**Linear Equations in One Variable**- In this chapter students will get to know about

**linear expressions in one variable**. A short revision of previously discussed topics is done in the beginning.

- An
**algebraic equation**is an**equality**involving**variables**having**equality sign**.

**Solving equations which have a linear expression on one side and numbers on the other side**is explained.

Students can practice the questions based on the same through exercise 2.1. Section 2.3 lays emphasis on

**Some Applications of linear equations in one variable**. A puzzle is given here to make the topic interesting for students. Several solved examples are discussed before moving to the next exercise. These examples are in the form of word problems.

The next topic is

**Solving equations having the variable on both sides**. In this section, students will learn how to solve such equations which have expressions with the

**variable**on both sides.

- The expressions forming
**equations**have to be simplified before it can be solved with usual methods. - Some equations may not even be
**linear**in the beginning, but they can be brought to**linear form**by applying suitable operations like**transposing**etc.

**some more applications**involving age, two digit numbers, dimensions of the given figure etc. the next part of the chapter deals with

**Reducing Equations to Simpler Form**.

- The utility of
**linear equations**is in their diverse applications such as:

b. Ages

c. Perimeters

d. Combination of currency notes

The last exercise is based on the same containing 7 questions. For a better understanding of this chapter and to increase the fluency of forming

**linear equations**, students must do a good practice of all solved and unsolved problems.

For conclusion, a recap of a few important points is discussed.

#### Page No 23:

#### Question 1:

Solve:

#### Answer:

*x* −
2 = 7

Transposing 2 to R.H.S, we obtain

*x* =
7 + 2 = 9

#### Page No 23:

#### Question 2:

Solve:

#### Answer:

*y* +
3 = 10

Transposing 3 to R.H.S, we obtain

*y* =
10 − 3 = 7

#### Page No 23:

#### Question 3:

Solve:

#### Answer:

6 = *z*
+ 2

Transposing 2 to L.H.S, we obtain

6 −
2 = *z*

*z* =
4

#### Page No 23:

#### Question 4:

Solve:

#### Answer:

Transposingto R.H.S, we obtain

#### Page No 23:

#### Question 5:

Solve:

#### Answer:

6*x*
= 12

Dividing both sides by 6, we obtain

*x* =
2

#### Page No 23:

#### Question 6:

Solve:

#### Answer:

Multiplying both sides by 5, we obtain

#### Page No 23:

#### Question 7:

Solve:

#### Answer:

Multiplying both sides by, we obtain

#### Page No 23:

#### Question 8:

Solve:

#### Answer:

Multiplying both sides by 1.5, we obtain

##### Video Solution for linear equations in one variable (Page: 23 , Q.No.: 8)

NCERT Solution for Class 8 maths - linear equations in one variable 23 , Question 8

#### Page No 23:

#### Question 9:

Solve:

#### Answer:

7*x*
− 9 = 16

Transposing 9 to R.H.S, we obtain

7*x*
= 16 + 9

7*x*
= 25

Dividing both sides by 7, we obtain

#### Page No 24:

#### Question 10:

Solve:

#### Answer:

14*y*
− 8 = 13

Transposing 8 to R.H.S, we obtain

14*y*
= 13 + 8

14*y*
= 21

Dividing both sides by 14, we obtain

#### Page No 24:

#### Question 11:

Solve:

#### Answer:

17 + 6*p*
= 9

Transposing 17 to R.H.S, we obtain

6*p*
= 9 − 17

6*p*
= −8

Dividing both sides by 6, we obtain

#### Page No 24:

#### Question 12:

Solve:

#### Answer:

Transposing 1 to R.H.S, we obtain

Multiplying both sides by 3, we obtain

#### Page No 28:

#### Question 1:

If you subtract from a number and multiply the result by, you get. What is the number?

#### Answer:

Let the
number be *x*. According to the question,

On multiplying both sides by 2, we obtain

On transposing to R.H.S, we obtain

Therefore, the number is .

#### Page No 28:

#### Question 2:

The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

#### Answer:

Let the breadth be *x* m. The length will be (2*x* + 2) m.

Perimeter of swimming pool = 2(*l + b*) = 154 m

2(2*x* + 2 + *x*) = 154

2(3*x* + 2) = 154

Dividing both sides by 2, we obtain

3*x *+ 2 = 77

On transposing 2 to R.H.S, we obtain

3*x* = 77 − 2

3*x* = 75

On dividing both sides by 3, we obtain

*x* = 25

2*x* + 2 = 2 × 25 + 2 = 52

Hence, the breadth and length of the pool are 25 m and 52 m respectively.

##### Video Solution for linear equations in one variable (Page: 28 , Q.No.: 2)

NCERT Solution for Class 8 maths - linear equations in one variable 28 , Question 2

#### Page No 28:

#### Question 3:

The base of an isosceles triangle is cm. The perimeter of the triangle is cm. What is the length of either of the remaining equal sides?

#### Answer:

Let the
length of equal sides be *x* cm.

Perimeter
= *x *cm *+ x *cm + Base =
cm

On transposing to R.H.S, we obtain

On dividing both sides by 2, we obtain

Therefore, the length of equal sides is cm.

#### Page No 28:

#### Question 4:

Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

#### Answer:

Let one
number be* x*. Therefore, the other number
will be *x*
+ 15.

According to the question,

*x
+ x* + 15 = 95

2*x*
+ 15 = 95

On transposing 15 to R.H.S, we obtain

2*x*
= 95 − 15

2*x*
= 80

On dividing both sides by 2, we obtain

*x* =
40

*x* +
15 = 40 + 15 = 55

Hence, the numbers are 40 and 55.

#### Page No 28:

#### Question 5:

Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

#### Answer:

Let
the common ratio between these numbers be *x*.
Therefore, the numbers will be 5*x*
and 3*x*
respectively.

Difference between these numbers = 18

5*x*
− 3*x*
= 18

2*x*
= 18

Dividing both sides by 2,

*x* =
9

First
number = 5*x*
= 5 × 9 = 45

Second
number = 3*x*
= 3 × 9 = 27

#### Page No 28:

#### Question 6:

Three consecutive integers add up to 51. What are these integers?

#### Answer:

Let
three consecutive integers be*
x*, *x*
+ 1, and* x*
+ 2.

Sum
of these numbers* = x+ x*
+ 1 + *x* +
2 = 51

3*x*
+ 3 = 51

On transposing 3 to R.H.S, we obtain

3*x*
= 51 − 3

3*x*
= 48

On dividing both sides by 3, we obtain

*x*
= 16

*x*
+ 1 = 17

*x*
+ 2 = 18

Hence, the consecutive integers are 16, 17, and 18.

#### Page No 28:

#### Question 7:

The sum of three consecutive multiples of 8 is 888. Find the multiples.

#### Answer:

Let the
three consecutive multiples of 8 be 8*x*, 8(*x* + 1), 8(*x*
+ 2).

Sum
of these numbers* =*
8*x *+ 8(*x*
+ 1) + 8(*x*
+ 2) = 888

8(*x*
+ *x* + 1 +*
x *+ 2) = 888

8(3*x*
+ 3) = 888

On dividing both sides by 8, we obtain

3*x*
+ 3 = 111

On transposing 3 to R.H.S, we obtain

3*x*
= 111 − 3

3*x*
= 108

On dividing both sides by 3, we obtain

*x*
= 36

First
multiple = 8*x = *8 × 36 = 288

Second
multiple = 8(*x* + 1) = 8 × (36 + 1) = 8 × 37
= 296

Third
multiple = 8(*x* + 2) = 8 × (36 + 2) = 8 × 38
= 304

Hence, the required numbers are 288, 296, and 304.

#### Page No 28:

#### Question 8:

Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

#### Answer:

Let three consecutive
integers be *x*,
*x* + 1, *x*
+ 2. According to the question,

2*x
*+ 3(*x*
+ 1) + 4(*x*
+ 2) = 74

2*x*
+ 3*x* + 3 +
4*x* + 8 =
74

9*x*
+ 11 = 74

On transposing 11 to R.H.S, we obtain

9*x*
= 74 − 11

9*x*
= 63

On dividing both sides by 9, we obtain

*x*
= 7

*x*
+ 1 = 7 + 1 = 8

*x*
+ 2 = 7 + 2 = 9

Hence, the numbers are 7, 8, and 9.

#### Page No 28:

#### Question 9:

The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

#### Answer:

Let
common ratio between Rahul’s age and Haroon’s age
be *x*.

Therefore,
age of Rahul and Haroon will be 5*x*
years and 7*x*
years respectively. After 4 years, the age of Rahul and Haroon will
be (5*x *+
4) years and (7*x *+
4) years respectively.

According to the given question, after 4 years, the sum of the ages of Rahul and Haroon is 56 years.

∴
(5*x *+
4 + 7*x *+
4) = 56

12*x
*+ 8 = 56

On transposing 8 to R.H.S, we obtain

12*x*
= 56 − 8

12*x*
= 48

On dividing both sides by 12, we obtain

*x*
= 4

Rahul’s
age = 5*x*
years = (5 × 4) years = 20 years

Haroon’s
age = 7*x*
years = (7 × 4) years = 28 years

#### Page No 28:

#### Question 10:

The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

#### Answer:

Let the
common ratio between the number of boys and numbers of girls be *x*.

Number of
boys = 7*x*

Number of
girls = 5*x*

According to the given question,

Number of boys = Number of girls + 8

∴
7*x* = 5*x* + 8

On
transposing 5*x* to L.H.S, we obtain

7*x*
− 5*x* = 8

2*x*
= 8

On dividing both sides by 2, we obtain

*x* =
4

Number of
boys = 7*x* = 7 × 4 = 28

Number of
girls = 5*x* = 5 × 4 = 20

Hence, total class strength = 28 + 20 = 48 students

#### Page No 28:

#### Question 11:

Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

#### Answer:

Let
Baichung’s father’s age be *x* years. Therefore,
Baichung’s age and Baichung’s grandfather’s age
will be (*x* − 29) years and (*x* + 26) years
respectively.

According to the given question, the sum of the ages of these 3 people is 135 years.

∴ *x
+ x* − 29 + *x* + 26 = 135

3*x*
− 3 = 135

On transposing 3 to R.H.S, we obtain

3*x*
= 135 + 3

3*x*
= 138

On dividing both sides by 3, we obtain

*x* =
46

Baichung’s
father’s age = *x* years = 46 years

Baichung’s
age = (*x* − 29) years = (46 − 29) years = 17 years

Baichung’s
grandfather’s age = (*x* + 26) years = (46 + 26) years =
72 years

#### Page No 28:

#### Question 12:

Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

#### Answer:

Let Ravi’s
present age be *x* years.

Fifteen years later, Ravi’s age = 4 × His present age

*x* +
15 = 4*x*

On
transposing *x* to R.H.S, we obtain

15 = 4*x*
− *x*

15 = 3*x*

On dividing both sides by 3, we obtain

5 = *x*

Hence, Ravi’s present age = 5 years

#### Page No 28:

#### Question 13:

A rational number is such that when you multiply it by and add to the product, you get. What is the number?

#### Answer:

Let the
number be *x*.

According to the given question,

On transposing to R.H.S, we obtain

On multiplying both sides by, we obtain

Hence, the rational number is.

#### Page No 28:

#### Question 14:

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4, 00,000. How many notes of each denomination does she have?

#### Answer:

Let the common ratio between the numbers of notes of different denominations be *x*. Therefore, numbers of Rs 100 notes, Rs 50 notes, and Rs 10 notes will be2*x*, 3*x*, and 5*x* respectively.

Amount of Rs 100 notes = Rs (100 × 2*x*) = Rs 200*x*

Amount of Rs 50 notes = Rs (50 × 3*x*)= Rs 150*x*

Amount of Rs 10 notes = Rs (10 × 5*x*) = Rs 50*x*

It is given that total amount is Rs 400000.

∴ 200*x* + 150*x* + 50*x* = 400000

⇒ 400*x* = 400000

On dividing both sides by 400, we obtain

*x* = 1000

Number of Rs 100 notes = 2*x* = 2 × 1000 = 2000

Number of Rs 50 notes = 3*x* = 3 × 1000 = 3000

Number of Rs 10 notes = 5*x* = 5 × 1000 = 5000

##### Video Solution for linear equations in one variable (Page: 28 , Q.No.: 14)

NCERT Solution for Class 8 maths - linear equations in one variable 28 , Question 14

#### Page No 28:

#### Question 15:

I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

#### Answer:

Let the
number of Rs 5 coins be *x*.

Number of
Rs 2 coins = 3 × Number of Rs 5 coins = 3*x*

Number of Re 1 coins = 160 − (Number of coins of Rs 5 and of Rs 2)

= 160 − (3*x* + *x*) = 160 − 4*x*

Amount of
Re 1 coins = Rs [1 × (160 − 4*x*)] = Rs (160 −
4*x*)

Amount of Rs 2 coins = Rs (2 × 3*x*)= Rs 6*x*

Amount of
Rs 5 coins = Rs (5 × *x*) = Rs 5*x*

It is given that the total amount is Rs 300.

∴
160 − 4*x* + 6*x* + 5*x* = 300

160 + 7*x*
= 300

On transposing 160 to R.H.S, we obtain

7*x*
= 300 − 160

7*x*
= 140

On dividing both sides by 7, we obtain

*x* =
20

Number of
Re 1 coins = 160 − 4*x* = 160 − 4 × 20 = 160 −
80 = 80

Number of
Rs 2 coins = 3*x* = 3 × 20 = 60

Number of
Rs 5 coins = *x* = 20

#### Page No 28:

#### Question 16:

The organizers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3000. Find the number of winners, if the total number of participants is 63.

#### Answer:

Let the
number of winners be *x*. Therefore, the number of participants
who did not win will be 63 − *x*.

Amount given to the winners = Rs (100 × *x*) = Rs 100*x*

Amount given to the participants who did not win = Rs [25(63 −
*x*)]

= Rs (1575 − 25*x*)

According to the given question,

100*x*
+ 1575 − 25*x *= 3000

On transposing 1575 to R.H.S, we obtain

75*x *=
3000 − 1575

75*x *=
1425

On dividing both sides by 75, we obtain

*x* =
19

Hence, number of winners = 19

#### Page No 30:

#### Question 1:

Solve
and check result: 3*x*
= 2*x*
+ 18

#### Answer:

3*x*
= 2*x* + 18

On
transposing 2*x* to L.H.S, we obtain

3*x*
− 2*x* = 18

*x* =
18

L.H.S = 3*x*
= 3 × 18 = 54

R.H.S = 2*x*
+ 18 = 2 × 18 + 18 = 36 + 18 = 54

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 2:

Solve
and check result: 5*t*
− 3 = 3*t*
− 5

#### Answer:

5*t*
− 3 = 3*t* − 5

On
transposing 3*t* to L.H.S and −3 to R.H.S, we obtain

5*t*
− 3*t* = −5 − (−3)

2*t*
= −2

On dividing both sides by 2, we obtain

*t* =
−1

L.H.S = 5*t*
− 3 = 5 × (−1) − 3 = −8

R.H.S = 3*t*
− 5 = 3 × (−1) − 5 = − 3 − 5 = −8

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 3:

Solve
and check result: 5*x*
+ 9 = 5 + 3*x*

#### Answer:

5*x*
+ 9 = 5 + 3*x*On transposing 3

*x*to L.H.S and 9 to R.H.S, we obtain

5*x*
− 3*x *= 5 − 9

2*x*
= −4

On dividing both sides by 2, we obtain

*x *=
−2

L.H.S = 5*x*
+ 9 = 5 × (−2) + 9 = −10 + 9 = −1

R.H.S = 5
+ 3*x* = 5 + 3 × (−2) = 5 − 6 = −1

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 4:

Solve
and check result: 4*z*
+ 3 = 6 + 2*z*

#### Answer:

4*z*
+ 3 = 6 + 2*z*

On
transposing 2*z* to L.H.S and 3 to R.H.S, we obtain

4*z*
− 2*z* = 6 − 3

2*z*
= 3

Dividing both sides by 2, we obtain

L.H.S = 4*z*
+ 3 = 4 ×
+ 3 = 6 + 3 = 9

R.H.S = 6
+ 2*z* = 6 + 2 ×
= 6 + 3 = 9

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 5:

Solve
and check result: 2*x*
− 1 = 14 − *x*

#### Answer:

2*x*
− 1 = 14 − *x*

Transposing
*x* to L.H.S and 1 to R.H.S, we obtain

2*x*
+ *x* = 14 + 1

3*x*
= 15

Dividing both sides by 3, we obtain

*x* =
5

L.H.S = 2*x*
− 1 = 2 × (5) − 1 = 10 − 1 = 9

R.H.S = 14
− *x* = 14 − 5 = 9

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 6:

Solve
and check result: 8*x*
+ 4 = 3(*x*
− 1) + 7

#### Answer:

8*x*
+ 4 = 3(*x *− 1) + 7

8*x*
+ 4 = 3*x* − 3 + 7

Transposing
3*x* to L.H.S and 4 to R.H.S, we obtain

8*x *−
3*x* = − 3 + 7 − 4

5*x*
= − 7 + 7

L.H.S = 8*x*
+ 4 = 8 × (0) + 4 = 4

R.H.S =
3(*x* − 1) + 7 = 3 (0 − 1) + 7 = − 3 + 7 = 4

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 7:

Solve and check result:

#### Answer:

Multiplying both sides by 5, we obtain

5*x*
= 4(*x* + 10)

5*x*
= 4*x* + 40

Transposing
4*x* to L.H.S, we obtain

5*x*
− 4*x* = 40

*x* =
40

L.H.S = *x*
= 40

R.H.S = =

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 8:

Solve and check result:

#### Answer:

Transposing to L.H.S and 1 to R.H.S, we obtain

Multiplying both sides by 5, we obtain

*x* =
10

L.H.S = =

R.H.S =

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 9:

Solve and check result:

#### Answer:

Transposing
*y* to L.H.S and
to
R.H.S, we obtain

Dividing both sides by 3, we obtain

L.H.S =

R.H.S =

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 30:

#### Question 10:

Solve and check result:

#### Answer:

Transposing
5*m* to L.H.S, we obtain

Dividing both sides by −2, we obtain

L.H.S =

R.H.S =

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

#### Page No 31:

#### Question 1:

Amina thinks of a number and subtracts from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

#### Answer:

Let the
number be *x*.

According to the given question,

=
3*x*

8*x*
− 20 = 3*x*

Transposing
3*x* to L.H.S and −20 to R.H.S, we obtain

8*x*
− 3*x* = 20

5*x*
= 20

Dividing both sides by 5, we obtain

*x* =
4

Hence, the number is 4.

#### Page No 31:

#### Question 2:

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

#### Answer:

Let the
numbers be *x* and 5*x*. According to the question,

21 + 5*x*
= 2(*x* + 21)

21 + 5*x
*= 2*x* + 42

Transposing
2*x* to L.H.S and 21 to R.H.S, we obtain

5*x *−
2*x* = 42 − 21

3*x*
= 21

Dividing both sides by 3, we obtain

*x *=
7

5*x*
= 5 × 7 = 35

Hence, the numbers are 7 and 35 respectively.

#### Page No 31:

#### Question 3:

Sum of the digits of a two digit number is 9. When we interchange the digits it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

#### Answer:

Let the
digits at tens place and ones place be *x* and 9 − *x*
respectively.

Therefore,
original number = 10*x* + (9 − *x*) = 9*x* + 9

On
interchanging the digits, the digits at ones place and tens place
will be *x* and 9 − *x* respectively.

Therefore,
new number after interchanging the digits = 10(9 − *x*) +
*x*

= 90 − 10*x* + *x*

= 90 − 9*x*

According to the given question,

New number = Original number + 27

90 −
9*x* = 9*x* + 9 + 27

90 −
9*x* = 9*x* + 36

Transposing
9*x* to R.H.S and 36 to L.H.S, we obtain

90 −
36 = 18*x*

54 = 18*x*

Dividing both sides by 18, we obtain

3 = *x
*and 9 − *x* = 6

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

Therefore,
the two-digit number is 9*x* + 9 = 9 × 3 + 9 = 36

#### Page No 31:

#### Question 4:

One of the two digits of a two digit number is three times the other digit. If you interchange the digit of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

#### Answer:

Let the
digits at tens place and ones place be *x* and 3*x*
respectively.

Therefore,
original number = 10*x* + 3*x* = 13*x*

On
interchanging the digits, the digits at ones place and tens place
will be* x* and 3*x* respectively.

Number
after interchanging = 10 × 3*x* + *x* = 30*x* +
*x* = 31*x*

According to the given question,

Original number + New number = 88

13*x*
+ 31*x* = 88

44*x*
= 88

Dividing both sides by 44, we obtain

*x* =
2

Therefore,
original number = 13*x *= 13 × 2 = 26

By
considering the tens place and ones place as 3*x* and *x*
respectively, the two-digit number obtained is 62.

Therefore, the two-digit number may be 26 or 62.

#### Page No 31:

#### Question 5:

Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of this mother’s present age. What are their present ages?

#### Answer:

Let Shobo’s age be *x* years. Therefore, his mother’s age will be 6*x* years.

According to the given question,

*x* + 5 = 2*x*

Transposing *x* to R.H.S, we obtain

5 = 2*x* − *x*

5 = *x*

6*x* = 6 × 5 = 30

Therefore, the present ages of Shobo and Shobo’s mother will be 5 years and 30 years respectively.

##### Video Solution for linear equations in one variable (Page: 31 , Q.No.: 5)

NCERT Solution for Class 8 maths - linear equations in one variable 31 , Question 5

#### Page No 31:

#### Question 6:

There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs 100 per metre it will cost the village panchayat Rs 75, 000 to fence the plot. What are the dimensions of the plot?

#### Answer:

Let the common ratio between the length and breadth of the rectangular plot be *x*. Hence, the length and breadth of the rectangular plot will be 11*x* m and 4*x* m respectively.

Perimeter of the plot = 2(Length + Breadth)

It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.

∴ 100 × Perimeter = 75000

100 × 30*x* = 75000

3000*x* = 75000

Dividing both sides by 3000, we obtain

*x* = 25

Length = 11*x* m = (11 × 25) m = 275 m

Breadth = 4*x* m = (4 × 25) m = 100 m

Hence, the dimensions of the plot are 275 m and 100 m respectively.

##### Video Solution for linear equations in one variable (Page: 31 , Q.No.: 6)

NCERT Solution for Class 8 maths - linear equations in one variable 31 , Question 6

#### Page No 31:

#### Question 7:

Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36660. How much trouser material did he buy?

#### Answer:

Let 2*x* m of trouser material and 3*x* m of shirt material
be bought by him.

Per metre selling price of trouser material = = Rs 100.80

Per metre selling price of shirt material = = Rs 55

Given that, total amount of selling = Rs 36660

100.80 × (2*x*) + 55 × (3*x*) = 36660

201.60*x *+ 165*x* = 36660

366.60*x***
= **36660

Dividing both sides by 366.60, we obtain

*x* =
100

Trouser
material = 2*x* m = (2 × 100) m = 200 m

#### Page No 32:

#### Question 8:

Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

#### Answer:

Let the number of deer be *x*.

Number of deer grazing in the field =

Number of deer drinking water from the pond = 9

Multiplying both sides by 8, we obtain

*x* = 72

Hence, the total number of deer in the herd is 72.

##### Video Solution for linear equations in one variable (Page: 32 , Q.No.: 8)

NCERT Solution for Class 8 maths - linear equations in one variable 32 , Question 8

#### Page No 32:

#### Question 9:

A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages

#### Answer:

Let the
granddaughter’s age be *x* years. Therefore, grandfather’s
age will be

10*x*
years.

According to the question,

Grandfather’s age = Granddaughter’s age + 54 years

10*x*
= *x* + 54

Transposing
*x* to L.H.S, we obtain

10*x*
− *x* = 54

9*x*
= 54

*x* =
6

Granddaughter’s
age* = x* years = 6 years

Grandfather’s
age = 10*x* years = (10 × 6) years = 60 years

#### Page No 32:

#### Question 10:

Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

#### Answer:

Let Aman’s
son’s age be *x* years. Therefore, Aman’s age will
be 3*x* years. Ten years ago, their age was (*x* −
10) years and (3*x* − 10) years respectively.

According to the question,

10 years ago, Aman’s age = 5 × Aman’s son’s age 10 years ago

3*x*
− 10 = 5(*x* − 10)

3*x*
− 10 = 5*x* − 50

Transposing
3*x* to R.H.S and 50 to L.H.S, we obtain

50 −
10 = 5*x* − 3*x*

40 = 2*x*

Dividing both sides by 2, we obtain

20 = *x*

Aman’s
son’s age = *x* years = 20 years

Aman’s
age = 3*x* years = (3 × 20) years = 60 years

#### Page No 33:

#### Question 1:

Solve
the linear equation**
**

#### Answer:

L.C.M. of the denominators, 2, 3, 4, and 5, is 60.

Multiplying both sides by 60, we obtain

⇒
30*x *− 12 = 20*x *+ 15 (Opening
the brackets)

⇒
30*x *− 20*x* = 15 + 12

⇒
10*x* = 27

⇒

#### Page No 33:

#### Question 2:

Solve the
linear equation**
**

#### Answer:

L.C.M. of the denominators, 2, 4, and 6, is 12.

Multiplying both sides by 12, we obtain

6*n*
− 9*n* + 10*n* = 252

⇒
7*n* = 252

#### Page No 33:

#### Question 3:

Solve
the linear equation**
**

#### Answer:

L.C.M. of the denominators, 2, 3, and 6, is 6.

Multiplying both sides by 6, we obtain

6*x*
+ 42 − 16*x *= 17 − 15*x*

⇒
6*x* − 16*x *+ 15*x *= 17 − 42

⇒
5*x* = −25

#### Page No 34:

#### Question 4:

Solve
the linear equation**
**

#### Answer:

L.C.M. of the denominators, 3 and 5, is 15.

Multiplying both sides by 15, we obtain

5(*x*
− 5) = 3(*x* − 3)

⇒
5*x* − 25 = 3*x* − 9 (Opening
the brackets)

⇒
5*x* − 3*x* = 25 − 9

⇒
2*x* = 16

#### Page No 34:

#### Question 5:

Solve the linear equation

#### Answer:

L.C.M. of the denominators, 3 and 4, is 12.

Multiplying both sides by 12, we obtain

3(3*t*
− 2) − 4(2*t* + 3) = 8 − 12*t*

⇒
9*t *− 6 − 8*t* − 12 = 8 −
12*t* (Opening the brackets)

⇒
9*t *− 8*t* + 12*t* = 8 + 6 + 12

⇒
13*t* = 26

#### Page No 34:

#### Question 6:

Solve the linear equation

#### Answer:

L.C.M. of the denominators, 2 and 3, is 6.

Multiplying both sides by 6, we obtain

6*m*
− 3(*m* − 1) = 6 − 2(*m* − 2)

⇒
6*m* − 3*m* + 3 = 6 − 2*m* + 4 (Opening
the brackets)

⇒
6*m* − 3*m* + 2*m* = 6 + 4 − 3

⇒
5*m* = 7

⇒

#### Page No 34:

#### Question 7:

Simplify and solve the linear equation

#### Answer:

3(*t*
− 3) = 5(2*t* + 1)

⇒
3*t* − 9 = 10*t* + 5 (Opening
the brackets)

⇒
−9 − 5 = 10*t* − 3*t*

⇒ −14 = 7t

#### Page No 34:

#### Question 8:

Simplify and solve the linear equation

#### Answer:

15(*y*
− 4) − 2(*y* − 9) + 5(*y* + 6) = 0

⇒
15*y *− 60 − 2*y* + 18 + 5*y* + 30 =
0 (Opening the brackets)

⇒
18*y *− 12 = 0

⇒
18*y *= 12

⇒

#### Page No 34:

#### Question 9:

Simplify and solve the linear equation

#### Answer:

3(5*z*
− 7) − 2(9*z* − 11) = 4(8*z* −
13)−17

⇒
15*z* − 21 − 18*z* + 22 = 32*z* − 52
− 17 (Opening the brackets)

⇒
−3*z* + 1 = 32*z* − 69

⇒
−3*z* − 32*z* = −69 − 1

⇒
−35*z* = −70

⇒

#### Page No 34:

#### Question 10:

Simplify and solve the linear equation

#### Answer:

0.25(4*f* − 3) = 0.05(10*f* − 9)

Multiplying both sides by 20, we obtain

5(4*f* − 3) = 10*f* − 9

⇒ 20*f *− 15 = 10*f* − 9 (Opening the brackets)

⇒ 20*f *− 10*f* = − 9 + 15

⇒ 10*f* = 6

⇒

##### Video Solution for linear equations in one variable (Page: 34 , Q.No.: 10)

NCERT Solution for Class 8 maths - linear equations in one variable 34 , Question 10

#### Page No 35:

#### Question 1:

Solve:
**
**

#### Answer:

On
multiplying both sides by 3*x*, we obtain

8*x*
− 3 = 6*x*

⇒
8*x* − 6*x* = 3

⇒
2*x* = 3

⇒

#### Page No 35:

#### Question 2:

Solve:
**
**

#### Answer:

Onmultiplying both sides by 7 − 6*x*, we obtain

9*x*
= 15(7 − 6*x*)

⇒
9*x* = 105 − 90*x*

⇒
9*x* + 90*x* = 105

⇒
99*x* = 105

⇒

#### Page No 35:

#### Question 3:

Solve:
**
**

#### Answer:

On
multiplying both sides by 9(*z* + 15), we obtain

9*z *=
4(*z* + 15)

⇒
9*z *= 4*z* + 60

⇒
9*z *− 4*z* = 60

⇒
5*z* = 60

⇒ *z*
= 12

#### Page No 35:

#### Question 4:

Solve:
**
**

#### Answer:

On
multiplying both sides by 5(2 − 6*y*), we obtain

5(3*y*
+ 4) = −2(2 − 6*y*)

⇒
15*y* + 20 = − 4 + 12*y*

⇒
15*y* − 12*y* = − 4 − 20

⇒
3*y* = −24

⇒ *y*
= −8

#### Page No 35:

#### Question 5:

Solve:
**
**

#### Answer:

On
multiplying both sides by 3(*y *+ 2), we obtain

3(7*y*
+ 4) = −4(*y* + 2)

⇒
21*y* + 12 = − 4*y* − 8

⇒
21*y* + 4*y* = − 8 − 12

⇒
25*y* = −20

⇒

#### Page No 35:

#### Question 6:

The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

#### Answer:

Let the common ratio between their ages be *x*. Therefore, Hari’s age and Harry’s age will be 5*x* years and 7*x* years respectively and four years later, their ages will be (5*x* + 4) years and (7*x* + 4) years respectively.

According to the situation given in the question,

Hari’s age = 5*x* years = (5 × 4) years = 20 years

Harry’s age = 7*x* years = (7 × 4) years = 28 years

Therefore, Hari’s age and Harry’s age are 20 years and 28 years respectively.

##### Video Solution for linear equations in one variable (Page: 35 , Q.No.: 6)

NCERT Solution for Class 8 maths - linear equations in one variable 35 , Question 6

#### Page No 35:

#### Question 7:

The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is. Find the rational number.

#### Answer:

Let the
numerator of the rational number be *x*. Therefore, its
denominator will

be *x*
+ 8.

The rational number will be. According to the question,

⇒
2(*x* + 17) = 3(*x* + 7)

⇒
2*x* + 34 = 3*x* + 21

⇒
34 − 21 = 3*x* − 2*x*

⇒13
= *x*

Numerator
of the rational number = *x* = 13

Denominator
of the rational number = *x* + 8 = 13 + 8 = 21

Rational number

View NCERT Solutions for all chapters of Class 8