# NCERT Solutions For Class 8 Maths Chapter 2

## NCERT Solutions Class 8 Maths Linear Equations in One Variable

### Ncert Solutions For Class 8 Maths Chapter 2 PDF Free Download

NCERT Solutions for Class 8 Maths Chapter 2 is provided here so that students can check for the solutions whenever they are facing difficulty while solving the NCERT questions from Linear Equations in one variable chapter. Class 8 is an important phase of a student’s life, in class 8 many new concepts are added.  Maths is a subject which requires students understanding and reasoning skills. Along with this, it also requires students to practice maths on a regular basis. Students of Class 8 are suggested to solve NCERT questions in order to practice the questions that are usually asked in the examination. From all the chapters, Linear Equations in one variable is an important topic for the students, and one needs to practice thoroughly to score well in the examination. The solutions for Linear Equations in One Variable are provided in a detailed manner, where one can find a step-by-step solution to all the questions of this chapter.

This chapter includes the following topics- Introduction. Solving Equations which have Linear Expressions on one Side and Numbers on the other Side, equations having the variable on both Sides, reducing equations to a simpler form, and equations reducible to the linear form. Here, NCERT class 8 maths solutions for Linear Equations in One Variable PDF is also provided for better understanding and clarification of the chapter.

Example:

Consider an algebraic equation 2x – 7 = 3

x is the variable

= is the equality symbol

2x – 3 = 7 is the equation

2x – 3 is the LHS (Left hand side)

7 is the RHS(Right hand side)

Therefore here the LHS = RHS

### NCERT Solutions Class 8 Maths Chapter 2 Exercises

Exercise 2.1

Question-1

Solve the linear equation $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

L.C.M. of the denominators, 2,3,4 and 5 is 60.

Multiplying both sides by 60, we obtain

$60 *(\frac{x}{2}-\frac{1}{5})= 60 * (\frac{x}{3}+\frac{1}{4})$

30x – 12= 20+ 15(opening the brackets)

30x– 20x = 15+12

10= 27

= $\frac{27}{10}$

Question-2

Solve the linear equation

$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$

$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$

L.C.M. of the denominators, 2, 4, and 6 is 12

Multiplying both sides by 12, we obtain

6n-9n+10n= 252

7= 252

$n=\frac{252}{7}$

n = 36

Question-3

Solve the linear equation

$x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

$x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

LCM of the denominators 2, 3, and 6 is 6.

Multiplying both sides by 6, we obtain

6x+42-16x = 17-15x

6x-16x+15x = 17-42

5= -25

$x=\frac{-25}{5}$

x= -5

Question-4

Solve the linear equation $\frac{x-5}{3}=\frac{x-3}{5}$

$\frac{x-5}{3}=\frac{x-3}{5}$

LCM of the denominators , 3 and 5 is 15.

Multiplying both the sides by 15, we obtain

5(x-5) = 3(x-3)

5x-25= 3x-9 (opening the brackets)

5x-3x = 25-9

2x = 16

$x=\frac{16}{2}$

x=8

Question- 5

Solve the linear equation

$\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

$\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

LCM of the denominators 3 and 4 is 12.

Multiplying both the sides by 12, we obtain

3(3t-2)-4(2t+3) =8-12t

9t-6-8t-12= 8-12t (opening the brackets)

9t-8t+12t= 8+6+12

12t= 26

t=2

Question-6

Solve the linear equation $m-\frac{m-1}{2}=1-\frac{m-2}{3}$

$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

LCM of the denominators, 2 and 3, is 6.

Multiplying both the sides by 6, we obtain

6m-3m (m-1) = 6-2(m-2)

6m-3m+3 = 6-2m+4 (opening the brackets)

6m-3m+2m = 6+4-3

5m = 7

$m=\frac{7}{5}$

Question-7

Simplify and solve the linear equation

3(t-3) = 5(2t+1)

3(t-3) = 5(2t+1)

3t-9 = 10t +5 (opening the brackets)

-9-5 = 10t-3t

-14= 7t

= -2

Question-8

Simplify and solve the linear equation

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 (opening the brackets)

18y-12 = 0

18y =12

$y=\frac{12}{18}=\frac{2}{3}$

Question-9

Simplify and solve the linear equation

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 (opening the bracket)

-3z+1 = 32z-69

-3z-32z =-69-1

-35z = -70

z = 2

Question-10

Simplify and solve the linear equation

0.25(4f-3) = 0.05(10f-9)

0.25(4f-3) = 0.05(10f-9)

$\frac{1}{4}\left ( 4f-3 \right )= \frac{1}{20}\left ( 10f-9 \right )$

Multiplying both the sides by 20, we obtain

5(4f-3) = 10f-9

20f-15 = 10f-9 (opening the brackets)

20f-10f = -9+15

10f = 6

$f = \frac{3}{5}=0.6$

Question-11

Solve: $\frac{8x-3}{3x}=2$

$\frac{8x-3}{3x}=2$

On multiplying both sides by 3x, we obtain

8x-3 =6x

8x-6x = 3

2x = 3

$x=\frac{3}{2}$

Question – 12

Solve: $\frac{9x}{7-6x}=15$

$\frac{9x}{7-6x}=15$

On multiplying both the sides by 7-6x, we obtain

9x = 15(7-6x)

9x=105-90x

9x+90x = 105

99x = 105

$x=\frac{105}{99}=\frac{35}{33}$

Question-13

Solve: $\frac{z}{z+15}=\frac{4}{9}$

$\frac{z}{z+15}=\frac{4}{9}$

On multiplying both the sides by 9(z+15), we obtain

9z=4(z+15)

9z=4z+60

9z-4z=60

5z=60

z=12

Question-14

Solve: $\frac{3y+4}{2-6y}=\frac{-2}{5}$

$\frac{3y+4}{2-6y}=\frac{-2}{5}$

On multiplying both the sides by 5(2-6y), we obtain

5(3y+4) = -2(2-6y)

15y + 20= -4+12y

15y-12y = -4-20

3y= -24

y =-8

Question-15

Solve: $\frac{7y+4}{y+2}=\frac{-4}{3}$

$\frac{7y+4}{y+2}=\frac{-4}{3}$

On multiplying both the sides by 3(y+2), we obtain

3(7y+4) = -4(y+2)

21y + 12= -4y-8

21y + 4y = -8-12

25y = -20

$y=-\frac{4}{5}$

Question-16

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.

Let the common ratio between their ages be x. Therefore, Hari’s ages and Harry’s ages will be 5x years and 7x years respectively and four years later, their ages will be (5x+4) years and (7x+4) years respectively.

According to the situation given in the question,

$\frac{5x+4}{7x+4}=\frac{3}{4}$

4(5x+4)=3(7x+4)

20x+16= 21x+12

16-12=21x-20x

x=4

Hari’s age = 5x years = (5×4) years = 20 years

Harry’s age = 7x years = (7×4) years = 28 years

Therefore, Hari’s age and Harry’s ages are 20years and 28 years respectively.

Question-17

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 3/2. Find the rational number.

Let the numerator of the rational number be x. Therefore, its denominator will be x+8.

The rational number will be $\frac{x}{x+8}$. According to the question,

$\frac{x+17}{x+8-1}=\frac{3}{2}$

$\frac{x+17}{x+7}=\frac{3}{2}$

2(x+17) = 3(x+7)

2x+34 = 3x+21

34-21 = 3x-2x

x=13

Numerator of the rational number = x=13

Denominator of the rational number = x+8=13+8=21

Rational number = $\frac{13}{21}$

Exercise 2.2

Question -1

Amina thinks of a number and substracts 5/2 from it. She multiplies the result by 8. The result now obtained in 3 times the same number she thought of. What is the number?

Let the number be x.

According to the given question,

$8\begin{pmatrix} x-\frac{5}{2} \end{pmatrix}=3x$

8x-20=3x

Transposing 3x to L.H.S and 20 to R.H.S, we obtained

8x-3x=20

5x=20

Dividing both sides by 5, we obtain

X=4

Hence the number is 4.

Question-2

A positive number is 5 times another number. If 21 is added to both the numbers, the one of the new numbers become twice the other new number. What are the numbers?

Let the number be x and 5x. according to the question,

21+5x=2(x+21)

21+5x=2x+42

Transposing 2x to L.H.S and 21 to R.H.S, we obtain

5x-2x=42-21

3x=21

Dividing both the sides by 3, we obtain

X=7

5x= 5×7= 35

Hence, the number are 7 and 35 respectively.

Question-3

Sum of the digits of a two digit number is 9. When we interchange the digit it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Let the digits are tens place and ones place be x and 9-x respectively.

Therefore, original number= 10x+(9-x)=9x+9

On interchanging  the digits, the digits at the ones place and tens place will be x and 9-x respectively.

Therefore new number afterinterchanging the digits = 10(9-x)+x

= 90-10x+x

= 90-9x

According to the given question ,

New number = original number + 27

90-9x=9x+9+27

90-9x=9x+36

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

90-36 = 18x

54 = 18x

Dividing both sides by 18, we obtain

3 = x and 9-x=6

Hence the digits at the tens and ones place of the number is 3 and 6 respectively.

Therefore the two digits number is 9x+9 = 9 x 3+9 =36

Question-4

One of the two digits of the two number is three times the other digit. If you interchange the digits of this two-digits number and add the resulting number to the original number, you  get 88. What is the original number?

Let the digits at tens place and once place be x and 3x respectively.

Therfore the original number = 10x+3x = 13x

On interchanging the digits , the digits at ones place and tens placewill be x and 3x respectively.

Number after interchanging  = 10 X 3x + x=31x

According to the given question,

Original number + new number = 88

13+ 31= 88

44= 88

Dividing both sides by 44, we obtain

= 2

therefore the original number = 13x = 13 X 2= 26

By considering the tens place and ones place as 3and x respectively, the two-digit number obtained is 62.

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 this mother’s present age. What are their present ages?

Let shobo’s age be years. Therefore, his mother’s age will be 6years.

According to the given question,

$After\: 5\: years,\: Shobo’s\: age\: =\: \frac{shobo’s\: mother’s\: present\: age }{3}$

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

+ 5 = 2x

Transposing to R.H.S, we obtain

5= 2x-x

5=x

6x= 6 X 5= 30

Therefore, the present age of Shobo and Shobo’s mother will be 5 years and 30 years respectively.

Question-6

A narrow rectangular plot is reserved for a school in village. The length and breadth of a plot are in the ratio 11:4. At the rate of Rs 100 per meter it will cost the village panchayat Rs 75,000 to fence the plot. What are the dimensions of the plot?

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 11m and 4xm respectively.

Perimeter of the plot = 2(length + breadth) = [2(11x + 4x)]m = 30m

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

100 x 30x = 75,000

3000x = 75,000

Dividing both sides by3000, we obtain

X =25

Length = 11x m = (11×25)m = 275m

Breadth = 4x m = (4×25)m = 100m

Hence, the dimension of the plot are 275m 100m respectively.

Question-7

For school uniform Hasan buys two kinds of cloth material, the shirt material cost him Rs50 per metre and trouser material that cost him Rs 90 per metre. For every 2 metre of the trouser material he buys 3 metres of the shirt material. He sells the material at 12% and 10% profit respectively. His total sales are Rs 36,660. How much trouser material did he buy?

Let 2m of trouser material and 3x m of the shirt material be bought by him.

Per metre selling price of the trouser material = $Rs 90+\frac{90\times 12}{100}= Rs 100.80$

Per metre selling price of trouser of shirt material = $Rs 50+\frac{50\times 10}{100} = Rs 55$

Given that, total amount of selling= Rs 36660

100.80 x (2x) + 55 x (3x) = Rs36660

201.60x + 165x = 36660

366.60x = 36660

Dividing both the sides by 366.60, we obtain

= 100

Trouser material = 2xm = (2×100)m = 200m

Question-8

Half of the herd of deer are gazing in the field and three fourth of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Let the number of deer be x.

Number of deer grazing in the field = $\frac{x}{2}$

Number of deer playing nearby = $\frac{3}{4}\times number\: of \: remaining \: deer$

= $\frac{3}{4}\times \begin{pmatrix} x-\frac{x}{2} \end{pmatrix}=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}$

Number of deer drinking water from the pond = 9

= $x- \frac{x}{2}+\frac{3x}{8} =9$

= $x- \frac{4x+3x}{8} =9$

= $x-\frac{7x}{8}=9$

= $\frac{x}{8}= 9$

Multiplying both the sides by 8, we obtain

X= 72

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

Question-9

A grandfather is ten times older than his granddaughter. He is also 54 year older than her.  Find their present age.

Let the daughter’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 L.H.S, we obtain

= 10x= 54

9= 54

X=6

Granddaughter’s age = x years = 6 years

Grandfather’s age = 10years = (10×6) years = 60 years

Question-10

Arman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages

Let Arman’s son’s age be x years. Therefore, Arman’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, Arman’s age = 5 x Arman’s son’s age 10 years ago

3– 10 = 5(x-10)

3x-10 = 5x-50

Transposing 3x to R.H.S and 50 to L.H.S, we obtain

50 – 10 = 5x-3x

40= 2x

Dividing both sides by 2, we obtain

20= x

Arman’s son’s age = years = 20 years

Arman’s age = 3x years= (3×20)years = 6  years.

Exercise 2.3

Question 1:

Solve and check. 3x = 2x + 18.

Solution:

3X = 2X + 18

Transpose 2x to L.H.S,

3x – 2x = 18   →   [ x = 18 ]

L.H.S :  3X   → (3 x 18) = 54

R.H.S :  2X + 18  → (2 X 18) + 18

→ 36 + 18 = 54

L.H.S = R.H.S

Hence proved.

Question 2:

Solve and check. 5x + 9 = 5 + 3x

Solution:

5x – 3x = 5 – 9

2x =  -4

Dividing 2 by both sides,

x  =  -2

L.H.S :  5X + 9 →  5 X  (-2) + 9

→  -10 + 9  =    -1

R.H.S :  5 + 3X  →  5 +  3 X (-2)

→ 5 – 6  = -1

L.H.S   =   R.H.S

Hence proved.

Question 3:

Solve and check. 4z + 3 = 6 + 2z.

Solution:

4z + 3 = 6 + 2z

Transpose 2z to left hand side   &  3 to right hand side.

4z – 2z   =   6 – 3

2z  =  3  →    z  = $\left ( 3\div 2 \right )$

L.H.S :    4z + 3    =    4 x $\left ( 3\div 2 \right )$+ 3

→  $\left ( 12\div 2 \right )$  + 3   →  6 + 3  =    9

R.H.S :     6 + 2z  = 6 + 2  x $\left ( 3\div 2 \right )$

→ 6 + $\left ( 6\div 2 \right )$

→ 6 + 3 =   9

L.H.S    =    R.H.S

Hence proved.

Question 4:

Solve and check. x = $\left ( 4\div 5 \right )$ (x  + 10)

Solution:

X =  $\left ( 4\div 5 \right )$ ( X + 10)

Multiply 5 on both sides

5x = 4(x + 10) → 5x = 4x + 40

Transpose 4x to L.H.S,

5X – 4X = 40.  →  X  =  40

L.H.S :        X   =  40

R.H.S :   $\left ( 4\div 5 \right )$ ( X + 10)    →   $\left ( 4\div 5 \right )$ ( 40 + 10)

→ $\left ( 4\div 5 \right )$ X  50 = 40

L.H.S    =   R.H.S

Hence proved.

Question 5:

Solve and check.  $2x / 3 + 1 = 7x /15 + 3$

Solution:

$2x / 3 + 1 = 7x /15 + 3$

Transpose 7x / 15   to  L.H.S

Transpose    1  to  R.H.S

2x / 3  –  7x / 15   =   3 – 1

5  X 2x – 7x / 15  = 2

3x / 15 = 2 →      x /5  = 2

→    x /5  = 2  multiply 5 on both sides  .

X = 10

L.H.S  = 2x / 3  + 1 =  2 x 10 / 3   + 1

=$2 \times 10 + 1 \times 3 \div 3$

= $23\div 3$

R.H.S  = $= 7x\div 15 + 3 = 7 \times 10 \div 15 + 3$

= $7 \times 2 \div 3 + 3 = 14\div 3 + 3$

=$14 + 3 \times 3 \div 3$

=$23 \div 3$

L.H.S  =  R.H.S

Hence proved.

Question 6:

Solve and check.  2y + 5/3 = 26 / 3 – y.

Solution:

2Y  + 5/3 = 26/3 – Y

Transpose y to L.H.S.

Transpose 5/3 to R.H.S.

2y + y  = 26/3  –  5/3

3y = 21/3    (i.e)  3y = 7

Divide 3 by both sides,

→    y = 7/3

L.H.S :

2Y = 5/3        =  2 X 7/3 + 5/3     = 14/3 + 5/3     =  19/3

R.H.S :

26/3 – y          =  26/3 – 7/3              =  19/3

L.H.S = R.H.S

Hence proved.

Question 7:

Solve and check. 3m = 5m – $\left ( 8\div 5 \right )$

Solution:

3m -5m = – $\left ( 8\div 5 \right )$

-2m =  – $\left ( 8\div 5 \right )$

Divide -2 on both sides,

m = $\left ( 4\div 5 \right )$

L.H.S :

3m = 3 x $\left ( 4\div 5 \right )$

=  $\left ( 12\div 5 \right )$

R.H.S :

= 5M – $\left ( 8\div 5 \right )$

= 5 X$\left ( 4\div 5 \right )$  –– $\left ( 8\div 5 \right )$

= $\left ( 12\div 5 \right )$

L.H.S = R.H.S

Hence proved.

Question 8:

Solve and check. 5g – 3 = 3g – 5

Solution:

Transpose 3t to L.H.S.

Transpose -3 to R.H.S.

5g – 3g = -5 – (-3)

2g = -2

Divide 2 on both sides,

g = -1

L.H.S :

5g – 3      =   5 x (-1) -3                 = -5-3       = -8

R.H.S :

3g – 5      = 3 x (-1) -5                    = -3 -5    = -8

L.H.S  = R.H.S

Hence proved

Exercise 2.4

Question 1:

solve x-3=6

x-3=6

Transporting 3 to the right hand side(R.H.S)  We get

X=6+3=9

Question 2:

solve : y+4=9

Transporting 4 to R.H.S we obtain

Y=9-4=5

Question 3:

solve : 7=z+3

Bring 3 to L.H.S ,we obtain

7-3=z

Z=4

Question 4:

Solve :  $\frac{3}{8}+x=\frac{19}{6}$

$\frac{3}{8}+x=\frac{19}{6}$

Bring $\frac{3}{8}$  to the R.H.S ,we obtain

X= $\frac{19}{6}-\frac{3}{8}$

X= $\frac{23}{8}$

Question 5:

Solve 3x=6

3x=6

Dividing both side by 3,we get

$\frac{3x}{3}=\frac{6}{3}$

x=2

Question 6:

Solve : $\frac{t}{2}=5$

$\frac{t}{2}=5$

Multiplying both side by 2 ,we get

$\frac{t}{2}*2=5*2$

T=10

Question 7:

Solve :

$\frac{2x}{5}=16$

Multiply both sides by $\frac{5}{2}$ ,we obtain

$\frac{5}{2}*\frac{2x}{5}=16*\frac{5}{2}$

X=40

Question 8:

Solve : 1.4= $\frac{y}{1.2}$

Solutions:

$\frac{y}{1.2}$

Multiply both side by 1.2 ,we obtain

1.4*1.2= $\frac{y}{1.2}*1.2$

Y=1.68

Question 9:

Solve: 5x-3=15

Bring 3 to R.H.S , we obtain

5x=15+3

5x=18

Dividing both sides by 5 ,we get

$\frac{5x}{5}=\frac{12}{5}$

X=  $\frac{12}{5}$

Question 10:

solve 12y-6=7

12y-6=7

Bring 6 to RHS ,We get

12y=7+6

12y=13

Divide both side by 12

$\frac{12y}{12}=\frac{13}{12}$

Y= $\frac{13}{12}$

Question 11:

Solve 13+5x=9

Tranposing 13 to RHS We get

5x=9-13

5x= -4

Divide 5 by both side we obtain

$\frac{5x}{5}=\frac{-4}{5}$

X =$\frac{-4}{5}$

Question 12:

Solve: $\frac{x}{4}+1=\frac{6}{15}$

$\frac{x}{4}+1=\frac{6}{15}$

Changing 1 to the RHS , we get

$\frac{x}{4}=\frac{6}{15}-1$

$\frac{x}{4}=\frac{6-15}{15}$

$\frac{x}{4}=\frac{-9}{15}$

Multiply either side by 4 we get

$\frac{x*4}{4}=\frac{-9*4}{15}$

X=$\frac{12}{5}$

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