Linear Equations In One Variable

EXERCISE-2.1

P1. Solve : x − 2 = 7

Sol. x − 2 = 7

Transposing 2 to R.H.S, we obtain

x = 7 + 2 = 9

P2. Solve : y + 3 = 10

Sol. y + 3 = 10

Transposing 3 to R.H.S, we obtain

y = 10 − 3 = 7

P3. Solve : 6 = z + 2

Sol. 6 = z + 2

Transposing 2 to L.H.S, we obtain

6 − 2 = z $\Rightarrow$ z = 4

P4. Solve : $\frac{3}{7}+x=\frac{17}{7}$

Sol. $\frac{3}{7}+x=\frac{17}{7}$

Transposing $\frac{3}{7}$ to R.H.S, we obtain

$x=\frac{17}{7}–\frac{3}{7}=\frac{14}{7}=2$

P5. Solve : 6x = 12

Sol. 6x = 12

Dividing both sides by 6, we obtain

$\frac{6x}{6}=\frac{12}{6}\Rightarrow$x = 2

P6. Solve : $\frac{t}{5}=10$

Sol. $\frac{t}{5}=10$

Multiplying both sides by 5, we obtain

$\frac{t}{5}\times 5=10\times 5$

P7. Solve : $\frac{2x}{3}=18$

Sol. $\frac{2x}{3}=18$

Multiplying both sides by $\frac{3}{2}$, we obtain

$\frac{2x}{3}=\frac{3}{2}=18\times \frac{3}{2}$

x = 27

P8. Solve : $1.6=\frac{y}{1.5}$

Sol. $1.6=\frac{\mathrm{y}}{1.5}$

Multiplying both sides by 1.5, we obtain

$1.6\times 1.5=\frac{\mathrm{y}}{1.5}\times 1.5$

2.4 = y

P9. Solve : 7x − 9 = 16

Sol. 7x − 9 = 16

Transposing 9 to R.H.S, we obtain

7x = 16 + 9

7x = 25

Dividing both sides by 7, we obtain

$\frac{7x}{7}=\frac{25}{7}$

$x=\frac{25}{7}$

P10. Solve : 14y − 8 = 13

Sol. 14y − 8 = 13

Transposing 8 to R.H.S, we obtain

14y = 13 + 8

14y = 21

Dividing both sides by 14, we obtain

$\frac{14y}{14}=\frac{21}{14}$

$y=\frac{3}{2}$

P11. Solve : 17 + 6p = 9

Sol. 17 + 6p = 9

Transposing 17 to R.H.S, we obtain

6p = 9 − 17

6p = −8

Dividing both sides by 6, we obtain

$\frac{6p}{6}=–\frac{8}{6}$

$p=–\frac{4}{3}$

P12. Solve : $\frac{x}{3}+1=\frac{7}{15}$

Sol. $\frac{x}{3}+1=\frac{7}{15}$

Transposing 1 to R.H.S, we obtain

$\frac{x}{3}=\frac{7}{15}–1$

$\frac{x}{3}=\frac{7–15}{15}$

$\frac{x}{3}=–\frac{8}{15}$

Multiplying both sides by 3, we obtain

$\frac{x}{3}\times 3=–\frac{8}{15}\times 3$

$x=–\frac{8}{5}$

EXERCISE-2.2

P1. If you subtract $\frac{1}{2}$ from a number and multiply the result by $\frac{1}{2}$ , you get $\frac{1}{8}$. What is the number?

Sol. Let the number be x.

According to the P,

$\left(x–\frac{1}{2}\right)\times \frac{1}{2}=\frac{1}{8}$

On multiplying both sides by 2, we obtain

$\left(x–\frac{1}{2}\right)\times \frac{1}{2}\times 2=\frac{1}{8}\times 2$

$x–\frac{1}{2}=\frac{1}{4}$

On transposing $\frac{1}{2}$ to R.H.S, we obtain

$x=\frac{1}{4}+\frac{1}{2}=\frac{1+2}{4}=\frac{3}{4}$

Therefore, the number is $\frac{3}{4}$.

P2. 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?

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

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

2(2x + 2 + x) = 154

2(3x + 2) = 154

Dividing both sides by 2, we obtain

$\frac{2(3x+2)}{2}=\frac{154}{2}$

3x + 2 = 77

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

3x = 77 − 2

3x = 75

On dividing both sides by 3, we obtain

$\frac{3x}{3}=\frac{75}{3}$

x = 25

2x + 2 = 2 × 25 + 2 = 52

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

P3. The base of an isosceles triangle is $\frac{4}{3}$cm. The perimeter of the triangle is $4\frac{2}{15}$cm. What is the length of either of the remaining equal sides?

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

Perimeter = x cm + x cm + Base = $4\frac{2}{15} \mathrm{cm}.$

$2x+\frac{4}{3}=\frac{62}{15}$

On transposing $\frac{4}{3}$ to R.H.S, we obtain

$2x=\frac{62}{15}–\frac{4}{3}$

$2x=\frac{62–4\times 5}{15}=\frac{62–20}{15}$

$2\mathrm{x}=\frac{42}{15}$

On dividing both sides by 2, we obtain

$\frac{2x}{2}=\frac{42}{15}\times \frac{1}{2}$$\Rightarrow$$x=\frac{7}{5}=1\frac{2}{5}$

Therefore, the length of equal sides is $1\frac{2}{5} \mathrm{cm}$.

P4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

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

According to the P,

x + x + 15 = 95

2x + 15 = 95

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

2x = 95 − 15

2x = 80

On dividing both sides by 2, we obtain

$\frac{2x}{2}=\frac{80}{2}$

x = 40

x + 15 = 40 + 15 = 55

Hence, the numbers are 40 and 55.

P5. Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?

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

Difference between these numbers = 18

5x − 3x = 18

2x = 18

Dividing both sides by 2,

$\frac{2x}{2}=\frac{18}{2}$

x = 9

First number = 5x = 5 × 9 = 45

Second number = 3x = 3 × 9 = 27

P6. Three consecutive integers add up to 51. What are these integers?

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

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

3x + 3 = 51

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

3x = 51 − 3

3x = 48

On dividing both sides by 3, we obtain

$\frac{3x}{3}=\frac{48}{3}$

x = 16

x + 1 = 17

x + 2 = 18

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

P7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

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

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

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

8(3x + 3) = 888

On dividing both sides by 8, we obtain

$\frac{8(3x+3)}{8}=\frac{888}{8}$

3x + 3 = 111

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

3x = 111 − 3

3x = 108

On dividing both sides by 3, we obtain

$\frac{3x}{3}=\frac{108}{3}$

x = 36

First multiple = 8x = 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.

P8. 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.

Sol. Let three consecutive integers be x, x + 1, x + 2. According to the P,

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

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

9x + 11 = 74

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

9x = 74 − 11

9x = 63

On dividing both sides by 9, we obtain

$\frac{9x}{9}=\frac{63}{9}$

x = 7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

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

P9. 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?

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

Therefore, 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.

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

$\therefore$ (5x + 4 + 7x + 4) = 56

12x + 8 = 56

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

12x = 56 − 8

12x = 48

On dividing both sides by 12, we obtain

$\frac{12\mathrm{x}}{12}=\frac{48}{12}$

x = 4

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

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

P10. 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?

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

Number of boys = 7x

Number of girls = 5x

According to the given P,

Number of boys = Number of girls + 8

$\therefore$ 7x = 5x + 8

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

7x − 5x = 8

2x = 8

On dividing both sides by 2, we obtain

$\frac{2x}{2}=\frac{8}{2}$

x = 4

Number of boys = 7x = 7 × 4 = 28

Number of girls = 5x = 5 × 4 = 20

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

P11. 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?

Sol. 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 P, the sum of the ages of these 3 people is 135 years.

$\therefore$ x + x − 29 + x + 26 = 135

3x − 3 = 135

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

3x = 135 + 3

3x = 138

On dividing both sides by 3, we obtain

$\frac{3x}{3}=\frac{138}{3}$

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

P12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Sol. Let Ravi’s present age be x years.

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

x + 15 = 4x

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

15 = 4x − x

15 = 3x

On dividing both sides by 3, we obtain

$\frac{15}{3}=\frac{3x}{3}$

5 = x

Hence, Ravi’s present age = 5 years

P13. A rational number is such that when you multiply it by $\frac{5}{2}$and add $\frac{2}{3}$ to the product, you get $–\frac{7}{2}$. What is the number?

Sol. Let the number be x.

According to the given P,

$\frac{5}{2}x+\frac{2}{3}=–\frac{7}{12}$

On transposing $\frac{2}{3}$ to R.H.S, we obtain

$\frac{5}{2}x=–\frac{7}{12}–\frac{2}{3}$

$\frac{5}{2}x=\frac{–7–(2\times 4)}{12}$

$\frac{5}{2}x=–\frac{15}{12}$

On multiplying both sides by $\frac{2}{5}$, we obtain

$x=–\frac{15}{12}\times \frac{2}{5}=–\frac{1}{2}$

Hence, the rational number is $–\frac{1}{2}$ .

P14. 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?

Sol. 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 be 2x, 3x, and 5x respectively.

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

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

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

It is given that total amount is Rs 400000.

$\therefore$ 200x + 150x + 50x = 400000

$\Rightarrow$ 400x = 400000

On dividing both sides by 400, we obtain x = 1000

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

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

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

P15. 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?

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

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

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

= 160 − (3x + x) = 160 − 4x

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

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

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

It is given that the total amount is Rs 300.

$\therefore$ 160 − 4x + 6x + 5x = 300

160 + 7x = 300

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

7x = 300 − 160

7x = 140

On dividing both sides by 7, we obtain

$\frac{7x}{7}=\frac{140}{7}$

x = 20

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

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

Number of Rs 5 coins = x = 20

P16. 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.

Sol. 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 100x

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

= Rs (1575 − 25x)

According to the given P,

100x + 1575 − 25x = 3000

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

75x = 3000 − 1575

75x = 1425

On dividing both sides by 75, we obtain

$\frac{75x}{75}=\frac{1425}{75}$

x = 19

Hence, number of winners = 19

EXERCISE-2.3

P1. Solve and check result : 3x = 2x + 18

Sol. 3x = 2x + 18

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

3x − 2x = 18

x = 18

L.H.S = 3x = 3 × 18 = 54

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

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

Hence, the result obtained above is correct.

P2. Solve and check result: 5t − 3 = 3t − 5

Sol. 5t − 3 = 3t − 5

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

5t − 3t = −5 − (−3)

2t = −2

On dividing both sides by 2, we obtain

t = −1

L.H.S = 5t − 3 = 5 × (−1) − 3 = −8

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

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

Hence, the result obtained above is correct.

P3. Solve and check result : 5x + 9 = 5 + 3x

Sol. 5x + 9 = 5 + 3x

On transposing 3x to L.H.S and 9 to R.H.S, we obtain

5x − 3x = 5 − 9

2x = −4

On dividing both sides by 2, we obtain

x = −2

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

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

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

Hence, the result obtained above is correct.

P4. Solve and check result: 4z + 3 = 6 + 2z

Sol. 4z + 3 = 6 + 2z

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

4z − 2z = 6 – 3 $\Rightarrow$ 2z = 3

Dividing both sides by 2, we obtain

$z=\frac{3}{2}$

L.H.S = 4z + 3 = 4 × $\left(\frac{3}{2}\right)$+3 = 6 + 3 = 9

R.H.S = 6 + 2z = 6 + 2 ×$\left(\frac{3}{2}\right)$ = 6 + 3 = 9

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

Hence, the result obtained above is correct.

P5. Solve and check result: 2x − 1 = 14 − x

Sol. 2x − 1 = 14 − x

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

2x + x = 14 + 1 $\Rightarrow$ 3x = 15

Dividing both sides by 3, we obtain x = 5

L.H.S = 2x − 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.

P6. Solve and check result : 8x + 4 = 3(x − 1) + 7

Sol. 8x + 4 = 3(x − 1) + 7

8x + 4 = 3x − 3 + 7

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

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

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

Hence, the result obtained above is correct.

P7. Solve and check result : $x=\frac{4}{5}(x+10)$

Sol. $x=\frac{4}{5}(x+10)$

Multiplying both sides by 5, we obtain

5x = 4(x + 10)

5x = 4x + 40

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

5x − 4x = 40

x = 40

L.H.S = x = 40

R.H.S = $\frac{4}{5}(x+10)$$=\frac{4}{5}(40+10)=\frac{4}{5}\times 50=40$

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

Hence, the result obtained above is correct.

P8. Solve and check result : $\frac{2x}{3}+1=\frac{7x}{15}+3$

Sol. $\frac{2x}{3}+1=\frac{7x}{15}+3$

Transposing $\frac{7x}{15}$ to L.H.S and 1 to R.H.S, we obtain

$\frac{2x}{3}–\frac{7x}{15}=3–1$

$\frac{5\times 2x–7x}{15}=2$

$\frac{3x}{15}=2\Rightarrow \frac{x}{5}=2$

Multiplying both sides by 5, we obtain

x = 10

L.H.S $=\frac{2x}{3}+1=\frac{2\times 10}{3}+1$

$=\frac{2\times 10+1\times 3}{3}=\frac{23}{3}$

R.H.S $=\frac{7x}{15}+3=\frac{7\times 10}{15}+3$

$=\frac{7\times 2}{3}+3=\frac{14}{3}+3$

$=\frac{14+3\times 3}{3}=\frac{23}{3}$

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

Hence, the result obtained above is correct.

P9. Solve and check result : $2y+\frac{5}{3}=\frac{26}{3}–y$

Sol. $2y+\frac{5}{3}=\frac{26}{3}–y$

Transposing y to L.H.S and $\frac{5}{3}$ to R.H.S, we obtain

$2y+y=\frac{26}{3}–\frac{5}{3}$$\Rightarrow 3y=\frac{21}{3}=7$

Dividing both sides by 3, we obtain

$y=\frac{7}{3}$

L.H.S $=2y+\frac{5}{3}$$=2\times \frac{7}{3}+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}=\frac{19}{3}$

R.H.S$=\frac{26}{3}–y=\frac{26}{3}–\frac{7}{3}=\frac{19}{3}$

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

Hence, the result obtained above is correct.

P10. Solve and check result : $3 \mathrm{m}=5 \mathrm{m}–\frac{8}{5}$

Sol. $3m=5m–\frac{8}{5}$

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

$3 \mathrm{m}–5 \mathrm{m}=–\frac{8}{5}$

$–2m=–\frac{8}{5}$

Dividing both sides by −2, we obtain

$m=\frac{4}{5}$

L.H.S = $3m=3\times \frac{4}{5}=\frac{12}{5}$

R.H.S$=5 \mathrm{m}–\frac{8}{5}=5\times \frac{4}{5}–\frac{8}{5}=\frac{12}{5}$

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

Hence, the result obtained above is correct.

EXERCISE-2.4

P1. Amina thinks of a number and subtracts $\frac{5}{2}$ 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?

Sol. Let the number be x.

According to the given P,

$8\left(x–\frac{5}{2}\right)=3x$ = 3x

8x − 20 = 3x

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

8x − 3x = 20

5x = 20

Dividing both sides by 5, we obtain

x = 4

Hence, the number is 4.

P2. 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?

Sol. Let the numbers be x and 5x. According to the P,

21 + 5x = 2(x + 21)

21 + 5x = 2x + 42

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

5x − 2x = 42 − 21

3x = 21

Dividing both sides by 3, we obtain

x = 7

5x = 5 × 7 = 35

Hence, the numbers are 7 and 35 respectively.

P3. 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?

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

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

= 90 − 9x

According to the given P,

New number = Original number + 27

90 − 9x = 9x + 9 + 27

90 − 9x = 9x + 36

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

90 − 36 = 18x

54 = 18x

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 9x + 9 = 9 × 3 + 9 = 36

P4. 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?

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

Therefore, original number

= 10x + 3x = 13x

On interchanging the digits, 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 P,

Original number + New number = 88

13x + 31x = 88

44x = 88

Dividing both sides by 44, we obtain

x = 2

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

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

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

P5. 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?

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

According to the given P,

After 5 years, Shobo;s age$=\frac{Shobo‘s mother‘s present age}{3}$

$x+5=\frac{6x}{3}$

$\Rightarrow$x + 5 = 2x

Transposing x to R.H.S, we obtain

5 = 2x – x

$\Rightarrow$ 5 = x

$\Rightarrow$ 6x = 6 × 5 = 30

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

P6. 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?

Sol. 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 11x m and 4x m respectively.

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

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

$\therefore$ 100 × Perimeter = 75000

100 × 30x = 75000

3000x = 75000

Dividing both sides by 3000, we obtain

x = 25

Length = 11x m = (11 × 25) m = 275 m

Breadth = 4x m = (4 × 25) m = 100 m

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

P7. 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?

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

Per metre selling price of trouser material = $=Rs\left(90+\frac{90\times 12}{100}\right)=Rs.100.80$

Per metre selling price of shirt material = $Rs\left(50+\frac{50\times 10}{100}\right)=Rs55$

Given that, total amount of selling = Rs 36660

100.80 × (2x) + 55 × (3x) = 36660

201.60x + 165x = 36660

366.60x = 36660

Dividing both sides by 366.60, we obtain

x = 100

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

P8. 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.

Sol. Let the number of deer be x.

Number of deer grazing in the field = $\frac{x}{2}$

Number of deer playing nearby

$=\frac{3}{4}\times$ Number of remaining deer

$=\frac{3}{4}\times \left(x–\frac{x}{2}\right)=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}$

Number of deer drinking water from the pond = 9

$x–\left(\frac{x}{2}+\frac{3x}{8}\right)=9$

$x–\left(\frac{4x+3x}{8}\right)=9$

$x–\frac{7x}{8}=9$

Multiplying both sides by 8, we obtain

x = 72

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

P9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages

Sol. Let the granddaughter’s age be x years.

Therefore, grandfather’s age will be 10x years.

According to the P,

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

10x = x + 54

Transposing x to L.H.S, we obtain

10x − x = 54

9x = 54

x = 6

Granddaughter’s age = x years = 6 years

Grandfather’s age = 10x years

= (10 × 6) years = 60 years

P10. 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.

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

According to the P,

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

3x − 10 = 5(x − 10)

3x − 10 = 5x − 50

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

50 − 10 = 5x − 3x

40 = 2x

Dividing both sides by 2, we obtain

20 = x

Aman’s son’s age = x years = 20 years

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

EXERCISE-2.5

P1. Solve the linear equation $\frac{x}{2}–\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

Sol. $\frac{x}{2}–\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

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

Multiplying both sides by 60, we obtain

$60\left(\frac{x}{2}–\frac{1}{5}\right)=60\left(\frac{x}{3}+\frac{1}{4}\right)$

$\Rightarrow$ 30x − 12 = 20x + 15 (Opening the brackets)

$\Rightarrow$ 30x − 20x = 15 + 12

$\Rightarrow$ 10x = 27

$\Rightarrow \mathrm{x}=\frac{27}{10}$

P2. Solve the linear equation $\frac{n}{2}–\frac{3n}{4}+\frac{5n}{6}=21$

Sol. $\frac{n}{2}–\frac{3n}{4}+\frac{5n}{6}=21$

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

Multiplying both sides by 12, we obtain

6n − 9n + 10n = 252

$\Rightarrow$ 7n = 252

$\Rightarrow \mathrm{n}=\frac{252}{7}$$\Rightarrow$ n = 36

P3. Solve the linear equation $x+7–\frac{8x}{3}=\frac{17}{6}–\frac{5x}{2}$

Sol. $x+7–\frac{8x}{3}=\frac{17}{6}–\frac{5x}{2}$

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

Multiplying both sides by 6, we obtain

6x + 42 − 16x = 17 − 15x

$\Rightarrow$ 6x − 16x + 15x = 17 − 42

$\Rightarrow$ 5x = −25

$\Rightarrow x=\frac{–25}{5}$

$\Rightarrow$x = −5

P4. Solve the linear equation $\frac{x–5}{3}=\frac{x–3}{5}$

Sol. $\frac{x–5}{3}=\frac{x–3}{5}$

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

Multiplying both sides by 15, we obtain

5(x − 5) = 3(x − 3)

$\Rightarrow$ 5x − 25 = 3x − 9 (Opening the brackets)

$\Rightarrow$ 5x − 3x = 25 – 9 $\Rightarrow$ 2x = 16

$\Rightarrow x=\frac{16}{2}$

$\Rightarrow$ x = 8

P5. Solve the linear equation $\frac{3t–2}{4}–\frac{2t+3}{3}=\frac{2}{3}–t$

Sol. $\frac{3t–2}{4}–\frac{2t+3}{3}=\frac{2}{3}–t$

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

Multiplying both sides by 12, we obtain

3(3t − 2) − 4(2t + 3) = 8 − 12t

$\Rightarrow$ 9t − 6 − 8t − 12 = 8 − 12t (Opening the brackets)

$\Rightarrow$ 9t − 8t + 12t = 8 + 6 + 12

$\Rightarrow$ 13t = 26

$\Rightarrow \mathrm{t}=\frac{26}{13}$

$\Rightarrow$ t = 2

P6. Solve the linear equation $\mathrm{m}–\frac{\mathrm{m}–1}{2}=1–\frac{\mathrm{m}–2}{3}$

Sol. $\mathrm{m}–\frac{\mathrm{m}–1}{2}=1–\frac{\mathrm{m}–2}{3}$

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

Multiplying both sides by 6, we obtain

6m − 3(m − 1) = 6 − 2(m − 2)

$\Rightarrow$ 6m − 3m + 3 = 6 − 2m + 4 (Opening the brackets)

$\Rightarrow$ 6m − 3m + 2m = 6 + 4 − 3

$\Rightarrow$ 5m = 7

$\Rightarrow \mathrm{m}=\frac{7}{5}$

P7. Simplify and solve the linear equation 3(t − 3) = 5(2t + 1)

Sol. 3(t − 3) = 5(2t + 1)

$\Rightarrow$ 3t − 9 = 10t + 5 (Opening the brackets)

$\Rightarrow$ −9 − 5 = 10t − 3t

$\Rightarrow$ −14 = 7t

$\Rightarrow \mathrm{t}=\frac{–14}{7}$

$\Rightarrow$ t = −2

P8. Simplify and solve the linear equation 15(y – 4) – 2 (y – 9) + 5 (y + 6) = 0

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

$\Rightarrow$ 15y − 60 − 2y + 18 + 5y + 30 = 0 (Opening the brackets)

$\Rightarrow$18y − 12 = 0

$\Rightarrow$ 18y = 12

$\Rightarrow y=\frac{12}{18}=\frac{2}{3}$

P9. Simplify and solve the linear equation 3(5z − 7) − 2(9z − 11) = 4(8z − 13)−17

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

$\Rightarrow$ 15z − 21 − 18z + 22 = 32z − 52 − 17 (Opening the brackets)

$\Rightarrow$ −3z + 1 = 32z − 69

$\Rightarrow$−3z − 32z = −69 − 1

$\Rightarrow$ −35z = −70

$\Rightarrow z=\frac{70}{35}=2$

P10. Simplify and solve the linear equation 0.25(4f − 3) = 0.05(10f − 9)

Sol. 0.25(4f − 3) = 0.05(10f − 9)

$\frac{1}{4}(4f–3)=\frac{1}{20}(10f–9)$

Multiplying both sides by 20, we obtain

5(4f − 3) = 10f − 9

$\Rightarrow$ 20f − 15 = 10f − 9 (Opening the brackets)

$\Rightarrow$ 20f − 10f = − 9 + 15 $\Rightarrow$ 10f = 6

$\Rightarrow \mathrm{f}=\frac{3}{5}=0.6$

EXERCISE-2.6

P1. Solve : $\frac{8x–3}{3x}=2$

Sol. $\frac{8x–3}{3x}=2$

On multiplying both sides by 3x, we obtain

8x − 3 = 6x

$\Rightarrow$ 8x − 6x = 3

$\Rightarrow$ 2x = 3

$\Rightarrow x=\frac{3}{2}$

P2.  Solve : $\frac{9x}{7–6x}=15$

Sol. $\frac{9x}{7–6x}=15$

On multiplying both sides by 7 − 6x, we obtain

9x = 15(7 − 6x)

$\Rightarrow$ 9x = 105 − 90x

$\Rightarrow$ 9x + 90x = 105 $\Rightarrow$ 99x = 105

$\Rightarrow x=\frac{105}{99}=\frac{35}{33}$

P3. Solve : $\frac{z}{z+15}=\frac{4}{9}$

Sol. $\frac{z}{z+15}=\frac{4}{9}$

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

9z = 4(z + 15)

$\Rightarrow$ 9z = 4z + 60    

$\Rightarrow$ 9z − 4z = 60

$\Rightarrow$ 5z = 60          

$\Rightarrow$ z = 12

P4. Solve : $\frac{3y+4}{2–6y}=\frac{–2}{5}$

Sol. $\frac{3y+4}{2–6y}=\frac{–2}{5}$

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

5(3y + 4) = −2(2 − 6y)

$\Rightarrow$ 15y + 20 = − 4 + 12y

$\Rightarrow$ 15y − 12y = − 4 − 20

$\Rightarrow$ 3y = −24

$\Rightarrow$ y = −8

P5. Solve : $\frac{7y+4}{y+2}=\frac{–4}{3}$

Sol. $\frac{7y+4}{y+2}=\frac{–4}{3}$

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

3(7y + 4) = −4(y + 2)

$\Rightarrow$ 21y + 12 = − 4y − 8

$\Rightarrow$ 21y + 4y = − 8 – 12    

$\Rightarrow$ 25y = −20

$\Rightarrow y=–\frac{4}{5}$

P6. 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.

Sol. Let the common ratio between their ages be x. Therefore, 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 P,

$\frac{5x+4}{7x+4}=\frac{3}{4}\Rightarrow$4(5x + 4) = 3(7x + 4)

$\Rightarrow$ 20x + 16 = 21x + 12

$\Rightarrow$ 16 – 12 = 21x – 20x

$\Rightarrow$ 4 = x

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

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

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

P7. 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 $\frac{3}{2}$. Find the rational number.

Sol. Let the numerator of the rational number be x. Therefore, its denominator will be x + 8.

The rational number will be $\frac{x}{x+8}$. According to the P,

$\frac{x+17}{x+8–1}=\frac{3}{2}$

$\Rightarrow \frac{x+17}{x+7}=\frac{3}{2}$

$\Rightarrow$ 2(x + 17) = 3(x + 7)

$\Rightarrow$ 2x + 34 = 3x + 21

$\Rightarrow$ 34 − 21 = 3x − 2x

$\Rightarrow$13 = x

Numerator of the rational number  = x = 13

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

Rational number$=\frac{13}{21}$