Kerala Syllabus Class 8 Mathematics - Unit 5 Solutions of Equations - Questions and Answers | Teacher Text 

Questions and Answers for Class 8 Mathematics - Unit 5 Solutions of Equations - Study Notes | Text Books Solution STD 8 - Maths: Unit 5 Solutions of Equations - Questions and Answers | ഗണിതം - സമവാക്യപരിഹാരം - ചോദ്യോത്തരങ്ങൾ 

എട്ടാം ക്ലാസ്സ്‌  Mathematics - Unit 5 Solutions of Equations എന്ന പാഠം ആസ്പദമാക്കി തയ്യാറാക്കിയ ചോദ്യോത്തരങ്ങള്‍. ഈ അധ്യായത്തിന്റെ Teachers Handbook ഡൗൺലോഡ് ചെയ്യാനുള്ള ലിങ്ക് ചോദ്യോത്തരങ്ങളുടെ അവസാനം നൽകിയിട്ടുണ്ട്. പുതിയ അപ്‌ഡേറ്റുകൾക്കായി ഞങ്ങളുടെ Telegram Channel ൽ ജോയിൻ ചെയ്യുക.

Kerala Syllabus STD 8 Maths: Unit 5 Solutions of Equations - Textbook Solutions
രണ്ടളവുകൾ തുല്യമാണെന്ന് കാണിക്കുന്ന പ്രസ്താവനയാണ് സമവാക്യം. ഇതിനെ സമീകരണം എന്നും പറയാറുണ്ട്. സമവാക്യങ്ങൾ ഉപയോഗിച്ച് ഗണിതപ്രശ്നങ്ങൾ ലളിതമാക്കാനും പരിഹരിക്കാനും സാധിക്കുന്നു.

♦ LESSON ANALYSIS 
Adding and subtracting
One day, Zuhra's mother asked Zuhra to open the money box and count it. How much money is in the money box, mother asked. "If you give me seven rupees more, I would have a round fifty", Zuhra said hopefully. Let's see how much money Zuhra had in her money box.
7 more rupees would make it 50, which means what she has is 7 less than 50.
50-7=43

♦ Textbook Activities (Textbook Page No: 79)

(1) "Six more marks and I would have got hundred out of hundred in my math exam”, Rajan thought sadly. How much did he score?
It will become 100 if 6 more are added. So his marks are 6 less than 100.
100 − 6 = 94
These are Rajan's marks.
Checking: 94 + 6 = 100

(2) "5 more years, and I would be 18. I can vote", Lissy calculated. How old is she now?
Lissy's present age: 18 − 5 = 13. 
Checking: 13+5=18.

3. What should be added to 123 to make it 321? 
Number to be added = 321 − 123 = 198 
Checking: 123 + 198 = 321

4. What should be subtracted from 432 to get 234?
Number to be subtracted = 432 − 234 = 198
Checking: 432 − 198 = 234

Another problem: 
Unni spent 8 rupees from what he got for Vishu to buy a pen. Now he has 42 rupees left. How much did he get for Vishu?
When it is decreased by 8 rupees, it becomes 42 rupees. So what he got is 8 more than 42. 
∴ Amount he get for vishu = 42 + 8 = 50

♦ Textbook Activities (Textbook Page No: 80)

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?
After removing 7 rotten bananas, there were 46. So the number of bananas in the bunch before was 7 more than 46.
∴ Number of bananas in the bunch = 46 + 7 = 53
Checking: 53 − 7 = 46

2. Vimala bought some things for 163 rupees, and now she has 337 rupees left. How much did she have at first?
After buying things for 163 rupees, there were 337 rupees left. So, the amount of money that was first in hand was 163 rupees more than 337.
∴ Amount of money at first = 163 + 337 = 500 rupees

3. What number becomes 321 on subtracting 123? 
The number before subtraction is 123 more than 321. 
Number to be subtracted = 321 + 123 = 444 
♦ Textbook Activities (Textbook Page No: 80)
Multiplication and division

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?
The manager's salary is 5 times the peon's salary. 
The manager's salary = 75000 rupees. So the peon's salary is ⅕ of 75000.
75000 × ⅕ = 75000 ÷ 5 = 15000 rupees
∴ The peon's monthly salary is 15000 rupees

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?
When the number of people who went on the trip is multiplied by 2500, it becomes 12,500 rupees. To find the number of people who went on the trip, we can simply divide 12,500 by 2500.
12500 ÷ 2500 = 5
∴ Number of people = 5

3. A number multiplied by 12 gives 756. What is the number?
When the number is multiplied by 12, it becomes 756. To find the number, we can simply divide 756 by 12.
Number: 756 ÷ 12 − 63

4. A number divided by 21 gives 756. What is the number?
To find the number, we can multiply 756 by 21. 
Number 756 x 21 = 15876

♦ Textbook Activities (Textbook Page No: 82, 83)
Different kinds of change

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?
The perimeter of the rectangle = 50 metres 
2 (largest side + shortest side) = 50
If the shortest side is 5 metres, then 2 shortest sides = 2 x 5 = 10 metres
2 largest sides = 50 − 10 = 40 metres
∴ Largest side 40 ÷ 2 = 20 metres.

2. When Anita and her friends 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?
The amount spent = 32 rupees
The price before getting a reduction of 3 rupees = 32 + 3 = 35 rupees
The price of 5 pens = 35 rupees
The price of 1 pen = 35 ÷ 5 = 7 rupees.

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
(i) Before adding 3 = 101 − 3 = 98
    If 2 times the number is 98, then the number is 98 ÷ 2 = 49
(ii) Before adding 2 = 101 − 2 = 99
    If 3 times the number is 99, then the number = 99 ÷ 3 = 33
(iii) Before subtracting 3 = 101 + 3 = 104 
    If 2 times is 104, number = 104 ÷ 2 = 52 
(iv) Before subtracting 2 = 101 + 2 = 103 
    If 3 times is 103, number = 103 ÷ 3 = 34.33

4. The sum of 6 times a number and 4 is 100. What is the number?
Before adding 4 = 100 − 4 = 96
When multiplied by 6 first, it became 96.
So, the original number is 96 ÷ 6 = 16.

5. 4 times the sum of a number and 6 gives 100. What is the number?
Before multiplying by 4 = 100 ÷ 4 = 25
When 6 was added first, it became 25, so the original number = 25 − 6 = 19.

♦ Textbook Activities (Textbook Page No: 84)
Multiple and part

1) The sum of 2 times a number and 7 times the same number is 27. What is the number?
When 2 times and 7 times of any number are added, it becomes 9 times. So, 9 times the number is 27, the number is ⅑ of 27. That is, number = 27 × ⅑ = 3.

2. 99 added to 12 times a number gives 21 times the number. What is the number?
21 times is 99 more than 12 times.
The difference between 21 times and 12 times is 99. 
9 times the number is 99.
Number = ⁹⁹⁄₉ = 11

♦ Textbook Activities (Textbook Page No: 85)

1. A third of a piece of rope cut away leaves 10 meters. What was the original length?
If ⅓ of the rope is cut off the remaining part
= 1 − ⅓ = ³⁄₃ − ⅓ − ⅔ part of the rope.
⅔ part of the total length of the rope = 10 metres.
Total length = ³⁄₂ part of 10 metres
= 10 × ³⁄₂ = 5 × 3 = 15 metres

2. Half the milk in a can and a third of the remaining was used up, and now there is 5 litres remaining. How many litres did it originally contain?
If half and one third of the remaining milk are combined = ½ + (½ + ⅓) = ½ + ⅙
= ³⁄₆ + ⅙ = ⁴⁄₆ = ⅔
Remaining amount of milk = 5 litres
Total milk amount (1 − ⅔) part = 5 litres
1 − ⅔ = ³⁄₃ − ⅔ = ⅓
So ⅓ part of the total milk amount is 5 litres.
Total amount of milk = 5 × 3 = 15 litres 
The can originally contained 15 litres of milk.
♦ Textbook Activities (Textbook Page No: 89)
Algebraic methods

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?
Let's denote the width as x.
Length = 2 x + 1
Given that the perimeter is 80 metres, we can write 2 (length + width) = 80
2 (2x + 1 + x) = 80
2 (3x + 1) = 80
6x + 2 = 80
6x = 80 − 2 = 78
x = ⁷⁸⁄₆ = 13
Width of the rectangle = 13 metres
Length of the rectangle = 2 × 13 + 1 = 27 metres 

2. When a hundred-rupee note was changed to twenty-rupee and ten-rupee notes, seven notes were got. How many of each? 
Number of 20 rupee notes = x
Their value = 20x
Number of 10-rupee notes = 7 − x
Their value 10 (7 − x)
Total value = 20x + 10 (7 − x)
∴ 20x + 10 (7 − x) = 100 
   20x + 70 −10x = 100 
   10x + 70 = 100
   10x = 100 −70
   10x = 30, x = ³⁰⁄₁₀ = 3
∴ Number of 20 rupee notes = 3 
Number of 10 rupee notes = 7 − 3 = 4.

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?
Let's take the price of a pen is x rupees.
The price of a book = x + 4
The price of a pencil = x − 2
The price of 5 books = 5 (x + 4) = 5x + 20
The price of 2 pens = 2x
The price of 3 pencils = 3(x − 2) = 3x − 6 
Total price = 5x + 20 + 2x + 3x − 6 = 10x + 14
∴ 10x + 14 = 54
   10x = 54 − 14, 10x = 40
    x = ⁴⁰⁄₁₀ = 4
The price of a pen = 4 rupees
The price of a book = 4 + 4 = 8 rupees 
The price of a pencil = 4 − 2 = 2 rupees 

4. A square containing four numbers in a calendar is marked, and the sum of these numbers is 80. What are the numbers?
Let's take the first number is x. 
x + x + 1 + x + 7 + x + 8 = 80
4x + 16 = 80
4x = 80 − 16 
4x = 64
x = ⁶⁴⁄₄ = 16
Numbers = 16, 17, 23, 24
♦ Textbook Activities (Textbook Page No: 92)

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? 
Let's take Appu's current age is x.
Mother's current age = 9x
After 9 years,
Appu's age = x + 9
Mother's age = 9x + 9
Mother's age is 3 times Appu's age; that is, 3 (x + 9) = 9x + 9, 3x + 27 = 9x + 9 
9x − 3x = 27 − 9, 6x = 18
x = ¹⁸⁄₆ = 3
Appu's current age = 3
Mother's current age = 3 × 9 = 27

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?
Let's take the number of boys x.
Number of girls = x
On the day 8 boys were absent, the number of boys = x − 8
The number of girls was 2 times the number of boys on that day.
2 (x − 8) = x, 2x − 16 = x, 2x − x = 16, x = 16
Number of boys = 16
Number of girls = 16

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? 
The number of students initially in the class = x 
Number of girls = 50% of x = x × ⁵⁰⁄₁₀₀ = ˣ⁄₂
When 10 more girls arrived 
Total students = x + 10
No. of girls = ˣ⁄₂ +10
Now the number of girls is 60%.
ˣ⁄₂ + 10/x + 10 = ⁶⁰⁄₁₀₀,  ˣ⁄₂ + 10/x + 10 = 0.6
ˣ⁄₂ + 10 = 0.6 (x + 10),  ˣ⁄₂ +10 = 0.6x + 6
0.5x + 10 = 0.6x + 6, 10 − 6 = 0.6x − 0.5x
4 = 0.1x, x = 4/0.1 = ⁴⁰⁄₁ = 40
So, the initial number of students was 40 

4. Another problem from folk-math. Some lotus flowers have bloomed in a pond. A flock birds sat on the flowers to rest. First, one bird sat on each flower, but one bird didn't have a flower to sit. Then two birds sat on each flower and there was one flower extra. How many lotuses and how many birds?
If the number of lotus flowers is x, then the number of birds is (x + 1)
When 2 birds sit on one lotus, one lotus will be extra. 
So, the number of birds is 2 (x − 1)
Number of lotus flowers = (x + 1)
That is,
  2(x − 1) = x + 1, 2x − 2 = x + 1
  2x− x = 1 + 2, x = 3
∴ Number of lotus flowers = 3
   Number of birds = 4