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)

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\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

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\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}

Q.14 (a) Construct 3 equations starting with x=2(b) Construct 3 equations starting with x=−2

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$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=2\\ \left(\mathrm{b}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=-2\end{array}$

Q.15 Set up equations and solve them to find the unknownnumbers in the following case:a Add 4 to eight times a number; you get 60.b One fifth of a number minus 4 gives 3.c If I take three fourths of a number and count up 3 more, I get 21.(d) When I subtracted 11 from twice a number, the resultwas 15.(e) Munna subtracts thrice the number of notebooks he has from 50,he finds the result to be 8.f Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.g Anwar thinks of a number. If he takes away 7 from 52 of the number,the result is 112.

Ans.

\begin{array}{l}\text{(a) Let the number be x}\text{.}\\ \text{8 times of this number = 8x}\\ \text{So, we get}\\ 8x+4=6\text{0}\\ 8x=\text{56}\\ \text{x=}\frac{56}{8}\\ Thus,\boxed{x=7}\\ \text{(b) Let the number be}x.\\ \text{One-fifth of this number=}\frac{x}{5}\\ \text{So, we get}\\ \frac{x}{5}-4=3\\ \frac{x}{5}=7\\ \boxed{x=35}\\ \text{(c) Let the number be}x.\\ \text{Three-fourth of this number}=\frac{3x}{4}\\ \text{So,}\text{we}\text{get}\\ \frac{3x}{4}+3=21\\ \frac{3x}{4}=18\\ 3x=72\\ x=\frac{72}{3}\\ Thus,\boxed{x=24}\\ (d)\text{Let the number be}x.\\ \text{So, we have}\\ \text{2x-11=15}\\ \text{2x=26}\\ \text{x=}\frac{26}{13}=13\\ Thus,\boxed{x=13}\\ (e)\text{Let the number be}x\\ \text{Thrice the number of books}=\text{3}x\\ \text{So, we get}\\ \text{50}-3x=\text{8}\\ -3x=8-50=-42\\ \text{Divide both sides by -3 to get}\\ \text{x=}\frac{-42}{-3}=14\\ Thus,\boxed{x=14}\\ (f)\text{Let the number be}x.\\ \text{We, have}\\ \frac{x+19}{5}=8\\ x+19=8\times 5=40\\ x=40-19=21\\ \text{Thus,}\boxed{x=21}\\ (g)\text{Let the number be}x.\\ \boxed{So,wehave}\\ \frac{5x}{2}-7=23\\ \frac{5x}{2}=23+7=30\\ 5x=30\times 2=60\\ x=\frac{60}{5}=12\\ \text{Thus},\boxed{x=12}\end{array}

Q.16 Solve the following:a The 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. What is thelowest score?b In an isosceles triangle, the base angles are equal.The vertex angle is 40°. What are the base angles of thetriangle? (Remember, the sum of three angles of a triangleis 180°)

Ans.

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{a}\right)\mathrm{The}\mathrm{teacher}\mathrm{tells}\mathrm{the}\mathrm{class}\mathrm{that}\mathrm{the}\mathrm{highest}\mathrm{marks}\\ \mathrm{obtained}\mathrm{by}\mathrm{a}\mathrm{student}\mathrm{in}\mathrm{her}\mathrm{class}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{lowest}\\ \mathrm{marks}\mathrm{plus}7.\mathrm{The}\mathrm{highest}\mathrm{score}\mathrm{is}87.\mathrm{What}\mathrm{is}\mathrm{the}\\ \mathrm{lowest}\mathrm{score}?\\ \left(\mathrm{b}\right)\mathrm{In}\mathrm{an}\mathrm{isosceles}\mathrm{triangle},\mathrm{the}\mathrm{base}\mathrm{angles}\mathrm{are}\mathrm{equal}.\\ \mathrm{The}\mathrm{vertex}\mathrm{angle}\mathrm{is}40°.\mathrm{What}\mathrm{are}\mathrm{the}\mathrm{base}\mathrm{angles}\mathrm{of}\mathrm{the}\\ \mathrm{triangle}?\; (\mathrm{Remember},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\\ \mathrm{is}180°)\end{array}$

Q.17 Solve the following:(i)Irfan says that he has 7 marbles more than five timesthe marbles Parmithas. Irfan has 37 marbles. How manymarbles does Parmit have?(ii)Laxmi‘s father is 49 year sold.He is 4 years older thanthree times Laxmi‘s age.What is Laxmi‘ sage?(iii) Maya, Madhura and Mohsina are friends studying in thesame class.In a class testin geography, Maya got 16out of 25. Madhura got 20. Their average score was 19.How much did Mohsina score?

$\begin{array}{l}\left(\mathrm{iv}\right)\mathrm{People}\mathrm{of}\mathrm{Sundargram}\mathrm{planted}\mathrm{a}\mathrm{total}\mathrm{of}102\mathrm{trees}\mathrm{in}\mathrm{the}\\ \mathrm{village}\mathrm{garden}.\mathrm{Some}\mathrm{of}\mathrm{the}\mathrm{trees}\mathrm{were}\mathrm{fruit}\mathrm{trees}.\mathrm{The}\\ \mathrm{number}\mathrm{of}\mathrm{non}-\mathrm{fruit}\mathrm{trees}\mathrm{were}\mathrm{two}\mathrm{more}\mathrm{than}\mathrm{three}\mathrm{times}\\ \mathrm{the}\mathrm{number}\mathrm{of}\mathrm{fruit}\mathrm{trees}.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{fruit}\\ \mathrm{trees}\mathrm{planted}?\end{array}$

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}\\ \text{5}m+\text{7}=\text{37}\\ 5m=37-7=30\\ 5m=30\\ m=\frac{30}{5}=6\\ \text{Therefore,}\text{}\text{Parmit}\text{has}\boxed{\text{6}\text{marbles}}\text{.}\\ \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+4=\text{49}\\ \text{3}y+\text{4}=\text{49}\\ 3y=49-4=45\\ 3y=45\\ y=\frac{45}{3}=15\\ \text{Therefore,}\text{laxmi’s}\text{age}\text{is}\boxed{\text{15}\text{years}}\text{.}\\ \text{(iii) Let the number of fruit trees be}x.\\ \text{So, we have}\\ \text{3}\times \text{Number of fruit trees}+2=\text{Number of non-fruit trees}\\ 3x+2=77\\ 3x=77–2=75\\ 3x=75\\ x=\frac{75}{3}=25\\ \text{Therefore,}\text{the}\text{number}\text{of}\text{fruit}\text{trees}\text{was}\boxed{25}.\end{array}

Q.18 Solve the following riddle.I am a number,Tell my identity!Take me seven times overAnd add a fifty!To reach a triple centuryYou still need forty

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{be}\mathrm{x}\mathrm{.}\\ \mathrm{Then}\mathrm{we}\mathrm{have}\\ (7\mathrm{x}+50)+40=300\\ 7\mathrm{x}+50+40=300\\ 7\mathrm{x}+90+300\\ 7\mathrm{x}=300-90\\ 7\mathrm{x}=210\\ \mathrm{x}=\frac{210}{7}=30.\\ \mathrm{Therefore},\mathrm{the}\mathrm{number}\mathrm{is}\boxed{\mathrm{30}}\mathrm{.}\end{array}$