NCERT Solutions For Class 7 Maths Chapter 4, Simple Equations, are given here in a simple and detailed way. Simple Equations, solutions can be extremely helpful for the students to clear all their doubts easily and understand the basics of this chapter. Students can use them as worksheets to prepare for their exams. To download pdf of Chapter 4 NCERT solutions for 7th Class, click on the link given above.

The NCERT solutions given here are very easy to understand such that students do not face any difficulties while solving questions based on NCERT book. The simple equations for class 7 worksheets are designed by our subject experts with respect to the CBSE syllabus (2018-2019) prescribed by the board itself.

BYJU’S also provide simple equations word problems for class 7, ncert notes, questions papers similar to sample papers and previous year question papers to get an idea of the types of questions asked from this chapter.

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

- NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations Exercise 4.1
- NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations Exercise 4.2
- NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations Exercise 4.3

### NCERT Class 7 Maths Solutions For Chapter 4 – Exercises 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.**

**x + 5 = 18 ( x = 1)****7x + 5 = 17 ( x = – 2)****7x + 5 = 19 ( x = 2)****4y – 3 = 11 (y = 1)****4y – 3 = 12 ( y = – 4)****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;**

**5m + 1 = 11****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**

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

**Ans.)**

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

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

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

**Ans.)**

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

### NCERT Class 7 Maths Solutions For Chapter 4 – Exercises 4.2

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

**a + 2 = 0****b – 3 = 0****p – 2 = 6****q + 4 = 2****m – 5 = – 8****n – 5 = 5****x + 5 = 5****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;**

**4x = 48****y/3 = 5****z/8 = 5****5m = 32****7n = 28****t/4 = 7/3****x/6 = 5/17****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;**

**4x – 5 = 48****6y + 9 = 19****40x = 160****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:**

**20x = 400****15y + 15 = 150****m/5 = 6****-3x/4 = 4****4m/5 = 7****5n = 25****4k + 10 = 0****4t = 0****3n = 12****4y – 12 = 0****3m + 9 = 0****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

### NCERT Class 7 Maths Solutions For Chapter 4 – Exercises 4.3

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

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

**The teacher noticed that in a Class of standard 7**^{th}, 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?**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 \(180^{\circ}\))****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:**

**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,***x*is the number of toffies with the Shami).**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***y*years old at present.)?**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**.

## Class 7 Maths NCERT Solutions Chapter 4 – Simple Equations

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

Simple Equations, chapter 4 from Class 7 Maths book covers important topics such as;

- What is the equation?
- Setting up an equation
- How to solve an equation?
- Solutions to equations and applications of equations to practical situations

For the students of class 7, it is prime important to learn the concepts of a simple equation. As simple equation forms the backbone of various polynomial equations. Thus, one needs to have a good grasp of this topic. This chapter also forms a major part in the examination, therefore, in order to perform exceptionally well in the Class 7 examination, students need to have a good practice of this topic. Simple equations topic is easy and scoring as one just needs to have a basic understanding of the fundamentals. With a lot of practice, students can easily get an in-depth knowledge of the simple equation concepts which will help them to enhance their learning.

Solving the questions on Simple equation can be fun, as the students get to know the different types of question and equation, which will help them to score good marks. For the students of class 7, it is essential to have a focus on understanding and learning rather than just by-hearing the concepts. These concepts could also be asked in the talent squad test or any competitive tests conducting in 2019, to test the logical and mathematical abilities of students. Therefore, students who are willing to appear for those tests can learn from online materials provided by BYJU’S to clear their concepts.

Get all chapters solutions for class 7 maths subject covered in NCERT here and also, download BYJU’S app to learn from videos and creative content.