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.

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

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Access Answers to NCERT Solutions for Class 7 Maths Chapter 4 â€“ 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 â€“ 4

= 8x = 60 â€“ 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) â€“ 4 = 3

By transposing â€“ 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) Ã— 5 = 7 Ã— 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 is 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 â€“ 3

= (3/4) x = 21 â€“ 3

= (3/4) x = 18

Multiply both sides by 4,

Then we get,

= (3x/4) Ã— 4 = 18 Ã— 4

= 3x = 72

Then,

Divide both sides 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 is x

Twice a number = 2x

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

= 2x â€“11 = 15

By transposing -11 from LHS to RHS it becomes 11

= 2x = 15 + 11

= 2x = 26

Then,

Divide both sides 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 is x

Thrice the number = 3x

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

= 50 â€“ 3x = 8

By transposing 50 from LHS to RHS it becomes â€“ 50

= â€“ 3x = 8 â€“ 50

= -3x = â€“ 42

Then,

Divide both sides by -3,

= (-3x/-3) = â€“ 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 is x

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

= (x + 19)/5 = 8

Multiply both sides by 5,

= ((x + 19)/5) Ã— 5 = 8 Ã— 5

= x + 19 = 40

Then,

By transposing 19 from LHS to RHS it becomes â€“ 19

= x = 40 â€“ 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 is x

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

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

= (5/2) x â€“ 7 = 23

By transposing -7 from LHS to RHS it becomes 7

= (5/2) x = 23 + 7

= (5/2) x = 30

Multiply both sides by 2,

= ((5/2) x) Ã— 2 = 30 Ã— 2

= 5x = 60

Then,

Divide both the sides 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 is 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 above given 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 â€“ 7

= 2x = 80

Now,

Divide both the sides 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 40Â°.

What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180Â°).

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 â€“ 40

= 2b = 140

Now,

Divide both the sides 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 Rahulâ€™s score is x

Then,

Sachin scored twice as many runs as Rahul is 2x

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

= Rahulâ€™s score + Sachinâ€™s score = 200 â€“ 2

= x + 2x = 198

= 3x = 198

Divide both the sides by 3,

= 3x/3 = 198/3

= x = 66

So, Rahulâ€™s score is 66

And Sachinâ€™s score is 2x = 2 Ã— 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 Parmitâ€™s marbles = m

From the question it is given that,

Then,

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

= 5 Ã— Number of Parmitâ€™s marbles + 7 = Total number of marbles Irfan has

= (5 Ã— m) + 7 = 37

= 5m + 7 = 37

By transposing 7 from LHS to RHS it becomes -7

= 5m = 37 â€“ 7

= 5m = 30

Divide both the sides by 5

= 5m/5 = 30/5

= m = 6

So, Permit has 6 marbles

(ii) Laxmiâ€™s father is 49 years old. He is 4 years older than three times Laxmiâ€™s age.

What is Laxmiâ€™s age?

Solution:-

Let Laxmiâ€™s age be = y years old

From the question it is given that,

Lakshmiâ€™s father is 4 years older than three times of her age

= 3 Ã— Laxmiâ€™s age + 4 = Age of Lakshmiâ€™s father

= (3 Ã— y) + 4 = 49

= 3y + 4 = 49

By transposing 4 from LHS to RHS it becomes -4

= 3y = 49 â€“ 4

= 3y = 45

Divide both the sides by 3

= 3y/3 = 45/3

= y = 15

So, Lakshmiâ€™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 Ã— 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 â€“ 2

= 3f = 75

Divide both the sides 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

To reach a triple century

You still need forty!

Solution:-

Let us assume the number is 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 â€“ 90

= 7x = 210

Divide both sides 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 â€“ Simple Equations

Exercise 4.1 Solutions

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Also, explore â€“Â

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7Â

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