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 Excercise: 2.1

Q1 Solve the following equations x – 2 = 7

Answer:

1. Transposing -2 to the RHS, we get

x = 7 + 2 = 9

Q2 Solve the following equation y + 3 = 10

Answer:

Transposing 3 to the RHS, we get

y = 10 - 3 = 7

y = 7

Q3 Solve the following equation 6 = z + 2

Answer:

Transposing 2 to the LHS, we get

6 - 2 = z => z= 4

Thus z = 4

Q4 Solve the following equations

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

Answer:

4. Transposing $\frac{3}{7}$ to the RHS, we get

$x = \frac{17}{7}-\frac{3}{7} = \frac{14}{7} = 2$

Thus x =2

Q5 Solve the following equation 6x = 12

Answer:

Dividing both sides by 6, we get

$x = \frac{12}{6} => 2$

Thus x = 2

Q6 Solve the following equation.

$\frac{t}{5} = 10$

Answer:

Multiplying both sides by 5, we get

$t = 10\times 5 = 50$

Thus t = 50

Q 7 Solve the following equations.

$\frac{2x}{3} = 18$

Answer:

$\frac{2x}{3} = 18$

Multiplying both sides by 3, we get

$2x = 18\times 3 = 54$

Now, dividing both sides by 2, we get

$x = \frac{54}{2} = 27$

Q8 Solve the following equation.

$1.6 = \frac{y}{1.5}$

Answer:

Multiplying both sides by 1.5, we get

$1.6\times 1.5 = y = 2.4$

Thus y = 2.4

Q9 Solve the following equation 7x – 9 = 16

Answer:

Transposing 9 to the RHS, we get

7x = 16 + 9 = 25

Now, dividing both sides by :

$x = 25\times \frac{1}{7} = \frac{25}{7}$

Q10 Solve the following equation 14y – 8 = 13

Answer:

Transposing -8 to the RHS, we get

14y = 13 + 8 = 21

Now dividing both sides by 14:

$y = \frac{21}{14} = \frac{3}{2}$

Q11 Solve the following equation 17 + 6p = 9

Answer:

At first, transposing 17 to the RHS:

6p = 9 - 17 = -8

Now, dividing both sides by 6, we get

$p = \frac{-8}{6} = \frac{-4}{3}$

Q12 Solve the following equations.

$\frac{x}{3} +1 = \frac{7}{15}$

Answer:

12.

$\frac{x}{3} + 1 = \frac{7}{15}$

Transposing 1 to the RHS, we get

$\frac{x}{3} = \frac{7}{15} -1 = \frac{7-15}{15} = \frac{-8}{15}$

Now multiplying both sides by 3, it becomes

$x = \frac{-8}{15}\times 3 = \frac{-8}{5}$

Solve the following equations. Q.12
Edit Q

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.2

Q1 If you subtract $\frac{1}{2}$ from a number and multiply the result by $\frac{1}{2}$ , you get $\frac{1}{8}$ . What is the number?

Answer:

Assume the number to be x.

Thus according to the question,

$\left ( x-\frac{1}{2} \right )\times \frac{1}{2} = \frac{1}{8}$

Multiplying both sides by 2, we get

$\left ( x-\frac{1}{2} \right ) = \frac{1}{8}\times 2 = \frac{1}{4}$

Now transposing $-\frac{1}{2}$ to the RHS, we get

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

Thus the number was x = 3/4

Q 2 The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth of the pool?

Answer:

Let the breadth of the pool be x m.

According to the question, the length of the pool = 2x + 2 m.

Perimeter of recatangle = 2(l + b) = 154 m .

i.e., 2(2x + 2 + x) = 154

2(3x + 2) = 154

Dividing both sides by 2, we get

3x + 2 = 77

Now transposing 2 to the RHS, we get

3x = 77 - 2 = 75

Dividing both sides by 3, we get

$x = \frac{75}{3} = 25$

Thus breadth of pool = 25 m

and lenght of the pool = 2 $\times$ 25 + 2 = 52 m

Q3 The base of an isosceles triangle is $\frac{4}{3}$ cm. The perimeter of the triangle is $4\frac{2}{15} cm$ . What is the length of either of the remaining equal sides?

Answer:

In isosceles triangles, we have 2 sides of equal length.

Given that its perimeter is $\frac{62}{15}$ cm.

Let's assume the length of the equal side is x cm.

Also,

$Perimeter = x + x + \frac{4}{3} = \frac{62}{15}$

Transposing $\frac{4}{3}$ to the RHS side it becomes,

$2x = \frac{62}{15}-\frac{4}{3} = \frac{62-20}{15} = \frac{42}{15} = \frac{14}{5}$

Now dividing both sides by 2, we get

$x = \frac{14}{5}\times \frac{1}{2} = \frac{7}{5}$

Hence, the length of the equal sides of the isosceles triangle is $\frac{7}{5}$ cm.

Q4 Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Answer:

Assume one number to be x. Then the other number will be x + 15.

Now it is given that sum of the two numbers is 95.

Thus quation becomes x + x + 15 = 95

or 2x +15 = 95

Transposing 15 to the RHS, it becomes

2x = 95 - 15 = 80

Dividng both sides by 2, we get

$x = \frac{80}{2} = 40$

Thus two numbers are 40 and 55

Q5 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Answer:

Given that numbers are in the ratio of 5:3, so we can assume numbers to be 5x and 3x.

Also, the difference between these numbers is 18, so the equation becomes

5x - 3x = 18

2x = 18

Dividing both sides by 2, we get

x = 9

Hence the two numbers are 5 $\times$ 9 = 45 and 3 $\times$ 9 = 27

Q6 Three consecutive integers add up to 51. What are these integers?

Answer:

Let the consecutive integers be x, x+1, x+2.

Sum of these integers is given to be 51.

Thus equation becomes : x + x + 1 + x + 2 = 51

3x + 3 = 51

Transposing 3 to the RHS,

3x = 48

Now dividing both sides by 3, we get

$x = \frac{48}{3} = 16$

Hence the consecutive integers are 16, 17, 18.

Q7 The sum of three consecutive multiples of 8 is 888. Find multiples.

Answer:

Let x be the multiple of 8.

Then the three consecutive integers (multiple of 8) are x, x+8, x+16.

Given their sum is 888,

thus the equation becomes: x + x + 8 + x + 16 = 888

or 3x + 24 = 888

Transposing 24 to the RHS

3x = 888 - 24 = 864

Now dividing both sides by 3, we get

$x = \frac{864}{3} = 288$

Thus the required consecutive integers are 288, 296, 304.

Q8 Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Answer:

Let the consecutive integers be x, x+1, x+2.

Now according to the question equation becomes,

2x + 3(x+1) + 4(x+2) = 74

or 2x + 3x + 3 + 4x + 8 = 74

or 9x + 11 = 74

Transposing 11 to the RHS we get,

9x = 74 - 11 = 63

Divinding both sides by 9.

$x = \frac{63}{9} = 7$

Therefore required consecutive integers are 7, 8, 9.

Q9 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Answer:

Let the current age of Rahul and Haroon be 5x and 7x respectively(Since there age ratio is given as 5:7).

Four years later their ages become 5x + 4 and 7x + 4 respectively.

According to the question,

5x + 4 +7x + 4 = 56

or 12x + 8 = 56

Transposing 8 to the RHS we get,

12x = 48

Dividing both sides by 12:

$x = \frac{48}{12} = 4$

Thus current age of Rahul = 5 $\times$ 4 = 20

Haroon = 7 $\times$ 4 = 28

Q10 The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Answer:

Let us assume the number of boys and the number of girls is 7x and 5x.

According to the given data in the question,

7x = 5x + 8

Transposing 5x to the LHS we get,

7x - 5x = 8

2x = 8

Dividing both sides by 2

x = 4

Hence number of boys = 7 $\times$ 4 = 28

and number of girls = 5 $\times$ 4 = 20

So total number of students = 28 + 20 = 48.

Q11 Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Answer:

Let the age of Baichung be x years.

Then according to question age of Baichung's father = x + 29 years

and age of Baichung's grandfather = x + 29 + 26 years

The sum of their ages is given 135 years. According to that, x + x + 29 + x + 29 + 26 = 135

3x + 84 = 135

Transposing 81 on the RHS we get,

3x = 135 - 84 = 51

Dividing both sides by 3

$x =\frac{51}{3} = 17$

Thus, the age of Baichung = 17 years

The age of Baichung's father = 17 + 29 = 46 years

The age of Baichung's grandfather = 46 + 26 = 72 years

Q12 Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Answer:

Let us assume Ravi's present age to be x years.

According to the question,

x + 15 = 4x

Transposing x to the RHS

15 = 3x

Now dividing both sides by 3, we get

$x = \frac{15}{3} = 5$

Hence Ravi's current age is 5 years.

Q13 A rational number is such that when you multiply it by $\frac{5}{2}$ and add $\frac{2}{3}$ to the product, you get $\frac{-7}{12}$ . What is the number?

Answer:

Let the rational number be x.

According to the question,

$x\times \frac{5}{2} + \frac{2}{3} = -\frac{7}{12}$

Transposing $\frac{2}{3}$ to the RHS:

$x\times \frac{5}{2} = -\frac{7}{12} - \frac{2}{3} = \frac{-15}{12} = \frac{-5}{4}$

Multiplication by 2 in both sides, we get

$5x = \frac{-5}{4}\times 2 = \frac{-5}{2}$

Dividing both sides by 5, we get

$x = \frac{-5}{2}\times \frac{1}{5} = \frac{-1}{2}$

Q14 Lakshmi is a cashier in a bank. She has currency notes of denominations ?100, ?50 and ?10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ?4,00,000. How many notes of each denomination does she have?

Answer:

Since the ratio of currency notes is 2:3:5. Therefore,

Let the number of currency notes of Rs.100, Rs.50 and Rs.10 be 2x, 3x, 5x respectively.

Hence according to the question equation becomes,

100 $\times$ 2x + 50 $\times$ 3x + 10 $\times$ 5x = 400000

or 200x + 150x + 50x = 400000

or 400x = 400000

Dividing both sides by 400 we get,

x = 1000 .

No. of Rs.100 notes = 2 $\times$ 1000 = 2000 notes

No. of Rs.50 notes = 3 $\times$ 1000 = 3000 notes

No. of Rs.10 notes = 5 $\times$ 1000 = 5000 notes

Q15 I have a total of Rs 300 in coins of denomination Rs1, Rs2 and Rs5. The number of Rs2 coins is 3 times the number of Rs5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Answer:

Let the no. of Rs.1 coin be y.

And the no. of Rs.5 coin be x.

Thus according to question no. of Rs.2 coin will be 3x.

Also, the total no. of coins = 160

This implies : y + x + 3x = 160

or y + 4x = 160

Transposing 4x to the RHS

y = 160 - 4x = No. of Rs.1 coin.

Now, it is given that total amount is Rs.300.

i.e., 1(160-4x) + 2(3x) + 5(x) = 300

or 160 - 4x + 6x + 5x = 300

or 160 + 7x = 300

Transposing 160 to the RHS :

7x = 300 - 160 = 140

Dividing both sides by 7:

$x = \frac{140}{7} = 20$

x = 20

Thus number of Rs.1 coin = 160 - 4x = 160 - 80 = 80

number of Rs.2 coin = 3x = 60

number of Rs.5 coin = x = 20

Q16 The organizers of an essay competition decide that a winner in the competition gets a prize of Rs100 and a participant who does not win gets a prize of Rs25. The total prize money distributed is Rs3,000. Find the number of winners, if the total number of participants is 63.

Answer:

Let the no. of winners be x.

Since total no. of participants is 63, thus no. of participants who does not win = 63 - x

According to the question, equation becomes:

x(100) + (63-x)(25) = 3000

or 100x + 1575 - 25x = 3000

or 75x + 1575 = 3000

Transposing 1575 to the RHS :

75x = 3000 - 1575

or 75x = 1425

Dividing both sides by 75, we get

$x = \frac{1425}{75} = 19$

x = 19

Therefore no. of winners = 19

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.3

Q1 Solve the following equations and check your results 3x = 2x + 18

Answer:

We have

3x = 2x + 18

Subtracting 2x from both sides

3x - 2x = 2x - 2x + 18

or x = 18

Check :- Put x = 18 in both LHS and the RHS.

LHS = 3x = 3(18) = 54

RHS = 2x + 18 = 2(18) + 18 = 36 + 18 = 54

Thus, LHS = RHS

Q2 Solve the following equation and check your results 5t – 3 = 3t – 5

Answer:

We have

5t - 3 = 3t - 5

Transposing 3t to the LHS and -3 to the RHS, we get:

5t - 3t = -5 + 3

or 2t = -2

Dividing both sides by 2 :

t = -1

Check :- Put t = -1 in the LHS we have,

5t - 3 = 5(-1) - 3 = -5 -3 = -8

Similarly put t = -1 in the RHS:

3t - 5 = 3(-1) - 5 = -3 - 5 = -8

Hence, LHS = RHS

Q3 Solve the following equation and check your result 5x + 9 = 5 + 3x

Answer:

We have

5x + 9 = 5 + 3x

Transposing 3x to the LHS and 9 to the RHS, we get:

5x - 3x = 5 - 9

or 2x = -4

Dividing both sides by 2 :

x = -2

Check :- Put x = - 2 in both LHS and RHS

LHS :- 5x + 9 = 5(-2) + 9 = -10 + 9 = -1

RHS :- 5 + 3x = 5 + 3(-2) = 5 - 6 = -1

Hence, LHS = RHS

Q4 Solve the following equation and check your result. 4z + 3 = 6 + 2z

Answer:

We have

4z + 3 = 6 +2z

Transposing 2z to the LHS and 3 to the RHS, we get:

4z - 2z = 6 - 3

or 2z = 3

DIviding both sides by 2,

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

Check :- Put $z = \frac{3}{2}$ in LHS as well as RHS, we have

LHS :- 4z + 3

$= 4\times \frac{3}{2} + 3 = 9$

RHS :- 6 + 2z

$= 6 + 2\times \frac{3}{2} = 6 + 3 = 9$

Thus, LHS = RHS

Q5 Solve the following equation and check your result 2x – 1 = 14 – x

Answer:

We have

2x - 1 = 14 - x

Transposing -x to the LHS and -1 to the RHS

2x + x = 14 + 1

or 3x = 15

Dividing both sides by 3, we get

x = 5

Check :- Put x = 5 in both the LHS and the RHS

LHS :- 2x - 1 = 2(5) - 1 = 10 - 1 = 9

RHS :- 14 - x = 14 - 5 = 9

Hence, LHS = RHS

Q6 Solve the following equation and check your result 8x + 4 = 3 (x – 1) + 7

Answer:

We have

8x + 4 = 3 (x - 1) + 7

or 8x + 4 = 3x - 3 + 7 = 3x + 4

Transposing 3x to the LHS and 4 to the RHS, we get

8x - 3x = 4 - 4 = 0

5x = 0

Dividing both sides by 5:

x = 0

Check :- Putting x = 0 in both LHS and RHS:

LHS :- 8x + 4 = 8(0) + 4 = 4

RHS :- 3(x-1) + 7 = 3(-1) + 7 = -3 + 7 = 4

Hence, LHS = RHS

Q7 Solve the following equations and check your results.

$x =\frac{4}{5}\times \left ( x+10 \right )$

Answer:

We have

$x = \frac{4}{5}\left ( x + 10 \right )$

$= \frac{4}{5}x + 8$

Transposing $\frac{4}{5}x$ to the left-hand side:

$x - \frac{4}{5}x = 8$

or $\frac{5x-4x}{5} = 8$

or $\frac{x}{5} = 8$

Multiplying both sides by 5

x = 40

Check :- Put x = 40 in both the LHS and the RHS.

LHS :- x = 40

RHS :- $\frac{4}{5}\left ( x + 10 \right ) = \frac{4}{5}\left ( 40 + 10 \right )$

$= \frac{4}{5}\left ( 50 \right ) = 40$

Thus, LHS = RHS

Q8 Solve the following equation and check your result.

$\frac{2x}{3}+1 = \frac{7x}{15}+3$

Answer:

We have

$\frac{2x}{3} + 1 = \frac{7x}{15} + 3$

Transposing $\frac{7x}{15}$ to the LHS and 1 to the RHS:

$\frac{2x}{3} - \frac{7x}{15} = 3 - 1 = 2$

or $\frac{10x - 7x}{15} = 2$

or $\frac{3x}{15} = 2$

or x = 10

Check :- Put x = 10 in both the LHS and the RHS

LHS :

$\frac{2x}{3} + 1 = \frac{2}{3}\times 10 + 1 = \frac{23}{3}$

RHS :

$\frac{7x}{15} + 3 = \frac{7}{15}\times 10 + 3 = \frac{14}{3} + 3 = \frac{23}{3}$

Thus, LHS = RHS

Q9 Solve the following equations and check your results.

$2y+\frac{5}{3} = \frac{26}{3} -y$

Answer:

We have

$2y+ \frac{5}{3} = \frac{26}{3} - y$

Transposing -y to the LHS and $\frac{5}{3}$ to the RHS, we get

$2y + y = \frac{26}{3}- \frac{5}{3} = \frac{21}{3} = 7$

or 3y = 7

Dividing both sides by 3:

$y = \frac{7}{3}$ i

Check :- Put $y = \frac{7}{3}$ in both the LHS and the RHS:

LHS :

$2y + \frac{5}{3} = 2\times \frac{7}{3} + \frac{5}{3} = \frac{19}{3}$

RHS :

$\frac{26}{3} - \frac{7}{3} = \frac{19}{3}$

Hence, LHS = RHS

Q10 Solve the following equation and check your result.

$3m = 5m - \frac{8}{5}$

Answer:

We have

$3m = 5m - \frac{8}{5}$

Transposing 5m to the LHS

$3m -5m = - \frac{8}{5}$

or $-2m = - \frac{8}{5}$

Dividing both sides by -2, we get

$m = \frac{4}{5}$

Check :- Put $m = \frac{4}{5}$ in the LHS and the RHS:

LHS :-

$3m = 3\times \left ( \frac{4}{5} \right ) = \frac{12}{5}$

RHS :-

$5m - \frac{8}{5} = 5 \times \frac{4}{5} - \frac{8}{5} = \frac{12}{5}$

Hence, LHS = RHS

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.4

Q1 Amina thinks of a number and subtracts $\frac{5}{2 }$ 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?

Answer:

Let us assume that the number Amina thought was x.

As per the question,

$\left ( x-\frac{5}{2} \right )\times8 = 3x$

or <img alt="8x - 20 = 3x" height="12"

src="https://lh4.googleusercontent.com/g0Gfsl_ge2fKco6rn9ZEhXpVr1BkhY7g5s2yjAtg4OjWqydAzAEPdyOwHCKUYbq_PVFmal7_YLGkw4gUP_kacubUU-v1-QkeST4hIkZNsICpg3kadvkMKZCjO95fcqGoCJXeBAo" style="margin-left: 0px; margin-top: 0px;" width="103" />

Transposing 3x to the LHS and -20 to the RHS, we get

8x - 3x = 20

or 5x = 20

Dividing both sides by 5.

x = 4

Therefore the number Amina thought was 4.

Q2 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?

Answer:

Let the No.1 = x

Then the second no. No.2 = 5x.

Now the question states that if we add 21 to both numbers one number becomes twice of the other.

i.e., 2(x + 21) = 5x + 21

or 2x + 42 = 5x + 21

Transposing 2x to the RHS and 21 to the LHS:

21 = 5x - 2x = 3x

or x = 7

Thus the two numbers are 7 and 35 . (Since No.2 = 5x)

Q3 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?

Answer:

Given that sum of the two digits is 9.

Let us assume the digit of units place be x.

Then the digit of tens place will be 9-x.

Thus the two digit number is 10(9-x) + x

Now if we reverse the digits, the number becomes 10x + (9-x).

As per the question,

10x + (9-x) = 10(9-x) + x + 27

or 9x + 9 = 90 - 10x + x + 27

or 9x + 9 = 117 - 9x

Transposing -9x to the LHS and 9 to the RHS:

9x + 9x = 117 - 9

or 18x = 108

x = 6

Thus two digit number is 36.

Q4 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?

Answer:

Let us assume that the units place of a two digit number is x.

Then as per the question, tens place of the number is 3x.

Thus the two digit number is 10(3x) + x.

If we interchange the digits the number becomes : 10x + 3x

According to the question,

10(3x) + x + 10x + 3x = 88

or 30x + 14x = 88

or 44x = 88

Dividing both sides by 44, we get

x = 2

Hence the two digit number is 10(6) + 2 = 62

Note that the two-digit number can also be found if we reverse the digits. So one more possible number is 26.

Q5 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?

Answer:

Let us assume that the Shobo's present age be x.

So, Shobo's mother's present age = 6x.

Shobo's age five years from now will be = x + 5.

It is given that Shobo’s age five years from now will be one-third of his mother’s present age.

So,

$x + 5 = \left ( \frac{1}{3} \right )\times6x = 2x$

or x + 5 = 2x

Transposing x to the RHS,

5 = 2x - x = x

Hence Shobo's present age is x = 5 years

and Shobo's mother's present age is 6x = 30 years

Q6 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 Rs100 per metre it will cost the village panchayat Rs75000 to fence the plot. What are the dimensions of the plot?

Answer:

Let the lenght and breadth of the plot be 11x and 4x repectively.

(Since they are in the ratio11:4)

We know that the fencing will be done on the boundary of the plot, so we need to calculate its perimeter.

Perimeter of rectangle = 2(lenght + breadth)

Total cost to fence the plot = Rs.75000 = Rs.100 $\times$ Perimeter of plot.

Thus equation becomes,

100 $\times$ 2(11x + 4x) = 75000

or 200(15x) = 75000

or 3000x = 75000

Dividing both sides by 3000:

x = 25

Hence lenght of plot is 11x = 275 m

and breadth of the plot is 4x = 100m

Q7 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs50 per meter and trouser material that costs him Rs90 per meter. For every 3 meters of the shirt material, he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs36,600. How much trouser material did he buy?

Answer:

Let the total shirt material that Hasan bought be x metre.

and the total trouser material be y metre.

As per the question,

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

Thus trouser material $y = \frac{2}{3}x$

Now we will make another equaion by using cost of the materials.

Profit for shirt is 12% i.e., selling price of shirt is

$=50 +\frac{12}{100}\times 50 = 50 + 6 = 56$

and selling price for trouser is

$=90 +\frac{10}{100}\times 90 = 90 + 9 = 99$

Hence equation becomes:

$x\left ( 56 \right ) + \frac{2}{3}x\left ( 99 \right ) = 36600$

or 56x + 66x = 36600

or 122x = 36600

Dividing both sides by 122:

x = 300

Thus trouser material that Hasan bought was = 200 meter

$\left ( \frac{2}{3}x = \frac{2}{3}\times300 = 200 \right )$

Q8 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.

Answer:

Let us assume the number of deer in the herd to be x.

Half of a herd of deer are grazing in the field i.e., $\frac{x}{2}$ .

Now half are remaining.

Three fourths of the remaining are playing nearby, implies :

$\frac{3}{4}\times \frac{x}{2} = \frac{3x}{8}$

Remaining number of deer will be :

$\frac{x}{2} - \frac{3x}{8} = \frac{4x - 3x}{8} = \frac{x}{8}$

Given that rest are 9, so

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

Hence number of deer in the herd = 72

Q9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Answer:

Let the age of granddaughter is x years.

So the age of grandfather is x + 54.

Also grandfather is ten times older than his granddaughter.

i.e., x + 54 = 10x

Transposing x to the RHS :

54 = 10x - x

or 54 = 9x

or x = 6

Therefore age of granddaughter = 6 years

and the age of grandfather = 6 + 54 = 60 years

Q10 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.

Answer:

Let the age of Aman's son is x years.

So the age of Aman will be 3x years.

Ten years ago, the age of Aman's son = x - 10 and Aman's age = 3x - 10.

So according to question,

5(x - 10) = 3x - 10

or 5x - 50 = 3x - 10

Transposing 3x to the LHS and -50 to the RHS:

5x - 3x = -10 + 50 = 40

or 2x = 40

or x = 20

So Aman's age = 3x = 60 years and

Aman's son's age = x = 20 years.

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.5

Q1 Solve the following linear equations.

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

Answer:

The detailed solution of the above-written question is here,

We have

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

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

or $\frac{3x - 2x}{6} = \frac{4 +5}{20}$

or $\frac{x}{6} = \frac{9}{20}$

Multiplying both sides by 6:

$x = \frac{9}{20}\times 6 = \frac{27}{10}$

Thus $x = \frac{27}{10}$

Q2 Solve the following linear equations.

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

Answer:

We have

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

Multiplying both sides 12.

(We multiplied both sides by 12 because it is the lowest common factor.)

6n - 9n + 10n = 252

or 7n = 252

or n = 36

Q3 Solve the following linear equations.

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

Answer:

We have

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

Multiplying both sides by 6, we get

6x + 42 - 16x = 17 - 15x

or -10x +42 = 17 - 15x

Transposing -15x to the LHS and 42 to the RHS :

-10x + 15x = 17 - 42

or 5x = -25

or x = -5

Q4 Solve the following linear equations.

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

Answer:

We have

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

Multiplying both sides by 15 . (because the LCM of denomenator is 15 )

5x - 25 = 3x - 9

Transposing 3x to the LHS and -25 to the RHS.

5x - 3x = -9 + 25

or 2x = 16

x = 8

Q5 Solve the following linear equations.

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

Answer:

We have

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

Multiplying both sides by 12, we get

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

or 9t - 6 - 8t - 12 = 8 - 12t

or t - 18 = 8 - 12t

Transposing -18 to the RHS and -12t to the LHS:

12t + t = 8 + 18 = 26

or 13t = 26

Thus, t = 2

Q6 Solve the following linear equation.

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

Answer:

We have

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

Multiplying both sides by 6 (as it is LCM of the denomenator.)

6m - 3(m - 1) = 6 - 2(m - 2)

or 6m - 3m + 3 = 6 - 2m + 4

or 3m + 3 = 10 - 2m

Transposing -2m to the LHS and 3 to the RHS:

3m + 2m = 10 - 3

or 5m = 7

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

Q7 Simplify and solve the following linear equation. 3(t – 3) = 5(2t + 1)

Answer:

Let us open the brackets :

LHS : 3(t - 3) = 3t - 9

RHS : 5(2t + 1) = 10t + 5

The equation is 3t - 9 = 10t + 5

or -9 = 10t - 3t + 5

or -9 - 5 = 7t

or -14 = 7t

Therefore t = -2

Q8 Simplify and solve the following linear equation . 15(y – 4) –2(y – 9) + 5(y + 6) = 0

Answer:

Let us open the brackets

Equation becomes: 15y - 60 - 2y + 18 + 5y + 30 = 0

or 18y - 12 = 0

or 18y = 12

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

Q9 Simplify and solve the following linear equation. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Answer:

Let us open the brackets.

LHS: 3(5z - 7) - 2(9z - 11)

= 15z - 21 - 18z + 22

= -3z + 1

RHS: 4(8z - 13) - 17

= 32z - 52 - 17

= 32z - 69

So the equation becomes : -3z + 1 = 32z - 69

or 69 + 1 = 32z + 3z

or 70 = 35z

z = 2

Q10 Simplify and solve the following linear equations. 0.25(4f – 3) = 0.05(10f – 9)

Answer:

We have

0.25(4f – 3) = 0.05(10f – 9)

Let us open the brackets:

f - 0.75 = 0.5f - 0.45

or f - 0.5f = - 0.45 + 0.75

or 0.5f = 0.30

Dividing both sides by 0.5, we get

f = 0.6

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.6

Q1 Solve the following equation.

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

Answer:

We observe that the equation is not a linear equation since LHS is not linear.

So we multiply 3x to both the sides to make it linear.

We get, 8x - 3 = 2 $\times$ 3x

or 8x - 3 = 6x

or 8x - 6x = 3

or 2x = 3

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

Q2 Solve the following equation.

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

Answer:

We will convert the given equation into a linear equation by multiplying

(7 - 6x) to both sides.

Equation becomes: 9x = 15(7 - 6x)

or 9x = 105 - 90x

or 99x = 105

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

Q3 Solve the following equation.

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

Answer:

We have

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

Cross-multiplication gives:

9z = 4z + 60

or 9z - 4z = 60

or 5z = 60

or z = 12

Q4 Solve the following equation.

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

Answer:

We have

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

By cross-multiplication we get:

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

or 15y + 20 = -4 + 12y

or 15y - 12y = -4 - 20

or 3y = -24

y = -8

Q5 Solve the following equation.

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

Answer:

We have

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

Cross-multiplication gives,

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

or 21y + 12 = -4y - 8

or 21y + 4y = -8 - 12

or 25y = -20

$y = \frac{-20}{25} = \frac{-4}{5}$

Q6 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.

Answer:

Let the present ages of Hari and Harry be 5x and 7x respectively.

(Since there are in the ratio of 5:7)

Four years from age of Hari will be = 5x + 4

and Harry's age will be = 7x + 4.

According to question four years from now, the ratio of their ages will be 3:4.

So the equation becomes :

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

Cross-multiplication gives:

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

or 20x + 16 = 21x + 12

or x = 4

Hence present age of Hari = 5x = 20 and

present age of Harry = 7x = 28

Q7 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.

Answer:

Let the numerator of the rational number be x.

Then denominator will be x + 8.

Further, according to question,

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

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

Cross-multiplication gives:

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

or 2x + 34 = 3x + 21

or x = 13

Hence the rational number is $\frac{13}{21}$ .

NCERT Class 8 Mathematics Solutions

Chapter 01 - Rational Numbers

Chapter 02 - Linear Equations in One Variable

Chapter 03 -Understanding Quadrilaterals

Chapter 04 - Practical Geometry

Chapter 05 - Data Handling

Chapter 06 - Squares and Square Roots

Chapter 07 - Cubes and Cube Roots

Chapter 08 - Comparing Quantities

Chapter 09 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Indirect proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

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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 Excercise: 2.1

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.2

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.3

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.4

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.5

NCERT solutions for class 8 maths chapter 2 Linear Equations in One Variable Excercise: 2.6

NCERT Class 8 Mathematics Solutions

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