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NCERT Solutions for Class 8: Chapter 2 - Linear Equations in One Variable

Check NCERT Solutions for Class 8 Chapter 2 - Linear Equations in One Variable. With this article, you can also download the PDF of this chapter.

Aug 1, 2019 16:07 IST
NCERT Solutions for Class 8: Chapter 2 - Linear Equations in One Variable

NCERT Solutions for Class 8 Chapter 2 - Linear Equations in One Variable is available here. This chapter is one of the most important chapters of CBSE Class 8. With this article, you can also download the PDF of this chapter. Download link is given at the end of this article.

NCERT Solutions for Class 8 Chapter 2 - Linear Equations in One Variable:

Chapter - 2

Linear Equations in One Variable

EXERCISE 2.1

Solve the following equations.

1. x – 2 = 7

Solutions:

x − 2 = 7

⇒x = 7 + 2 (by transposing 2 to R.H.S.)

⇒x= 9

2. y + 3 = 10

Solutions:

y + 3 = 10

⇒y = 10 − 3  (by transposing 3 to R.H.S.)

⇒y= 7.

3. 6 = z + 2

Solutions:

6 = z + 2

⇒6 − 2 = z

⇒z = 4

4. (3/7) + x = 17/7

Solutions:

(3/7) + x = 17/7

⇒x = (17/7) – (3/7) [By transposing 3/7 to R.H.S]

⇒x= 14/7

⇒x= 2

5. 6x = 12

Solutions:

6x = 12

⇒6x/6 = 12/6 [On dividing both sides by 6]

⇒x = 2

6. t/5 = 10

Solutions:

t/5 = 10

⇒(t/5) x 5 = 10 x 5 [On multiplying both sides by 5]

⇒t = 50

7. 2x/3 = 18

Solutions:

2x/3 = 18

⇒(2x/3) x (3/2) = 18 x (3/2) [On multiplying both sides by 3/2]

x = 27

8. 1.6 = y/1.5

Solutions:

1.6 = y/1.5

⇒1.6 x 1.5 = (y/1.5) x 1.5 [On multiplying both sides by 1.5]

⇒2.4 = y

9. 7x – 9 = 16

Solutions:

⇒7x − 9 = 16

⇒7x = 16 + 9

⇒7x = 25

⇒7x/7 = 25/7 [On dividing both sides by 7]

⇒x = 25/7

10. 14y – 8 = 13

Solutions:

14y − 8 = 13

⇒14y = 13 + 8

⇒14y = 21

⇒14y/14 = 21/14 [On dividing both sides by 14]

⇒y = 3/2

11. 17 + 6p = 9

Solutions:

17 + 6p = 9

⇒6p = 9 − 17

⇒6p = −8

⇒6p/6 = - 8/6 [On dividing both sides by 6]

p = - 4/3

12. (x/3) +1 = 7/15

Solutions:

(x/3) +1 = 7/15

x/3 = (7/15) – 1

x/3 = 7-15/15

x/3 = - 8/15

⇒(x/3) x 3 = (- 8/15) x 3 [On multiplying both sides by 3]

x = - 8/5

EXERCISE 2.2

1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?

Solutions:

Suppose the number be x.

Now, according to the question,

[x – (1/2)] x 1/2 = 1/8

⇒[x – (1/2)] x (1/2) x 2 = (1/8) x 2 [On multiplying both sides by 2]

⇒x - 1/2 = 1/4

⇒x = (1/4) + (1/2)

⇒x = 1+2/4 = 3/4

Hence, the number is 3/4.

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?

Solutions:

Suppose the breadth be a m.

According to the questions, the length will be (2a + 2) m.

Now, perimeter of swimming pool = 2(length + breadth) = 154 m

2(2a + 2 + a) = 154

⇒2(3a + 2) = 154

Dividing both sides by 2, we obtain

⇒2 (3a + 2)/2 = 154/2

⇒3a + 2 = 77

⇒3a = 77 − 2

⇒3a = 75

⇒3a/3 = 75/3 [On dividing both sides by 3]

a = 25

On substituting a = 25, we have 2a+ 2 = 2 x 25 + 2 = 52

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

3. The Base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4(2/15) cm. what is the length if either if the remaining equal sides?

Solutions:

Given perimeter of the triangle = 62/15

Suppose the length of equal sides beacm.

Perimeter = Length of two equal sides + Base = (a cm + a cm) + Base = 62/15

⇒2a + (4/3) = 62/15

⇒2a = (62/15) – (4/3)

⇒2a = (62 – 20)/15 = (62 - 20)/15

⇒2a = 42/15

⇒2a/2 = (42/15) x (1/2) [On dividing both sides by 2]

a= 7/5 cm.

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

Solutions:

Suppose one number is a. So, the other number will be a + 15

a + a + 15 = 95

⇒2a + 15 = 95

⇒2a = 95 − 15

⇒2a = 80

⇒2a/2 = 80/2

a = 40

Now, a + 15 = 40 + 15 = 55

Therefore the numbers are 40 and 55.

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

Solutions:

Suppose the common ratio between numbers be x. So, the numbers are 5x and 3x respectively.

Now, the difference between these numbers = 18

⇒5x − 3x = 18

⇒2x = 18

On dividing both sides by 2, we have

⇒2x/2 = 18/2

⇒x = 9

So, the first number = 5x = 5 x 9 = 45

Second number = 3x = 3 x 9 = 27

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

Solutions:

Suppose the three consecutive integers are a, a + 1 and a + 2

Now, sum of these numbers = a+ a + 1 + a + 2

a+ a + 1 + a + 2= 51

⇒3a + 3 = 51

⇒3a = 51 − 3

⇒3a = 48

⇒3x/3 = 48/3 [On dividing both sides by 3]

a = 16

Now, a + 1 = 17 and a + 2 = 18

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

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

Solutions:

Suppose the three consecutive multiples of 8 are 8x, 8(x + 1), 8(x + 2).

Now, sum of these numbers = 8x + 8(x + 1) + 8(x + 2) = 888

⇒8(x + x + 1 + x + 2) = 888

⇒8(3x + 3) = 888

⇒8(3x + 3)/8 = 888/8 [On dividing both sides by 8]

⇒3x + 3 = 111

⇒3x = 111 − 3

⇒3x = 108

⇒3x/3 = 108/3  [On dividing both sides by 3]

⇒x = 36

So, the first multiple of 8 = 8x = 8 x 36 = 288

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

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

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

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.

Solutions:

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

⇒2x + 3(x + 1) + 4(x + 2) = 74

⇒2x + 3x + 3 + 4x + 8 = 74

⇒9x + 11 = 74

⇒9= 74 − 11

⇒9x = 63

⇒9x/9 = 63/9 [On dividing both sides by 9]

⇒x = 7

First integer = 7

Second integer =x + 1 = 7 + 1 = 8

Third integer = x + 2 = 7 + 2 = 9

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?

Solutions:

Suppose the common ratio between Rahul’s age and Haroon’s age is x.

So, the age of Rahul and Haroon will be 5x years and 7x years respectively.

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

Now according to the questions

⇒(5x + 4 + 7x + 4) = 56

⇒12x + 8 = 56

⇒12x = 56 − 8

⇒12x = 48

⇒12x/12 = 48/12 [On dividing both sides by 12]

⇒x = 4

So, Rahul’s age = 5x= (5 × 4) = 20 years and Haroon’s age = 7x = (7 × 4) = 28 years.

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?

Solutions:

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

So, the number of boys = 7x and the number of girls = 5x

According to the question,

Number of boys = Number of girls + 8

⇒7x = 5x + 8

⇒7x − 5x = 8

⇒2x = 8

⇒2x/2 = 8/2

⇒x = 4

So, the number of boys = 7x = 7 × 4 = 28 and the number of girls = 5x = 5 × 4 = 20

Total number of students in the class= 28 + 20 = 48 students

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?

Solutions:

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

⇒3x − 3 = 135

⇒3x = 135 + 3

⇒3x = 138

⇒3x/3 = 138/3 [On dividing both sides by 3]

⇒x = 46

Therefore, Baichung’s father’s age = x = 46 years

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

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

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

Solutions:

Suppose Ravi’s present age be x years.

After fifteen years, Ravi’s age = 4 × His present age

⇒x + 15 = 4x

⇒15 = 4x − x

⇒15 = 3x

⇒15/3 = 3x/3 [On dividing both sides by 3]

⇒5 = x

Hence, Ravi’s present age = 5 years

13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?

Solutions:

Let the number be x.

According to the given question,

(5/2) x + (2/3) = - 7/12

⇒(5/2) x = (- 7/12) - (2/3)

⇒(5/2) x = [-7 - (2 X 4)]/12

⇒(5/2) x = - 15/12

x = (- 15/12) X (2/5) [On multiplying both sides by 2/5]

x= - 1/2

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?

Solutions:

Suppose the common ratio between the numbers of notes of different denominations be x.

So the numbers of Rs 100 notes, Rs 50 notes, and Rs 10 notes will be2x, 3x, and 5x respectively.

Now, the value of Rs 100 notes = Rs (100 x 2x) = Rs 200x

The values of Rs 50 notes = Rs (50 x 3x) = Rs 150x

The values of Rs 10 notes = Rs (10 x 5x) = Rs 50x

According to the question,

200x + 150x + 50x = 400000

⇒ 400x = 400000

x = 1000 [On dividing both sides by 400]

So the number of Rs 100 notes = 2x = 2 x 1000 = 2000

The number of Rs 50 notes = 3x = 3 x 1000 = 3000

The number of Rs 10 notes = 5x = 5 x 1000 = 5000

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?

Solutions:

Suppose the number of 5 rupees coins are x.

According to the question,

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

Also, Number of Rs 1 coins = 160 − (Number of coins of Rs 5 and of Rs 2)= 160 − (3x + x) = 160 − 4x

Now, value of Re 1 coins = Rs [1 X (160 − 4x)] = Rs (160 − 4x)

Value of Rs 2 coins = Rs (2 x 3x) = Rs 6x

Values of Rs 5 coins = Rs (5 xx) = Rs 5x

According to the question, 160 − 4x + 6x + 5x = 300

⇒160 + 7x = 300

⇒7x = 300 − 160

⇒7x = 140

⇒7x/7 = 140/7 [On dividing both sides by 7]

⇒x = 20

Number of Rs 5 coins = x = 20

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

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

16. The organisers 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 3,000. Find the number of winners, if the total number of participants is 63.

Solutions:

Suppose the number of winners are x. So, the number of participants who did not win will be 63 − x.

Now, the amount given to the winners = Rs (100 xx) = Rs 100x

Also,  given to the participants who will not win = Rs [25(63 − x)]= Rs (1575 − 25x)

According to the given question,

100x + 1575 − 25x = 3000

⇒75x = 3000 − 1575

⇒75x = 1425

⇒75x/78 = 1425/75 [On dividing both sides by 75]

x = 19

So the number of winners = 19.

EXERCISE 2.3

Solve the following equations and check your results.

1. 3x = 2x + 18

Solutions:

3x = 2x + 18

⇒3x − 2x = 18

x = 18

Now, L.H.S = 3x = 3 × 18 = 54

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

Here, L.H.S. = R.H.S.

So, the result is correct.

2.5t – 3 = 3t – 5

Solutions:

5t − 3 = 3t − 5

⇒5t − 3t = −5 − (−3)

2t = −2

t = −1[On dividing both sides by 2]

L.H.S = 5t − 3 = [5 × (−1)] − 3 = −8

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

As,L.H.S. = R.H.S.

So, the result is correct.

3.5x + 9 = 5 + 3x

Solutions:

5x + 9 = 5 + 3x

5x − 3x = 5 − 9

⇒2x = −4

x = −2 [On dividing both sides by 2]

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

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

As, L.H.S. = R.H.S.

So, the result is correct.

4.4z + 3 = 6 + 2z

Solutions:

4z + 3 = 6 + 2z

⇒4z − 2z =6 − 3

⇒2z = 3

⇒z = 3/2[Dividing both sides by 2]

L.H.S. = 4z + 3 = [4 × (3/2)] + 3 = 6 + 3 = 9

R.H.S. = 6 + 2z = 6 + [2 × (3/2)] = 6 + 3 = 9

As, L.H.S. = R.H.S.

So, the result is correct.

5.2x 1= 14–x

Solutions:

2x − 1 = 14 − x

⇒2x + x = 14 + 1

⇒3x = 15

x = 5 [Dividing both sides by 3]

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

R.H.S. = 14 − x = 14 − 5 = 9

As, L.H.S. = R.H.S.

So, the result is correct.

6.8x + 4 = 3 (x – 1) + 7

Solutions:

8x + 4 =3(x − 1) + 7

⇒8x + 4 =3x − 3 + 7

⇒8x − 3x = − 3 + 7 − 4

⇒5x =− 7 + 7

x = 0

L.H.S = 8x + 4 = [8 × (0)] + 4 = 4

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

As, L.H.S. = R.H.S.

So, the result is correct.

7. x = 4/5 (x+10)

Solutions:

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

⇒5x = 4(x + 10) [Multiplying both sides by 5]

⇒5x = 4x + 40

⇒5x − 4x = 40

x = 40

L.H.S. = x = 40

R.H.S = (4/5) (x + 10) =(4/5) (40 + 10) = (4/5) x 50 = 40

As, L.H.S. = R.H.S.

So, the result is correct.

8.(2x/3) + 1 =(7x/15) + 3

Solutions:

(2x/3) + 1 = (7x/15) +3

Transposing 7x/15 to L.H.S and 1 to R.H.S, we obtain

⇒(2x/3) - (7x/15) = 3 - 1

⇒(10x - 7x)/15 = 2

3x/15 = 2

x/5 = 2

x = 10 [Multiplying both sides by 5]

L.H.S = (2x/3) + 1 = [(2 x 10)/3] + 1 = (20 + 3)/3 = 23/2

R.H.S = (7x/15) + 3 = (7x 10)/15 + 3 = 14/3 + 3 = (14 + 9)/3= 23/3.

As, L.H.S. = R.H.S.

So, the result is correct.

9. 2y + (5/3) = (26/3) – y

Solutions:

2y +(5/3) = (26/3) - y

⇒2y + y = (26/3) - (5/3)

⇒3y = 21/3 = 7

y = 7/3 [Dividing both sides by 3]

L.H.S. = 2y + (5/3) = [2 x (7/3)] + (5/3) = (14/3) + (5/3) = 19/3

R.H.S. =(26/3) - y = (26/3) - (7/3) = 19/3

Now, L.H.S. = R.H.S.

So, the result is correct.

10.3m = 5m - (8/5)

Solutions:

3m = 5m - (8/5)

⇒3m - 5m = - 8/5

- 2m = - 8/5

m = 4/5 [Dividing both sides by −2]

L.H.S = 3m = 3 x (4/5) = 12/5

R.H.S = 5m - (8/5) = [5 x(4/5)] - (8/5) = 12/5

As, L.H.S. = R.H.S.

So, the result is correct.

EXERCISE 2.4

1. Amina thinks of a number and subtracts 5/2 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?

Solutions:

Suppose the number is x.

According to thequestion,

8 [x - (5/2)] = 3x

⇒8x − 20 = 3x

⇒8x − 3x = 20

⇒5x = 20

⇒x = 4 [Dividing both sides by 5]

So, the number is 4.

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?

Solutions:

Suppose the numbers are x and 5x.

According to the question,

21 + 5x = 2(x + 21)

⇒21 + 5x = 2x + 42

⇒5x − 2x = 42 − 21

⇒3x = 21

⇒x = 7 [Dividing both sides by 3]

⇒5x = 5 x 7 = 35

So, the numbers are 7 and 35.

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?

Solutions:

Suppose the digits at tens place and ones place are x and 9 − x respectively.

Now, original number = 10x + (9 − x) = 9x + 9

As we will interchange the digits, the digits at ones place and tens place will be x and 9 − x respectively.

Now, new number after interchanging the digits = 10(9 − x) + x= 90 − 10x + x = 90 − 9x

According to the question,

New number = Original number + 27

⇒90 − 9x = 9x + 9 + 27

⇒90 − 9x = 9x + 36

⇒90 − 36 = 18x

⇒54 = 18x

⇒3 = x and 9 − x = 6

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

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

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

Solutions:

Suppose the digits at tens place and ones place are x and 3x respectively.

Now the original number = 10x + 3x = 13x

If we interchange the digits then the digits at ones place and tens place will be x and 3x respectively.

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

According to the given question,

Original number + New number = 88

⇒13x + 31x = 88

⇒44x = 88

⇒x = 2 [Dividing both sides by 44]

So the original number = 13x = 13 × 2 = 26

Now the digitat tens place and ones place as 3x and x respectively, the two-digit number obtained is 62.

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

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 his mother’s present age. What are their present ages?

Solutions:

Suppose, Shobo’s age is x years.

So, the mother’s age will be 6x years.

According to the given question,

Shobo's age after 5 years= (Shobo's Mother's Present age)/3

⇒x + 5 = 6x/3

⇒x + 5 = 2x

⇒5 = 2x − x

⇒5 = x

⇒6x = 6 × 5 = 30

So, the present ages of Shobo and Shobo’s mother are 5 years and 30 years respectively.

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 Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Solutions:

Suppose the common ratio between the length and breadth of the rectangular plot isx.

So, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.

Now, perimeter of the plot = 2(Length + Breadth) = [2 (11x + 4x)] m = 30x m

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

So, 100 × Perimeter = 75000

⇒100 × 30x = 75000

⇒3000x = 75000

⇒x = 25 [Dividing both sides by 3000]

Therefore, Length = 11x m = (11 × 25) m = 275 m

Breadth = 4x m = (4 × 25) m = 100 m

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 36,660. How much trouser material did he buy?

Solutions:

Suppose he brought 2x m cloth for trouser and 3x m cloth for shirt.

Selling price of shirt’scloth(per metre) = Rs. {50 + [(50 X 10)100]} = Rs 55

Selling price of trouser’scloth(per metre) = Rs. {90 + [(90 x 12)100]} = Rs 100.80

Now, total amount of selling = Rs. 36660

100.80 × (2x) + 55 × (3x) = 36660

⇒201.60x + 165x = 36660

⇒366.60x = 36660

x = 100

Length of trouser’scloth = 2x m = (2 × 100) m = 200 m

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.

Solutions:

Given, number of deer drinking water from the pond = 9

Suppose the number of deer are x.

So, the number of deer grazing in the field = x/2

Number of deer playing nearby = (3/4) x Number of remaining deer= (3/4) x [x - (x/2)] = (3/4) x (x/2) = 3x/8

According to the question,

x - [(x/2) + (3x/8)] = 9

⇒x - [(4x + 3x) 8] = 9

⇒x - (7x/8) = 9

⇒x/8 = 9

⇒x = 72

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

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

Solutions:

Suppose, the granddaughter’s age be x years. So, grandfather’s age will be 10x years.

Now, according to the question,

10x = x + 54

⇒10x − x = 54

⇒9x = 54

⇒x = 6

So, granddaughter’s age = x years = 6 years and grandfather’s age = 10x years = (10 × 6) years = 60 years

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.

Solutions:

Suppose Aman’s son’s age is x years.

So, Aman’s age will be 3x years.

According to the question,

(Aman’s age 10 years ago) = 5 × (Aman’s son’s age 10 years ago)

⇒3x − 10 = 5(x − 10)

⇒3x − 10 = 5x − 50

⇒50 − 10 = 5x − 3x

⇒40 = 2x

⇒20 = x [Dividing both sides by 2]

So, Aman’s son’s age = x years = 20 years and Aman’s age = 3x years = (3 × 20) years = 60 years.

EXERCISE 2.5

Solve the following linear equations.

Solutions:

1. (x/2) - (1/5) = (x/3) + (1/4)

⇒60 [(x/2) - (1/5)] = 60 [(x/3) + (1/4)] {multiplying by 60 on both sides}

⇒ 30x − 12 = 20x + 15    

⇒ 30x − 20x = 15 + 12

⇒ 10x = 27

⇒ x = 27/10

2. (n/2) - (3n/4) + (5n/6) = 21

⇒6n − 9n + 10n = 252 [Multiplying 12 on both sides]

⇒ 7n = 252

⇒n = 252/7

⇒n = 36

3. x + 7 - (8x/3) = (17/6) - (5x/2)

⇒[3(x + 7) - 8 x]/3 = [17 - 15x]/6

⇒2(3x + 21 - 8x) = (17 - 15x)

⇒ 6 x + 42 - 16 x = 17 - 15x

⇒ 6x − 16x + 15x = 17 − 42

⇒ 5x = −25

⇒ x = -25/5

⇒ x = - 5

4. (x-5)/3 = (x - 3)/5

⇒5(x − 5) = 3(x − 3)

⇒ 5x − 25 = 3x − 9

⇒ 5x − 3x = 25 − 9

⇒ 2x = 16

⇒ x = 16/2

⇒ x = 8

5. [(3t - 2)/4] - [(2t + 3)/3] = (2/3) - t

⇒3(3t − 2) − 4(2t + 3) = 8 − 12t

⇒ 8 − 12t = 9t − 6 − 8t − 12

⇒ 9t − 8t + 12t = 8 + 6 + 12

⇒ 13t = 26

t = 26/13

t = 2

6.m - [(m-1)/2] = 1 - [(m - 2)/3]

⇒ (2m - m + 1)/2 = (3 - m + 2)/3

⇒ 6m − 3m + 3 = 6 − 2m + 4        

⇒ 6m − 3m + 2m = 6 + 4 − 3

⇒ 5m = 7

⇒ m = 7/5

Simplify and solve the following linear equations.

7. 3(t – 3) = 5(2t + 1)

Solutions:

3(t − 3) = 5(2t + 1)

⇒ 3t − 9 = 10t + 5

⇒ 10t − 3t = − 9 − 5

⇒  7t = − 14

t = - 14/7

t = - 2

8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0

Solutions:

15(y − 4) − 2(y − 9) + 5(y + 6) = 0

⇒ 15y − 60 − 2y + 18 + 5y + 30 = 0

⇒ 18y − 12 = 0

⇒ 18y = 12

⇒ y = 12/18 = 2/3

9. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solutions:

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

⇒ 15z − 21 − 18z + 22 = 32z − 52 − 17

⇒ − 3z + 1 = 32z − 69

⇒ − 3z − 32z = − 69 − 1

⇒ − 35z = − 70

z = 70/35

z = 2

10. 0.25(4f – 3) = 0.05(10f – 9)

Solutions:

0.25(4f − 3) = 0.05(10f − 9)

⇒1/4 (4f - 3) = 1/20 (10f - 9)

⇒5(4f − 3) = 10f − 9 [Multiplying both sides by 20]

⇒ 20f − 15 = 10f − 9

⇒ 20f − 10f = − 9 + 15

⇒ 10f = 6

f = 3/5 = 0.6

EXERCISE 2.6

Solve the following equations.

Solutions:

1. (8x - 3)/3x = 2

⇒8x − 3 = 6x [On multiplying both sides by 3x]

⇒ 8x − 6x = 3

⇒ 2x = 3

⇒ x = 3/2

2. 9x/(7 - 6x) = 15

⇒9x = 15(7 − 6x) [On multiplying both sides by (7 − 6x)]

⇒ 9x = 105 − 90x

⇒ 9x + 90x = 105

⇒ 99x = 105

⇒ x = 105/99 = 35/33

3. z/z + 15 = 4/9

⇒9z = 4 (z + 15) [On multiplying both sides by 9(z + 15)]

⇒ 9z = 4z + 60

⇒ 9z − 4z = 60

⇒ 5z = 60

⇒ z = 12

4. (3y + 4)/(2 - 6y) = - 2/5

⇒5 (3y + 4) = −2 (2 − 6y) [On multiplying both sides by 5(2 − 6y)]

⇒ 15y + 20 = − 4 + 12y

⇒ 15y − 12y = − 4 − 20

⇒ 3y = −24

⇒ y = −8

5. (7y + 4)/(y + 2) = - 4/3

⇒3(7y + 4) = −4(y + 2) [ On multiplying both sides by 3(y + 2)]

⇒ 21y + 12 = − 4y − 8

⇒ 21y + 4y = − 8 − 12

⇒ 25y = −20

⇒ y = - 4/5

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.

Solutions:

Suppose the common ratio between their ages be x. So, Hari’s age and Harry’s age will be 5x years and 7x years respectively

 and four years later, their ages will be (5x + 4) years and (7x + 4) years respectively.

According to the situation given in the question,

(5x + 4)/(7x + 4)= 3/4

⇒ 4 (5x + 4) = 3 (7x + 4)

⇒ 20x + 16 = 21x + 12

⇒ 16 – 12 = 21x - 20x

⇒ 4 = x

It means, Hari’s age = 5x years = (5 × 4) years = 20 years

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

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 3/2. Find the rational number.

Solutions:

Suppose the numerator of the rational number isx. So, its denominator will be x + 8.

Now, the rational number is x/(x + 8).

Now, according to the question,

(x + 17)/(x + 8 - 1) = 3/2

⇒ (x + 17)/(x + 7) = 3/2

⇒ 2(x + 17) = 3(x + 7)

⇒ 2x + 34 = 3x + 21

⇒ 34 − 21 = 3x − 2x

⇒13 = x

Therefore, numerator of the rational number = x = 13

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

Now, rational number = 13/21.