Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions Class 7 Maths Chapter 4 Simple Equations

Last updated date: 17th Jul 2024
Total views: 845.4k
Views today: 14.45k

## NCERT Solutions for Maths Chapter 4 Simple Equations Class 7- FREE PDF Download

NCERT Chapter 4 of Maths, Simple Equations Class 7, introduces students to the fundamental concept of forming and solving equations. According to the latest CBSE Class 7 Maths Syllabus, this chapter is essential as it lays the groundwork for understanding algebraic expressions and their applications. It explains how to set up equations from given statements and solve NCERT Solutions for Maths Class 7  to find the value of unknown variables.

Table of Content
1. NCERT Solutions for Maths Chapter 4 Simple Equations Class 7- FREE PDF Download
2. Glance on Maths Chapter 4 Class 7 - Simple Equations
3. Access Exercise wise NCERT Solutions for Chapter 4 Maths Class 7
4. Exercises Under NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations
5. Access NCERT Solutions for Class 7 Maths Chapter 4 – Simple Equations
5.1Exercise 4.1
5.2Exercise 4.2
5.3Exercise 4.3
6. Overview of Deleted Syllabus for CBSE Class 7 Maths Simple Equations
7. Class 7 Maths Chapter 4: Exercises Breakdown
8. Other Study Material for CBSE Class 7 Maths Chapter 4
9. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs

Students should focus on understanding the process of balancing equations and practising various problems to build confidence. Vedantu's NCERT Solutions provides detailed explanations and step-by-step methods to solve each problem, ensuring that students grasp the concepts thoroughly. Mastering these skills will help students in solving more complex algebraic problems in higher classes.

## Glance on Maths Chapter 4 Class 7 - Simple Equations

• NCERT Solutions for Maths Chapter 4 Simple Equations explains the concept of simple equations and their applications.

• A simple equation is a mathematical statement that shows the equality of two expressions with an unknown variable.

• Forming equations involves translating word problems into mathematical statements using variables.

• Solving equations means finding the value of the unknown variable that makes the equation true.

• Balancing equations refers to performing the same mathematical operation on both sides to maintain equality.

• The addition method involves adding the same number to both sides of the equation to isolate the variable.

• The subtraction method involves subtracting the same number from both sides of the equation to isolate the variable.

• Multiplication and division methods involve multiplying or dividing both sides of the equation by the same number to solve for the variable.

• This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 4 - Simple Equations Class 7, which you can download as PDFs.

• There are three exercises (14 fully solved questions) in NCERT Class 7 Maths Chapter 4 solutions pdf.

## Access Exercise wise NCERT Solutions for Chapter 4 Maths Class 7

 Current Syllabus Exercises of Class 7 Maths Chapter 4 NCERT Solutions of Class 7 Maths Simple Equations Exercise 4.1 NCERT Solutions of Class 7 Maths Simple Equations Exercise 4.2 NCERT Solutions of Class 7 Maths Simple Equations Exercise 4.3

## Exercises Under 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:

 S.No. Equation Value Say, whether the Equation is satisfied. (Yes / No) (i) ${\text{x}} + 3 = 0$ ${\text{x}} = 3$ (ii) ${\text{x}} + 3 = 0$ ${\text{x}} = 0$ (iii) ${\text{x}} + 3 = 0$ ${\text{x}} = - 3$ (iv) ${\text{x}} - 7 = 1$ ${\text{x}} = 7$ (v) ${\text{x}} - 7 = 1$ ${\text{x}} = 8$ (vi) $5{\text{x}} = 25$ ${\text{x}} = 0$ (vii) $5{\text{x}} = 25$ ${\text{x}} = 5$ (viii) $5{\text{x}} = 25$ ${\text{x}} = - 5$ (ix) $\dfrac{{\text{m}}}{3} = 2$ ${\text{m}} = - 6$ (x) $\dfrac{{\text{m}}}{3} = 2$ ${\text{m}} = 0$ (xi) $\dfrac{{\text{m}}}{3} = 2$ ${\text{m}} = 6$

Ans:

 (i) ${\text{x}} + 3 = 0$ ${\text{x}} = 3$ No (ii) ${\text{x}} + 3 = 0$ ${\text{x}} = 0$ No (iii) ${\text{x}} + 3 = 0$ ${\text{x}} = - 3$ Yes (iv) ${\text{x}} - 7 = 1$ ${\text{x}} = 7$ No (v) ${\text{x}} - 7 = 1$ ${\text{x}} = 8$ Yes (vi) $5{\text{x}} = 25$ ${\text{x}} = 0$ No (vii) $5{\text{x}} = 25$ ${\text{x}} = 5$ Yes (viii) $5{\text{x}} = 25$ ${\text{x}} = - 5$ No (ix) $\dfrac{{\text{m}}}{3} = 2$ ${\text{m}} = - 6$ No (x) $\dfrac{{\text{m}}}{3} = 2$ ${\text{m}} = 0$ No (xi) $\dfrac{{\text{m}}}{3} = 2$ ${\text{m}} = 6$ Yes

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:

${\text{x}} - 1 + 1 = 0 + 1$

$\Rightarrow {\text{x}} = 1$

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

$\Rightarrow {\text{x}} = 6$

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:

${\text{y}} - 4 + 4 = - 7 + 4$

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

f) ${y - 4 = 4}$

Ans:

${\text{y}} - 4 + 4 = 4 + 4$

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

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

 Chapter Dropped Topics Simple Equations Exercise 4.3 4.6 From solution to equation

## Class 7 Maths Chapter 4: Exercises Breakdown

 Exercise Number of Questions Exercise 4.1 6 Questions & Solutions Exercise 4.2 3 Questions & Solutions Exercise 4.3 4 Questions & Solutions

## 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.

## Other Study Material for CBSE Class 7 Maths Chapter 4

 S. No Important Links for Chapter 4 Simple Equations 1 Class 7 Simple Equations Important Questions 2 Class 7 Simple Equations Revision Notes 3 Class 7 Simple Equations NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 7 Maths Chapter 4 Simple Equations

1. If Both Sides of an Equation are Divided by the same (Non-Qero) Quantity, the Equality.

• Does not change.

• Changes.

• May or may not change.

• None of these.

• Does not change.

2. If 2x – 3 = 5 , then

• x = 1

• x = -1

• x = 4

• x = - 4

• x = 4

3. How do you Avail of the Study Materials for NCERT Solutions for Class 7 Maths?

NCERT Solutions for Class 7 Maths and its study material per topic are available on Vedantu’s portal. They are available in pdf format and you can download them on your phone, computers, laptop, and any other device. The teachers have crafted the NCERT Solutions Maths in step-to-step method and they are self-explanatory. The students can revise and solve many questions to master the topic.

4. What are the Important Topics Covered in NCERT Solutions Class 7 Maths Chapter 4?

Vedantu provides solutions of Class 7 Maths Chapter 4 in its portal covering all the concepts in a precise as well as a detailed manner. The content is prepared by Maths subject matter experts and is easy to grasp. The syllabus and exercises are completely set according to the CBSE guidelines. A few of the important topics covered under Chapter 7 are given below :

• Constants and Variables

• L.H.S and R.H.S

• Equations.

5. How Many Questions are there in NCERT Solutions Class 7 Maths Chapter 4  Simple equations?

There are a variety of questions found in the NCERT Solutions for  Class 7 Maths, Chapter 4, Simple Equations. This chapter consists of four exercises. In Exercise 4.1, there are a total of six questions, Exercise 4.2 has four questions, Exercise 4.3 has four questions and lastly Exercise 4.4 has  four questions. These questions are available on the Vedantu portal, which can be accessed easily as it is free of cost. You can download NCERT Solutions of Class 7 Maths Chapter 4 for future use and can learn anytime.

6. In Chapter 4 Simple Equations of Class 7 Maths, is there any theory asked?

No, in Chapter 4 Simple Equations there is no theory asked. The series of questions asked from this Chapter is mainly practical. Even if a theory comes, it won’t be in long format, it could be either one-liner statements or one-word type. So, your main focus should be on solving, not on learning the theory. Numericals are of more weightage in Simple Equations. Practise is the best solution to have a grip of the topics in an accurate way.  Since the numericals are to be solved accurately the concepts have to be cleared.

7. How to remember the difficult formulae of simple equations?

Simple Equations holds maximum marks in the examination. If you want to remember the difficult formulas, then try to understand them. Solve a question and analyse it by yourself. If you mug up all the formulas, then you won’t gain anything. Start with knowing how the formulas have been structured from step one. This way you cannot forget and if you forget you will know the method of derivation. When you learn in this method then you will never forget it.

8. What should I do to solve the numerical problems of Chapter 4 of Class 7 Maths quickly?

To solve numerical problems quickly, there is only one solution, which is ‘practice’. Make numbers of your daily friends, and try to play with them regularly. Devote an ample number of hours to create a friendship with them. Don’t just think priorly that numerical problems are tough, instead be confident that you can easily solve them. Also, you can have a look at Vedantu’s solutions available on the Vedantu app or website, where numerical problems are solved in simple steps.

10. How to introduce simple equations in NCERT Class 7 Maths Chapter 4 solutions pdf?

Simple equations can be introduced by explaining that they are mathematical statements showing the equality of two expressions. Start with real-life examples, such as balancing weights on a scale, to illustrate the concept. Use simple word problems to demonstrate how to form equations using variables. Highlight the importance of finding the value of the unknown variable to make the equation true. Practice basic examples to build a strong foundation.

11. What is the transposing method in NCERT Class 7 Maths Chapter 4 Solutions pdf?

The transposing method involves moving a term from one side of an equation to the other by changing its sign. For example, if you have x+5=12, you can transpose 5 to the other side, resulting in x=12−5. This method helps in isolating the variable to solve the equation. It is crucial to maintain the balance of the equation while transposing terms. Practicing this method simplifies solving equations in ncert class 7 maths chapter 4 solutions pdf.

12. What is algebra in class 7 math chapter 4?

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. In class 7 math chapter 4, students learn to use variables to represent numbers and write algebraic expressions. Algebra involves operations such as addition, subtraction, multiplication, and division with variables. It includes forming and solving equations, understanding patterns, and working with expressions. Algebra provides tools to solve problems involving unknown values.

13. What is a coefficient in simple equations for class 7 pdf?

A coefficient is a numerical factor that multiplies a variable in an algebraic expression. For example, in the expression 3x, 3 is the coefficient of the variable x. Coefficients can be positive, negative, or fractional numbers. They indicate how many times the variable is being multiplied. Understanding coefficients in simple equations for class 7 pdf is important for simplifying and solving algebraic expressions. Recognizing coefficients helps in performing algebraic operations correctly.

14. What is the formula for integers in chapter 4 class 7 maths?

There isn't a single formula for integers in chapter 4 class 7 maths, but students learn various properties and operations involving integers. Key concepts include addition, subtraction, multiplication, and division of integers. For example, $\left ( -a \right )+\left ( -b \right )= -\left ( a+b \right )$ illustrates the addition of negative integers. Understanding the rules for combining positive and negative numbers is essential. Students also learn about the distributive property and the use of integers in equations.

15. What is the formula for integers in Class 7 Chapter 4?

As mentioned, there isn't a specific formula for integers. Instead, students focus on the rules and properties governing integer operations. These include addition, subtraction, multiplication, and division rules for positive and negative numbers. For instance, multiplying two negative integers results in a positive product. Practicing these operations helps students handle integers effectively in various mathematical contexts. Understanding these rules is crucial for solving problems involving integers in Class 7 Chapter 4.