NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

• Exercise 4.1 focuses on the basics of forming simple equations from given statements. Students learn to translate word problems into algebraic equations by identifying the unknown variables and defining them clearly. This exercise includes various problems that require setting up equations from everyday situations, helping students understand how to convert real-life scenarios into mathematical expressions. By practising these problems, students get a strong foundation in recognizing and formulating equations.

• Exercise 4.2 deals with solving simple equations using the balancing method. This exercise teaches students how to manipulate equations by performing the same operations on both sides to maintain equality. Problems in this exercise require adding, subtracting, multiplying, or dividing both sides of the equation to isolate the variable. Additionally, students practice checking their solutions by substituting the values back into the original equations to ensure their answers are correct. This step-by-step approach helps reinforce the concept of maintaining balance in equations.

• Exercise 4.3 emphasizes solving equations that involve more complex operations. Students encounter problems that require multiple steps to isolate the variable, including combining like terms and using the distributive property. This exercise builds on the skills learned in previous exercises, challenging students to apply their understanding of balancing equations to solve more intricate problems. By mastering these problems, students develop confidence in handling more sophisticated algebraic equations.

Access NCERT Solutions for Class 7 Maths Chapter 4 – Simple Equations

Exercise 4.1

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 $

Exercise 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$

Exercise 4.3

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) Anwar 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. What is the Lowest score?

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:

(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?

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.

(ii) 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.

(iii) People of Sundargram 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?

Ans: Let the number of planted fruit trees be x.

From given question we can say that, the number of non-fruit trees = 3x+2 and that is equal to 77

$ \Rightarrow 3x + 2\, = \,77$

$\begin{gathered} \Rightarrow 3x\, = \,77 - 2 \hfill \\ \Rightarrow 3x\, = \,75 \hfill \\ \Rightarrow x\, = \,\frac{{75}}{3} \hfill \\ \Rightarrow x\, = \,25 \hfill \\ \end{gathered} $

$\therefore $The number of planted fruit trees in 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 $

Overview of Deleted Syllabus for CBSE Class 7 Maths Simple Equations

Class 7 Maths Chapter 4: Exercises Breakdown

Conclusion

NCERT Chapter 4 - Simple Equations is a fundamental part of Class 7 Maths, helping students understand how to form and solve equations. It is crucial to focus on the basics of setting up equations from word problems and mastering the balancing method to solve them. By practicing these concepts, students can build a strong foundation for more advanced algebraic topics. In previous year's question papers, around 3 to 4 questions were asked from this chapter. NCERT Class 7 Maths Chapter 4 Solutions pdf provides detailed explanations and step-by-step guidance to help students excel in their exams and understand the concepts thoroughly.

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