# NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.4) Exercise 2.4

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**CBSE NCERT Solutions for Class 8 Maths Chapter 2 – Exercise 2.4**

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**Important Topics under NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.4) Exercise 2.4**

Class 8 Maths Chapter 2 Exercise 2.4, includes questions based on some applications of Linear Equations in One Variable. This exercise primarily covers linear equations with variables on both sides. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, greatly reduces students’ burden as they attempt to solve these questions. The problems presented in this exercise have clear solutions in the Extramarks provided NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, are written by knowledgeable subject experts with consideration for students’ level of comprehension in Class 8. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, are written using the most recent curriculum that the CBSE has approved for use in Class 8. As a result, students no longer need to search for reliable solutions elsewhere or worry about the syllabus. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, are a one-stop destination for Class 8 students to find thorough, trustworthy solutions to all of the exercises’ questions.

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**Importance of Linear Equations in One Variable**

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**Exercise 2.4 NCERT Solutions Class 8 Maths Chapter 2**

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**NCERT Solutions Class 8 Math Chapter 2 All Exercises**

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**Q.1 **

$\begin{array}{l}\mathrm{Amina}\mathrm{thinks}\mathrm{of}\mathrm{a}\mathrm{number}\mathrm{and}\mathrm{subtracts}\frac{5}{2}\mathrm{from}\mathrm{it}.\mathrm{She}\\ \mathrm{multiplies}\mathrm{the}\mathrm{result}\mathrm{by}8.\mathrm{The}\mathrm{result}\mathrm{now}\mathrm{obtained}\mathrm{is}\\ 3\mathrm{times}\mathrm{the}\mathrm{same}\mathrm{number}\mathrm{she}\mathrm{thought}\mathrm{of}.\mathrm{What}\mathrm{is}\mathrm{the}\\ \mathrm{number}?\end{array}$

**Ans**

Let the number be *x*.

According to the given question,

$8(\mathrm{x}-\frac{5}{2})=3\mathrm{x}$

8*x* − 20 = 3*x*

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

8*x* − 3*x* = 20

5*x* = 20

On dividing both sides by 5, we get

*x* = 4

Hence, the number is 4.

**Q.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? What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

**Ans**

Let the numbers be x and 5x.

$\begin{array}{l}\text{According to the question},\text{the equation becomes}\\ \text{21}+\text{5}\mathrm{x}=\text{2}(\mathrm{x}+\text{21})\\ \Rightarrow \text{21}+\text{5}\mathrm{x}=\text{2}\mathrm{x}+\text{42}\\ \\ \text{Transposing 2}\mathrm{x}\text{to L}.\text{H}.\text{S and 21 to R}.\text{H}.\text{S},\text{we obtain}\\ \Rightarrow \text{5}\mathrm{x}-\text{2}\mathrm{x}=\text{42}-\text{21}\\ \text{3}\mathrm{x}=\text{21}\\ \\ \text{Dividing both sides by 3},\text{we obtain}\\ \mathrm{x}=\text{7}\\ \text{5}\mathrm{x}=\text{5}\times \text{7}=\text{35}\\ \\ \text{Hence},\text{the numbers are 7 and 35 respectively}.\end{array}$

**Q.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?

**Ans**

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

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

= 90 − 9*x*

According to the given question,

New number = Original number + 27

90 − 9*x* = 9*x* + 9 + 27

90 − 9*x* = 9*x* + 36

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

90 − 36 = 18*x*

54 = 18*x*

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 9*x* + 9

= 9 × 3 + 9

= 36

**Q.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?

**Ans**

$\begin{array}{l}\text{Let the digits at tens place and ones place be \u2018}\mathrm{x}\u2018\text{and \u20183}\mathrm{x}\u2018.\\ \therefore \mathrm{the}\text{original number}=\text{1}0\mathrm{x}+\text{3}\mathrm{x}=\text{13}\mathrm{x}\\ \\ \text{On interchanging the digits},\text{the digits at ones place will be \u2018x\u2019and}\\ \text{tens place will be \u20183}\mathrm{x}\u2018.\\ \\ \therefore \text{New number}=\text{1}0\times \text{3}\mathrm{x}+\mathrm{x}=\text{3}0\mathrm{x}+\mathrm{x}=\text{31}\mathrm{x}\\ \\ \text{According to the given question},\mathrm{the}\text{}\mathrm{equation}\text{\hspace{0.17em}}\mathrm{becomes}\\ \text{8813}\mathrm{x}+\text{31}\mathrm{x}=\text{88}\\ \Rightarrow \text{44}\mathrm{x}=\text{88}\end{array}$

$\begin{array}{l}\text{Dividing both sides by 44},\text{we obtain}\\ \Rightarrow \frac{\text{44}\mathrm{x}}{44}=\frac{\text{88}}{44}\\ \Rightarrow \mathrm{x}=\text{2}\\ \\ \therefore ,\text{original number}=\text{13}\mathrm{x}=\text{13}\times \text{2}=\text{26}\\ \\ \text{Hence},\text{the two}-\text{digit number may be 26 or 62}.\end{array}$

**Q.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?

**Ans**

Let Shobo’s age be *x* years.

His mother’s age will be 6*x* years.

According to the given question, the equation becomes

:

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

Multiplying both sides by 3

$\begin{array}{l}(\mathrm{x}+5)\times 3=\frac{6\mathrm{x}}{3}\times 3\\ \Rightarrow 3\mathrm{x}+15=6\mathrm{x}\\ \text{Transposing 3x to R.H.S,we get}\\ \Rightarrow \text{15}=\text{6x}-\text{3x}\\ \Rightarrow \text{15}=3\mathrm{x}\\ \text{Dividing both sides by 3, we get}\\ \Rightarrow \frac{\text{15}}{3}=\frac{3\mathrm{x}}{3}\\ \Rightarrow \mathrm{x}=5\\ \\ \therefore 6\mathrm{x}=6\times 5=30\end{array}$

Therefore, the present ages of Shobo and his mother will be 5 years and 30 years.

**Q.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 ₹100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?

**Ans**

Let the ratio between the length and breadth of the rectangular plot be *x*.

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

Perimeter of the plot = 2(Length + Breadth)

$\begin{array}{l}=\left[2(11\mathrm{x}+4\mathrm{x})\right]\mathrm{m}\\ =30\mathrm{x}\end{array}$

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

100 × Perimeter = 75000

100 × 30*x* = 75000

3000*x* = 75000

Dividing both sides by 3000, we obtain

\frac{\text{3}000x}{3000}=\text{}\frac{\text{75}000}{3000}

*x* = 25

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

and Breadth = 4*x* m = (4 × 25) m = 100 m

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

**Q.7 **Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy?

**Ans**

Let the trouser and shirt material bought by Hasan be 2x and 3x .

The selling price of trouser material is

$\begin{array}{l}=\text{\hspace{0.17em}}\u20b9(90+\frac{90\times 12}{100})\\ =\text{\hspace{0.17em}}\u20b9100.80\end{array}$

The selling price of Shirt material is

$\begin{array}{l}=\text{\hspace{0.17em}}\u20b9\left(50+\frac{50\times 10}{100}\right)\\ =\text{\hspace{0.17em}}\u20b955\end{array}$

Total sale =36,600

$\begin{array}{l}\text{1}00.\text{8}0\times \left(\text{2}\mathrm{x}\right)+\text{55}\times \left(\text{3}\mathrm{x}\right)=\text{36,60}0\\ \text{2}0\text{1}.\text{6}0\mathrm{x}+\text{165}\mathrm{x}=\text{36,60}0\\ \text{366}.\text{6}0\mathrm{x}=\text{3660}0\\ \text{Dividing both sides by 366}.\text{6}0,\text{we obtain}\\ \Rightarrow \frac{\text{366}.\text{6}0\mathrm{x}}{366.60}=\frac{\text{3660}0}{366.60}\\ \Rightarrow \mathrm{x}=99.836\\ \Rightarrow \text{x}=\text{1}00\\ \\ \therefore \text{Trouser material}=\text{2}\mathrm{x}\text{m}\\ \text{}=(\text{2}\times \text{1}00)\text{m}\\ \text{}=\text{2}00\text{m}\end{array}$

**Q.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.

**Ans**

Let the number of deer be x.

Number of deer grazing in the field = x / 2

Remaining deer= x – (x / 2)

Therefore, number of deer playing nearby is

$\begin{array}{l}=\frac{3}{4}\times (\mathrm{x}-\frac{\mathrm{x}}{2})\\ =\frac{3}{4}\times \frac{\mathrm{x}}{2}\\ =\frac{3\mathrm{x}}{8}\end{array}$

The rest 9 are drinking water from the pond, so the equation becomes:

$\begin{array}{l}\mathrm{x}-(\frac{\mathrm{x}}{2}+\frac{3\mathrm{x}}{8})=9\\ \Rightarrow \mathrm{x}-\left(\frac{4\mathrm{x}+3\mathrm{x}}{8}\right)=9\\ \Rightarrow \mathrm{x}-\frac{7\mathrm{x}}{8}=9\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\Rightarrow \frac{8\mathrm{x}-7\mathrm{x}}{8}=9\\ \Rightarrow \frac{\mathrm{x}}{8}=9\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{Mutiplying both sides by 8,we get}\\ \Rightarrow \frac{\mathrm{x}}{8}\times 8=9\times 8\\ \Rightarrow \mathrm{x}=72\end{array}$

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

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

**Ans**

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

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

According to the question, the equation becomes:

10*x* = *x* + 54

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

10*x* − *x* = 54

9*x* = 54

*x* = 6

Granddaughter’s age* = x* years = 6 years

Grandfather’s age = 10*x* years

= (10 × 6) years

= 60 years

Therefore, the grandfather’s age is 60 years and granddaughter’s age is 6 years.

**Q.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.

**Ans**

Let Aman’s son’s age be *x* years.

Therefore, Aman’s age will be 3*x* years.

Ten years ago, their age was (*x* − 10) years and (3*x* − 10) years.

According to the question, the equation is:

$\begin{array}{l}3\mathrm{x}-10=5(\mathrm{x}-10)\\ \Rightarrow 3\mathrm{x}-10=5\mathrm{x}-50\\ \mathrm{On}\text{}\mathrm{transposing}\text{5x to L.H.S and 10 to R.H.S,we get}\\ \Rightarrow 3\mathrm{x}-5\mathrm{x}=10-50\\ \Rightarrow -2\mathrm{x}=-40\\ \Rightarrow 2\mathrm{x}=40\end{array}$

$\begin{array}{l}\text{On dividing both the sides by 2,we get}\\ \Rightarrow \frac{2\mathrm{x}}{2}=\frac{40}{2}\\ \Rightarrow \mathrm{x}=20\end{array}$

Therefore,

Aman’s son’s age is 20 years and

Aman’s age = 3 *× *20 years = 60 years.

## FAQs (Frequently Asked Questions)

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### 2. Are the concepts under the NCERT Class 8 Maths Chapter 2 Exercise 2.4 clarified by the NCERT solutions?

For students in Class 8, NCERT solutions are crucial for exam preparation. In order to prepare for the annual exams, students can use learning modules like the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4. Students can develop the ability to respond to any question of a similar nature that appears in the exam by becoming more familiar with the steps and procedure.

### 3. When preparing for the board exams, how should students use the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4?

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