NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations

1. Complete the last column of the table:

Ans:

2. Check whether the value given in the brackets is a solution to the given equation or not:

(a) $ {n + 5 = 19(n = 1)}$ 

Ans: Putting $ {\text{n}} = 1$ in the above equation, we get:

$  1 + 5 = 19 $ 

$ \Rightarrow 6 = 19 $  

$ \therefore 6 \ne 19$ 

$ \because {\text{n}} = 1$  is not a solution to the given equation.

(b) $ {7n + 5 = 19(n =  - 2)}$ 

Ans: Putting $ {\text{n}} =  - 2$  in the above equation, we get:

$  7( - 2) + 5 = 19 $ 

$ \Rightarrow  - 14 + 5 = 19 $  

$ \Rightarrow  - 9 = 19 $ 

$ \therefore  - 9 \ne 19 $ 

$ \because {\text{n}} =  - 2$  is not a solution to the given equation.

(c) $ {7n + 5 = 19(n = 2)}$ 

Ans: Putting $ {\text{n}} = 2$  in the above equation, we get:

$ 7(2) + 5 = 19 $ 

$  \Rightarrow 14 + 5 = 19 $

$  \Rightarrow 19 = 19 $

$ \because $ The above relation holds to be true, $ {\text{n}} = 2$  is a solution to the given equation.

(d) $ {4p - 3 = 13(p = 1)}$ 

Ans: Putting $ {\text{p}} = 1$ in the above equation, we get:

$  4(1) - 3 = 13 $ 

$    \Rightarrow 4 - 3 = 13 $ 

$  \Rightarrow 1 = 13 $

$   \because 1 \ne 13 $

$ \therefore {\text{p}} = 1$  is not a solution to the given equation.

(e) $ {4p - 3 = 13(p =  - 4)}$ 

Ans: Putting $ {\text{p}} =  - 4$  in the above equation, we get:

$   4( - 4) - 3 = 13 $ 

$    \Rightarrow  - 16 - 3 = 13 $ 

$    \Rightarrow  - 19 = 13 $

$   \because  - 19 \ne 13 $

$ \therefore {\text{p}} =  - 4$ is not a solution to the given equation.

(f) $ {4p - 3 = 13(p = 0)}$

Ans: Putting $ {\text{p}} = 0$  in the above equation, we get:

$  4(0) - 3 = 13 $ 

$  \Rightarrow 0 - 3 = 13 $ 

$  \Rightarrow  - 3 = 13 $ 

$ \because  - 3 \ne 13 $  

$ \therefore {\text{p}} = 0$ is not a solution to the given equation.

3. Solve the following equations by trial and error method:

(i) $ {5p + 2 = 17}$ 

Ans: Putting $ {\text{p}} =  - 3$  in L.H.S. $ 5( - 3) + 2 =  - 15 + 2 =  - 13$ 

$ \because  - 13 \ne 17$ $ \therefore {\text{p}} =  - 3$  is not the solution.

Putting $ {\text{p}} =  - 2$  in L.H.S. $ 5( - 2) + 2 =  - 10 + 2 =  - 8$ 

$ \because  - 8 \ne 17$  $ \therefore {\text{p}} =  - 2$  is not the solution.

Putting $ {\text{p}} =  - 1$  in L.H.S. $ 5( - 1) + 2 =  - 5 + 2 =  - 3$ 

$ \because  - 3 \ne 17$ $ \therefore {\text{p}} =  - 1$  is not the solution.

Putting $ {\text{p}} = 0$  in L.H.S. $ 5(0) + 2 = 0 + 2 = 2$ 

$ \because 2 \ne 17$ $ \therefore {\text{p}} = 0$  is not the solution. Putting $ {\text{p}} = 1$  in L.H.S. $ 5(1) + 2 = 5 + 2 = 7$ 

$ \because 7 \ne 17$  $ \therefore {\text{p}} = 1$  is not the solution.

Putting $ {\text{p}} = 2$  in L.H.S. $ 5(2) + 2 = 10 + 2 = 12$ 

$ \because 12 \ne 17$ $ \therefore {\text{p}} = 2$  is not the solution. Putting $ {\text{p}} = 3$  in L.H.S. $ 5(3) + 2 = 15 + 2 = 17$ 

$ \because 17 = 17$ $ \therefore {\text{p}} = 3$  is the solution.

(ii) $ {3m - 14 = 4}$

Ans: Putting $ {\text{m}} =  - 2$  in L.H.S. $ 3( - 2) - 14 =  - 6 - 14 =  - 20$ 

$ \because  - 20 \ne 4$  $ \therefore {\text{m}} =  - 2$  is not the solution.

Putting $ {\text{m}} =  - 1$  in L.H.S. $ 3( - 1) - 14 =  - 3 - 14 =  - 17$ 

$ \because  - 17 \ne 4$  $ \therefore {\text{m}} =  - 1$  is not the solution.

Putting $ {\text{m}} = 0$  in L.H.S. $ 3(0) - 14 = 0 - 14 =  - 14$ 

$ \because  - 14 \ne 4$ $ \therefore {\text{m}} = 0$  is not the solution.

Putting $ {\text{m}} = 1$  in L.H.S. $ 3(1) - 14 = 3 - 14 =  - 11$ 

$ \because  - 11 \ne 4$  $ \therefore {\text{m}} = 1$  is not the solution.

Putting $ {\text{m}} = 2$  in L.H.S. $ 3(2) - 14 = 6 - 14 =  - 8$ 

$ \because  - 8 \ne 4$ ∴m=2 is not the solution.

Putting $ {\text{m}} = 3$  in L.H.S. $ 3(3) - 14 = 9 - 14 =  - 5$ 

$ \because  - 5 \ne 4$  $ \therefore {\text{m}} = 3$ is not the solution.

Putting $ {\text{m}} = 4$  in L.H.S. $ 3(4) - 14 = 12 - 14 =  - 2$ 

$ \because  - 2 \ne 4$ $ \therefore {\text{m}} = 4$  is not the solution.

Putting $ {\text{m}} = 5$  in L.H.S. $ 3(5) - 14 = 15 - 14 =  - 1$ 

$ \because 1 \ne 4$ $ \therefore {\text{m}} = 5$  is not the solution.

Putting $ {\text{m}} = 6$ in L.H.S. $ 3(6) - 14 = 18 - 14 = 4$ 

$ \because 4 = 4$ $ \therefore {\text{m}} = 6$ is the solution.

4. Write the equations for the following statements:

(i) The sum of numbers x and 4 is 9.

Ans: $ {\text{x}} + 4 = 9$

(ii) 2 subtracted from y is 8.

Ans: $ {\text{y}} - 2 = 8$

(iii) Ten times a is 70.

Ans: $ 10{\text{a}} = 70$

(iv) The number b divided by 5 gives 6.

Ans: $ \dfrac{{\text{b}}}{5} = 6$

(v) Three-fourth of t is 15.

Ans: $ \dfrac{{\text{3}}}{4}{\text{t}} = 15$

(vi) Seven times m plus 7 gets you 77.

Ans: $ 7{\text{m}} + 7 = 77$

(vii) One-fourth of a number x minus 4 gives 4.

Ans: $ \dfrac{{\text{1}}}{4}{\text{x}} - 4 = 4$

(viii) If you take away 6 from 6 times y, you get 60.

Ans: $ 6{\text{y}} - 6 = 60$

(ix) If you add 3 to one-third of z, you get 30.

Ans: $ \dfrac{1}{3}{\text{z}} + 3 = 30$

5. Write the following equations in statement form:

(i) $ {p + 4 = 15}$ 

Ans: Sum of numbers p and 4 gives you 15.

(ii) $ {m - 7 = 3}$ 

Ans: 7 subtracted from m gives number 3.

(iii) $ {2m = 7}$ 

Ans: Two times m equals to 7.

(iv) $ \dfrac{{m}}{{5}}{ = 3}$ 

Ans: One fifth of number m equals 5.

(v) $ \dfrac{{{3m}}}{{5}}{ = 6}$ 

Ans: Three fifth of m gives 6.

(vi) $ {3p  +  4  =  25}$ 

Ans: 4 added to three times p gives 25.

(vii) $ {4p - 2  = 18}$ 

Ans: 2 subtracted from four times p equals to 18.

(viii) $ \dfrac{{p}}{{2}}{ + 2 = 8}$ 

Ans: 2 added to half of p gives 8.

6. 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. (Tale m to be the number of Parmit’s marbles.)

Ans: Let m be the number of Parmit’s marbles:

$ \therefore 5{\text{m}} + 7 = 37$

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.)

Ans: Let age of Laxmi be y:

$ \therefore 3{\text{y}} + 4 = 49$

iii) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)

Ans: Let the lowest score in class be l:

$ \therefore 2{\text{l}} + 7 = 87$

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°.)

Ans: Let the base angle of isosceles triangle be b:

⇒ the vertex angle is 2b

$ \therefore {\text{b}} + {\text{b}} + 2{\text{b}} = 180^\circ $ 

$  \Rightarrow 4{\text{b}} = 180^\circ $

Class –VII Mathematics (Ex. 4.2)

1. Give first the step you will use to separate the variable and then solve the equations:

a) $ {x - 1 = 0}$ 

Ans:  

$   {\text{x}} - 1 + 1 = 0 + 1 $ 

$   \Rightarrow {\text{x}} = 1  $  

(Adding 1 both sides)

b) $ {x + 1 = 0}$ 

Ans:

$  {\text{x}} + 1 - 1 = 0 - 1 $ 

$  \Rightarrow {\text{x}} =  - 1 $  

(Subtracting 1 from both sides)

c) $ {x - 1 = 5}$ 

Ans:
$   {\text{x}} + 1 + 1 = 5 + 1 $ 

$   \Rightarrow {\text{x}} = 6  $  

(Adding 1 both sides)

d) $ {x + 6 = 2}$ 

Ans:

$   {\text{x}} + 6 - 6 = 2 - 6 $ 

$    \Rightarrow {\text{x}} =  - 4 $  

(Subtracting 6 from both sides)

e) $ {y - 4 =  - 7}$ 

Ans:

$   {\text{y}} - 4 + 4 =  - 7 + 4 $ 

$    \Rightarrow {\text{y}} =  - 3 $  

(Adding 4 both sides)

f) $ {y - 4 = 4}$ 

Ans:

$   {\text{y}} - 4 + 4 = 4 + 4 $

$    \Rightarrow {\text{y}} = 8 $  

(Adding 4 both sides)

g) $ \operatorname{y}  + 4 = 4$ 

Ans:

$   {\text{y}} + 4 - 4 = 4 - 4 $ 

$    \Rightarrow {\text{y}} = 0 $  

(Subtracting 4 from both sides)

h) $ \operatorname{y}  + 4 =  - 4$ 

Ans:

$   {\text{y}} + 4 - 4 =  - 4 - 4 $ 

$    \Rightarrow {\text{y}} =  - 8 $  

(Subtracting 4 from both sides)

2. Give first the step you will use to separate the variable and then solve the equations

a) $ 3l = 42$ 

Ans:

$   \dfrac{{3{\text{l}}}}{3} = \dfrac{{42}}{3} $ 

$    \Rightarrow {\text{l}} = 14 $  

(Dividing both sides by 3)

b) $ \dfrac{b}{2} = 6$ 

Ans:

$ \dfrac{{\text{b}}}{2} \times 2 = 6 \times 2 $ 

$ \Rightarrow {\text{b}} = 12 $  

(Multiplying both sides by 2)

c) $ \dfrac{p}{7} = 4$ 

Ans:

$ \dfrac{{\text{p}}}{7} \times 7 = 4 \times 7 $ 

$  \Rightarrow {\text{p}} = 28 $  

(Multiplying both sides by 7)

d) $ 4{\text{x}} = 25$ 

Ans:

$   \dfrac{{{\text{4x}}}}{4} = \dfrac{{25}}{4} $ 

$   \Rightarrow {\text{x}} = \dfrac{{25}}{4} $  

(Dividing both sides by 4)

e) $ 8{\text{y}} = 36$ 

Ans:

$   \dfrac{{8{\text{y}}}}{8} = \dfrac{{36}}{8} $ 

$    \Rightarrow {\text{y}} = \dfrac{9}{2} $  

(Dividing both sides by 8)

f) $ \dfrac{{\text{z}}}{3} = \dfrac{5}{4}$ 

Ans:

$  \dfrac{{\text{z}}}{3} \times 3 = \dfrac{5}{4} \times 3 $ 

$  \Rightarrow {\text{z}} = \dfrac{{15}}{4} $  

(Multiplying both sides by 3)

g) $ \dfrac{a}{5} = \dfrac{7}{{15}}$ 

Ans:

$  \dfrac{{\text{a}}}{5} \times 5 = \dfrac{7}{{15}} \times 5 $ 

$    \Rightarrow {\text{a}} = \dfrac{7}{3} $  

(Multiplying both sides by 5)

h) $ 20t =  - 10$ 

Ans:

$  \dfrac{{20{\text{t}}}}{{20}} = \dfrac{{ - 10}}{{20}} $ 

$ \Rightarrow {\text{t}} = \dfrac{{ - 1}}{2} $  

(Dividing both sides by 20)

3. Give first the step you will use to separate the variable and then solve the equations

(a) $ {3n - 2 = 46}$ 

Ans: $ 3{\text{n}} - 2 + 2 = 46 + 2$ (Adding 2 both sides)

$  \Rightarrow 3{\text{n}} = 48$

$  \Rightarrow \dfrac{{3{\text{n}}}}{3} = \dfrac{{48}}{3}$ (Dividing both sides by 3)

$  \Rightarrow {\text{n}} = 16$

(b) $ {5m + 7 = 17}$ 

Ans: $ 5{\text{m}} + 7 - 7 = 17 - 7$ (Subtracting 7 both sides)

$  \Rightarrow 5{\text{m}} = 10$ 

$  \Rightarrow \dfrac{{5{\text{m}}}}{5} = \dfrac{{10}}{5}$ (Dividing 5 both sides)

$  \Rightarrow {\text{m}} = 2$

(c) $ \dfrac{{{20p}}}{{3}}{ = 40}$ 

Ans: $ \dfrac{{20{\text{p}}}}{3} \times 3 = 40 \times 3$ (Multiplying both sides by 3)

$  \Rightarrow 20{\text{p}} = 120$ 

$  \Rightarrow \dfrac{{20p}}{{20}} = \dfrac{{120}}{{20}}$ (Dividing both sides by 20)

$  \Rightarrow {\text{p}} = 6$

(d) $ \dfrac{{{3p}}}{{{10}}}{ = 6}$ 

Ans: $ \dfrac{{3{\text{p}}}}{{10}} \times 10 = 6 \times 10$ (Multiplying both sides by 10)

$  \Rightarrow 3{\text{p}} = 60$ 

$  \Rightarrow \dfrac{{3{\text{p}}}}{3} = \dfrac{{60}}{3}$ (Dividing both sides by 3)

$  \Rightarrow {\text{p}} = 20$

4. Solve the following equation:

(a) $ {10p = 100}$ 

Ans: $ \dfrac{{10p}}{{10}} = \dfrac{{100}}{{10}}$ 

$  \Rightarrow {\text{p}} = 10$

(b) $ {10p + 10 = 100}$ 

Ans: $ 10{\text{p}} + 10 - 10 = 100 - 10$
$  \Rightarrow 10{\text{p}} = 90 $ 

$    \Rightarrow \dfrac{{10{\text{p}}}}{{10}} = \dfrac{{90}}{{10}} $ 

$  \Rightarrow {\text{p}} = 9 $

(c) $ \dfrac{{p}}{{4}}{ = 5}$ 

Ans: $ \dfrac{{\text{p}}}{4} \times 4 = 5 \times 4$ 

$  \Rightarrow {\text{p}} = 20$

(d) $ \dfrac{{{ - p}}}{{3}}{ = 5}$ 

Ans: $ \dfrac{{ - {\text{p}}}}{3} \times 3 = 5 \times 3$

$  \Rightarrow  - {\text{p}} = 15 $

$  \Rightarrow {\text{p}} =  - 15 $

(e) $ \dfrac{{{3p}}}{{4}}{ = 6}$ 

Ans: $ \dfrac{{{\text{3p}}}}{4} \times 4 = 6 \times 4$ 

$  \Rightarrow 3{\text{p}} = 24$ 

$  \Rightarrow \dfrac{{{\text{3p}}}}{3} = \dfrac{{24}}{3}$ 

$  \Rightarrow {\text{p}} = 8$

(f) $ {3s =  - 9}$ 

Ans: $ \dfrac{{3{\text{s}}}}{3} = \dfrac{{ - 9}}{3}$ 

$  \Rightarrow {\text{s}} =  - 3$

(g) $ {3s + 12 = 0}$ 

Ans: $ 3{\text{s}} + 12 - 12 = 0 - 12$ 

$  \Rightarrow 3{\text{s}} =  - 12$ 

$  \Rightarrow \dfrac{{3{\text{s}}}}{3} = \dfrac{{ - 12}}{3}$ 

$  \Rightarrow {\text{s}} =  - 4$

(h) $ {3s = 0}$ 

$  \Rightarrow \dfrac{{{\text{3s}}}}{3} = \dfrac{0}{3}$ 

$  \Rightarrow {\text{s}} = 0$

(i) $ {2q = 6}$ 

Ans: $ \dfrac{{2{\text{q}}}}{2} = \dfrac{6}{2}$ 

$  \Rightarrow {\text{q}} = 3$

(j) $ {\mathbf{2q - 6 = 0}}$ 

Ans: $ 2{\text{q}} - 6 + 6 = 0 + 6$ 

$  \Rightarrow 2{\text{q}} = 6$ 

$  \Rightarrow \dfrac{{2{\text{q}}}}{2} = \dfrac{6}{2}$ 

$  \Rightarrow {\text{q}} = 3$

(k) $ {\mathbf{2q + 6 = 0}}$ 

Ans: $ 2{\text{q}} + 6 - 6 = 0 - 6$ 

$  \Rightarrow 2{\text{q}} =  - 6$ 

$  \Rightarrow \dfrac{{2{\text{q}}}}{2} = \dfrac{{ - 6}}{2}$ 

$  \Rightarrow {\text{q}} =  - 3$

(l) $ {\mathbf{2q + 6 = 12}}$ 

Ans: $ 2{\text{q}} + 6 - 6 = 12 - 6$ 

$  \Rightarrow 2{\text{q}} = 6$ 

$  \Rightarrow \dfrac{{2{\text{q}}}}{2} = \dfrac{6}{2}$ 

$  \Rightarrow {\text{q}} = 3$

Class –VII Mathematics (Ex. 4.3)

1. Solve the following equations:

(a) $ {\mathbf{2y + }}\dfrac{{\mathbf{5}}}{{\mathbf{2}}}{\mathbf{ = }}\dfrac{{{\mathbf{37}}}}{{\mathbf{2}}}$ 

Ans: $ 2{\text{y}} + \dfrac{5}{2} = \dfrac{{37}}{2}$ 

$  \Rightarrow 2{\text{y}} = \dfrac{{37}}{2} - \dfrac{5}{2}$ 

$  \Rightarrow 2y = \dfrac{{37 - 5}}{2}$ 

$  \Rightarrow 2y = \dfrac{{32}}{2}$ 

$  \Rightarrow 2y = 16$ 

$  \Rightarrow y = \dfrac{{16}}{2}$ 

$  \Rightarrow y = 8$

(b) $ {\mathbf{5t + 28 = 10}}$ 

Ans: $  \Rightarrow 5{\text{t}} = 10 - 28$ 

$  \Rightarrow 5{\text{t}} =  - 18$ 

$  \Rightarrow {\text{t}} = \dfrac{{ - 18}}{5}$

(c) $ \dfrac{{\mathbf{a}}}{{\mathbf{5}}}{\mathbf{ + 3 = 2}}$ 

Ans: $ \dfrac{{\text{a}}}{5} + 3 = 2$ 

$  \Rightarrow \dfrac{{\text{a}}}{5} = 2 - 3$ 

$  \Rightarrow \dfrac{{\text{a}}}{5} =  - 1$ 

$  \Rightarrow {\text{a}} =  - 1 \times 5$ 

$  \Rightarrow {\text{a}} =  - 5$

(d) $ \dfrac{{\mathbf{q}}}{{\mathbf{4}}}{\mathbf{ + 7 = 5}}$ 

Ans: $  \Rightarrow \dfrac{{\text{q}}}{4} = 5 - 7$ 

$  \Rightarrow \dfrac{{\text{q}}}{4} =  - 2$ 

$  \Rightarrow {\text{q}} =  - 2 \times 4$ 

$  \Rightarrow {\text{q}} =  - 8$

(e) $ \dfrac{{\mathbf{5}}}{{\mathbf{2}}}{\mathbf{x = 10}}$ 

Ans: $ \dfrac{5}{2}{\text{x}} = 10$ 

$  \Rightarrow 5{\text{x}} = 10 \times 2$ 

$  \Rightarrow 5{\text{x}} = 20$ 

$  \Rightarrow {\text{x}} = \dfrac{{20}}{5}$ 

$  \Rightarrow {\text{x}} = 4$

(f) $ \dfrac{{\mathbf{5}}}{{\mathbf{2}}}{\mathbf{x = }}\dfrac{{{\mathbf{25}}}}{{\mathbf{4}}}$ 

Ans: $ \dfrac{5}{2}{\text{x}} = \dfrac{{25}}{4}$ 

$  \Rightarrow 5{\text{x}} = \dfrac{{25}}{4} \times 2$ 

$  \Rightarrow 5{\text{x}} = 25 \times 2$ 

$  \Rightarrow 5{\text{x}} = 50$

(g) $ {\mathbf{7m + }}\dfrac{{{\mathbf{19}}}}{{\mathbf{2}}}{\mathbf{ = 13}}$ 

Ans: $ 7{\text{m}} + \dfrac{{19}}{2} = 13$ 

$  \Rightarrow 7{\text{m}} = 13 - \dfrac{{19}}{2}$ 

$  \Rightarrow 7{\text{m}} = \dfrac{{26 - 19}}{2}$ 

$  \Rightarrow 7{\text{m}} = \dfrac{7}{2}$ 

$  \Rightarrow {\text{m}} = \dfrac{7}{2} \times \dfrac{1}{7}$ 

$  \Rightarrow {\text{m}} = \dfrac{1}{2}$

(h) $ {\mathbf{6z + 10 =  - 2}}$ 

Ans: $ 6z + 10 =  - 2$ 

$  \Rightarrow 6z =  - 2 - 10$ 

$  \Rightarrow 6z =  - 12$ 

$  \Rightarrow z = \dfrac{{ - 12}}{6}$ 

$  \Rightarrow z =  - 2$

(i) $\dfrac{{3l}}{2} = \dfrac{2}{3}$

Ans: $\dfrac{{3l}}{2} = \dfrac{2}{3}$

$  \Rightarrow 3l = \dfrac{2}{3} \times 2 $

$ \Rightarrow 3l = \dfrac{4}{3} $ 

$ \Rightarrow l = \dfrac{4}{3} \times \dfrac{1}{3} $ 

$ \Rightarrow l = \dfrac{4}{9} $

(j) $\dfrac{{{2b}}}{{3}} - {5 = 3}$

Ans: $\dfrac{{2b}}{3} - 5 = 3$

$ \Rightarrow \dfrac{{2b}}{3} = 3 + 5 $ 

$\Rightarrow \dfrac{{2b}}{3} = 8 $ 

$  \Rightarrow 2b = 8 \times 3 $ 

$   \Rightarrow 2b = 24 $ 

$  \Rightarrow b = 12 $

2. Solve the following equations:

(a) ${2(x + 4) = 12}$

Ans: $2(x + 4) = 12$

$ \Rightarrow x + 4 = \dfrac{{12}}{2} $ 

$ \Rightarrow x + 4 = 6 $ 

$  \Rightarrow x = 6 - 4 $ 

$ \Rightarrow x = 2 $

(b) ${3(n} - {5) = 21}$

Ans: $3(n - 5) = 21$

$  \Rightarrow n - 5 = \dfrac{{21}}{3} $ 

$   \Rightarrow n - 5 = 7 $ 

$ \Rightarrow n = 7 + 5 $ 

$ \Rightarrow n = 12 $

(c) ${3(n} - {5) = } - {21}$

Ans: $ 3(n - 5) =  - 21$

$    \Rightarrow n - 5 = \dfrac{{ - 21}}{3} $

$\Rightarrow n - 5 =  - 7 $ 

$  \Rightarrow n =  - 7 + 5 $ 

$   \Rightarrow n =  - 2 $

(d) ${3 - 2(2 - y) = 7}$

Ans: $3 - 2(2 - y) = 7$

$  \Rightarrow  - 2(2 - y) = 7 - 3 $

$ \Rightarrow (2 - y) = \dfrac{4}{{ - 2}} $ 

$  \Rightarrow 2 - y =  - 2 $

$ \Rightarrow  - y =  - 2 - 2 $ 

$ \Rightarrow  - y =  - 4 $

$  \Rightarrow y = 4 $

(e) ${ - 4(2 - x) = 9}$

Ans: $ - 4(2 - x) = 9$

$  \Rightarrow 2 - x = \dfrac{9}{{ - 4}} $ 

$ \Rightarrow  - x = \dfrac{{ - 9}}{4} - 2 $ 

$  \Rightarrow  - x = \dfrac{{ - 9 - 8}}{4} $ 

$  \Rightarrow  - x = \dfrac{{ - 17}}{4} $ 

$  \Rightarrow x = \dfrac{{17}}{4} $

(f) ${4(2 - x) = 9}$

Ans: $4(2 - x) = 9$

$   \Rightarrow 2 - x = \dfrac{9}{4} $ 

$  \Rightarrow  - x = \dfrac{9}{4} - 2 $ 

$ \Rightarrow  - x = 2\dfrac{1}{4} - 2 $ 

$ \Rightarrow  - x = \dfrac{1}{4} $ 

$   \Rightarrow x =  - \dfrac{1}{4} $

(g) ${4 + 5(p - 1) = 34}$

Ans: $4 + 5(p - 1) = 34$

$   \Rightarrow 5(p - 1) = 34 - 4 $ 

$   \Rightarrow 5(p - 1) = 30 $ 

$  \Rightarrow p - 1 = \dfrac{{30}}{5} $ 

$  \Rightarrow p - 1 = 6 $ 

$  \Rightarrow p = 6 + 1 $ 

$ \Rightarrow p = 7 $

(h) ${34 - 5(p - 1) = 4}$$$

Ans: $34 - 5(p - 1) = 4$

$    \Rightarrow  - 5(p - 1) = 4 - 34 $ 

$ \Rightarrow  - 5(p - 1) =  - 30 $ 

$ \Rightarrow p - 1 = \dfrac{{ - 30}}{{ - 5}} $ 

$ \Rightarrow p - 1 = 6 $ 

$ \Rightarrow p = 6 + 1 $ 

$ \Rightarrow p = 7 $

3. Solve the following equations:

(a) ${4 = 5(p - 2)}$

Ans: $4 = 5(p - 2)$

$  \Rightarrow \dfrac{4}{5} = p - 2 $ 

$  \Rightarrow \dfrac{4}{5} + 2 = p $ 

$  \Rightarrow \dfrac{{4 + 10}}{5} = p $ 

$ \Rightarrow \dfrac{{14}}{5} = p $

(b) ${ - 4 = 5(p - 2)}$

Ans: $ - 4 = 5(p - 2)$

$   \Rightarrow \dfrac{{ - 4}}{5} = (p - 2) $ 

$ \Rightarrow \dfrac{{ - 4}}{5} + 2 = p $ 

$\Rightarrow \dfrac{{ - 4 + 10}}{5} = p $ 

$  \Rightarrow \dfrac{6}{5} = p $

(c) ${ - 16 =  - 5(2 - p)}$

Ans: $ - 16 =  - 5(2 - p)$ 

$ \Rightarrow \dfrac{{ - 16}}{{ - 5}} = (2 - p) $ 

$ \Rightarrow \dfrac{{16}}{5} - 2 = p $ 

$  \Rightarrow \dfrac{{16 - 10}}{5} = p $ 

$  \Rightarrow \dfrac{6}{5} = p $

(d) ${10 = 4 + 3(t + 2)}$

Ans: $10 = 4 + 3(t + 2)$

$   \Rightarrow 10 - 4 = 3(t + 2) $ 

$ \Rightarrow \dfrac{6}{3} = t + 2 $ 

$   \Rightarrow 2 = t + 2 $ 

$  \Rightarrow 2 - 2 = t $ 

$ \Rightarrow t = 0 $

(e) ${28 = 4 + 3(t + 5)}$

Ans: $28 = 4 + 3(t + 5)$

$  \Rightarrow 28 - 4 = 3(t + 5) $ 

$ \Rightarrow \dfrac{{24}}{3} = t + 5 $ 

$ \Rightarrow 8 - 5 = t $ 

$  \Rightarrow 3 = t $

(f) ${0 = 16 + 4(m - 6)}$

Ans: $0 = 16 + 4(m - 6)$

$   \Rightarrow 0 - 16 = 4(m - 6) $ 

$  \Rightarrow \dfrac{{ - 16}}{4} = m - 6 $ 

$\Rightarrow  - 4 + 6 = m $ 

$ \Rightarrow 2 = m $

4. (a) Construct 3 equations starting with ${x = 2}$

(i) $x = 2$

Adding 5 to both sides, we get:

$  x + 5 = 2 + 5 $ 

$ \Rightarrow x + 5 = 7 $

(ii) $x = 2$

Multiplying both sides by 11, we get:

$11x = 22$

(iii) $x = 2$

Multiplying both sides by 11, we get:

$11x = 22$

Adding 5 to both sides, we get:

$11x + 5 = 27$

(b) Construct 3 equations starting with ${x =  - 2}$

(i) $x =  - 2$

Adding 3 to both sides, we get:

$x + 3 = 1$

Dividing both sides by 5, we get

$\dfrac{{x + 3}}{5} = \dfrac{1}{5}$

(ii) $x =  - 2$

Adding 2 to both sides, we get:

$x + 2 = 0$

Multiplying both side by 100, we get:

$  100(x + 2) = 100 \times 0 $ 

$ \Rightarrow 100x + 200 = 0 $

(iii) $x =  - 2$

Subtracting 4 from both sides, we get:

$  x - 4 =  - 2 - 4 $

$  \Rightarrow x - 4 =  - 6 $

Class-VII Mathematics (Ex 4.4)

1. Setup equations and solve them to find the unknown numbers in the following cases:

(a) Add 4 to eight times a number; you get 60.

Ans: Let the number be x, we get the equation:

$  8x + 4 = 60 $ 

$ \Rightarrow 8x = 60 - 4 $ 

$ \Rightarrow 8x = 56 $ 

$   \Rightarrow x = \dfrac{{56}}{8} $ 

$   \Rightarrow x = 7 $

(b) One fifth of a number minus 4 gives 3.

Ans: Let the number be x, we get the equation:

$  \dfrac{x}{5} - 4 = 3 $ 

$\Rightarrow \dfrac{x}{5} = 3 + 4 $ 

$\Rightarrow \dfrac{x}{5} = 7 $

$\Rightarrow x = 7 \times 5 $ 

$\Rightarrow x = 35 $

(c) If I take three-fourth of a number and add 3 to it, I get 21.

Ans: Let the number be x, we get the equation:

$  \dfrac{3}{4}x + 3 = 21 $ 

$\Rightarrow \dfrac{3}{4}x = 21 - 3 $ 

$ \Rightarrow \dfrac{3}{4}x = 18 $ 

$ \Rightarrow x = 18 \times \dfrac{4}{3} $ 

$  \Rightarrow x = 24 $

(d) When I subtracted 11 from twice a number, the result was 15.

Ans: Let the number be x, we get the equation:

$  2x - 11 = 15 $ 

$  \Rightarrow 2x = 15 + 11 $ 

$\Rightarrow 2x = 26 $ 

$ \Rightarrow x = \dfrac{{26}}{2} $ 

$ \Rightarrow x = 13 $

(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

Ans: Let the number of notebooks be x, we get the equation:

$  50 - 3x = 8 $ 

$  \Rightarrow  - 3x = 8 - 50 $ 

$  \Rightarrow  - 3x =  - 42 $ 

$   \Rightarrow x = \dfrac{{ - 42}}{{ - 3}} $ 

$  \Rightarrow x = 14 $

(f) Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.

Ans: Let the number be x, we get the following equation:

$  \dfrac{{x + 19}}{5} = 8 $ 

$  \Rightarrow x + 19 = 8 \times 5 $ 

$  \Rightarrow x + 19 = 40 $ 

$   \Rightarrow x = 40 - 19 $ 

$   \Rightarrow x = 21 $

(g) Answer thinks of a number. If he takes away ${7}$ from $\dfrac{{5}}{{2}}$of the number, the result is $\dfrac{{{11}}}{{2}}$.

Ans: Let the number be x, we get the equation:

$  \dfrac{{5x}}{2} - 7 = \dfrac{{11}}{2} $ 

$\Rightarrow \dfrac{{5x}}{2} = \dfrac{{11}}{2} + 7 $ 

$  \Rightarrow \dfrac{{5x}}{2} = \dfrac{{11 + 14}}{2} $ 

$ \Rightarrow \dfrac{{5x}}{2} = \dfrac{{25}}{2} $ 

$  \Rightarrow 5x = 25 $ 

$ \Rightarrow x = \dfrac{{25}}{5} $ 

$ \Rightarrow x = 5 $

2. Solve the following:

(a) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87

Ans: Let the lowest score in class be x:

$ \therefore 2{\text{x}} + 7 = 87$

$   \Rightarrow 2x = 87 - 7 $

$ \Rightarrow 2x = 80 $ 

$\Rightarrow x = \dfrac{{80}}{2} $

$ \Rightarrow x = 40 $

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°.)

Ans: Let one of the base angles be x.

$  \therefore x + x + 40^\circ  = 180^\circ  $

$ \Rightarrow 2x = 180^\circ  - 40^\circ  $ 

$  \Rightarrow 2x = 140^\circ  $ 

$  \Rightarrow x = \dfrac{{140^\circ }}{2} $ 

$ \Rightarrow x = 70^\circ  $

(c) Sachin scored twice as many as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

Ans: Let Rahul’s score be x.

$ \Rightarrow $Sachin’s score is 2x.

$  \therefore x + 2x = 200 - 2 $

$  \Rightarrow 3x = 198 $

$ \Rightarrow x = \dfrac{{198}}{3} $

$  \Rightarrow x = 66 $

$  \therefore 2x = 132 $

Rahul’ score is 66

Sachin’s score is 132.

3. Solve the following:

(a) 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?

Ans: Let the number of marbles Parmit has, be x

$  \therefore 5x + 7 = 37 $

$   \Rightarrow 5x = 37 - 7 $

$\Rightarrow 5x = 30 $

$  \Rightarrow x = \dfrac{{30}}{5} $ 

$  \Rightarrow x = 6 $

Hence, Parmit has a total of 6 marbles.

(b) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

Ans: Let the age of Laxmi be x,

$ \Rightarrow 3x + 4 = 49 $ 

$ \Rightarrow 3x = 49 - 4 $ 

$  \Rightarrow 3x = 45 $ 

$ \Rightarrow x = \dfrac{{45}}{3} $ 

$ \Rightarrow x = 15 $

Hence, Laxmi is 15 years old.

(c) People of Sundergram planted a total of 102 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?

Ans: Let the number of fruit trees be x,

$\therefore $number of non-fruit trees is 3x+2

$  \Rightarrow x + 3x + 2 = 102 $

$ \Rightarrow 4x + 2 = 102 $ 

$  \Rightarrow 4x = 102 - 2 $ 

$  \Rightarrow 4x = 100 $ 

$ \Rightarrow x = \dfrac{{100}}{4} $ 

$  \Rightarrow x = 25 $

Hence, number of fruit trees planted is 25.

4. 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!

Ans: Let the number be x,

“Take me seven times over, And add a fifty”

$ \Rightarrow 7x + 50$

“To reach a triple century, You still need forty!”

$ \Rightarrow $Result of the equation is :

$300 - 40 = 260$

Hence the final equation becomes:

$7x + 50 = 260$

Solving for the equation:

$7x+50=260$

$\Rightarrow 7x=260-50$

$\Rightarrow 7x=210$

$ \Rightarrow x = \dfrac{{210}}{7} $ 

$ \Rightarrow x = 30 $

NCERT Solutions for Class 7 Maths Chapter 4  Free PDF Download

NCERT Solutions for Class 7 Maths Chapter 4 Exercises

Facts

• A variable takes on different numerical values whereas a constant has a fixed value.

• An equation is a statement of a variable in which two expressions in the variable should have equal value.

• An equation remains unchanged if its L.H.S and R.H.S are interchanged.

• Transposing a number means moving it to the other side.

• The equations remain unchanged when we:

• Add the same number to both sides.

• Subtract the same number from both sides.

• Multiply both sides by the same number.

• Divide both sides by the same number.

• When we transpose a number from one side of the equation to the other side its sign is changed.

Variable

As we just read that variable is not fixed which means the numerical value of the variable changes. These variables are denoted by letters of the alphabet such as l, m, n, p, q, r, s, u, v, x, y, z, etc. When we perform operations like addition, subtraction, multiplication, and division on variables then expressions are formed.

• The value of an expression depends upon the chosen value of the variable. If there is only one term in an expression then it is called a monomial expression.

• If there are two terms in an expression then it is a binomial expression.

• If there are three terms in an expression then it is a trinomial expression.

• A polynomial expression is an expression that has four terms.

Note: A polynomial expression can have many terms but none of the terms can have a negative exponent for any variable.

An Equation

An equation is a mathematical statement on a variable where two expressions on either side of the equality sign should have equal value. At least one of the expressions must contain the variable.

Note: An equation does not change when the expression on the left-hand side and on the right-hand side is interchanged.

In an equation, there is always an equality sign between two expressions.

Example:  Write the following statements in the form of equations.

1. The difference of five times x and 11 is 28.

2. One-fourth of a member minus 8 is 18.

Solution:

i) We have, five times x that is 5x.

Difference of 5x – 11 is 5x -11

∴ 5x – 11 = 28

Thus, the required equation is: 5x – 11 = 28

ii) Let the number be x.

∴ One-fourth of x is ¼ (x).

Now, one-fourth of x minus 8 is ¼(x) – 8.

Thus, the required equation is: ¼  (x) – 8 = 18

Let us see one more example, which will help you with the Exercise 4.1 of NCERT Solutions Chapter 4.

Example: Write a statement for the equations: 2x – 5 = 15

Solution: 2x – 5 = 15

Taking away 5 from twice a number is 15.

Solving an Equation

An equation is like a weighing balance having equal weights on both of its pans such that the arm of the balance is exactly horizontal. The horizontal arm is not disturbed if the same amount of weights are added to both pans (or the same amount of weights are removed from both the pans).

We use this principle when we solve an equation. The equality sign between the L.H.S and R.H.S corresponds to the horizontal beam (arm) of a balance.

An equation remains undisturbed or unchanged:

1. If L.H.S and R.H.S are interchanged.

2. To both the sides, if the same number is added. 

3. From both the sides, if the same number is subtracted.

4. When both L.H.S and R.H.S are multiplied by the same number.

5. If both L.H.S and R.H.S  are divided by the same number.

To understand the concept  better, let us try an example.  This will help you with Exercise 4.2 of NCERT Solutions Chapter 4.

Example: Solve 5x – 3 = 12

Solution: 5x – 3 = 12

Adding 3 to both sides, we have

5x – 3 + 3 = 12 + 3

5x  = 15

Dividing both side by 5, we have: 5x/5 = 15/5

x = 3, which is the required solution.

Note: For checking the corrections of the answer, we substitute the value of the variable in the given equation.

i.e., L.H.S = (5 x 3) - 3 = 15 - 3 = 12 = R.H.S

or   L.H.S = R.H.S

Example: Solve ½ (x) + 5 = 65

Solution: ½ (x) + 5 = 65

Subtracting 5 from both sides, we have

½ (x) + 5 - 5 = 65 - 5

½ (x) = 60

Multiplying both sides by 2, we have

[½ (x)] x 2 = 60 x 2 or x = 120

x = 120 is the required solution.

Forming an Equation

So far we have learned to solve an equation. Now, we shall form (or construct) the equation when its solution (root) is given. Let us know the following successive steps:

1. Start with x = 9

2. Multiply both side by 3,

3x = 27

3. Subtract 2 from both sides,

3x -2 = 27 - 2

3x - 2 = 25, which is an equation.

Note: For a given equation, you get one solution; but for a given solution one can make many equations.

Let us understand this with some more examples so that you can solve the Exercise 4.3 of NCERT Solutions Chapter 4.

Example: Solve 5 ( x - 3 ) = 25

Solution: 5 (x - 3) = 25

(or) x - 3   = 25/5    (Dividing both side by 5)

(or) x - 3   =  5

(or) x =  5 + 3        (Transposing -3 to R.H.S)

          =  8

Thus, x = 8 is the required solution.

Example: Solve 3 (x + 1)/2 = 18

Solution: 3 ( x + 1)/2 = 18

(or) (x + 1)/2 = 18/3 (Dividing both sides by 3)

(or) (x + 1)/2 = 6

(or) x/2 = (6 - 1)/2  (Transposing  1 to R.H.S)

(or) x = (12 - 1)/2 = 11/2

∴ x = 11/2 is the required solution.

Application of Simple Equations to Practical Situations

Let us understand this with some more examples so that you can solve the Exercise 4.4 of NCERT Solutions Chapter 4.

Example: The sum of five times a number and 18 is 63. Find the number.

Solution: Let the required number be ‘ x ‘.

5 times the number = 5x

∴ According to the condition, we have

5x + 18 = 63

5x = 63 – 18 (Transposing 18 from L.H.S to R.H.S)

5x = 45

(or) Dividing both sides by 5, we have

5x/5 = 45/5

x = 9

Thus, the required number = 9

Significances of NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations

The significance of NCERT Solutions for Class 7 Maths Chapter 4, "Simple Equations," is paramount in the realm of mathematics education. These solutions play a pivotal role in helping students grasp the fundamental concepts of solving equations, a skill that forms the basis of more complex mathematical concepts in later grades. They provide step-by-step explanations and worked-out examples, ensuring that students understand the underlying principles and techniques required to solve equations confidently. Moreover, these solutions offer clarity and practice, enabling students to build a strong foundation in mathematics. Beyond acade mic excellence, they nurture problem-solving skills, critical thinking, and logical reasoning, which are valuable assets both inside and outside the classroom. In essence, NCERT Solutions for this chapter empower students to navigate the world of mathematics with confidence and competence.

Conclusion 

NCERT Solutions for Class 7 Maths Chapter 4, "Simple Equations," are indispensable resources in the journey of mathematical learning. These solutions provide a structured and comprehensive approach to understanding the essential concepts of equation-solving. They offer step-by-step guidance and practical examples that aid students in developing a strong foundation in mathematical problem-solving. Beyond the classroom, these solutions promote critical thinking and logical reasoning skills, which are invaluable in various aspects of life. Moreover, they enhance students' confidence and competence in tackling mathematical challenges. In essence, NCERT Solutions for this chapter serve as a catalyst for mathematical proficiency, equipping students with essential skills for their academic journey and beyond.