NCERT Solutions for Class 7 Maths Exercise 4.4 Chapter 4 Simple Equations

NCERT Solutions for Class 7 Maths Exercise 4.4 Chapter 4 Simple Equations in simple PDF are available here. Application of simple equations to practical situations is the only topic covered in this exercise of NCERT Solutions for Class 7 Maths Chapter 4. The method is first to form equations corresponding to such situations and then to solve these equations, thereby giving solutions to the problems. Students are suggested to try solving the questions from NCERT Solutions for Class 7 Maths Chapter 4 and strengthen their knowledge.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations – Exercise 4.4

 

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ncert solution class 7 maths chapter 4 exercise 4.4
ncert solution class 7 maths chapter 4 exercise 4.4
ncert solution class 7 maths chapter 4 exercise 4.4
ncert solution class 7 maths chapter 4 exercise 4.4
ncert solution class 7 maths chapter 4 exercise 4.4
ncert solution class 7 maths chapter 4 exercise 4.4

 

Access Answers to NCERT Solutions for Class 7 Maths Chapter 4 – Simple Equations Exercise 4.4

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.

Solution:-

Let us assume the required number be x

Eight times a number = 8x

The given above statement can be written in the equation form as,

= 8x + 4 = 60

By transposing 4 from LHS to RHS it becomes – 4

= 8x = 60 – 4

= 8x = 56

Divide both side by 8,

Then we get,

= (8x/8) = 56/8

= x = 7

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

Solution:-

Let us assume the required number be x

One-fifth of a number = (1/5) x = x/5

The given above statement can be written in the equation form as,

= (x/5) – 4 = 3

By transposing – 4 from LHS to RHS it becomes 4

= x/5 = 3 + 4

= x/5 = 7

Multiply both side by 5,

Then we get,

= (x/5) × 5 = 7 × 5

= x = 35

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

Solution:-

Let us assume the required number be x

Three-fourths of a number = (3/4) x

The given above statement can be written in the equation form as,

= (3/4) x + 3 = 21

By transposing 3 from LHS to RHS it becomes – 3

= (3/4) x = 21 – 3

= (3/4) x = 18

Multiply both side by 4,

Then we get,

= (3x/4) × 4 = 18 × 4

= 3x = 72

Then,

Divide both side by 3,

= (3x/3) = 72/3

= x = 24

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

Solution:-

Let us assume the required number be x

Twice a number = 2x

The given above statement can be written in the equation form as,

= 2x –11 = 15

By transposing -11 from LHS to RHS it becomes 11

= 2x = 15 + 11

= 2x = 26

Then,

Divide both side by 2,

= (2x/2) = 26/2

= x = 13

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

Solution:-

Let us assume the required number be x

Thrice the number = 3x

The given above statement can be written in the equation form as,

= 50 – 3x = 8

By transposing 50 from LHS to RHS it becomes – 50

= – 3x = 8 – 50

= -3x = – 42

Then,

Divide both side by -3,

= (-3x/-3) = – 42/-3

= x = 14

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

Solution:-

Let us assume the required number be x

The given above statement can be written in the equation form as,

= (x + 19)/5 = 8

Multiply both side by 5,

= ((x + 19)/5) × 5 = 8 × 5

= x + 19 = 40

Then,

By transposing 19 from LHS to RHS it becomes – 19

= x = 40 – 19

= x = 21

(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.

Solution:-

Let us assume the required number be x

5/2 of the number = (5/2) x

The given above statement can be written in the equation form as,

= (5/2) x – 7 = 23

By transposing -7 from LHS to RHS it becomes 7

= (5/2) x = 23 + 7

= (5/2) x = 30

Multiply both side by 2,

= ((5/2) x) × 2 = 30 × 2

= 5x = 60

Then,

Divide both the side by 5

= 5x/5 = 60/5

= x = 12

2. Solve the following:

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

Solution:-

Let us assume the lowest score be x

From the question it is given that,

The highest score is = 87

Highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7

5/2 of the number = (5/2) x

The given above statement can be written in the equation form as,

Then,

= 2x + 7 = Highest score

= 2x + 7 = 87

By transposing 7 from LHS to RHS it becomes -7

= 2x = 87 – 7

= 2x = 80

Now,

Divide both the side by 2

= 2x/2 = 80/2

= x = 40

Hence, the lowest score is 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°).

Solution:-

From the question it is given that,

We know that, the sum of angles of a triangle is 180o

Let base angle be b

Then,

= b + b + 40o = 180o

= 2b + 40 = 180o

By transposing 40 from LHS to RHS it becomes -40

= 2b = 180 – 40

= 2b = 140

Now,

Divide both the side by 2

= 2b/2 = 140/2

= b = 70o

Hence, 70o is the base angle of an isosceles triangle.

(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?

Solution:-

Let us assume Rahul’s score be x

Then,

Sachin scored twice as many runs as Rahul is 2x

Together, their runs fell two short of a double century,

= Rahul’s score + Sachin’s score = 200 – 2

= x + 2x = 198

= 3x = 198

Divide both the side by 3,

= 3x/3 = 198/3

= x = 66

So, Rahul’s score is 66

And Sachin’s score is 2x = 2 × 66 = 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?

Solution:-

Let us assume number of Parmit’s marbles = m

From the question it is given that,

Then,

Irfan has 7 marbles more than five times the marbles Parmit has

= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having

= (5 × m) + 7 = 37

= 5m + 7 = 37

By transposing 7 from LHS to RHS it becomes -7

= 5m = 37 – 7

= 5m = 30

Divide both the side by 5

= 5m/5 = 30/5

= m = 6

So, Permit has 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?

Solution:-

Let Laxmi’s age to be = y years old

From the question it is given that,

Lakshmi’s father is 4 years older than three times of her age

= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father

= (3 × y) + 4 = 49

= 3y + 4 = 49

By transposing 4 from LHS to RHS it becomes -4

= 3y = 49 – 4

= 3y = 45

Divide both the side by 3

= 3y/3 = 45/3

= y = 15

So, Lakshmi’s age is 15 years.

(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?

Solution:-

Let the number of fruit tress be f.

From the question it is given that,

3 × number of fruit trees + 2 = number of non-fruit trees

= 3f + 2 = 77

By transposing 2 from LHS to RHS it becomes -2

=3f = 77 – 2

= 3f = 75

Divide both the side by 3

= 3f/3 = 75/3

= f = 25

So, number of fruit tree was 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!

Solution:-

Let us assume the number be x.

Take me seven times over and add a fifty = 7x + 50

To reach a triple century you still need forty = (7x + 50) + 40 = 300

= 7x + 50 + 40 = 300

= 7x + 90 = 300

By transposing 90 from LHS to RHS it becomes -90

= 7x = 300 – 90

= 7x = 210

Divide both side by 7

= 7x/7 = 210/7

= x = 30

Hence the number is 30.


Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 4 – Simple Equations

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

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