**Class 7 Maths NCERT Solutions Chapter 4 – Simple Equations comprises of the 4 Exercises, **

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.1

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.2

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.3

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.4

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.1

**1. Complete the last column of the table:**

**(i) No **Substitute x=3 in the equation x + 3 = 0

3 + 3 = 9 ≠ 0

**(ii) No **Substitute x=0 in the equation x + 3 = 0

0 + 3 = 3 ≠ 0

**(iii) Yes **Substitute x=-3 in the equation x + 3 = 0

-3+ 3 = 0

**(iv) No **Substitute x=7 in the equation x – 7 = 1

7 – 7 = 0 ≠ 1

**(v) Yes **Substitute x=8 in the equation x – 7 = 1

8- 7 = 1

**(vi) No **Substitute x = 0 in the equation 5x = 25

5(0) = 0 ≠ 25

**(vii) Yes **Substitute x = 5 in the equation 5x = 25

5(5) = 25

**(viii) No **Substitute x = -5 in the equation 5x = 25

5(-5) = -25 ≠ 25

**(ix) No **Substitute m = -6 in the equation = 2

= -2 ≠ 2

**(x) No **Substitute m = 0 in the equation = 2

= 0 ≠ 2

**(xi) Yes **Substitute m = 6 in the equation = 2

= 2

**2. Check whether the value given in the brackets is a solution to the given equation or not:**

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

1 + 5 = 6 ≠ 19

**Therefore n = 1 is not a solution.**

b) 7n + 5 = 19 (n = – 2)

7(-2) + 5 = -14 + 5

= -9 ≠ 19

**Therefore n = -2 is not a solution.**

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

7(2) + 5 = 14 + 5 = 19

**Therefore n = 2 is a solution.**

d) 4p – 3 = 13 (p = 1)

4(1) – 3 = 4 – 3 = 1 ≠ 13

**Therefore p = 1 is not a solution**

e) 4p – 3 = 13 (p = – 4)

4(-4) – 3 = -16 – 3

= -19 ≠ 13

**Therefore p = -4 is not a solution**

f) 4p – 3 = 13 (p = 0)

4(0) – 3 = 0 – 3

= -3 ≠ 13

**Therefore p = 0 is not a solution**

**3. Solve the following equations by trial and error method**

i) 5p + 2 = 17

p = 0

5(0) + 2 = 0 + 2

= 2 ≠ 17

p = 1

5(1) + 2 = 5 + 2

= 7 ≠ 17

p = -1

5(-1) + 2 = -5 + 2

= -3 ≠ 17

p = 2

5(2) + 2 = 10 + 2

= 12 ≠ 17

p = -2

5(-2) + 2 = -10 +2

= -8 ≠ 17

p = 3

5(3) + 2 = 15 + 2

= 17

**Therefore p=3 is a solution of the equation.**

ii) 3m – 14 = 4

m = 0

3(0) – 14 = 0 – 14

= -14 ≠ 4

m = 1

3(1) – 14 = 3 – 14

= -11 ≠ 4

m = -1

3(-1) – 14 = -3 -14

= – 17

m = 2

3(2) – 14 = 6 – 14

= -8 ≠ 4

.

.

.

.

m = 6

3(6) – 14 = 18 – 14

= 4

**Therefore m = 6 is a solution of the equation **

**4. Write equations for the following statements**

(i) The sum of numbers x and 4 is 9.

Sum of x and 4 is x + 4

The sum is 9

**Therefore the equation is x + 4 = 9**

(ii) 2 subtracted from y is 8.

2 subtracted from is y – 2

The given difference is 8

**Therefore the equation is y – 2 = 8**

(iii) Ten times a is 70

Ten times a is 10 x a i.e 10a

The given product is 70

**Therefore the equation is 10a = 70**

(iv) The number b divided by 5 gives 6.

Number b divided by 5 is

The division gives quotient 6

**Therefore the equation is ** ** = 6**

(v) Three-fourth of t is 15.

Three-fourth of t is t

**Therefore the equation is t**** = 15**

(vi) Seven times m plus 7 gets you 77

Seven times m is 7 x m i.e 7m

Seven times m plus 7 is 7m + 7

The sum yields 77

**Therefore the equation is 7m + 7 = 77**

(vii) One-fourth of a number x minus 4 gives 4.

One-fourth of a number x is

One-fourth of a number x minus 4 is – 4

The difference yields 4

**Therefore the equation is **** – 4 = 4**

(viii) If you take away 6 from 6 times y, you get 60.

6 times y is 6 x y i.e 6y

Taking away 6 from 6 times y is 6y – 6

The difference gives 60

**Therefore the equation is 6y – 6 = 60**

(ix) If you add 3 to one-third of z, you get 30.

One-third of z is z

Add 3 to one-third of z is z + 3

The sum gives 30

**Therefore the equation is ** **z + 3 = 30**

**5. Write the following equations in statement forms:**

i) p + 4 = 15

**The sum of numbers p and 4 is 15**

ii) m – 7 = 3

**7 subtracted from m is 3**

iii) 2m = 7

**Two times m is 7**

iv) = 3

**The number m divided by 5 gives 3**

v) = 6

**Three times a number m divided by 5 gives 6**

vi) 3p + 4 = 25

**Three times p plus 4 gets 25**

vii) 4p – 2 = 18

**If you take 2 away from 4 times p, you get 18**

viii) + 2 = 8

**If you add 2 to half of a number p you get 8**

**6. Set up an equation in the following cases:**

i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)

**Solution:**

Number of marbles Parmit has = m

Five times Parmit marbels = 5m

Irfan has 7 marbles more than five times the marbles Parmit has i.e 5m + 7

Number of marbles Irfan has = 37

**Therefore, the equation is 5m + 7 = 37**

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.)

**Solution:**

Laxmi’s age = y years

Three times Laxmi’s age = 3y

Laxmi’s father is 4 years older than three times Laxmi’s age i.e 3y + 4

Laxmi’s father’s age = 49 years

**Therefore, the equation is 3y + 4 = 49**

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.)

**Solution:**

Lowest score = l

Twice the lowest marks = 2l

The highest marks obtained by a student in class is twice the lowest marks plus 7 i.e 2l + 7

The highest score = 87

**Therefore the equation is 2l +7 = 873**

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:**

Let the angles of the Isosceles triangle be b^{o}, b^{o} and v^{o }Where b^{o} is the base angles and v^{o } is the vertex angle

The vertex angle is twice either base angle i.e v^{o} = 2b^{o }Sum of angles of a triangle = 180^{o }i.e b^{o} + b^{o} + v^{o} = 180^{o }i.e b^{o} + b^{o }+ 2b^{o} = 180^{o} (by substituting the value of v^{o})

**Therefore the equation is 3b ^{o }= 180**

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.2

**1. Give first the step you will use to separate the variable and then solve the equation.**

a) x – 1 = 0

Add 1 to both sides of the equation

x – 1 + 1 = 0 + 1

**Hence x = 1**

b) x + 1 = 0

Subtract 1 from both sides of the equation

x + 1 – 1 = 0 – 1

**Hence x = -1**

c) x – 1 = 5

Add 1 to both sides of the equation

x – 1 + 1 = 5 + 1

**Hence x = 6**

d) x + 6 = 2

Subtract 6 from both sides of the equation

x + 6 – 6 = 2 – 6

**Hence x = -4**

e) y – 4 = -7

Add 4 to both sides of the equation

y – 4 + 4 = -7 + 4

**Hence y = -3**

f) y – 4 = 4

Add 4 to both sides of the equation

y – 4 +4 = 4 + 4

**Hence y = 8**

g) y + 4 = 4

Subtract 4 from both sides of the equation

y + 4 – 4 = 4 – 4

**Hence y = 0**

h) y + 4 = – 4

Subtract 4 from both sides of the equation

y + 4 – 4 = -4 – 4

**Hence y = -8**

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

a) 3l = 42

Dividing both sides by 3 we get,

=

**Hence l = 14**

b) = 6

Multiplying both sides by 2 we get,

(2) = 6(2).

**Hence b = 12**

c) = 4

Multiplying both sides by 7 we get,

(7) = 4 (7)

**Hence p = 28**

d) 4x = 25

Dividing both sides by 4 we get,

=

**Hence x = **

e) 8y = 36

Dividing both sides by 8 we get,

=

**Hence y = ** ** = **

f) =

Multiplying both sides by 3 we get,

x (3) = x (3)

**Hence z = **

g) =

Multiplying both sides by 5 we get,

x (5) = x (5)

**Hence a = **

h) 20t = -10

Dividing both sides by 20 we get,

=

**Hence t = **

**3. Give the steps you will use to separate the variable and then solve the equation**

a) 3n – 2 = 46

Adding 2 to both sides we get,

3n – 2 + 2 = 46 + 2

3n = 48

Dividing both sides by 3 we get,

=

**Hence n = 16**

b) 5m + 7 = 17

Subtracting 7 from both sides we get,

5m + 7 – 7 = 17 – 7

5m = 10

Dividing both sides by 5 we get,

=

**Hence m = 2**

c) = 40

Multiplying both sides by 3 we get,

(3) = 40(3)

20p = 40 (3)

Dividing both sides by 20 we get,

=

**Hence p = 6 **

d) = 6

Multiplying both sides by 10 we get,

(10) = 6(10)

3p = 60

Dividing both sides by 3 we get,

=

**Hence p = 20**

**4. Solve the following equations: **

(a) 10p = 100

Dividing both sides by 10 we get,

=

**Hence p = 10**

(b) 10p + 10 = 10

Subtracting 10 from both sides we get,

10p + 10 – 10 = 10 – 10

10p = 0

Dividing both sides by 10 we get,

=

**Hence p = 0**

(c) = 5

Multiplying both sides by 4 we get,

(4) = 5(4)

**Hence p = 20**

(d) = 5

Multiplying both sides by -3 (so that p becomes positive)we get,

(-3) = 5(-3)

**Hence p = -15**

(e) = 6

Multiplying both sides by 4 we get,

(4) = 6(4)

3p = 24

Dividing both sides by 3 we get,

=

**Hence p = 8**

(f) 3s = -9

Dividing both sides by 3 we get,

=

**Hence s = -3**

(g) 3s + 12 = 0

Subtracting 12 from both sides we get,

3s + 12 – 12 = 0 – 12

3s = -12

Dividing both sides by 3 we get,

=

**Hence s = -4 **

(h) 3s = 0

Dividing both sides by 3 we get,

=

**Hence s = 0**

(i) 2q = 6

Dividing both sides by 2 we get,

=

**Hence q = 3**

(j) 2q – 6 = 0

Adding 6 to both sides we get,

2q -6 + 6 = 0 + 6

2q = 6

Dividing both sides by 2 we get,

=

**Hence q = 3**

(k) 2q + 6 = 0

Subtracting 6 from both sides we get,

2q + 6 – 6 = 0 – 6

2q = -6

Dividing both sides by 2 we get,

=

**Hence q = -3**

(l) 2q + 6 = 12

Subtracting 6 from both sides we get,

2q +6 – 6 = 12 – 6

2q = 6

Dividing both sides by 2 we get,

=

**Hence q = 3**

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.3

**1. Solve the following equations**

a) 2y + =

Transposing (+ ) to RHS we get,

2y = –

2y = = 16

Divide both sides by 2

=

**Hence y = 8**

b)5t + 28 = 10

Transposing (+ 28) to RHS we get,

5t = 10 – 28

5t = -18

Divide both sides by 5

=

**Hence t = **

c) + 3 = 2

Transposing (+3) to RHS we get,

= 2 – 3

= -1

Multiplying both sides by 5

(5) = -1 (5)

**Hence a = -5**

d) + 7 = 5

Transposing (+7) to RHS we get,

= 5 – 7

= – 2

Multiply both sides by 4

(4) = – 2(4)

**Hence q = -8**

e) x = -5

Multiply both sides by 2

(2) = -5(2)

5x = -10

Divide both sides by 5

=

**Hence x = -2**

f) =

Multiply both sides by 2

(2) = (2)

5x = =

Divide both sides by 5

= = .

**Hence x = **

g) 7m + = 13

Transposing to RHS we get,

7m = 13 –

7m = –

7m =

Divide both sides by 7

7m = .

**Hence m = **

h) 6z + 10 = -2

Transposing (+10) to RHS we get,

6z = -2 -10

6z = -12

Divide both sides by 6

=

**Hence z = -2**

i) =

Multiply both sides by

. = .

**Hence l = **

j) – 5 = 3

Transposing (-5) to RHS we get,

= 3 + 5

= 8

Multiply both sides by

. = 8.

**Hence b = 12**

**2. Solve the following equations:**

a) 2(x + 4) = 12

Divide both sides by 2, so as to remove the brackets in the LHS,we get,

x + 4 =

x + 4 = 6

Transposing (+4) to RHS we get,

x = 6 – 4

**x = 2**

b) 3(n – 5) = 21

Divide both sides by 3, so as to remove the brackets in the LHS,we get,

n – 5 =

n – 5 = 7

Transposing (-5) to RHS

n = 7 + 5

**Hence n = 12**

c) 3(n – 5) = – 21

Divide both sides by 3, so as to remove the brackets in the LHS,we get,

n – 5 =

n – 5 = -7

Transposing (-5) to RHS

n = -7 + 5

**Hence n = -2**

d) – 4(2 + x) = 8

Divide both sides by -4, so as to remove the brackets in the LHS,we get,

2 + x =

2 + x = -2

Transposing (+2) to RHS|

x = -2 – 2

**Hence x = -4**

e) 4(2 – x) = 8

Divide both sides by 4, so as to remove the brackets in the LHS,we get,

2 – x =

2 – x = 2

Transposing (+2) to RHS

-x = 2 – 2

-x = 0

Multiplying -1 (to make x positive) to both sides we get,

**Hence x = 0**

**3. Solve the following equations:**

a) 4 = 5(p – 2)

The above equation can also be written as 5(p – 2) = 4

Thus, divide both sides by 5, so as to remove the brackets in the LHS,we get,

p – 2 =

Transposing (-2) to RHS we get,

p = + 2

p = +

**Hence p = **

b) – 4 = 5(p – 2)

The above equation can also be written as 5(p – 2) = -4

Thus, divide both sides by 5, so as to remove the brackets in the LHS,we get,

p – 2 =

Transposing (-2) to RHS we get,

p = + 2

p = +

**Hence p = **

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

The above equation can also be written as 4 + 3(t + 2) = 16

Transposing (+4) to RHS

3(t + 2) = 16 – 4

3(t + 2) = 12

Divide both sides by 3, so as to remove the brackets in the LHS,we get,

t + 2 =

t + 2 = 4

Transposing (+2) to RHS we get,

t = 4 – 2

**Hence t = 2**

d) 4 + 5(p – 1) =34

Transposing (+4) to RHS

5(p – 1) = 34 – 4

5(p – 1) = 30

Divide both sides by 5, so as to remove the brackets in the LHS,we get,

p – 1 =

p – 1 = 6

Transposing (-1) to RHS we get,

p = 6 + 1

**Hence p = 7**

e) 0 = 16 + 4(m – 6)

The above equation can also be written as 16 + 4(m – 6) = 0

Transposing (+16) to RHS

4(m – 6) = -16

Divide both sides by 4, so as to remove the brackets in the LHS,we get,

m – 6 =

m – 6 = -4

Transposing (-6) to RHS we get,

m = -4 + 6

**Hence m = 2**

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

(i) x = 2

Multiply both sides by 4

**4x = 8**

(ii) x = 2

Add 3 to both sides

**x + 3 = 5**

(iii) x = 2

Subtract 11 from both sides

x – 11 = 2 – 11

**x – 11 = -9 **

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

Follow the same steps as the previous question and construct 3 different equations.

You can multiply, divide, add or subtract any number as you wish.

### NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.4

**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

Let, the number be ‘x’

Eight times a number is 8x

Add 4 to eight times a number is 8x + 4

You get 60

i.e 8x + 4 = 60

8x = 60 – 4 (Transposing +4 to RHS)

8x = 56

**Therefore x = 7 (Dividing both sides by 8)**

b) One-fifth of a number minus 4 gives 3.

Let, the number be ‘x’

One-fifth of a number is

Subtracting 4 from it i.e – 4

The result gives 3

i.e – 4 = 3

– = 3

= 3 (Taking LCM)

x – 20 = 15 (Multiplying both sides by 5)

x = 15 + 20 (Transposing -20 to RHS)

**Therefore x = 35**

c) If I take three-fourths of a number and add 3 to it, I get 21.

Let, the number be ‘x’

Three-fourths of a number is

Add 3 to it + 3

The result gives 21

i.e + 3 = 21

+ = 21 (Taking LCM)

= 21

3x + 12 = 84 (Multiplying both sides by 4)

3x = 84 – 12 (Transposing +12 to RHS)

3x = 72

**Therefore x = 24 (Dividing both sides by 3)**

d) When I subtracted 11 from twice a number, the result was 15

Let, the number be ‘x’

Twice a number is 2x

I subtracted 11 from it, 2x – 11

Result was 15

i.e 2x – 11 = 15

2x = 15 + 11 (Transposing -11 to RHS)

2x = 26

**Therefore x = 13 (Dividing both sides by 2)**

e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

Let the number of notebooks = x

Thrice the number of notebooks = 3x

Munna subtracts thrice the number of notebooks he has from 50 ,

i.e 50 – 3x

The result is 8

i e 50 – 3x = 8

50 – 8 = 3x (Transposing +8 to LHS and -3x to RHS)

The above equation can be written as 3x = 42

**Therefore x = 14 (Dividing both sides by 3)**

f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.

Let the number = x

Add 19 to it; x + 19

Divide the sum by 5 ;

The result is 8

Therefore = 8

Multiply both sides by 5, we get,

x + 19 = 40

x = 40 – 19 (Transposing +19 to RHS)

**Therefore x = 21 **

g) Anwar thinks of a number. If he takes away 7 from of the number, the result is .

Let the number = x

of the number =

Take away 7 from of the number; – 7

Result is

Therefore – 7 =

– = (Taking LCM)

=

5x – 14 = 11 (Multiplying both sides by 2)

5x = 11 + 14 (Transposing -14 to RHS)

5x = 25

**Therefore x = 5 (Dividing both sides by 5)**

**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 marks plus 7. The highest score is 87. What is the lowest score?

**Solution:**

Let lowest mark = x

Given highest score equals 87 and it is twice the lowest mark plus 7

87 = 2x + 7

The above equation can be written as 2x + 7 = 87

2x = 87 – 7 (Transposing +7 to RHS)

2x = 80

x = 40 (Dividing both sides by 2)

**Therefore the lowest score is 40**

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°).

**Solution:**

Given:

Vertex angle = 40°

Triangle is an isosceles triangle

Let the base angles = b^{o }Therefore, b^{o }+ b^{o}+ 40^{o} = 180^{o} (Sum of three angles of a triangle is 180°)

2b^{o} + 40^{o} = 180^{o }2b^{o} = 180^{o} – 40^{o} (Transposing +40 to RHS)

2b^{o} = 140^{o }b^{o} = 70^{o} (Dividing 2 on both sides)

**The base angles of the triangle is 70 ^{o}**

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:**

Let Rahul’s score = x

Therefore Sachin’s score = 2x

Two short of a double century is 200 – 2 = 198

Since together their runs fell two short of a double century

2x + x = 198

3x = 198

**x = 66 which is Rahul’s score. **

**Sachin’s score is 2 x 66 = 132**

**3. Solve the following:**

(i) 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?

**Solution:**

Let number of marbles Parmit has = x

Number of marbles Irfan has = 5x + 7

i.e 5x + 7 = 37

5x = 37 – 7 (Transposing +7 to RHS)

5x = 30

**x = 6 is the number of marbles Parmit has.**

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

**Solution:**

Let Laxmi’s age = x

Laxmi’s father’s age is 3x + 4 = 49

3x = 49 – 4 (Transposing +4 to RHS)

3x = 45

**x = 15 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 trees planted if the number of non-fruit trees planted was 77?

**Solution:**

Let the number of fruit trees = x

Number of non-fruit trees planted = 77 (Given)

The number of non-fruit trees were two more than three times the number of fruit trees , therefore

77 = 3x + 2

i.e 3x + 2 = 77

3x = 77 – 2 (Transposing +2 to RHS)

3x = 75

**x = 25 is the number of fruit trees planted.**

**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:**

Let the number = x

Take the number seven times over and add a fifty i.e 7x + 50

To reach a triple century you still need forty i.e 300 = 7x + 50 + 40

7x + 90 = 300

7x = 300 – 90 (Transposing +90 to RHS)

7x = 210

**x = 30 is the number**

**This Chapter contains the Exercises relating to the following topics , which are discussed in Chapter 4 – Simple Equations Class 7 NCERT book : –**

- 4.1 A MIND-READING GAME!
- 4.2 SETTING UP OF AN EQUATION
- 4.3 REVIEW OF WHAT WE KNOW
- 4.4 WHAT EQUATION IS?
- 4.4.1 Solving an Equation
- 4.5 MORE EQUATIONS
- 4.6 FROM SOLUTION TO EQUATION
- 4.7 APPLICATIONS OF SIMPLE EQUATIONS TO PRACTICAL SITUATIONS

**Download NCERT Solutions for Class 7 Maths Chapter 4 – Simple Equations**