# Excercise 2.4 Linear Equations in One Variable- NCERT Solutions Class 8

Linear Equations in One Variable

## Chapter 2 Ex.2.4 Question 1

Amina thinks of a number and subtracts \(\begin{align}\frac{5}{2}\end{align}\) 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?

**Solution**

**Video Solution**

**What is known?**

i) \(\begin{align}\frac{5}{2}\end{align}\) is subtracted from a number

ii) Then multiplied by \(8\)

iii) Result is \(3\) times the actual number

**What is unknown?**

Number thought of by Amina

**Reasoning: **

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

**Steps:**

Let the number be \(x\).

According to the given question,

\[\begin{align}{8\left( {x - \frac{5}{2}} \right) = 3x} \\{\,\,\,\,8x - 20 = 3x}\end{align}\]

Transposing \(3x\) to LHS and \(−20\) to RHS, we obtain

\[\begin{align}{8x - 3x = {\text{ }}20} \\{\,\,\,\,\,\,\,\,\,\,\,5x = 20}\end{align}\]

Dividing both sides by \(5\), we obtain

\[x = 4\]

Hence, the number is \(4\).

## Chapter 2 Ex.2.4 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?

**Solution**

**Video Solution**

**What is known?**

i) A positive number is \(5\) times another number

ii) \(21\) is added to both the numbers

iii) Then one of the new numbers becomes twice the other new number

**What is unknown?**

Numbers

**Reasoning: **

Assume one positive number to be variable then use the conditions to form a linear equation.

**Steps:**

Let the numbers be \(x\) and \(5x\). According to the question,

\[\begin{align}21 + 5x &= 2\left( {x{\text{ }} + {\text{ }}21} \right) \\21 + 5x &= 2x{\text{ }} + {\text{ }}42\end{align}\]

Transposing \(2x\) to LHS and \(21\) to RHS, we obtain

\[\begin{align}5x - 2x &= 42 - 21 \\3x &= 21\end{align}\]

Dividing both sides by \(3\), we obtain

\[x = 7\]

First number is \(\,x=7\)

Second number is \(5x = {\text{ }}5 \times 7 = 35\)

Hence, the numbers are \(7\) and \(35\) respectively.

## Chapter 2 Ex.2.4 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?

**Solution**

**Video Solution**

**What is known?**

i) Sum of the digits of a two-digit number is \(9\)

ii) Interchanging the digits result in a new number greater than the original number by \(27\)

**What is unknown?**

Number

**Reasoning: **

Assume one of the digits of two-digit as variable then use other conditions and form a linear equation.

**Steps:**

Let the digits at tens place and ones place be *\(x\)* and \(9{\text{ }} - {\text{ }}x\) respectively.

Therefore, original number \( = {\text{ }}10x{\text{ }} + {\text{ }}\left( {9{\text{ }} - {\text{ }}x} \right){\text{ }} = {\text{ }}9x{\text{ }} + {\text{ }}9\)

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:

\[\begin{align}&= 10\left( {9 - {\text{ }}x} \right) + x \\&= 90 - 10x + x \\&= 90 - 9x\end{align}\]

According to the given question,

New number \(=\) Original number \(+\; 27\)

\[\begin{align}90 - 9x &= 9x + 9 + 27 \\90 - 9x &= 9x + 36\\\end{align} \]

Transposing \(9x\) to RHS and \(36\) to LHS, we obtain

\[\begin{align}90 - 36 = 18x \\54 = 18x\end{align}\]

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 \(9x + 9\)

\[\begin{align}&= {\text{ }}9{\text{ }} \times {\text{ }}3{\text{ }} + {\text{ }}9{\text{ }} \\& = {\text{ }}36 \\\end{align}\]

## Chapter 2 Ex.2.4 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?

**Solution**

**Video Solution**

**What is known?**

i) One of the two digits of a two-digit number is three times the other digit

ii) Interchanging the digit of this two-digit number and adding the resulting number to the original number results in \(88\).

**What is unknown?**

Original number

**Reasoning: **

Assume one of the digits of two-digit as variable then use other conditions and form a linear equation.

**Steps:**

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 \( = 10 \times 3x + x = 30x + x = 31x\)

According to the given question,

Original number \(+\) New number \(=\) \(88\)

\[\begin{align}{13x + 31x = 88} \\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,44x = 88}\end{align}\]

Dividing both sides by \(44\), we obtain

\[x = 2\]

Therefore, original number \( = {\text{ }}13x = 13 \times 2 = 26\)

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 \(62.\)

## Chapter 2 Ex.2.4 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 his mother’s present age. What are their present ages?

**Solution**

**Video Solution**

**What is known?**

i) Shobo’s mother’s present age is six times Shobo’s present age

ii) Shobo’s age five years from now will be one third of his mother’s present age

**What is unknown?**

Present age of Shobo’s and his mother’s.

**Reasoning: **

Assume Shobo’s age as variable then his mother’s age will be six time of his age. Use second condition and form a linear equation.

**Steps:**

Let Shobo’s age be \( x\) years. Therefore, his mother’s age will be \(6x\) years.

According to the given question,

After \(5\) years Shobo's age

\(\begin{align} &= \frac{{{\rm{Shobo's}}\,\,{\rm{mother's}}\,\,{\rm{present}}\,\,{\rm{age}}}}{{\rm{3}}}\end{align}\)

\(\begin{align} x + 5 &= \frac{{6x}}{3}\\ x + 5 &= 2x\end{align}\)

Transposing *\(x\)* to RHS, we obtain

\[\begin{align}5&= 2x - x \\5&= x\end{align}\]

Shobo’s age is \(x = 5\)

Shobo’s mother’s age is \(6x=6\times 5=30\)

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

## Chapter 2 Ex.2.4 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 \(\rm Rs \;100\) per metre it will cost the village panchayat \(\rm{Rs} \,75,000\) to fence the plot. What are the dimensions of the plot?

**Solution**

**Video Solution**

**What is known?**

i) The length and breadth of the plot are in the ratio \(11:4.\)

ii) At the rate \(\rm Rs \;100\) per metre it will cost Rs \(75,000\) to fence the plot

**What is unknown?**

Dimensions of the plot

**Reasoning: **

Use the ratio condition and express length and breadth in terms of a variable and form linear equation.

**Steps:**

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 \(11x\rm{ \;m}\) and \(4x\rm\; m\) respectively.

Perimeter of the plot

\[\begin{align} &={\rm{ 2}}\left( {\rm{Length + Breadth}} \right)\\ &= {\rm{[}}2(11x + 4x){\rm{]m}}\\\ &= 30x\,{\rm{m}}\end{align}\]

It is given that the cost of fencing the plot at the rate of \(\rm Rs \;100\) per metre is \(\rm{Rs} \,75,000\).

\[\begin{align}\therefore\, 100 \times {\rm{Perimeter}}\,\,\, &= 75000\\ 100{\rm{ }} \times {\rm{ }}30x &= {\rm{ }}75000\\ \,\,\,\,\,\,\,\,\,3000x{\rm{ }} &= {\rm{ }}75000\end{align}\]

Dividing both sides by \(3000\), we obtain

\[x = 25\]

\[\begin{align}\,\,\,{\rm{Length}}\, &= 11x\,{\rm{m}}\\ &= \left( {11 \times 25} \right){\rm{m}}\\ & = {\rm{ }}275{\rm{m}}\\{\rm{Breadth}} &= {\rm{ }}4x\,\,{\rm{m}}\\ &= \left( {4 \times 25} \right){\rm{m}}\\ & = {\rm{ }}100\,{\rm{m}}\end{align}\]

Hence, the dimensions of the plot are \(275 \,\rm{m}\) and \(100\;\rm{m}\) respectively.

## Chapter 2 Ex.2.4 Question 7

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

**Solution**

**Video Solution**

**What is known?**

i) Shirt material that costs him \(\rm{Rs} \;50\) per meter and trouser material that costs him \(\rm{Rs} \;90 \) per meter

ii) For every \(2\) meters of the trouser material, \(3\) meters of the shirt material is purchased.

iii)Selling of the materials takes place at \(12\%\) and \(10\%\) profit respectively

iv) Total sale is \(\rm{Rs} \;36660\)

**What is unknown?**

How much trouser material is being purchased.

**Reasoning: **

Trouser material and shirt material are purchased in the ratio of \(2:3\). Use the ratio condition and express clothes material purchased in the form of variable and form the linear equation.

**Steps:**

Let \(2x\;\rm{m}\) of trouser material and \(3x\) m of shirt material be bought by him.

Per metre selling price of trouser material

\[\begin{align} &= {\rm{Rs}}\,\,\,\left( {90 + \frac{{90 \times 12}}{{100}}} \right)\\ &= {\rm{Rs}}\,\,100.80\end{align}\]

Per metre selling price of shirt material

\[\begin{align} &= {\rm{Rs}}\,\,\left( {50 + \frac{{50 \times 10}}{{100}}} \right)\\ &= {\rm{Rs}}\,\,\,55\end{align}\]

Given that, total amount of selling = \(\rm{Rs} \;36660\)

\[\begin{align}100.80 \! \times \! \left( {2x} \right) \! + \! 55 \! \times \! \left( {3x} \right) &= \! {\rm{ }}36660\\201.60x \! + \! 165x &= \! 36660\\366.60x &= \! 36660\end{align}\]

Dividing both sides by \(366.60\), we obtain

\[x{\text{ }} = {\text{ }}100\]

Trouser material

\[\begin{align}&= 2x\,{\rm{m}}\\&= \left( {2{\rm{ }} \times {\rm{ }}100} \right)\,{\rm{m}}\\ &= 200\,{\rm{m}}\end{align}\]

## Chapter 2 Ex.2.4 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.

**Solution**

**Video Solution**

**What is known?**

i) Half of the herd of deer are grazing in the field

ii) Three-fourths of the remaining are playing nearby

iii)The rest \(9\) are drinking water from the pond

**What is unknown?**

The number of deer in the herd

**Reasoning: **

Assume total number of deer as a variable now use other conditions to form linear equation.

**Steps:**

Let the number of deer be \(x.\)

Number of deer grazing in the field \(\begin{align}=\frac{x}{2}\end{align}\)

Number of deer playing nearby

\[\begin{align} &= \frac{3}{4} \times \begin{Bmatrix} \text{ Number of} \\ \text{ remaining deer} \end{Bmatrix} \\&= \frac{3}{4} \times \left( {x - \frac{x}{2}} \right)\\&= \frac{3}{4} \times \frac{x}{2}\\&= \frac{{3x}}{8}\end{align}\]

Number of deer drinking water from the pond \(=9\)

\[\begin{align}x - \left( {\frac{x}{2} + \frac{{3x}}{8}} \right)&= 9\\x - \left( {\frac{{4x + 3x}}{8}} \right)&= 9\\x - \frac{{7x}}{8}&= 9\\\frac{x}{8} &= 9\end{align}\]

Multiplying both sides by \(8\), we obtain

\[x{\text{ }} = {\text{ }}72\]

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

## Chapter 2 Ex.2.4 Question 9

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

**Solution**

**Video Solution**

**What is known?**

i) Grandfather is ten times older than his granddaughter

ii) also \(54\) years older than her

**What is unknown?**

Present age of grandfather and granddaughter.

**Reasoning: **

Assume granddaughter age as a variable then grandfather’s age will be \(10\) time his granddaughter. Use second condition to form linear equation.

**Steps:**

Let the granddaughter’s age be \(x\) years. Therefore, grandfather’s age will be \(10x\) years.

According to the question,

Grandfather’s age \(=\) Granddaughter’s age \(+\) \(54\) years

\[10x = x + 54\]

Transposing \(x\) to LHS, we obtain

\[\begin{align}10x - x &= 54\\9x &= 54\\x &= 6\end{align}\]

Grand daughter’s age

\[\begin{align} &= x\,\rm{years}\\&= 6\rm{ years}\end{align}\]

Grand father’s age

\[\begin{align}&=10x\,{\rm{years }}\\ &= \left( {10{\rm{ }} \times {\rm{ }}6} \right){\rm{ years }}\\ &= 60 \,{\rm{ years}}\end{align}\]

## Chapter 2 Ex.2.4 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.

**Solution**

**Video Solution**

**What is known?**

i) Aman’s age is three times his son’s age

ii) Ten years ago he was five times his son’s age

**What is unknown?**

Present age of Aman and his son.

**Reasoning: **

Assume Aman’s son’s age as a variable now use other conditions and form a linear equation.

**Steps:**

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 question,

\(10\) years ago, Aman’s age \( =5\times \) Aman’s son’s age \(10\) years ago

\[\begin{align}3x - 10&= 5\left( {x - 10} \right) \\3x - 10&= 5x - 50 \\\end{align} \]

Transposing \(3x\) to RHS and \(50\) to LHS, we obtain

\[\begin{align}50 - 10&= 5x - 3x \\40 &= 2x \\\end{align}\]

Dividing both sides by \(2\), we obtain

\[20{\text{ }} = {\text{ }}x\]

Aman's sons age

\[\begin{align} &= x\;{\rm{ years}}\\ &= 20\;{\rm{ years}}\end{align}\]

Aman's age

\[\begin{align} &= 3x\;{\rm{ years}}\\ &= {\rm{ }}\left( {3{\rm{ }} \times {\rm{ }}20} \right){\rm{ years}}\\ &= {\rm{ }}60\;{\rm{ years}}\end{align}\]