Vedanta Excel in Mathematics Book 5 Final (2078) NEW

Time, Money, Bill and Budget

10.5 Multiplication and division of time

Let's study the following examples and learn about the multiplication and
division of units of time.

Example 8: A water pump can ϔill a tank completely in 1 h 20 min. How
long does the pump take to ϔill 4 such tanks?

Solution

1 h 20 min 80 min = 60 min + 20 min
×4 = 1 h 20 min

4 h 80 min 4 h + 1 h 20 min = 5 h 20 min

5 h 20 min

Hence, the pump takes 5 h 20 min to ill 4 such tanks.

Example 9: The total teaching hours in 6 days of a week is 31 h 30 min and
there are equal teaching hours in each day. Find the teaching
hours in each day.

Solution

Here, the teaching hours in each day = 31 h 30 min ÷ 6

6 )31 h 30 min) 5 h 15 min

–30
1 h 30 min

+ 60 min
90 min

–6 min
30 min
30 min
0

Hence, the teaching hours in each day is 5 h 15 min.

EXERCISE 10.2
Section A - Classwork
1. Let's add and regroup into the higher units.

a) 30 min + 40 min =

b) 40 min + 50 min =

c) 50 s + 30 s =

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d) 25 days + 10 days = 35 days = 1 month 5 days

e) 20 days + 20 days =

f) 7 months + 8 months =

g) 9 months + 5 months =
2. Let's convert into the lower units and subtract.

a) 1 h – 10 min = 60 min – 10 min =

b) 1 h – 25 min =

c) 1 m – 20 s =

d) 1 year – 7 months =

e) 1 month – 12 days =
Section B

3. Let's add or subtract.

a) 2 h 30 min + 1 h 35 min b) 3 h 45 min + 2 h 40 min

c) 4 h 20 min – 1 h 30 min d) 6 h 15 min – 3 h 45 min

e) 2 years 7 months + 1 year 6 months

f) 4 years 9 months + 3 years 10 months

g) 5 years 3 months – 2 years 4 months

h) 8 years 6 months – 4 years 9 months

i) 4 months 20 days + 5 months 25 days

j) 11 months 10 days – 6 months 15 days
4. Let's add or subtract and write the time using a. m. or p. m.

a) 7 : 40 a. m. + 1 h 35 min b) 4 : 30 p. m. + 2 h 50 min

c) 9 : 45 a. m. + 3 h 25 min d) 11 : 50 p.m. + 4 h 40 min

e) 8 : 15 a. m. – 2 h 30 min f) 5 : 20 p. m. – 3 h 45 min

g) 1 : 10 p. m. – 4 h 50 min h) 6 : 30 p. m. – 10 h 30 min

5. a) Inter-house basketball match started at 2 : 45 p. m. and it was of 2 h 45 min.
At what time was the match over?

b) Mathematics class started at 11 : 30 a. m. and the class was over after
45 minutes. At what time was the class over?

c) Buses arrive in every 45 minutes at a bus park. The irst bus arrived at
6 : 30 a. m. At what time will the next bus arrive?

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d) The duration of a movie show was of 2 h 50 min. If it was over at 1 : 20 p. m.,
at what time was it started?

e) Because of the bad weather a plane was delayed by 1 h 25 min to take off
from Lukla airport. If it took off at 1 : 05 p. m., what was its actual take off
schedule?

6. a) Mrs. Shakya travelled 1 h 20 min by a taxi to reach a bus park. Then she
travelled 3 h 55 min by a bus and arrived her home town. Calculate the time
she spent in travelling.

b) A maths teacher teaches 2 h 15 min before the tif in break and 1 h 30 min
after the tif in break everyday. Find her/his teaching hours in a day.

c) Kamala completed her Maths and Science homework in 1 h 30 min. If she
completed Science homework in 50 minutes, how long did she take to
complete Maths homework?

d) The working hours of a man in a factory is 7 h 30 min everyday in two
shifts, morning and evening. If he works 4 h 45 min in the morning shift,
how long does he work in the evening shift?

7. a) A maths class starts at 12 : 30 p. m. and gets over at 1 : 15 p. m. How long is
the class conducted?

b) A plane takes off at 12 : 50 p. m. from Tribhuvan airport. It is landed at
Dhangadhi airport at 2 : 05 p. m. How long is the light?

c) A T20 cricket match started at 10 : 45 a. m. and it was over at 2 : 15 p. m.
How long was the match played?

d) On July 1st the sun rose at 5 : 15 a. m. and set at 6 : 55 p. m. How long was
the sun above the horizon?

8. a) The departure timetable of the buses from Kathmandu to different places is
given below. Workout the following problems.

Places Bus - A Bus - B Bus - C

Itahari 6 : 15 a. m. 7 : 30 a. m. 8 : 45 a. m.

Pokhara 6 : 10 a. m. 7 : 15 a. m. 8 : 05 a. m.

Nepalganj 1 : 45 p. m. 2 : 30 p. m. 3 : 30 p. m.

a) If the bus - A arrived Itahari at 4 : 10 p. m., how long did it travel?

b) If the bus - B takes 5 h 30 min to arive at Pokhara, at what time does it
arrive?

c) If the bus - C arrived Nepalganj at 6 : 15 a. m. next day, how long did it
travel?

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9. a) Mrs. Mahato borrowed a loan from a bank on 15 Kartik 2076 and paid on
25 Chaitra 2077 B. S. How long did she use the money?

b) The construction of a health post building was started on 18 Baisakh 2077
and completed on 27 Chaitra 2077 B. S. How long did it take to complete the
construction?

c) Let's subtract your date of birth from today's date. Then, ind how old you
are today.

10. a) A water pump can ill a tank completely in 1 h 30 min. How long does the
pump take to ill 3 such tanks?

b) Each class period of a school is of 40 minutes long.

(i) Find the duration of 4 periods in hours and minutes.

(ii) If the irst period starts at 9 : 45 a. m., at what time does the fourth
period start?

c) Daily working hours in a noodle factory in Nepal is 8 h 45 min. Find the
working hours of the workers in 6 days.

d) Each terminal session of a school is of 65 days. How many months and days
are there in 3 terminal sessions of the school?

11. a) There are 7 equal durations of class periods in a school each day. If the total
school hours in a day is 5 h 15 min, ind the duration of each period.

(Hint: 5 h 15 min = 5 × 60 min + 15 min) = 315 min and 315 min ÷ 7)

b) The total school hours from Sunday to Thursday in a school is 32 h 30 min.
Find the daily school hours of the school from Sunday to Thursday.

c) How many weeks and days are there in 1 year (365 days)?

d) A school has 210 school days in a year. If these school days are equally
divided into 3 terminal sessions, how many months and days are there in
each terminal session?

It's your time - Project work!

12. a) Let's prepare your school routine of your class from Sunday to Friday on a
chart paper.

b) How many school hours does your school have everyday? Also, calculate
the school hours of your school in a week.

c) How much time do you spend while coming to school and going back to

your home. (i) in 1 day? (ii) in 6 days?

d) How much time do you spend on watching TV?

(i) in 1 day? (ii) in 1 week? (iii) in 1 month?

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e) Write the date of birth of your any ive friends. Then, ind how old they are
today.

f) Write the date of birth of your parents. Then, ind how old they are today.

10.6 Money - Looking back

Classwork - Exercise

1. Let's say and write the answers as quickly as possible.

a) Re 1 = paisa (p) b) Rs 2 = paisa (p)

c) Rs 2.25 = paisa (p) d) Rs 4.50 = paisa (p)

e) 300 p = rupees (Rs) f) 500 p = rupees (Rs)

g) 175 p = rupees (Rs) h) 74o p = rupees (Rs)

10.7 Conversion of money

In our Nepali currency system, we use rupees (Rs) and
paisa (p). Rupees is higher unit and paisa is lower unit of
money.

Re 1 = 100 p, Rs 2 = 200 p, … and so on.

We use this relation between rupees and paisa to convert rupees to paisa or
paisa to rupees.

× 100

Rupees Paisa

÷ 100

(i) We should multiply rupees by 100 to convert into paisa.
(ii) We should divide paisa by 100 to convert into rupees.
We usually express rupees and paisa in the decimal of rupees.
For example: Rs 7 and 65 p is Rs 7.65.
Now, let's study the given examples and learn more about the conversion of
money.

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Example 1: Convert Rs 4 and 50 p a) into paisa b) into rupees

Solution It's easier!
a) Rs 4 and 50 p = 4 × 100 p + 50 p Re 1 = 100 p
So, Rs 4 = 4 × 100 p = 400p
= 400 p + 50 p

= 450 p

b) Rs 4 and 50 p = Rs 4 + Rs 50 I got it!
100
100 p = Re 1
1
= Rs 4 + Rs 0.50 1p = Rs 100 and

= Rs 4.50 50p = Rs 50 = Rs 0.50
100

10.8 Addition and subtraction of money

Usually, we express rupees and paisa in the decimal of rupees. Then, we use
the general rules of addition and subtraction of decimal numbers.

Example 2: Add Rs 72 and 45 p + Rs 30 and 60 p.

Solution Rs p
Rs 72 and 45 p = Rs 72.45 72 45
Rs 30 and 60 p = + Rs 30.60 + 30 60
102 105
Rs 103.05 103 05
= Rs 103.05

Example 3: Subtract Rs 96 and 15 p – Rs 58 and 50 p.

Solution Rs p
Rs 96 and 15 p = Rs 96.15 96 15
Rs 58 and 50 p = – Rs 58.50 – 58 50
37 65
Rs 37.65 = Rs 37.65

10.9 Multiplication and division of money

Let's study the following examples and learn about the multiplication and
division of money. In this case, we use the general rules of multiplication and
division of decimal of rupees.

Example 4: Mr. Thapa bought 3 kg of vegetables at Rs 50.40 per kg and
2 kg of fruits at Rs 90.75 per kg. How much money did he spend
altogether?

Solution

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Cost of 3 kg of vegetables = 3 × Rs 50.40 = Rs 151.20

Cost of 2 kg of fruits = 2 × Rs 90.75 = + Rs 181.50

Rs 332.70

So, he spent Rs 332.70 altogether.

Example 5: A fruit seller bought oranges at Rs 80.50 per kg and sold at
Rs 90.25 per kg. How much proϔit did she make in selling
20 kg of oranges?

Solution

Cost price of 1 kg of oranges = Rs 80.50

Selling price of 1 kg of oranges = Rs 90.25

Pro it in selling 1 kg of oranges = selling price – cost price

= Rs 90.25 – Rs 80.50

= Rs 9.75

Now, pro it in selling 20 kg of oranges = 20 × Rs 9.75 = Rs 195.00

So, she made a pro it of Rs 195 in selling 20 kg of oranges.

Example 6: A stationer makes a proϔit of Rs 51.60 by selling 1 dozen of
exercise books. How much proϔit does he make in 1 exercise
book?

Solution

Pro it of 1 dozen of exercise books = Rs 51.60

Pro it of 12 exercise books = Rs 51.60

Pro it of 1 exercise book = Rs 51.60 ÷ 12 = Rs 4.30

Hence, he makes a pro it of Rs 4.30 in 1 exercise books.

EXERCISE 10.3
Section A - Classwork

1. Let's convert into rupees or paisa as quickly as possible.

a) Re 1 = p b) Rs 3 = p c) Rs 4.25 = p

d) Rs 5 = p e) 400 p = f) 700 p =

g) 150 p = h) 255 p = i) 375 p =

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2. Let's say and write in the decimal of rupees.

a) Rs 2 and 15 p = b) Rs 3 and 40 p =

c) Rs 4 and 68 p = d) Rs 7 and 99 p =

Section B
3. Let's add or subtract.

a) Rs 15.20 + Rs 12.90 b) Rs 36.45 + Rs 54.75

c) Rs 48.15 – Rs 18.30 d) Rs 120.35 – Rs 60.50

e) Rs 70 and 80 p + Rs 115 and 60 p f) Rs 295 and 55p + Rs 769 and 85 p

g) Rs 450 and 40 p – Rs 125 and 75 p h) Rs 930 and 5 p – Rs 580 and 10 p

4. Let's multiply or divide.

a) 4 × Rs 5.30 b) 6 × Rs 8.25 c) 5 × Rs 9.40
f) Rs 60.45 ÷ 3
d) 12 × Rs 10.75 e) 15 × Rs 18.20 i) Rs 240.40 ÷ 8

g) Rs 80.50 ÷ 5 h) Rs 150.75 ÷ 9

5. a) You had a plate of Mo:Mo for Rs 110.50 and an ice-cream for Rs 45.75 in a
restaurant.

(i) How much money did you pay to clear the bill?

(ii) If you gave a note of Rs 500 to pay the bill, how many rupees did the
shopkeeper return you?

b) Mother bought 2 kg of vegetables at Rs 45.60 per kg and 3 kg of fruits at
Rs 80.50 per kg.

(i) How much money did she spend altogether?

(ii) If she gave a note of Rs 1000 to the shopkeeper, how much money did the
shopkeeper return her?

c) Mr. Pandey earns Rs 750.50 everyday and he spends Rs 420.80 to run his
family.

(i) How much money does he save in 1 day?

(ii) How much money does he save in 1 week?

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6. a) A fruit seller bought apples at Rs 110.60 per kg and sold at Rs 120.40
per kg. How much pro it did she make in selling 15 kg of apples?

b) A grocer purchased eggs at the rate of Rs 250.25 per crate (30 eggs)
and sold at Rs 280.50 per crate, How much pro it did he make in selling
12 crates of eggs?

c) Mrs. Priyar is a stationer. She makes a pro it of Rs 41.40 by selling 1 dozen
of colour pencil. How much pro it does she make in 1 colour pencil?

d) If the cost of 5 l of petrol is Rs 552.50, ind the cost of 1 l of petrol.

It's your time - Project work!
7. a) Let's copy the table given below in a chart paper. Write the names of

greater and smaller units of money of these countries. For example, rupees
is greater and paisa is smaller units of money in Nepal. You can visit the
available website such as www.google.com to ind the information.

Countries USA UK India Malaysia Israel Qatar
Greater units

Smaller units

b) Find today's exchange rate of the currencies of these countries with Nepali
currency. Then, calculate the values of the following currencies.

(i) 10 (ii) 10 Ringgit (iii) 10 Rial

(iv) 10 pound (v) 10 Shekel (vi) 10 dollar

10.10 Bill

When we buy things from a shop the shopkeeper
write out the details (name of items, quantities,
rates, amounts, total amount, etc.) on a piece of
paper and gives us for the payment. This piece
of paper is called a bill.

Let's study the given example and learn to
prepare a bill.

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Example 1: Father purchased 10 kg of rice at Rs 90.50 per kg, 2 kg of sugar
at Rs 85 per kg, and 3 l, of cooking oil at Rs 180 per litre from
a grocery store. Prepare a bill received by him.

Solution

Maitidevi Grocery Store

Rajabas - Sunsari

Bill No. 0342 Date: 2077/08/17

Customer Name and address: Dhanapati Limbu, Koshi Tappu

S. No. Particulars Quantity Rate (Rs) Amount (Rs)

1 Rice 10 kg Rs 90.50 905.00

2 Sugar 2 kg Rs 85.00 170.00
3 Cooking oil 3 l Rs 180.00 540.00
Sold by: Badri Rai
Total 1615.00

10.11 Budget

A budget is a description of income of a family or organisation from different
sources and a plan of how it will be spent over a period of time. A budget may
help to balance between income and expenditure and manges saving.

Example 2: The annual earning of a family from different sources and

the planning of expenses on different titles are given below.

Prepare an annual budget of the family.

Sources and income Titles and expenditure

Service Rs 2,50,000 Food and cloths Rs 1,10,000

Business Rs 1,80,000 Education Rs 1,20,000

Farming Rs 90,000 Taxes Rs 25,000

Insurance Rs 50,000

Solution

Income Expenditure)

Sources Amounts (Rs) Titles Amount (Rs)

Services 2,50,000 Food and cloths 1,10,000
1,20,000
Business 1,80,000 Education
25,000
Farming 90,000 Taxes 50,000
Insurance 3,05,000

Total 5,20,000 Total

Saving = Rs 5,20,000 – Rs 3,05,000 = Rs 2,15,000

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EXERCISE 10.4

Section A - Classwork

1. Let's read the bill given below. Then, say and write the answer of the
questions.

Janaki Stationery

Ram Chowk, Janakpur

Bill No. 0252 Date: 2077/10/18

Customer Name and address: Rekha Gupta Address: Wakil Tol, Dhanusha

S. No. Particulars Quantity Rate (Rs) Amount (Rs)

1 Exercise books 8 Rs 40.00 320.00

2 Gel pens 3 Rs 25.00 75.00

3 Colour pencils 6 Rs 10.00 60.00

Sold by: Manish Chaudhari Total 455.00

a) What is the name of the shop?

b) Who is the customer?

c) What is the rate of cost of exercise book?

d) How many gel pens did the customer buy?

e) What amount did the customer pay for colour pencils?

f) What is the total amount of the bill?

Section B

2. Let's prepare the bills given by a shopkeeper to customers while
purchasing the following items.

a) 5 kg of wheat lour at Rs 56 per kg b) 12 exercise books at Rs 50 each
3 kg of sugar at Rs 80 per kg 2 paint boxes at Rs 75 each
2 kg of pulses at Rs 120 per kg 3 drawing books at Rs 60 each
2 l mustard oil at Rs 250 per litre 4 pencils at Rs 10 each

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c) On your birthday, you and your family had 3 plates Mo:Mo at Rs 115 per
plate, 1 cake for Rs 450, 4 glasses of fresh juice at Rs 75 per glass, and 2 ice-
cream at Rs 60 per piece in a restaurant. Prepare a bill given to you for the
payment.

3. a) Let's copy the annual budget of Mr. Gurung given below. Then answer the
following questions.

Income Expenditure
Titles Amount (Rs)
Sources Amounts (Rs)

Vegetable farming 1,05,000 Food and 90,000
cloths

Poultry 1,50,000 Education 1,10,000

Fishery 1,80,000 Insurance 45,000

Miscellaneous 25,000

Total Total

(i) Calculate his total annual income and expenditure.

(ii) How much money is he planning to save?

b) The annual income of Mrs. Sangita Thakuri from different sources and
expenditures on different titles are given below. Prepare her annual budget.

Sources and income Titles and expenditure

Goat farming Rs 1,25,000 Food and cloths Rs 85,000

Dairy Rs 1,10,000 Education Rs 1,20,000

Bee Keeping Rs 75,000 Health Insurance Rs 35,000

Taxes Rs 15,000

How much money is she planning to save in a year?

It's your time - Project work!

4. a) Let's make groups of your friends. Visit a few number of shops in your
surroundings and collect the sample of some bills.

b) Let's discuss with your parents about the monthly income and expenditure
of your family. Then, prepare an annual budget of your family. How much
money are you planning to save for your family in a year?

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Unit Algebra - Algebraic Expressions

11

11.1 Constant and variable - Looking back

Classwork - Exercise

1. Let's say and write the values of letters as quickly as possible.

a) x is the number of students in your class. So, x =

b) x is the number of provinces in Nepal. So, x =

c) x is the number of districts in Nepal. So, x =

d) y is the number of teachers in your school. So, y =

e) y is the number of maths teacher in your school. So, y =

2. Let's say and write whether the letters represent 'constant' or 'variable'.

a) x represents the ages of your friends. x is a

b) x represents the number of sides of a triangle. x is a

c) y represents the natural numbers less than 10. y is a

d) y represents the natural numbers between 8 and 10. y is a

In this way, the number 1, 2, 3, 4, 5, … always represent a ixed number of objects.
For example, 5 always represents ive number of objects, 8 always represents
8 number of objects, and so on. So, the numbers are called constants.

On the other hand, the letters x, y, z, a, b, p, q, … do not represent a ixed or
constant number of objects. We can use these letters to represent any number
of objects. For example, when x represents the number of provinces of Nepal,
its value is 7. But when it represents the number of districts of Nepal, its
value is 77. So, these letters are called variables. The word 'variable' means
something that can vary or change.

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11.2 Operation on constant and variable
Let's study the following illustrations and learn about the operations (addition,
subtraction, multiplication, and division) between constant and variables.
How many marbles are there altogether in bags on each skateboard?

333 xx y yy5

3+3+3 x+x y+y+y+5
3 is added 3 times x is added 2 times 3 times y and 5 more
3×3=9 2 × x = 2x 3 × y + 5 = 3y + 5
Similarly,

a is added to 1 = a + 1 a and a + 1 are variables and 1 is constant.

4 is subtracted from p = p – 4 p and p – 4 are variables and 4 is constant.

The sum of x and 3 is divided by 2= x+ 3 x and x+3 are variables. 3 and 2
2 2
are constants.

11.3 Algebraic term and expression

Algebraic terms : x, y, 1, 2, 5, 2x, 3a, 4xy, … are terms. A term can be a number,

a variable, or a number and variable combined by multiplication or division.

3x , 5xy , ab, 2pqr, … are a few more examples of terms.
2 4

Algebraic expressions

5x is a term. But 5 + x , 5 – x, x + 5, or x – 5 are expressions.

Similarly, x + y, x – y, a + b + c, p – q + 3, x +y – 1, … are a few more examples
of algebraic expressions. 2

In this way, an algebraic expression is a collection of terms separated by
addition or subtraction signs.

11.4 Types of algebraic expressions

Depending on the number of terms of an algebraic expression, it may be
monomial, binomial, trinomial, or polynomial expression.

Monomial expression

x, 3x, 2xy, 4abc, … are monomial expressions. A monomial expression has only
one term.

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Algebra - Algebraic Expressions

Binomial expression
x + 2, x – y, 2a + 3b, … are binomial expressions. A binomial expression has
two terms.
Trinomial expression
x + y – 3, a + b + c, 2p – 3q – 1, … are trinomial expressions. A trinomial
expression has three terms.
Polynomial expression
x + y + z – 4, ab + bc – ca + abc – 2, … are polynomial expression. A polynomial
expression has more than three terms.

It's interesting!
5xyz is a monomial, 5 + xyz is a binomial,
5 + x + yz is a trinomial and 5 + x + y + z
is a polynomial expressions!

11.5 Evaluation of algebraic expression
When x = 2, the value of x + 3 = 2 + 3 = 5.
When y = 3, the value of y – 7 = 3 × 3 – 7 = 9 – 7 = 2
When l = 4 and b = 3, the value of l × b = 4 × 3 = 12
Thus, we can ind the value of an algebraic expression by replacing the variables
with number. It is called evaluation of algebraic expression.

EXERCISE 11.1
Section A - Classwork

1. Let's list the constants and variables separately in the table.

x, 2, 2x, 5 Constants
3y, 8, 5p

x , 2 , a + b Variables
2 3

2. Let's say and write whether these letters represent constants or variables.

a) x is an odd number less than 5. x is a
b) y is an odd number between 2 and 4. y is a

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c) p is the number of students in your class. p is a

d) q is the height of the students in your school. q is a

3. Let's say and write the correct answers in the blank spaces.

a) In 3x, constant is variables are ,

b) In 5y, constant is variables are ,

c) In x + 7, constant is variables are ,

d) In 3y + 2, constants are variables are , ,

4. Let's say and write the correct answers in the blank spaces.

a) Cost of 1 book is Rs x, cost of 3 books is

b) Cost of 1 pencil is Rs 9, cost of y number of pencils is

c) Cost of 2 kg of rice is Rs x, cost of 1 kg of rice is

d) 2 marbles are added to x number of marbles, total marbles are

e) 3 pens are taken away from y number of pens, pens are left.

5. Let's say and write the terms and the types of expressions.

a) 4xy Terms Types of expressions
b) 4 + xy
a)

c) 4 + x + y b)
d) 4 + x + y + z c)
d)

6. Let's say and write the values of the expressions quickly.

a) If x = 2, then (i) x + 2 = (ii) 2x = (iii) x
2

b) If y = 4, then (i) y – 1 = (ii) 3y = (iii) y
2

c) If x = 2 and y = 1, then (i) x + y = (ii) x – y =

d) If a = 5 and b = 3, then (i) ab = (ii) a =
b
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Section B
7. Let's answer the following questions.

a) De ine the meaning of constant and variable? Give one example of each.

b) What is a term of an algebraic expression? Give any two examples of terms.

c) What is an algebraic expressions? Give any two examples of algebraic
expressions.

d) What is the meaning of evaluation of an algebraic expression?

e) What type of expression is 6pqr? Let's make a binomial, a trinomial and a
polynomial expression of your own from this expression.

8. Let's use these operations and make algebraic expressions.

a) The sum of x and 2y. b) The difference of 3a and b.

c) The product of 2p and 3q. d) The sum of x and y divided by 2.

e) 2 times the sum of l and b. f) 3 times the difference of m and n.

9. a) Sahayata is x years old now. How old will she be after 5 years?

b) Dakshes is y years old now. How old was he 2 years ago?

c) Sunayana is 7 years younger than Bishwant. If Bishwant is x years old now,
how old is Sunayana?

d) Shreyasha's father is 4 times older than her. If Shreyasha is y years old, how
old is her father?

e) A number is 5 more than another number. If the smaller number is x, what
is the bigger number?

f) A number is 4 less than another number. If the bigger number is p, what is
the smaller number?

g) A number is 2 times and 3 more than another number. If the smaller
number is x, what is the bigger number?

10. Let's evaluate the given algebraic expressions.

a) If x = 3, evaluate (i) 2x + 1 (ii) 3x – 2 (iii) x + 5 (iv) 6 – 1
2 x
3xy
b) If x = 4 and y = 2, evaluate (i) x + 3y (ii) 3x – 2y (iii) x+y (iv) 4
3

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c) The area of rectangle = l × b. Find the area of rectangle in sq. cm.

(i) l = 5 cm, b = 2 cm (ii) l = 6 cm, b = 4 cm (iii) l = 10 cm, b = 7 cm

d) The perimeter of a rectangle = 2(l + b). Find the perimeter in cm.

(i) l = 4 cm, b = 3 cm (ii) l = 7.5 cm, b = 5 cm (iii) l = 8.4 cm, b = 4.6 cm

e) The volume of cuboid = l × b × h. Find the volume in cubic cm.

(i) l = 5 cm, b = 4 cm, h = 2 cm (ii) l = 6 cm, b = 5 cm, h = 3 cm

(iii) l = 40 cm, b = 5.4 cm, h = 4 cm

11. a) If x = 2y, express (i) x + 2y (ii) 5y – x in terms of y and ind the values of
these expressions when y = 2.

b) If b = 4a, express (i) 6a – b (ii) a + b in terms of a and evaluate these
expressions when a = 3.

12. Perimeter of each plane shape = total sum of the length of all sides. If
p = 2, ind the perimeters of these shapes.

a) b) p + 3 c) 2p

p 3p p 2p
p+1
p+1

p 3p p

It's your time - Project work!

13. a) Let the number of students in your class is x. Write two algebraic
expressions to represent the number of girls and the number of boys in
your class.

b) Let's count the number of students in your class.

(i) Let the number of boys be x. Write an algebraic expression to represent
the number of girls.

(ii) Let the number of girls be y. Write an algebraic expression to represent
the number of boys.

c) (i) Let's write any two monomial expressions using x and y.

(ii) Let's write any two binomial expressions using x and y.

(iii) Let's write any two trinomial expressions using x and y.

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11.6 Coefficient, base, power, and exponent of algebraic term
Coef icient and base

Let's study the following illustrations and learn about the coef icient and base
of algebraic terms.

Let's add a variable x two times.

x + x = 2x x is multiplied by 2. So, 2 is the coef icient.
x is a variable and it is the base.

Let's add a variable term y three times.

y + y + y = 3y y is multiplied by 3. So, 3 is the coef icient.
y is a variable and it is the base.

Similarly,

In a + a + a + a = 4a, coef icient is 4 and base is a.

In p + p + p + p + p = 5p, coef icient is 5 and base is p.

Thus, the constant number which is used to multiply a variable is called a
coef icient and the variable is called the base.

Power and exponent

Now, let's learn about the power and exponent of an algebraic term from the
following illustrations.

22 × 2 is square of 2 = 2 exponent We read 22 as
base '2 squared.'

power

xx × x is square of x= 2 exponent We read x2 as
base 'x squared.'

power

55 × 5 × 5 is cube of 5 = 3 exponent We read 53 as
base '5 cubed.'

power We read y3 as
'y cubed.'
yy × y × y is cube of y = 3 exponent
base

power

aa × a × a × a = 4 exponent We read a4 as
base 'a to the power 4.'

power

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Similarly, I got it!
x × x × x × x × x = x5 I read x5 as x to the power 5!!
p × p × p × p × p × p = p6 I read p6 as p to the power 6!!

In this way, a power is a product of repeated multiplication of the same base.
The exponent of a power is the number of times the base is multiplied.

Remember, power and exponent do not have the same meaning. They are
different. For example, x3 is the power of x and it has two parts: base and
exponent. x is the base and 3 is the exponent of the power x3.

Furthermore, Very simple!
3y2 is read as '3y squared'. I read 7a7 as
2x3 is read as '2x cubed'. '7a to the power 7'!!
5b4 is read as '5b to the power 4, ... and so on.

11.7 Like and unlike terms

Let's study the following illustrations and investigate the idea about like and
unlike algebraic terms.

2 pencils and 3 pencils are the same (like) things. Let's
replace pencils by x. Then, 2x and 3x are like terms.

2 pencils and 3 rulers are the different (unlike) things. Let's
replace pencils by x and rulers by y. Then, 2x and 3y are unlike
things. So, 2x and 3y are unlike terms.

Again, let's learn more about the like and unlike terms from
the following examples.
3a and 4a are like terms, but 3a and 4b are unlike terms.
x2 and 2x2 are like terms, but x2 and 2y2 are unlike terms.
2a3 and 5a3 are like terms, 2a3 and 5a2 are unlike terms.
4pq and 6pq are like terms, but 4pq and 6qr are unlike terms.
3x2y and x2y are like terms, but 3x2y and xy2 are unlike terms.

Thus, like algebraic terms have the same bases and the equal exponents.
However, unlike algebraic terms have either the different basses or the
different exponents or both.

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EXERCISE 11.2
Section A - Classwork

1. Let's tell and write the coef icient, base, power and exponent of each of
the following algebraic terms.

Terms Coef icients Bases Powers Exponents

a) x 1 x x1 1

b) 4y3 4 y y3 3

c) 2x

d) p2

e) x3

f) 3y4

g) 5y5

2. Let's say and write the products in the power forms and express in words.

a) 3 × 3 × 3 × 3 = 34 3 to the power 4.

b) 2 × 2 × 2 =

c) p × p =

d) a × a × a =

e) x × x × x × x =

f) y × y × y × y × y =
3. Let's express these algebraic terms in words.

a) 4x2 is 4x squared b) 2a2 is

c) 5p3 is d) 3y4 is

e) 6m5 is f) 10x3 is

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4. Let's list the like and unlike terms separately.

Terms Like terms Unlike terms

a) 4x, 4y, x, z

b) 3a, b, 3c, 5b

c) y2, 2y, 3y2, y3

d) p2, q2, 6p2, q

e) xy, yz, xz, 4xy

f) a2b, ab2, ab, 2a2b

11.8 Addition and subtraction of monomial expressions
Addition and subtraction of unlike terms

Let's add x and y o It is x + y x+y

Let's add 2a and 3b o It is 2a + 3b a a + b b b

Let's subtract 2y from x o It is x – 2y x – y y

Let's subtract b from 2b2 o It is 2b2 – b b2 b2 – b

Thus, we cannot get a single total or a single difference by adding or subtracting
unlike terms. We simply write them in the form of addition or subtraction.

Addition and subtraction of like terms

Let's add x and x o It is x + x = 2x x+x=x x

Let's add y2 and 2y2 o It is y2 + 2y2 = 3y2 y2 + y2 y2 = y2 y2 y2

Let's subtract 2a from 3a o It is 3a – 2a = a a a a = a

Let's subtract x2 from 4x2 o It is 4x2–x2=3x2 x2 x2 x2 x2 = x2 x2 x2

Now, let's investigate the rules of addition and subtraction of like terms.

2x + 3x = (2 + 3)x = 5x x x + x x x = x x x x x

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3y + y = (3 + 1)y = 4y y y y +y =y y y y
5a – 2a = (5 – 2)a = 3a
4p2 – 3p2 = (4 – 3)p2 = p2 a a a a a =a a a

p2 p2 p2 p2 = p2

Thus, when we add like terms, we simply add the coef icients of the given
bases. When we subtract the like terms, we simply subtract the coef icient
of the given bases.

11.9 Addition and subtraction of polynomial expressions

Let's study the following examples and learn about the addition and subtraction
of polynomial expressions.

Example 1: Add x + 2y + 2 and 2x + 3y + 1
Solution

x + 2y + 2 x + 2y + 2 + 2x + 3y + 1 I understood!
2x + 3y + 1 = x + 2x + 2y + 3y + 2 + 1 Like terms x and 2x, 2y and 3y, 2
3x + 5y + 3 = 3x + 5y + 3 and 1 are separately added!!

Example 2: Subtract 2x2 – y2 from 4x2 – 3y2.
Solution

4x2 – 3y2 4x2 – 3y2 – (2x2 – y2) I got it!
± 2x2 y2 = 4x2 – 3y2 – 2x2 + y2 The like terms are
= 4x2 – 2x2 – 3y2 + y2 separately subtracted!!
2x2 – 2y2

Thus, when we add or subtract polynomial expressions, we should arrange the

like terms in the same column. Then, the coef icients of like terms are added or

subtracted. 4x2 – 3y2
–+ 2x2 +– y2
Furthermore, in the case of subtraction,
when we subtract (+) term it is changed to

(–) term and (–) term is changed to (+) term.

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Example 3: Simplify 9x2 + 4x – 2 – 2x2 – x + 6.
Solution
9x2 + 4x – 2 – 2x2 – x + 6

= 9x2 – 2x2 + 4x – x – 2 + 6

= 7x2 + 3x + 4

Example 4: What should be added to x + 1 to get 3x + 4?
Solution

3x + 4 Let's think, what should be added to 3 to get 7?
± x ±1
It is 4. Because 7 – 3 = 4. Similarly, what should be
2x + 3 added to x + 1 to get 3x + 4 is also 3x + 4 – (x + 1)

EXERCISE 11.3
Section A - Classwork

1. Let's say and write the total number of letters in two bags.

a) x2 + x2 b) x x + x c) yy + y y
x2 x x yy y

2x2 + x2 = 3x2 + = +=

d) a2 + a2 e) pp33pp33 + p3 f) x2x2 + xx2x22xx2 2
a2a2 a2

+= += +=

2. Let's subtract the letter-cards which are crossed and taken away.

a) a2 a2 a2 a2 a2 b) y y y

5a2 – 2a2 = 3a2 –=

c) x x x x d) x2 x2 x2 x2 x2 x2

–= –=
e) y2 y2 y2 y2 y2 f) b3 b3 b3 b3 b3 b3 b3

–= –=
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3. Let's say and write the sums or differences quickly.

a) x + 2x = b) x2 + 2x2 = c) 3y + 2y =

d) 3y2 + 2y2 = e) 4x + 2x = f) 4x3 + 2x3 =

g) 2a – a = h) 2a2 – a2 = i) 7p – 3p =

j) 7p2 – 3p2 = k) 8m – 5m = l) 8m3 – 5m3 =

4. Let's add, then say and write the sums quickly.

a) x + 2 and x + 3 = b) x2 + 2 and x2 + 3 =

c) y + 1 and y + 4 = d) y2 + 1 and y2 + 4 =

e) a + b and a + b = f) a2 + b2 and a2 + b2 =

5. Let's subtract, then say and write the differences quickly. It's easy!
a) From 3x + 4 subtract x + 2 = 3x – x = 2x
4–2 =2
b) From 6x + 5 subtract 2x + 1 =

c) From 5y + 4 subtract 4y + 3 =

d) From 7y + 6 subtract 5y + 4 =

6. Let's say and write the correct terms in the blank spaces.

a) 2x + = 6x b) + 3y = 5y c) a2 + = 3a2

d) 5x – = 2x e) – 2y = 4y f) 8a2 – = 5a2

7. Let's write your like terms to get the given sums or differences.

a) + = 5x b) + = 7x2 c) + = 8y

d) – = 2a e) – = 3a2 f) – = y

Section B

8. Let's add or subtract.

a) 2x + 3x b) x2 + x2 c) y + 5y d) 2y2 + 2y2

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e) ab + ab f) 3xy + 4xy g) p2q2 + p2q2 h) a2b2 + 2a2b2
i) 4x – x j) 3x2 – 2x2 k) 7y – 2y l) 5y2 – 3y2
m) 6xy – 3xy n) 2ab – ab o) 9a2b2 – 2a2b2 p) 8p2q2 – 5p2q2

9. Let's add and ind the sums. b) x2 + 1 and x2 + 3
a) x + 3 and x + 2 d) 3y2 + 2 and y2 + 1
c) 2y + 4 and y + 6 f) 2a2 + b2 and 3a2 + 2b2
e) 4a + b and a + 3b h) 5p2 + 7 and 3p2 + 4
g) 3p + 9 and 2p + 5 j) 3a + 2b + 5 and 2a + 4b + 2
i) 2x + y + 4 and x + 3y + 3

10. Let's subtract and ind the differences.

a) x + 2 from 3x + 5 b) 2x2 + 1 from 5x2 + 3

c) 3y + 4 from 4y + 7 d) y2 + 3 from 3y2 + 6
e) 2a – 3 from 7a – 9 f) p2 – 2 from 4p2 – 3
g) x + y + 1 from 3x + 2y + 4 h) 2a + 3b + 4 from 4a + 6b + 5
i) p + q + r from 2p + 3q – 2r j) x – y – z from 2x – 2y – 2z

11. Let's simplify.

a) x + 3 + x – 1 b) 2x2 – 4 + x2 + 2 c) 3y + 5 – y – 2
f) 2x2 + 3y2 – x2 – 4y2
d) 2x + y + 3x + 2y e) a2 – 2b2 + a2 – b2 h) 5a2 – 3a – 2 – 2a2 + a – 5

g) 3x2 + 2x + 3 + x2 + x + 1

12. a) What should be added to 3x to get 5x?
b) What should be added to 2a2 to get 6a2?
c) What should be added to p + 1 to get 2p + 2?
d) What should be added to 2x + 3 to get 4x + 5?
e) What should be subtracted from 5a to get 3a?
f) What should be subtracted from 7p2 to get 4p2?
g) What should be subtracted from 3x + 4 to get x + 2?

13. Let's ind the perimeters of the following plane shapes.

a) (x + 2) cm b) (2x + 1) cm c) (x + 1) cm
4 cm
5 cm
5 cm
3 cm
(x + 2) cm

(x + 3) cm (2x + 1) cm 2 cm

If x = 3, ind the perimeters of the above given plane igures.

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It's your time - Project work!

14. a) Let write any ive pairs of like terms and ind the sum and difference of
each pair.

b) Let's write any ive pairs of unlike terms and ind the sum and difference of
each pair.

11.10 Multiplication of algebraic expressions
Multiplication of monomial expressions

Let's study the following illustrations and investigate the rules of multiplication
of algebraic terms.

22 o 1 2 2 = 4 unit squares 32 o 1 2 3 3 = 9 unit squares
34 4 5 6

2 = 21 × 21 = 21 + 1 = 22 789 = 31 × 31 = 31 + 1 = 32
3

1234

42 o 5 6 7 8 4 = 16 unit squares x2 o x = x1 × x1
9 10 11 12 = x1 + 1 = x2
= 41 × 41 = 41 + 1 = 42
13 14 15 16 x

4

Similarly, a × a = a1 × a1 = a1 + 1 = a2, y × y = y2, p × p = p2, … and so on.

Again,

23 o 2 = 8 unit cubes 33 o 3 = 27 unit cubes

2 = 21 × 21 × 21 = 31 × 31 × 31
2 = 21 + 1 + 1 = 23 3 3 = 31 + 1 + 1 = 33

Similarly, x × x × x = x1 × x1 × x1 = x1+1+1 = x3, y × y × y = y1+1+1 = y3,
a × a × a = a1+1+1 = a3, … and so on.

Furthermore, I got it.
2x × x = 2x × 1x = (2 × 1)x1+1 = 2x2 Coefϔicients are multiplied
3y × 4y = (3 × 4)y1+1 = 12y2, … and so on. and the exponents of the
same bases are added.

Thus, when we multiply monomial algebraic expressions, we should multiply
the coef icients of the bases and the exponents of the same base are added.

Now, let's learn more about the multiplication of monomial expressions from
the following examples.

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Example 1: Multiply a) 4a2 × 2a b) 3ab × ab c) 2x × 3y

Solution 4 × 2 = 8 and the exponents of the same
a) 4a2 × 2a = 8a2+1 = 8a3 base are added!

b) 3ab × ab = 3a1+1 b1+1 = 3a2b2 3 × 1 = 3 and the exponents of the same
c) 2x × 3y = 6xy bases a and b are separately added!

2 × 3 = 6, x and y are unlike terms.
So, x × y = xy!

11.11 Multiplication of binomial expressions

Let's study the following examples and learn the rules of multiplication of
binomial expressions.

Example 2: Find the product of

a) (x + 2) × (x + 1) b) (a + b) × (a – b) c) (p – q) × (2p – 3q)

Solution

a) (x + 2) × (x + 1) Each term of x + 1 is multiplied by each
= x(x + 1) + 2(x + 1) term of x + 2.
= x2 + x + 2x + 2

= x2 + 3x + 2

b) (a + b) × (a – b) Each term of a – b is multiplied by each
= a(a – b) + b(a – b) term of a + b.
= a2 – ab + ab – b2

= a2 – b2

Sign rules:

c) (p – q) × (2p – 3q) (+) × (+) = +

= p(2p – 3q) – q(2p – 3q) (+) × (–) = –

= 2p2 – 3pq – 2pq + 3q2 (–) × (+) = –

= 2p2 – 5pq + 3q2 (–) × (–) = +

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EXERCISE 11.4
Section A - Classwork

1. Let's say and write the answer as quickly as possible.

a) x + x = and x + x = x+x =x x
b) y + y = and y + y =
x × x = x2 x

x

c) a + a + a = and a + a + a = x+x+x=x x x
d) p + p + p = and p + p + p =
x × x × x = x3 x
x
x

2. Let's say and write the products as quickly as possible.

a) 2x × x = b) 3y × y = c) 2a × 3a =

d) 4p × 5p = e) 2x2 × x = f) 3y2 × 4y =

g) 4a × 2a2 = h) 5b × 3b2 = i) 2p2 × 5p =

3. Look at these targets. Investigate the ideas and complete the other targets

as quickly as possible. 2x

x x a ay y
2x

x2 a yy x

3y y a 2a a 2a

y a2 a2 3a
3a
a
4. Let's say and write the product as quickly as possible.

a) 2(x + 1) = b) –2(x + 1) = c) –2(x – 1) =

d) x(x + 1) = e) –x(x + 1) = f) –x(x – 1) =

g) a(a + 2) = h) –a(a + 2) = i) –a(a – 2) =

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Section B
5. Let's ind the area of the squares and volume of the cubes.

a) x b) a c) y d) 2x

x a y 2x

Area = x × x = x2

e) x f) p g) 2m h) 3x

xx p p 2m 2m 3x 3x

Volume = x × x × x = x3

6. Let's multiply and ind the products.

a) 2x × 3x b) 3y × 4y c) 2a × 4a d) (–2) × 3p
e) (–2x) × 5x f) (–x) × (–x) g) (–2y) × (–y) h) (–3a) × 5b
i) 3x × (–3y) j) (–2p) × (–6q) k) 3b × b2 l) 5x2 × (–x)

7. Let's multiply and ind the products.

a) 2(x + 2) b) –2(x + 2) c) –2(x – 2) d) x(x + 3)
e) –x(x + 3) f) –x(x – 3) g) 2a(a + 4) h) –2a(a – 4)
i) 3y(y – 2) j) x(x + y) k) x(x – y) l) –x(x – y)

8. Let's multiply and simplify the products. c) p(p – 1) + 2(p – 1)
a) x(x + 1) + 2(x + 1) b) y(y + 2) + 1(y + 2) f) a(a + b) – b (a + b)
d) a(a + 3) – 1(a + 3) e) x(x – y) – y(x – y)

9. Let's multiply and simplify the products.

a) (x + 1) (x + 2) b) (x + 2) (x + 3) c) (a + 3) (a + 4)

d) (y + 3) (y + 5) e) (a + b) (a + b) f) (a + b) (a – b)

g) (a – b) (a – b) h) (p – 2) (p – 2) i) (x – 3) (2x + 1)

j) (2y – 3) (y – 2) k) (2p – q) (p + q) l) (x – y) (x – 2y)

10. Let's ind the area of the squares and rectangles in cm2.

a) b) c) d)
3x cm
3x cm (x+2)cm
3x cm

(a+1)cm
(a+3)cm
(x+2)cm
2x cm

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11. a) A square ground is 5a m long, ind its area in m2.
b) A square garden is (x + 3) m long, ind its area in m2.
c) A rectangular park is 4a m long and 3a m broad, ind its area in m2.
d) A rectangular ield is (x + 5) m long and (x + 3) m broad, ind its area in m2.
e) A cubical block is 4x cm long, ind its volume in cm3.
f) A cubical tank is 5x m high, ind its volume in m3.

It's your time - Project work!

12. a) Let's write any ive pairs of monomial expressions and ind the product of
each pair of expressions.

b) Let's write any three pairs of binomial expressions and ind the product of
each pair of expressions.

11.12 Division of monomial expressions

Let's study the following examples and learn the rules of division of monomial
algebraic expressions.

Example 1: Divide a) 6x2 by 3x b) 10a3b3 by 2ab

Solution = 6x2 b) 10a3b3 ÷ 2ab = 10a3b3
a) 6x2 ÷ 3x 3x 2ab

= 2 ×x× x = 5 × a ×a×a×b × b × b
3×x 2×a×b
6 10

= 2x = 5 × a × a × b × b = 5a2b2

11.13 Simplification by removing brackets

Let's study the given examples and learn how to remove the brackets while
simplifying the expressions inside the brackets.

Example 2: Simplify a) 4x + 3(x – 2) b) 5a – 2b –2(a – b)
Solution
a) 4x + 3(x – 2) = 4x + 3x – 6 I got it!
3(x – 2) = 3 × x – 3 × 2
= 7x – 6
= 3x – 6

b) 5a – 2b – 2(a – b) = 5a– 2b – 2a + 2b It's easier!
–2(a – b) = –2 × a – 2 × (– b)
= 5a – 2a – 2b + 2b
= 3a = –2a + 2b

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EXERCISE 11.5
Section A - Classwork
1. Let's say and write the quotient as quickly as possible.

a) x ÷ x = b) x2 ÷ x = c) x3 ÷ x =

d) a2 ÷ a2 = e) a3 ÷ a2 = f) 2y ÷ y =

g) 3y2 ÷ y = h) 2p3 ÷ p = i) 3b3 ÷ b2 =

2. Let's say and write the correct answers quickly.

a) x × x = , then x2 ÷ x =

b) 2a × a = , then 2a2 ÷ 2a = and 2a2 ÷ a =

c) y × y2 = , then y3 ÷ y = and y3 ÷ y2 =

d) 3p × p2 = , then 3p3 ÷ 3p = and 3p3 ÷ p2 =

Section B

3. Let's divide and ind the quotients.

a) 4x2 ÷ 2x b) 6y2 ÷ 3y c) 6y2 ÷ 2y d) 8a2 ÷ 4

e) 8a2 ÷ 4a f) 8a2 ÷ 4a2 g) 8a2 ÷ 2a h) 9x2 ÷ 3

i) 9x2 ÷ 3x j) 9x2 ÷ 3x2 k) 12p3 ÷ 3p l) 12p3 ÷ 4p2

m) 12p3 ÷ 6p3 n) 15x2y2 ÷ 5xy o) 15x2y2 ÷ 3xy p) 14a3b3 ÷ 2ab

q) 14a3b3 ÷ 7ab r) 14a3b3 ÷ 2a2b2 s) 14a3b3 ÷ 7a2b2 t) 14a3b3 ÷ 7a3b3

4. Let's remove the brackets and simplify.

a) 2(x + 3) b) 2(x – 3) c) –3(x + 2) d) –3(x – 2)

e) a(a + 5) f) –a(a – 5) g) 2y(2y – 3) h) –3y(3 – 2y)

i) 5x + 2(x + 2) j) 5x – 2(x + 2) k) 5x – 2(x – 2) l) p + 2q + 2(p – q)

m) a + b – (a – b) n) 3x + 3y – 3(x – y) o) 4x + 4y – 3(x + y)

5. a) From the sum of 2x and 3, subtract x + 2.
b) From the difference of 3a and 4, subtract a + 3.
c) From the sum of x and y, subtract the difference of x and y.
d) From the difference of a and b, subtract the sum of a and b.

It's your time - Project work!

6. Let's write any ive pairs of monomial expressions with the same bases in each
pair.
a) Find the product of each pair of expressions.
b) Divide the product of each pair of expressions by each term of the expressions.

"

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Unit Algebra - Equation

12

12.1 Open mathematical sentence (or statement) - review

'4 is added to 5 is 9', '3 subtracted from 7 is 4', … are mathematical sentences.
A mathematical sentence contains mathematical operations.

4 is added to 5 is 9 ‘ 4 + 5= 9, which is true.

3 subtracted from 7 is 4 ‘ 7 – 3 = 4, which is true.
Such true mathematical sentences are closed sentences (or closed
statements).

On the other hand,
x is added to 5 is 9 ‘ x + 5 = 9, we cannot say it is true or false.
y is subtracted from 7 is 4 ‘ 7 – y = 4, we cannot say it is true or false.

Such mathematical sentences for which we cannot say whether they are true
or false are open mathematical sentences.

We make open mathematical sentences by using variables.

12.2 Equation - review

x + 5 = 9 is an open mathematical sentence. It is an equation.

7 – y = 4 is an open mathematical sentence. It is an equation.

3 × a = 6 is an open mathematical sentence. It is an equation.

x = 3 is an open mathematical sentence and it is an equation.
2

Thus, an equation is on open mathematical sentence that has two equal sides
separated by an equal (=) sign.

We write left hand side of an equation as L.H.S. and the right hand side as
R.H.S.

We can compare an equation to a pan balance, where the equal sign works as
the balance point.

5 =5 7+2 = 9 x+3 = 8 2y = 10

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Algebra - Equation

12.3 Solving equation

Let's take an equation x + 2 = 5 and replace the variable x by different values.

When x = 1, then 1 + 2 = 5, it is not true.

When x = 2, then 2 + 2 = 6, it is not true.

When x = 3, then 3 + 2 = 5, it is true.
Thus, the equation x + 2 = 5 is true only for x = 3. Therefore, 3 is solution of
the equation.

Similarly,

If x + 1 = 2, then solution is x = 1. If x – 1 = 3, the solution is x = 4.

If 2 × x = 6, the solution is x = 3, … and so on.

We use the following facts while solving equations.
Fact I

When we add an equal number to both sides of an equation, the sums is equal.
For example, if x = 7, then, x + 3 = 7 + 3.

x =7 x+3 7 x+3 = 7+3

Fact II

When we subtract an equal number from both sides of an equation, the
differences is equal. For example, if x = 5, then, x – 2 = 5 – 2.

x =5 x–2 5 x–2 = 5–2

Fact III

When we multiply both sides of an equation by an equal number, the products
x x
is equal. For example, if 2 = 3, then, 2 × 2 = 3 × 2.

x =3 x × 2 3 x × 2 = 3×2
2 2 2

Fact IV

When we divide both sides of an equation by an equal number, the quotient is
2x 6
equal. For example, if 2x = 6, then, 2 = 2

2x = 6 2x 2x 6
26 2= 2

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Algebra - Algebraic Equation

Now, let's investigate the processes of solving equations from the following
illustrations.
a) Solve: x + 3 = 5

xxx

x+3=5 x+3–3=5–3 x=2
b) Solve: x – 2 = 3

x–2 x–2 x

x–2=3 x–2+2=3+2 x =5
c) Solve: 2x = 6 x x

xx x

2x = 6 2x = 6 x=3
2 2 x
x
d) Solve: 2 = 3

x x
2 x2
2

x = 3 2 × x = 2 × 3 x=6
2 2

Now, let's use the above rules to solve the equations in the given examples.

Example 1: Solve a) x + 4 = 9 b) x – 5 = 2 c) 3x = 15 d) x = 4
Solution: b) x – 5 = 2 3
a) x + 4 = 9
or, x – 5 + 5 = 2 + 5
or, x + 4 – 4 = 9 – 4
or, x = 7
or, x = 5

Transposition method Transposition method
x+4 =9 x–5 =2

or, x = 9 – 4 or, x = 2 + 5
or, x = 5 or, x = 7

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Algebra - Equation

c) 3x = 15 d) x =4
or, 3
or, 13x = 155 or, x
31 31 3 × 3 =4×3

or, x = 5 x = 12

Transposition method Transposition method

3x= 15 x =4
3
or, x= 155 or, x = 4 × 3
31
or, x = 5
or, x = 12
2x
Example 2: Solve a) 3x – 2 = 10 b) 3 + 1 = 3
Solution:

a) 3x – 2 = 10 Checking the solution

or, 3x = 10 + 2 –2 is transposed to R. H. S. 3x – 2 = 10

or, 3x = 12 or, 3 × 4 – 2 = 10

or, x = 124 3 is transposed to divide R. H. S. or, 12 – 2 = 10
31 or, 10 = 10

or, x = 4 ? 4 is the correct solution.

b) 2x + 1 = 3
3
2x
or, 3 =3–1 1 is transposed to R. H. S. Checking the solution

or, 2x =2 3 is transposed to 2x + 1 =3
3 multiply R. H. S. 3

or, 2x = 2 × 3 or, 2×3 +1 =3
3
or, 2x = 6 or, 2 + 1 = 3

or, x = 63 2 is transposed to or, 3 = 3
21 divide R. H. S. ? 3 is the correct solution.

or, x = 3

Example 3: Solve a) 3x – 2 =2 b) 1.5x + 3 = 9
2

Solution: Cross-multiplication

a) 3x – 2 =2 3x – 2 = 2
2 2 1

or, 3x – 2 = 2 × 2 or, 1 × (3x – 2) = 2 × 2

or, 3x – 2 = 4 Multiplicationofnumeratorsanddenominator

or, 3x = 4 + 2 of L.H.S. and R.H.S. in the directions of
arrow-heads is cross-multiplication.

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or, 3x = 6 Algebra - Algebraic Equation

or, x = 62 Checking the solution
31 1.5x + 3 = 9
or, x = 2
or, 1.5 × 4 + 3 = 9
b) 1.5x + 3 = 9 or, 6.0 + 3 = 9
or, 9 = 9
or, 1.5x = 9 – 3 3 is transposed to R. H. S. ? 4 is the correct solution.
or, 1.5x = 6

or, 315x =6 1.5 = 15
102 10

or, 3x =6
2
or, 3x = 6 × 2 By cross-multiplication

or, 3x = 12

or, x = 124 3 is transposed to
31 divide R. H. S.

or, x = 4

EXERCISE 12.1
Section A - Classwork

1. Let's say and tick the correct answers.
a) The sum of 3 and 5 is 8. It is (closed/open) mathematical sentence.
b) The sum of x and 4 is 7. It is (closed/open) mathematical sentence.
c) The difference of y and 2 is 6. It is (closed/open) mathematical sentence.
d) 5 times 10 is 50. It is (closed/open) mathematical sentence.
e) 5 times x is 20. It is (closed/open) mathematical sentence.

2. Let's list out the equations separately. Equations are
x – 4, x – 4 = 6, y + 1 = 9,

y + 1, 3a, 3a = 3, x , x = 4
2 2
3. Let's say and write the answers as quickly as possible.

a) What should be added to 2 to get 5? x + 2 = 5, so, x =

b) From which number is 3 subtracted to get 4? x – 3 = 4, so, x =

c) 2 times of which number is equal to 10? 2 × x = 10, so, x =
so, x =
d) What should be divided by 2 to get 7? x = 7,
2

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Algebra - Equation

4. Let's say and write the correct number inside each circle. Then, say and
write the value of letters.

a) + 3 = 8 o x + 3 = 8 So, x =

b) – 4 = 6 o y – 4 = 6 So, y =

c) 3 × = 12 o 3a = 12 So, a =

d) =2 o x = 2 So, x =
4 4

5. Let's say and write the values of letters quickly.

a) x + 1 = 2, x = b) x – 2 = 1, x = c) a + 2 = 4, a =

d) 2x = 12, x = e) 3y = 15, y = f) x = 2 , x=
3

6. The solution of each equation is given. Let's make equations.

Equations Solutions Equations Solutions
a) x + 1 = 3 6
b) x + 4 = 2 e) 2x = 5
c) a – 2 = 7 5
d) a – 5 = 9 f) 3x =
4
g) x =
2 d) y + 5 = 8
h) 2y + 3 = y + 7
h) y = d) a – 5 = 4
3 h) 2a – 3 = a + 3
d) 4p = 8
Section B h) 5x + 4 = 14

Let's solve the equations.

7. a) x + 1 = 5 b) x + 2 = 3 c) y + 3 = 8
g) 2x + 2 = x + 4
e) a + 4 = 6 f) 2a + 1 = a + 3 c) p – 3 = 7
g) 2x – 5 = x + 1
8. a) x – 1 = 2 b) x – 4 = 1 c) 2y = 6
g) 4a – 3 = 13
e) y – 2 = 9 f) 2y – 1 = y + 2
9. a) 2x = 2 b) 3a = 3

e) 2x – 1 = 7 f) 3y + 2 = 8

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Algebra - Algebraic Equation

10. a) x = 1 b) a = 2 c) y = 3 d) 2x = 2
2 3 4 3

e) 3a = 6 f) 2y =4 g) 3x = 9 h) 5x = 10
2 5 4 3

11. a) x + 1 = 3 b) x – 1 = 2 c) y + 2 = 4 d) a – 3 = 0
2 3 2 3

e) x – 2 = 0 f) 2a + 4 = 6 g) 3x – 1=5 h) 2x + 3 = 7
4 3 2 5

12. a) x + 1 = 1 b) x–2 = 2 c) y–3 =0 d) 2y – 4 = 0
2 3 2 3

e) x + 2 + 1= 2 f) a–2 –1= 1 g) y + 3 + 2 = 4 h) 2x + 3 – 2 = 1
2 3 2 5

13. a) 0.5x = 1 b) 0.5x = 2 c) 1.5x = 3 d) 1.5x = 6

e) 0.5x + 1 = 2 f) 0.5x – 1 = 2 g) 1.5x + 1 = 4 h) 1.5x – 2 = 4

14. Let's solve these equations. Then, check whether the solutions are correct
or not.

a) x + 4 = 10 b) x – 5 = 7 c) 2x + 3 = x + 8 d) 3x – 1 = x + 5

e) 3a + 2 = a + 8 f) 2a – 2 = 4 g) 3y + 1 = 10 h) 2x – 3 =3
3 2 3

15. Let's make equations from these balances, then solve them. Also, check
whether the solutions are correct or not.

a) x b) x x c) x x x x

16. The lengths of the straight line segments are given. Let's ind the unknown
part of each line segments by making and solving equations.

a) x cm 4 cm b) x cm 2x cm 2 cm
11 cm
9 cm

17. The perimeters of the plane igures are given. Let's ind the lengths of
unknown sides of the igures.
a) b) c) 3x cm

5 cm 5 cm
3 cm
2x cm

2x cm
2x cm

x cm x cm 3x cm

perimeter = 14 cm perimeter = 11 cm perimeter = 20 cm

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Algebra - Equation

It's your time - Project work!
18. a) Let's make your own ive different equations.

(i) Solve your equations.

(ii) Check whether the solutions of the equations are correct or not.

b) Let's make any three different equations which have the solutions 1, 2 and
3 respectively. Check whether these solution are correct or not for your
equations.

12.4 Use of equation

We can ind unknown numbers or quantities in our real life situations by
making equations and solving them. Let's follow the given steps to ind the
unknown numbers or quantities asked in word problems.

At irst, let's consider the unknown number or quantity as a variable such as
x, y, a, …

Then, translate the word problem into mathematical expression according
to the given condition. It is the required equation.

Now, let's solve the equation and ind the unknown number or quantity.

Now, let's study the following examples and learn the processes of solving
word problems by using equations.

Example 1: The sum of two numbers is 18 and one of the numbers is 11.

Find the other number.

Solution I got it!
The given condition is-
Let the other number be x. 'sum of two numbers is 18'.
Then, x + 11 = 18

or, x = 18 – 11 = 7 ? x + 11 = 18!!

Therefore, the required number is 7.

Example 2: The difference between two numbers is 9. If the smaller

number is 15, ϔind the greater number.

Solution

Let the greater number be x. It's very easy!
Then, x – 15 = 9 Difference of x and 15 means
or, x = 9 + 15 = 24 x – 15 and the difference is
9 means x – 15 = 9

Therefore, the required greater number is 24.

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Algebra - Algebraic Equation

Example 3: The area of a rectangular ϔloor is 63 m2 and it is 9 m long. Find

its breadth.

Solution

Let the breadth of the rectangle be x m. I recall!

Now, length × breadth = area of the loor Area of rectangle = l × b

Then, 9 × x = 63

or, x = 637 = 7 m
91

Therefore, the breadth of the loor is 7 m.

Example 4: There are 30 students in a class and the number of girls are 4

more than the number of boys. Find the number of girls and

boys.

Solution

Let the number of boys be x.

Then, the number of girls = x + 4 I got it!
Now, x + (x + 4) = 30 4 more than x is x + 4.
or, 2x + 4 = 30 The total of x and x + 4 is
or, 2x = 30 – 4 x + (x + 4) which is 30!
? x + (x + 4) = 30

or, 2x = 26

or, x = 26 13 = 13
21

Therefore, number of boys = 13.

The number of girls = x + 4 = 13 + 4 = 17.

Example 5: The number of marbles with Daffy Duck is double than the
number of marbles with Donald Duck. If they have 45 marbles
in total, how many marbles does each have?

Solution

Let the number of marbles with Donald Duck is x.
Then, the number of marbles with Daffy Duck = 2x
Now, x + 2x = 45
or, 3x = 45

or, x = 45 15 = 15
31
Therefore, the number of marbles with Donald Duck = x = 15 marbles.

The number of marbles with Daffy Duck = 2x = 2 × 15 = 30 marbles.

Example 6: If 1 part of the distance between Kathmandu and Pokhara is
4

50 km, ϔind the distance between these two places?

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Algebra - Equation

Solution

Let the distance between Kathmandu and Pokhara be x km.

Then, 1 of x = 50
4

or, 1 × x = 50
or, 4 x = 50

4
or, x = 50 × 4 = 200

Therefore, the distance between these two places is 200 km.

EXERCISE 12.2
Section A - Classwork

1. Let's say and write the values of letters as quickly as possible.

a) The sum of x and 4 is 9. x=

b) The difference of y and 3 is 5. y=

c) The product of 4 and a is 28, a=

d) The quotient of p divided by 3 is 10. p =

e) One-quarter of x is 6. x=

2. Let's make equations. Then, say and write the values of variables.

a) The sum of a and 5 is 12. a + 5 = 12 a=

b) The sum of x and 6 is 10.

c) The difference of y and 2 is 7.

d) 3 times p is 15.

e) The quotient of x divided by 2 is 6.

Section B

3. a) The sum of x and 7 is 18, ind x. b) The difference of y and 5 is 15, ind y.

c) The product of 6 and a is 54, ind a. d) The quotient of p divided by 4 is 6, ind p.

e) The double of x is 14, ind x. f) One-third of y is 9, ind y.

g) x increased by 4 is 16, ind x. h) y decreased by 8 is 10, ind y.

i) p is more than 15 by 3, ind p. j) 7 is less than x by 5, ind x.

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Algebra - Algebraic Equation

Let's solve these word problems by making algebraic equations.

4. a) The sum of two numbers is 27 and one of them is 15. Find the other number.

b) The difference of two numbers is 9 and the smaller number is 18. Find the
bigger number.

c) The product of two numbers is 72 and one of them is 8. Find the other
number.

d) When a number is divided by 5, the quotient is 10, ind the number.

e) If 1 of a number is 12, ind the number.
2

f) If one-quarter of a number is 7, ind the number.

5. a) There are 35 students in a class and 17 of them are boys. Find the number
of girls.

b) Mrs. Upadhyaya earns Rs 25,000 in a month. She earns Rs 9,500 from
vegetable farming and the rest from poultry farming. How much does she
earn from poultry farming?

6. a) There are 18 girls in a class which is 4 more than the number of boys. Find
the number of boys.

b) Anjolina has Rs 9 less than Anamol. If Anjolina has Rs 25, how much money
does Anamol have?

c) Sita is 3 years older than Geeta. If Sita is 17 years old, how old is Geeta?

d) When 5 more new students joined the class, there were 38 students in the
class. How many students were there at the beginning?

e) Bishwant had some money. When he spent Rs 150, he had Rs 125 left. How
much money did he have at the beginning?

7. a) The area of a rectangle is 24 cm2 and its length is 6 cm. Find its breadth.

b) A rectangular loor of a room is 8 m wide and the area of the loor is 80 m2.
Find the length of the room.

c) The perimeter of a square garden is 100 m. Find the length of the garden.

d) The perimeter of a rectangular compound is 110 m and it is 30 m long. Find
the breadth of the compound.

8. a) The number of sweets with Mickey Mouse is twice as much as that of Bugs
Bunny. If they have 36 sweets altogether, how many sweets does each have?

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Algebra - Equation

b) The number of girls in a school is double than that of boys. If there are 360
students in total, ind the number of boys and girls.

c) Age of father is 2 times and 10 years more than the of his daughter. If the
father is 40 years old now, how old is the daughter?

d) Pradeep has 3 times and Rs 50 more money than that of Santosh. If Pradeep
has Rs 230, how much money does Santosh have?

9. a) If 1 of the distance between Bharatpur and Hetauda is 38 km, ind the
2

distance between these two places.

b) One-third of the distance between Kohalpur and Attariya is 55 km. Find the
distance between these two places.

c) 1 part of the capacity of a water tank is 500 litres. Find the full capacity of
4

the tank.

d) Mother spends Rs 6,000 every month for her children's education, which is
1
5 part of her monthly income. Find her monthly income.

It's your time - Project work!

10. a) Let's measure the length and breadth of your mathematics book using a
30-cm ruler. By how many centimetres is the length longer than the
breadth? Let's use the answer and do the following problems.

(i) Let the length of the book be x cm. Now, make an equation and ind the
value of x. (Hint: x – … = breadth)

(ii) Let the breadth of the book be x cm. Now, make an equation and ind the
value of x. (Hint: x + … = length)

b) There are x number of students in your class who like pizza and the rest

like Mo:Mo. Let's count the number of students who like Mo:Mo and the
total number of students. Then make an equation and ind x.

c) There are y number of lady teachers in your school. Let's count the number

of gents teachers and the total number of teachers. Then make an equation
and ind the value of y.

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Unit The Metric Measurement System

13

13.1 Measurement of length and distance - Looking back

Classwork - Exercise

1. Let's say and write the correct answers as quickly as possible.

a) 1 cm = mm, 2 cm = mm, 4.5 cm = mm
cm
b) 1 m = cm, 3 m = cm, 2.5 m = m
cm
c) 1 km = m, 4 km = m, 3.2 km = m
km
d) 1 mm = 0.1 cm, 2 mm = cm, 46 mm =

e) 1 cm = 0.01 m, 3 cm = m, 65 cm =

f) 1 m = 0.001 km, 5 m = km, 500 m =

We use rulers, measuring tapes, milometers,

etc. to measure lengths and distances.
These instruments are called the standard
instruments. Millimetre (mm), centimetre

(cm), meter (m), kilometre (km) are the
standard units of measurement of lengths and
distances in Metric Measurement System.

13.2 Conversion of units of length

Let's review the relationship between the units of measurement of length. It is
useful to convert higher to lower units or lower to higher units of lengths.

1 cm = 10 mm cm × 10 mm and mm ÷ 10 cm
1 m = 100 cm m × 100 cm and cm ÷ 100 m
1 km = 1000 m km × 1000 m and m ÷ 1000 km

Now, let's study the following examples and learn about the conversion of
different units of lengths.

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The Metric Measurement System

Example 1: Convert a) 4 cm 6 mm into mm b) 7 cm 5 mm into cm
Solution

a) 4 cm 6 mm b) 7 cm 5 mm
= 4 × 10 mm + 6 mm
= 40 mm + 6 mm 1 cm = 10 mm = 7 cm + 5 cm 1 mm= 1 cm
4 cm = 4 × 10 mm 10 5
5 mm= 10 cm
= 40 mm = 7 cm + 0.5 cm
= 0.5 cm

= 46 mm = 7.5 cm

Example 2: Convert a) 2 m 75 cm into cm b) 5 m 60 cm into m.

Solution

a) 2 m 75 cm b) 5 m 60 cm 100 cm = 1 m

= 2 × 100 cm + 75 cm 1m = 100 cm = 5 m + 60 m 60 cm = 60 cm
2m = 2 × 100 cm 100 100
= 0.60 cm
= 200 cm + 75 cm = 200 cm = 5 m + 0.60 cm
= 0.6 m

= 275 cm = 5.6 m

Example 3: Convert a) 3 km 250 m into m b) 6 km 400 m into km.

Solution

a) 3 km 250 m b) 6 km 400 m

1 km = 1000 m 400 1000 m = 1 km
3 km = 3 × 1000m 1000 400
= 3 × 1000 m + 250 m = 6 km + km 400 m = 1000 km

= 3000 m + 250 m = 3000 m = 6 km + 0.400 km = 0.400 km
= 0.4 km

= 3250 m = 6.4 km

EXERCISE 13.1
Section A - Classwork

1. Let's convert higher to lower or lower to higher units.

a) 1 cm = mm, 6 cm = mm, 7.2 cm = mm
cm
b) 1 m = cm, 4 m = cm, 3.8 m = m
cm
c) 1 km = m, 3 km = m, 1.25 km =

d) 1 mm = cm, 5 mm = cm, 74 mm =

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The Metric Measurement System

e) 1 cm = m, 6 cm = m, 48 cm = m

f) 1 m = km, 32 m = km, 672 m = km

2. Let's say and write the measurements pointed by arrows.

cm

3. Let's say and write the answers as quickly as possible.

a) How many cm and mm are in 45 mm? cm mm
cm
b) How many m and cm are in 320 cm? m m
mm
c) How many km and m are in 2250 m? km cm
m
d) How many cm and mm are in 5.8 cm? cm

e) How many m and cm are in 1.6 m? m

f) How many km and m are in 3.5 km? km

Section B

4. Let's convert the units of length as indicated:

a) 3 cm 4 mm (into mm) b) 8 cm 5 mm (into mm)

c) 5 cm 7 mm (into cm) d) 9 cm 3 mm (into cm)

e) 1 m 50 cm (into cm) f) 3 m 25 cm (into cm)

g) 2m 18 cm (into m) h) 4 m 70 cm (into m)

i) 1 km 350 m (into m) j) 2 km 200m (into m)

k) 4 km 675 m (into km) l) 7 km 800 m (into km)

5. Let's convert the following units into the decimal of higher units.

a) 1 mm, 4 mm, 25 mm, 60 mm, 136 mm (into cm)

b) 1 cm, 5 cm, 48 cm, 80 cm, 264 cm (into m)

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The Metric Measurement System

6. Let's convert the following units into the lower units.
a) 0.2 cm, 0.5 cm, 3.8 cm, 4.0 cm, 12.2 cm (into mm)
b) 0.01 m, 0.07 m, 0.25 m, 0.7 m, 3.5 m (into cm)
c) 0.001 km, 0.009 km, 0.036 km, 0.06 km, 0.8 km (into m)
It's your time - Project work!

7. Let's use a 15 cm - ruler and ind:

a) thickness of your math book in cm and mm.
b) length of your middle inger in cm and mm.
8. Let's use a measuring tape and ind:
a) your waist in cm and in mm
b) length and breadth of your classroom in metres (m)
9. Ask your parents or teachers and estimate:
a) distance between your house and school in km,
b) a place which is about 5 km away from your school,
c) distance between your house and the nearest hospital or health post.
10. Let's cut two paper strips from a chart paper; one for a 15 cm and the other for
a 30 - cm paper rulers. Then make these two paper rulers.

13.3 Addition and subtraction of lengths

While adding or subtracting lengths, we should add or subtract the units
separately. Alternatively, we write the given lengths in the decimal of higher
units. Then, we perform the addition or subtraction of decimals.

Let's study the following examples and learn about the processes of addition
and subtraction of lengths.

Example 1: A rubber is 15 cm 6 mm long and it is stretched by
4 cm 8 mm. Find the length of the stretched rubber.

Solution

Here, the length of the stretched rubber = 15 cm 6 mm + 4 cm 8 mm

15 cm 6 mm Another process

+ 4 cm 8 mm 15 cm 6 mm = 15 . 6 cm
19 cm 14 mm
4 cm 8 mm = + 4 . 8 cm
= 20 cm 4 mm 20 . 4 cm

Therefore, the length of the stretched rubber is 20 cm 4 mm or 20.4 cm.

vedanta Excel in Mathematics - Book 5 196
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The Metric Measurement System

Example 2: The total length of an electric pole is 15 m 20 cm. If the
underground part of the pole is 1 m 25 cm long, ϔind the
height of the pole above the ground.

Solution

Here, the height of the pole above the ground = 15 m 20 cm – 1 m 25 cm

15 m 20 cm Another process
– 1 m 25 cm
15 m 20 cm = 15 . 20 m
13 m 95 cm 1 m 25 cm = – 1 . 25 cm

13 . 95 cm

Therefore, the height of the pole above the ground is 13 m 95 cm or 13.95 m.

13.4 Multiplication and division of lengths

Now, let's study the following examples and learn about the processes of
multiplication and division of lengths.

Example 3: A book is 1 cm 8 mm thick. Find the height made by 9 such

books placed one above another.

Solution Another process
Here, the required height = 9 × 1 cm 8 mm
1 cm 8 mm
1 cm 8 mm = 1.8 cm

× 9 4 cm 8 mm = × 9

9 cm 72 mm 16.2 cm

16 cm 2 mm

Hence, the required height is 16 cm, 2 mm or 16.2 cm.

Example 4: A local bus travels 44 km 700 m in its 6 trips between two

villages. How much distance does it travel in 1 trip?

Solution

Here, the distance travelled in 1 trip = 44 km 700 m ÷ 6

6 44 km 700 m 7 km 450 Another process
– 42 44 km 700 m = 44.7 km
2 km 700 m
2700 m 6 44.7 km 7.45 km
– 24 – 42
300 27
– 24
30

– 300 – 30

00

Hence, the distance travelled by the bus in 1 trip is 7 km 450 m or 7.45 km.

197 vedanta Excel in Mathematics - Book 5

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

The Metric Measurement System

EXERCISE 13.2
Section A - Classwork

1. Let's add and regroup into the higher units.

a) 4 mm + 6 mm = mm = cm

b) 7 mm + 8 mm = mm = cm mm

c) 30 cm + 70 cm = cm = m

d) 60 cm + 80 cm = cm = m cm

e) 200 m + 800 m = m= km

f) 500 m + 900 m = m= km m

2. Let's convert into the lower units and subtract.

a) 1 cm – 6 mm = mm – 6 mm = mm

b) 1 m – 70 cm = cm – 70 cm = cm

c) 1 km – 500 m = m – 500 m = m

3. Let's multiply. Say and write the products in the decimal of higher units.

a) 4 × 6 mm = mm = cm

b) 3 × 50 cm = cm = m

c) 5 × 700 m = m= km

4. Let's convert into lower units and divide.

a) 1 cm ÷ 5 = mm ÷ 5 = mm
b) 2 cm ÷ 4 = mm ÷ 4 = mm
c) 1 m ÷ 10 = cm ÷ 10 = cm
d) 3 m ÷ 6 = cm ÷ 6 = cm
e) 1 km ÷ 5 = m÷5 = m
Section B

5. Let's add or subtract. b) 15 m 40 cm + 10 m 80 cm
a) 18 cm 9 mm + 12 cm 6 mm d) 25 cm 5 mm – 13 cm 8 mm
c) 9 km 550 m + 5 km 875 m f) 10 km 250 m – 4 km 500 m
e) 30 m 25 cm – 16 m 60 cm 198

vedanta Excel in Mathematics - Book 5 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur