Ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable in English | Class 8 Chapter 2 maths exercise 2.4, 2.5, 2.6

Ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable class 8th maths. From class 8 chapter 2 maths in english. Here in ncert solutions for maths chapter 2 class 8 we given all questions solutions in detail. NCERT Solution of class 8 subject math chapter 2 Linear Equations in One Variable get all questions answers, notes, pdf like format. In this ncert maths class 8 chapter 2 students will get to know the concept of Pair of Linear Equations in Two Variables. chapter 2 maths class 8 explains how to represent a situation algebraically and graphically. Students will explore the methods of solving ncert maths class 8 chapter 2 the pair of the linear equation through Graphical Method. This chapter describes the Algebraic Method, Elimination Method, Cross-Multiplication Method, Substitution Method respectively. Students must practice this chapter to master the method of solving the linear equations. The exercises present in the chapter should be dealt with utmost sincerity if one wants to score well in the examinations.

Students can practice the questions based on the same through class 8th maths ncert solutions for class 8 maths chapter 2 exercise 2.4, ncert solutions for class 8 maths chapter 2 exercise 2.5, ncert solutions for class 8 maths chapter 2 exercise 2.6 and many more. These examples are in the form of word problems.
The next topic is Solving equations having the variable on both sides. In this section, students will learn how to solve such equations which have expressions with the variable on both sides. class 8th maths

Here we solve ncert solutions for ncert maths class 8 chapter 2 solution From class 8 maths chapter 2 pdf concepts all solutions with easy method with expert solutions. chapter 2 maths class 8 help students in their study, home work and preparing for exam. Soon we provide CBSE chapter 2 maths class 8 solutions English Medium. NCERT solutions for class 8 maths chapter 2 pdf Linear Equations in One Variable PDF here. You can download ncert english book class 8 from official NCERT website or Click HERE.

ncert solutions for class 8 maths chapter 2
Linear Equations in One Variable

ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable in English

NCERT solutions for class 8 maths chapter 2 pdf – Questions Answer Click Here

Ncert Solutions for Class 8 Maths Chapter 2 EXERCISE 2.4
class 8th maths

1. Amina thinks of a number and subtracts \(\displaystyle \frac{5}{2}\) 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’s Amina thought the number is = x
She subtract \(\displaystyle \frac{5}{2}\) from x 
\(\displaystyle =x-\frac{5}{2}\)
\(\displaystyle =\left( {\frac{{2x-5}}{2}} \right)\) (by L.C.M.)
now multiplies by 8
\(\displaystyle \begin{array}{l}=\left( {\frac{{2x-5}}{2}} \right)\times 8\\=4\times (2x-5)\\=8x-20\end{array}\)
According to the question –
\(\displaystyle 8x-20=3x\)
\(\displaystyle 8x-3x=20\) (by transposing)
5x = 20
\(\displaystyle x=\frac{{20}}{5}\)
x = 4
That number which Amina think is 4 answer.

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?
Sol.: Let’s second number is = x
and then first number is = 5x
adding 21 in both numbers –
then second number is = x + 21
and first number is = 5x + 21
According to the question – 
5x + 21 = 2 × (x + 21)
5x + 21 = 2x + 42
5x – 2x = 42 – 21 (by transposing)
3x = 21
\(\displaystyle x=\frac{{21}}{3}\)
x = 7
So, the second number = x = 7
and first number is = 5x
= 5 × 7
= 35

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?
Sol.: Let’s one’s of two-digit number is = x
and ten’s of two-digit number is = 9 – x
then two-digit number is = 10 × (9 – x) + x (bcoz place value of one’s multiply by 1 and place value of ten’s multiply by 10)
= 90 – 10x + x
so the two-digit number is = 90 – 9x
After interchange the digits –
Now one’s of two-digit number is = 9 – x
and ten’s of two-digit number is = x
then the two-digit number is = 10 × x + 1 × (9 – x)
= 10x + 9 – x
= 9x + 9
According to the question – 
90 – 9x + 27 = 9x + 9
– 9x – 9x = 9 – 27 – 90 (by transposing)
– 18x = – 117 + 9
– 18x = – 108 
\(\displaystyle x=\frac{{-108}}{{-18}}\)
x = 6
So the one’s of two digit numbers is x = 6
and ten’s of two digit number is 9 – x
= 9 – 6 = 3
Then two-digit number is 63 or 36 answer.

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?
Sol.: Let’s one’s of two-digit number is = x
and ten’s of two-digit number is = 3x
then two-digit number is = (10 × 3x) + x (bcoz place value of one’s multiply by 1 and place value of ten’s multiply by 10)
= 30x + x
= 31x
After interchange the digits –
Now one’s of two-digit number is = 3x
and ten’s of two-digit number is = x
then the two-digit number is = (10 × x) + 3x
= 10x + 3x
= 13x
According to the question – 
31x + 13x = 88
44x = 88
\(\displaystyle x=\frac{{88}}{{44}}\)
x = 2
So, one’s of two-digit number is = x = 2
and ten’s of two-digit number is = 3x
= 3 × 2
= 6
So the two-digit number is = 26 answer.

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?
Sol.: Let’s Shobo’s present age is = x
and Shobo’s mother age is = 6x
After 5 years –
Shobo age is = x + 5
According to the question –
\(\displaystyle x+5=\frac{1}{3}\times 6x\)
x + 5 = 2x
or 2x = x + 5
2x – x = 5 (by transposing)
x = 5 answer

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 rs 100 per metre it will cost the village panchayat rs 75000 to fence the plot. What are the dimensions of the plot?
Sol.: Let’s the length of rectangular plot is = 11x
and the breadth of the rectangular plot is = 4x
Bcoz perimeter of the rectangle is = 2 (Length + Breadth)
= 2 (11x + 4x)
= 2 × 15x
= 30x
So, cost of fence the plot = 100 × 30x
= 3000x rs
According to the question –
3000x = 75000
\(\displaystyle x=\frac{{75000}}{{3000}}\)
x = 25
So the length of rectangular plot is = 11x
= 11 × 25
= 275 metres answer
and the breadth of rectangular plot is = 4x
= 4 × 25 
= 100 metres answer

7. 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 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?
Sol.: Let’s total length of shirt cloth is = 3x metre
and total length of trouser cloth is = 2x metre
Total cost price of shirt cloth is = 3x × 50 = 150 rs
and total cost price of trouser cloth is = 2x × 90 = 180 rs
Suppose cost price of shirt cloth is = 100 rs
profit = 12% or 12 rs
so, selling price of shirt cloth is = 100 + 12 = 112 rs
By unitary rule –
Bcoz cost price of shirt cloth 100 rs then selling price = 112 rs
∴ cost price of shirt cloth 1 rs then selling price = \(\displaystyle \frac{{112}}{{100}}\)
∴ cost price of shirt cloth 150x rs then selling price = \(\displaystyle \frac{{112}}{{100}}\times 150x\) 
= 168x

Same as this,
Suppose cost price of trouser cloth is = 100 rs
profit = 10% or 10 rs
so, selling price of trouser cloth is = 100 + 10 = 110 rs
By unitary rule –
Bcoz cost price of trouser cloth 100 rs then selling price = 110 rs
∴ cost price of trouser cloth 1 rs then selling price = \(\displaystyle \frac{{110}}{{100}}\)
∴ cost price of trouser cloth 180x rs then selling price = \(\displaystyle \frac{{110}}{{100}}\times 180x\) 
= 198x
So, total selling price is = 36600 rs
168x + 198x = 36600
366x = 36600
\(\displaystyle \frac{{36600}}{{366}}\)
x = 100
So, total length of trouser cloth is = 2x
= 2 × 100
= 200 metres

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.
Sol.: Let’s total number of deer is = x
Half of the total grazing in the field = \(\displaystyle \frac{x}{2}\)
Three fourth of the remaining half is in playing = \(\displaystyle \frac{x}{2}\times \frac{1}{4}=\frac{x}{8}\)
Drinking water = 9
According to the question –
\(\displaystyle \begin{array}{l}\frac{x}{2}+\frac{3x}{8}+9=x\\\frac{{4x+3x+72}}{8}=x\end{array}\)
4x + 3x + 72 = 8x    (by cross multiplication)
7x – 8x = – 72     (by transposing)
– x = – 72
x = 72
So, the total number of deer is x = 72

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Sol.: Let’s granddaughter age is = x years
and grandfather age is = 10x years
According to the question –
10x – x = 54
9x = 54
\(\displaystyle x=\frac{{54}}{9}\)
x = 6
So granddaughter age x = 6 years
and grandfather age = 10x
= 10 × 6
= 60 years

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.
Sol.: Let’s Aman son’s age is = x years
and Aman’s age is = 3x years
Before 10 years –
Aman son’s age is = x – 10 years
and Aman’s age is = 3x – 10 years
According to the question –
5 (x – 10) = 3x – 10
5x – 50 = 3x – 10
5x – 3x = – 10 + 50 (by transposing)
– 2x = – 40
\(\displaystyle x=\frac{{-40}}{{-2}}\)
x = 20
So, Aman son’s age = x = 20 years
and Aman’s age = 3x
= 3 × 20
= 60 Years

Ncert Solutions for Class 8 Maths Chapter 2 EXERCISE 2.5
class 8th maths

Solve the following linear equations.

1. \(\displaystyle \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\) 1. \(\displaystyle \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\) 3. \(\displaystyle x+7-\frac{{8x}}{3}=\frac{{17}}{6}-\frac{{5x}}{2}\)
4. \(\displaystyle \frac{{x-5}}{3}=\frac{{x-3}}{5}\) 5. \(\displaystyle \frac{{3t-2}}{4}-\frac{{2t+3}}{3}=\frac{2}{3}-t\) 6. \(\displaystyle m-\frac{{m-1}}{2}=1-\frac{{m-2}}{3}\)

Solutions : 

1. \(\displaystyle \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Sol.: \(\displaystyle \frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
\(\displaystyle \frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5}\) (by transposing)
\(\displaystyle \begin{array}{l}\frac{{3x-2x}}{6}=\frac{{5+4}}{{20}}\\\frac{x}{6}=\frac{9}{{20}}\end{array}\)
20x = 9 × 6 (by cross multiplication)
20x = 54
\(\displaystyle \begin{array}{l}x=\frac{{54}}{{20}}\\x=\frac{{27}}{{10}}\end{array}\)

2. \(\displaystyle \frac{n}{2}-\frac{{3n}}{4}+\frac{{5n}}{6}=21\)
Sol.: \(\displaystyle \frac{n}{2}-\frac{{3n}}{4}+\frac{{5n}}{6}=21\)
\(\displaystyle \begin{array}{l}\frac{{6n-9n+10n}}{{12}}=21\\\frac{{16n-9n}}{{12}}=21\\\frac{{7n}}{{12}}=21\end{array}\)
\(\displaystyle 7n=12\times 21\) (by cross multiplication)
\(\displaystyle \begin{array}{l}7n=252\\n=\frac{{252}}{7}\\n=36\end{array}\)

3. \(\displaystyle x+7-\frac{{8x}}{3}=\frac{{17}}{6}-\frac{{5x}}{2}\)
Sol.: \(\displaystyle x+7-\frac{{8x}}{3}=\frac{{17}}{6}-\frac{{5x}}{2}\)
\(\displaystyle x-\frac{{8x}}{3}+\frac{{5x}}{2}=\frac{{17}}{6}-7\) (by transposing)
\(\displaystyle \frac{{6x-16x+15x}}{6}=\frac{{17-42}}{6}\) (by L.C.M.)
\(\displaystyle \frac{{21x-16x}}{6}=\frac{{-25}}{6}\)
5x = – 25 (denominator is same)
\(\displaystyle x=\frac{{-25}}{5}\)
x = – 5 answer

4. \(\displaystyle \frac{{x-5}}{3}=\frac{{x-3}}{5}\)Sol.: \(\displaystyle \frac{{x-5}}{3}=\frac{{x-3}}{5}\)
\(\displaystyle 5(x-5)=3(x-3)\) (by cross multiplication)
\(\displaystyle 5x-3x=-9+25\) (by transposing)
2x = 16
\(\displaystyle x=\frac{{16}}{2}\)
x = 8 answer

5. \(\displaystyle \frac{{3t-2}}{4}-\frac{{2t+3}}{3}=\frac{2}{3}-t\)
Sol.: \(\displaystyle \frac{{3t-2}}{4}-\frac{{2t+3}}{3}=\frac{2}{3}-t\)
\(\displaystyle \frac{{3\times \left( {3t-2} \right)-4\times (2t+3)}}{{12}}=\frac{{2-3t}}{3}\)
\(\displaystyle \frac{{9t-6-8t-12}}{{12}}=\frac{{2-3t}}{3}\)     
\(\displaystyle \frac{{t-18}}{{12}}=\frac{{2-3t}}{3}\)
3 × (t – 18) = 12 × (2 – 3t) (by cross multiplication)
3t – 54 = 24 – 36t
3t + 36t = 24 + 54 (by transposing)
39t = 78
\(\displaystyle t=\frac{{78}}{{39}}\)
t = 2 answer

6. \(\displaystyle m-\frac{{m-1}}{2}=1-\frac{{m-2}}{3}\)
Sol.: \(\displaystyle m-\frac{{m-1}}{2}=1-\frac{{m-2}}{3}\)
\(\displaystyle \begin{array}{l}\frac{{2m-(m-1)}}{2}=\frac{{3-(m-2)}}{3}\\\frac{{2m-m+1}}{2}=\frac{{3-m+2}}{3}\end{array}\)
3 (2m – m + 1) = 2 (3 – m + 2) (by cross multiplication)
6m – 3m + 3 = 6 – 2m + 4
6m – 3m + 2m = 4 + 6 – 3
5m = 7
\(\displaystyle m=\frac{7}{5}\)

Simplify and solve the following linear equations.

7. 3 (t -3) = 5 (2t + 1)          8. 15 (y -4) – 2 (y – 9) + 5 (y + 6) = 0
9. 3 (5z – 7) – 2 (9z – 11) = 4 (8z – 13) – 17
10. 0.25 (4f – 3) = 0.05 (10f – 9)

Solution :

7. 3 (t -3) = 5 (2t + 1)
Sol.: 3 (t -3) = 5 (2t + 1)
3t – 9 = 10t + 5
3t – 10t = 5 + 9
– 7t = 14
\(\displaystyle t=-\frac{{14}}{7}\)
t = – 2 answer

8. 15 (y -4) – 2 (y – 9) + 5 (y + 6) = 0
15 (y -4) – 2 (y – 9) + 5 (y + 6) = 0
15y – 60 – 2y + 18 + 5y + 30 = 0
15y – 2y + 5y = 60 – 18 – 30 (by transposing)
20y – 2y = 60 – 48
18y = 12
\(\displaystyle \begin{array}{l}y=\frac{{12}}{{18}}\\y=\frac{2}{3}\end{array}\)

9. 3 (5z – 7) – 2 (9z – 11) = 4 (8z – 13) – 17
3 (5z – 7) – 2 (9z – 11) = 4 (8z – 13) – 17
15z – 21 – 18z + 22 = 32z – 52 – 17
15z – 18z – 32z = 21 – 22 – 17 – 52 (by transposing)
15z – 50z = – 91 + 21
-35z = – 70
\(\displaystyle z=\frac{{-70}}{{-35}}\)
z = 2

10. 0.25 (4𝑓 – 3) = 0.05 (10𝑓 – 9)
1.00𝑓 – 0.75 = 0.5𝑓 – 0.45
1.00𝑓 – 0.5𝑓 = 0.75 – 0.45
0.05𝑓 = 0.30
\(\displaystyle z=\frac{{0.30}}{{0.05}}\)
𝑓 = 6

Ncert Solutions for Class 8 Maths Chapter 2 EXERCISE 2.6
class 8th maths

Solve the following equations.

1. \(\displaystyle \frac{{8x-3}}{{3x}}=2\) 2. \(\displaystyle \frac{{9x}}{{7-6x}}=15\) 3. \(\displaystyle \frac{z}{{z+15}}=\frac{4}{9}\)
4. \(\displaystyle \frac{{3y+4}}{{2-6y}}=\frac{{-2}}{5}\)          5. \(\displaystyle \frac{{7y+4}}{{y+2}}=\frac{{-4}}{3}\)  

Solution :

1. \(\displaystyle \frac{{8x-3}}{{3x}}=2\)
Sol.: \(\displaystyle \frac{{8x-3}}{{3x}}=2\)\(\displaystyle \frac{{8x-3}}{{3x}}=\frac{2}{1}\)8𝑥 – 3 = 2 × 3𝑥 (by cross multiplication)
8𝑥 – 3 = 6𝑥
8𝑥 – 6𝑥 = 3 (by transposing)
2𝑥 = 3
\(\displaystyle x=\frac{3}{2}\) 

2. \(\displaystyle \frac{{9x}}{{7-6x}}=15\)
Sol.: \(\displaystyle \frac{{9x}}{{7-6x}}=15\)\(\displaystyle \frac{{9x}}{{7-6x}}=\frac{{15}}{1}\)9𝑥 = 15 (7 – 6𝑥)
9𝑥 = 105 – 90𝑥 (by cross multiplication)
9𝑥 + 90𝑥 = 105 (by transposing)
99𝑥 = 105
\(\displaystyle \begin{array}{l}x=\frac{{-105}}{{99}}\\x=\frac{{-35}}{{33}}\end{array}\)

3. \(\displaystyle \frac{z}{{z+15}}=\frac{4}{9}\)
Sol.: \(\displaystyle \frac{z}{{z+15}}=\frac{4}{9}\)
9 × 𝑧 = 4 (𝑧 + 15) (by cross multiplication)
9𝑧 = 4𝑦 + 60
9𝑧 – 4𝑧 = 60 (by transposing)
5𝑧 = 60
\(\displaystyle z=\frac{{60}}{5}\)
𝑧 = 12

4. \(\displaystyle \frac{{3y+4}}{{2-6y}}=\frac{{-2}}{5}\)         
Sol.: \(\displaystyle \frac{{3y+4}}{{2-6y}}=\frac{{-2}}{5}\)
5 (3𝑦 + 4) = – 2 (2 – 6𝑦) (by cross multiplication)
15𝑦 + 20 = – 4 + 12𝑦
15𝑦 – 12𝑦 = – 4 – 20 (by transposing)
3𝑦 = – 24
\(\displaystyle y=\frac{{-24}}{3}\)
𝑦 = – 8

5. \(\displaystyle \frac{{7y+4}}{{y+2}}=\frac{{-4}}{3}\)
Sol.: \(\displaystyle \frac{{7y+4}}{{y+2}}=\frac{{-4}}{3}\)
3 (7𝑦 + 4) = – 4 (𝑦 + 2) (by cross multiplication)
21𝑦 + 12 = – 4𝑦 – 8
21𝑦 + 4𝑦 = – 8 – 12 (by transposing)
25𝑦 = – 20
\(\displaystyle \begin{array}{l}y=\frac{{-20}}{{25}}\\y=\frac{{-4}}{5}\end{array}\)

6. 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’s Hari present age = 5𝑥
and Harry present age = 7𝑥
After 4 year
Hari age = 5𝑥 + 4
Harry age = 7𝑥 + 4
According to the question –
\(\displaystyle \frac{{5x+4}}{{7x+4}}=\frac{3}{4}\)
3 (7𝑥 + 4) = 4 (5𝑥 + 4) (by cross multiplication)
21𝑥 + 12 = 20𝑥 + 16
21𝑥 – 20𝑥 = 16 – 12 (by transposing)
𝑥 = 4
So, Hari present age = 5𝑥 = 5 × 4 = 20 years
and Harry present age = 7𝑥 = 7 × 4 = 28 years

7. 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 \(\displaystyle \frac{3}{2}\). Find the rational number.
Sol.: Let’s numerator of a rational number = 𝑥
and denominator of rational number = 𝑥 + 8
according to the question –
\(\displaystyle \frac{{x+17}}{{x+8-1}}=\frac{3}{2}\)
3 (𝑥 + 8 – 1) = 2 (𝑥 + 17) (by cross multiplication)
3𝑥 + 24 – 3 = 2𝑥 + 34
3𝑥 – 2𝑥 = 34 + 3 – 24 (by transposing)
𝑥 = 13
So, numerator of a rational number 𝑥 = 13
and denominator of rational number = 𝑥 + 8 = 13 + 8 = 21
So, that rational number is  \(\displaystyle \frac{{13}}{{21}}\) answer.

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