NCERT Solutions Class 7 Maths Chapter 4 Simple Equations

4.0NCERT Questions with Solutions for Class 7 Maths Chapter 4 - Detailed Solutions

Exercise: 4.1

• Complete the last column of the table.

Sol.

• Check whether the value given in the brackets is a solution to the given equation or not: (a) $\mathrm{n}+5=19(\mathrm{n}=1)$ (b) $7\mathrm{n}+5=19(\mathrm{n}=-2)$ (c) $7\mathrm{n}+5=19(\mathrm{n}=2)$ (d) $4p-3=13(p=1)$ (e) $4\mathrm{p}-3=13(\mathrm{p}=-4)$ (f) $4p-3=13(p=0)$ Sol. (a) $n+5=19(n=1)$ Putting $\mathrm{n}=1$ in L.H.S., $1+5=6$ $\because$ L.H.S. $\ne$ R.H.S., $\therefore \mathrm{n}=1$ is not the solution of the given equation. (b) $7\mathrm{n}+5=19(\mathrm{n}=-2)$ Putting $\mathrm{n}=-2$ in L.H.S., $7(-2)+5=-14+5=-9$ $\because$ L.H.S. $\ne$ R.H.S., $\therefore \mathrm{n}=-2$ is not the solution of the given equation. (c) $7\mathrm{n}+5=19(\mathrm{n}=2)$ Putting $\mathrm{n}=2$ in L.H.S., $7(2)+5=14+5=19$ $\because$ L.H.S. $=$ R.H.S., $\therefore \mathrm{n}=2$ is the solution of the given equation. (d) $4\mathrm{p}-3=13(\mathrm{p}=1)$ Putting $p=1$ in L.H.S., $4(1)-3=4-3=1$ $\because$ L.H.S. $=$ R.H.S., $\therefore \mathrm{p}=1$ is not the solution of the given equation. (e) $4p-3=13(p=-4)$ Putting $\mathrm{p}=-4$ in L.H.S., $4(-4)-3=-16-3=-19$ $\because$ L.H.S. $\ne$ R.H.S., $\therefore \mathrm{p}=-4$ is not the solution of the given equation. (f) $4p-3=13(p=0)$ Putting $\mathrm{p}=0$ in L.H.S., $4(0)-3=0-3=-3$ $\because$ L.H.S. $\ne$ R.H.S., $\therefore \mathrm{p}=0$ is not the solution of the given equation. 3
• Solve the following equations by trial and error method: (i) $5p+2=17$ (ii) $3m-14=4$ Sol. (i) $5\mathrm{p}+2=17$ Putting $\mathrm{p}=-3$ in L.H.S. $5(-3)+2=-15+2=-13$ $\because -13\ne 17$ Therefore, $p=-3$ is not the solution. Putting $\mathrm{p}=-2$ in L.H.S. $5(-2)+2=-10+2=-8$ $\because -8\ne 17$ Therefore, $\mathrm{p}=-2$ is not the solution. Putting $p=-1$ in L.H.S. $5(-1)+2=-5+2=-3$ $\because -3\ne 17$ Therefore, $\mathrm{p}=-1$ is not the solution. Putting $\mathrm{p}=0$ in L.H.S. 5(0) $+2=0+2=2$ $\because 2\ne 17$ Therefore, $\mathrm{p}=0$ is not the solution. Putting $\mathrm{p}=1$ in L.H.S. $5(1)+2=5+2=7$ $\because 7\ne 17$ Therefore, $\mathrm{p}=1$ is not the solution. Putting $p=2$ in L.H.S. $5(2)+2=10+2=12$ $\because 12\ne 17$ Therefore, $p=2$ is not the solution. Putting $\mathrm{p}=3$ in L.H.S. $5(3)+2=15+2=17$ $\because 17=17$, Therefore, $\mathrm{p}=3$ is the solution. (ii) $3m-14=4$ Putting $\mathrm{m}=-2$ in L.H.S. $3(-2)-14=-6-14=-20$ $\because -20\ne 4$ Therefore, $m=-2$ is not the solution. Putting $\mathrm{m}=-1$ in L.H.S. $3(-1)-14=-3-14=-17$ $\because -17\ne 4$ Therefore, $\mathrm{m}=-1$ is not the solution. Putting $\mathrm{m}=0$ in L.H.S. 3(0) $-14=0-14=-14$ $\because -14\ne 4$ Therefore, $\mathrm{m}=0$ is not the solution. Putting $\mathrm{m}=1\mathrm{in}$ L.H.S. $3(1)-14=3-14=-11$ $\because -11\ne 4$ Therefore, $\mathrm{m}=1$ is not the solution. Putting $\mathrm{m}=2$ in L.H.S. $3(2)-14=6-14=-8$ $\because -8\ne 4$ Therefore, $\mathrm{m}=2$ is not the solution. Putting $\mathrm{m}=3$ in L.H.S. $3(3)-14=9-14=-5$ $\because -5\ne 4$ Therefore, $\mathrm{m}=3$ is not the solution. Putting $\mathrm{m}=4$ in L.H.S. $3(4)-14=12-14=-2$ $\because -2\ne 4$ Therefore, $\mathrm{m}=4$ is not the solution. Putting $\mathrm{m}=5$ in L.H.S. $3(5)-14=15-14=1$ $\because 1\ne 4$ Therefore, $m=5$ is not the solution. Putting $\mathrm{m}=6$ in L.H.S. $3(6)-14=18-14=4$ $\because 4=4$ Therefore, $\mathrm{m}=6$ is the solution.
• Write equations for the following statements: (i) The sum of numbers $x$ and 4 is 9 . (ii) 2 subtracted from y 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 x minus 4 gives 4 . (viii) If you take away 6 from 6 times $y$, you get 60. (ix) If you add 3 to one-third of $z$, you get 30 . Sol. (i) $\mathrm{x}+4=9$ (ii) $y-2=8$ (iii) $10\mathrm{a}=70$ (iv) $\frac{b}{5}=6$ (v) $\frac{3}{4}t=15$ (vi) $7\mathrm{m}+7=77$ (vii) $\frac{x}{4}-4=4$ (viii) $6y-6=60$ Sol. (i) The sum of numbers p and 4 is 15 . (ii) 7 subtracted from $m$ is 3 . (iii) Two times $m$ is 7 . (iv) The number $m$ is divided by 5 gives 3 . (v) Three-fifth of the number $m$ is 6 . (vi) Three times p plus 4 gets 25 . (vii) If you take away 2 from 4 times $p$, you get 18. (viii) If you added 2 to half of $p$, you get 8 .
• Set up an equation in the following cases : (i) Irfan says that he has 7 marbles more than five times 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 than three times Laxmi's age. (Take Laxmi's age to be y years.) (iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be $\ell$ ). (iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be $b$ in degrees. Remember that the sum of angles of a triangle is 180 degrees). Sol. (i) Let $m$ be the number of Parmit's marbles. $\therefore 5\mathrm{m}+7=37$ (ii) Let the age of Laxmi be y years. $\therefore 3y+4=49$ (iii) Let the lowest score be $\ell$. $\therefore 2\ell +7=87$ (iv) Let the base angle of the isosceles triangle be b , so vertex angle $=2\mathrm{b}$. $\therefore 2\mathrm{b}+\mathrm{b}+\mathrm{b}=180$ (angle sum property of triangle) $4\mathrm{b}=180^{\circ }$

Exercise : 4.2

• Give the first step you will use to separate the variable and then solve the equation (a) $x-1=0$ (b) $\mathrm{x}+1=0$ (c) $x-1=5$ (d) $x+6=2$ (e) $y-4=-7$ (f) $y-4=4$ (g) $y+4=4$ (h) $y+4=-4$ Sol. (a) $\mathrm{x}-\mathrm{l}=0$ [Adding 1 both sides] $\Rightarrow \mathrm{x}-1+1=0+1$ $\Rightarrow \mathrm{x}=1$ (b) $\mathrm{x}+\mathrm{l}=0$ [Subtracting 1 both sides] $\Rightarrow \mathrm{x}+1-1=0-1$ $\Rightarrow \mathrm{x}=-1$ (c) $\mathrm{x}-1=5$ [Adding 1 both sides] $\Rightarrow \mathrm{x}-1+1=5+1$ $\Rightarrow \mathrm{x}=6$ (d) $x+6=2$ [Subtracting 6 both sides] $\Rightarrow \mathrm{x}+6-6=2-6$ $\Rightarrow \mathrm{x}=-4$ (e) $y-4=-7$ [Adding 4 both sides] $\Rightarrow \mathrm{y}-4+4=-7+4$ $\Rightarrow \mathrm{y}=-3$ (f) $y-4=4$ [Adding 4 both sides] $\Rightarrow \mathrm{y}-4+4=4+4$ $\Rightarrow y=8$ (g) $y+4=4$ [Subtracting 4 both sides] $\Rightarrow \mathrm{y}+4-4=4-4$ $\Rightarrow \mathrm{y}=0$ (h) $y+4=-4$ [Subtracting 4 both sides] $\Rightarrow y+4-4=-4-4$ $\Rightarrow y=-8$
• Give the first step you will use to separate the variable and then solve the equation : (a) $3\ell =42$ (b) $\frac{b}{2}=6$ (c) $\frac{p}{7}=4$ (d) $4x=25$ (e) $8y=36$ (f) $\frac{\mathrm{Z}}{3}=\frac{5}{4}$ (g) $\frac{a}{5}=\frac{7}{15}$ (h) $20\mathrm{t}=-10$ Sol. (a) $3\ell =42$ [Dividing both sides by 3] $\Rightarrow \frac{3\ell }{3}=\frac{42}{3}$ $\Rightarrow \ell =14$ (b) $\frac{b}{2}=6$ [Multiplying both sides by 2] $\Rightarrow \frac{\mathrm{b}}{2}\times 2=6\times 2$ $\Rightarrow \mathrm{b}=12$ (c) $\frac{p}{7}=4$ [Multiplying both sides by 7] $\Rightarrow \frac{\mathrm{p}}{7}\times 7=4\times 7$ $\Rightarrow \mathrm{p}=28$ (d) $4\mathrm{x}=25$ [Dividing both sides by 4] $\Rightarrow \frac{4\mathrm{x}}{4}=\frac{25}{4}$ $\Rightarrow \mathrm{x}=\frac{25}{4}$ (e) $8y=36$ [Dividing both sides by 8] $\Rightarrow \frac{8y}{8}=\frac{36}{8}$ $\Rightarrow \mathrm{y}=\frac{9}{2}$ (f) $\frac{\mathrm{z}}{3}=\frac{5}{4}$ [Multiplying both sides by 3] $\Rightarrow \frac{\mathrm{z}}{3}\times 3=\frac{5}{4}\times 3$ $\Rightarrow \mathrm{z}=\frac{15}{4}$ (g) $\frac{a}{5}=\frac{7}{15}$ [Multiplying both sides by 5] $\Rightarrow \frac{\mathrm{a}}{5}\times 5=\frac{7}{15}\times 5$ $\Rightarrow \mathrm{a}=\frac{7}{3}$ (h) $20\mathrm{t}=10$ [Dividing both sides by 20] $\Rightarrow \frac{20\mathrm{t}}{20}=\frac{-10}{20}$ $\Rightarrow \mathrm{t}=\frac{-1}{2}$
• Give the steps you will use to separate the variable and then solve the equation : (a) $3n-2=46$ (b) $5\mathrm{m}+7=17$ (c) $\frac{20p}{3}=40$ (d) $\frac{3p}{10}=6$ Sol. (a) $3n-2=46$ Step I: [Adding 2 both sides] $3n-2+2=46+2$ $\Rightarrow 3\mathrm{n}=48$ Step II: $\frac{5\mathrm{m}}{5}=\frac{10}{5}$ [Dividing both sides by 3 ] $\frac{3n}{3}=\frac{48}{3}$ $\Rightarrow \mathrm{n}=16$ (b) $5\mathrm{m}+7=17$ Step I: [Subtracting 7 both sides] $5m+7-7=17-7$ $\Rightarrow 5\mathrm{m}=10$ Step II: [Dividing both sides by 5] $\frac{5m}{5}=\frac{10}{5}$ $\Rightarrow \mathrm{m}=2$ (c) $\frac{20\mathrm{p}}{3}=40$ Step I: [Multiplying both sides by 3] $\frac{20\mathrm{p}}{3}\times 3=40\times 3$ $\Rightarrow 20\mathrm{p}=120$ Step II: [Dividing both sides by 20] $\frac{20\mathrm{p}}{20}=\frac{120}{20}$ $\Rightarrow \mathrm{p}=6$ (d) $\frac{3p}{10}=6$ Step I: [Multiplying both sides by 10] $\frac{3p}{10}\times 10=6\times 10$ $\Rightarrow 3\mathrm{p}=60$ Step II: [Dividing both sides by 3] $\frac{3p}{3}=\frac{60}{3}$ $\Rightarrow \mathrm{p}=20$
• Solve the following equations : (a) $10\mathrm{p}=100$ (b) $10\mathrm{p}+10=100$ (c) $\frac{\mathrm{p}}{4}=5$ (d) $\frac{-p}{3}=5$ (e) $\frac{3p}{4}=6$ (f) $3s=-9$ (g) $3s+12=0$ (h) $3s=0$ (i) $2\mathrm{q}=6$ (j) $2q-6=0$ (k) $2q+6=0$ (l) $2q+6=12$ Sol. (a) $10p=100$ [Dividing both sides by 10] $\Rightarrow \frac{10\mathrm{p}}{10}=\frac{100}{10}$ $\Rightarrow \mathrm{p}=10$ (b) $10\mathrm{p}+10=100$ [Subtracting 10 on both sides] $\Rightarrow 10\mathrm{p}+10-10=100-10$ $\Rightarrow 10\mathrm{p}=90$ [Dividing both sides by 10] $\Rightarrow \frac{10\mathrm{p}}{10}=\frac{90}{10}$ $\Rightarrow \mathrm{p}=9$ (c) $\frac{p}{4}=5$ [Multiplying both sides by 4] $\Rightarrow \frac{\mathrm{p}}{4}\times 4=5\times 4$ $\Rightarrow \mathrm{p}=20$ (d) $\frac{-\mathrm{p}}{3}=5$ [Multiplying both sides by -3 ] $\Rightarrow \frac{-\mathrm{p}}{3}\times (-3)=5\times (-3)$ $\Rightarrow \mathrm{p}=-15$ (e) $\frac{3p}{4}=6$ [Multiplying both sides by 4] $\Rightarrow \frac{3\mathrm{p}}{4}\times 4=6\times 4$ $\Rightarrow 3\mathrm{p}=24$ [Dividing both sides by 3] $\Rightarrow \frac{3\mathrm{p}}{4}$ $\Rightarrow \mathrm{p}=8$ (f) $3\mathrm{s}=-9$ [Dividing both sides by 3] $\Rightarrow \frac{3\mathrm{s}}{3}=\frac{-9}{3}$ $\Rightarrow \mathrm{s}=-3$ (g) $3\mathrm{s}+12=0$ [Subtracting 12 on both sides] $\Rightarrow 3\mathrm{s}+12-12=0-12$ $\Rightarrow 3\mathrm{s}=-12$ [Dividing both sides by 3] $\Rightarrow \frac{3\mathrm{s}}{3}=\frac{-12}{3}$ $\Rightarrow \mathrm{s}=-4$ (h) $3\mathrm{s}=0$ [Dividing both sides by 3] $\Rightarrow \frac{3\mathrm{s}}{3}=\frac{0}{3}$ $\Rightarrow \mathrm{s}=0$ (i) $2q=6$ [Dividing both sides by 2] $\Rightarrow \frac{2\mathrm{q}}{2}=\frac{6}{2}$ $\Rightarrow \mathrm{q}=3$ (j) $2q-6=0$ [Adding both sides 6] $\Rightarrow 2q-6+6=0+6$ $\Rightarrow 2\mathrm{q}=6$ [Dividing both sides by 2] $\Rightarrow \frac{2\mathrm{q}}{2}=\frac{6}{2}$ $\Rightarrow \mathrm{q}=3$ (k) $2\mathrm{q}+6=0$ [Subtracting 6 on both sides] $\Rightarrow 2q+6-6=0-6$ $\Rightarrow 2\mathrm{q}=-6$ [Dividing both sides by 2] $\Rightarrow \frac{2\mathrm{q}}{2}=\frac{-6}{2}$ $\Rightarrow \mathrm{q}=-3$ (l) $2q+6=12$ [Subtracting 6 on both sides] $\Rightarrow 2q+6-6=12-6$ $\Rightarrow 2\mathrm{q}=6$ [Dividing both sides by 2] $\Rightarrow \frac{2\mathrm{q}}{2}=\frac{6}{2}$ $\Rightarrow \mathrm{q}=3$

Exercise : 4.3

• Solve the following equations: (a) $2y+\frac{5}{2}=\frac{37}{2}$ (b) $5\mathrm{t}+28=10$ (c) $\frac{a}{5}+3=2$ (d) $\frac{q}{4}+7=5$ (e) $\frac{5}{2}x=-5$ (f) $\frac{5}{2}x=\frac{25}{4}$ (g) $7\mathrm{m}+\frac{19}{2}=13$ (h) $6z+10=-2$ (i) $\frac{3\ell }{2}=\frac{2}{3}$ (j) $\frac{2\mathrm{b}}{3}-5=3$ Sol. (a) $2y+\frac{5}{2}=\frac{37}{2}\Rightarrow 2y=\frac{32}{2}$ $\Rightarrow 2\mathrm{y}=\frac{37-5}{2}\Rightarrow 2\mathrm{y}=\frac{37}{2}-\frac{5}{2}$ $\Rightarrow 2\mathrm{y}=\frac{37-5}{2}16\Rightarrow \mathrm{y}=\frac{16}{2}\Rightarrow \mathrm{y}=8$ (b) $5\mathrm{t}+28=10\Rightarrow 5\mathrm{t}=10-28$ $\Rightarrow 5\mathrm{t}=-18\Rightarrow \mathrm{t}=\frac{-18}{5}$ (c) $\frac{\mathrm{a}}{5}+3=2\Rightarrow \frac{\mathrm{a}}{5}=2-3\Rightarrow \frac{\mathrm{a}}{5}=-1$ $\Rightarrow \mathrm{a}=-1\times 5\Rightarrow \mathrm{a}=-5$ (d) $\frac{\mathrm{q}}{4}+7=5\Rightarrow \frac{\mathrm{q}}{4}=5-7\Rightarrow \frac{\mathrm{q}}{4}=-2$ $\Rightarrow \mathrm{q}=-2\times 4\Rightarrow \mathrm{q}=-8$ (e) $\frac{5}{2}x=-5\Rightarrow 5x=2\times -5\Rightarrow x=-2$ (f) $\frac{5}{2}x=\frac{25}{4}\Rightarrow 5x=\frac{25}{4}\times 2\Rightarrow 5x=\frac{25}{4}$ $\Rightarrow \mathrm{x}=\frac{25}{2\times 5}\Rightarrow \mathrm{x}=\frac{5}{2}$ (g) $7\mathrm{m}+\frac{19}{2}=13\Rightarrow 7\mathrm{m}=13-\frac{19}{2}$ $\Rightarrow 7\mathrm{m}=\frac{7}{2}\Rightarrow \mathrm{m}=\frac{7}{2\times 7}$ $\Rightarrow \mathrm{m}=\frac{1}{2}$ (h) $6\mathrm{z}+10=-2\Rightarrow 6\mathrm{z}=-2-10$ $\Rightarrow 6\mathrm{z}=-12\Rightarrow \mathrm{z}=\frac{-12}{6}\Rightarrow \mathrm{z}=-2$ (i) $\frac{3\ell }{2}=\frac{2}{3}\Rightarrow 3\ell =\frac{2}{3}\times 2\Rightarrow 3\ell =\frac{4}{3}$ $\Rightarrow \ell =\frac{4}{3\times 3}\Rightarrow \ell =\frac{4}{9}$ (j) $\frac{2\mathrm{b}}{3}-5=3\Rightarrow \frac{2\mathrm{b}}{3}=3+5\Rightarrow \frac{2\mathrm{b}}{3}=8$ $\Rightarrow 2\mathrm{b}=8\times 3\Rightarrow 2\mathrm{b}=24\Rightarrow \mathrm{b}=\frac{24}{2}$ $\Rightarrow \mathrm{b}=12$
• Solve the following equations : (a) $2(x+4)=12$ (b) $3(\mathrm{n}-5)=21$ (c) $3(\mathrm{n}-5)=-21$ (d) $-4(2+x)=8$ (e) $4(2-x)=8$ Sol. (a) $2(x+4)=12\Rightarrow x+4=\frac{12}{2}$ $\Rightarrow x+4=6\Rightarrow x=6-4\Rightarrow x=2$ (b) $3(\mathrm{n}-5)=21\Rightarrow \mathrm{n}-5=\frac{21}{3}\Rightarrow \mathrm{n}-5=7$ $\Rightarrow \mathrm{n}=7+5\Rightarrow \mathrm{n}=12$ (c) $3(\mathrm{n}-5)=-21\Rightarrow \mathrm{n}-5=\frac{-21}{3}$ $\Rightarrow \mathrm{n}-5=-7\Rightarrow \mathrm{n}=-7+5$ $\Rightarrow \mathrm{n}=-2$ (d) $-4(2+x)=8$ $\Rightarrow 2+\mathrm{x}=\frac{8}{-4}$ $\Rightarrow 2+\mathrm{x}=-2$ $\Rightarrow \mathrm{x}=-2-2$ $\Rightarrow \mathrm{x}=-4$ (e) $4(2-\mathrm{x})=9\Rightarrow 4\times 2-\mathrm{x}\times (-4)=9$ $\Rightarrow 8+4\mathrm{x}=9\Rightarrow 4\mathrm{x}=9-8$ $\Rightarrow 4\mathrm{x}=1\Rightarrow \mathrm{x}=\frac{1}{4}$
• Solve the following equations: (a) $4=5(p-2)$ (b) $-4=5(p-2)$ (c) $16=4+3(t+2)$ (d) $4+5(p-1)=34$ (e) $0=16+4(m-6)$ Sol. (a) $4=5(p-2)\Rightarrow 4=5\times p-5\times 2$ $\Rightarrow 4=5\mathrm{p}-10\Rightarrow 5\mathrm{p}=4+10$ $\Rightarrow 5\mathrm{p}=14\Rightarrow \mathrm{p}=\frac{14}{5}$ (b) $-4=5(\mathrm{p}-2)\Rightarrow -4=5\times \mathrm{p}-5\times 2$ $\Rightarrow -4=5\mathrm{p}-10\Rightarrow 5\mathrm{p}-10=-4$ $\Rightarrow 5\mathrm{p}=-4+10\Rightarrow 5\mathrm{p}=6$ $\Rightarrow \mathrm{p}=\frac{6}{5}$ (c) $16=4+3(\mathrm{t}+2)$ $\Rightarrow 16=4+3\mathrm{t}+6$ $\Rightarrow 3\mathrm{t}+10=16$ $\Rightarrow 3\mathrm{t}=6$ $\Rightarrow t=\frac{6}{3}=2$ (d) $4+5(p-1)=34$ $\Rightarrow 4+5\mathrm{p}-5=34$ $\Rightarrow 5\mathrm{p}-1=34$ $\Rightarrow 5\mathrm{p}=35$ $\Rightarrow \mathrm{p}=\frac{35}{5}=7$ (e) $0=16+4(\mathrm{m}-6)\Rightarrow 0-16=4(\mathrm{m}-6)$ $\Rightarrow -16=4(\mathrm{m}-6)\Rightarrow \frac{-16}{4}=\mathrm{m}-6$ $\Rightarrow -4=\mathrm{m}-6\Rightarrow -4+6=\mathrm{m}$ $\Rightarrow 2=\mathrm{m}\Rightarrow \mathrm{m}=2$
• (a) Construct 3 equations starting with x $=2$ (b) Construct 3 equations starting with x = -2 Sol. (a) 3 equations starting with $\mathrm{x}=2$. (i) $\mathrm{x}=2$ Multiplying both sides by 10 , $10\mathrm{x}=20$ Adding 2 both sides $10\mathrm{x}+2=20+2$ $\Rightarrow 10\mathrm{x}+2=22$ (ii) $\mathrm{x}=2$ Multiplying both sides by 5 $5\mathrm{x}=10$ Subtracting 3 from both sides $5x-3=10-3$ $\Rightarrow 5\mathrm{x}-3=7$ (iii) $x=2$ Dividing both sides by 5 $\frac{x}{5}=\frac{2}{5}$ (b) 3 equations starting with $\mathrm{x}=-2$. (i) $\mathrm{x}=-2$ Multiplying both sides by 3 $3x=-6$ (ii) $\mathrm{x}=-2$ Multiplying both sides by 3 $3x=-6$ Adding 7 to both sides $3x+7=-6+7$ $\Rightarrow 3\mathrm{x}+7=1$ (iii) $\mathrm{x}=-2$ Multiplying both sides by 3 $3x=-6$ Adding 10 to both sides $3\mathrm{x}+10=-6+10$ $\Rightarrow 3\mathrm{x}+10=4$

Exercise : 4.4

• Set up equations and solve them to find the unknown number in the following cases : (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-fourth of a number and add 3 to it, I get 21. (d) When I subtracted 11 from twice a number, the result was 15. (e) Munna subtracts thrice the number of notebooks he has, from 50 , find 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 $\frac{5}{2}$ of the number, the result is $\frac{11}{2}$. Sol. (a) Let the number be $x$. According to the question, $8x+4=60$ $\Rightarrow 8\mathrm{x}=60-4\Rightarrow 8\mathrm{x}=56$ $\Rightarrow \mathrm{x}=\frac{56}{8}\Rightarrow \mathrm{x}=7$ (b) Let the number be $y$. $\Rightarrow \mathrm{y}=7\times 5\Rightarrow \mathrm{y}=35$ (c) Let the number be z . According to the question, $\frac{3}{4}z+3=21$ $\Rightarrow \frac{3}{4}\mathrm{z}=21-3\Rightarrow \frac{3}{4}\mathrm{z}=18$ $\Rightarrow 3\mathrm{z}=18\times 4\Rightarrow 3\mathrm{z}=72$ $\Rightarrow \mathrm{z}=\frac{72}{3}\Rightarrow \mathrm{z}=24$ (d) Let the number be x . According to the question, $2\mathrm{x}-11=15$ $\Rightarrow 2\mathrm{x}=15+11\Rightarrow 2\mathrm{x}=26$ $\Rightarrow \mathrm{x}=\frac{26}{2}\Rightarrow \mathrm{x}=13$ (e) Let the number be $m$. According to the question, $50-3\mathrm{m}=8$ $\Rightarrow -3\mathrm{m}=8-50\Rightarrow -3\mathrm{m}=-42$ $\Rightarrow \mathrm{m}=\frac{-42}{-3}\Rightarrow \mathrm{m}=14$ (f) Let the number be $n$. According to the question, $\frac{\mathrm{n}+19}{5}=8$ $\Rightarrow \mathrm{n}+19=8\times 5\Rightarrow \mathrm{n}+19=40$ $\Rightarrow \mathrm{n}=40-19$ $\Rightarrow \mathrm{n}=21$ (g) Let the number be x . According to the question, $\frac{5}{2}x-7=\frac{11}{2}$ $\Rightarrow \frac{5}{2}\mathrm{x}=\frac{11}{2}+7\Rightarrow \frac{5}{2}\mathrm{x}=\frac{11+14}{2}$ $\Rightarrow \frac{5}{2}\mathrm{x}=\frac{25}{2}\Rightarrow 5\mathrm{x}=\frac{25\times 2}{2}$ $\Rightarrow 5\mathrm{x}=25$ $\Rightarrow \mathrm{x}=\frac{25}{5}\Rightarrow \mathrm{x}=5$
• Solve the following: (a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87 . What is the lowest score? (b) In an isosceles triangle, the base angles are equal. The vertex angle is $40^{\circ }$. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180^{\circ }$ ). (c) Sachin scored twice as many runs as Rahul. Together, their runs fell two shorts of a double century. How many runs did each one score? Sol. (a) Let the lowest marks be $y$. According to the question, $2\mathrm{y}+7=87$ $\Rightarrow 2\mathrm{y}=87-7\Rightarrow 2\mathrm{y}=80$ $\Rightarrow \mathrm{y}=\frac{80}{2}\Rightarrow \mathrm{y}=40$ Thus, the lowest score is 40 . (b) Let the base angle of the triangle be b . Given, $\mathrm{a}=40^{\circ },\mathrm{b}=\mathrm{c}$ Since, $\mathrm{a}+\mathrm{b}+\mathrm{c}=180^{\circ }$ [Angle sum property of a triangle] $\Rightarrow 40^{\circ }+\mathrm{b}+\mathrm{b}=180^{\circ }$ $\Rightarrow 40^{\circ }+2\mathrm{b}=180^{\circ }$ $\Rightarrow 2\mathrm{b}=180^{\circ }-40^{\circ }$ $\Rightarrow 2\mathrm{b}=140^{\circ }$ $\Rightarrow \mathrm{b}=\frac{140^{\circ }}{2}$

• $\Rightarrow \mathrm{b}=70^{\circ }$ Thus, the base angles of the isosceles triangle are $70^{\circ }$ each. (c) Let the score of Rahul be x runs and Sachin's score is 2 x . According to the question, $\mathrm{x}+2\mathrm{x}=198$ $\Rightarrow 3\mathrm{x}=198$ $\Rightarrow \mathrm{x}=\frac{198}{3}$ $\Rightarrow \mathrm{x}=66$ Thus, Rahul's score $=66$ runs And Sachin's score $=2\times 66=132$ runs.
• Solve the following: (i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have? (ii) Laxmi's father is 49 years old. He is 4 years older than three times Laxmi's age. What is Laxmi's age? (iii) People of Sundargam planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77? Sol. (i) Let the number of marbles Parmit has be $m$. According to the question, $5\mathrm{m}+7=37$ $\Rightarrow 5\mathrm{m}=37-7\Rightarrow 5\mathrm{m}=30$ $\Rightarrow \mathrm{m}=\frac{30}{5}\Rightarrow \mathrm{m}=6$ Thus, Parmit has 6 marbles. (ii) Let the age of Laxmi be y years. Then her father's age $=(3y+4)=49$ $\Rightarrow 3\mathrm{y}=49-4\Rightarrow 3\mathrm{y}=45$ $\Rightarrow \mathrm{y}=\frac{45}{3}\Rightarrow \mathrm{y}=15$ Thus, the age of Laxmi is 15 years. (iii) Let the number of fruit trees be $t$. Then the number of non-fruits tree $=3t+2$ According to the question, $3\mathrm{t}+2=77$ $\Rightarrow 3\mathrm{t}=77-2$ $\Rightarrow 3\mathrm{t}=75$ $\Rightarrow \mathrm{t}=\frac{75}{3}\Rightarrow \mathrm{t}=25$ Thus, the number of fruit trees are 25.
• Solve the following riddle : I am a number, Tell my identity! Take me seven times over and add a fifty! To reach a triple century, you still need forty! Sol. Let the number be $n$. According to the question, $7\mathrm{n}+50+40=$ 300 $\Rightarrow 7\mathrm{n}+90=300$ $\Rightarrow 7\mathrm{n}=300-90$ $\Rightarrow 7\mathrm{n}=210$ $\Rightarrow \mathrm{n}=\frac{210}{7}\Rightarrow \mathrm{n}=30$ Thus, the required number is 30 .

5.0Importance of Practice NCERT Solutions Class 7 Maths Chapter 4 Simple Equations

• The concepts presented in Chapter 4 of the NCERT Solutions for Class 7 Maths offer an initial basis for enhancing the mathematical skills needed for examinations.
• The exercises and examples in the NCERT solutions on simple equations simplify complex concepts and make them easy to understand. 
• The exercise questions cover the basics, formulas, and principles of simple equations, which helps students understand how to solve problems.
• The abilities students gain from preparing this chapter enable them to handle increasingly complex problems in the future, which is essential for upcoming exams.