Linear Equations in One Variable-NCERT Solutions

Class VIII Math
NCERT SOLUTIONS For Liner Equation in one Variable

Exercise 2.1

Solve the following equations:

Question 1:

x - 2 = 7

Solution:

Given, x – 2 = 7

By adding 2 to both sides, we get

x – 2 + 2 = 7 + 2

⇒ x = 7 + 2

⇒ x = 9

Question 2:

y + 3 = 10

Solution:

Given,y + 3 = 10

By subtracting 3 from both sides, we get

y + 3 – 3 = 10 – 3

⇒ y = 10 – 3

⇒ y = 7

Question 3:

6 = z + 2

Solution:

Given,6 = z + 2

By substracting 2 from both sides, we get

6 – 2 = z + 2 – 2

⇒4 = z + 2 – 2

⇒ 4 = z

⇒ z = 4

Question 4:

Solution:

Given,

By subtracting

from both sides, we get

Question 5:

6x = 12

Solution:

Given,6x = 12

After dividing both sides by 6, we get

Question 6:

Solution:

Given,

After multiplying both sides by 5, we get

Question 7:

Solution:

Given,

By multiplying both sides with 3, we get

After dividing both sides by 2, we get

Question 8:

Solution:

Given,

After multiplying both sides by 1.5, we get

Question 9:

7x – 9 = 16

Solution:

Given,7x – 9 = 16

By adding 9 to both sides, we get

After dividing both sides by 7, we get

Question 10:

Solution:

Given,14y – 18 = 13

By adding 8 to both sides, we get

14y – 8 + 8 = 13 + 8

⇒ 14y = 21

After dividing both sides by 14, we get

Question 11:

17 + 6p = 9

Solution:

Given, 17 + 6p = 9

After subtracting 17 from both sides, we get

17 + 6p – 17 = 9 – 17

⇒ 6p = –8

After dividing both sides by 6, we get

Question 12:

Solution:

Given,

By transposing 1 from LHS to RHS, we get

After multiplying both sides with 3, we get

Exercise 2.2

Question 1:

If you subtract

from a number and multiply the result by

, you get

. What is the number?

Solution:

Let the number is m.

As per question we have

After dividing both sides by

we get

By transposing

to RHS, we get

Question 2:

The perimeter of a rectangular swimming pool is 154m. Its length is 2m more than twice its breadth.

What are the length and the breadth of the pool?

Solution:

Given, perimeter of the rectangular swimming pool = 154m

Length is 2 metre more than the breadth.

Let the breadth of the swimming pool = a metre

Therefore, as per question, length of the swimming pool = (2a + 2) metre

We know that,

perimeter of rectangle = 2 (length + breadth)

Therefore, 154m = 2[(2a + 2) + a]

⇒ 154m = 2(2a + 2 + a )

⇒ 154m = 2 (3a + 2)

⇒ 154m = 6a + 4

By subtracting 4 from both sides, we get

154m – 4 = 6a + 4 – 4

⇒ 150m = 6a

After dividing both sides by 6, we get

Since, length = (2a + 2)m

Therefore, by substituting the value of breadth (a), we get

(2 x 25 + 2)m= (50 + 2)m = 52m

Thus, length of the given pool = 52m

And breadth = 25m

The base of an isosceles triangle is

The perimeter of the triangle is

What is the length of either of the remaining equal sides?

Solution:

Given, Base of the isosceles triangle

Isosceles triangles have two sides equal.

We know that perimeter of an isosceles triangle

= Sum of two equal sides + third side

Let the length of equal sides of the given isosceles triangle = a

And length of unequal side = b

Therefore, perimeter = 2a + b

Therefore,

By subtracting

from both sides, we get

After dividing both sides by 2, we get

Thus,

lenght of either of remaining equal sides

Question 4:

Sum of two numbers is 95. If one

exceeds the other by 15, find the numbers.

Solution:

Let one number is ‘a’.

Therefore,

according to question second number = a + 15

Now, as given, sum of two numbers = 95

Therefore,

a + a + 15 = 95

⇒ 2a + 15 = 95

By subtracting 15 from both sides, we get

2a + 15 – 15 = 95 – 15

⇒ 2a = 95 – 15

⇒ 2a = 80

After dividing both sides by 2, we get

Now, since second number = a + 15

Therefore, by substituting the value of ‘a’, we get

The second number = 40 + 15 = 55

Thus, first number = 40 and second number = 55

Question 5:

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

Solution:

Given, two numbers are in the ratio of 5:3

Their difference = 18

Let first number = 5x and second number = 3x

As per question, 5x – 3x = 18

⇒ 2x = 18

After dividing both sides by 2, we get

Since, first number = 5x and second number = 3x

Thus, by substituting the value of x, in both the nubmers, we get

First number = 5x = 5 X 9 = 45

Ans second number = 3x = 3 X 9 = 27

Thus,

first number = 45 and second number = 27

Question 6:

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

Solution:

Let first integer = a

Therefore, second consecutive integer = a + 1

And, third consecutive integer = a + 2

Since,

sum of the given three consecutive number = 51

Therefore,

a + (a +1) + (a + 2) = 51

⇒ a + a + 1 + a + 2 = 51

⇒ a + a + a + 1 + 2 = 51

⇒ 3a + 1 + 2 = 51

⇒ 3a + 3 = 51

By subtracting 3 from both sides, we get

3a + 3 – 3 = 51 – 3

⇒ 3a = 51 – 3

⇒ 3a = 48

After dividing both sides by 3, we get

Now, second consecutive number = a + 1

Therefore, by substituting the value of ‘a’, we get

Second consecutive number = 16 + 1 = 17

Similarly, after substituting the value of ‘a’ in third consecutive number, we get

Third consecutive number = a + 2 = 16 + 2 = 18

Thus, three required consecutive numbers are 16, 17 and 18

Question 7:

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

Solution:

Let first multiple of 8 = 8a

Therefore, second consecutive multiple of 8 = 8 ( a + 1)

And, third consecutive multiple of 8 = 8 ( a + 2 )

Since,

sum of three consecutive multiples of 8 = 888 as given in the question

Therefore, 8a + [8 (a + 1)] + [8 (a + 2)] = 888

⇒ 8a + (8a + 8) + (8a + 16) = 888

⇒ 8a + 8a + 8 + 8a + 16 = 888

⇒ 8a + 8a + 8a + 8 + 16 = 888

⇒ 24a + 24 = 888

By transposing 24 to RHS, we get

⇒ 24a = 888 – 24

⇒ 24a = 864

After dividing both sides by 24, we get

Now, since first multiple of 8 = 8a,

Second multiple of 8 = 8(a + 1)

And third multiple of 8 = 8 (a + 2)

Thus, by substituting the value of a we get

First multiple of 8 = 8a = 8 x 36 = 288

Second multiple of 8

= 8 (a + 1) = 8 (36 + 1) = 8 x 37 = 296

Third multiple of 8

= 8 (a + 2) = 8 (36 + 2) = 8 x 38 = 304

Thus, three required consecutive multiples of 8

= 288, 296 and 304.

Question 8:

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.

Solution:

Let first integer = a

Therefore, second consecutive integer = a + 1

And, third consecutive integer = a + 2

Since, according to question, by taking in increasing number, first integer is multiplied by 2, second consecutive integer is multiplied by 3 and third consecutive integer is multiplied by 4. And sum of all the three integers = 74

Therefore,

(First integer x 2) + (second consecutive integer

x 3) + (third consecutive integer x 4) = 74

⇒ (a ×2) + [(a + 1) × 3] + [(a + 2) × 4] = 74

⇒ 2a + (3a + 3) = (4a + 8) = 74

⇒ 2a + 3a + 3 + 4a + 8 = 74

⇒ 2a + 3a + 4a + 3 + 8 = 74

⇒ 9a + 11 = 74

By subtracting 11 from both sides, we get

9a + 11 – 11 = 74 – 11

⇒9a = 63

After dividing both sides by 9, we get

By substituting the value of ‘a’ the rest two consecutive integers can be obtained.

Second consecutive integer = a + 1 = 7 + 1 = 8

Third consecutive integer = a + 2 = 7 + 2 = 9

Thus, three required consecutive integers are 7, 8 and 9.

Question 9

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?

Solution:

Let the age of Rahul is 5x and the age of Haroon is 7x

After 4 years,

The age of Rahul = 5x + 4

The age of Haroon = 7x + 4

According to question, the sum of age of Rahul and Harron after 4 years = 56 years

Therefore, (5x + 4_ + (7x + 4) = 56

⇒ 5x + 4 + 7x + 4 = 56

⇒ 5x + 7x + 4 + 4 = 56

⇒ 12x + 8 = 56

After transposing 8 to RHS, we get

12x = 56 – 8

⇒21x = 48

After dividing both sides by 12, we get

Since, present age of Rahul = 5x

Therefore, after substituting the value of x, we get

Present age of Rahul = 5x = 5 × 4 = 20 years

Similarly, present age of Haroon = 7x = 7 × 4 = 28 years

Therefore, present age of Rahul = 20 years

And present age of Haroon = 28 years

Question 10:

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?

Solution:

Let the numbe of boys = 7x and number of girls = 5x

As per question,

the number of boys –8 = Number of girls

Therefore, 7x – 8 = 5x

After transposing 7x to RHS, we get

–8 = 5x – 7x

⇒ –8 = –2x

⇒ 8 = 2x

After dividing both sides by 2, we get

The strength of class = Number of boys + Number of girls

⇒ Strenght of class = 7x + 5x = 12x

By substituting the value of x, we get

Strength of class = 12x = 12 × 4 = 48

Thus strenght of class = 48

Question 11:

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?

Solution:

Let the age of Baichung’s father = m year

Since, Baichung’s father is 26 year younger than Baichung’s grand father

Therefore, Age of Baichung’s grandfather = Age of Baichung’s father + 26 = m + 26 year

Since, Baichung’s father is 29 year older than Baichung

Therefore, Age of Baichung = Age of Baichung’s father – 29 = m – 29 year

Now, we have

Age of Baichung = m – 29 year

Age of Baichung’s father = m year

Age of Bahichung’s grandfather = m + 26 year

According to question, the sum of ages of all the three = 135 year

Therefore,

Age of Baichung + Age of Baichung’s father + Age of Baichung’s grandfather = 135

⇒ (m – 29) + m + (m + 26) = 135 year

⇒ m – 29 + m + m + 26 = 135 year

⇒ m + m + m – 29 + 26 = 135 year

⇒ 3m – 3 = 135 year

After transposing – 3 to the RHS, we get

3m = 135 year + 3

⇒ 3m = 138 year

After dividing both sides by 3, we get

This means age of Baichung’s father = 46 year

Now, by substituting the value of m, we can calculate the age of Baichung and the age of Baichung’s grandfather.

Therefore,

Age of Baichung = m – 29 = 46 – 29 = 17 year

Age of Baichung’s grandfather

= m + 26 = 46 + 26 = 72 year

Thus,

Age of Baichung = 17 year

Age of Baichung’s father = 46 year

Age of Baichung’s grandfather = 72 year

Question 12:

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

Solution:

Let the present age of Ravi = m year

Age of Ravi after 15 years = m + 15 year

According to question, age of Ravi will be four times of his present age.

i.e. Age of Ravi after 15 year = 4 x present age of Ravi

⇒ m + 15 = 4 x m

⇒ m + 15 = 4m

After transposing m to RHS, we get

15 = 4m – 3

⇒ 15 = 3m

After dividing both sides by 3, we get

Question 8:

Thus, Ravi’s present age = 5 year

Question 13:

A rational number is such that when you multiply it by

and add

to the product, you get

What is the number?

Solution:

Let the rational number

According to question,

By transposing

to RHS, we get

After dividing both sides by

, we get

Thus, the rational number is

Question 14:

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000.

How many notes of each denomination does she have?

Solution: Given

Ratio of currency notes of denominations Rs 100, Rs 50 and Rs 10 = 2:3:5

Thus,

Let the number of notes of denomination of Rs 100 = 2x

Let the number of notes of denomination of Rs = 3x

Let the number of notes of denomination of Rs 10 = 5x

Therefore,

Value of notes of denomination of

Rs 100 = 2x × 100 = 200x

Value of notes of denomination of

Rs 50 = 3x × 50 = 150x

Value of notes of denomination of

Rs 10 = 5x × 10 = 50x

Therefore, According to question

Total cash = Rs4,00,000 = 200x + 150x + 50x

⇒ Rs 4,00,000 = 400x

After dividing both sides by 400, we get

Now, since the number of notes of denomination of

Rs 100 = 2x

Therefore, number of notes of denomination of

Rs 100 = 2 ⇒ 1000 = 2000

Similarly,

Number of notes of denomination of

Rs 50 = 3x = 3 × 1000 = 3000

Number of notes of denomination of

Rs 10 = 5x = 5 × 1000 = 5000

Thus, number of notes of Rs 100 = 2000

Number of notes of Rs 50 = 3000

Number of notes of Rs 10 = 5000

Question 15:

I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Solution:

Given, total value of Rs = Rs 300

Total number of coins = 160

Coins of denomination = Re 1, Rs 2 and Rs 5

Number of Rs 2 coins = 3 x number Rs 5 coins

Let the number of coins of Rs 5 = m

Since, the number coins of Rs 2 is 3 times of the number of coins of Rs 5

Therefore, number of coins of Rs 2 = m x 3 = 3m

Now, Number of coins of Re 1 = Total number of coins – (Number of Rs 5 coins + Number of Rs 2 coins)

Therefore,

Number of coins of Re 1 = 160 – (m + 3m) = 160 – 4m

Total Rs = (Re 1 x Number of Re 1 coins) + (Rs 2 x Number of Rs 2 coins) + (Rs 5 x Number of Rs 5 coins)

⇒ 300 = [1 x (160 – 4m)] + (2 x 3m) + (5 x m)

⇒ 300 = (160 – 4m) + 6m + 5m

⇒ 300 = 160 – 4m + 6m + 5m

⇒ 300 = 160 – 4m + 11m

⇒ 300 = 160 + 7m

After transposing 160 to LHS, we get

300 – 160 = 7m

⇒ 140 = 7m

After dividing both sides by 7, we get

Thus, number of coins of Rs 5 = 20

Now, since, number of coins of Re 1 = 160 – 4m

Thus, by substituting the value of m, we get

Number of coins of

Rs 1 = 160 – (4 x 20) = 160 ndash; 80 = 80

Now, number coins of Rs 2 = 3m

Thus, by substituting the value of m, we get

Number of coins of Rs 2 = 3m = 3 x 20 = 60

Therefore,

Number of coins of Re 1 = 80

Number of coins of Rs 2 = 60

Number of coins of Rs 5 = 20

Question 16:

The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000.

Find the number of winners, if the total number of participants is 63.

Solution:

Given, Total number participants = 63

Total prize money distributed = Rs 3000

Winner gets a prize of Rs 100

Who does not win gets a prize of Rs 25

Number of winners = ?

Let the number of winners = m

Since,

Number of winners + Number of losers = Total number of participants

Or, m + Number of losers = 63

By transposing ‘m’ to RHS, we get

Number of losers = 63 – m

Now,

Total Prize money distributed to winners

= Number of winners X prize money distributed to each winner = m X 100 = 100m

Total prize money distributed to losers

= Number of losers X prize money distributed to each loser

= (63 – m) X 25 = (63 x 25) – 25m = 1575 – 25m

Now,

Total Prize money of winners + Total Prize money of losers = Total prize money

By substituting the total prize money distributed to winners and total prize money distributed to losers, we get

100m + 1575 – 25m = 3000

⇒ 100m – 25m + 1575 = 3000

By transposing 1575 to RHS, we get

100m – 25m = 3000 – 1575

⇒75m = 1425

After dividing both sides by 75, we get

Thus, number of winners = 19

Exercise 2.3

Solve the following equations and check your results

Question 1:

3x = 2x + 18

Solution:

We have 3x = 12x + 18

By transposing 2x to LHS, we get

3x – 2x = 18

⇒ x = 18

Check

Given question is 3x = 2x + 18

By substituting the value of x in the LHS of the given question,

we get LHS = 3x = 3 × 18 = 54

By substituting the value of x in RHS we get

RHS = 2x + 18 = 2 × 18 + 18 = 36 + 18 = 54

Since, LHS = RHS, thus the value of x satisfies the given equation

Question 2:

5t – 3 = 3t – 5

Solution:

We have 5t – 3 = 3t – 5

By transposing 3t to LHS we get

5t – 3 – 3t = –5

By transposing –3 to RHS, we get

5t – 3t = –5 + 3

⇒ 2t = –2

After dividing both sides by 2, we get

Check

Given question is 5t – 3 = 3t – 5

By substituting the value of t in LHS, we get

LHS = 5t – 3 = 5 × (–1) – 3 = –5 –3 = –8

By substituting the valueo f t in RHS we get

RHS = 3t – 5 = 3 × (–1) – 5 = –3 –5 = –8

Since, LHS = RHS, thus the obtained value of ‘t’ satisfies the equation

Question 3:

5x + 9 = 5 + 3x

Solution:

Given,5x + 9 = 5 + 3x

By transposing 3x to LHS, we get

5x + 9 – 3x = 5

⇒ 5x + 9 = 5 + 3x

By transposing 9 to RHS, we get

5x + 9 = 5 + 3x

⇒ 2x = – 4

Aftre dividing both sides by 2, we get

Check

Given question is 5x + 9 = 5 + 3x

By substituting the value of x in LHS, we get

LHS = 5x + 9 = 5 × (–2) + 9 = –10 + 9 = –1

By substituting the value of x in RHS we get

RHS = 5 + 3x = 5 + 3 × (–2) = 5 – 6 = – 1

Since LHS = RHS, thus the obtained value of x satisfies the given equation

Question 4:

4z + 3 = 6 + 2z

Solution:

Given,4z + 3 = 6 + 2z

By transposing 2x to LHS, we get

4z + 3 = 6 + 2z = 6

By transposing 3 to RHS, we get

4z – 2z = 6 – 3

⇒ 2z = 6 – 3

⇒ 2z = 3

After dividing both sides by 2, we get

Check

Given question is 4z + 3 = 6 + 2z

By substituting the value of z in LHS, we get

By substituting the value of z in RHS we get

Since, LHS = RHS thus obtained value of z satisfies the equation

Question 5:

2x – 1 = 14 – x

Solution:

Given,2x – 1 = 14 – x

By transposing –x to LHS, we get

2x – 1 + x = 14

By transposing –1 to RHS, we get

2x + x = 14 + 1

⇒ 3x = 15

After dividing both sides by 3, we get

Check

Given question is 2x – 1 = 14 – x

By substituting the value of x in LHS, we get

LHS = 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9

By substituting the value of x in RHS we get

RHS = 14 – x = 14 – 5 = 9

Thus, LHS = RHS proved

Solve the following equations and check your results

Question 6:

8x + 4 = 3(x – 1) + 7

Solution:

Given,8x + 4 = 3(x – 1) + 7

By removing bracket from RHS, we get

8x + 4 = 3(x – 1) + 7

By transposing 3x to LHS, we get

8x + 4 = 3(x – 1) + 7

⇒ 8x – 3x + 4 = 4

By transposing 4 to RHS, we get

8x – 3x = 4 – 4

⇒ 5x = 0

After dividing both sides by 5, we get

Check

Given question is 8x + 4 = 3(x – 1) +7

By substituting the value of x in LHS, we get

LHS = 8x + 4 = 8 X 0 + 4 = 0 + 4 = 4

By substituting the value of x in RHS, we get

RHS = 3(x – 1) + 7 = 3(0 – 1) + 7 = –3 + 7 = 4

Thus, LHS = RHS proved

Question 7:

Solution:

Given,

By transposing

to LHS, we get

After multipying both sides with 5, we get

Check

Given question is

By substituting the value of x in LHS, we get

LHS = x = 40

By substituting the value of x in RHS we get

RHS

Thus, LHS = RHS proved

Question 8:

Solution:

Given,

By transposing

to LHS, we get

By transposing 1 to RHS, we get

After multiplying both sides with 5, we get

Check

Given question is

By substituting the value of x in LHS, we get

By substituting the value of x in RHS, we get

Thus, LHS = RHS

Question 9:

Solution:

Given,

By transposing –y to LHS, we get

After dividing both sides by 3, we get

Check

Given question is

By substituting the value of y in LHS, we get

By substituting the value of y in RHS, we get

Thus, LHS = RHS proved

Question 10:

Solution:

Given,

By transposing 5m to LHS, we get

(Negative sign of both sides become positive)

After dividing both sides by 2, we get

Check

Given question is

By substituting the value of m in LHS, we get

By substituting the value of m in RHS, we get

Thus, LHS = RHS proved

Exercise 2.4

Solution of Questions from 1 to 5

Question 1:

Amina thinks of a number and subtracts

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?

Solution:

Let the number thought by Anamika = a

According to question,

After removing bracket, we get

After transposing 3a to LHS, we get

8a – 20 – 3a = 0 ⇒ 8a – 3a – 20 = 0

By transposing – 20 to RHS, we get

⇒ 8a – 3a = 20 ⇒ 5a = 20

After dividing both sides by 5, we get

Question 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?

Solution:

Let the given positive number = a

Therefore,

another number which is 5 times of it = 5a

Now, after adding 21 to both of the number,

First number = a + 21

Second number = 5a + 21

According to question, one new number becomes twice of the other new number

Therefore,

Second number = 2 x first number

i.e. 5a + 21 = 2 (a + 21)

⇒ 5a + 21 = 2a + 42

By transposing ‘2a’ to LHS, we get

⇒ 5a + 21 – 2a = 42

Now, after transposing 21 to RHS, we get

⇒ 5a – 2a = 42 – 21

⇒ 3a = 21

After dividing both sides by 3, we get

Therefore, another number 5a = 5 x 7 = 35

Thus, required numbers are 7 and 35

Question 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?

Solution:

Let the number at ones place of two digit number = a

According to question, the sum of digits of given two digit number = 9

i.e. Digit at tens place + digit at ones place = 9

Or, Digit at tens place + a = 9

By transposing ‘a’ to RHS, we get

Digit at tens place = 9 – a

Thus, the number = 10(9 – a) + a

After interchange of digit, the number = 10a + (9 – a)

Since, number obtained after interchange of digit is greater than the original number by 27

Therefore, New number – 27 = Original number

Here, we have original number = 10(9 – a) + a

And, new number = 10a + (9 – a)

⇒ 10a + (9 – a) – 27 = 10 (9 – a) + a

⇒ 10a + 9 – a – 27 = 90 – 10a + a

⇒ 10a – a + 9 – 27 = 90 – 9a

⇒ 9a – 18 = 90 – 9a

By transposing 18 to RHS, we get

9a = 90 – 9a + 18

By transposing – 9a to LHS, we get

9a + 9a = 90 + 18

⇒18 a = 108

After dividing both sides by 18, we get

Since, digit at tens place = 9 – a

Thus, by substituting the value of a, we get

The number at tens place = 9 – a = 9 – 6 = 3

Thus, number at tens place = 3

And number at ones place = a = 6

Thus, the number = 36

Question 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?

Solution:

Let one of the digit , which is at ones place, of a two digit number = a

Therefore, other digit of the two digit number = 3a

Therefore, number = (10 x 3a) + a = 30a + a = 31a

After interchange, the digit = 10a + 3a = 13a

Now,

since sum of the original and resulting number = 88

Therefore, 31a + 13a = 88

⇒ 44 a = 88

Now, after dividing both sides by 44, we get

By substituting the value of ’a’in original number we get

Since, original number = 31a = 31 x 2 = 62

Thus, the number = 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 his mother’s present age. What are their present ages?

Solution:

Let the present age of Shobo = a

Therefore, Shobo’s mother’s present age = 6a

After five years, the age of Shobo = a + 5

As per question,

By transposing ‘a’ to LHS, we get

2a – a = 5 ⇒ a = 5

Thus, present age of Shobo = 5 year

Since, present age of Shobo’s mother = 6a

Thus, present age of Shobo’s mother = 6 × 5 = 30 year

Therefore,

Present age of Shobo = 5 year and present age of Shobo’s mother = 30 year

Question 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 Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Solution:

Cost of fence per meter = Rs 100

Total cost of fence the plot = Rs 75000

Given, ratio of length and breadth of the rectangular plot = 11:4

Let the length of the plot = 11x

And the breadth of the plot = 4x

∵ Rs 100 is the cost to fence the plot of 1 meter

∴Rs 1 will be cost to fence the plot of

meter

∴Rs 75000 will be the cost to fence the plot of

Since, fence is around the plot.

Thus, perimeter of plot = lenght of fence = 750m

We know that Perimeter = 2(lenght + breadth)

⇒ 750m = 2(11x + 4x) ⇒ 750m = 2(15x)

⇒ 750m = 30x

After dividing both sides by 30, we get

By substituting the value of x the lenght and breadth can be calculated

Therefore, lenght = 11x = 11 X 25 = 275m

And breadth = 4x = 4 X 25 = 100m

Thus, length = 275m and Breadth = 100m

Question 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?

Solution:

Given,

Rate of shirt material = Rs 50 per meter

Rate of trouser material = Rs 90 per meter

Profit on shirt material = 12%

Profit on trouser material = 10%

Total sale price = Rs 36600.00

Since profit on the cost price of shirting material = 12%

Therefore,

sale price of shirting material = cost price + 12% of cost price

Therefore, sale price of shirt material = Rs 56.00

Since, profit on the cost price of trouser material = 10%

Therefore, sale price of trouser material = cost price of trouser material + 10% of cost price

Therefore, sale price of trouser material=Rs 99.00

Now, since Hasan buys 3m of shirt material for every 2m of trouser material

Thus, let he buys 3x m of shirting material and 2x m of trouser material

Total sale price

= Total sale price of shirt matreial + Total sale price of trouser material

⇒ Rs 36600 = 3x × Rs 56 + 2x × Rs 99

⇒ Rs 36600 = Rs168x + Rs 198x

⇒ Rs 36600 = Rs 366x

After dividing both sides by 366 we get

Since, purchase of trouser material = 2x

Thus, after substituting the value of x, we get

Purchase of trouser material = 2 X 100 = 200m

Thus, Hasan buys 200m of trouser material.

Question 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.

Solution:

Let the total number of deer = x

Since, half of the herd are grazing in the field

Thus, number of deer grazing in the field

Since,

th 4 of remaining of the half are playing nearby

Thus, number of deer playing nearby

Number of deer drinking water = 9

(as given in question)

Now, Total number of deer

= No. of deer grazing + No. of deer playing

+ No. of deer playing

Now, after transposing

to LHS, we get

Now, after multiplying both sides by 8, we get

Thus, total number of deer in the herd = 72

Question 9:

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

Solution:

Let the age of granddaughter = x year

As given, the age of grandfather = 10 × age of his granddaughter

Therefore, age of grandfather = 10x

Now, again given, age of grandfather = Age of granddaughter + 54

Thus, age of grandfather = x + 54

Since, age of grandfather will equal in all given conditions

Therefore, 10x = x + 54

Now, by transposing x to LHS, we get

10x – x = 54 ⇒ 9x = 54

Now, after dividing both sides by 9, we get

Thus, age of granddaughter = 6 year

And age of grandfather = 6 × 10 = 60 year

Question 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.

Solution:

Let the age of Aman’s son = x year

Since, as given Aman’s age = 3 × age of his son

Therefore, age of Aman = 3x year

Ten years ago,

Present age of Aman – 10 year = (present age of his son – 10) × 5

3x – 10 = (x – 10) × 5

⇒ 3x – 10 = 5x – 50

By transposing 5x to LHS and –10 to RHS, we get

3x – 5x = – 50 + 10 ⇒ –2x = –40

After canceling negative sign both sies, we get

⇒ 2x = 40

Now, after dividing both sides by 2, we get

Thus, present age of Aman’s son=20 year

And present age of Aman=20 year X 3=60 year

Exercise 2.5

Solution of NCERT Exercise from Question 1 to 5

Solve the following linear equations:

Question 1:

Solution:

Given,

After transposing

to LHS and

to RHS, we get

After multiplying both sides with 6, we get

Question 2:

Solution:

Given,

After multiplying both sides by 12, we get

Now after dividing both sides by 7, we get

Question 3:

Solution:

After transposing 7 to RHS and

to LHS, we get

After multiplying both sides with 6, we get

After dividing both side by 5, we get

Question 4:

Solution:

After multiplying both sides with 3, we get

After multiplying both sides with 5, we get

After removing of bracket from both sides we get

5x – 25 = 3x – 9

After transposing 3x to LHS and –25 to RHS, we get

5x – 3x = –9 + 25 ⇒ 2x = 16

After dividing both sides by 2, we get

Question 5:

Solution:

Given,

After transposing –t to LHS, we get

After removing of brackers, we get

After multiplying both sides by 12, we get

After transposing –18 to RHS, we get

By dividing both sides by 13, we get

Question 6:

Solution:

Given,

After transposing

to LHS, we get

Now, after multiplying both sides with 6, we get

After transposing –1 to RHS, we get

5m = 6 + 1 = 7

After dividing both sides with 5, we get

Simplify and solve the following linear equations:

Question 7:

3(t – 3) = 5(2t + 1)

Solution:

Given,3(t – 3) = 5(2t + 1)

⇒ 3t – 9 = 10t + 5

After transposing, –9 to RHS and 10t to LHS, we get

3t – 10t = 5 + 9 ⇒ –7t = 14

After dividing both sides by 7, we get

Question 8:

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Solution:

Given,15(y – 4) – 2(y – 9) + 5(y + 6) = 0

After removing brackets, we get

15y – 60 – 2y + 18 + 5y + 30 = 0

After arranging the above equation, we get

15y – 60 – 2y + 18 + 5y + 30 = 0

After arranging the above equation, we get

15y – 2y + 5y – 60 + 18 + 30 = 0

⇒ 18y – 12 = 0

After transposing –12 to RHS, we get

18y = 12

After dividing both sides by 18, we get

Question 9:

3(5z – 7) –2(9z – 11) = 4(8z – 13) – 17

Solution:

Given 3(5z – 7) –2(9z – 11) = 4(8z – 13) – 17

After removing brackets, we get

15z – 21 – 18z + 22 = 32z – 52 – 17

⇒ 15z – 18z – 21 + 22 = 32z – 69

⇒ –3z + 1 = 32z – 69

After tranposing 1 to RHS and 32z to LHS, we get

–3z – 32z = – 69 – 1

⇒ –35z = –70

After dividing both sides by 35, we get

Question 10:

0.25(4f – 3) = 0.05(10f – 9)

Solution:

Given, 0.25(4f – 3) = 0.05(10f – 9)

After removing brackets, we get

f – 0.75 = 0.5f – 0.45

After transposing 0.5f to LHS and – 0.75 to RHS, we get

f – 0.5f = –0.45 + 0.75

0.5f = 0.3

After dividing both sides by 0.5 we get

Exercise 2.6

Solve the following equations:

Question 1:

Solution:

Given,

After multiplying both sides with 3x, we get

After transposing 8x to RHS, we get

–3 = 6x – 8x

⇒ – 3 = –2x

⇒ 3 = 2x

After dividing both sides by 2, we get

Question 2:

Solution:

Given,

By multiplying both sides with 7 – 6x, we get

After transposing –90x to LHS, we get

9x + 90x = 105

⇒ 99x = 105

After dividing both side by 99, we get

Question 3:

Solution:

Given,

After multiplying both sides by z + 15, we get

After transposing,

to LHS, we get

After multiplying both sides by 9, we get

After dividing both sides by 5, we get

Question 4:

Solution:

Given,

After muliplying both sides with 2 – 6y, we get

After transposing 4 to RHS and

to LHS, we get

After multiplying both sides with 5, we get

After dividing both sides by 3, we get

Question 5:

Solution:

Given,

After multiplying y + 2 we get

After transposing

to LHS, and 4 RHS, we get

After multiplying both sides with 3, we get

After dividing both sides by 25, we get

Question 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.

Solution:

Let the present age of Hari = 5x

And the present age of Harry = 7x

After four years from now

Age of Hari will be = 5x + 4

and age of Harry will be = 7x + 4

As given the ratio of their age after 4 years = 3:4

Therefore,

Now after multiplying both sides by 7x + 4, we get

After transposing

to LHS and 4 to RHS, we get

After multiplying both sides by 4, we get

Since, present age of Hari = 5x

Therefore, after substituting the value of x, we get

Present age of hari = 5x = 5 X 4 = 20 year

Since, present age of Harry = 7x

Thus, after substituting the value of x, we get

Present age of Harry = 7x = 7 X 4 = 28 year

Thus, present age of Hari = 20 year

And present age of Harry = 28 year

Question 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

Find the rational number.