Math5

Exercise 10.3

1. In the given figures, volume of each cube is 1 cu. cm. Count the number of
cubes to find the volume of the given solids:

2. Find the volume of the following cuboids:

3. Find the volume of the following cubes:
Oasis School Mathematics Book-5 201

4. Find the volume of the following cuboids:
a. length = 2.5 cm, breadth = 2 cm, height = 2 cm
b. length = 6 cm, breadth = 4.5 cm, height = 2 cm
c. length = 5 cm, breadth = 3.5 cm, height = 4 cm
d. length = 4 cm, breadth = 3.5 cm, height = 3 cm
e. length = 4.2 cm, breadth = 3.2 cm, height = 2 cm
f. length = 4.8 cm, breadth = 4.2 cm, height = 3.5 cm

5. Find the volume of the following cubes having side:
a. 3 cm b. 5cm c. 4.5 cm d. 3.6 cm e. 4.8 cm

6. Count the number of cubes having volume 1 cu.cm each along length, breadth
and height to find l, b and h of the given figures and then find the volume:

7. a. A room is 4 m long, 3.5 m wide and 2.5 m high. Find the volume of the room.
b. Find the volume of the cubical box having each side of 2.5 m.

1. Consult your teacher 2. a) 24cm2 b) 40cm3 c) 96cm3 d) 200 cm3 3. a) 27cm3
b) 15.625cm3 c) 42.875 cm3 d) 8cm3 4. a) 10cm3 b) 54cm3 c) 70cm3 d) 42 cm3
e) 26.88 cm3 f) 70.56 cm3 5. a) 27cm3 b) 125 cm3 c) 91.125cm3 d) 46.656cm3
e) 110.592 cm3 6. a) 20cm3 b) 40cm3 c) 12cm3 d) 20cm3 7.a) 35cm3 b) 15.625 cm3

202 Oasis School Mathematics Book-5

Worksheet

On a graph paper, draw the figure containing full square and half square.

Figure (ii) Figure (iii)

Figure (i)

Figure (vi)

Figure (iv) Figure (v)

Figure (vii)

Number of Number of 1 Number of corners of Area counting by
corners on the corners enclosed 2 square and half
the figure + Number of
Figure figure by the figure square
corners enclosed by the

figure – 1

1 ×9+5-1=9+4 6 + 1 × 5 = 8 1
2 2 22
(i) 9 (Red dots) 5 (Green dots) = 81

2

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Oasis School Mathematics Book-5 203

Objective Questions

Colour the correct alternatives:
1. Polygon having 7 sides is called

pentagon hexagon heptagon

2. If the length and breadth of a rectangle are 6cm and 5cm respectively, then its
perimeter is

22cm 11cm 30cm

3. A side of a square is 6cm. Its area is equal to

36cm² 24cm² 6cm²

4. Which of the following relations is not true?

Area of square = l² perimeter of rectangle = 4l Area of rectangle = l × b

5. Area of given figure is equal to

5cm² 2.5cm² 10cm²
6. 1 litre is equal to

10ml 100ml 1000ml

7. The area of shaded part in the given figure is 10cm

8cm

48cm³ 72cm³ 120cm³ 12cm 6cm

8. Volume of a cuboid having length 6cm, breadth 4cm and height 3cm is

72cm² 36cm² 144cm²

A

9. The length of the fencing material required to fence the 15m 17m
given triangular field thrice is

52m 104m 156m B 20m C

Number of correct answers

204 Oasis School Mathematics Book-5

Unit Test

Full marks: 15

Attempt all the questions:

1. Find the perimeter of the given figure. 2×2=4

(a) 2.5cm 3.5cm (b) 2.5cm 3cm
3cm
6cm 5cm 4cm
6cm
6.5cm

2. a. Three sides of a triangle are 6.2cm, 5.3cm and 7cm. Find its perimeter. 2
b. Find the perimeter of a rectangle whose length and breadth are 5cm and

6.5cm 1

3. a. Find the area of a rectangle whose length is 7cm and breadth is 6cm. 1
b. Find the area of the given square. 1

6cm

4. Find the area of the shaded part in the given figure. 2
2
12cm
6cm

3cm 10cm

5. Find the volume of the given cuboid.

8cm

10cm 6cm

6. Find the volume of a cube having each side of 5cm. 2

Oasis School Mathematics Book-5 205

UNIT Chart, Bar Graph,

11 Temperature, Ordered
Pairs and Co-ordinates

12 Estimated Teaching Hours: 15
93

6

Contents • Chart
• Bar graph
• Temperature
• Order pairs
• Co-ordinates

Expected Learning Outcomes

Upon completion of this unit, students will be able
to develop the following competencies:

• To study the given chart and find the information from the
given chart

• To study the bar graph and obtain the information from the bar
graph

• To draw the bar graph from the given information
• To read the thermometer of Celsius and Fahrenheit scale
• To write two different sets in order pair
• To write the co-ordinates of the given point
• To plot the given points on the graph sheet

Materials Required : Graph sheet, thermometer, models of bar graph, etc.

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Chart

A chart gives the numerical information. Let's study the given chart.
Number of students in different sectors are shown in the given chart.

Boarders Day Boarders Day Scholars

Boys Girls Total Boys Girls Total Boys Girls Total

124 118 242 105 110 215 340 365 705

Study the chart above properly and answer the following questions.

 In which sector the number of students is maximum?

 In which sector the number of students is minimum?

 What is the number of day boarders? I can answer all the

 What is the total number of girls in the school? questions.

 What is the total number of boys in the school?

 What is the number of day scholars?

Bar graph

Data can be represented by using different types of figures. A figure gives
complete information about the data.
A bar graph consists of bars which are vertical with equal width. The gap
between two bars should be same. To get numerical information, note the
height of the vertical bar corresponding to the reading on the left.
Marks obtained by 5 students of class V is shown below on a bar graph.

Amit Anisha Manish Asmita Ranjan
Oasis School Mathematics Book-5 207

Study the graph above properly and answer the following questions.
Which student got the highest marks?
Which student got the lowest marks?
How much did Anisha score?
How much did Asmita score?
What is the highest mark of the class?

Manish got the highest marks. Asmita got the lowest marks.

Anisha got 20 marks. Asmita got 15 marks.

Highest marks in the class Oh! Its easy to study the bar graph.
is 30.

Class Assignment

The highest marks in different subjects in the second terminal examinations are
given below.

Subject English Nepali Maths Social Studies Science
70 90
Marks 60 80 90

Show the information in the given bar graph.

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Exercise 11.1

1. The given chart shows the number of students in different classes:

Class I II III IV V

Number

of
students 15 25 40 27 23 50 18 19 37 23 16 39 31 12
Boys
Girls
Total
Boys
Girls
Total
Boys
Girls
Total
Boys
Girls
Total
Boys
Girls
Total

43

Study the chart above properly and answer the questions given below:

a. What is the total number of students in

i. class I? ii. class II? iii. class III? iv. class IV? v. class V?

b. What is the total number of students in the school?

c. Which class has the maximum number of boys?

d. Which class has the highest number of students?

e. What is the total number of boys in the school?

f. What is the total number of girls in the school?

2. The given bar graph shows the favourite fruits of class V students. Study the
bar graph properly and answer the following questions:

Number of students

Banana Grapes Orange Pear Mango

a. Which is the most popular fruit in the class?
b. Which is the least popular fruit in the class?
c. How many students like banana?
d. How many more students like oranges than pears?
e. How many students are there in class V?

Oasis School Mathematics Book-5 209

3. The given table shows the marks obtained by Arya in different subjects.
Draw bar graph to represent the information:

Subjects English Nepali Mathematics Science Social
Studies

Marks 35 30 40 30 20

4. The given table shows the number of students present in class V in a week.
Draw the bar graph to show the information:

Days Sunday Monday Tuesday Wednesday Thursday Friday
Number of 28
present students 30 32 25 30 35

Temperature Answers: Consult your teacher

Temperature means hotness or coldness of an object. The degree of hotness or
coldness of an object is expressed by its temperature.

The instrument which is used to measure the temperature is a thermometer.

Celcius scale thermometre

Fahrenheit scale
thermometre

The first figure is a thermometer with Celsius scale and the second is
thermometer with Fahrenheit scale.
The reading of the first thermometer is 35°C.
The reading of the second thermometer is 97°F.

Exercise 11.2

1. Write down the temperature shown by the following thermometers with Celsius scale.
a.

b.

c.

d.

e.

210 Oasis School Mathematics Book-5

2. Write down the temperature shown by the following thermometers with
Fahrenheit scale.

a.

b.

c.

d.

3. Use the thermometer of Fahrenheit scale to measure :
a. your temperature b. temperature of your friend

4. From a daily magazine, collect the record of maximum and minimum
temperature of Dhankuta, Kathmandu, Pokhara, Birendranagar and Dipayal.
Answer the following questions:
a. Which place is the hottest place?
b. Which place is the coldest place?
c. What is the difference between the maximum and minimum
temperature of Kathmandu?
d. What is the difference between the maximum and minimum
temperature of Pokhara?
Answers: Consult your teacher

Activity
Use Celsius scale thermometer to record the temperature at 9 am, 10 am,
11 am, 12 noon, 1 pm, 2 pm and 3 pm and observe the rise and fall in the
temperature.

Do you know?
Normal temperature of human body is 98.6°F.
Water freezes at 0°C or 32°F.
Water boils at 100°C or 212°F.

Oasis School Mathematics Book-5 211

Order pairs

Let’s take any two sets
A = {Nepal, India, China} B = {Kathmandu, New Delhi, Beijing}
A is a set of countries.
B is a set of capitals.
The countries and capitals can be written in the ordered pairs.
{(Nepal, Kathmandu), (India, New Delhi), (China, Beijing)}
Here are three pairs, first element of the pair represents country and the second
element represents capital. Such pairs are called ordered pairs.

Class Assignment

1. Let,
A = {Kailali, Kanchanpur, Bardiya}
B = {Dhangadi, Mahendranagar, Gulariya}
Make the ordered pair representing a district and its headquarters.
{(.................., ..................), (.................., ..................), (.................., ..................)}
2. Let A = {2, 3, 4}
B = {4, 9, 16}
Make the ordered pair representing a number and its square.
Order pair = { }

Co-ordinates

In the figure below, OX represents X-axis and OY represents Y-axis. While finding

the position of a point we have to count the division along X-axis and the division

along Y-axis. Count horizontally on X-axis and vertically on Y-axis .

Position of point ‘A’ counting horizontally from the left = 1 unit

Position of ‘A’ counting vertically from the bottom = 1 unit

\ position of point A = (1,1)

i.e. Co-ordinates of point A = (1,1)

For the position of point B, K
Counting horizontally from the left = 3 units
Counting vertically from the bottom = 2 units N

\ Position of point B = (3,2) P L
i.e. Co-ordinates of point B = (3,2) M
Similarly, co-ordinates of C = (3,4)

Co-ordinates of D = (5,3)

Now, we can also find the co-ordinates of other points also.

212 Oasis School Mathematics Book-5

Let's complete the given table.

Points Co-ordinates Points Co-ordinates
E J
F K
G L
H M
I N

Remember !

• Thenumberofdivisioncountedhorizontally(alongX-axis)representsX-coordinate.
The number of division counted vertically (along Y-axis) represents Y-coordinates.

Co-ordinates of point E = (6,4) I understand, first element of the pair is
X- co-ordinate of E = 6 X-co-ordinates and second element of the
Y- co-ordinate of E = 4
pair is Y-co-ordinate.

Exercise 11.3

1. The following sets give the numbers and their squares. Express them in ordered pairs:
A = {1, 2, 3, 4} B = {1, 4, 9, 16}

2. Write the co-ordinates of each of the following points:

3. Plot the following points in the graph paper:

a. A(2,3) b. B(4,7) c. C(3,4) d. D(4,5)
h. H(6,5)
e. E(7,2) f. F(5,3) g. G(1,4)

Answers: Consult your teacher

Oasis School Mathematics Book-5 213

Objective Questions

Colour the correct alternatives:
A. The given chart shows the number of students of different Houses.

Red House Blue House Green House Yellow House

Boys Girls Total Boys Girls Total Boys Girls Total Boys Girls Total

92 88 180 79 91 170 93 96 189 86 92 178

1. The number of boys in Red House is

92 93 86

2. The number of girls in Yellow House is 92
88 96

3. Which House has the highest number of students?

Red Green Yellow

4. How many more students are there in Red House than in Yellow House?

10 9 2

B. The given bar graph shows the marks obtained by Dorje in four different

subjects. 50

5. Marks obtained by Dorje in Maths is 40

30

35 45 40

20

10

0 Nepali English Maths Science

6. In which subject did he get the lowest marks?

Nepali Science English

7. How many more marks he got in Maths than in Science?

5 10 20

Number of correct answers

214 Oasis School Mathematics Book-5

Unit Test Full marks: 14

Attempt all the questions:

1. The given chart shows the number of students in different sections of class V.
Read the given chart properly and answer the questions given below: 4

Section A Section B Section C Section D

Boys Girls Total Boys Girls Total Boys Girls Total Boys Girls Total

22 13 35 18 16 34 14 21 35 15 21 36

a. Which section has the highest number of students?
b. Which two sections have equal number of students?
c. What is the total number of boys in class V?
d. What is the total number of girls in class V?

2. The given table shows the maximum temperature on June 15, 2015 of five
different cities.

City Biratnagar Kathmandu Pokhara Nepalgunj Dhangadhi
Maximum 35° 30° 30° 40° 40°
temperature

Represent the above information in the bar graph. 4

3. Write down the reading of the given thermometers. 2

4. Plot the following points on the graph paper. 4
A(2, 3), B(5, 1), C(3, 7), D(5, 4)
E(1, 2), F(7, 3), 4(8, 1), H(6, 6)

Oasis School Mathematics Book-5 215

UNIT Bills and Budget

12 and Sets

12 Estimated Teaching Hours: 15
93

6

Contents • Bill
• Budget
• Sets

Expected Learning Outcomes

Upon completion of this unit, students will be able
to develop the following competencies:

• To prepare the bill
• To read the bill and extract information from the bill
• To get the information from the budget
• To know the meaning of sets
• To represent the set in different ways

Materials Required : Sample of bill, Chart of household budget, etc.
216 Oasis School Mathematics Book-5

Bills

When we go to buy something from a shop, the shopkeeper gives us a bill. It
has the details about the quantity and price of the thing that we buy.
Look at the given bill properly and learn to prepare it.

Atharai Books Distributor Remember !
Anamnagar, Kathmandu
A bill has
Mr/ Mrs/ Ms: Dayakanta Mishra Date: 067 / 10 / 25 • name and address of

1. Oasis School Mathematics 15 Rs 210 Rs 3,150 shop or store
V • bill number
• date of issue of the bill
2. Oasis School Science V 20 Rs 195 Rs 3,900 • quantity of each item
• rate of each item
3. Exercise book 25 Rs 30 Rs 750 • total cost of all items
• signature of
4. Pen 45 Rs 35 Rs 1,575
salesperson

Total Rs 9,375

................................
Salesman

Budget

A budget shows information about the income and expenditure under different
headings.

The monthly expenditure of a household is given below.

Income Expenditure

Title Rs Title Rs

Salary Rs 15,000 Food Rs 12000

Business Rs 25,000 Education Rs 6000

Clothes Rs 14000

Miscellaneous Rs 4000

Rs 40,000 Rs 36,000

Oasis School Mathematics Book-5 217

From the budget information above, we should be able to answer the following
questions.
• What is the income of the household from the salary?
• What is the total income of the household?
• What is the total expenditure of the household?
• What is the expenditure of the household on food?

I can answer all the questions. Income from salary = Rs. 15000
Total income = Rs. 40,000

Total expenditure = Rs. 36,000
Expenditure on food = Rs. 12000

Exercise 12.1

1. Bikalpa purchased the following items from Manisha Books and Stationery
Butawal. Prepare the bill given to him by the shopkeeper:

Rs. 10 Rs. 35 Rs. 150 Rs. 45

2. Aadhya bought the following items from Arun Stores Nepalgunj. Prepare the
bill the shopkeeper issued her.
a. 6 kg rice at the rate of Rs 60 per kg.
b. 3 kg sugar at the rate of Rs 65 per kg.
c. 1 kg tea at the rate of Rs 350 per kg.
d. 2 litres oil at the rate of Rs 120 per litre.
e. 5 soaps at the rate of Rs 35 per soap.

218 Oasis School Mathematics Book-5

3. Ankit , Anishma and Anasuya went to Kids Fun Corner:

• Ankit ordered one plate

vegetable Mo : Mo and a

cup of tea. KIDS FUN CORNER
• Anishma ordered one plate Menu

chicken Mo : Mo and a Vegetable Mo:Mo - Rs 65 per plate
coffee.
• Anasuya ordered a tandoori Chicken Mo:Mo - Rs 80 per plate
chicken and a tea.
Prepare the bill. Tandoori chicken - Rs 220

Tea - Rs 10

Coffee - Rs 25

4. Monthly budget of a household is given below:

Income Expenditure
Rs
Title Title Rs
Salary Rs 22,000
Rent Rs 25000 Food Rs 18,000
Business Rs 30,000
Education Rs 15,000

Colthes Rs 16,000

Miscellaneous Rs 10,000

Total Rs 77,000/- Total Rs 59,000

Study the budget above and answer the following questions:
a. What is the total income of the household?
b. What is the total expenditure of the household?
c. What is the income of the household from the business?
d. What is the expenditure of the household on clothes?
e. What is the monthly saving of the household?
f. On which title does the household spend the least?

Answers: Consult your teacher

Oasis School Mathematics Book-5 219

Sets
Review:
The meaning of set is a ‘group’ or ‘aggregate’. We can say that collection of
objects is a set like a collection of books, a collection of flowers, a collection of
odd numbers etc.

It is the collection of flowers.
It is a set of flowers.

It is the collection of fruits.
It is a set of fruits.

It is the collection of animals.
It is the set of animals.

In each collection, there are objects of the same nature. Hence a set is the
“collection of well defined objects.” The objects of the set are called its members
or elements.
Example :
Identify which of the following collections are sets?
a. collection of birds.
b. collection of intelligent students
c. collection of honest people of Nepal
d. collection of vowel letters
Solution:
a. The collection of birds is a well defined collection, because we can specify

the birds. So it is a set.
b. The collection of intelligent students is not a well-defined collection

because we can’t specify who is intelligent. So it is not a set.
c. The collection of honest people is not a well-defined collection because

we cannot specify who is honest.
\ The collection of honest people is not a set.
d. The collection of vowel letters is a well defined collection because we can

specify the vowel letters.
\ The collection of vowel letters is a set.

220 Oasis School Mathematics Book-5

Representation of set:

The sets are generally denoted by capital letters A, B,
C, D..... Its elements are denoted by small letters a, b,
c, d.....

A set can be represented by pictures of its objects in a
circle, rectangle, square etc.

Sometimes, it is difficult to represent a set by means of pictures. The other
methods to represent sets are:

a. Listing method (tabulation method)

b. Description method

c. Set builder method

Listing method:
In this method, all the elements of the set are listed within curly braces { }
separating each element by commas.

A = {potato, tomato, cauliflower, cabbage}

A set of vegetables.

B = {elephant, tiger, rhino, lion}

A set of wild animals.

Description method:

In this method, we describe the type of members of the set by descriptive
phrase.
For example:

A = {a set of even numbers less than 10}

B = {a set of vowel letters}

Set builder method:

In this method, the elements or members of a set are represented by a variable
and the common properties of elements are described by the variables.

A = {x : x is a prime number less than 10}

It is read as, A is the set of element x such that x is a prime number less than 10.

B = {y : y is a factor of 6}

It is read as 'B is the set of element y such that y is a factor of 6.

A = {1, 3, 5, 7, 9} A = {a set of odd A = {x : x is an odd

numbers less than 10.} number less than 10}

Listing method Description method Set builder method

Oasis School Mathematics Book-5 221

Member of a set:

Let A = {a, b, c, d, e}

It is a set of first five English letters.

Here, ‘a’ is a member of A.

‘a’ belongs to A.

i.e. ‘a’ ∈ A

‘f’ is not a member of A

‘f’ does not belong to A.

f ∉A.

Example :

Express the given set in:

i. listing method Baisakh, Jestha, Asar,
ii. description method Shrawan, Bhadra
iii. set builder method

Solution:

i. Expressing the above set in listing method,

A = {Baisakh, Jestha, Asar, Shrawan, Bhadra}

ii. Expressing the above set in description method,

A = {a set of first five Nepali months}

iii. Expressing the above set in set builder method,

A = {x:x is first five Nepali months}

Exercise 12.2

1. Identify whether the following collections are sets or not:

a. collection of animals

b. collection of even numbers from 1 to 10

c. collection of honest students of the class
d. collection of tall students of the class

e. collection of planets

f. collection good cricket players

2. Cross the odd element of the following sets and write the name of the set of rest

of the elements:

a. b. c. d.

1 5 Sunday, Tuesday a b e
6 7 9 Friday, Magh i o u

222 Oasis School Mathematics Book-5

3. List the elements of the following sets within braces by using listing method:
a. The set of first four days of the week.
b. The set of first five English letters.
c. The set of even numbers from 1 to10.
d. The set of prime numbers from 1 to 10.
e. The set of any three fruits.

4. Rewrite the following sets in description as well as set builder method:

a. A = {a, b, c, d} b. B = {2, 4, 6, 8}

c. A = {cow, buffalo, goat} d. B = {2, 3, 5, 7}

e. C = {1, 2, 3, 4, 5, 6, 7, 8, 9}

5. List the following sets:
a. A = {x : x is an odd number less than 10}
b. B = {x : x is the factor of 8}
c. C = {x : x is a square number less than 30}
d. A = {x : x is the first five multiples of 5}

6. Rewrite the given sets by listing, description as well as set builder method:

7. Write ∈ or ∉ in the blanks:
a. A = {a, e, i, o, u}
a.......A, b.......A, c.......A, e.......A, o.......A, u.......A
b. B = {2, 4, 6, 8, 10, 12}
2.......B, 4.......B, 7.......B, 9.......B
c. C = {x:x is a prime number less than 20}
2.......C, 6.......C, 11.......C, 5.......C, 10.......C
d. D is a set of first five multiples of 5
3.......D, 5.......D, 10.......D, 14.......D

Answers: Consult your teacher

Oasis School Mathematics Book-5 223

Types of sets:
Depending upon the number of elements or members of sets, the following
are its types.
Empty set or Null set

A set having no element is called an empty set or null set. The symbol for this
type of set is f or { }.
e.g: A set of boys in Padma Kanya College
Unit set

A set having only one element is called unit set. It is also known as singleton set.
e.g: A set of the highest peak of the world
A set of capital city of Nepal
In both sets there is only one element, they are unit sets.
Finite set

A set having a finite number of elements is called a finite set. e.g.: ‘A’ = {2, 4, 6, 8, 10}
In set ‘A’ there are finite elements.
\ A is a finite set.
Infinite set:

A set having an unlimited number of elements is called an infinite set. Its
elements are countless.
Example, A is a set of even numbers
A = {2, 4, 6, 8, 10, 12, ....}
There are unlimited even numbers. So A is an infinite set.
Example :

Identify whether the following sets are null, unit, finite or infinite.
i. A = {2, 4, 6, 8}
ii. B = A set of odd numbers
iii. C = A set of capital city of Nepal
iv. D = A set of odd numbers which are multiples of 2
Solution:
i. A = {2, 4, 6, 8}
Number of elements in A = 4
Since ‘A’ has a finite number of elements, A is a finite set.
ii. B = a set of odd numbers.
B = {1, 3, 5, 7, 9......}

224 Oasis School Mathematics Book-5

There are unlimited odd numbers.
\ B is an infinite set
iii. C = A set of capital city of Nepal.
C = {Kathmandu}
There is only one element in set C. So C is a unit set.
iii. D = A set of odd numbers which are multiples of 2.
None of the odd numbers is multiples of 2.
\ D is a null set.

Exercise 12.3

1. Identify whether the following are unit or null sets:
a. A = φ b. B = {3}
c. A = a set of prime numbers from 4 to 6
d. C = a set of people 10 ft tall.

2. Identify whether the following sets are finite or infinite:
a. A = {2, 3, 5, 7}
b. B = {1, 2, 3, 4,......}
c. A = A set of first five English months
d. C = A set of even numbers
e. D = A set of odd numbers less than 10
f. A = A set of multiples of 5

Answers: Consult your teacher

Set relation:
Depending upon the nature and the number of elements in two sets, the
following are their relations.
Equal sets
Let A = { a, b, c, d, e}
B = {a, b, c, d, e}
In both sets ‘A’ and ‘B’ the number of elements is 5.
In both sets all elements are same.
\ A and B are equal sets.
Hence, two sets ‘A’ and ‘B’ are said to be equal sets if they have same elements
and equal number of elements.

Oasis School Mathematics Book-5 225

Equivalent sets
Let A = {1, 2, 3, 4}

B = {a, b, c, d}

In both sets ‘A’ and ‘B’, the number of elements is 4. So ‘A and ‘B’ are equivalent
sets.
Hence two sets ’A’ and ‘B’ are said to be equivalent sets if they have equal
number of elements.
Overlapping sets
Let, A = {a, b, c, d, e}
B = {a, e, i, o, u}
Here, two elements ‘a’ and ‘e’ belong to both sets. Such sets are overlapping
sets.
Hence, two sets ‘A’ and ‘B’ are said to be overlapping sets if they have at least
one common element.
Disjoint sets
Let A = {1, 3, 5, 7, 9}
B = { 2, 4, 6, 8, 10}
In both sets, no element is common. ‘A’ and ‘B’ are disjoint sets. Hence, two sets
‘A’ and ‘B’ are said to be disjoint sets if they do not have a common element.

Example :

Identify whether the following sets are equal or equivalent.
a. A = {1, 2, 3, 4, 5} and B = {a, e, i, o, u}
b. A = {2, 4, 6, 8} and B = {8, 6, 4, 2}
Solution:
a. Number of elements in set A = 5
Number of elements in set B = 5
Since the number of elements in both sets are equal A and B are equivalent

sets.
b. In both sets A and B, elements are same and the number of elements is also

same.
\ A and B are equal sets.

226 Oasis School Mathematics Book-5

Example :
Identify whether the following are overlapping or disjoint sets.
a. A = {5, 6, 7, 8,}, B = {1, 2, 3, 4}
b. A = {a, b, c, d, e}, B = {a, e, i, o, u}
Solution:
a. Here, A = {5, 6, 7, 8} and B = {1, 2, 3, 4}. None of the elements from the two sets is

common.
\ A and B are disjoint sets.
b. Here, A = {a, b, c, d, e}, B = {a, e, i, o, u}
Two elements 'a' and 'e' are common in both sets.
\ A and B are overlapping sets.

Exercise 12.4

1. State whether the following are equal or equivalent sets:
a. A = {10, 15, 20, 25, 30} and B = {30, 20, 10, 15, 25}
b. X = {Lily, Rose, Sunflower, Marigold}, Y = {1, 2, 3, 4}
c. P = a set of even numbers less than 9, Q = {2, 4, 6, 8}
d. C = {tiger, elephant, rhino, lion}, D = {cow, buffalo, goat, dog}

2. State whether the following are overlapping or disjoint sets:
a. A = {p, q, r, s, t}, B = {s, t, u, v, w}
b. A = {1, 4, 9, 16, 25}, B = {1, 8, 27, 64, 125}
c. P = {11, 13, 15, 17, 19}, Q = {10, 12, 14, 16, 18}
d. X = {x : x is an even number less than10}
Y = {x : x is a prime number less than 10}
e. P = a set of multiples of 3 less than 20
Q = a set of multiples of 7 less than 20
f. A = a set of factors of 12
B = a set of factors of 18

Answers: Consult your teacher

Oasis School Mathematics Book-5 227

Objective Questions

Colour the correct alternatives:

A. Income and expenditure of a household is given in the table,

Income Expenditure

Business Rs 80,000 Food - Rs 25,000

Clothes - Rs 20,000

Education - Rs. 20,000

1. The total expenditure of the household is

45,000 65,000 25,000

2. The total saving of the household is

15,000 20,000 25,000

3. Onwhichtitlethehouseholdhasmaximumexpenditure?

Food Cloths Education

4. Which of the following collection is not a set?

Collection of first Collection of clever Collection of even
5 English letters students of class V numbers from 1 to 10

5. Among the given elements, the odd one is

Football Rose Basketball

6. A set of capital city of Nepal is

null set unit set infinite set
infinite set
7. Set of multiples of 2 is unit set
null set

Number of correct answers
228 Oasis School Mathematics Book-5

Unit Test Full marks: 15

Attempt all the questions:

1. Lata bought some goods from Arun Store Basundhara, Kathmandu.

Prepare the bill. 4

She bought 2 toothpaste tubes Price list
2 soaps
2 leaves of bread Tooth paste - Rs. 80
and 5 kg rice
Soap - Rs. 35

Bread - Rs. 65

Rice - Rs. 115 per kg

Sugar - Rs. 85 per kg

2. Monthly income and expenditure of Yam Nath Adhikari is given below:

Income Expenditure
Salary : Rs. 50,000 Food: Rs. 20,000
House rent: Rs. 30,000 Cloths: Rs. 24,000
Business: Rs. 80,000 Education: Rs. 16,000
Miscellaneous: Rs. 20,000
Total Rs. 1,60,000 Total : Rs. 80,000

Study the above table and answer the questions given below: 5
a. What is the total income of Yam Nath Adhikari?
b. What is the total expenditure of Yam Nath Adhikari?
c. What is his income from business?
d. On which title he spends most money?
e. What is his main source of income?

3. Write the following set of elements in listing method. 4
a. A set of even numbers less than 10
b. A set of prime numbers less than 10
c. A set of vowel sounds
d. A set of first three English month

4. Identify whether the given sets are null or unit set. 2
a. A set of capital city of Nepal
b. A set of men having height 9 ft.

Oasis School Mathematics Book-5 229

UNIT

13 Algebra

12 Estimated Teaching Hours: 10
93

6

Contents • Constant and variables
• Algebraic terms and expressions
• Types of algebraic expressions
• Factors and co-efficients
• Like and unlike terms
• Addition and subtraction of algebraic

expressions
• Multiplication of algebraic expressions
• Division of algebraic expressions
• Algebraic equations

Expected Learning Outcomes

Upon completion of this unit, students will be able
to develop the following competencies:

• To identify constants and variables
• To form algebraic expressions from the given statement
• To identify the type of algebraic expressions
• To identify the factor co-efficient and power from the given

term
• To identify the like and unlike terms
• To add and subtract the algebraic expressions
• To multiply and divide the algebraic expressions
• To divide the polynomials by monomials
• To solve the algebraic equation

Materials Required : Flash cards, chart paper, A4 size paper, etc.

230 Oasis School Mathematics Book-5

Review

Algebra is the generalised form of arithmetic. In arithmetic, we use the numbers
1, 2, 3, 4..., etc. which have fixed value. In algebra, quantities are represented
by the symbols a, b, c, d,.... x, y, z etc. which may have any value.

Constants and variables:
Constant: The numbers 1, 2, 3, 4, 5..., etc. can be used in algebra to represent
different values of quantities. These numbers are called constants.
Variables: The letters like a, b, c, .....x, y, z can be used in algebra to represent
different values of quantities. These letters are called variables.
Height of Mount Everest is a constant.
Temperature of Kathmandu is a variable.

Algebraic terms and expressions:

x, y, 2x, 5y, 6z are algebraic terms. If two (5x² + 3xy) expression
or more algebraic terms are connected by term term
arithmetic operations, such expression is
called an algebraic expression.

x + y, 3x + 5, 6y + z, etc. are algebraic expressions.

Formation of algebraic expressions:
The arithmetic operations can be used to connect variables and constants. See
some examples of algebraic expressions.

Example : Use (+) sign in case of ‘more than’,
‘is added to’ and ‘increase by’.
5 more than x means x + 5
y is added to z means y + z Use (-) sign in the case of ‘less than’ ‘is
Increase x by 8 means x + 8 subtracted from’ and is ‘decreased by’.
Again,
6 less than x means x - 6
Decrease z by 2 means z - 2
7 is subtracted from x means x - 7
2 is decreased by y means y -2

Oasis School Mathematics Book-5 231

Again, Use (×) sign in case of product and
5 times x means 5x ‘multiplied by’.
The product of 6 and y means 6y
z is multiplied by 8 means 8z Use (÷) sign in case ‘quotient of’ and
‘is divided by’.
Again, 4mmeaenasnsy5 x4

Quotient of x by
y is divided by 5

Types of algebraic expressions:
Monomial: An algebraic expression having only one term is called monomial.
5, 4x, 6y, etc. are monomials.
Binomial : An algebraic expression having two terms is called binomial. Eg.

(x + 2), (x + y), (3x + 4), etc. are binomials.
Trinomial: An algebraic expression having three terms is called trinomial. Eg.
(x + y + z), (2x + y + 2), (3x - y + 2), etc. are trinomials.

Evaluation of terms or algebraic expression:

If there is numerical value of each of the literal, the total value of algebraic
expression or a term can be calculated. Example,
If x = 2, y = 3 and z = 4, then the value of 2x - 3y + 4z is
2x - 3y + 4z = 2 × 2 - 3 × 3 + 4 × 4
= 4 - 9 + 16
= 20 - 9
= 11

Factors and co-efficients:

We know that 5 × 7 = 35

5 and 7 are the factors of 35.

x × y = xy ‘x’ and ‘y’ are the factors of xy.

a × b × c = abc ‘a’, ‘b’ and ‘c’ are the factors of abc.

2 × x × y = 2xy 2, x and y are the factors of 2xy.

232 Oasis School Mathematics Book-5

Again,
2 × x = 2x
2 and x are the factors of 2x.
2 is the co-efficient of x.
‘x’ is the literal co-efficient of 2.
Again,
x × y × z = xyz
‘x’, ‘y’ and ‘z’ are the factors of xyz.
Literal co-efficient of yz = x
Co-efficient of xyz = 1

Base and power:

In x², power of x is 2.
2 is the power and x is the base.
In y5, 5 is the power and y is the base.
In,

Class Assignment

1. Complete the given table

Term Factors Co-efficient Literal co-efficient of x
Power Co-efficient
3xy

5xy²z³

xy5z6

2. Complete the given table

Term Base
6x³
7x8
10x²
5x7

Oasis School Mathematics Book-5 233

Exercise 13.1

1. State whether the following letters are constants or variables:
a. ‘x’ denotes the number of fingers in human body.
b. ‘y’ denotes the temperature of Kathmandu.
c. ‘a’ denotes the number of sides of a quadrilateral.
d. ‘b’ denotes the number of public vehicles in Kathmandu.

2. Identify whether the given are an algebraic term or an expression:

a. 2x + 3y b. 5x c. x + y + z d. x e. 3

f. 4x + y + z g. x -y + z h. 2a + 3b + c i. y

3. Write the expression for each of the following statements:

a. 1 more than x b. z is decreased by 5

c. 2 less than y d. 3 times z

e. x is increased by 3 f. z is decreased by 5

g. x is decreased by 10 h. 5 times of y

i. z is multiplied by 3 j. twice of x is subtracted by 2

k. 7 is added to three times of x l. 5 is added to six times of y

m. 3 times of x is added to 2 times of y

n. 4 is added to 5 times of x o. the quotient of x to 3 is added to 6

4. State whether the following expressions are monomial, binomial or trinomial:

a. 5xy b. 3x + 2y c. a + b + c

d. 2x + 3y e. x + y + 2z f. 3a

g. 5x² - 2xy h. x² + 2xy + y² i. 2p + 3q - 4r

j. 4c k. a - 4c l. 2m + 3n + 5c

5. If a = 1, b = 2, c = 3, d = 0, x = 4, y = 5 and z = 6, find the value of:

a. 3ab b. 2a + 5b c. a + b +c

d. 2x - y + z e. 5x + 6y - 2z f. 3x - y + z

g. 2a + b - c h. 3a + b - 4d i. a + b + c + d

j. 4xyz - 2abcd k. (2x + y - 2z) × a l. (2a + b) ÷ x

m. (3c + 2d) ÷ (x + y) n. a² + b² o. x² + xy + y²

234 Oasis School Mathematics Book-5

6. a If l = 3 cm, b = 2 cm, find the value of P, if P = 2(l + b).
b. If l = 5 cm, b = 4 cm, find the value of A, if A = l × b.
c. If l = 3 cm, b = 4 cm and h = 5 cm, find the value of V, if V = l × b × h.
d. If l = 3 cm, b = 2 cm and h = 4 cm, find the value of S, if S = 2(lb + lh + bh).

7. If x = 2 cm, y = 4 cm and z = 5 cm, find the total length of the given line segments:

8. Name the co-efficient, base and power of:

a. 3x² b. 5x7 c. 10x³ d. 7y6 e. 6z³

9. If x × y × z = xyz:
a. What are the factors of xyz?
b. What is the numerical co-efficient of xyz?
c. What is the literal co-efficient of x?
d. What is the literal co-efficient of z?

1. Consult your teacher, 2. Consult your teacher, 3. Consult your teacher,
4. Consult your teacher 5. a) 6 b) 12 c) 6 d) 9 e) 38 f) 13 g) 1 h) 5 i) 6 j) 480 k) 1
l) 1 m) 1 n) 5 o) 61 6. a) 10 cm b) 20 cm2 c) 60 cm2 d) 52cm2 7. a) 10 cm
b) 21 cm 7. Consult your teacher, 8. Consult your teacher, 9. Consult your teacher.

Like and unlike terms:

x², 5x², 7x² are like terms.
3ab, 7ab, 10ab are like terms. 3 oranges, 5 oranges, 6 oranges are like terms
4y, 8y, 10y are like terms.
Hence, the terms with same variables with the same power or exponent are
called like terms.
5a², 2b², 3x² are unlike terms 2ab, 4y², 6xy are unlike terms
Hence, the terms which have different variables or same variables with
different power or exponent are called unlike terms.

Oasis School Mathematics Book-5 235

Addition and subtraction of algebraic expressions

Addition and subtraction of algebraic expressions can be done only if the like
terms are like.

While adding or subtracting algebraic expressions, we should just add or
subtract the co-efficient of like terms.

Example :

6a + 4a + 3a Again, Again,
= (6 + 4 + 3)a - 5x - 4x - 2x 2y - 7y + 9y
= 13a = -(5 + 4 + 2)x = (2 - 7 + 9)y
= -11x = 4y
Way of addition:

Addition of algebraic expression can be done in two methods

a. Vertical arrangement method b. Horizontal arrangement method
Vertical arrangement method:
Example :

Add (3a + 4b + 5c) and (2a + b + 6c)

Here, a + 4b + 5c • Write each expression in separate rows in
such a way that like terms are arranged one
below the other in column.
+ 2a + b + 6c
• Add the like terms in each column.
5a + 5b + 11c

Horizontal arrangement method:

Let's see an example,

Add: (5x - 2y + 6z) and (2x + 5y - 2z)

Solution: • Write the given expressions
(5x - 2y + 6z) + (2x + 5y - 2z) horizontally.
= 5x - 2y + 6z + 2x + 5y - 2z
= 7x + 3y + 4z • Collect the like terms and simplify.

Ways of subtraction:

As in addition, subtraction of two algebraic expressions can also be done in
two methods

a. Vertical arrangement method b. Horizontal arrangement method

236 Oasis School Mathematics Book-5

Vertical arrangement method:

Subtract: (4x - 3y + 6z), from (5x + y - 3z)

Here, • Rewrite the given expressions in two
5x + y - 3z lines such away that lower lines is the
expression to be subtracted and like
4x - 3y + 6z terms of both expressions are one below
the other.
(–) (+) (–)

x + 4y - 9z

• Change the sign of each term of the lower
line.

• Combine the terms column wise.

Horizontal arrangement method:

Let's see an example,

Subtract: (5a - 2b - 3c) from (6a + 2b - 2c)

Here, • Enclose the expression to be subtracted
(6a + 2b - 2c) - (5a - 2b - 3c) in the brackets with a minus (-) sign.
= 6a + 2b - 2c - 5a + 2b + 3c
= a + 4b + c • Open the bracket with minus sign.

• Combine the like terms.

Exercise 13.2

1. Add or Subtract:

a. 5x + 3x b. 7a - 4a c. 8b + 12b d. 8b - 5b

e. 3x + 2x + 5x f. 5a² + 2a² g. 3mn + 6mn + 2mn h. 11mn - 7mn

i. 7a² - 5a² j. 4p - 8p k. 2xy + 5xy + xy l. 5x² + 2x² + x²

m. n² + 3n² + 2n²

2. Simplify:

a. 6x - 3x - 5x b. 7a² - 6a² - 2a² c. 5ab - 2ab + 3ab
f. n² - 3n² + 4n²
d. 3xy + 5xy - 2xy e. 4mn - 2mn + 6mn i. 3a²b + 4a²b + 5a²b

g. - 5y - 6y + 8y - y + 2y h. 3x - x - 2x + 8x - 7x

3. a.

Find AB + BC + AC.

Oasis School Mathematics Book-5 237

b.

Find PQ + QR + RS + PS.

4. Add the following:

a. 5a + 2b b. 2a + 7b c. 4x - 3y d. x + 2y + 3z
+ 8a - 6b + 6a + 3b + 8x + 6y + 2x + y + 2z

5. Add:
a. 2a + 3b + 5c and a - 2b + 3c
b. x² - xy + 2y² abd 2x² - 5xy + 3y²
c. 2x² + 5 - 3x and 3x - 2x² + 8
d. 5a + 4b + 6c and - 3c + 2b - 8a
e. -2m + 3n + 4p and 2m + 3n - 2p
f. 4xy - 3yz + 2zx, 2xy + 5yz - 3zx and xy + yz + zx
g. a² - ab + 2b², 4a² + 3ab + b² and 5a² - 2ab + b²
h. 3m² + 2n² - 4mn, m² - mn + n² and m² + mn + n²
i. 4xy - 7yz + 8zx, 3xy + 2yz - 4zx and 5xy - 16yz - 12zx
j. 8x - 6y - 8z, 5x + 3y + 6z and 5x - 3y + 5z

6. Subtract:

a. 2x + 5 from 3x + 7 b. 2a + b + c from 3a + 2b + 2c

c. 2x + y from 5x + 6y d. m + n - 2p from 4m - 2n + 3p

e. 8x + 6y - 4z from 9x + 8y + 2z f. 3a² - 2a + 4 from 4a² - a + 6

g. 5a² + 3ab + 5b² from 3a² + 2ab + 8b² h. 4x³ - 6x² - 2xy from 2x³ + 5x²

i. 3x² + 2xy + 5y² from x² - xy - 2y² j. 4m² + 6mn - n² from 2m² + 3mn - n²

238 Oasis School Mathematics Book-5

7. Simplify: b. 5x² + 6xy + 2y² - 3x² + 2xy - y²
a. 3x + 4y - 2x - 3y d. a²b + b²c + c²a - 2a²b - 3c²a + 4b²c
c. 5x² - 2xy + 3y² - x² + 6xy - 2y²

8. a. What should be added to 3x - 4y + 6z to make 7x - y + 4z?
b. By how much 4x - y + z is greater than x - 2y + 3z?
c. From what 2a + 3b - c be subtracted to get 3a - b -c?
d. By how much x² + xy + y² is less than 2x² - xy?
e. What should be subtracted from 3m² - 4n² to make 2m² - mn + 6n²?
f. Take a²b - b²c + 2c²a from 2a²b - 6b²c + c²a.

9. If x = a + 3b - 4c, y = 2a - 5b + c and z = a + b + c, find the value of x + y + z.

10. Find the perimeter of the given figures:
a. b.

1. Consult your teacher 2. Consult your teacher 3. Consult your teacher
4. Consult your teacher 5. Consult your teacher 6. Consult your teacher
7. Consult your teacher 8. a) 4x+3y-2z b) 3x+y-2z c) 5a + 2b – 2c
d) x2–2xy–y2 e) m2+mn–10n2 f) a2b–5b2c–c2a 9. 4a–b–2c
10. a) 10a+b+6c b) 10x+3z

Oasis School Mathematics Book-5 239

Multiplication of algebraic expressions

Let's consider an algebraic term 3x². If x is multiplied two times it

In 3x², is x2 if x is multiplied 5 times
it is x5.

3 is the co-efficient, x is the base and 2 is the power.

Again, Note:
x × x = x² 1a1 = a
a × a × a× a = a4 1x1 = x
b × b × b × b × b = b5

Remember !
• (+) × (+) = (+), (+) × (-) = (-), (-) × (+) = (-), (-) × (-) = (+)

Let's be clear with some examples.

While multiplying the algebraic expressions the co-efficients of the terms are
multiplied and the power of the base are added.

Example :

Multiply: Multiply the co-efficient and add the power of like terms.
3x² × 5x7
= 3 × 5x²+7 (-) × (+) = (-)
= 15x9
Again, Class Assignment
(-6a3) × (5a²)
= - 6 × 5 a3+2 Multiply:
= -30a5 x7 × x5 =
Again, a10 × a6 =
(7y) × (-3y³) 15a³ × 2a6 =
= 7 × (-3) y1+3 (-3a7) × (-4a6) =
= -21 y4 2a9 × 13a7 =
(-5y³) × 7y9 =
13y7 × (-5y8) =

240 Oasis School Mathematics Book-5

Multiply:

x³y² × x5y7 =

x7y9 × x²y² =

3x²y6 × 2x4y =

7x³y × (-2x5y4) =

13x²y4 × (-2x4y²) =

12x4b3c × (-3x7y²) =

Multiplication of unlike terms Power of like base
While multiplying the unlike terms, co-efficient is added and unlike
of the terms are multiplied and the base remains base remains same.
same.

3x × 5y² = 3 × 5 xy²

= 15 xy²

6x³ × 5x²y² = 6 × 5 x³+2y²

= - 6 × 5 x5y²

= - 30x5y²

Again, (-3xy²) × (5x²y³) (+) × (-) = (-)
= - 3 × 5x1+2 y2+3 x1 × x² = x³, y² × y³ = y5
= - 15x³y5

Exercise 13.3

1. Multiply: b. a × a × a × a c. n × n × n
a. m×m e. y × y × y × y × y × y f. z × z × z × z × z × z × z
d. x × x × x × x × x × x h. 6b × 2b × 4b i. 5c × 2c × 3c × c
g. 2a×3a×5a

2. Find the product of:

a. 2x × 3x² b. 5x² × 6x³ c. 3a² × 5a³
f. (- 2a) × (5a²)
d. 6a³ × 2a4 e. 3a² × 5a³ × 2a i. 5b³ × (-4b³)

g. 4a³ × (-2a4) h. 5a² × 3a³

j. 5z³ × (2z²) k. 3x² × (-2x) × 5x²

Oasis School Mathematics Book-5 241

3. Find the product of:

a. 3a²b × 5a³b² b. 3xy² × 4x²y c. xy × x²y³
f. (-2x³y7) × (-7x4y³)
d. x³y4 × 2x7y³ e. 7x4y7 × (-4x³y²)

g. 5x3y7 × (-3x4y8) h. (-10x²y4) × (-x³y7)

Answers: Consult your teacher.

Multiplication of polynomials by monomials and binomials

While multiplying polynomials by monomial, multiply each term of the
polynomial by monomial.

Example : [(+) × (-) = (-)]

Multiply: 2a²b³ (3a4b² - a³b) It is the multiplication of
Solution: binomial by binomial
2a²b³ (3a4b² - a³b)
= 2a²b³ × 3a4b² - 2a²b³ × a³b
= 2 × 3a²+4 b3+2 - 2 × 1a2+3 b3+1
= 6a6b5 - 2a5b4
Example :

Multiply: (5x + 2y) (2x² - 3xy)

Solution: -15x²y + 4x²y = -11x²y

(5x + 2y) (2x² - 3xy)

= 5x(2x² - 3xy) + 2y (2x² - 3xy)

= 5x × 2x² - 5x × 3xy + 2y × 2x² - 2y × 3xy
= 5 × x1+2 - 5 × 3x1+1y + 2 × 2x²y - 2 × 3xy1+1
= 10x³ - 15x²y + 4x²y - 6xy²
= 10x³ - 11x²y - 6xy²

Alternative method: Multiplying by 2x²
5x + 2y Multiplying by -3xy
2x² - 3xy
10x³ + 4x²y
- 15x²y - 6xy²
10x³ - 11x²y - 6xy²

242 Oasis School Mathematics Book-5

Again, Look at one example:
Multiply: (3a + 4b) × (a - 2b)
Solution:
(3a + 4b) ( a - 2b)
= 3a² - 6ab + 4ab - 8b²
= 3a² - 2ab - 8b²

Alternative method:

3a + 4b

× a - 2b Multiplying by a
3a² + 4ab Multiplying by -2b
- 6ab - 8b² Adding
3a² - 2ab - 8b²

Example : l = 2x + y

Find the area of the given rectangle: b=x+y
Solution:
Given, l = 2x + y
b = x + y
Area of a rectangle = l × b
= (2x+y) (x+y)
= 2x² + 2xy + xy + y²
= 2x² + 3xy + y²

Exercise 13.4

1. Multiply:

a. 3xy(2x + 3y) b. ab(3a² - 4b²) c. a²b² (5a - 6b)
f. x²y²z²(x - y - z)
d. abc(a + 2b - c) e. 2x²yz(3x - 2y + z) i. x²y²z(3x - 4xy + 2z)

g. mn²(3m² - 4n³ + 6) h. -pq2r³(2p² - 3q² + 6r²)

j. -5c²d(2c + 3d + 4)

2. Multiply: b. (x - 3) (x + 4) c. (a + 5) ( a - 2)
a. (a + 2) (a+3) e. (y + 2) (y - 4) f. (2x + 3) (3x + 1)
d. (x - 6) (x - 2) h. (2x - 1) (3x + 1) i. (5x - 2) (7x - 3)
g. (3x - 5) (2x + 3) k. (a + 2b) (2a + 3b) l. (x - 3y) (2x + y)
j. (x + y) (2x + y) n. (2x + 3y) (x - 2y) o. (x - y) (2x - y)
m. (4x - y) (x - y)

Oasis School Mathematics Book-5 243

3. Using the formula A = l × b, find the area of the following figures:

a. l = x + 2 b. l = 2x + 3

b = 2x + 3 b=x+1

c. l = 3x - 2 d. l = m+n
b = m+n
b = 2x + 1

4. Using the formula V = l×b×h, find the volume of the given figures:

a. b.

h=x h=y

l=x+2 b=x-2 b=x+y

Answers: Consult your teacher l = 2x + y

Division of algebraic expression

Let's learn the division of algebraic expression from the given examples:

a5 ÷ a² a5 a×a×a×a×a Short cut way:
a² a×a a5 ÷ a² = a5-² = a³
= = = a³ x6 ÷ x4 = x6-4 = x²

x6 ÷ x4

= x6 = x×x×x×x×x×x = x² I understand, while dividing the two
x4 x×x×x×x terms having like base, we have to

subtract the power of the divisor from

the power of the dividend of the same

12x6 base.
3x²
Again, 12x6 ÷ 3x² =
= 4x6-2
I understand, in these cases we
have to divide the co-efficient and

= 4x4 subtract the power of divisor from
the power of dividend of the same

base.

244 Oasis School Mathematics Book-5

Again,

24a³b5 ÷ 4a²b³ 24 ÷ 4 = 6 Class Assignment
a³ ÷ a² = a Divide:
Here, 24a³b5 b5 ÷ b³ = b² x15 ÷ x6 =
4a²b³ x20 ÷ x4 =
15x10 ÷ 3x7 =
20x5 ÷ 5x² =
= 6a²-2b5-3 24a10 ÷ 3a4 =
20x5y6 ÷ 5x²y² =
= 6ab² 25x6y7 ÷ 5x³y4 =

Division of polynomials by monomials Splitting each term

While dividing polynomials by Remember:
monomials, we divide each term of the • x0 = 1, y0 = 1
polynomials by monomials.

Example :

Divide: (15a³b² - 10a²b²) ÷ 5ab

Solution:

(15a³b² - 10a²b²) ÷ 5ab

15a³b² - 10a²b²
= 5ab

= 15a³b² – 10a²b²
5ab 5ab

= 3a³-1b2-1 - 2a²-1b²-1
= 3a²b - 2ab

Example :

Divide: 12x4y³ - 8x²y + 4x²y³ by 2x²y
Solution:
12x4y³ - 8x²y + 4x²y³ by 2x²y

= 12x4y³ - 8x²y + 4x²y³
2x²y

= 12x4y³ – 8x²y + 4x²y³
2x²y 2x²y 2x²y

= 6x4-2y3-1 - 4x2-2y1-1 + 2x2-2y3-1
= 6x²y² - 4x0y0 + 2x0y²
= 6x²y2 - 4 + 2y²

Note:
(-) ÷ (-) = (+), (+) ÷ (+) = (+), (+) ÷ (-) = (-), (-) ÷ (+) = (-)

Oasis School Mathematics Book-5 245

Exercise 13.5

1. Divide: b. y7 ÷ y4 c. x8 ÷ x² d. y6 ÷ y4
a. x6 ÷x² f. 12y5 ÷ 3y² g. 20z6 ÷ 5z4 h. 24a9 ÷ 8a³
e. 15x6 ÷ 3x²

2. Divide: b. p7q² ÷ p²q³ c. a9b7c4 ÷ a4b³c
a. x4y³ ÷ x²y² e. 4x²y7 ÷ 2xy f. 8x6y5z³ ÷ 2x²y³z²
d. x7y³z9 ÷ x²y³z² h. -20a7b²c5 ÷ 4a³bc i. -15xy³z ÷ 5xyz
g. (-15a5b7c) ÷ (-5a²b²c²)
j. 27p6q²r ÷ (-3p³qr)

3. Divide:

a. (x³ +x²) ÷ x b. (a4 - a²) ÷ a c. (4a³ - 3a²)÷ a

d. (8a³ - 4a²) ÷ 2a e. (4x4 - 8x²) ÷ 2x² f. (9x7 - 3x5) ÷ 3x³

g. (12y7 - 9y6) ÷ 3y4 h. (15x³y² - 9xy) ÷ 3xy

i. (12x7y³ - 6x²y²) ÷ 6x²y j. (12x4y³ - 8xy + 4x²y³) ÷ 2xy

k. 21a²b + 7a³b² - 14a²b² ÷ (-7a²b) l. (20a³b² - 10ab)÷ (-5ab)

m. (20p³q² - 25p²q³) ÷ 5p²q²

4. a. What should be multiplied by 5x²y² to make 15x4y³?
b. What should be multiplied by 5xy to make (10x³y4 - 5x²y²)?
c. Find the quotient of (12m³n² - 6mn) divided by 6m.

Simplification of algebraic expressions

Look at these examples properly and get the idea of simplification of algebraic
expressions.

Simplify: (5xy + 3y²) - 2x(3x + 2y)

Solution:

(5xy + 3y²) - 6x² - 4xy - 2x × 3x = - 6x²
= 5xy + 3y² - 6x² - 4xy -2x × 2y = - 4xy
= xy + 3y² - 6x²

Look at one more example. -4x × x = -4x2
Simplify: -4x (-2) = 8x
3x² - 4x(x-2)
= 3x² - 4x² + 8x
= -x² + 8x

246 Oasis School Mathematics Book-5

Exercise 13.6

1. Simplify: b. 4a + (6a - 4b) c. 2x + (7x + 3y)
a. 5x + (2x - 3y) e. (3x - 4y) + 3y f. (x - 2y) - 3y
d. (5x + 2y) + 7y h. 3a - (2a - 3b) i. a - (2a - 5b)
g. 4x - (x - 2y) k. (a - b) + (a + 2b) l. (3a + 5b) + (a - 2b)
j. (3a - 4b) + (5a + 3b) n. (3a - 5b) - (a + 2b) o. (2a - b) - (a + b)
m. (2p - 5q) - (3p - q)
p. (2p - 5q) + (3p - q) c. (m -2n) + 2(m - n)
f. (4x - y) - 3(x + y)
2. Simplify: i. (2x² + xy) + x(x + y)
a. (a + b) + 2(a + 2b) b. (x- y) + 3(x -y)
d. (2p + q) - 3(p - 2q) e. (x + 3y) - 2(x - y)
g. - 3(x + 2y) + 2(3x - y) h. (4m - n) - 3(m + n)
j. (5x² - 3xy) - x(2x - 3y) k. (p² - q²) - 3p(p - q)

Algebraic equations

An equation is a mathematical statement equating

two quantities. ‘=’ stands for equal to
x + 5 = 9 is an equation.

(x + 5) and 9 are two quantities.

In an equation, the value of the terms on the left hand
side is equal to the terms on the right hand side.

An equation is like a weighing balance having equal
weight on each pan.

Solution of an equation:

Let's take an example,

x + 2 = 6.

We have to find the value of x, which will satisfy the equation.

When, x = 1,

1+2=6 To solve an equation, we have to
3 = 6 (which is not true) determine the value of variable that will
When, x = 2,
make the equation true.

2+2=6

4 = 6 (which is not true)

Oasis School Mathematics Book-5 247

When, x = 3

3 + 2 = 6,

5 = 6 (which is not true) This method of solving equation is
When, x = 4, known as Trial and Error method.

4+2=6

6 = 6 (which is true)

\ x = 4 is the solution to this equation.

If x + 2 = 6, x = ? means, what should be added to 2 to make 6?

Example :

Determine, if 5 is the solution to the equation 2x + 3 = 13

Solution:

Given equation, 2x + 3 = 13

When x = 5,

Left hand side = 2 × 5 + 3

= 10 + 3 = 13

Right hand side = 13

When x = 5, Left hand side = Right hand side

\ x = 5 is the solution to the given equation.

Exercise 13.7

1. Determine by substitution, whether:

a. 3 is the solution to 2x - 1 = 5

b. 2 is the solution to 3x + 1 = 5

c. 5 is the solution to x + 2 = 7

2. Replace the variable by 1, 2 and 3 to solve the given equations:

a. x + 3 = 6 b. 2x + 3 = 7 c. 3x - 1 = 2

d. 4x + 3 = 11 e. 3x - 5 = 1

3. Replace the by suitable number to make the given statement true:

a. + 3 = 4 b. 2 × - 3 = 3 c. 3 × + 5 = 8

d. 2 × + 1 = 3 e. 4 × + 1 = 5

248 Oasis School Mathematics Book-5

Solving of the equation using the principle of balance

While solving any equation, we have to use the following facts.

i. If same number is added to both sides of an equation, the equation remains true.

Example :

x-2=5 Adding 2 on both sides Shortcut way
or x - 2 + 2 = 5 + 2 x–2=5
or, x = 5 + 2
or x = 7 or, x = 2

ii. If the same number is subtracted from both sides
of an equation the equation remains true.

Example :

y+5=7 Subtracting 5 from both sides Shortcut way
or y + 5 - 5 = 7 - 5 y+5=7
\ y = 2 or, y = 7 – 5
or, y = 2

iii. If the same number is multiplied on both sides of an equation, the
equation remains true.

Example :

x3 = 2 Shortcut way
x Multiplying both sides by 3 x
or, 3 × 3 = 2 × 3 3 =2

or, x = 6 or, x = 2 × 3

iv. If the same number divides both sides of an equation or, x = 6
(except zero), the equation remains true.

Example :

3x = 12 Shortcut way

or, 3x = 12 Dividing both sides by3 3x = 12
3 3
or, x= 12
or, x = 4 3

Example : or, x = 4

Solve: 3x - 3 = 6
4

Solution:

3x – 3 = 6
4

Oasis School Mathematics Book-5 249

or, 3x – 3 + 3 = 6 + 3 Adding 3 on both sides Shortcut way:
4
3x 3x - 3 = 6
4 4
or, =9
3x
or, 3x × 4 = 9 × 4 Multiplying both sides by4 4 = 6 + 3
4
3x =9
or, 3x = 36 o4r, 3x = 9 × 4

or, 3x = 36 Dividing both sides by 3 or, x = 9 × 4
3 3 3

or, x = 12 I understand! or, x = 12

3x - 3 = 6 can be written as 3x = 6 + 3 directly.
4 4

Exercise 13.8

1. Solve the following: c. z - 2 = 7 d. x - 4 = 7 e. x - 8 = 9
a. y - 2 = 6 b. x - 1 = 5

2. Solve: c. y + 5 = 8 d. x + 1 = 6 e. z + 2 = 5
a. x + 2 = 4 b. y + 3 = 7

3. Solve: y z x y
x 3 2 5 3
a. 4 = 2 b. = 2 c. = 1 d. = 1 e. = 1

4. Solve:

a. 3x = 15 b. 2x = 6 c. 5y = 20 d. 4y = 12 e. 7z = 14

5. Solve: b. 3x - 1 = 8 c. 2x + 9 = 13 d. 2z + 5 = 13
a. 2x + 5 = 9 f. 4z + 1 = 13 g. 5 + 3y = 20 h. 5x + 3 = 23
e. 3y - 2 = 10

6. Solve: c. 2x - 3 = x + 2
a. 3x + 2 = 2x + 1 b. 5x + 3 = 4x + 5
d. 3x - 1 = x + 1 e. 6x - 7 = 4x + 5

7. Solve:

250 Oasis School Mathematics Book-5