\mathrm{use}\mathrm{to}\mathrm{separate}\mathrm{the}\mathrm{variable}\\ \mathrm{and}\mathrm{then}\mathrm{solve}\mathrm{the}\mathrm{equation}:\\ \left(\mathrm{a}\right)\mathrm{x}–1=0\left(\mathrm{b}\right)\mathrm{x}+1=0\left(\mathrm{c}\right)\mathrm{x}–1=5\left(\mathrm{d}\right)\mathrm{x}+6=2\\ \left(\mathrm{e}\right)\mathrm{y}–4=–7\left(\mathrm{f}\right)\mathrm{y}–4=4\left(\mathrm{g}\right)\mathrm{y}+4=4\left(\mathrm{h}\right)\mathrm{y}+4=–4\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)x-\text{1}=0\\ \text{Add 1 to both sides to get}\\ \boxed{x=1}\\ \left(\text{b}\right)\text{x}+\text{1}=0\\ \text{Subtract 1 from both sides to get}\\ \boxed{\text{x}=-\text{1}}\\ \left(\text{c}\right)\text{x}-\text{1}=\text{5}\\ \text{Add 1 to both sides to get}\\ \boxed{x=6}\\ \left(\text{d}\right)\text{x}+\text{6}=\text{2}\\ \text{Subtract 6 from both sides to get}\\ \boxed{\text{x}=-\text{4}}\\ \left(\text{e}\right)\text{y}-\text{4}=-\text{7}\\ \text{Add 4 to both sides to get}\\ \boxed{\text{y}=-3}\\ \left(\text{f}\right)\text{y}-\text{4}=\text{4}\\ \text{Add 4 to both sides to get}\\ \boxed{y=0}\\ \left(\text{g}\right)\text{y}+\text{4}=\text{4}\\ \text{Subtract 4 from both sides to get}\\ \boxed{y=0}\\ \left(\text{h}\right)\text{y}+\text{4}=-\text{4}\\ \text{Subtract 4 from both sides to get}\\ \boxed{y=-8}\end{array}

Q.2

$\begin{array}{l}\mathrm{Give}\mathrm{first}\mathrm{the}\mathrm{step}\mathrm{you}\mathrm{will}\mathrm{use}\mathrm{to}\mathrm{separate}\mathrm{the}\mathrm{variable}\\ \mathrm{and}\mathrm{then}\mathrm{solve}\mathrm{the}\mathrm{equation}:\\ \left(\mathrm{a}\right)3\mathrm{l}=42\left(\mathrm{b}\right)\frac{\mathrm{b}}{2}=6\left(\mathrm{c}\right)\frac{\mathrm{p}}{7}=4\left(\mathrm{d}\right)4\mathrm{x}=25\\ \left(\mathrm{e}\right)8\mathrm{y}=36\left(\mathrm{f}\right)\frac{\mathrm{z}}{3}=\frac{5}{4}\left(\mathrm{g}\right)\frac{\mathrm{a}}{5}=\frac{7}{15}\left(\mathrm{h}\right)2\mathrm{t}=–10\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)3l=42\\ \text{Divide both sides by 3 to get}\\ l=\frac{42}{3}=\boxed{14}\\ \left(\text{b}\right)\frac{b}{2}=6\\ \text{Multiply both sides by 3 to get}\\ \boxed{b=12}\\ \left(\text{c}\right)\frac{p}{7}=\text{4}\\ \text{Multiply both sides by 7 to get}\\ \boxed{p=28}\\ \left(\text{d}\right)\text{4x}=\text{25}\\ \text{Divide both sides by 4 to get}\\ \boxed{x=\frac{25}{4}}\\ \left(\text{e}\right)\text{8y}=36\\ \text{Divide both sides by 8 to get}\\ y=\frac{36}{8}=\boxed{\frac{9}{2}}\\ \left(\text{f}\right)\frac{z}{3}=\frac{5}{4}\\ \text{Multiply both sides by 3 to get}\\ \boxed{z=\frac{15}{4}}\\ \left(\text{g}\right)\frac{a}{5}=\frac{7}{15}\\ \text{Multiply both sides by 5 to get}\\ \text{a=}\frac{7\times 5}{15}=\frac{7\times \cancel{5}}{3\times \cancel{5}}=\boxed{\frac{7}{3}}\\ \left(\text{h}\right)2t=-\text{10}\\ \text{Divide both sides by 2 to get}\\ \text{t=}\frac{-10}{2}=\frac{-5\times \cancel{2}}{\cancel{2}}=\boxed{-5}\end{array}$

Q.3

$\begin{array}{l}\mathrm{Give}\mathrm{the}\mathrm{steps}\mathrm{you}\mathrm{will}\mathrm{use}\mathrm{to}\mathrm{separate}\mathrm{the}\mathrm{variable}\mathrm{and}\\ \mathrm{then}\mathrm{solve}\mathrm{the}\mathrm{equation}:\\ (\mathrm{a})\; 3\mathrm{n}-2\; =\; 46(\mathrm{b})\; 5\mathrm{m}+7\; =\; 17\\ \left(\mathrm{c}\right)\frac{20\mathrm{p}}{\mathrm{3}}=\; 40(\mathrm{d}) \frac{3\mathrm{p}}{\mathrm{10}}\mathrm{=\; 6}\end{array}$

Ans.

\begin{array}{l}\text{(a)}3n-2=46\\ \text{Add 2 to both sides to get}\\ 3n=48\\ \text{Now, divide both sides by 3 to get}\\ \boxed{n=16}\\ \text{(b)}5m+7=17\\ \text{Subtract 7 from both sides to get}\\ 5m=10\\ \text{Now, divide both sides by 5 to get}\\ \boxed{m=2}\\ \text{(c)}\frac{20p}{3}=40\\ \text{Multiply both sides by 3 to get}\\ 20p=120\\ \text{Now divide both sides by 20, to get}\\ p=\frac{120}{20}=\frac{\cancel{20}\times 3}{\cancel{20}}\\ \text{Thus},\boxed{p=3}\\ \text{(d)}\frac{3p}{10}=6\\ \text{Multiply both sides by 10 to get}\\ 3p=60\\ \text{Now divide both sides by 3, to get}\\ p=\frac{60}{3}=\frac{20\times \cancel{3}}{\cancel{3}}\\ \text{Thus},\boxed{p=20}\end{array}

Q.4

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}:\\ \left(\mathrm{a}\right)10\mathrm{p}=100\left(\mathrm{b}\right)10\mathrm{p}+10=100\left(\mathrm{c}\right)\frac{\mathrm{p}}{4}=5\\ \left(\mathrm{d}\right)\frac{\mathrm{p}}{3}=5\left(\mathrm{e}\right)\frac{3\mathrm{p}}{4}=6\left(\mathrm{f}\right)\mathrm{\hspace{0.17em}}3\mathrm{s}=–9\\ \\ \left(\mathrm{g}\right)\mathrm{\hspace{0.17em}}3\mathrm{s}+12=0\left(\mathrm{h}\right)\mathrm{\hspace{0.17em}}3\mathrm{s}=0\left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}2\mathrm{q}=6\\ \\ \left(\mathrm{j}\right)\mathrm{\hspace{0.17em}}2\mathrm{q}–6=0\left(\mathrm{k}\right)\mathrm{\hspace{0.17em}}2\mathrm{q}+6=0\left(\mathrm{l}\right)\mathrm{\hspace{0.17em}}2\mathrm{q}+6=12\end{array}$

Ans.

\begin{array}{l}(a)10p=100\\ p=\frac{100}{10}=\boxed{10}\\ \text{(b)}10p+10=100\\ 10p=90\\ p=\frac{90}{10}=\boxed{9}\\ \text{(c)}\frac{p}{4}=5\\ p=5\times 4=\boxed{20}\\ \text{(d)}\frac{p}{3}=5\\ p=5\times 3=\boxed{15}\\ (e)\frac{3p}{4}=6\\ 3p=6\times 4=24\\ p=\frac{24}{3}=\boxed{8}\\ (f)3s=-9\\ s=\frac{-9}{3}=\boxed{-3}\\ (g)3s+12=0\\ 3s=-12\\ s=\frac{-12}{3}=\boxed{-4}\\ (h)3s=0\\ s=\frac{0}{3}=\boxed{0}\\ (i)2q=6\\ q=\frac{6}{2}=\boxed{3}\\ (j)2q-6=0\\ 2q=6\\ q=\frac{6}{2}=\boxed{3}\\ (k)2q+6=0\\ 2q=-6\\ q=\frac{-6}{2}=\boxed{-3}\\ (l)2q+6=12\\ 2q=12-6\\ 2q=6\\ q=\frac{6}{3}=\boxed{2}\end{array}