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

Simple equations are an important part of algebra and help students learn how to find the value of unknown numbers using basic operations like addition, subtraction, multiplication, and division. This chapter introduces students to the concept of forming and solving simple equations in a step-by-step way, making it easier to understand. The NCERT Solutions provided for this chapter give students a thorough understanding of each type of problem and help them build a strong foundation in algebra.

Latest: NCERT Class 7 Maths: Chapterwise Important Formulas with Examples and Points

This Story also Contains

1. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Points
2. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation
3. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation (Exercise)
4. Simple Equations Class 6 Maths Chapter 4 -Topics
5. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - Points to Remember
6. NCERT Solutions for Class 7 Maths Chapter Wise

These solutions are created by subject experts at Careers360 to make learning easier for students. The explanations are clear and simple, helping students understand the method behind solving equations. To access all the solutions for all the chapters, students can visit the NCERT Solutions for Class 7 Maths.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Points

Variable: Something that can vary, its value is not fixed, such as x, y, z, p, q, r, etc.

Equation: It is a mathematical statement that uses an equal sign(=) to show that the value of the expression to the left of the sign (LHS) is equal to the value of the expression to the right of the sign (RHS).

Interchangeability: If the LHS and RHS are interchanged, the equation remains the same.

Balanced equation: the value of LHS remains equal to the value of the RHS. If 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.

Solution of an equation: The value of the variable for which the equation is satisfied ( for that variable, LHS = RHS ).

Transposing: Transposing a number has the same effect as adding the same number to both sides or subtracting the same number from both sides of the equation.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation

Download PDF

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation (Exercise)

1. Complete the last column of the table.

Answer:

The table is shown below:-

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

(a) n + 5 = 19 (n = 1)

(b) 7n + 5 = 19 (n = – 2)

(c) 7n + 5 = 19 (n = 2)

(d) 4p – 3 = 13 (p = 1)

(e) 4p – 3 = 13 (p = – 4)

(f) 4p – 3 = 13 (p = 0)

Answer:

(a) Put n = 1 in the equation, we have :

n + 5 = 15

or 1 + 5 = 15

or 6 _$\ne$ 15

Thus, n = 1 is not a solution.

(b) Put n = - 2, we have :

7n + 5 = 19

or 7(-2) + 5 = - 14 + 5 = - 9 _$\ne$ 19.

So, n = - 2 is not a solution to the given equation.

(c) Put n = 2, we have :

7n + 5 = 19

or 7(2) + 5 = 14 + 5 = 19 = R.H.S

Thus, n = 2 is the solution for the given equation.

(d) Put p = 1 , we have:

4p - 3 = 13

or 4(1) - 3 = 1 _$\ne$ 13 .

Thus, p = 1 is not a solution.

(e) Put p = - 4, we get :

4p - 3 = 13

or 4(1) - 3 = 1 _$\ne$ 13 .

Thus, p = 1 is not a solution.

(f) Put p = 0 , we get :

4p - 3 = 13

or 4(0) - 3 = - 3 _$\ne$ 13 .

Thus, p = 0 is not a solution.

3. Solve the following equations by the trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4

Answer:

(i) Put p = 1,

We have: $5(1)+2=7\ne 17$

Put p = 2,

We have: $5(2)+2=12\ne 17$

Put p = 3,

we have: $5(3)+2=17=17$

Thus, the solution is p = 3.

(ii) Put m = 4,

we have : $3(4)-14=-2\ne 4$

Put m = 5,

we have: $3(5)-14=1\ne 4$

Now, put m = 6,

we have: $3(6)-14=4=4$

Thus, m = 6 is the solution.

4. 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-fourths 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.

Answer:

The equations are given below :

(i) $x+4=9$

(ii) $y-2=8$

(iii) $10a=70$

(iv) $\frac{b}{5}=6$

(v) $\frac{3}{4}t=15$

(vi) $7m+7=77$

(vii) $\frac{x}{4}-4=4$

(viii) $6y-6=60$

(ix) $\frac{z}{3}+3=30$

5. Write the following equations in statement form:

(i) p + 4 = 15

(ii) m – 7 = 3

(iii) 2m = 7

(iv) m /5 = 3

(v) 3 m/5 = 6

(vi) 3p + 4 = 25

(vii) 4p – 2 = 18

(viii) p /2 + 2 = 8

Answer:

(i) Add 4 to the number p, and we get 15.

(ii) Subtract 7 from m to get 3.

(iii) Twice the number m is 7.

(iv) One-fifth of m is 3.

(v) Three-fifths of m is 6.

(vi) 4 is added to thrice the number p to get 25.

(vii) 2 is subtracted from the product of 4 times p to get 18.

(viii) When 2 is added to half of the number p, we get 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. (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 are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
(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.)

Answer:

(a) Let the Parmit's marbles be m.

Then, according to the question, we have: $5m+7=37$

or $5m=30$

(b) Let the age of Laxmi be y years.

Then we have : $3y+4=49$

or $3y=45$

(c) Let the lowest marks be l, then :

$2l+7=87$

or $2l=80$

(d) Let the base angle of the triangle be b degrees.

Then, according to the question, we have:

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

or $3b={180}^{\circ }$

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

(a) x – 1 = 0

(b) x + 1 = 0

(c) x – 1 = 5

(d) x + 6 = 2

(e) y – 4 = – 7

(f) y – 4 = 4

(g) y + 4 = 4

(h) y + 4 = – 4

Answer:

(a) Add 1 to both sides, we have:

$x=1$

(b) Transposing 1 to the RHS, we have:

$x=-1$

(c) Transposing -1 to the RHS, we have:

$x=6$

(d) Transposing 6 to the RHS, we get:

$x=-4$

(e) Transposing - 4 to the RHS, we have:

$y=-3$

(f) Transposing - 4 to the RHS, we get:

$y=8$

(g) Transposing 4 to the RHS, we get:

$y=0$

(h) Transposing 4 to the RHS, we get:

$y=-8$

2. Give the first step you will use to separate the variable, and then solve the equation:

(a) 3l = 42

(b) b / 2 = 6

(c) p /7 = 4

(d) 4x = 25

(e) 8y = 36

(f) z/ 3 = 5 /4

(g) a /5 =7/ 15

(h) 20t = – 10

Answer:

(a) Divide both sides by 3, and we get:

$l=14$

(b) Multiply both sides by 2, we get:

$b=12$

(c) Multiply both sides by 7, we get:

$p=28$

(d) Divide both sides by 4, and we get:

$x=\frac{25}{4}$

(e) Divide both sides by 8, we get:

$y=\frac{9}{2}$

(f) Multiply both sides by 3, we get:

$z=\frac{15}{4}$

(g) Multiply both sides by 5, we get:

$a=\frac{7}{3}$

(h) Divide both sides by 20, and we get:

$t=-\frac{1}{2}$

3. Give the steps you will use to separate the variable and then solve the equation:

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p / 3 = 40

(d) 3p/10 = 6

Answer:

(a) We have 3n – 2 = 46.

Transposing - 2 to the RHS, we have:

$3n=46+2=48$

or $n=16$

(b) We have 5m + 7 = 17

Transposing 7 to the RHS, we have:

$5m=17-7=10$

or $m=2$

(c) We have 20p / 3 = 40

Multiply both sides by $\frac{3}{20}$ :

$\frac{20p}{3}\times \frac{3}{20}=40\times \frac{3}{20}$

or $p=6$

(d) We have 3p/10 = 6

Multiply both sides by $\frac{10}{3}$ :

$\frac{3p}{10}\times \frac{10}{3}=6\times \frac{10}{3}$

or $p=20$

4. Solve the following equations:

(a) 10p = 100

(b) 10p + 10 = 100

(c) p /4 = 5

(d) – p/3 = 5

(e) 3 p/4 = 6

(f) 3s = –9

(g) 3s + 12 = 0

(h) 3s = 0

(i) 2q = 6

(j) 2q – 6 = 0

(k) 2q + 6 = 0

(l) 2q + 6 = 12

Answer:

(a) Divide both sides by 10, we get:

$p=10$

(b) Transposing 10 to the RHS, we get:

$10p=100-10=90$

Now, dividing both sides by 10 gives: $p=9$

(c) Multiplying both sides by 4, we have:

$p=20$

(d) Multiplying both sides by - 3, we have:

$p=15$

(e) Multiplying both sides by $\frac{4}{3}$ , we have:

$p=8$

(f) Dividing both sides by 3, we have:

$s=-3$

(g) Transposing 12 to the RHS and then dividing both sides by 3, we have:

$s=-4$

(h) Dividing both sides by 3, we get:

$s=0$

(i) Dividing both sides by 2, we get:

$q=3$

(j) Transposing - 6 to the RHS and then dividing both sides by 2, we get:

$q=3$

(k) Transposing 6 to the RHS and then dividing both sides by 2, we get:

$q=-3$

(l) Transposing 6 to the RHS and then dividing both sides by 2, we get:

$q=3$

1. Set up equations and solve them to find the unknown numbers 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-fourths 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, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.

Answer:

Let the number in each case be n.

(a) According to the question: $8n+4=60$

or $8n=60-4=56$

or $n=7$

(b) We have :

$\frac{n}{5}-4=3$

or $\frac{n}{5}=7$

or $n=35$

(c) The equation is :

$\frac{3n}{4}+3=21$

or $\frac{3n}{4}=18$

or $n=24$

(d) We have :

$2n-11=15$

or $2n=26$

or $n=13$

(e) The equation is :

$50-3n=8$

or $3n=42$

or $n=14$

(f) We have :

$\frac{n+19}{5}=8$

or $n+19=40$

or $n=21$

(g) We have :

$\frac{5n}{2}-7=23$

or $\frac{5n}{2}=30$

or $n=12$

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?
(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 the three angles of a triangle is ${180}^{\circ }$ ).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

Answer:

(a) Let the lowest score be l.

Then a,ccording to the question, we have :

$2l+7=87$

or $2l=80$

or $l=40$

Thus, the lowest marks are 40.

(b) Let the base angle of a triangle be $\mathrm{\Theta }$.

Then, according to the question, we get :

$\mathrm{\Theta }+\mathrm{\Theta }+{40}^{\circ }={180}^{\circ }$

or $2\mathrm{\Theta }+{40}^{\circ }={180}^{\circ }$

or $2\mathrm{\Theta }={140}^{\circ }$

or $\mathrm{\Theta }={70}^{\circ }$

(c) Let the runs scored by Rahul be x. Then runs by Sachin is 2x.

Further, it is given that their runs fell two short of a double century.

Thus we have : $x+2x=198$

or $3x=198$

or $x=66$ .

Hence, Rahul is 66, and the runs scored by Sachin are 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?
(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 Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was 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?

Answer:

(a) Let the number of Parmit's marbles be n.

Then, according to the question, we have :

$5n+7=37$

or $5n=37-7$

or $5n=30$

or $n=6$

(b) Let the age of Laxmi be x.

Then, according to the question, we have :

$3x+4=49$

or $3x=49-4=45$

or $x=15$

(c) Let the number of fruit trees planted be z.

Then, according to the question, we have :

$3z+2=77$

or $3z=77-2=75$

or $z=25$

4. Solve the following riddle:
I am a number,
Tell me my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!

Answer:

Let the number be x.

According to the question ,the equation is :

$7x+50+40=300$

or $7x+90=300$

or $7x=300-90=210$

or $x=30$

Hence, the number is 30.

Simple Equations Class 6 Maths Chapter 4 -Topics

• A Mind-Reading Game
• Setting Up an Equation
• Review Of What We Know
• What is an Equation?
• More Equations
• From Solution To Equation
• Application Of Simple Equations To Practical Situations

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - Points to Remember

Variable: Something that can vary, its value is not fixed, such as x, y, z, p, q, r, etc.

Equation: It is a mathematical statement that uses an equal sign(=) to show that the value of the expression to the left of the sign (LHS) is equal to the value of the expression to the right of the sign (RHS).

Balancing an equation:

Whatever operations are performed, the value of LHS remains equal to the value of the RHS.

LHS = RHS

(LHS ± a) = (RHS ± a)

(LHS × b) = (RHS × b)

(LHS ÷ c) = (RHS ÷ c)

Solution of an Equation: For a variable 'x', if LHS(x) = RHS(x), then 'x' is a solution.

NCERT Solutions for Class 7 Maths Chapter Wise

NCERT Solutions for Class 7 Subject Wise

Practising problems is important to score well in exams. The NCERT solutions are helpful in practising homework problems. To access the solutions for all the chapters in each subject of Class 7, check the links given below.

Students can also check NCERT Books and NCERT Syllabus here:

• NCERT Syllabus Class 7 Maths
• NCERT Books Class 7
• NCERT Syllabus Class 7