NCERT Solutions Class 7 Maths Chapter 4

Q.1 Complete the last column of the table.

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Q.2 Check whether the value given in the brackets is a solutionto the given equation or not:(a) n+5=19(n=1) (b) 7n+5=19(n=–2)(c) 7n+5=19(n=2) (d) 4p–3=13(p=1)(e) 4p–3=13(p=–4) (f)4p–3=13(p=0)

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 Solve the following equations by trial and error method:(i) 5p+2=17 (ii) 3m–14=4

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

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\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

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

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\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 Set up an equation in the following cases:i Irfan says that he has 7 marbles more than fivetimes the marbles Parmit has. Irfan has 37 marbles.(Take m to be the number of Parmit’s marbles.(ii) Laxmi’s father is 49 years old. He is 4 years older thanthree times Laxmi’s age.(Take Laxmi’s age to be y years.)iiiThe teacher tells the class that the highest marksobtained by a student in her class is twice the lowestmarks plus 7.The highest score is 87.(Take the lowest score to be l).iv In an isosceles triangle, the vertex angle is twice eitherbase angle.(Let the base angle be b in degrees. Rememberthat the sum of angles of a triangle is 180​ degrees).

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\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}

Q.7 Give first the step you will use to separate the variableand then solve the equation:(a) x–1=0 (b) x+1=0 (c) x–1=5 (d) x+6=2(e) y–4=–7 (f) y–4=4 (g) y+4=4 (h) y+4=–4

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\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.8 Give first the step you will use to separate the variableand then solve the equation:(a) 3l=42 (b)b2=6 (c)p7=4 (d)4x=25(e) 8y=36 (f)z3=54 (g)a5=715 (h)2t=–10

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

Q.9 Give the steps you will use to separate the variable andthen solve the equation:(a) 3n-2 = 46 (b) 5m+7 = 17(c) 20p3 = 40 (d) 3p10 = 6

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\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.10 Solve the following equations:(a)10p=100(b)10p+10=100(c)p4=5(d)p3=5(e)3p4=6(f) 3s=–9(g) 3s+12=0(h) 3s=0(i) 2q=6(j) 2q–6=0(k) 2q+6=0(l) 2q+6=12

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}

Q.11 Solve the following equations:(a) 2y+52=372 (b) 5t+28=10 (c)a5+3=2(d) q4+7=5 (e) 52x=10 (f) 52x=254(g) 7m+192=13 (h) 6z+10=−2 (i) 3l2=23(j) 2b3−5=3

Ans.

\begin{array}{l}\left(\text{a}\right)\text{2}y+\frac{5}{2}=\frac{37}{2}\\ 2y=\frac{37}{2}-\frac{5}{2}=\frac{37-5}{2}=\frac{32}{2}=16\\ 2y=16\\ y=\frac{16}{2}=8\\ Thus,\boxed{y=8}\\ \left(\text{b}\right)\text{5t}+\text{28}=\text{1}0\\ 5t=10-28\\ 5t=-18\\ \boxed{t=\frac{-18}{5}}\\ \left(\text{c}\right)\frac{\text{a}}{5}+3=2\\ \frac{a}{5}=2-3\\ \frac{a}{5}=-1\\ \boxed{a=-5}\\ \left(\text{d}\right)\frac{q}{4}+7=5\\ \frac{q}{4}=5-7\\ \frac{q}{4}=-2\\ \boxed{q=-8}\\ \left(\text{e}\right)\frac{5}{2}x=-10\\ 5x=-10\times 2\\ x=\frac{-10\times 2}{5}=\frac{-\cancel{5}\times 2\times 2}{\cancel{5}}=-4\\ Thus,\boxed{x=-4}\\ \left(\text{f}\right)\frac{5}{2}x=\frac{25}{4}\\ 5x=\frac{25\times 2}{4}\\ x=\frac{\cancel{5}\times 5\times \cancel{2}}{\cancel{2}\times 2\times \cancel{5}}\\ Thus,\boxed{x=\frac{5}{2}}\\ \left(\text{g}\right)7m+\frac{19}{2}=13\\ 7m=13-\frac{19}{2}=\frac{26-19}{2}=\frac{7}{2}\\ 7m=\frac{7}{2}\\ m=\frac{\cancel{7}}{2\times \cancel{7}}\\ Thus,\boxed{m=\frac{1}{2}}\\ \left(\text{h}\right)\text{6z}+\text{1}0=-\text{2}\\ 6z=-2-10\\ 6z=-12\\ z=\frac{-12}{6}=\frac{-2\times \cancel{6}}{\cancel{6}}0\\ Thus,\boxed{z=-2}\\ \left(\text{i}\right)\frac{3l}{2}=\frac{2}{3}\\ 3l=\frac{4}{3}\\ \boxed{l=\frac{4}{9}}\\ \left(\text{j}\right)\frac{2b}{3}-5=3\\ \frac{2b}{3}=8\\ 2b=24\\ b=\frac{24}{2}=\frac{\cancel{2}\times 12}{\cancel{2}}\\ Thus,\boxed{b=12}\end{array}

Q.12 Solve the following equations:(a) 2(x+4)=12 (b) 3(n−5)=21 (c) 3(n−5)=21(d) 3−2(2−y)=7 (e)−4(2−x)=9 (f) 4(2−x)=9(g) 4+5(p−1)=34 (h)34−5(p−1)=4

Ans.

\begin{array}{l}\text{(a})2(x+4)=12\\ \text{Divide both sides by 2 to get}\\ x+4=6\\ \text{Thus,}\boxed{x=2}\\ \left(\text{b}\right)3(n-5)=-21\\ \text{Divide both sides by 3 to get}\\ n-5=-\text{7}\\ \text{Thus,}\boxed{n=-2}\\ \left(\text{c}\right)3(n-5)=-\text{21}\\ \text{Divide both sides by 2 to get}\\ n-5=-7\\ Thus,\boxed{n=-12}\\ \left(\text{d}\right)\text{3}-\text{2}(\text{2}-y)=\text{7}\\ -2\left(2-y\right)=4\\ \text{Divide both sides by}-\text{2 to get}\\ 2-y=-2\\ \text{Multiply both sides by}-\text{1 to get}\\ y-2=2\\ \text{Thus,}\boxed{y=4}\\ \left(\text{e}\right)-4(2-x)=\text{9}\\ \text{Divide both sides by}-\text{4 to get}\\ \text{2}-x=\frac{-9}{4}\\ \text{Multiply both sides by}-\text{1 to get}\\ x-2=\frac{9}{4}\\ x=\frac{9}{4}+2=\frac{9+8}{4}\\ Thus,\boxed{x=\frac{17}{4}}\\ \left(\text{f}\right)4(2-x)=\text{9}\\ \text{Divide both sides by 4 to get}\\ \text{2}-x=\frac{9}{4}\\ \text{Multiply both sides by -1 to get}\\ \text{x}-2=\frac{-9}{4}\\ x=\frac{-9}{4}+2=\frac{-9+8}{4}\\ \text{Thus},\boxed{x=\frac{-1}{4}}\\ \left(\text{g}\right)4+5(p-1)=34\\ 5\left(p-1\right)=30\\ \text{Divide both sides by 5 to get}\\ \text{p}-1=6\\ p=6+1=7\\ \text{Thus},\boxed{p=7}\\ \left(\text{h}\right)34-5(p-1)=4\\ -5\left(p-1\right)=-30\\ \text{Divide both sides by}-\text{5 to get}\\ p-1=6\\ p=6+1=7\\ \text{Thus,}\boxed{p=7}\end{array}

Q.13 Solve the following equations.(a) 4=5 (p−2) (b)−4=5 (p−2) (c)−16=−5(2−p)(d) 10=4+3 (t+2) (e) 28=4+3 (t+5)(f) 0=16+4 (m−6)

Ans.

\begin{array}{l}\left(\text{a}\right)\text{4}=\text{5}(p-\text{2})\\ \text{Divide both sides by 5 to get}\\ \frac{4}{5}=p-2\\ p=\frac{4}{5}+2=\frac{4+10}{5}\\ \text{Thus},\boxed{p=\frac{14}{5}}\\ \left(\text{b}\right)-\text{4}=\text{5}(p-2)\\ \text{Divide both sides by 5 to get}\\ \frac{-4}{5}=p-2\\ p=\frac{-4}{5}+2=\frac{-4+10}{5}\\ \text{Thus},\boxed{p=\frac{6}{5}}\\ \left(\text{c}\right)-16=-5(2-p)\\ \text{Divide both sides by}-\text{5 to get}\\ \frac{16}{5}=2-p\\ \text{Multiply both sides by}-\text{1 to get}\\ \frac{-\text{16}}{5}=p-2\\ p=\frac{-16}{5}+2=\frac{-16+10}{5}\\ \text{Thus},\boxed{p=\frac{-6}{5}}\\ \left(\text{d}\right)\text{1}0=\text{4}+\text{3}(t+2)\\ 6=3\left(t+2\right)\\ \text{Divide both sides by 3 to get}\\ 2=t+2\\ \text{Thus,}\boxed{t=0}\\ \left(\text{e}\right)\text{28}=\text{4}+\text{3}(t+\text{5})\\ 24=3\left(t+5\right)\\ \text{Divide both sides by 3 to get}\\ 8=t+5\\ \text{Thus,}\boxed{t=3}\\ \left(\text{f}\right)0=\text{16}+\text{4}(m-\text{6})\\ -16=4\left(m-6\right)\\ \text{Divide both sides by 4 to get}\\ -4=m-6\\ \text{Thus,}\boxed{m=2}\end{array}