# Ncert Solutions For Class 8 Maths Ex 2.1

## Ncert Solutions For Class 8 Maths Chapter 2 Ex 2.1

Question-1

Solve the linear equation x215=x3+14$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

x215=x3+14$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

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

Multiplying both sides by 60, we obtain

60(x215)=60(x3+14)$60 *(\frac{x}{2}-\frac{1}{5})= 60 * (\frac{x}{3}+\frac{1}{4})$

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

30x– 20x = 15+12

10= 27

= 2710$\frac{27}{10}$

Question-2

Solve the linear equation

n23n4+5n6=21$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$

n23n4+5n6=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=2527$n=\frac{252}{7}$

n = 36

Question-3

Solve the linear equation

x+78x3=1765x2$x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

x+78x3=1765x2$x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

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=255$x=\frac{-25}{5}$

x= -5

Question-4

Solve the linear equation x53=x35$\frac{x-5}{3}=\frac{x-3}{5}$

x53=x35$\frac{x-5}{3}=\frac{x-3}{5}$

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=162$x=\frac{16}{2}$

x=8

Question- 5

Solve the linear equation

3t242t+33=23t$\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

3t242t+33=23t$\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

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 mm12=1m23$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

mm12=1m23$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

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=75$m=\frac{7}{5}$

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=1218=23$y=\frac{12}{18}=\frac{2}{3}$

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)

14(4f3)=120(10f9)$\frac{1}{4}\left ( 4f-3 \right )= \frac{1}{20}\left ( 10f-9 \right )$

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=35=0.6$f = \frac{3}{5}=0.6$

Question-11

Solve: 8x33x=2$\frac{8x-3}{3x}=2$

8x33x=2$\frac{8x-3}{3x}=2$

On multiplying both sides by 3x, we obtain

8x-3 =6x

8x-6x = 3

2x = 3

x=32$x=\frac{3}{2}$

Question – 12

Solve: 9x76x=15$\frac{9x}{7-6x}=15$

9x76x=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=10599=3533$x=\frac{105}{99}=\frac{35}{33}$

Question-13

Solve: zz+15=49$\frac{z}{z+15}=\frac{4}{9}$

zz+15=49$\frac{z}{z+15}=\frac{4}{9}$

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: 3y+426y=25$\frac{3y+4}{2-6y}=\frac{-2}{5}$

3y+426y=25$\frac{3y+4}{2-6y}=\frac{-2}{5}$

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: 7y+4y+2=43$\frac{7y+4}{y+2}=\frac{-4}{3}$

7y+4y+2=43$\frac{7y+4}{y+2}=\frac{-4}{3}$

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=45$y=-\frac{4}{5}$

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,

5x+47x+4=34$\frac{5x+4}{7x+4}=\frac{3}{4}$

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 xx+8$\frac{x}{x+8}$. According to the question,

x+17x+81=32$\frac{x+17}{x+8-1}=\frac{3}{2}$ x+17x+7=32$\frac{x+17}{x+7}=\frac{3}{2}$

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 = 1321$\frac{13}{21}$