Prime Math 6 - Final rabi_final

Worked Out Examples.

Example 1: Find the profit or loss if cost price = Rs.350 and selling price
Solution: = Rs.335.
Here, Cost price (C.P.) = Rs.350.
Selling price (S.P.) = Rs.335
C.P. > S.P. So, there is loss.
\ Loss = C.P. – S.P.

= Rs.350 – Rs.335 = Rs.15

Example 2: Find the profit or loss percent when C.P. = Rs.50 and S.P. = Rs.75.
Solution: Here, C.P. = Rs.50

S.P. = Rs.75

S.P. > C.P. So, there is profit.

\ Profit = S.P. - C.P.

= Rs.75 – Rs.50 = Rs.25

Now, profit percent = Profit x 100%
C.P.

= 25 x 100% 2 = 50%.
50

Example 3: A shopkeeper bought 2 dozen pens at Rs.25 each and sold the pens

at Rs.28 each. Find the profit or loss amount. Also the result in

percent.

Solution: Here,

The shopkeeper bought 2 dozen pens at Rs.25 each

Number of Pens = 2 × 12 [1 dozen = 12 pieces]

= 24
C.P. of 24 pens = Rs.25 × 24 = Rs.600.

Selling price of 1 pen is Rs.28.

S.P. of 24 pens = Rs.28 × 24 = Rs.672

Here, S.P. > C.P.

Therefore, profit amount = S.P. – C.P.

= Rs.672 – Rs.600 = Rs.72

Now, Profit 72
C.P. 600
Profit percent = × 100% = × 100% = 12%.

Prime Mathematics Book - 6 195

Exercise 9.1

1. Find the profit in the following cases.

a. C.P. = Rs 35 , S.P. = Rs.42

b. C.P. = Rs 273 , S.P. = Rs.292

c. C.P. = Rs 4275 , S.P. = Rs.4480

2. Find the loss in the following cases.

a. C.P. = Rs. 75 , S.P. = Rs. 62

b. C.P. = Rs. 634 , S.P. = Rs. 527

c. C.P. = Rs. 8350 , S.P. = Rs. 7590

3. Find the profit percent in the following cases.

a. C.P. = Rs. 75 , S.P. = Rs. 90

b. C.P. = Rs. 625 , S.P. = Rs. 675

c. C.P. = Rs. 1280 , S.P. = Rs. 1408

4. Find the loss percent in the following cases.

a. C.P. = Rs. 150 , S.P. = Rs. 135

b. C.P. = Rs. 750 , S.P. = Rs. 735

c. C.P. = Rs. 728 , S.P. = Rs. 800

5. (a) Dolma bought a watch for Rs.625 and sold it for Rs.550. Find the profit or
loss amount made by her.

(b) Mr. Khan bought a T.V. for Rs.5250 and sold it for Rs.6300. Find his profit
or loss percent.

(c) Mr. Yadav bought 2 dozen of copies for Rs.250. He sold a dozen for Rs.150
and another dozen for Rs.175. Calculate his profit or loss percent.

(d) Mr. Khanal bought 2 pens for Rs.80. He sold one pen for Rs.30 and another
for Rs.42. Calculate his profit or loss percent.

(e) A fruit seller purchased 20kg of oranges at the rate of Rs.35 per kg. If he
could not sell 2 kg of oranges due to rotten, find his profit or loss percent
after selling the good oranges at the rate of Rs. 42 per kg.

(f) Mr. Gurung bought a calculator for Rs.750 and a watch for Rs.825. He sold
the calculator for Rs.653 and the watch for Rs.985. What percent profit
or loss did he make in all?

196 Prime Mathematics Book - 6

Mixed problems on profit and loss.

Anu! Recall the formula for the
profit and loss.

Profit = S.P. – C.P.
Loss = C.P. – S.P.

From the formula, Profit = S.P. – C.P., We can write

S.P. = Profit + C.P. C.P. = S.P. – Profit

From the formula, Loss = C.P. – S.P., We can write

S.P. = C.P. – Loss C.P. = S.P. + Loss

Work Out Examples.

Example 1: If C.P. = Rs.75 and profit = Rs.12, find the S.P.

Solution: Here,

C.P. = Rs.75 and profit = Rs.12

S.P. = C.P. + Profit

= Rs.75 + Rs.12

\ S.P. = Rs.87

Example 2: If C.P. = Rs.120 and loss percent = 5%, find the S.P.

Solution: Here, Alternate process

C.P. = Rs.120 and loss percent = 5%, Here, C.P. = Rs. 120

Loss percent = Loss x 100% loss = 5 %
C.P.
Loss We have
120
or, 5% = x 100% S.P. = C.P. - loss

or, 5 x 120 = Loss x 100 = C.P. - 5% of C.P.

6 = 120 - 160502G0 120 6
5 × 120 = 120 -
\ Loss = 1002 =6

Now, S.P. = C.P. – Loss = 114

= 120 – 6 = Rs.114 \ S.P. = Rs. 114

Prime Mathematics Book - 6 197

Example 3: A fruit seller bought 12kg of mangoes at the rate of Rs.45 per kg.
He got a loss of Rs.40. At what amount did he sell all the mangoes?

Solution: Here,
Total mangoes = 12kg.
C.P. of 1 kg of mangoes = Rs.45
Therefore, total C.P. = Rs.45 x 12

= Rs.540
\ Loss = Rs.40
Now, S.P. = C.P. – Loss

= Rs.540 – Rs.40
= Rs.500
\ He sold all the mangoes for Rs.500.

Exercise 9.2

1. Find the S.P. in the following cases.

a. C.P. = Rs.210 Profit = Rs.43

b. C.P. = Rs.575 Loss = Rs.40

c. C.P. = Rs.1025 Profit = Rs.105

2. Find the C.P. in the following cases:

a) S.P. = Rs.95, profit = Rs. 15

b) S.P. = Rs.350, loss = Rs. 45

c) S.P. = Rs. 2140, profit = Rs.260

3. Find the profit or loss amount in the following cases:

a) C.P. = Rs.80, profit percent = 5%

b) C.P. = Rs.370, loss percent = 20%

c) C.P. = Rs.140, profit percent = 15%

4. a) A fruit seller bought 4 Kgs. of oranges at the rate of Rs.50 per kg. At what
price should he sell the oranges so that he makes the profit of Rs.30?

b) A shopkeeper sold a watch for Rs. 2650 and made a loss of Rs.125. At
what price did he purchase the watch?

198 Prime Mathematics Book - 6

c) Mr. Shrestha bought a shirt for Rs.450. At what price did he sell it so that
his loss was Rs. 75?

d) Mr. Rai bought 2 dozen pens at the rate of Rs.40 each. He made a loss of
Rs. 48 after selling all the pens. At what price did he sell each pen?
Find it.

e) Mrs. Gupta buys a sari for Rs. 700 and sells at 15% profit
(i) How much is her profit amount?
(ii) How much is the selling price of the sari?

f) Mr. Mathema bought a computer set for Rs. 15500 and sold at a loss
of 12%.
(i) Find his loss amount.
(ii) Find the selling price of the computer set.

Unit Revision Test I

1. Find the profit or loss if
a) C.P. = Rs.285 and S.P. = Rs. 315.
b) S.P. = Rs.2017 and C.P. = Rs. 2117

2. Find the profit or loss percent if
a) C.P. = Rs.500 and S.P. = Rs. 575
b) S.P. = Rs.540 and C.P. = Rs. 600

3. A man bought 30 kg. of apples at the rate of Rs. 40 per kg. If he could not sell
5 kg. of apples due to rotten, find his profit or loss percent after selling the
good apples at the rate of Rs. 50 per kg.

4. Find the S.P. or C.P. if
a) C.P. = Rs. 420 and Profit = Rs. 80
b) S.P. = Rs. 3240 and loss = Rs.125

5. Mr. Chaudhary bought a T.V. set for Rs. 17500 and sold at a profit of 14%.
a) Find his profit of 14%
b) Find the selling price of the T.V. set.

Prime Mathematics Book - 6 199

Estimated periods – 5

Objectives

At the end of this unit, the students will be able to:
use the idea of direct proportions
find the value of certain number of object when value or cost of one object is
known.
find the value or cost of one object when the value or cost of certain number of
objects is known.

Rs. 30

Rs. 400

Teaching Materials Activities

Menu, price list e.t.c. It is better to:
discusss about use of direct proportions
for solving unitary method problems.
let the students solve the related problems
by group discussion.

Unitary Method
Raj went to buy 4 pens. He asked with the shopkeeper about the price of 4 pens.
But the shopkeeper answered in this way: If the cost of a pen is Rs. 33, please find
the price of 4 pens and pay to me. He thought for a while and answered that the
cost of 4 pens is Rs. 142. So, here as Raj knows the cost of 4 pens. Thus, with the
help of cost of unit number of pen, we can determine the cost of any number of
pens. This method is known as 'Unitary Method.’

Examples1: If the cost of 1 chair is Rs. 450, find the cost of 7 such chairs.

Solution: The cost of 1 chair is Rs. 450. As the cost of 7 chairs is
\ The cost of 7 chairs is Rs. 450 × 7 more than the cost of 1
= Rs. 3150 chair, so we multiply]

Hence, the cost of seven chairs is Rs. 3150.

Examples 2: If the cost of 15 apples is Rs. 375, then find the cost of 1 apple.

Solution: The cost of 15 apples is Rs. 375

Then, the cost of 1 apple is Rs. 375 = Rs. 25
15

[As the cost of 1 apple is less than the cost of 15 apples, so division

work is done.]

Hence, the cost of 1 apple is Rs. 25

Examples 3: The cost of 20 bananas is Rs. 600. Find the cost of 45 bananas.

Solution: The cost of 20 bananas is Rs. 600

\ The cost of 1 banana is Rs. 600 = Rs. 30
20

Then, the cost of 45 bananas is Rs. 30 × 45

= Rs. 450

Hence, the cost of 45 bananas is Rs. 450.

Prime Mathematics Book - 6 201

Examples 4: 16 men can accomplish a construction in 18 day. How many man
can accomplish the same construction in 12 days?

Solution: In 18 day a construction can be accomplished by 16 men.

\ In 1 day the construction can be accomplished by 16 × 18 men.

\ In 12 days the construction can be accomplished by 16 × 18 men
12

= 24 men
Therefore, 24 men can accomplish the construction in 12 days.

Note: More men, less days it takes, less number of men more days
it takes to do a work. We say number of men and number of
days vary indirectly.

Exercise 10

1. Solve the following problems.
(a) If the total wages of 7 workers is Rs. 2450, then find the wages of 1
worker.
(b) The cost of 8 kgs of sugar is Rs. 720. Find the cost of 1 kg of sugar.
(c) If the bus fare of a person is Rs. 450, find the fare of 9 persons.
(d) The cost of 1 kg of ghee is Rs. 280 find the cost of 6 kgs of ghee.
(e) By 1 litre of petrol, a car can travel for 12 km. How far will the car travel
for 4 litres of petrol?

2. Solve the following problems:
(a) 24 apples cost Rs. 192. What will be the cost of 15 apples?
(b) 6m of cloth cost of Rs. 1440. How many metres of cloth can be purchased
for Rs. 1920?
(c) A car can travel 480km in 6 hrs. What time is required to cover a distance
of 320km?
(d) If 20 men can finish a piece of work in 30 day, how long will 24 men take
to do it?
(e) The cost of 15 books is Rs. 3150. What is the cost of 25 books?

202 Prime Mathematics Book - 6

3. Solve the following problems:
(a) If 5 watches cost Rs. 3625, what is the cost of 8 watches?
(b) Four dozen pencils cost Rs. 216. What is the cost of 9 pencils?
(c) 10 men can build a wall in 24 days. How many men are required to build
the wall in 30 days?
(d) 150 men have enough food for 40 days. How long the food will last for
200 men?
(e) A man can finish a piece of work in 24 days working 8 hours per day. In
how many days will he finish the work working 6 hours per day?

Unit Revision Test I

1. Sove the following problems:
a) If the cost of 5 kgs of sugar is Rs. 450, find the cost of 1 kg of sugar.
b) The cost of a pencil is Rs. 8.25. Find the cost of a dozen of pencils.
c) With 7 litres of petrol a bike can travel 315 km. What is the milage (km/litre)
of the bike ?

2. Solve the following problems.
a) The cost of 16 books is Rs. 3440. What is the cost of 25 such books?
b) 8 m of cloth cost Rs. 2080. How many meters of cloth can be brought for
Rs. 3120?
c) 20 masons can complete a wall in 30 days. How many masons can
complete the same wall in 12 days?

Prime Mathematics Book - 6 203

Estimated periods – 6

Objectives

At the end of this unit, the students will be able to:
understand what interest is.
calculate simple interest using unitary method.
calculate simple interest using formula.

Teaching Materials Activities

Current interest rates of different banks It is better to:
and finance companies, sample of loan discuss about interest, process of
agreements.
providing loan.
discuss about the use of unitary method
to find interest.
show, how the formula I = PTR is
obtained.
100

Simple interest

Suppose, a man took a loan from a bank. The money borrowed is called the principal
and is represented by ‘P’. The man has to return the borrowed money with some
extra money to the bank in lieu to the use of the money. The extra money paid by
the man to the bank is called interest. It is represented by ‘I’. The money is borrowed
for a certain period of time. This certain period of time is called time. It is represented
by ‘T’. The interest paid for Rs. 100 for one year is known as rate percent per annum
or simply rate. If the rate is 12%, it means interest for Rs. 100 for 1 year is Rs. 12.
Similarly, if the rate is 15% means, the interest for every Rs. 100 borrowed for 1 year
is Rs. 15. At the end of the period, the man has to pay principal together with
interest which is known as Amount. It is represented by ‘A’. thus,

Amount (A) = Principal (P) + Interest (I)

Calculation of simple interest by using unitary method

Example1: Raj deposited Rs. 5000 in a bank at the rate of 8%. Find the simple
interest he will get after 3 years.

Solution: The rate is 8% per annum means
\
Interest on Rs. 100 for 1 year is Rs. 8

Interest on Re. 1 for 1 year is Rs. 8/100

Interest on Rs. 5000 for 3 years is Rs. 8 × 5000 × 3
100

= Rs. 1200

Hence, Raj will get Rs. 1200 as simple interest.

Example2: Calculate the simple interest on Rs. 4000 at the rate of 5%
Solution: for 321 years. Also find the amount.

The rate of interest is 5% per annum means

Interest on Rs. 100 for 1 year is Rs. 5

\ Interest on Re. 1 for 1 year is Rs. 5/100

Prime Mathematics Book - 6 205

Interest on Rs. 4000 for 7 years is Rs. 5 × 4000 × 7
2 100 2
= Rs. 700

Thus, the simple interest is Rs. 700.

Now, Amount = P + I = Rs 4000 + Rs 700 = Rs 4700

Example3: Find the simple interest on Rs. 6000 at the rate of 6% for 9
Solution:
months. Remember:

Time (T) = 9 months = 9 years = 3 years 1 year = 12 months
12 4 1 year = 52 weeks
1 year = 365 days
The rate is 6% means Time (T) should be

Interest on Rs. 100 for 1 year is Rs. 6

Interest on Re. 1 for 1 year is Rs. 6/100 in years.

Interest on Rs. 6000 for 3 years is Rs. 6 × 6000 × 3
4 100 4

= Rs. 270

Hence, the interest is Rs. 270.

Calculation of simple interest by using formula:

To calculate the simple interest on Rs. P at the rate of R% per annum for
T years.

The rate is R% p.a. means

Interest on Rs.100 for 1 year is Rs. R

\ Interest on Re. 1 for 1 year is Rs. R
100

\ Interest on Rs. P for T years is Rs. R × P × T
100

Hence, I = PTR Note:
100 Principal (P) is also called borrowed

We can obtain

P = 100 ×I sum, lent sum, sum invested,
T× R deposited sum etc.

R = 100 × I Amount (A) is also called the sum to
PT

And T = 100 × I pay all, sum to clear the debt etc.
PR

206 Prime Mathematics Book - 6

Example 4: Find the simple interest on Rs 3500 for 2 years at 14% per annum.

Solution: Here, Principal (P) = Rs 3500

Time (T) = 2 years

Rate (R) = 14%

\ Simple Interest (I) = P×T×R
100

= 3500 × 2 × 14
100

= 980

\ Simple Interest (I) = Rs. 980

Example 5: What sum invested at 16% p.a. for 8 months will yield a simple
interest of Rs 1280 ?

Solution: Here, Principal (P) = ?

Time (T) = 8 months =182 years = 2 years
3
Rate (R) = 16% p.a.

\ Simple Interest (I) = Rs. 1280

We have, P = I×100 = 1280 × 100 = 1280 × 100 × 3
T×R 2 × 16
2 × 16
3

= 12000

\ Principal (P) = Rs. 12000

Example 6: In what time will Rs 10,000 yield simple interest of Rs 4800 at
Solution: 12% P.a. ?

Here, Principal (P) = Rs. 10,000

Rate (R) = 12% p.a.

Simple Interest (I) = Rs. 4800, Time (T) = ?

We have, T = I×100 = 4800 × 100 12= 4
P×R 10,000 ×

\ Time (T) = 4 years

Prime Mathematics Book - 6 207

Example 7: At what rate will a sum of Rs.12000 yield a simple interest of
Rs. 5400 in 3 years ?

Solution: Here, Principal (P) = Rs. 12,000

Rate (R) = ?

Simple Interest (I) = Rs. 4800, Time (T) = 3 years

We have, R = I×100 = 5400 × 100 = 15
P×T 12,000 ×
3

\ Rate (R) = 15 %

Exercise 11

1. Using the unitary method, calculate the simple interest on

(a) Rs. 4500 for 5 years at 7% per annum.
(b) Rs. 5500 for 4 years at 321 % per annum.
(c) Rs. 6000 for 6 years at 8% per annum.
(d) Rs. 9000 for 521 years at 6% p.a.

2. Use the unitary method and find the simple interest on

(a) Rs. 3500 for 9 months at 5% p.a.
(b) Rs. 4800 for 6 months at 7% p.a.
(c) Rs. 5000 for 2 years 6 months at 8% p.a.

3. Using the formula for simple interest, calculate the simple interest on
(a) Rs. 4800 for 5 years at 412 % p.a.
(b) Rs. 7500 for 3 years at 5% p.a.

(c) Rs. 8000 for 421 years at 6% p.a.
(d) Rs. 10000 for 6 months at 8% p.a.

(e) Rs. 9500 for 9 months at 5% p.a.

208 Prime Mathematics Book - 6

(f) Rs. 8500 for 1 year 4 months at 6% p.a.

(g) Rs. 6500 for 2 years 6 months at 3% p.a.

(h) Rs. 2500 for 3 years 9 months at 4% p.a.

4. (a) What sum invested at 14% p.a. for 3 years will yield simple interest of
Rs 1029 ?

(b) In what time will Rs 2500 produce simple interest of Rs. 5000 at 8% per
annum ?

(c) Sangdup took a loan of Rs. 50,000 from a bank and cleared the debt paying
Rs. 54000 after 8 months. What was the rate ?

(d) Mrs. Sherchan borrowed Rs. 1,75,000 from Mrs. Maharjan to buy a scooter
at 12% p.a. simple interest for 3 years. how much she has to pay to clear
the debt after 3 years ?

Unit Revision Test I

1. Calculate the simple interest using unitary method.

a) P = Rs. 1500, T = 3 years, R = 5% p.a.

a) P = Rs. 1,50,000, T = 2 years, R = 12% p.a.

2. Calculate the Simple Interest using formula.

a) P = Rs. 12000, T = 4 years, R = 8% p.a.

a) P = Rs. 80,000, T = 1 year 6 months, R = 15% p.a.

3. a) If P = Rs. 9000, T = 5 years and S.I. = Rs 5400, find R.

b) If = A = Rs. 2112, S.I. = Rs. 512, R = 8%, find P and hence find T.

Prime Mathematics Book - 6 209

Estimated periods – 8

Objectives

At the end of this unit, the students will be able to:

define statistics
know the methods of collecting data present the data in frequency
distribution table.
represent the single charractered data in simple bar graph.
study the given table and bar graph.

No. of students Y The number of students of different classes.

350 V VI VII VIII IX
300 classes
250
200
150
100

50

X X

Teaching Materials Activities

Graphs, grids, different data presented in It is better to:
tables and bar graphs. discuss about the word statistics, its
definition and meanings.
involve the students in groups to collect
simple data available in the class, and
shool
show different tables and graph and
discuss about the data.

Statistics

Every day we come across a lot of informations in the form of facts, numerical
figures, tables, graphs e.t.c. which may be related to cricket batting/ bowling
averages, profit/loss of a company, tempereatures of various parts of country, polling
results, expenditures in various sectos of the budget of a fiscal year e.t.c. These
facts or figures which are numerical or otherwise, collected for a definite purpose
are called data. Data is the plural form of the Latin word datum.

We utilise data in one form or the other to extract meaningful information
from it. The extraction of meaningful information from such data is studied in a
branch of mathematics called statistics.

The word ‘statistics’ appears to have been derived from the Latin word
‘status’ meaning ‘a (political) state’. In its origin, statistics was simply the collection
of data on different aspects of the life of people, useful to the state like military
status, production of crops in the state, land owned by the people etc. Over the
lapse of time, however its scope has been broadened.

Statistics deals with collection, presentation/organization, analysis and
interpretation of the data. Thus four stages in statistics are

(i) Collection of data

(ii) Presentation or organization of data

(iii) Analysis of data

(iv) Interpretation of data.

In this chapter we will discuss about the organization/ presentation of data.

Presentation / organization of data:

After the collection of data, the next step in statistics is the organization / presentation
of data i.e. presenting them in a form which is meaningful, easily understood and
gives its main featured at a glance, under which comes arranging the data, presenting
in tables, diagrams, graphs etc. To some extent the secondary data are well presented
but primary data are row and need to present well.

The marks collected by a teacher in mathematics of 20 students in a terminal
examination is as follows.

80, 85, 90, 75, 82, 98, 68, 100, 95, 78, 99,

56, 88, 96, 79, 97, 64, 50, 89

Prime Mathematics Book - 6 211

Each mark in the above collection is called as observation. The data collected in the
original form is called a raw data or ungrouped data. So, the data collected above
is a raw data.
To represent the data, we arrange them whether in ascending order or in descending
order.

Let's arrange the data in ascending order.

50, 56, 64, 68, 75, 78, 79, 80, 82, 85, 88,

89, 90, 92, 95, 96, 97, 98, 99, 100
Now, it is easy to analysis the arranged data. We can say that the highest mark is
100 and the lowest mark is 50.

Frequency and frequency table:
Let's consider the marks scored by 30 students.
80, 90, 98, 85, 98, 80, 85, 95, 80, 90,
90. 94, 80, 100, 90, 94, 94, 80, 85, 80,
85, 100, 90, 94, 94, 94, 100, 98, 98, 94

Let's represent the data given in a table.
i. Arrange the data either in ascending order or in descending order.
ii. The number of repetitions of a particular data is called the frequency of the

data. For example, 100 has occurred 3 times, so the frequency of 100 is 3.
iii. The representation of this data in a table showing the frequency of each

observation in called frequency distribution table or simply frequency table.
iv. If the data is raw, it is easy to construct frequency table with tally marks. If

the frequency of an observation is 1, we use a short varied line called a stroke.
If the same observation occurs again, mark another stroke against the observation.
We do the same work till the same observation occurs for four times. For the fifth
time we draw a diagonal stroke, making a bundle of 5. If the same observation occurs
again, start with new stroke.

212 Prime Mathematics Book - 6

Marks obtained Tally marks Frequency
80 Exercise 12.1 6
85 4
90 5
94 7
95 1
98 4
100 3

Total = 30

1. The marks obtained by 20 students in an examination is given below.
70, 80, 90, 70, 70, 80, 90, 95, 100, 95, 90, 95, 70, 70, 90, 95, 80, 85,
95, 90
Construct a frequency distribution table by using the tally marks.

2. The number of children of 30 families are given below.
1, 2, 6, 4, 2, 3, 2, 3, 2, 2, 5, 2, 3, 4, 1, 1, 2, 3, 3, 2, 3, 1, 6, 1, 2, 3,
1, 2, 1, 1
Construct a frequency distribution table by using tally marks.

3. The weight of 15 teachers are as follows (in kg)
65, 70, 65, 70, 75, 80, 60, 60, 70, 70, 65, 60, 65, 70, 85
Construct of frequency table by using tally marks.

4. A dice is thrown 30 times add the outcome are noted as follows.
1, 2, 6, 4, 2, 3, 4, 1, 2, 2, 2, 3, 6, 3, 2, 1, 6, 5, 2, 2, 3, 6, 6, 2, 3, 5,
1, 1, 2, 6,
Prepare a frequency table.

5. The height (in cm) of 28 students are as follow.
146, 148, 160, 146, 160, 150, 152, 148, 146, 160, 148, 150, 152, 146,
146, 148, 152, 152, 146, 146, 150, 148, 160, 146, 150, 150, 150
Prepare a frequency table.

Prime Mathematics Book - 6 213

Bar Graph

To understand easily and quickly we represent the collected data through graphs,
diagrams or chart. Representing the data by bars is called bar graph. The height of
the bar represents the frequency of the corresponding data. The following things
should be noted while drawing the bar graph.

i. Width of the bars should be equal.

ii. The distance between the bars should be the same.

iii. A title of the bar graph should be given.

iv. A suitable scale should be chosen to represent the whole data. The scale is
shown a long the y-axis.

Examples:-

1. The number of students of different classes of a school is given below.

Class V VI VII VIII IX X

No. of students 350 200 300 250 100 50

The data can be represented as

No. of students Y The number of students of different classes.

350 V VI VII VIII IX
300 classes
250
200
150
100

50

X X

214 Prime Mathematics Book - 6

2. Read the bar graph shown below.
Y Number of students absent in a class on different days

No. of absent students 6

5

4

3

2

1

X1 Sun Mon Tue Wed Thur Fri X
Y1 Days

Answer the following questions.

i. On which day, least number of students is absent? What is the number of absent
students?

ii. What is the number of students absent on Thursday?

iii. On which day, maximum number of students is absent? What is it?

iv. What is the total number of students absent in a week?

Solution:
i. The least number of students is absent on Thursday and it is 1.
ii. The number of students absent on Thursday is 4?
iii. The maximum number of students is absent on Sunday and it is 6.
iv. The total number of students absent in a week is 21.

Exercise 12.2

1. The number of students passed in S.L.C. in different years of a school is as
follow.

Years (B.S.) 2063 2064 2065 2066 2067 2068
200 350 250 400 300
No. of students 150

Represent it is a bar graph.

Prime Mathematics Book - 6 215

2. The data shows the number of students passed in different subjects in a
terminal examination of a class.

Subjects Science Nepali Maths English Computer
70 110 120 130
No. of students 80

Draw a bar graph showing the above information.
3. The data shows the number of cars produced by a company in 5 years.

Years (in A.D.) 2006 2007 2008 2009 2010 2011

No. of cars (in thousands) 8 10 12 13 16 19

Present in a bar graph.

4. The number of buses crossing a check-post on 5 consecutive days is given
below. Draw a bar graph to represent the data.

Days Sunday Monday Tuesday Wednesday Thursday

No. of buses 100 150 120 140 130

5. Read the bar graph and answer the following questions.

45 The number of students passed in S.L.C. in different years
of a school.
40

No. of students 35
30
25
20
15
10

5

2064 2065 2066 2067 2068
classes

i. What is the information given by this graph?

ii. What is the number of students passed in S.L.C. examination in 2066?

iii. In which year only 15 students passed S.L.C. examination?

iv. In which year maximum number of students passed S.L.C. examination and what

is it?

v. By above bar graph, what is the total number of students passed in S.L.C.

examination?

216 Prime Mathematics Book - 6

Unit Revision Test

1. The blood groups of 30 students of a class are recorded as follows:

A,B,O,O,AB,O,A,O,B,A,O,B,A,O,O,A,AB,O,A,A,O,O,AB,B,A,O,B,O,B,A
Represent the data in a frequency distribution table. Which is the most common
and which is the rarest(r) blood group among these students ?

2. The value of p up to 40 digits is given below:

3.1415926535897932384626433832795028841971

i) Make a frequency distribution table of the digits form 0 to 9 after
the decimal point.

ii) What are the most and the least occuring digits ?

3. Given below are the sets won by different political parties in the polling
out comes of an election.

Political Parties A B C D E F

Seats won 80 56 35 29 10 40

Draw a bar graph to represent the polling result.

4. Study the given bar diagram and answer the following questions.

No. of students Birth months of 40 students of class VI
7
6
5
4
3
2
1

0 jan feb mar apr may jun july aug sep oct nov dec
Months of birth

(i) How many students were born in the month of September ? 217
(ii) In which month was the maximum and minimum students born ?

Prime Mathematics Book - 6

Estimated periods – 29

Objectives

At the end of this unit, the students will be able to:
identify the constants and variables.
classify the algebraic expressions according to the number of terms.
find the numerical values of the given algebraic expressions by substituting the
values of variables.
perform the four fundamental mathematical operations
on algebraic expressions.

3x = 12
\x=4

Teaching Materials Activities

Chart of algebraic expressions, tiles, local It is better to:
materials, etc. revise the concept of variable, constant,
coefficient, base, power, terms and
algebraic expressions before starting the
lesson.
discuss the way of writing the algebraic
expression of simple varbel problems.
discuss the activities of four fundamental
mathematical operations on algebraic
expressions by using tiles.
drill the problems of algebraic
expressions.

Historical fact:

Diophantus (Probably about 250 A.D. Alexandria, Greek) is often regarded as the
father of algebra. He was the first to take steps towards the algebraic notation. But
the word ‘algebra’ is originated from ‘Hisab al-jabr Wal- muqa-balah’ a book by
Arabian mathematician Mohammed ibn-Musa al-khuwarizmi (about 825 A.D.)

Revision Exercise:
1. Identify whether the following are variables or constants.

(a) x represents even numbers between 4 and 15.
(b) y represents the number of people in Kathmandu valley.
(c) z represents the difference of 19 and 8.

2. Identify whether the following are terms or expression.

(a) 2x (b) -4xy2 (c) a + 2b

(d) 5xy2z (e) 3x + a + 2b

3. Identify whether the following are like or unlike terms.

(a) 4x and -3x (b) 3a2b and 2ab2 (c) -5xy and 2xy

(d) 5xz and 3x2y (e) 2ab and 2b (f) 12y2 and y2

4. Identify the coefficient, base and power in the following.

(a) 7x2 (b) -3yn (c) 2ax3 (d) 9x7

5. Write the following in algebraic expressions.

(a) x is added to 5
(b) 2 is subtracted from y
(c) 7x is added to the product of 2x and 3y
(d) The sum of 3a and 5b is multiplied by 2ab

6. If x = 2 and y = 5, find the value of the following expression.

(a) 2xy2 (b) 3x3 - 2y

(c) 5x2 + 2xy - y2 (d) 3(7y - x2)

Constant and variables

In algebra, we can use letter or symbol to represent the numbers. Let's study the
following statements.

(a) x represents the counting numbers more than 6 but less than 12. Then the
value of x may be any one of 7, 8, 9, 10, 11.

Prime Mathematics Book - 6 219

(b) y represents the odd numbers between 10 and 16. Then the value of y may be
any one among 11, 13, 15.

(c) z represents the prime number between 20 and 28. Then the value of z is
only 23.

In the above statements, x and y have more than one values. But z has only one
value.

So, the letter or symbol which represents more than one values is called variable.
In the above statements, x and y are the variables. The letter or symbol which always
represent only one value is called constant. In the above statements, z is a constant.

Exercise 13.1

1. Identify whether the following are variables or constants.
(a) x represents the number of teachers of your school.
(b) y represents the number of students of Nepal in different year.
(c) z represents the prime number between 24 and 30.
(d) a represents the whole number more than 9 but less than 14.
(e) b represents the number of districts of Janakpur Zone.

2. ‘a’ represents a counting numbers more than 20 but less than 22 and ‘b’
represents only 20.

(a) Are a and b variable or constant? (b) What is the sum of a and b?

(c) In a and b, which one is bigger? (d) What is the difference of a and b?

3. Write the possible values of x and y in the following conditions.
(a) x represents the counting numbers between 8 and 12.
(b) y represents the counting numbers between 12 and 14.
(c) x represents the odd numbers more than 20 but less than 30.
(d) y represents the sum of 7 and 10.
(e) x represents the product of 8 and 14.

4. Identify that x and y are variables or constants in Q. No. 3.

220 Prime Mathematics Book - 6

Algebraic expressions

The constants and variables with different power or variables only with different
powers are connected by using any one mathematical fundamental operation. Then
the result obtained is called an algebraic expression.

For example, the sum of 5 and y is 5 + y. In 5 + y, 5 and y are connected by '+' sign.
So, 5 + y is an algebraic expression.

Similarly, other examples of algebraic expressions are 3a2b, 9a + 7b, 3a2 + 2ab - 5b2,
etc.

Algebraic terms:
The parts of an algebraic expression seperated by + or - signs are called the algebraic
terms. For example, 3a - 2b is an algebraic expression. In this expression, 3a and
2b are its two parts. Because 3a and 2b are connected by '-' sign. So, 3a and 2b are
the terms.

Similarly, in an expression 5x2 - 3xy + 7y2, 5x2, 3xy and 7y2 are the terms.

Types of algebraic expression:
There are following algebraic expressions.
Monomial, Binomial, Trinomial and multinomial expressions.

Monomial expression: An algebraic expression which contains only one term is called
monomial expression. For example, 2a, 5b2, 3x2yz, etc. These are also expression.
So, one term Itself is an monomial expression.

Binomial expression: An algebraic expression which contains two terms is called
binomial expression. For example, a + 2b, 3x + 2y2, etc.

Trinomial expression: An algebraic expression which contains three terms is called
trinomial expression. For example, 2a + b - c, 3a2b - 7bc2 + 9c3, etc.

Multinomial expression: An algebraic expression which contains more than three
terms is called multinomial expression. For example, x + 3y - 2z + 7,
a3 + 2a2 + 3a - 2b + 7, etc.

Prime Mathematics Book - 6 221

Coefficient, base and power of an algebraic terms:
Let us consider a variable x and add it four times itself. Then, x + x + x + x = 4x.
Here, 4 represents the repetation of x four times. So, 4 is called the coefficient of
x.
What is the coefficient of x in a term x?
If we consider a term bx where x is the variable, b is called the coefficient of x.
In 7y, 7 is called numeral coefficient of y.
In ay, a is called literal coefficient of y.
What is the coefficient of x in the terms -8x and 4ax?
Let us consider a variable y and multiply it three times itself. Then, y × y × y = y3.
In y3, the numeral 3 is called the power of the variable y and y is called the base.

3x4 Power

Coefficient Base

Mohan! What 3 is called coefficient of x4.
are 3, x and 4 in x is called the base; 4 is
called the power of x.
3x4 called ?

Exercise 13.2

1. Write the terms and the number of terms in the following expressions.

(a) 2a (b) 3x - 4y
(c) 3x2y - 2xy2 + 4y3 (d) 9xy3 - 17x3y

2. Identify the types of expression in the following.

(a) 3xy (b) 2x + 3y

(c) a2 + 2ab + b2 (d) -9x2y

(e) a + b + c + d - 7 (f) a3 + 3a2b - 3ab2 + b3

(g) 2x3y - 3xy3 (h) 9x - 3b + 7y + d - 8

222 Prime Mathematics Book - 6

3. Write the coefficient, base and power in the following terms.

(a) 3x2 (b) -x3 (c) 2x5
(d) 3ax
(e) 4 ay2
3

4. Write the term having.
(a) Base = x, power = 3 and coefficient = 2
(b) Base = y, power = 2 and coefficient = 5
(c) Base = x, power = 5 and coefficient = -1
(d) Base = x, power = 2 and coefficient = 2a
(e) Base = z, power = 7 and coefficient = -b

5. Write two examples of the following algebraic expressions.

(a) Monomial (b) Binomial (c) Trinomial

6. Rewrite the following statements in algebraic expressions.

(a) The sum of thrice x and 7.
(b) A boy had 7 pencils. He lost x pencils. Now, how many pencils are left

with him?
(c) Two times the product of x and y is increased by 2a.

(d) There were x trees in a garden. y of them died due to diseases. Now,
how many trees are left in the garden?

(e) z is added to the product of y and 6.

7. Match the following.

A B
(a) Five times the sum of a and b.
(b) Two times the product of x and y is increased by z. (i) 2x - 3y
(c) The difference of a and product of x and y.
(d) Subtract the sum of a and b from the product of x and y. (ii) x/y + z
(e) Addition of z to the quotient of x divided by y.
(f) The quotient of the sum of 5x and 7y divided by 3. (iii) 5(a + b)

(iv) 2(x + y)

(v) 2(x y) + z

(vi) xy - (a + b)

(vii) xy - a

(viii) 5x + 7y
3

Prime Mathematics Book - 6 223

8. The age of Kalaman is x years. Rewrite in the algebraic expressions.
(a) How old was he 4 years ago?
(b) How old will he be 3 years hence?
(c) If his mother's age is three times his age, how old is she?

9. Rewrite the statements from the following algebraic expressions.

(a) 3(x + y) + 5 (b) 2x + y (c) 2(x - y) - 4

(d) 3x + y (e) 3x - 2y (f) xy - (x + y)
2
x + y
(g) 2 - 9

Like and unlike terms:
Let's study the following terms.
(a) 3a, -4a, 5a, -7a are like terms.

(b) X2, 3x2, 3ax2, -x2, -2x2 are like terms.

(c) 3xy, -7xy, 9axy are like terms.

(d) 2x, 3x2, 3x3, -y2 are unlike terms.
(e) x3, -4x5, -x, x7 are unlike terms.

(f) 2zy, 2xy2, 3x2y are unlike terms.

From the above examples, it is clear that
the terms are like if the terms have the same base with

same power of the variable.
The terms are unlike if the terms have
the different variables or different power of the same

variable (base).

224 Prime Mathematics Book - 6

Evaluation of terms or expressions:
When we substitute the variable of a term or expression with number(s), the value
of the term or expression is obtained after use of mathematical fundamental
operations. It is called the evaluation of a term or expression.
For example, if x = 3 and y = 2, find the value of 2x2 + 3yx.

Here, 2x2 + 3yx
= 2(3)2 + 3(2) (3)

= 2 × 9 + 18

= 18 + 18

= 36

Example: If a = 2 and b = -3, find the value of 3a3 + 2ab2 - b3.
Here, 3a3 + 2ab2 - b3
= 3(2)3 + 2(2)(-3)2 - (-3)3
= 3 × 8 + 4 × 9 - (-27)
= 24 + 36 + 27
= 87

Exercise 13.3

1. Identify the like and unlike terms in the following.

(a) 2x, -3x, 7x (b) 2ab, 6ab, -3ab
(c) 3x, 2x2, 3x3 (d) 2ab, 3a2b, 3ab2

(e) 5y2, -9y2, 7y2 (f) x3y, xy2, x2y2

2. If x = 3, find the value of: (b) 3x + 7
(a) 2x2 - 4 (d) x4 - 3x2 + 2x
(c) 4x3 - 2x2 + 3x

Prime Mathematics Book - 6 225

3. If x = 3 and y = 4, find the value of:

(a) 2x + y (b) 4x- 2y (c) 3x2 + 2xy

(d) 2(x + y) (e) x - y

4

4. Find the value of the following expressions when x = 4, y = 3 and z = -2.

(a) 2x + 3z (b) (x + y)2 - 2z

(c) 2xy - 3z (d) 2x2 - 4y + z

5. If a = 3, b = 4, c = 2 and d = -2, find the value of:

(a) bc + ad (b) 2bc - 3ab (c) 3b2c - 2ad

(d) b+c (e) ac - bd
2 7

6. Write the perimeter of the following figures in the algebraic expressions.
Also find the numerical value of the perimeter if a = 3, b = 2 and c = 4.

(a) b b
a c (b) (c)

b+c 2a
3c

a 2(b+c)

(d) (e)

b a
a

7. If x = 3cm, find the length of the following line segments.

(a) (b)
x 2cm
3x cm 4cm

(c) x cm (d) (2x-4)cm
(2x+4)cm (3x-2)cm 3 cm

226 Prime Mathematics Book - 6

Addition and subtraction of algebraic terms

In the addition or subtraction of algebraic terms, we should add or subtract the
coefficients of like terms only. We can't add or subtract the coefficient of unlike
terms.

For examples: a + a = 2a, 2b + 3b = 5b, 2x + 3x + 6x = 11x

2x2 + 5x2 + 2x = 7x2 + 2x, 2x + 3y + 4x = 6x + 3y

7a - 3a = 4a, 8x2 - 3x2 = 5x2, 4x2 - 2x - 3x2 = x2 - 2x

Maya! Can you add 4x and 7x? Yes, sir!

Why do you add? Because, 4x and
Very good! Maya. 7x are like terms.

Sir!

Exercise 13.4

1. Add or subtract the following expressions.

(a) 2x + 3x (b) 4a + 3a + 7a
(d) 9x2 - 3x2
(c) 4x2 + 3x2 + 2x2 (f) 7xyz + 3xyz + 2x2yz
(e) 9ab + 2ab + 3ab

(g) 19x3 - 12x3 (h) 8x2y - 3x2y

2. Simplify:

(a) 7x + 8x + 5x (b) 2a2 + 3a + 9a2 + 4a
(c) 5x - 2x + 7x - 3x (d) 5p + 2q + 7q - 2p
(e) 2ab + 3bc - ab - 2bc (f) 12xy2 - 8x2y - 5xy2 - 2x2y

3. If a = 3cm, find the perimeter of the following figures.

(a) (b) (4a+2)cm (c) (4a+2) cm

a cm
(a+2) cm
2a cm
3a cm
(5a+1) cm
3a cm 2a+1cm (2a-2) cm
Prime Mathematics Book - 6
227

4. (a) What should be added to 3x to get 7x?
(b) What should be added to 5a to get 12a?
(c) What should be subtracted from 15xy to get 4xy?
(d) What should be subtracted from 5a to get 5a - 3b?
(e) By how much is 4xy greater than xy?
(f) By how much is 8xy less than 10xy?

Addition and subtraction of algebraic expressions:
In the case of addition or subtraction of two or more algebraic expressions having
two or more than two terms, we should arrange the like terms in the same column.
Then we add or subtract their coefficient. We can add or subtract the terms of the
algebraic expressions in vertical and horizontal arrangement.

Study and learn the following examples:

Example 1: Add: 2x2 + 4xy - 3y2 and 3x2 - xy + 5y2
Solution: Addition by vertical arrangement

2x2 + 4xy - 3y2 - Arrange the like terms in the same
3x2 - xy + 5y2 column.
5x2 + 3xy + 2y2
- Add or subtract the coefficient of
like terms.

2x2 + 3x2 = 5x2, 4xy - xy = 3xy, -3y2
+ 5y2 = 2y2

Addition by horizontal arrangement

2x2 + 4xy - 3y2 + 3x2 - xy + 5y2 Group the like terms together
= 2x2 + 3x2 + 4xy - xy - 3y2 + 5y2 and then performed the
= 5x2 + 3xy + 2y2 addition and subtraction.

Example 2: Add: 3a2 + 2ab, 4ab + b2 and 2a2 + 3ab + 3b2.
Solution: Here,
3a2 + 2ab

4ab + b2
2a2 + 3ab + 3b2
5a2 + 9ab + 4b2

228 Prime Mathematics Book - 6

Example 3: Subtract 2xy - 3yz + 7 from 5xy + 2yz - 2.

Solution: Subtraction by vertical arrangement

5xy + 2yz - 2 - Arrange the like terms in same column.

2xy - 3yz + 7 - Change the sign of each term of the
(-) (+) (-) second expression.

3xy + 5yz - 9 - Perform the addition or subtraction.

Subtraction by horizontal arrangement
(5xy + 2yz - 2) - (2xy - 3yz + 7)
= 5xy + 2yz - 2 - 2xy + 3yz - 7
= 5xy - 2xy + 2yz + 3yz - 2 - 7
= 3xy + 5yz - 9

Example 4: What should be added to 7x + 5y - 2z to get 11x + 3y + 9z?

Solution:

To solve this problem, let's think first.
What should be added to 9 to get 15?
6 is the difference of 15 and 9. That means,
15 - 9 = 6.
It is very good idea. So, we use this idea in
algebra also.

To solve this problem, we Now,
should subtract 11x + 3y + 9z

7x + 5y - 2z from (-)7x(+-) 5y(+- )2z
11x + 3y + 9z. 4x - 2y + 11z

\ The required expression to be added is 4x - 2y + 11z.

Prime Mathematics Book - 6 229

Example 5: What should be subtracted from 9a - 2b + 4bc to get 4a + 5b - bc?

Solution:

To solve this problem, let's think first.
What should be subtracted from 10 to get 4?
It's definitely 6. Oh! It's 10 - 4 = 6.
It is very good idea. So, we can use this idea in
algebra also.

To solve this problem, we Now,
should subtract 4a + 5b - bc 9a - 2b + 4bc
from 9a - 2b + 4bc.
(-)4a (+-)5b(-+)bc
5a - 7b + 5bc

\ The required expression to be subtracted is 5a - 7b + 5bc.

Exercise 13.5

1. Add:

(a) 5x + 3 and 7x + 9
(b) 2ab + 4bc and 5ab - 2bc
(c) 3a2 + 2b2 and 4a2 - 5b2
(d) 7a2 + 2ab + b2 and 2a2 + 3ab + 4b2
(e) 9x2 - 4y2 - 7 and 3x2 + 2y2 + 3
(f) 8x3 + 5x2 + 7x - 5 and 7x3 - 2x2 + 3x + 15
(g) 3x + 2y, 4x + 7y and 5x + 6y
(h) 2a + b, 4a + 2c, 2b + 6c and 3a + 9b
(i) 3x2 + 5x + 7, 11x2 - 2x + 6 and 9x - 2x2 - 3
(j) 4x - 5y and 2x - 10y
(k) 9x2 - 4xy and 5x2 - 4xy

230 Prime Mathematics Book - 6

2. Subtract:

(a) 2a + 4 from 7a + 9
(b) 4x + 5 from 11x + 7
(c) 2x + 5y from 6x + 9y
(d) 7p - 2q from 18p + 5q
(e) 3a2 + 5b2 from 9a2 + 7b2
(f) 2x + 3y - 7 from 9x + 6y + 8
(g) 2x2 + 3x - 2 from 5x2 - 2x + 9
(h) 3a3 + 4b3 + c3 from 5a3 - 2b3 + 4c3
(i) 5x3 + 2x2 - 4x from 7x3 + 9x2 + 11x + 7

3. Simplify: (b) 2x - 7 -(x + 9)
(d) 9x - 2y - (2x - 7y)
(a) 4a + 3b - 2a + 7b (f) 2a + 3b + 2c - (5a + 4b + 3c)
(c) 5a - 3 - (2a - 8)
(e) 11a - (3a + 4b) + (5a - 2b)

4. (a) What should be added to 3x + 4y to get 7x + 9y?
(b) What should be added to 4a + 3b - 7 to get 9a + 11b + 6?
(c) What should be added to 9x - 2y + z to get 4x + 6y + 3z?

5. (a) What should be subtracted from 7a + 5 to get 4a + 2?
(b) What should be subtracted from 11x + 7y to get 5x + 6y?
(c) What should be subtracted from 9a + 11b - 2c to get 4a + 7b - 6c?

6. (a) If x = 4a + 2b and y = 6a + 5b, find x + y and x - y.
(b) If a = 4x - 3 and b = 7x + 5, find 2a + b and 2b - a.
(c) If x = a + b and y = 2a - b, find 3x + 2y and 2x - 3y.

7. If x = 2cm, y = 3cm and z = 1cm, calculate A 5z+3
the perimeter of the given D ABC. 2x 3y+2

B C

8. If x = 4cm, find the total B
length of the given line A C D

segment. 2x x+3 3x+1

Prime Mathematics Book - 6 231

Multiplication of algebraic expression

Multiplication of 2x and 5x is 2x × 5x
= 2 × 5 × x1 × x1
= 10x1+1 = 10x2

Multiplication of Anima! What do you
2ab2 and -5a3b is 2ab2 × -(5a3b) understand about the
= 2 × (-5) × a × a3 × b2 × b multiplication of algebraic
= -10a1+3b2+1 terms from the above
= -10a4b3
examples?

Form the above examples, I
understand that in multiplication of

algebraic terms, we multiply the
coefficients of the terms and add

the power of the same base.

In the case of multiplication of algebraic expressions, the coefficients of the terms
are multiplied and the power of the same base (variable) are added.

Multiplication of monomial algebraic expressions:

We know that the monomial algebraic expression means the expression which contains
only one term. So, when we multiply two or more monomial algebraic expressions,
first we multiply the coefficients of all given expressions and add the power of the
same bases (variables) of the given expressions.

Let's learn it from the following examples.

Example 1: Multiply 6x3 and 8x5. 6 × 8 = 48 (coefficient are
Solution: Here, multiplied) x3 × x5 = x3+5 = x8
6x3 × 8x5 (power of the same base are
= 6 × 8 × x3 × x5
added.)

= 48x3 + 5 = 48x8

232 Prime Mathematics Book - 6

Example 2: Multiply 2a2b, 5b2 and - 1 ac
Solution: 6

Here, I remembered it!
2a2b × 5b2 × (-16 )ac
= 2 × 5 × (-61 ) × a2 × a × b × b2 × c (+) × (-) = (-)
= -35 a2+1b1+2c = -35 a3b3c So, 2 × 5 × -61 = -53
a2 × a = a2+1 = a3
b2 × b = b2+1 = b3

Exercise 13.6

1. Express the following without multiplication sign.

(a) 2a × b (b) 3x × 2y (c) 5a × -b (d) 1 × xy (e) 0 × abc

2. Multiply: (b) 4 × (-6b) (c) 7a × 2b (d) ab × 4c
(a) 3 × 4a (f) 4p × (-2q) (g) 5x3 × (3x4) (h) 8m3 × 2n
(e) 3x × 5x (j) -3x2y × 2xy3
(i) 7y × 3y3

3. Multiply: 1
3
(a) 2x × x × 3x2 (b) 3y × -(2y2) × 5y3 (c) y × 6y2z × 2x

(d) 1 ab × 15b2c × 1 c2d (e) 3 a × 6bc2 × 1 ab2c3
5 3 2 4

4. Area of a rectangle is length (l) × breadth (b). Find the area of the following
rectangles. Also calculate the actual area, if x = 2cm, y = 3cm and z = 4cm.

(a) (b) (c)

xz 2x

2y 3z
x

5. Volume of a cuboid is length (l) × breadth (b) × height (h). Find the volume

of the following cuboids. Also calculate the actual volume if x = 3cm,

y = 4cm and z = 2cm. (b) (c)
(a)

x x+3 2x 2z
y z 3y
233
z
y-1

Prime Mathematics Book - 6

6. (a) If x = 3b, y = 2b2 and b = 2, find 2x × y.
(b) If p = 2x, q = 3x and x = 3, find pxq2.

Multiplication of algebraic expressions by monomial expressions.

Multiply (2a + b) by 3a.
The multiplication of (2a + b) by 3a is (2a + b) × 3a
= 2a × 3a + b × 3a
= 6a2 + 3ab

Adarsh! What do you understand
about this multiplication?

I understand that each
term of the expression (2a + b) is

multiplied by 3a.

Very good! Adarsh.

Adarsh! Can you May I try, Sir!
multiply (2a2 + 3a) by

4ab?

Ok! (2a2 + 3a) × 4ab

Sir! My solution = 2a2 × 4ab + 3a × 4ab
is correct or = 2 × 4 × a2 ×a × b + 3 × 4 × a × a × b
not? = 8a3b + 12a2b

Adarsh! Your solution is absolutely
correct. You have understood very

well.Well done!

From the above conversation, it is clear that when we multiply the algebraic
expression by monomial expression then we should multiply each terms of the
expression separately by the given monomial expression. This property is called
distributive property.

234 Prime Mathematics Book - 6

Worked Out Examples.

Example 1: Multiply: (4a2 - 2b2) by 3ab.
Solution:
Here, Each term of (4a2 - 2b2) is
Example 2: (4a2 - 2b2) × 3ab separately multiplied by
Solution: = 4a2 × 3ab - 2b2 × 3ab
3ab.

= 4 × 3 × a2 × a × b - 2 × 3 × b2 × b × a

= 12a3b - 6b3a

= 12a3b - 6ab3

Multiply: (3x2y2 + 4xy - 5y3) by 8x3y.

Here,

(3x2y2 + 4xy - 5y3) × 8x3y

= 8x3y (3x2y2 + 4xy - 5y3)

= 3x2y2 × 8x3y + 4xy × 8x3y - 5y3 × 8x3y

= 24x2+3 y2+1 + 32x1+3 y1+1 - 40x3 y3+1

= 24x5y3 + 32x4y2 - 40x3y4

Example 3: If x = p - 4 and y = 2p2, find the value of xy.
Solution:
Here, It is very easy !
x = p - 4 and y = 2p2 Just substituting the values
Now, xy = (p - 4) × 2p2
= p × 2p2 - 4 × 2p2 of x = p - 4 and
= 2p3 - 8p2 y = 2p2 in xy and multiply as

above example.

1. Multiply: Exercise 13.7
(a) (a + b) by 2ab (b) (x + 3y) by 2x2

(c) (2x - 4) by 3xy (d) (2a2 - 3ab) by -5ab2

(e) (7x + 5y) by -3xy (f) (5x2 - 4y2) by 3xy

(g) (4p2 - 9q2) by 3pq2 (h) (9b + 7c) by 5bc

(i) (12a2 + 17bc) by 3abc

Prime Mathematics Book - 6 235

2. Multiply: (b) (x2 - xy + y2) by 5xy
(a) (a + 2b - 3c) by 4abc (d) (x2 - 2x + 5) by 3x
(c) (3a2 - 5ab + 4b2) by -2ab (f) (2a3 + 3a2b - 4b2 - 8) by -3ab2
(e) (x3 - 4x2 + 5x - 7) by 3x

3. Area of a rectangle is length (l) × breadth (b). Find the area of the following
rectangles. Also calculate the actual area if a = 2 and b = 3.

(a) (b) (c) 2a
b 3b

(a+b) (2a+3b) (3a+b)

4. (a) If a = 2x + 3 and b = 2y, find the product of a and b.

(b) If m = 4x - y and n = 2xy, find the product of m and n.

(c) If x = 3a + 2b and y = 4ab, find the value of x × y.

Multiplication of algebraic expressions by binomial expression

In case of multiplication of an algebraic expression by a binomial expression, each
term of the algebraic expression is separately multiplied by each term of the binomial
expression. Then the product is simplified. Let's learn this process in the following
worked out examples.

Worked Out Examples.

Example 1: Multiply: (2a + 3b) by (a - 2b)
Solution: Here,
(2a + 3b) by (a - 2b)
= a(2a + 3b) - 2b(2a + 3b)
= 2a2 + 3ab - 4ab - 6b2
= 2a2 - ab - 6b2

236 Prime Mathematics Book - 6

Example 2: Multiply: (3x + 4y - 3) by (x - y)
Solution: Here,
(3x + 4y - 3) by (x - y)
= x (3x + 4y - 3) - y(3x + 4y - 3)
= 3x2 + 4xy - 3x - 3xy - 4y2 - 3y
= 3x2 + xy - 3x - 4y2 - 3y

Example 3: If a = 2x2 + 3xy - 2y2 and b = x + 2y, find the product of a and b.
Solution: Here,
a = 2x2 + 3xy - 2y2 and b = x + 2y
The product of a and b is = a × b

= (2x2 + 3xy - 2y2) × (x + 2y)
= x(2x2 + 3xy - 2y2) + 2y(2x2 + 3xy - 2y2)
= 2x3 + 3x2y - 2xy2 + 4x2y + 6xy2 - 4y3
= 2x3 + 7x2y + 4xy2 - 4y3

1. Multiply: Exercise 13.8
(a)(x + 2) by (x + 3)
(b) (2x - 3) by (x + 4) (c)(y - 3) by (2y + 1)
(d)(2x + y) by (x - y)
(e) (4a - b) by (2a + 3b) (f)(4x - y) by (4x + y)

2. Multiply: (b) (x2 - 2x + 5) by (x - 3)
(d) (a2 + a + 1) by (a - 1)
(a) (x2 + x + 2) by (x + 3)
(c) (2x2 + 5x - 7) by (2x + 1)

3. (a) If x = 2a2 + 5a + 3 and y = a + 3, find the product of x and y.

(b) If x = a + 4 and y = a - 4, find the product of x and y.
(c) If p = x + y, q = x - y and r = x2 + y2, find pq + r.

Prime Mathematics Book - 6 237

Division of algebraic expressions

Study and learn the following.

Let's take the area of a rectangle of which length is 3x untis be 6xy sq. units. To find
the breadth of the rectangle, we apply the formula for the area of the rectangle.

Area of the rectangle = length × breadth

or, 6xy = 3x × breadth

or, 6xy = 3x × breadth [dividing both sides by 3x]
3x 3x

or, 2 × 3 × x × y = breadth
3 × x

\ breath = 2y units

Hence, the breadth of the rectangle is 2y units.

From the above example, it is clear that while dividing an algebraic term by another
algebraic term, we should divide the coefficient of dividend by the coefficient of
divisor and cancel (subtract) the same base (variables) of the dividend by the divisor.

Pinki! Can you divide First I find the Yes, I can.
18x2y by 6xy? factors of 18 and 6. May I try, sir!
Then I write x2y and
Ok! xy as x × x × y and My this process is
x × y. Then after I correct or not, sir!
Pinki! Your process is absolutely cancel common
correct. Very good. factors of the
numerator and
denominator. Divide 18x2y by 6xy.

= 18x2y
6xy

= 2 × 3 × 3 × x × x × y
2 × 3 × x × y

= 3x

238 Prime Mathematics Book - 6

Worked Out Examples

Example 1: Divide 15x3 by 3x2.
Solution:
Here,

15x3 ÷ 3x2

= 15x3
3x2
3 × 5 × x × x × x
= 3 × x × x [\15 = 3 × 5, x3 = x × x × x and x2 = x × x]

= 5x

Example 2: Divide 21a3b2 by 7a2b.
Solution:
Here,

21a3b2 ÷ 7a2b

= 21a3b2
7a2b
3 × 7 × a × a × a × b × b
= 7 × a × a × b

= 3ab

Division of a binomial expression by a monomial expression:

Study and learn the following:

Divide (8x3 + 6x2) by 2x. Alternative process
2x ) 8x3 + 6x2 ( 4x2 + 3x
Here, (8x3 + 6x2) ÷ 2x
-8x3
= 8x3 + 6x2 6x2
2x -6x2
0
= 8x3 + 6x2
2x 2x \ Quotient = 4x2 + 3x

= 2×2×2×x×x×x + 2 × 3 × x × x
2×x 2 × x

= 4x2 + 3x

Prime Mathematics Book - 6 239

From the above example, it is clear that
while dividing a binomial expression by a monomial
expression, we should divide separately each term of

dividend by divisor.

Pemba! Can you divide Yes sir! I can.
(15a3b2 - 12a2b2) by 3a2b ?
Sir! My process is
Ok! Show your process. correct or not?

Pemba! Your process is
absolutely correct.

The process of pemba

(15a3b2 - 12a2b2) ÷ 3a2b

= 15a3b2 - 12a2b2
3a2b

= 15a3b2 - 12a2b2
3a2b 3a2b

= 3 × 5 × a × a × a × b × b - 2 × 2 × 3 × a × a × b × b
3 × a × a × b 3 × a × a × b

= 5ab - 4b

Worked Out Examples

Example 1: Divide (21x4 + 14x2) by 7x
Solution:
Here,

(21x4 + 14x2) ÷ 7x Alternative process
7x ) 21x4 + 14x2 ( 3x3 + 2x
= 21x4 + 14x2
7x -21x4
21x4 14x2
= 7x + 14x2 -14x2
7x 0

= 3 × 7 ×x × x × x × x + 2 × 7 × x × x \ Quotient = 3x3 + 2x
7 × x 7 × x

= 3x3 + 2x

240 Prime Mathematics Book - 6

Example 2: Divide (2p2q - 6pq2) by 2pq.
Solution:
Here, Alternative process
2pq) 2p2q - 6pq2 (p - 3q
(2p2q - 6pq2) ÷ 2pq
-2p2q
= 2p2q - 6pq2 - 6pq2
2pq +- 6pq2
0
= 2p2q - 6pq2
2pq 2pq \ Quotient = p - 3q

= 2×p×p×q - 2×3×p×q×q
2×p×q 2×p×q

= p - 3q

Exercise 13.9

1. Divide: (b) 8x2 by 2x (c) 12a3 by 4a (d) 10m4 by 5m2
(a) 5x by x (f) 18b4 by 6b2 (g) 7a3 by a (h) -21p3 by 7p2
(e) -15p3 by 5p2

2. Divide: (b) 7x2y by xy (c) 12m3n2 by 4mn
(a) 6a2b by 3ab (f) 24m7n5 by -4m3n3
(i) 121x3y2z by 11x2y
(d) 35x3y2 by 5x2y2 (e) -40x3y4 by 8x2y3

(g) 45a3b4 by 9a2b2 (h) -42p3q2 by 7p2q

(j) 48a3b2c4 by (-4a2bc2)

3. Divide: (b) (2x3 - 3x2) by x2
(a) (a3 + a2) by a (d) (10x3 + 15x2) by 5x
(c) (4a3 + 6a4) by 2a2 (f) (2x3y - x2y2) by x2y
(e) (21x4 - 28x2) by 7x2 (h) (72x2y2z - 18x2y3z2) by 6x2yz
(g) (12x2y3 + 20x3y2) by 4x2y

4. (a) Area of a rectangle = 20x2y sq. units and it's length = 5xy unit. Find it's
breadth.

(b) The product of two algebraic terms is 12x4y2 + 15x2y3. If one of the
expressions is 3x2y2, find the other expression.

(c) Divide the product of 3xy and (2x2 + 5xy2) by 6x2y.

(d) Divide the sum of 4a3y and 10a2y3 by 2ay2.

Prime Mathematics Book - 6 241

Equation and Inequality
Mathematical statements:

x + 5 = 17 9 + 7 = 16

If x = 12, x + 5 = 17 is true. 9 + 7 = 16 is already true.

The sum of 9 and 4 is 13, i.e. 9 + 4 = 13
The difference of 10 and 6 is 4, i.e. 10 - 6 = 4
The product of 9 and 4 is 36, i.e. 9 × 4 = 36
The quotient of 12 divided by 4 is 3, i.e. 12 ÷ 4 = 3

All the above statements contain the mathematical operations like addition
(+), subtraction (-), multiplication (×) and division (÷).

Similarly, the sum of x and 5 is 12, i.e. x + 5 = 12.

The product of x and 4 is 20, i.e. x × 4 = 20

All the above statements are called mathematical statements.

A mathematical statement may be true or false. For example, the sum of 9 and 6
is 15, which is true statement.

The quotient of 20 divided by 5 is 6, which is false statements. From the above
examples, it is clear that the same statement can not be true or false at the same
time.

Open mathematical statements
Study and learn the following:
Let's consider a mathematical statement, the sum of x and 9 is 15.
It is written mathematically as x + 9 = 15.
We can not say whether x + 9 = 15 is a true or false statement before replacing x
by any numeral value.
If x is replaced by 6 in x + 9 = 15, 6 + 9 = 15 or 15 = 15, which is true.

If x is replaced by 5 is x + 9 = 15, 5 + 9 = 15 or 14 = 15, which is false.

242 Prime Mathematics Book - 6

Let's study the examples:

Open statement True statement False statement

y-3=7 10 - 3 = 7 12 - 3 = 7

x Î {odd numbers} 3 Î {odd number} 4 Î {odd numbers}

x is less than 9. 4 is less than 9 12 is less than 9

From the above examples, it is clear that such mathematical statements which can
not be predicated as true or false statements before replacing the numeral value
of the variable are known as open mathematical statements.

Exercise 13.10

1. Identify whether the following mathematical statements are true, false or

open statements.

(a) The sum of 3 and 9 is 12. (b) 7 is a prime number.

(c) The product of 7 and 5 is 40. (d) y + 7 = 19

(e) x Î {even number} (f) 3x is always less than 12.

(g) The product of y and 7 is 56. (h) 1190 is exactly divisible by 25.

(i) If x = 5, 2x - 3 = 7

2. Fill the correct numeral in so that the given statement can be true.

(a) The sum of and 14 is 20. (b) is more than 8 by 4.

(c) The difference of 12 and is 3. (d) is an odd number.

(e) is a whole number between 17 and 19. (f) The product of and 9 is 63.

(g) is an odd number and exactly divides 20. (h) ÷ 9 = 12

(i) The sum of and 1 is the first even square number.

3. Write the all possible values of the variables (letters) in the following open
mathematical statements so, that each statement is true.

(a) x + 8 = 13 (b) y - 4 = 9 (c) 3a = 24
x = ____ y = ____ A = ____

(d) x divides 21 exactly. (e) y represents the prime number
x = ____ between 9 and 20.
y = ____

(f) 2 of x = 12 (g) x ÷ 4 = 7 (h) 19 - y = 7
3
x = ____ x = ____ y = ____

Prime Mathematics Book - 6 243

Equation

Suppose, a boy has some mangoes (x mangoes). His friend give him 5 mangoes, then
he has 12 mangoes altogether. This statement can be written as x + 5 = 12. It is an
open mathematical statement containing equal (=) sign. This statement is true by
only one value (fixed value) of x. such type of open mathematical statement is called
an equation.

What value of x be Let me think sir. When I
substitute in x + 5 substitute 7 in place of x,
then x + 5 becomes 12. Is it
gives 12 ?
true ? sir!
Yes! It is true.

Now, x + 5 becomes 7 + 5 = 12.
So, x = 7 is the solution of x + 5 = 12.

The other examples of equations are x - 5 = 7, 2y + 4 = 12, 3x/2 = 15 etc.
Let's take an equation x + 9 = 27.
In the equation x + 9 = 27, x + 9 is an expression on left side of equal (=) sign. So,
it is called the expression of left hand side (L.H.S.). Similarly, 27 is the expression
on right side of equal (=) sign. So, it is called expression of right hand side (R.H.S.).

Solution of an equation
Let's take a set of numbers i.e. {1, 2, 3, 4, …………} and an equation y + 2
= 5. Now, substitute the place of y by each element of the given set.
When, y = 1, then y + 2 = 5 becomes 1 + 2 = 5 or 3 = 5, it is false.
When, y = 2, then y + 2 = 5 becomes 2 + 2 = 5 or 4 = 5, it is false.
When, y = 3, then y + 2 = 5 becomes 3 + 2 = 5 or 5 = 5, it is true.
When, y = 4, then y + 2 = 5 becomes 4 + 2 = 5 or 6 = 5, it is false.
From the above example, it is clear that y + 2 = 5 is true by substituting
y = 3 only.

244 Prime Mathematics Book - 6