NCERT Solutions | Class 7 Maths Chapter 4

NCERT Solutions | Class 7 Maths Chapter 4 | Simple Equations 

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

CBSE Solutions | Maths Class 7

Check the below NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Pdf free download. NCERT Solutions Class 7 Maths  were prepared based on the latest exam pattern. We have Provided Simple Equations Class 7 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 7 Maths

NCERT Solutions Class 7 Maths
Book: National Council of Educational Research and Training (NCERT)
Board: Central Board of Secondary Education (CBSE)
Class: 7th
Subject: Maths
Chapter: 4
Chapters Name: Simple Equations
Medium: English

Simple Equations | Class 7 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 7 Maths Chapter 4 Simple Equations to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 7 Maths Chapte 4 Simple Equations Exercise 4.1

Ex 4.1 Class 7 Maths Question 1.

Complete the last column of the table.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 1

Solution:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 2

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 3

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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 4
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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 5
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:

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

Solution:
The equations for the given statements are :

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

Ex 4.1 Class 7 Maths Question 5.

Write the following equations in statement forms :

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

Solution:
The statements for the given equations are :

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

NCERT Solutions for Class 7 Maths Chapte 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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 7

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 8

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 9

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 10

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 11

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 12

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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 13

(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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 14

(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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 15

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 16

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 17

(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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 18

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 19

(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.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 21
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 22
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 23

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 24
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 25
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 26
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 27

NCERT Solutions for Class 7 Maths Chapte 4 Simple Equations Exercise 4.3

Ex 4.3 Class 7 Maths Question 1.

Solve the following equations :
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 28

Solution:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 29
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 30
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 31
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 32
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 33

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 34
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 35

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 36
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 37
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 38

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,
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 39

(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

NCERT Solutions for Class 7 Maths Chapte 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
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 40

(b) Let x be the required number. Then, the required equation is \(\frac { x }{ 5 } \) – 4 = 3.
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 41

(c) Let y be the required number. Then, the required equation be \(\frac { 3y }{ 4 } \) + 3 = 21
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 42

(d) Let the required number be m. Then, the required equation is 2m -11 = 15
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 43

(e) Let Munna have x notebooks. Then, the required equation is 50 – 3x = 8
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 44

(f) Let the number be x. Then, the required equation is
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 45

(g) Let the number be n. Then, the required equation is
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 46
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 47

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 48

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 49

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:

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations 50

NCERT Class 7 Maths

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