Class 8 Maths Chapter 5 Solutions of Equations Questions and Answers Kerala Syllabus

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SCERT Class 8 Maths Chapter 5 Solutions Solutions of Equations

Class 8 Kerala Syllabus Maths Solutions Chapter 5 Solutions of Equations Questions and Answers

Solutions of Equations Class 8 Questions and Answers Kerala Syllabus

Adding and Subtracting (Page No. 79)

Question 1.
“Six more marks and I would have gotten a hundred out of a hundred in my math exam”, Rajan thought sadly. How much did he score?
Answer:
The mark is 6 less than 100.
Present mark is 94.

Question 2.
“5 more years, and I would be 18. I can vote”, Lissy calculated. How old is she now?
Answer:
The present age is 5 less than 18.
Her current age is 13 years.

Question 3.
What should be added to 123 to make it 321?
Answer:
Number to be added is 321 – 123 = 198

Question 4.
What should be subtracted from 432 to get 234?
Answer:
Number to be subtracted is 432 – 234 = 198

Textbook Page No. 80

Question 1.
Gopalan bought a bunch of bananas. 7 of the bananas had begun to rot, and he removed them. Now there are 46. How many bananas were there in the bunch?
Answer:
The number of good fruits is 46.
Including spoiled fruits total is 46 + 7 = 53.

Question 2.
Vimala bought some things for 163 rupees, and now she has 337 rupees left. How much did she have at first?
Answer:
The total amount in hand is the sum of the balance amount and the amount for purchasing goods.
It is 337 + 163 = 500 rupees

Question 3.
What number becomes 321 on subtracting 123?
Answer:
The subtraction should be done from the number 123 more than 321.
Number is = 321 + 123 = 444

Multiplication and Division (Page No. 80)

Question 1.
The salary of the manager of an office is five times the salary of a peon. The manager gets 75000 rupees a month. How much does the peon get?
Answer:
Peon’s salary is \(\frac {1}{5}\) of Manger’s salary
Peon’s salary is \(\frac {1}{5}\) × 75000 = 15000 rupees

Question 2.
Some friends went for a trip and decided to divide equally among them the 12500 rupees they had spent. Each had to pay 2500 rupees. How many were there in the group?
Answer:
We have to find out the number of 2500s that make 12500.
This will be the number of members
It is 12500 ÷ 2500
Number of members = \(\frac {12500}{2500}\) = 5

Question 3.
A number multiplied by 12 gives 756. What is the number?
Answer:
12 times a number is 756.
Number is \(\frac {1}{12}\) part of 756
Number = \(\frac {1}{12}\) × 756 = 63

Question 4.
A number divided by 21 gives 756. What is the number?
Answer:
\(\frac {1}{21}\) part of a number is 756.
The number is 21 times 756.
Number = 21 × 756 = 15876

Different Kinds of Change (Page No. 82, 83)

Question 1.
The perimeter of a rectangle is 50 metres and the length of one side is 5 metres. What is the length of the other side?
Answer:
Sum of two adjacent sides is 50 ÷ 2 = 25
One side is 5.
Other side 25 – 5 = 20 cm

Question 2.
When Anita and her friend bought 5 pens together, they got a reduction of 3 rupees in the price. The total cost was 32 rupees. Had they bought the pens individually, how much would it have cost each?
Answer:
If there is no reduction in price, the price of 5 pens is 35 rupees
Price of a pen = 35 ÷ 5 = 7 rupees

Question 3.
In each of the problems below, the result of doing some operations on a number is given. Find each number:
(i) Three added to two times the number gives 101
(ii) Two added to three times the number gives 101
(iii) Three subtracted from two times the number gives 101
(iv) Two subtracted from three times the number gives 101
Answer:
(i) Without adding three, twice the number = 101 – 3 = 98
Number is \(\frac {1}{2}\) part of 98.
Number = \(\frac {1}{2}\) × 98 = 49

(ii) Three times the number 101 – 2 = 99
Number is \(\frac {1}{3}\) of 99.
Number = \(\frac {1}{3}\) × 99 = 33

(iii) Twice the number = 101 + 3 = 104
Number is \(\frac {1}{2}\) of 104.
Number is \(\frac {1}{2}\) × 104 = 52

(iv) 3 times the number is 103.
Number is \(\frac {103}{3}\)

Question 4.
The sum of 6 times a number and 4 is 100. What is the number?
Answer:
If 4 is not added, 6 times the number = 100 – 4 = 96
Number is \(\frac {1}{6}\) of 96.
Number = \(\frac {1}{6}\) × 96 = 16

Question 5.
4 times the sum of a number and 6 gives 100. What is the number?
Answer:
6 added to \(\frac {1}{4}\) part of 100.
6 added to the number is 25
Number is 25 – 6 = 19.

Multiple and Part (Page No. 84)

Question 1.
The sum of 2 times a number and 7 times the same number is 27. What is the number?
Answer:
When 2 times a number and 7 times that number are added, we get 9 times that number.
9 times the number is 27.
Number = \(\frac {27}{9}\) = 3

Question 2.
99 added to 12 times a number gives 21 times the number. What is the number?
Answer:
When 12 times a number is subtracted from 21 times that number, we get 99.
9 times the number is 99.
\(\frac {1}{9}\) of 99 is 11.
Number = \(\frac {1}{9}\) × 99 = 11

Textbook Page No. 85

Question 1.
A third of a piece of rope cut away leaves 10 metres. What was the original length?
Answer:
When \(\frac {1}{3}\) is removed results \(\frac {2}{3}\) of length.
It is 10 m.
Length of rope is 10 × \(\frac {3}{2}\) = 15

Question 2.
Half the milk in a can and a third of the remaining was used up, and now there are 5 litres remaining. How many litres did it originally contain?
Answer:
Half of the amount remains \(\frac {1}{3}\) of this is \(\frac {1}{6}\) of the first amount
Used, \(\frac{1}{2}+\frac{1}{6}\). This is \(\frac {4}{6}\) part
Amount remains \(\frac {2}{6}\) part.
It is 5 L.
Amount in the beginning is 5 times \(\frac {6}{2}\).
It is 5 × \(\frac {6}{2}\) = 15L

Algebraic Methods (Page No. 89)

Question 1.
The perimeter of a rectangle is 80 metres, and its length is one metre more than twice the width. What are its length and width?
Answer:
Sum of length and breadth 40
Width = x, length = 2x + 1
∴ x + 2x + 1 = 40
⇒ 3x + 1 = 40
⇒ 3x = 39
⇒ x = 13
∴ Width = 13, Length = 2 × 13 + 1 = 27 cm

Question 2.
When a hundred-rupee note was changed to twenty-rupee and ten-rupee notes, seven notes were obtained. How many of each?
Answer:
Number of 20 rupee notes x, ten rupee notes 7 – x.
∴ 20x + 10(7 – x) = 100
⇒ 20x + 70 – 10x = 100
⇒ 20x – 10x + 70 = 100
⇒ 10x + 70 = 100
⇒ 10x = 100 – 70
⇒ 10x = 30
⇒ x = 3
∴ Number of 20 rupee notes = 3
∴ Number of 10 rupee notes = 4

Question 3.
The price of a book is 4 rupees more than the price of a pen. The price of a pencil is 2 rupees less than the price of this pen. A person bought 5 books, 2 pens, and 3 pencils and paid 54 rupees for them. What is the price of each?
Answer:
The price of a book and a pencil can be written as follows.
Price of pen = x
price of book = x + 4
price of pencil = x – 2
∴ 5(x + 4) + 2x + 3(x – 2) = 54
⇒ 5x + 20 + 2x + 3x – 6 = 54
⇒ 10x + 14 = 54
⇒ 10x = 40
⇒ x = 4
Price of pen 4 rupees, price of book 8, price of pencil 2.

Question 4.
A square containing four numbers in a calendar is marked, and the sum of these numbers is 80. What are the numbers?
Answer:

∴ x + x + 1 + x + 7 + x + 8 = 80
⇒ 4x + 16 = 80
⇒ 4x = 64
⇒ x = 16
Numbers are 16, 17, 23, 24.

Textbook Page No. 92

Question 1.
The age of Appu’s mother is now 9 times that of Appu. After nine years, it will be three times Appu’s age. What are their ages now?
Answer:
If Appu’s age is x, then the mother’s age is 9x
After 9 years ages are x + 9, 9x + 9
∴ (x + 9) × 3 = 9x + 9
⇒ 3x + 27 = 9x + 9
⇒ 18 = 6x
⇒ x = 3
∴ Appu’s age = 3, mother’s age = 27

Question 2.
A class has the same number of girls and boys. On a day when eight boys were absent, the number of girls was twice the number of boys. What are the number of girls and the number of boys?
Answer:
Let the number of boys be x, number of boys x.
When number of boys is x – 8, x = 2(x – 8)
⇒ x = 2(x – 8)
⇒ x = 16

Question 3.
In a class of girls and boys, 50% of the children are girls. When 10 more girls were admitted to this class, this became 60%. How many children were in the class at first?
Answer:
50% denotes the half.
Let the number of boys and the number of girls be x.
There are 2x students.
When 10 girls are added, the number of girls becomes x + 10
∴ x + 10 = 2x × \(\frac {60}{100}\)
⇒ x + 10 = \(\frac {6x}{5}\)
⇒ x = 50
There are 100 students in the class.

Question 4.
Another problem from folk-math. Some lotus flowers have bloomed in a pond. A flock of birds sat on the flowers to rest. First, one bird sat on each flower, but one bird didn’t have a flower to sit on. Then two birds sat on each flower, and there was one flower extra. How many lotuses and how many birds?
Answer:
Let x be the number of lotus flowers.
Number of birds = x + 1
∴ x + 1 = 2(x – 1)
⇒ x = 3
Number of lotus flowers 3, number of birds 4.

Class 8 Maths Chapter 5 Kerala Syllabus Solutions of Equations Questions and Answers

Class 8 Maths Solutions of Equations Questions and Answers

Question 1.
One third of a number is 10 less than the number. Which of the following is the number?
(a) 30
(b) 15
(c) 10
(d) 27
Answer:
\(\frac {2}{3}\) of the number is 10.
Number is \(\frac {3}{2}\) times 10.
Number = 10 × \(\frac {3}{2}\) = 15

Question 2.
The sum of two numbers is 39, difference is 15. Numbers are
(a) 17, 22
(b) 26, 13
(c) 27, 12
(d) 28, 11
Answer:
x, x + 15
∴ 2x + 15 = 39
⇒ 2x = 4
⇒ x = 12
Numbers = 12, 27

Question 3.
Two sides of a triangle are equal. Small angle is 30°. What is one of the equal angles?
(a) 100°
(b) 75°
(c) 70°
(d) 50°
Answer:
Angle sum = 180°
The smallest is 30°.
The sum of the equal angles is 150°
Equal angle is 75°.

Question 4.
When 3 is added to \(\frac {1}{4}\)th of a number gives 15. Find the number.
Answer:
\(\frac {1}{4}\)th of the number is 12.
Number is 4 × 12 = 48

Question 5.
When \(\frac {2}{3}\) of a number is subtracted from a number, we get 13. What is the number?
Answer:
\(\frac {1}{3}\) of the number is 13.
Number is 3 × 13 = 39

Question 6.
The sum of two numbers is 60. One number is four times the other. Find the number.
Answer:
When the number and four times the number are added, we get five times the number.
Number is \(\frac {1}{5}\) part of 60.
The number is 12.

Question 7.
The sum of two consecutive natural numbers is 63.
(a) If the smaller is x, then what is the other number?
(b) Write the equations and find the numbers.
Answer:
(a) x, x + 1 are the numbers.
(b) 2x + 1 = 63
⇒ 2x = 62
⇒ x = 31
Numbers are 31, 32.

Question 8.
Ammu’s mother is four times her mother’s age. After 20 years, mother’s age will be twice Ammu’s age.
(a) If Ammu’s age is x, then write the equation.
(b) Find the age of Ammu and her mother.
Answer:
(a) Ammu’s age x, Mothers age 4x
(b) 20 years hence
Ammu’s age = x + 20
mothers age = 4x + 20
∴ 4x + 20 = 2(x + 20)
⇒ 4x + 20 = 2x + 40
⇒ x = 10
Ammu’s age is 10, Mother’s age is 40.

Question 9.
A man’s age after 15 years will be twice his age before 15 years.
(a) If the present age is x, then write the equation.
(b) Find the present age?
Answer:
(a) Age after 15 years is x + 15
Age before 15 years is x – 15
∴ 2(x – 15) = x + 15
⇒ 2x – 30 = x + 15
⇒ x = 45
(b) Present age = 45 years

Question 10.
The perimeter of a rectangle is 50 cm. Length is 5 cm more than breadth.
(a) What is length + breadth?
(b) If x is the length, then write the equation.
(c) Find the sides.
Answer:
(a) 25 cm
(b) x + x + 5 = 25
⇒ 2x + 5 = 25
⇒ 2x = 20
⇒ x = 10
(c) Sides are 10 cm, 15 cm

Question 11.
The sum of two numbers is 90. The large number is 15 more than twice the small number. Write the equations and find the numbers.
Answer:
Smaller number = x
Larger number = 2x + 15
∴ x + 2x + 15 = 90
⇒ 3x = 75
⇒ x = 25
Numbers are 25, 65.

Question 12.
The sum of the digits of a two-digit number is 10. On reversing the digits, it is found that the number decreases by 36. Find the number.
Answer:
Digit in ten’s place is x.
Digit in one’s place is 10 – x.
Number is 10x + 10 – x
When the digits are reversed number becomes 10(10 – x) + x
It is given that 10(10 – x) + x = 10x + 10 – x – 36
⇒ 100 – 10x + x = 9x – 26
⇒ 100 – 9x = 9x – 26
⇒ 126 = 18x
⇒ x = 7
∴ Number is 73

Question 13.
Length of a bridge is 60 metres longer than \(\frac {1}{3}\) of its length. What is the length of the bridge?
Answer:
If x is the length of the bridge.
∴ \(\frac {2x}{3}\) = 60
⇒ x = 90

Question 14.
In a class of 52 students, the number of boys is \(\frac {6}{7}\) of the number of girls. How many are boys?
Answer:
Number of boys = x
Number of girls = 52 – x
∴ x = \(\frac {6}{7}\) × (52 – x)
⇒ 7x = 6(52 – x)
⇒ 7x = 312 – 6x
⇒ 13x = 312
⇒ x = 24
∴ Boys = 24, Girls = 28

Question 15.
The sum of three consecutive odd numbers is 63.
(a) If the smallest among them is x, then what are the other numbers?
(b) Find the numbers.
Answer:
(a) x, x + 2, x + 4 are the numbers.
(b) 3x + 6 = 63
⇒ 3x = 57
⇒ x = 19
∴ Numbers are 19, 21, 23.

Class 8 Maths Chapter 5 Notes Kerala Syllabus Solutions of Equations

→ Price of 5 pens and 3 pencils is 34 rupees. The price of 3 pens and 5 pencils is 30 rupees. What is the total price of a pen and a pencil together?

→ 64 rupees are needed for 8 pens and 8 pencils. So 8 rupees is needed for a pen and a pencil.

→ 5 pens and 3 pencils cost 34 rupees. 5 pens 5 pencils cost 40 rupees. Note the increased amount is for 2 pencils.

→ The price of 10 pens and 10 pencils is 80 rupees.

→ Three times \(\frac {1}{3}\) of a number is that number itself. So six times \(\frac {1}{3}\) of that number is two times that number.

→ Eight times \(\frac {1}{4}\) of a number is twice that number.

→ What is the sum of eight times \(\frac {1}{4}\) of a number and six times \(\frac {1}{3}\) of that number?

→ Let x be a number. Three times x is 18. This can be written as an equation. 3x = 18 is the equation \(\frac {1}{3}\) of a number is 18. What is the number?
That means, three times that number is 18.
3x = 18
x = \(\frac {1}{3}\) × 18
x = 6

→ 1 more than two times a number is 5.

→ If 1 is not added, two times the number is 4. The number is 2.

Problems that can be solved by simple logic are common in our daily lives. But some problems cannot be solved by simple logic. In mathematics, we use algebra to solve certain problems. The word algebra is derived from the Arabic word Al-Jabr, which means to join together.

Adding and Subtracting
Finding solutions in the form of equations,
x + a = b, x – a = b
Here, only addition and subtraction are needed to find the solutions. The reverse process is the easiest method.

Worksheet – 1

Question 1.
Babu has 600 rupees with him. He went to a shop and purchased some goods. Now he had 301 rupees left with him. How much did he spend?
Answer:
600 – 301 = 299

Question 2.
If we add 54 to a number, we will get 250. What is the number?
Answer:
250 – 54 = 196

Question 3.
Malini had 78 stamps with her. She gave some of them to her friend. Now she has 46 stamps with her. How many stamps did she give to her friend?
Answer:
78 – 46 = 32

Question 4.
When 164 is subtracted from a number, it makes it 235. What is the number?
Answer:
235 + 164 = 399

Multiplication and Division
Finding solutions in the form \(\frac {x}{a}\) = b and ax = b.
Here, only multiplication and division are needed to find the solution.
The reverse process is the easiest method here also.

Worksheet – 2

Question 1.
The price of 4 shirts is 720 rupees. What is the price of one shirt?
Answer:
720 ÷ 4 = 180

Question 2.
The age of Raju’s father is 9 times that of Raju. If the father’s age is 36 years, what is Raju’s age?
Answer:
36 ÷ 9 = 4

Question 3.
A number when multiplied by 14 gives 1008. What is the number?
Answer:
1008 ÷ 14 = 72

Question 4.
A number when divided by 41 gives 215. What is the number?
Answer:
215 × 41 = 8815

Different Kinds of Change
Here we discuss the changes in the form
ax + b = c, ax – b = c, \(\frac {x}{a}\) + b = c
First, we will adopt the method of reverse process, then only the algebraic method.

For example: One added to three times a number gives 100. What is the number?
Here, the last mathematical calculation done on a number is adding ‘1’.
So first we will subtract ‘1’ from the result 100.
So 100 – 1 = 99.
Then the next calculation is multiplying by three, so we will divide 99 by 3.
So the number = \(\frac {99}{3}\) = 33

Worksheet – 3

Question 1.
The double of 3 less than a number is 97. What is the number?
Answer:
3 less than, so add 3
97 + 3 = 100
Double, so half = \(\frac {100}{2}\) = 50

Question 2.
Two less than 3 times a number is 100. What is the number?
Answer:
2 less, so add 2
100 + 2 = 102
3 times, so divide by 3
\(\frac {102}{3}\) = 34

Question 3.
10 is added to half of a number, giving 100. What is the number?
Answer:
10 added, so subtract 10
100 – 10 = 90
Half of a number, so double it.
90 × 2 = 180

Question 4.
Alia has 150 rupees, which is 8 rupees more than twice what Adhil has. How much has Adhil?
Answer:
8 rupees more, so subtract 8
150 – 8 = 142
twice, so half of 142
\(\frac {142}{2}\) = 71

Multiple and Part
If 3 times a number and 2 times the same number added together make 100. What is the number?
Answer:
3 times any number and 2 times the same number make 5 times the number.
That means 5 times the number is 100.
So the number is \(\frac {1}{5}\) of 100.
Number = \(\frac {100}{5}\) = 20

Algebraic Methods
Now we will discuss the algebraic method of solving these word problems. Here, we will take the unknown number as x, then form an equation. Solving is based on the reverse process itself.

For example, the perimeter of a rectangle is 46 cm. The length is 5 cm more than the breadth. What are its length and breadth?
Here, the length is given as 5 more than the breadth.
Breadth is completely unknown.
So it is better to take breadth as ‘x’, which means length is x + 5
∴ Perimeter = 2(length + breadth)
⇒ 2(x + 5 + x) = 46
⇒ x + 5 + x = 23
⇒ 2x + 5 = 23
⇒ 2x = 23 – 5
⇒ 2x = 18
⇒ x = 9
∴ Length = 9 + 5 = 14

Worksheet – 4

Question 1.
Raju has 75 rupees more than Rani. If both of them together have 725 rupees, what are the amounts with Raju and Rani?
Answer:
Rani has ‘x’ rupees
Raju has 75 + x rupees
∴ x + 75 + x = 725
⇒ 2x + 75 = 725
⇒ 2x = 725 – 75
⇒ 2x = 650
⇒ x = 325
Raju has 325 + 75 = 400 rupees

Question 2.
The second angle of a triangle is double the first. The third angle is 40 degrees less than the first. Find the three angles.
Answer:
First angle = x
Second angle = x + x = 2x
Third angle = x – 40
∴ x + 2x + x – 40 = 180°
⇒ 4x – 40 = 180
⇒ 4x = 180 + 40
⇒ 4x = 220
⇒ x = 55°