# NCERT Solutions For Class 7 Maths Chapter 4

## NCERT Solutions Class 7 Maths Simple Equations

NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations are given here in a simple and detailed way. These NCERT Solutions for chapter 4 of class 7 maths can be extremely helpful for the students to clear all their doubts easily and understand the basics of this chapter in a better and detailed way.

NCERT class 7 maths Simple Equations chapter 4 solutions given here are very easily understandable so that students does not face any difficulties regarding any of the solutions. The NCERT solutions for class 7 maths Simple Equations PDF is also available here that the students can download and study.

### NCERT Solutions For Class 7 Maths Chapter 4 Exercises

Exercise 4.1

Q-1) Complete the last column of the table.

 Sr. No. Equation Value Say, whether the equation is satisfied. (Yes/No) 1 a + 4 = 0 a = 4 – 2 a + 5 = 0 a = 0 – 3 a + 4 = 0 a = – 4 – 4 a – 8 = 1 a = 8 – 5 a – 8 = 1 a = 9 – 6 5a = 25 a = 0 – 7 5a = 25 a = 5 – 8 5a = 25 a = – 5 – 9 x/3 = 3 x = – 9 – 10 x/3 = 3 x = 0 – 11 x/3 = 3 x = 9 –

Ans.)

1. a + 4 = 0, a = 4

L.H.S = a + 4

By putting, a = 4;

$$L.H.S = 4 + 4 = 8 \neq R.H.S$$

Therefore, “No”, the equation is not satisfied.

2. a + 5 = 0, a = 0

L.H.S = a + 5

By putting, a = 0;

$$L.H.S = 0 + 5 = 5 \neq R.H.S$$

Therefore, “No”, the equation is not satisfied.

3. a + 4 = 0, a = – 4

L.H.S = a + 4

By putting, a = – 4;

L.H.S = – 4 + 4 = 0 = R.H.S

Therefore, “Yes”, the equation is satisfied.

4. a – 8 = 1, a = 8

L.H.S = a – 8

By putting, a = 8;

$$L.H.S = 8 – 8 = 0 \neq R.H.S$$

Therefore, “No”, the equation is not satisfied.

5. a – 8 = 1, a = 9

L.H.S = a – 8

By putting, a = 8;

L.H.S = 9 – 8 = 1 = R.H.S

Therefore, “Yes”, the equation is satisfied.

6. 5a = 25, a = 0,

L.H.S = 5a

By putting, a = 0;

$$L.H.S = 5*0 =0 \neq R.H.S$$

Therefore, “No”, the equation is not satisfied.

7. 5a = 25, a = 5,

L.H.S = 5a

By putting, a = 5;

L.H.S = 5*5 = 25 = R.H.S

Therefore, “Yes”, the equation is satisfied.

8. 5a = 25, a = – 5,

L.H.S = 5a

By putting, a = – 5;

$$L.H.S = 5*(-5) =-25 \neq R.H.S$$

Therefore, “No”, the equation is not satisfied.

9. x/3 = 3, x = – 9,

L.H.S = x/3,

By putting, x = – 9;

$$L.H.S = -9/3 = -3 \neq R.H.S$$

Therefore, “No”, the equation is not satisfied.

10. x/3 = 3, x = 0,

L.H.S = x/3,

By putting, x = 0;

$$L.H.S = 0/3 = 0 \neq R.H.S$$

Therefore, “No”, the equation is not satisfied.

11. x/3 = 3, x = 9,

L.H.S = x/3,

By putting, x = 9;

L.H.S = 9/3 = 3 = R.H.S

Therefore, “Yes”, the equation is satisfied.

Q-2) Verify whether the value in the brackets is the solution of the given equation.

1. x + 5 = 18 ( x = 1)
2. 7x + 5 = 17 ( x = – 2)
3. 7x + 5 = 19 ( x = 2)
4. 4y – 3 = 11 (y = 1)
5. 4y – 3 = 12 ( y = – 4)
6. 4y – 3 = 14 ( y = 0)

Ans.)

1) Here, x + 5 is  H.S, 18 is R.H.S and x = 1  ( given data)

L.H.S = x + 5,

By putting, x = 1,

$$L.H.S = 1 + 5 = 6 \neq R.H.S$$

As, $$L.H.S \neq R.H.S$$ , so x = 1 is not a solution.

2) Here, 7x + 5 is  H.S, 17 is R.H.S and x = – 2  ( given data)

L.H.S = 7x + 5,

By putting, x = – 2,

$$L.H.S = 7(-2) + 5 = -9 \neq L.H.S$$

As, $$L.H.S \neq R.H.S$$ , so x = – 2 is not a solution.

3) Here, 7x + 5 is  H.S, 19 is R.H.S and x = 2  ( given data)

L.H.S = 7x + 5,

By putting, x = 2,

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

As, L.H.S = R.H.S, so x = 2 is a solution.

4) Here, 4y – 3 is  H.S, 11 is R.H.S and y = 1  ( given data)

L.H.S = 4y – 3,

By putting, y = 1,

$$L.H.S = 4(1) – 3 = 1 \neq L.H.S$$

As, $$L.H.S \neq R.H.S$$, so y = 1 is not a solution.

5) Here, 4y – 3 is  H.S, 12 is R.H.S and y = – 4  ( given data)

L.H.S = 4y – 3,

By putting, x = – 4,

$$L.H.S = 4(-4) – 3 = -19 \neq R.H.S$$

As, $$L.H.S \neq R.H.S$$, so y = – 4 is not a solution.

6) Here, 4y – 3 is  H.S, 14 is R.H.S and y = 0  ( given data)

L.H.S = 4y – 3,

By putting, x = 0,

$$L.H.S = 4(0) – 3 = -3 \neq R.H.S$$

As, $$L.H.S \neq R.H.S$$, so y = 0 is not a solution.

Q-3) Solve the equation given below by trial and error method;

1. 5m + 1 = 11
2. 3p – 15 = 6

Ans.)

1) 5m + 1 = L.H.S

By putting, m = 0,

$$5(0) + 1 = 1 \neq 11$$

By putting, m = 1,

$$5(1) + 1 = 2 \neq 11$$

By putting, m = 2,

5(2) + 1 = 11 = R.H.S

2) 3p – 15 = L.H.S

By putting, p = 5,

$$3(5) – 15 = 0 \neq 6$$

By putting, p = 6,

$$3(6) – 15 = 3 \neq 6$$

By putting, p = 7,

3(7) – 15 = 6 = R.H.S

Q-4) Make the questions of the statements given below

1. The sum of numbers and 5 is 10.
2. 6 subtracted from is 9.
3. Fifteen times xis 45.
4. The number divided by 8 gives 2.
5. One-third of is 9.
6. Ten times plus 5 gives you 49.
7. Two-third of a number minus 5 gives 7.
8. If you take away 7 from 7 times w, you get 56.
9. If you add 4 to One-fourth of v, you get 35.

Ans.)

1. + 5 = 10
2. q– 6 = 9
3. 15x= 45
4. m/8 = 2
5. 1/3n= 9
6. 10+ 5 = 49
7. 2/3h– 5 = 7
8. 7w– 7 = 56
9. 1/4v+ 4 = 35

Q-5) Write the equations given below in the statement form;

1. x + 6 = 8
2. y – 9 = 10
3. 4k = 14
4. 1/3x = 9
5. 1/5p = 8
6. 1/2x + 6 = 9
7. 4t + 3 = 27
8. 3t – 4 = 25

Ans.)

1. The sum of and 6 is 8.
2. 9 subtracted from is 10.
3. Four times is 14.
4. One-third of is 9.
5. One-fifth of is 8.
6. When 6 is added to the half of a number x, it gives 9.
7. Four times of a number t, when added to 3, gives 27.
8. When 4 is subtracted from three times of number t, gives 25.

Exercise 4.2

Q-1) Mention the first step you will use to separate the variable and then solve the equations:

1. a + 2 = 0
2. b – 3 = 0
3. p – 2 = 6
4. q + 4 = 2
5. m – 5 = – 8
6. n – 5 = 5
7. x + 5 = 5
8. k + 8 = – 8

Ans.)

1) a + 2 = 0

On Subtracting 2 from L.H.S and R.H.S of the above equation, we get

a + 2 – 2 = 0 – 2

a = – 2

2) b – 3 = 0

On adding 3 to L.H.S and R.H.S of the above equation, we get

b – 3 + 3 = 0 + 3

b = 3

3) p – 2 = 6

On adding 2 to L.H.S and R.H.S of the above equation, we get

p – 2 + 2 = 6 + 2

p = 8

4) q + 4 = 2

On Subtracting 4 from L.H.S and R.H.S of the above equation, we get

q + 4 – 4 = 2 – 4

q = – 2

5) m – 5 = – 8

On adding 5 to L.H.S and R.H.S of the above equation, we get

m – 5 + 5 = – 8 + 5

m = – 3

6) n – 5 = 5

On adding 5 to L.H.S and R.H.S of the above equation, we get

n – 5 + 5 = 5 + 5

n = 10

7) x + 5 = 5

On Subtracting 5 from L.H.S and R.H.S of the above equation, we get

x + 5 – 5 = 5 – 5

x = 0

8) k + 8 = – 8

On Subtracting 8 from L.H.S and R.H.S of the above equation, we get

k + 8 – 8 = – 8 – 8

k = – 16

Q-2) Mention the first step you have to use to separate the variable and then solve the given equations;

1. 4x = 48
2. y/3 = 5
3. z/8 = 5
4. 5m = 32
5. 7n = 28
6. t/4 = 7/3
7. x/6 = 5/17
8. 40m = – 5

Ans.)

1) 4x = 48

On dividing L.H.S and R.H.S by 4 in the above equation, we get

4x/4 = 48/4

x = 12

2) y/3 = 5

On multiplying L.H.S and R.H.S by 3 in the above equation, we get

(y*3)/3 = 5*3

y = 15

3) z/8 = 5

On multiplying L.H.S and R.H.S by 8 in the above equation, we get

(z*8)/8 = 5*8

z = 40

4) 5m = 32

On dividing L.H.S and R.H.S by 5 in the above equation, we get

5m/5 = 32/5

m = 32/5

5) 7n = 28

On dividing L.H.S and R.H.S by 7 in the above equation, we get

7n/7 = 28/7

n = 4

6) t/4 = 7/3

On multiplying L.H.S and R.H.S by 4 in the above equation, we get

(t*4)/4 = (7*4)/3

t = 28/3

7) x/6 = 5/17

On multiplying L.H.S and R.H.S by 6 in the above equation, we get

(x*6)/6 = (5*6)/17

x = 30/17

8) 40m = – 5

On dividing L.H.S and R.H.S by 40 in the above equation, we get

40m/40 = (- 5)/40

m = – 1/8

Q-3) Mention the first step you have to use to separate the variable and then solve the given equations;

1. 4x – 5 = 48
2. 6y + 9 = 19
3. 40x = 160
4. 5n/8 = 8

Ans.)

1) 4x – 5 = 48

On adding 5 to L.H.S and R.H.S of the above equation, we get

4x – 5 + 5 = 48 + 5

4x = 53

On dividing L.H.S and R.H.S by 4 in the above equation, we get

4x/4 = 53/4

x = 53/4

2) 6y + 9 = 19

On subtracting 9 from L.H.S and R.H.S of the above equation, we get

6y + 9 – 9 = 19 – 9

6y = 10

On dividing L.H.S and R.H.S by 6 in the above equation, we get

4x/4 = 53/4

6y/6 = 10/6

y = 5/3

3) 40x = 160

On dividing L.H.S and R.H.S by 40 in the above equation, we get

40x/40 = 160/40

x = 4

4) 5n/8 = 8

On multiplying L.H.S and R.H.S by 8 of in the above equation, we get

(5*n*8)/8 = 8*8

5n = 64

On dividing L.H.S and R.H.S by 5 in the above equation, we get

5n/5 = 64/5

n = 64/5

Q-4) Solve the equations given below:

1. 20x = 400
2. 15y + 15 = 150
3. m/5 = 6
4. -3x/4 = 4
5. 4m/5 = 7
6. 5n = 25
7. 4k + 10 = 0
8. 4t = 0
9. 3n = 12
10. 4y – 12 = 0
11. 3m + 9 = 0
12. 3m + 9 = 18

Ans.)

1) 20x = 400

On dividing L.H.S and R.H.S by 20 in the above equation, we get

20x/20 = 400/20

x = 20

2) 15y + 15 = 150

On subtracting 15 from L.H.S and R.H.S of the above equation, we get

15y + 15 – 15 = 150 – 15

15y = 135

On dividing L.H.S and R.H.S by 15 in the above equation, we get

15y/15 = 135/15

y = 9

3) m/5 = 6

On multiplying L.H.S and R.H.S by 5 in the above equation, we get

(m*5)/5 = 6*5

m = 30

4) -3x/4 = 4

On multiplying L.H.S and R.H.S by 4 of in the above equation, we get

(-3x*4)/4 = 4*4

– 3x = 16

On dividing L.H.S and R.H.S by (- 3) in the above equation, we get

(-3x) / (-3) = 16/(-3)

x = – 16/3

5) 4m/5 = 7

On multiplying L.H.S and R.H.S by 5 in the above equation, we get

(4m*5)/5 = 7*5

4m = 35

On dividing L.H.S and R.H.S by 4 in the above equation, we get

(4m)/4 = 35/4

m = 35/4

6) 5n = 25

On dividing L.H.S and R.H.S by 5 in the above equation, we get

5n/5 = 25/5

n = 5

7) 4k + 10 = 0

On subtracting 10 from L.H.S and R.H.S of the above equation, we get

4k + 10 – 10 = 0 – 10

4k = – 10

On dividing L.H.S and R.H.S by 4 in the above equation, we get

(4k)/4 = (-10)/4

k = – 10/4

8) 4t = 0

On dividing L.H.S and R.H.S by 4 in the above equation, we get

4t/4 = 0/4

t = 0

9) 3n = 12

On dividing L.H.S and R.H.S by 3 in the above equation, we get

3n/3 = 12/3

n = 4

10) 4y – 12 = 0

On adding 12 to L.H.S and R.H.S of the above equation, we get

4y – 12 + 12 = 0 + 12

4y = 12

On dividing L.H.S and R.H.S by 4 in the above equation, we get

4y/4 = 12/4

y = 3

11) 3m + 9 = 0

On subtracting 9 from L.H.S and R.H.S of the above equation, we get

3m + 9 – 9 = 0 – 9

3m = – 9

On dividing L.H.S and R.H.S by 3 in the above equation, we get

3m/3 = (-9)/3

m = – 3

12) 3m + 9 = 18

On subtracting 9 from L.H.S and R.H.S of the above equation, we get

3m + 9 – 9 = 18 – 9

3m = 9

On dividing L.H.S and R.H.S by 3 in the above equation, we get

3m/3 = 9/3

m = 3

Exercise 4.4

Q-1) Set up the equations and solve them to find the unknown numbers in the cases given below:

1. If you add 6 to five times a number, it gives 46.
2. Two-third of a number minus 5 gives 11.
3. If you take one-third of a number and add 4 to it, gives 45.
4. When person X subtracts 12 from thrice of a number, it gives 18.
5. When Jenny subtracts twice the number of pens she has from 40, she gets 16.
6. Virat guesses a number. If he adds 18 to that number and then divides the sum by 6, he gets answer 7.
7. Ami guesses a number. If she subtracts 8 from two-third of a number, she gets 6.

Ans.)

1) Say, m is the required number.

Therefore, five times a number = 5m

Thus, from the given statement,

5x + 6 = 46

5x = 46 – 6

5x = 40

x = 40/5

x = 8

2) Say, n is the required number.

Therefore, Two-third of a number = 2n/3

Thus, from the given statement,

2n/3 – 5 = 11

2n/3 = 11 + 5

2n/3 = 16

2n = 16*3

2n = 48

n = 48/2

n = 24

3) Say, n is the required number.

Therefore, One-third of a number = n/3

Thus, from the given statement,

n/3 + 4 = 45

n/3 = 45 – 4

n/3 = 41

n = 41*3

n = 123

4) Say, k is the required number.

Therefore, Thrice of a number = 3k

Thus, from the given statement,

3k – 12 = 18

3k = 18 + 12

3k = 30

k = 30/3

k = 10

5) Let jenny has x numbers of pen,

Therefore, Twice the number of pens = 2x

Thus, from the given statement,

40 – 2x = 16

40 = 2x + 16

40 – 16 = 2x

2x = 40 – 46

2x = 24

x = 12

6) Say, Virat guesses a number ‘a’.

Thus, from the given statement,

(a + 18) / 6 = 7

a + 18 = 7*6

a + 18 = 42

a = 42 – 18

a = 24

7) Say, ami guesses a number ‘a’.

Therefore, Two-third of a number = 2a/3

Thus, from the given statement,

2a/3 – 8 = 6

2a/3 = 6 + 8

2a/3 = 14

2a = 14*3

2a = 42,

a = 21

Q-2) Solve the following:

1. The teacher noticed that in a Class of standard 7th, the highest marks obtained in the class is thrice the lowest marks plus 4. The highest mark is obtained by Karan and is 91. If Kishan has got the lowest score, what is his mark?
2. In an isosceles triangle, all the base angles are equal. Vertex angle is given as $$50^{\circ}$$. Find the measure of base angle of an isosceles triangle.( Sum of all the interior angles of a triangle is $$180^{\circ}$$)
3. Virat scored thrice the number of runs scored by Rohit. The sum of their scores fall 4 short of a triple century. Find the individual scores of Virat and Rohit.

Ans.)

1) Say, Kishan scored ‘m’ marks (lowest marks)

Karan scored 91 marks (highest marks)

From the statement given in the question,

3m + 4 = highest marks

3m + 4 = 91

3m = 91 – 4

3m = 87

m = 87/3

m = 29

Thus, Kishan scored 29 marks.

2) Let, the base angle be $$x^{\circ}$$.

Vertex angle = $$50^{\circ}$$

Sum of all the interior angles of a triangle is $$180^{\circ}$$

x + x + $$50^{\circ}$$ = $$180^{\circ}$$

2x = $$180^{\circ}$$ – $$50^{\circ}$$

2x = $$130^{\circ}$$

x = $$130^{\circ}$$/2

x = $$65^{\circ}$$

3) Let Rohit’s scored ‘m’ runs

Virat scored thrice runs then that of Rohit   = 3m

Here, the sum of their scores falls 4 short of triple century.

So, 3m + m = 300 – 4

4m = 296

m = 296/4

m = 74

Rohit score = m = 74

Virat’s score = 3m = 3(74) = 232

So, Virat scored 232 runs and Rohit scored 74 runs.

Q-3) Solve the following:

1. Umesh says that he has 5 toffies more than 6 times the number of toffies Shami has. Umesh has 47 toffies. Calculate the number of toffies that Shami has? (Say, xis the number of toffies with the Shami).
2. John’s father is 59 years old. He is 3 years older than 4 times John’s age. What is the age of John (Say, John is yyears old at present.)?
3. Trees were planted by the people of Bengaluru city in the ‘Lalbagh Garden’. Some trees are fruit trees while some were not. The number of fruit trees is three less than four times the number of non- fruit trees. Calculate the number of non-fruit trees planted if the number of fruit trees planted was 65.

Ans.)

1) Let Shami has x toffies.

Therefore, Six times the toffies of shami = 6x

Here, Umesh says that he has 5 toffies more than 6 times the toffies Shami has.

So, 6x + 5 = 47

6x = 47 – 5

6x = 42

x = 42/6

x = 7

So, Umesh has 7 toffies.

2) Let John be y years old.

Therefore, 4 times John’s age = 4y

Here, John is 3 years older than 4 times John’s age.

So, 4y + 3 = 59

4y = 59 – 3

4y = 56

y = 14

Thus, John is 14 years old.

3) Let the m be the number of non-fruit trees.

Therefore, 4 times the number of non-fruit tress is = 4m

Number of fruit trees planted = 65.

Here, the number of fruit trees is three less than four times the number of non- fruit trees.

4m – 3 = 65

4m = 65 + 3

4m = 68

m = 68/4

m = 17

Thus, the number of non-fruit trees is 17.

Q-4) Solve the riddle given below:

“I am a Number,

Tell me my identity!

Take me nine times over

And subtract forty!

To reach a Double century

You still need sixty!

Ans.)  Let the required number be p.

As per the given riddle

(9p – 40) + 60 = 200

(9p – 40) = 200 – 60

(9p – 40) = 140

9p = 140 + 40

9p = 180

p = 180/9

p = 20

So, the identity of the number is 20.

For the students of class 7, it is prime important to learn the concepts of Simple equation. As simple equation forms the backbone of various polynomial equation, thus one needs to have a good grasp of this topic. Simple equation forms a major part in the examination for the students of class 7. In order to perform exceptionally well in the examination for class 7, students need to have a good understanding of the topic. Simple equation are easy and scoring topic, as one just needs to have a basic understanding of the topic. Students need to have a thorough understanding of the topic to excel in the examination. With a lot of practice students can easily get the in-depth knowledge of the concepts, which will help them to enhance their learning.

Simple equation basically consist an equation in one variable, which leads to give a particular value. Thus on equation the Left Hand side and the Right hand side of the equation, one can find the value of the unknown. It is to be noted that for a given unknown ‘n’, one needs to have n number of equation in order to find the value of n unknown.

Solving question on Simple equation can be fun, as the students get to know the different types of question and equation, which help them to score good marks. For the students of class 7, it is essential to have a focus on understanding to learning rather than just mugging up the concepts. As this topic hold importance for the students in the classes ahead, so it is essential for them to practice the concepts of simple equation.

Simple equation involves the equation forming, equation solving, problem solving etc. This helps the students to have develop the conceptual knowledge, reasoning skills, as well as develop logical skills. This helps students to encourage them in learning maths as well as to excel in the examination.

The above given are the Solution of class 7 Mathematics Simple Equation which needs to be practiced before the examination.