# Excercise 2.2 Linear Equations in One Variable- NCERT Solutions Class 8

## Chapter 2 Ex.2.2 Question 1

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

### Solution

Reasoning:

Forming a linear equation for the given problem statement and solving it will lead to the solution.

What is known?

(i) \begin{align}\frac{1}{2}\end{align} is subtracted from a number.

(ii) Result is multiplied by\begin{align}\frac{1}{2}\end{align}

(iii) Answer is \begin{align}\frac{1}{8}\end{align}

What is unknown?

The number

Steps:

Let the number be $$x$$

Then,

(i) \begin{align}\frac{1}{2}\end{align} is subtracted from a number \begin{align} \to x - \frac{1}{2}\end{align}

(ii) Result is multiplied by \begin{align}\frac{1}{2} \to \frac{1}{2}(x - \frac{1}{2})\end{align}

(iii) Answer is \begin{align} \frac{1}{8}\to \frac{1}{2}(x - \frac{1}{2}) = \frac{1}{8}\end{align}

Steps:

\begin{align}\frac{1}{2}(x - \frac{1}{2}) = \frac{1}{8} \qquad \\ \begin{Bmatrix}\text{Multiplying with} \\ \text{ 2 on both sides}\end{Bmatrix} \end{align}

\begin{align} x - \frac{1}{2} &= \frac{1}{8} \times 2\,\\x &= \frac{1}{4} + \frac{1}{2}\\x &= \frac{1}{4} + \frac{2}{4}\\x &= \frac{3}{4}\end{align}

The number is \begin{align}\frac{3}{4}\end{align}

## Chapter 2 Ex.2.2 Question 2

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

### Solution

What is known?

(i) Perimeter of a rectangular swimming pool is $$154\,\rm{m}$$.

(ii) Its length is $$2\,\rm{m}$$ more than twice its breadth.

What is unknown?

Length and breadth of the pool.

Reasoning:

Form a linear equation by using the formula for the perimeter of a rectangle. Assume either the breadth or the length to be a variable.

Steps:

Let the breadth of swimming pool be $$x \,\rm{m}$$.

Therefore, the length of the swimming pool will be $$(2x + 2)\,\rm{m}$$

Perimeter of rectangular swimming pool: $$2(\text{Length + Breadth})$$

Therefore, $$2(x + 2x + 2) = 154$$

\begin{align}3x + 2 &= \frac{{154}}{2}\\3x + 2 &= 77\\3x &= 77 - 2\\3x &= 75\\x &= \frac{{75}}{3}\\x &= 25\\{\rm{breadth}} &= 25\,{\rm{m}}\\{\rm{length}} &= 2x + 2\\ &= 2(25) + 2\\ &= 52\,{\rm{m}}\end{align}

Length of the pool $$= 52 \,\rm{m}$$

Breadth of the pool $$= 25 \;\rm{m}$$

## Chapter 2 Ex.2.2 Question 3

The base of an isosceles triangle is \begin{align}\frac{4}{3}\,\rm{cm}\end{align}. The perimeter of the triangle is \begin{align}4\frac{2}{{15}}\, \rm{cm}\end{align}. What is the length of either of the remaining equal sides?

### Solution

What is known?

(i) Base of an isosceles triangle \begin{align} = \frac{4}{3}{\rm{cm}}\end{align}

(ii) Perimeter of the triangle\begin{align} = 4\frac{2}{{15}}{\rm{cm}}\end{align}

What is unknown?

Length of sides of the equal sides of isosceles triangle.

Reasoning:

In an isosceles triangle, two sides of the triangle are equal. The value of one side is given. We can assume any one of the two sides to be a variable and form a linear equation using the following formula for the perimeter of any triangle:

Perimeter of a Triangle $$=$$ Sum of the Lengths of all Three Sides

Steps:

Let the length of either of equal sides be $$x \,\rm cm$$.

(i) Base of an isosceles triangle $$= \frac{4}{3}{\rm{cm}}$$

(ii) Perimeter of the triangle\begin{align}\to x + x + \frac{4}{3} = 4\frac{2}{{15}}\end{align}[adding all these sides]

\begin{align}2x + \frac{4}{3} &= 4\frac{2}{{15}}\\2x &= \frac{{62}}{{15}} - \frac{4}{3}\\2x &= \frac{{42}}{{15}}\\x &= \frac{{42}}{{15}} \times \frac{1}{2}\\x &= \frac{{21}}{{15}}\\x &= \frac{7}{5}\\\end{align}

Length of either of the equal side of isosceles triangle is\begin{align}\frac{7}{5}{\rm{cm}}\end{align}

## Chapter 2 Ex.2.2 Question 4

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

### Solution

What is known?

(i) Some of two numbers is $$95$$.

(ii) One exceeds the other by $$15$$.

What is unknown?

Numbers

Reasoning:

Assume any one number to be a variable and form a linear equation by using the relationships between the two numbers mentioned in the problem statement.

Steps:

Let one of the number be $$x$$.

Then the other number➔ $$(95-{\rm{ }}x)$$ (as sum of them is $$95$$).

If we assume $$x$$ to be larger among two of them.

$$x - (95 - x) = 15$$ (as one exceeds other by $$15$$)

\begin{align}x - 95 + x &= 15\\2x &= 110\\x &= \frac{{110}}{2}\\x &= 55\end{align}

Then other number➔ $$95 - x = 95 - 55 = 40$$

The numbers are $$40$$ and $$55$$.

## Chapter 2 Ex.2.2 Question 5

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

### Solution

What is known?

(i) Two number are in ratio $$5:3$$.

(ii) They differ by $$18$$.

What is unknown?

Number

Reasoning:

Assume any one number to be a variable and form a linear equation by using the relationships between the two numbers as mentioned in the problem statement. Any one relationship will allow you to express one number in terms of the other number (assumed to be a variable first). Then, the second relationship will allow you to form a linear equation.

Steps:

Two number are in ratio $$5:3$$

$$\therefore$$The number are $$5x$$ and $$3x$$

The two numbers differ by $$18$$:

$5x - 3x = 18$

\begin{align}2x &= 18\\x &= 9\end{align}

First number is $$5x = 5 \times 9 = 45$$

Second number is $$3x = 3 \times 9 = 27$$

The numbers will be $$45$$ and $$27$$.

## Chapter 2 Ex.2.2 Question 6

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

### Solution

What is known?

Three consecutive integers add up to $$51$$.

What is unknown?

Integers.

Reasoning:

Assume one integer to be a variable. Then use the property of consecutive integers and the given condition to form a linear equation.

Steps:

Let the first integer be $$x$$.

Then next consecutive two integers are $$x + 1\;{\rm{ and }}\;\,x + 2$$

$$x + (x + 1) + (x + 2) = 51$$ [$$\therefore$$ sum of the $$3$$ consecutive integers is $$51$$]

\begin{align}3x + 3 &= 51\\3x &= 51 - 3\\3x &= 48\\x &= \frac{{48}}{3}\\x &= 16\\x + 1 &= 17\\x + 2 &= 18\end{align}

Three consecutive integers are $$16$$, $$17$$ and $$18$$.

## Chapter 2 Ex.2.2 Question 7

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

### Solution

What is known?

Sum of three consecutive multiples of $$8$$ is $$888$$.

What is unknown?

Multiples.

Reasoning:

Assume one multiple to be a variable. Then use the property of consecutive multiples to form a linear equation.

Steps:

Let the first multiple be $$x$$

Then second consecutive multiple of $$8$$ ➔ $$x + 8$$

Third consecutive multiple of $$8$$ ➔$$x + 8 + 8 = {\rm{ }}x + 16$$

Sum of three consecutive multiples is $$888.$$

$$\therefore x + (x + 8) + (x + 16) = 888$$

\begin{align}3x + 24 &= 888\\3x &= 888 - 24\\3x &= 864\\x &= \frac{{864}}{3}\\x &= 288\\x + 8 &= 296\\x + 16 &= 304\end{align}

The multiples are $$288$$, $$296$$ and $$304$$.

## Chapter 2 Ex.2.2 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.

### Solution

What is known?

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 $$74$$.

What is unknown?

Number.

Reasoning:

Assume one integer to be a variable. Then use the property of consecutive integers and the given condition to form a linear equation.

Steps:

Let the first integer be $$x$$

Next two consecutive integers are $$x + 1\;{\rm{ and }}\; x + 2$$

They are taken in increasing order and multiplied by $$2$$,$$3$$ and $$4$$ respectively.

$\therefore x \times 2 = 2x$

\begin{align}\left( {x + 1} \right) \times 3 = 3x + 3\\\left( {x + 2} \right) \times 4 = 4x + 8\end{align}

They add up to $$74$$

$\therefore \,\,2x + 3x + 3 + 4x + 8 = 74$

\begin{align}9x + 11 &= 74\\9x &= 74 - 11\\9x &= 63\\x &= \frac{{63}}{9}\\x &= 7\end{align}

Then $x + 1 = 8$

$x + 2 = 9$

The numbers are $$7$$, $$8$$ and $$9$$.

## Chapter 2 Ex.2.2 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?

### Solution

What is known?

(i) Ages of Rahul and Haroon are in ratio $$5 :7$$

(ii) Four years later, sum of their ages will be $$56$$ years.

What is unknown?

Present ages of Rahul and Haroon.

Reasoning:

Assume the age of either Rahul or Haroon as a variable. Use the first condition to express the ages in terms of the variable. Then use the second condition to form a linear equation.

Steps:

Ages of Rahul and Haroon are in ratio $$5 :7$$

Present ages of Rahul and Haroon are $$5x$$ and $$7x$$ respectively.

Four years later, sum of their ages will be $$56$$ years.

Four years later, age of Rahul $$= 5x + 4$$

Four years later, age of Haroon $$= 7x + 4$$

Sum ➔$$5x + 4 + 7x + 4 = 56$$

\begin{align}12x + 8 &= 56\\12x &= 56 - 8\\12x &= 48\\x &= \frac{{48}}{{12}}\\x &= 4\end{align}

Present age of Rahul $$= 5x = 5 \times 4 = 20$$

Present age of Haroon $$= 7x = 7 \times 4 = 28$$

Present age of Rahul and Haroon are $$20$$ and $$28$$ years respectively.

## Chapter 2 Ex.2.2 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?

### Solution

What is known?

(i) Number of boys and girls in a class are in ratio $$7:5$$

(ii) Number of boys is $$8$$ more than the number of girls.

What is unknown?

Total class strength.

Reasoning:

Use the first condition to express the numbers in terms of the variable. Use the second condition to form a linear equation.

Steps:

Number of boys and girls in a class are in ratio $$7:5$$

$$\therefore$$ Number of boys in class$$\, = 7x$$

Number of girls in class $$= 5x$$

Number of boys is $$8$$ more than number of girls.

$\therefore 7x = 5x + 8$

\begin{align}7x - 5x = 8\\2x = 8\\x = 4\end{align}

Number of boys in class $$= 7x = 7 \times 4 = 28$$

Number of girls in class $$= 5x = 5 \times 4 = 20$$

Total class strength

\begin{align}&=\text{ number of boys} \\ & \quad +\text{ number of girls}\\&= 28 + 20\\ &= 48\end{align}

Total class strength is $$48$$ students.

## Chapter 2 Ex.2.2 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?

### Solution

What is known?

(i) Baichung’s father is $$26$$ years younger than Baichung’s grandfather and $$29$$ years older than Baichung

(ii) Sum of the ages of all the three is $$135$$ years.

What is unknown?

Ages of Baichung, Baichung’s father and Baichung’s grandfather.

Reasoning:

Assume Baichung’s age to be a variable. Then use the first condition to express the ages of Baichung’s father and grandfather. Finally, use the last second condition to form a linear equation.

Steps:

Let the age of Baichung be $$x$$ years.

$$\therefore$$ Age of Baichung’s father $$= {\text{ }}x + 29$$ years as he is $$29$$ years older than Baichung.

Age of Baichung’s grandfather $$=$$ age of Baichung father $$+$$ $$26$$ years

[$$\therefore$$ since Baichung’s father is $$26$$ years younger than Baichung’s grandfather]
$➔ x + 29 + 26$

$= x + 55\,{\rm{Years}}$

Sum of ages of all the three is $$135$$ years.

$$\therefore$$  $$x + x + 29 + x + 55 = 135$$

Ages: Baichung father grandfathers.

\begin{align}3x + 84 &= 135\\3x &= 135 - 84\\3x &= 51\\x &= \frac{{51}}{3}\\x &= 17\end{align}

Age of Baichung is $$17$$ years

Age of Baichung’s father is $$x + 29 = 17 + 29 = 46$$ years

Age of Baichung’ ‘s grandfathers is $$x + 55 = 17 + 55 = 72$$ years

## Chapter 2 Ex.2.2 Question 12

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

### Solution

What is known?

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

What is unknown?

Ravi’s present age

Reasoning:

Assume Ravi’s age to be a variable. Use the first and second condition to express his age after $$15$$ years in the form of a linear equation.

Steps:

Let the present age of Ravi be $$x$$ years.

$$15$$ years from now, Ravi’s age will be $$4$$ times his present age

\begin{align}x + 15 &= 4x\\x - 4x &= - 15\\ - 3x &= - 15\\x &= \frac{{ - 15}}{{ - 3}}\\x &= 5\end{align}

$$\therefore$$ Ravi’s present age is $$5$$ years

## Chapter 2 Ex.2.2 Question 13

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

### Solution

What is known?

A rational number is such that when multiply it by\begin{align}\frac{5}{2}\end{align} and add\begin{align}\frac{2}{3}\end{align}  to the product, answer is \begin{align}\frac{-7}{12}\end{align}.

What is unknown?

Numbers.

Reasoning:

Assume the rational number to be a variable. Apply the conditions sequentially to form a linear equation.

Steps:

Let the rational number be $$x$$

Multiply by \begin{align}\frac{5}{2} \to x \times \frac{5}{2} = \frac{{5x}}{2}\end{align}

Add  \begin{align}\frac{2}{3}\end{align}in the product  \begin{align}= \frac{{5x}}{2} + \frac{2}{3}\end{align}

Answer is \begin{align}➔ \frac{{ - 7}}{{12}}\end{align}

\begin{align} \therefore \quad \frac{{5x}}{2} + \frac{2}{3} &= \frac{{ - 7}}{{12}}\\\frac{{5x}}{2} &= \frac{{ - 7}}{{12}} - \frac{2}{3}\\\frac{{5x}}{2} &= \frac{{ - 7 - 8}}{{12}}\\\frac{{5x}}{2} &= \frac{{ - 15}}{{12}}\\x &= \frac{{ - 15}}{{12}} \times \frac{2}{5}\\x &= \frac{{ - 3}}{6}\\x &= \frac{{ - 1}}{2}\end{align}

The number is  \begin{align}- \frac{1}{2}.\end{align}

## Chapter 2 Ex.2.2 Question 14

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

### Solution

What is known?

(i) Cashier Lakshmi has currency notes of denominaton $$\rm{Rs}.\,100$$, $$\rm{Rs}.\,50$$, and $$\rm{Rs}.\,10$$.

(ii) Ratio of the number of notes is $$2:3:5$$.

(iii) Total cash with Lakshmi is $$\rm{Rs}.\, 4,00,000$$

What is unknown?

How many notes of each denomination.

Reasoning:

Use the ratio condition and express the number of notes of different denominations in terms of a variable. Use the second condition to form a linear equation.

Steps:

(i) Lakshmi has currency notes of denomination $$\rm{Rs}.\,100$$$$\rm{Rs}.\,50$$, $$\rm{Rs}.\,10$$.

(ii) Number of notes are in ratio $$2:3:5$$, therefore number of notes is $$2x,{\rm{ }}3x$$ and $$5x$$

 Denomination Number of notes Total $$\rm{Rs}.\,100$$ $$2x$$ $$200x$$ $$\rm{Rs}.\,50$$ $$3x$$ $$150x$$ $$\rm{Rs}.\,10$$ $$5x$$ $$50x$$

\begin{align}200x + 150x + 50x &= 400000 \\400x &= 400000 \\x &= \frac{400000}{400} \\x &= 1000 \\\end{align}

 Denomination Number of notes $$\rm{Rs}.\,100$$ $$2x = 2 \times 1000 = 2000$$ $$\rm{Rs}.\,50$$ $$3x = 3 \times 1000 = 3000$$ $$\rm{Rs}.\,10$$ $$5x = 5 \times 1000 = 5000$$

## Chapter 2 Ex.2.2 Question 15

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

### Solution

What is known?

(i) Total amount $$= \rm{Rs}.\, 300$$

(ii) Coin denominations are $$\rm{Rs.} 1$$, $$\rm{Rs}.\,2$$ and $$\rm{Rs}. 5$$

(iii) Number of $$\rm{Rs}.\,2$$ coins are $$3$$ times $$\rm{Rs.} \,5$$ coins.

(iv) Total coins ➔$$160$$

What is unknown?

How many coins of each denominations are there.

Reasoning:

Assume the number of $$\rm{Rs}.\,5$$ coin to be a variable. Use the second condition to obtain the number of coins for $$\rm{Rs}.\,2$$ in terms of the variable. Use the third condition to find the number of coin for $$\rm{Rs}.1$$. Finally, use the first condition to form a linear equation.

Steps:

Let the number of $$\rm{Rs} \,5$$ coins be $$x.$$

Then the number of $$\rm{Rs}.2$$ coins is $$3x.$$

Number of $$\rm{Rs.}1$$ coin is $$= 160 - (x + 3x)$$ [$$\therefore$$as total coins are 160]

$= 160 - 4x$

 Denomination Number of coins Amount $$\rm{Rs}.1$$ $$160-4x$$ $$160-4x$$ $$\rm{Rs}.2$$ $$3x$$ $$6x$$ $$\rm{Rs}.5$$ $$x$$ $$5x$$

$\to 160 - 4x + 6x + 5x = 300$

\begin{align}160 + 7x &= 300 \\7x &= 300 - 160 \\7x &= 140 \\x &= \frac{{140}}{7} \\x &= 20 \\\end{align}

 Denomination Number of coins $$\rm{Rs}.1$$ \begin{align} 160 - 4x &= 160 - 80 \\&= 80 \end{align} $$\rm{Rs}.2$$ $$3x = 3 \times 20 = 60$$ $$\rm{Rs}.5$$ $$x = 20$$

## Chapter 2 Ex.2.2 Question 16

The organizers of an essay competition decide that a winner in the competition gets a prize of $$\rm{Rs.}100$$ and a participant who does not win gets a prize of $$\rm{Rs.} \,25$$. The total prize money distributed is $$\rm{Rs}.\,3,000$$. Find the number of winners, if the total number of participants is $$63$$.

### Solution

What is known?

(i) Winner of essay competition gets a prize of $$\rm{Rs} \,100$$

(ii) Participant who does not win gets a prize of $$\rm{Rs} \,25$$

(iii) Total prize of money distributed is $$\rm{Rs.}\, 3000$$

(iv) Total number of participants is $$63$$.

What is unknown?

Number of winners.

Reasoning:

Assume the total number of winners is $$x$$. As total number of participants is $$63$$, number of participants who did not win will be $$63 - x$$. Now use first, second, and third conditions to form a linear equation.

Steps:

Let the number of number of winners be $$x$$

Then the number of participants who do not win ➔$$63- x$$ [as the total participants is $$63$$]

Total prize money $$= \rm{Rs}. 3000$$

\begin{align}\begin{bmatrix} \text{Number of winners}\!\times\! 100 \\ +\! \text{Number of participants } \\ \text{who do not win}\! \times \!25 \end{bmatrix}\!\!=\! 3000 \end{align}

\begin{align} x \!\times \!100 \!+\! (63 \!-\! x)25 &= 3000 \\100x + 1575 - 25x &= 3000 \\75x + 1575 &= 3000 \\75x &= 3000\! - \!1575 \\x &= \frac{{1425}}{{75}} \\x &= 19 \\\end{align}

Number of winners is $$19$$

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