NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

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:

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°

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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