NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (Ex 4.1) Exercise 4.1

Q.1

$\begin{array}{l}\mathrm{Complete}\mathrm{the}\mathrm{last}\mathrm{column}\mathrm{of}\mathrm{the}\mathrm{table}.\end{array}$

Ans.

Q.2

$\begin{array}{l}\mathrm{Check}\mathrm{whether}\mathrm{the}\mathrm{value}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{brackets}\mathrm{is}\mathrm{a}\mathrm{solution}\\ \mathrm{to}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{or}\mathrm{not}:\\ \left(\mathrm{a}\right)\mathrm{n}+5=19(\mathrm{n}=1)\left(\mathrm{b}\right)7\mathrm{n}+5=19(\mathrm{n}=–2)\\ \left(\mathrm{c}\right)7\mathrm{n}+5=19(\mathrm{n}=2)\left(\mathrm{d}\right)4\mathrm{p}–3=13(\mathrm{p}=1)\\ \left(\mathrm{e}\right)4\mathrm{p}–3=13(\mathrm{p}=–4)\left(\mathrm{f}\right)4\mathrm{p}–3=13(\mathrm{p}=0)\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\text{n}+\text{5}=\text{19}(\text{n}=\text{1})\\ \text{Put n}=\text{1 in L}\text{.H}\text{.S to get}\\ \text{n+5}=1+5=6\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{b}\right)\text{7n}+\text{5}=\text{19}(\text{n}=-\text{2})\\ \text{Put n}=-\text{2 in L}\text{.H}\text{.S to get}\\ \text{7n+5}=7\left(-2\right)+5=-14+5=-9\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{c}\right)\text{7n}+\text{5}=\text{19}(\text{n}=\text{2})\\ \text{Put n}=\text{2 in L}\text{.H}\text{.S to get}\\ \text{7n+5}=7\left(2\right)+5=14+5=19=\text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{d}\right)\text{4p}-\text{3}=\text{13}(\text{p}=\text{1})\\ \text{Put p}=1\text{in L}\text{.H}\text{.S to get}\\ \text{4p}-3=4\left(1\right)-3=4-3=1\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{e}\right)\text{4p}-\text{3}=\text{13}(\text{p}=-\text{4})\\ \text{Put p}=-\text{4 in L}\text{.H}\text{.S to get}\\ \text{4p}-3=4\left(-4\right)-3=-16-3=-19\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\\ \left(\text{f}\right)\text{4p}-\text{3}=\text{13}(\text{p}=0)\\ \text{Put p}=\text{0 in L}\text{.H}\text{.S to get}\\ \text{4p}-3=4\left(0\right)-3=0-3=-3\ne \text{R}\text{.H}\text{.S}\\ \text{So, the given value in the bracket is not a solution to the}\\ \text{given equation}\text{.}\end{array}

Q.3

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}\mathrm{by}\mathrm{trial}\mathrm{and}\mathrm{error}\mathrm{method}:\\ \left(\mathrm{i}\right)5\mathrm{p}+2=17\left(\mathrm{ii}\right)3\mathrm{m}–14=4\end{array}$

Ans.

\begin{array}{l}\left(\text{i}\right)\text{5p}+\text{2}=\text{17}\\ \text{Put p}=\text{1 to L}\text{.H}\text{.S to get}\\ \text{5}\left(1\right)\text{+2}=\text{5+2}=\text{7}\ne \text{R}\text{.H}\text{.S}\\ \text{Put p}=\text{2 to L}\text{.H}\text{.S to get}\\ \text{5}\left(2\right)\text{+2}=\text{10+2}=\text{12}\ne \text{R}\text{.H}\text{.S}\\ \text{Put p}=\text{3 to L}\text{.H}\text{.S to get}\\ \text{5}\left(3\right)\text{+2}=\text{15+2}=\text{17}=\text{R}\text{.H}\text{.S}\\ \text{Thus, p}=\text{3 is a solution to the given equation}\\ \left(\text{ii}\right)\text{3m}-\text{14}=\text{4}\\ \text{Put m}=\text{4 in L}\text{.H}\text{.S to get}\\ \text{3}\left(4\right)-14=12-14=-2\ne \text{R}\text{.H}\text{.S}\\ \text{Put m}=\text{5 in L}\text{.H}\text{.S to get}\\ \text{3}\left(5\right)-14=15-14=-1\ne \text{R}\text{.H}\text{.S}\\ \text{Put m}=\text{6 in L}\text{.H}\text{.S to get}\\ \text{3}\left(4\right)-14=18-14=4=\text{R}\text{.H}\text{.S}\\ \text{Thus, m}=\text{6 is a solution to the given equation}\end{array}

Q.4

$\begin{array}{l}\mathrm{Write}\mathrm{equations}\mathrm{for}\mathrm{the}\mathrm{following}\mathrm{statements}:\\ \left(\mathrm{i}\right)\mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{numbers}\mathrm{x}\mathrm{and}4\mathrm{is}9.\\ \left(\mathrm{ii}\right)\mathrm{The}\mathrm{difference}\mathrm{between}\mathrm{y}\mathrm{and}2\mathrm{is}8.\\ \left(\mathrm{iii}\right)\mathrm{Ten}\mathrm{times}\mathrm{a}\mathrm{is}70.\\ \left(\mathrm{iv}\right)\mathrm{The}\mathrm{number}\mathrm{b}\mathrm{divided}\mathrm{by}5\mathrm{gives}6.\\ \left(\mathrm{v}\right)\mathrm{Three}\mathrm{fourth}\mathrm{of}\mathrm{t}\mathrm{is}15.\\ \left(\mathrm{vi}\right)\mathrm{Seven}\mathrm{times}\mathrm{m}\mathrm{plus}7\mathrm{gets}\mathrm{you}77.\\ \left(\mathrm{vii}\right)\mathrm{One}\mathrm{fourth}\mathrm{of}\mathrm{a}\mathrm{number}\mathrm{minus}4\mathrm{gives}4.\\ \left(\mathrm{viii}\right)\mathrm{If}\mathrm{you}\mathrm{take}\mathrm{away}6\mathrm{from}6\mathrm{times}\mathrm{y},\mathrm{you}\mathrm{get}60.\\ \left(\mathrm{ix}\right)\mathrm{If}\mathrm{you}\mathrm{add}3\mathrm{toone}\mathrm{third}\mathrm{of}\mathrm{z},\mathrm{you}\mathrm{get}30.\end{array}$

Ans.

\begin{array}{l}\left(\text{i}\right)x+4=9\\ (\text{ii)}y-2=8\\ \text{(iii)}10a=70\\ \text{(iv)}\frac{b}{5}=6\\ \text{(v)}\frac{3}{4}t=15\\ \left(\text{vi}\right)7m+7=77\\ \left(\text{vii}\right)\frac{x}{4}-4=4\\ \left(\text{viii}\right)6y-6=60\\ \left(\text{ix}\right)\frac{z}{3}+3=30\end{array}

Q.5

$\begin{array}{l}\mathrm{Write}\mathrm{the}\mathrm{following}\mathrm{equations}\mathrm{in}\mathrm{statement}\mathrm{forms}:\\ \left(\mathrm{i}\right)\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{p}+4=15\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{m}–7=3\\ \left(\mathrm{iii}\right)\mathrm{\hspace{0.17em}\hspace{0.17em}}2\mathrm{m}=7\\ \left(\mathrm{iv}\right)\frac{\mathrm{m}}{5}=3\\ \left(\mathrm{v}\right)\hspace{0.17em}\frac{3}{5}\mathrm{m}=6\\ \left(\mathrm{vi}\right)3\mathrm{p}+4=25\\ \left(\mathrm{vii}\right)4\mathrm{p}–2=18\\ \left(\mathrm{viii}\right)\frac{\mathrm{p}}{2}+2=8\end{array}$

Ans.

\begin{array}{l}(i)\text{The sum of p and 4 is 15}\text{.}\\ \text{(ii) 7 subtracted from m is 3}\text{.}\\ \text{(iii) Twice of a number m is 7}\text{.}\\ \text{(iv) One-fifth of m is 3}\text{.}\\ \text{(v) Three-fifth of m is 6}\text{.}\\ \text{(vi) Three times of a number p, when add to 4 gives 25}\text{.}\\ \text{(vii) When 2 is subtracted from four times of a number p,}\\ \text{it gives 18}\text{.}\\ \text{(viii) When 2 is added to half of a number p, it gives 8}\text{.}\end{array}

Q.6

$\begin{array}{l}\mathrm{Set}\mathrm{up}\mathrm{an}\mathrm{equation}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{cases}:\\ \left(\mathrm{i}\right)\mathrm{Irfan}\mathrm{says}\mathrm{that}\mathrm{he}\mathrm{has}7\mathrm{marbles}\mathrm{more}\mathrm{than}\mathrm{five}\\ \mathrm{times}\mathrm{the}\mathrm{marbles}\mathrm{Parmit}\mathrm{has}.\mathrm{Irfan}\mathrm{has}37\mathrm{marbles}.\\ (\mathrm{Take}\mathrm{m}\mathrm{to}\mathrm{be}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{Parmit}’\mathrm{s}\mathrm{marbles}.\\ \left(\mathrm{ii}\right)\mathrm{Laxmi}’\mathrm{s}\mathrm{father}\mathrm{is}49\mathrm{years}\mathrm{old}.\mathrm{He}\mathrm{is}4\mathrm{years}\mathrm{older}\mathrm{than}\\ \mathrm{three}\mathrm{times}\mathrm{Laxmi}’\mathrm{s}\mathrm{age}.(\mathrm{Take}\mathrm{Laxmi}’\mathrm{s}\mathrm{age}\mathrm{to}\mathrm{be}\mathrm{y}\mathrm{years}.)\\ \left(\mathrm{iii}\right)\mathrm{The}\mathrm{teacher}\mathrm{tells}\mathrm{the}\mathrm{class}\mathrm{that}\mathrm{the}\mathrm{highest}\mathrm{marks}\\ \mathrm{obtained}\mathrm{by}\mathrm{a}\mathrm{student}\mathrm{in}\mathrm{her}\mathrm{class}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{lowest}\\ \mathrm{marks}\mathrm{plus}7.\mathrm{The}\mathrm{highest}\mathrm{score}\mathrm{is}87.\\ (\mathrm{Take}\mathrm{the}\mathrm{lowest}\mathrm{score}\mathrm{to}\mathrm{be}\mathrm{l}).\\ \left(\mathrm{iv}\right)\mathrm{In}\mathrm{an}\mathrm{isosceles}\mathrm{triangle},\mathrm{the}\mathrm{vertex}\mathrm{angle}\mathrm{is}\mathrm{twice}\mathrm{either}\\ \mathrm{base}\mathrm{angle}.(\mathrm{Let}\mathrm{the}\mathrm{base}\mathrm{angle}\mathrm{be}\mathrm{b}\mathrm{in}\mathrm{degrees}.\mathrm{Remember}\\ \mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\mathrm{degrees}).\end{array}$

Ans.

\begin{array}{l}\text{(i) Let Parmit has}m\text{marbles}\text{.}\\ \text{Then, according to the question, we have}\\ \text{5}\times \text{Number of marbles Parmit has +7}=\text{Number of marbles}\\ \text{Irfan has}\\ \text{5}\times m+\text{7}=\text{37}\\ \text{So, we get}\\ \boxed{\text{5}m+\text{7}=\text{37}}\\ \text{(ii) Let Laxmi be}y\text{years old}\\ \text{Then, according to the question, we have}\\ \text{3}\times \text{Laxmi’s age}+\text{4}=\text{Laxmi’s father age}\\ \text{3}\times y\text{+4}=\text{49}\\ \boxed{\text{3}y+\text{4}=\text{49}}\\ \text{(iii) Let the lowest marks be}l\text{.}\\ \text{Then, according to the question, we have}\\ 2\times \text{lowest marks}+\text{7}=\text{Highest marks}\\ \text{2}\times l+\text{7}=\text{87}\\ \boxed{2l+7=87}\\ \text{(iv) An isoceles triangle has two angles equal}\text{.}\\ \text{Let the base angle be}b\text{.}\\ \text{Then, according to the question, we have}\\ b+b+2b=\text{180}°\\ \boxed{4b=180°}\end{array}