Chapter 4 Simple Equations

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1**

**Question 1.**

Complete the last column of the table.

**Solution:
**

**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 (n = – 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:**

**(a) n + 5 = 19 (n = 1)**

L.H.S. = n + 5 = 1 + 5 | when n = 1 = 5

R.H.S. = 19

∵ L.H.S. ≠ R.H.S.

∴ n = 1 is not a solution to the given equation n + 5 = 19.

**(b) 7n + 5 = 19 (n = – 2)**

L.H.S. = 7n + 5 = 7(- 2) + 5 | when n = – 2 = – 14 + 5 = – 9

R.H.S. = 19

∵ L.H.S. ≠ R.H.S.

∴ n = – 2 is not a solution to the given equation 7n + 5 = 19.

**(c) 7n + 5 = 19 (n = 2)**

L.H.S. = In + 5 = 7(2) + 5 | when n = 2 = 14 + 5 = 19 = R.H.S.

∴ n = 2 is a solution to the given equation 7n + 5 = 19.

**(d) 4p – 3 = 13 (p = 1)**

L.H.S. = 4p – 3 = 4(1) – 3 | when p = 1 = 4 – 3 = 1

R.H.S. = 13

∵ L.H.S. ≠ R.H.S.

∴ p = 1 is not a solution to the given equation 4p – 3 = 13.

**(e) 4p – 3 = 13 (p = – 4)**

L.H.S. = 4p – 3 = 4(- 4) – 3 , | when p = – 4 = – 16 – 3 = – 19

R.H.S. = 13

∵ L.H.S. ≠ R.H.S.

∴ p = – 4 is not a solution to the given equation

4p – 3 = 13.

**(f) 4p – 3 = 13 (p = 0)**

L.H.S. = 4 (p) – 3 = 4(0) – 3 | when p = 0 = 0 – 3 = – 3

R.H.S. = 13

∵ L.H.S. ≠ R.H.S.

∴ p = 0 is not a solution to the given equation 4p – 3 = 13.

**Question 3.**

Solve the following equations by trial and error method.

- 5p + 2 = 17
- 3m – 14 = 4.

**Solution:**

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

So, p = 3 is the solution of the given equation 5p + 2 = 17.

**(ii) 3m – 14 = 4
**

So, m = 6 is the solution of the given equation 3m – 14 = 4.

**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 oft 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:**

- x + 4 = 9
- y – 2 = 8
- 10 a = 70
- b ÷ 5 = 6
- 34 × t = 15
- 7m + 7 = 77
- 14 × x – 4 = 4
- 6y – 6 = 60
- 13 × z + 3 = 30

**Question 5.**

Write the following equations in statement forms:

- p + 4 = 15
- m – 7 = 3
- 2m = 7
- m5 = 3
- 3m5 = 6
- 3p + 4 = 25
- 4p – 2 = 18
- p2 + 2 = 8.

**Solution:**

- The sum of p and 4 is 15.
- 7 subtracted from m is 3.
- Twice a number m is 7.
- One-fifth of a number m is 3.
- Three-fifth of a number m is 6.
- Three times a number p, when added to 4, gives 25.
- 2 subtracted from four times a number p is 18.
- Add 2 to half of a number p to get 8.

**Question 6.**

Set up an equation in the following cases:

- Irfan says that he has 7 marbles more than five times the marbles Permit has. Irfan has 37 marbles. (Take m to be the number of Permit’s marbles.)
- 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.)
- The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
- 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 triangle is 180 degrees).

**Solution:
(i)** Let the number of marbles with Parmit be m.

Then, 7 added to 5 times mis 5m + 7

It is given that 7 marbles more than five times the marble is 37. Thus, the equation obtained is 5m + 7 = 37.

**(ii)** Let Laxmi’s age be y years. Then, 4 added to 3 times y is 3y + 4

It is given that the father is 4 years older than 3 times Laxmi’s age. His age is 49years.

Then, we have the following equation : 3y + 4 = 49

**(iii)** Let the lowest marks be l. Then, twice the lowest marks plus 7 is 2l +7

It is given that, the highest marks 87 obtained by a student is twice the lowest marks plus 7.

So, we have the following equation : 2l + 7 = 87

**(iv)** Let the base angle be b. Then, the vertex angle = 2b.

Since, sum of the angles of a triangle is 180°

∴ b + b + 2b = 180°

⇒ 4b = 180°

which is the required equation.

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2**

**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

**(e)** y – 4 = – 7

**(f)** y – 4 = 4

**(g)** y + 4 = 4

**(h)** y + 4 = – 4

**Solution:
(a)** The given equation is x – 1 = 0

Add 1 to both sides,

x – 1 + 1 = 0 + 1 ⇒ x = 1

It is the required solution.

**Check.**Put the solution x = 1 back into the equation.

L.H.S. = x – 1 = 1 = 1 – 0 = R.H.S.

The solution is thus checked for its correctness.

**(b)** The given equation is x + 1 = 0

Subtract 1 from both sides, x + 1 – 1 = 0 – 1 ⇒ x = – 1

It is the required solution.

**Check.** Put the solution x = – 1 back into the equation.

L.H.S. = x + 1 = (-1)+1

= 0 = R.H.S.

The solution is thus checked for its correctness.

**(c)** The given equation is

x – 1 = 5

Add 1 to both sides,

x + 1 – 1 = 5 + 1 ⇒ x = 6

It is the required solution

**Check.** Put the solution x = 6 back into the equation.

L.H.S. = x – 1 = 6 – 1 = 5 = R.H.S.

The solution is thus checked for its correctness.

**(d)** The given equation is x + 6 = 2

Subtract 6 from both sides, x + 6 – 6 = 2 – 6 ⇒ x = – 4

It is the required solution.

**Check.** Put the solution x = – 4 back into the equation.

L.H.S. = x + 6 = – 4 + 6 = 2 = R.H.S.

The solution is thus checked for its correctness.

**(e)** The given equation is y – 4 = – 7

Add 4 to both sides, y – 4 + 4 = – 7 + 4 ⇒ y = – 3

It is the required solution.

**Check.** Put the solution

L.H.S. = y – 4 = – 3 – 4 = – 7 = R.H.S.

The solution is thus checked for its correctness.

**(f)** The given equation is y – 4 = 4

Add 4 to both sides,

y – 4 + 4 = 4 + 4 ⇒ y = 8

It is the required solution.

**Check.** Put the solution y = 8 back into the equation.

L.H.S. = y – 4 = 8 – 4 = 4 = R.H.S.

The solution is thus checked for its correctness.

**(g)** The given equation is y + 4 = 4

Subtract 4 from both sides, y + 4 – 4 = 4 – 4 ⇒ y = 0

It is the required solution.

**Check.** Put the solution y = 0 back into the equation.

L.H.S. =y + 4 = 0 + 4 = 4 = R.H.S.

The solution is thus checked for its correctness.

**(h)** The given equation is y + 4 = – 4

Subtract 4 from both sides, y + 4 – 4 = – 4 – 4 ⇒ y = -8

It is the required solution.

**Check.** Put the solution y = – 8 back into the equation.

L.H.S. = y + 4 = -8 + 4 = – 4 = R.H.S.

The solution is thus checked for its correctness.

**Question 2.**

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

**(a)** 3l = 42

**(b)** b2 = 6

**(c)** p7 = 4

**(d)** 4x = 25

**(e)** 8y = 36

**(f)** z3 = 54

**(g)** a5 = 715

**(h)** 20t = – 10

**Solution:
**

**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)** 20p3 = 40

**(d)** 3p10 = 6

**Solution:
**

**Question 4.**

Solve the following equations:

**Solution:
**

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3**

January 16, 2019 by Dattu

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3.

- Simple Equations Class 7 Ex 4.1
- Simple Equations Class 7 Ex 4.2
- Simple Equations Class 7 Ex 4.4
- Simple Equations Class 7 MCQ

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Simple Equations |

Exercise |
Ex 4.3 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3**

**Question 1.**

Solve the following equations:

**Solution:
**

**Question 2.**

Solve the following equations:

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

**(6)** 3 (n – 5) = 21

**(c)** 3 (n – 5) = -21

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

**(e)** 4 (2 – x) = 8

**Solution:
**

**Question 3.**

Solve the /bilowing 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:
**

**Question 4.**

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

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

**Solution:**

**(a) 1.** Start with x = 2

Multiply both sides by 3, 3x = 6

Subtract 2 from both sides, 3x – 2 = 4 …(1)

**2.** Start with x = 2

Multiply both sides by 4, 4x = 8

Add 5 to both sides, 4x + 5 = 13 …(2)

**3.** Start with x = 2 Multiply both sides by 5 5x = 10

Subtract 1 from both sides, 5x – 1 = 9 …(3)

**(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, x2 = -1

Add 3 to both sides, x2 + 3 = 2

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4**

**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)** 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 52 of the number, the result is 23.

**Solution:
**

**Question 2.**

Solve the following:

**(a)** The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest mark 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:
**

**Question 3.**

Solve the following:

- 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?
- Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
- The people of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was 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:
**

**Question 4.**

Solve the following riddle :

I am a number, Tell me my identity!

Take me seven times over And add a fifty!

To reach a triple century You still need forty!

**Solution:**

Let ‘x’ be the number,

Then, according to the question, we get (x × 7) + 50 = 300 – 40

7x + 50 = 260

7x = 210

x = 2107 = 30

So, the number is 30.