Contents

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Simple Equations |

Exercise |
Ex 4.1, Ex 4.2, Ex 4.3, Ex 4.4. |

Number of Questions Solved |
18 |

Category |
NCERT Solutions |

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

### Chapter 4 Simple Equations Exercise 4.1

**Ex 4.1 Class 7 Maths Question 1.**

Complete the last column of the table.

**Solution:
**

**Ex 4.1 Class 7 Maths Question 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 (w = -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)

**Solution:
**

**Ex 4.1 Class 7 Maths Question 3.**

Solve the following equations by trial and error method:

**(i)** 5p + 2 = 17 .

**(ii)** 3m – 14 = 4

**Solution:**

**(i)** Let us evaluate the L.H.S. and R.H.S. of the given equation for some values of p and continue to given new values till the L.H.S. becomes equal to the R.H.S.

The given equation is 5p + 2 = 17. We have,

L.H.S. = 5p+ 2 and R.H.S. = 17

Clearly, L.H.S. = R.H.S. for p = 3. Hence, p = 3 is the solution of the given equation.

**(ii)** Let us evaluate the L.H.S. and R.H.S. of the given equation for some values of m and continue to given new values till the L.H.S. becomes equal to the R.H.S.

The given equation is 3m – 14 = 4, that is, 14 subtracted from 3 times m gives 4.

So, we substitute values which gives 3m > 14.

We have, L.H.S. = 3m – 14 and R.H.S. = 4

Clearly, L.H.S. = R.H.S. for m = 6. Hence, m = 6 is the solution of the given equation.

**Ex 4.1 Class 7 Maths Question 4.**

Write equations for the following statements:

- The sum of numbers x and 4 is 9.
- 2 subtracted from y is 8.
- Ten times a is 70.
- The number b divided by 5 gives 6.
- Three-fourth of t is 15.
- Seven times m plus 7 gets you 77.
- One-fourth of a number x minus 4 gives 4.
- If you take away 6 from 6 times y, you get 60.
- If you add 3 to one-third of z, you get 30.

**Solution:**

The equations for the given statements are :

- x + 4 = 9
- y – 2 = 8
- 10 α = 70
- \(\frac { b }{ 5 } \) = 6
- \(\frac { 3 }{ 4 } \) t = 15
- 7m + 7 = 77
- \(\frac { 1 }{ 4 } \) x – 4 = 4
- 6y – 6 = 60
- \(\frac { 1 }{ 3 } \) z + 3 = 30

**Ex 4.1 Class 7 Maths Question 5.**

Write the following equations in statement forms :

- p + 4 = 15
- m – 7 = 3
- 2m = 7
- \(\frac { m }{ 5 } \) = 3
- \(\frac { 3m }{ 5 } \)= 6
- 3p + 4 = 25
- 4p – 2 = 18
- \(\frac { p }{ 2 } \) + 2 = 8

**Solution:
**The statements for the given equations are :

- The sum of numbers p and 4 is 15.
- The difference of m and 7 is 3.
- Two times m is 7.
- The number m divided by 5 gives 3.
- Three time’s m divided by 5 gives 6.
- Three times p plus 4 gives 25.
- Four times p minus 2 gives 18.
- p divided by 2 plus 2 gives 8.

**Ex 4.1 Class 7 Maths Question 6.**

Set up an equation in the following cases :

**(i)** Irfan says that he has 7 marbles more than five times the marbles Parxnit 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 is 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).

**Solution:
**

**(i)**Let Parmit has m marbles. Then, five times the marbles Parmit has = 5m

Irfan has 7 marbles more than five times the marbles Parmit has = 5m + 7,

i.e., Irfan has (5m + 7) marbles

But it is given that Irfan has 37 marbles. Therefore, 5m + 7 = 37

**(ii)** Let the age of Laxmi be y years.

Then, three times Laxmi’s age = 3y years Laxmi’s father is 4 years older than three times Laxmi’s age,

i.e., Age of Laxmi’s father = (3y + 4) years But it is given that Laxmi’s father is 49 years old Therefore, 3y + 4 = 49

**(iii)** Let the lowest marks be l.

Then, twice the lowest marks = 21

Highest score obtained by a student in her class is twice the lowest marks plus 7, i.e.,

Highest score = 2l + 7

But this is given to be 87

Therefore, 2l + 7 = 87

**(iv)** Let the base angle be b in degrees. Then the vertex angle is 2b in degrees.

∵ Sum of the angles of a triangle is 180 degrees.

∴ 2 b + b + b = 180° or 4 b = 180°

### Chapter 4 Simple Equations Exercise 4.2

**Ex 4.2 Class 7 Maths Question 1.**

Give first the 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

**Solution:**

**(a)** We have, x – 1 = 0

In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift -1. This can be done by adding 1 to both sides of the given equation.

x – 1 + 1 = 0 + 1 [ Adding 1 to both sides ]

x = 1 [ ∵ -1 + 1 = 0 and 0 + 1 = 1 ]

**(b)** We have, x + 1 = 0

In order to get x by itself on the L.H.S., we need to shift 1. This can be done by subtracting 1 from both sides of the given equation.

x + 1 – 1 = 0 – 1 [Subtracting 1 from both sides]

⇒ x = -1 [∵ 1 – 1 = 0, 0 – 1 = -1 ]

So, x = -1 is the solution of the given equation.

**(c)** We have, x – 1 = 5

In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift -1. This can be done by adding 1 to both sides of the given equation.

**(d)** We have, x + 6 = 2

In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift 6. This can be done by subtracting 6 from both sides of the given equation.

**(e)** We have, y – 4 = -7

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S.,we need to shift -4. This can be done by adding 4 to both sides of the given equation.

**(f)** We have, y – 4 = 4

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift -4. This can be done by adding 4 to both sides of the given equation.

**(g)** We have, y + 4 = 4

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift 4. This can be done by subtracting 4 from both sides of the given equation.

**(h)** We have, y + 4 = -4

In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift 4. This can be done by subtracting 4 from both sides of the given equation.

**Ex 4.2 Class 7 Maths Question 2.**

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

**(a)** 3l = 42

**(b)** \(\frac { b }{ 2 } \) = 6

**(c)** \(\frac { p }{ 7 } \) = 4

**(d)** 4x = 25

**(e)** 8y = 36

**(f)** \(\frac { z }{ 3 } \) = \(\frac { 5 }{ 4 } \)

**(g)** \(\frac { a }{ 5 } \) = \(\frac { 7 }{ 15 } \)

**(h)** 20t = -10

**Solution:**

**(a)** We have, 3l = 42

In order to solve this equation, we have to get l by itself on the L.H.S. For this, 3 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 3.

**(b)** We have, \(\frac { b }{ 2 } \) = 6

In order to solve this equation, we have to get b by itself on the L.H.S. To get b by itself on L.H.S., we have to remove 2 from L.H.S. This can be done by multiplying both sides of the equation by 2. Thus, we have

**(c)** We have, \(\frac { p }{ 7 } \) = 4

In order to solve this equation, we have to get p by itself on the L.H.S. To get p by itself on L.H.S., we have to remove 7 from L.H.S. This can be done by multiplying both sides of the equation by 7. Thus, we have

**(d)** We have, 4x = 25

In order to solve this equation, we have to get x by itself on the L.H.S. For this, 4 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 4.

**(e)** We have, 8y = 36

In order to solve this equation, we have to get y by itself on the L.H.S. For this, 8 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 8.

**(f)** We have, \(\frac { z }{ 3 } \) = \(\frac { 5 }{ 4 } \)

In order to solve this equation,we have to get z by itself on the L.H.S. To get z by itself on L.H.S., we have to remove 3 from L.H.S. This can be done by multiplying both sides of the equation by 3.

Thus, we have

**(g)** We have, \(\frac { a }{ 5 } \) = \(\frac { 7 }{ 15 } \)

In order to solve this equation, we have to get a by itself on the L.H.S. To get a by itself on L.H.S., we have to remove 5 from L.H.S. This can be done by multiplying both sides of the equation by 5.

**(h)** We have, 20t = -10

In order to solve this equation, we have to get t by itself on the L.H.S. For this, 20 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 20.

**Ex 4.2 Class 7 Maths Question 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)** \(\frac { 20p }{ 3 } \) = 40

**(d)** \(\frac { 3p }{ 10 } \) = 6

**Solution:
**

**Ex 4.2 Class 7 Maths Question 4.**

Solve the following equations :

**(a)** 10p = 100

**(b)** 10p + 10 = 100

**(c)** \(\frac { p }{ 4 } \) = 5

**(d)** \(\frac { -p }{ 3 } \) = 5

**(e)** \(\frac { 3p }{ 4 } \) = 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

**Solution:**

### Chapter 4 Simple Equations Exercise 4.3

**Ex 4.3 Class 7 Maths Question 1.**

Solve the following equations :

**Solution:**

**Ex 4.3 Class 7 Maths Question 2.**

Solve the following equations :

**(a)** 2 (x + 4) = 12

**(b)** 3 (n – 5) = 21

**(c)** 3 (n – 5) = -21

**(d)** -4 (2 + x) = 8

**(e)** 4 (2 – x) = 8

**Solution:
**

**Ex 4.3 Class 7 Maths Question 3.**

Solve the following equations :

**(a)** 4 = 5 (p – 2)

**(b)** -4 = 5 (p – 2)

**(c)** 16 = 4 + 3 (t + 2)

**(d)** 4 + 5 (p – 1) = 34

**(e)** 0 = 16 + 4 (m – 6)

**Solution:**

**Ex 4.3 Class 7 Maths Question 4.**

**(a)** Construct 3 equations starting with x = 2.

**(b)** Construct 3 equations starting with x = -2.

**Solution:**

**(a) First equation :**

Start with x = 2

Multiply both sides by 3, 3x = 6

Add 2 to both sides, 3x + 2 = 8

**Second equation :**

Start with x = 2

Multiply both sides by -3, -3x, = -6

Add 8 to both sides, 8 – 3x = 2.

**Third equation :**

Start with x = 2

Divide both sides by 5, \(\frac { x }{ 5 } \) = \(\frac { 2 }{ 5 } \)

Subtract 2 from both sides,

**(b) First equation :**

Start with x = -2

Multiply both sides by 2, 2x = -4

Subtract 3 from both sides, 2x -3 = -7

**Second equation :**

Start with x = -2

Multiply both sides by – 5, – 5x = 10

Add 10 to both sides, 10 – 5x = 20

**Third equation :**

Start with x = -2

Divide both sides by 2, \(\frac { x }{ 2 } \) = -1

Add 3 to both sides, \(\frac { x }{ 2 } \) + 3 = 2

### Chapter 4 Simple Equations Exercise 4.4

**Ex 4.4 Class 7 Maths Question 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 \(\frac { 5 }{ 2 } \) of the number, the result is 23.

**Solution:**

**(a)** Let x be the required number. Then, the required equation is

**(b)** Let x be the required number. Then, the required equation is \(\frac { x }{ 5 } \) – 4 = 3.

**(c)** Let y be the required number. Then, the required equation be \(\frac { 3y }{ 4 } \) + 3 = 21

**(d)** Let the required number be m. Then, the required equation is 2m -11 = 15

**(e)** Let Munna have x notebooks. Then, the required equation is 50 – 3x = 8

**(f)** Let the number be x. Then, the required equation is

**(g)** Let the number be n. Then, the required equation is

**Ex 4.4 Class 7 Maths Question 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?

**(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°). (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:
**

**Ex 4.4 Class 7 Maths Question 3.**

Solve the following :

**(i)** Irfan says that he has 7 marbles more than five times the marbles Pannit 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 were two more than three times the number of fruit trees. What was the number of fruit tree planted if the number of non-fruit trees planted was 77?

**Solution:
**

**Ex 4.4 Class 7 Maths Question 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:
**

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