NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable Ex 2.2

Q.1

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

Ans

$\begin{array}{l}\text{Let the number be x}.\\ \mathrm{}\text{According to the question the equation becomes},\\ (\mathrm{x}-\frac{1}{2})\times \frac{1}{2}=\frac{1}{8}\\ \text{On multiplying both the sides by 2,we get}\\ \Rightarrow (\mathrm{x}-\frac{1}{2})\times \frac{1}{2}\times 2=\frac{1}{8}\times 2\\ \Rightarrow \mathrm{x}-\frac{1}{2}=\frac{1}{4}\\ \text{On transposing}\frac{\text{1}}{\text{2}}\text{to R.H.S, we get}\\ \Rightarrow \mathrm{x}=\frac{1}{4}+\frac{1}{2}\\ \Rightarrow \mathrm{x}=\frac{1+2}{4}\\ \Rightarrow \mathrm{x}=\frac{3}{4}\end{array}$

Q.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?

Ans

The perimeter of a rectangular swimming pool is 154 m.
Let the breadth be x m. The length will be (2x + 2) m.
According to the question the equation becomes,

$\begin{array}{l}\text{2}\left(\text{2}x+\text{2}+x\right)=\text{154}\\ \Rightarrow \text{2}\left(\text{3}x+\text{2}\right)=\text{154}\\ \text{Dividing both sides by 2, we obtain}\\ \Rightarrow \frac{\text{2}\left(\text{3}x+\text{2}\right)}{2}=\frac{\text{154}}{2}\\ \Rightarrow x+\text{2}=\text{77}\\ \text{On transposing 2 to R.H.S, we obtain}\\ \Rightarrow \text{3}x=\text{77}-\text{2}\\ \Rightarrow \text{3}x=\text{75}\\ \text{On dividing both sides by 3, we obtain}\\ \frac{\text{3}x}{3}=\frac{\text{75}}{3}\\ x=\text{25}\\ \therefore \text{2}x+\text{2}=\text{2}\times \text{25}+\text{2}=\text{52}\end{array}$

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

Q.3

$\begin{array}{l}\mathrm{The}\mathrm{base}\mathrm{of}\mathrm{an}\mathrm{is}\mathrm{osceles}\mathrm{triangleis}\frac{4}{3}\mathrm{cm}.\mathrm{The}\mathrm{perimeter}\\ \mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}4\frac{2}{15}\mathrm{cm}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{either}\\ \mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{equal}\mathrm{sides}?\end{array}$

Ans

$\begin{array}{l}\text{Let the length of equal sides be x cm.}\\ \text{Perimeter =xcm+xcm+Base=4}\frac{\text{2}}{\text{15}}\text{cm}\\ \Rightarrow 2\mathrm{x}+\frac{4}{3}=\frac{62}{15}\\ \text{On transposing}\frac{\text{4}}{\text{3}}\text{to R.H.S, we obtain}\\ \Rightarrow 2\mathrm{x}=\frac{62}{15}-\frac{4}{3}\\ \Rightarrow 2\mathrm{x}=\frac{62-20}{15}\\ \Rightarrow 2\mathrm{x}=\frac{42}{15}\\ \text{On dividing both sides by 2, we obtain}\\ \Rightarrow \frac{2\mathrm{x}}{2}=\frac{42}{15\times 2}\\ \Rightarrow \mathrm{x}=\frac{7}{5}\\ \text{Therefore, the length of equal sides is}\frac{\text{7}}{\text{5}}\text{cm or 1}\frac{\text{2}}{\text{5}}\text{.}\end{array}$

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

Ans

Let one number be x.

Therefore, the other number will be x + 15.

According to the question,

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 get

x = 40

Therefore, x+15 = 40+15 = 55

Hence the numbers are 40 and 55.

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

Ans

Let the ratio between these numbers be x.
Therefore, the numbers will be 5x and 3x respectively.
Difference between these numbers = 18
According, to the question the equation becomes
5x − 3x = 18
2x = 18

$\begin{array}{l}\text{Dividing both sides by 2,we get}\\ \frac{2\mathrm{x}}{2}=\frac{18}{2}\\ \Rightarrow \mathrm{x}=9\end{array}$

Therefore, the first number is 5x=5×9=45 and,
The second number is 3x=3×9=27.

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

Ans

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

$\begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3,\mathrm{we}\mathrm{get}\\ \frac{3\mathrm{x}}{3}=\frac{48}{3}\\ \Rightarrow \mathrm{x}=16\\ \therefore \mathrm{x}+1=17\\ \mathrm{x}+2=18\end{array}$

Hence, the integers are 16, 17 and 18.

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

Ans

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

\begin{array}{l}\text{On dividing both sides by 8, we get}\\ \Rightarrow \frac{8(3x+3)}{8}=\frac{888}{8}\\ \Rightarrow 3x+3=111\\ \text{On transposing 3 to RHS,we get}\\ \Rightarrow 3x=111-3\\ \Rightarrow 3x=108\\ \text{On dividing both sides by 3,we get}\\ \Rightarrow \frac{3x}{3}=\frac{108}{3}\\ \Rightarrow x=36\end{array}

Therefore, 8x = 8×36 = 288
8(x+1) = 8(36+1) = 8×37 = 296
8(x+2) = 8(36+2) = 8×38 = 304
Hence, the numbers are 288,296 and 304.

Q.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.

Ans

Let 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

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

9x = 74 − 11

9x = 63

\begin{array}{l}\text{On dividing both sides by 9,we get}\\ \Rightarrow \frac{9x}{9}=\frac{63}{9}\\ \Rightarrow x=7\\ \therefore x+1=8\\ x+2=9\\ \text{Hence,the numbers are 7, 8 and 9}\text{.}\end{array}

Q.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?

Ans

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

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

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

According to the given question, the equation becomes,

(5x + 4 + 7x + 4) = 56

12x + 8 = 56

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

12x = 56 − 8

12x = 48

$$\begin{gathered} \begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}12,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{12\mathrm{x}}{12}=\frac{48}{12}\\ \Rightarrow \mathrm{x}=4\\ \therefore \mathrm{Rahul}‘\mathrm{s}\mathrm{age}=5\mathrm{x}=5\times 4=20\mathrm{years} \\ \mathrm{Haroon}‘\mathrm{s}\mathrm{age}\mathrm{is}=7\mathrm{x}\mathrm{years}=7\times 4=28\mathrm{years}\end{array} \end{gathered}$$

Q.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?

Ans

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

Then, number of boys = 7x

Number of girls = 5x

According to the given question,

Number of boys = Number of girls + 8

7x = 5x + 8

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

7x − 5x = 8

2x = 8

$\begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{2x}{2}=\frac{8}{2}\\ \Rightarrow x=4\\ \therefore \mathrm{Number}\mathrm{of}\mathrm{boys}=7\mathrm{x}=7\times 4=28\\ \mathrm{Number}\mathrm{of}\mathrm{girls}=\mathrm{x}=5\times 4=20\\ \mathrm{Hence},\mathrm{the}\mathrm{total}\mathrm{strength}\mathrm{of}\mathrm{class}=28+20=48\mathrm{students}\end{array}$

Q.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?

Ans

Let Baichung’s father’s age be x years.

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

According to the given question, the equation becomes:

x + x − 29 + x + 26 = 135

3x − 3 = 135

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

3x = 135 + 3

3x = 138

$$\begin{gathered} \begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{3\mathrm{x}}{3}=\frac{138}{3}\\ \Rightarrow \mathrm{x}=46\\ \\ \therefore \mathrm{Baichung}‘\mathrm{s}\mathrm{father}‘\mathrm{s}\mathrm{age}=\mathrm{x}\mathrm{years}=46\mathrm{years}\\ \mathrm{Baichung}‘\mathrm{s}\mathrm{age}=(\mathrm{x}-29)\mathrm{years}=(46-29)=17\mathrm{years}\\ \mathrm{Baichung}‘\mathrm{s}\mathrm{grandfather}‘\mathrm{s}\mathrm{age}=(\mathrm{x}+26)\mathrm{years}\\ =(46+26)\mathrm{years} \\ =72\mathrm{years}\end{array} \end{gathered}$$

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

Ans

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

$\begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{15}{3}=\frac{3\mathrm{x}}{3}\\ \Rightarrow \mathrm{x}=5\\ \\ \therefore \mathrm{Ravi}‘\mathrm{s}\mathrm{present}\mathrm{age}=5\mathrm{years}\end{array}$

Q.13

$\begin{array}{l}\mathrm{A}\mathrm{rational}\mathrm{number}\mathrm{is}\mathrm{such}\mathrm{that}\mathrm{when}\mathrm{you}\mathrm{multiply}\mathrm{it}\mathrm{by}\frac{5}{2}\\ \mathrm{and}\mathrm{add}\frac{2}{3}\mathrm{to}\mathrm{the}\mathrm{product},\mathrm{you}\mathrm{get}\frac{-7}{12}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{number}?\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{be}\mathrm{x}.\\ \text{According to the question,the equation becomes}\\ \frac{5}{2}\mathrm{x}+\frac{2}{3}=-\frac{7}{12}\\ \text{On transposing}\frac{2}{3}\text{to R.H.S,}\text{we get}\\ \Rightarrow \frac{5}{2}\mathrm{x}=-\frac{7}{12}-\frac{2}{3}\\ \Rightarrow \frac{5}{2}\mathrm{x}=\frac{-7-8}{12}\\ \Rightarrow \frac{5}{2}\mathrm{x}=\frac{-15}{12}\\ \text{On multiplying both sides by}\frac{2}{5},\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{5\mathrm{x}}{2}\times \frac{2}{5}=\frac{-15}{12}\times \frac{2}{5}\\ \Rightarrow \mathrm{x}=-\frac{1}{2}\\ \text{Hence,}\text{the number is}-\frac{1}{2}.\end{array}$

Q.14 Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?

Ans

Let the ratio between the numbers of notes of different denominations be x.

Therefore, numbers of ₹ 100 notes, ₹ 50 notes, and ₹ 10 notes will be 2x, 3x, and 5x respectively.

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

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

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

Total cash = ₹ 400000.

Therefore,

200x + 150x + 50x = 400000

⇒ 400x = 400000

On dividing both sides by 400, we obtain

x = 1000

Therefore,

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

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

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

Q.15 I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of Rs 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Ans

Let the number of ₹ 5 coins be x.

Since the number of ₹ 2 coins is 3 times the number of ₹ 5 coins, so the number of ₹ 2 coins = 3x

Therefore, the number of ₹1 coins = 160 − (Number of coins of ₹ 5 and of ₹ 2)

= 160 − (3x + x) = 160 − 4x

Now, Amount of ₹ 1 coins = ₹ [1 × (160 − 4x)] = ₹ (160 − 4x)

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

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

Total amount is ₹300.

Therefore, 160 − 4x + 6x + 5x =300

160 + 7x = 300

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

7x = 300 – 160 = 140

On dividing both sides by 7, we get

x = 20No. of ₹ 1 coins = 160 – 4x = 160 – 4×20 = 80

No. of ₹ 2 coins = 3x = 3×20 = 60

No. of ₹ 5 coins = x = 20

Therefore, number of ₹ 1 coins is 80, ₹ 2 coins is 60 and ₹ 5 coins is 20.

Q.16 The organisers of an essay competition decide that a winner of the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.

Ans

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 = ₹ (100 × x) = ₹ 100x

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

= ₹ (1575 − 25x)

According to the given question,

100x + 1575 − 25x = 3000

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

75x = 3000 − 1575

75x = 1425

$\begin{array}{l}\text{On dividing both sides by 75},\text{we obtain}\\ \Rightarrow \frac{75\mathrm{x}}{75}=\frac{1425}{75}\\ \Rightarrow \mathrm{x}=19\\ \\ \mathrm{Hence},\text{the number of winners}=19\end{array}$