Vedanta Excel in Mathematics Book 9 Final (2078)

Commission and Taxation

b) Mrs. Anjali Subba is a medical doctor in a government hospital. Her monthly salary is
Rs 36,900 and she receives one month's salary as festival bonus. 10% of her salary is
deducted as citizen investment trust. How much income tax does she pay in a year?

c) After deducting 10% provident fund a married person draws Rs 40,500 salary per
month and one month's salary as festival bonus, the person pays Rs 14,500 annually
as the premium of his/her insurance. Calculate the annual income tax paid by the
person.

d) Mr. Sayad Sharma an unmarried employee of a UN Project draws monthly salary
of Rs 51,000 after deducting 10% salary in his provident fund and 5% in citizen
investment trust. He also receive the Dashain bonus of one month's salary. He pays
Rs 22,000 annually as the premium of his life insurance. How much income tax does
he pay in a year?

Project work

4. a) Let's ask the monthly salary of your Mathematics, Science, and English teachers.
Then, complete the table given below.

Name of Marital Monthly Festival Provident Insurance
teachers status Salary allowance fund

Now, calculate the annual income tax paid by each subject teacher.

b) If your parents are involved in any government or private service, let's ask their
monthly salary, bonus, provident fund, insurance, etc. Then, calculate the income
tax paid by them in a year.

3.5 Dividend

A dividend is a payment made by a corporation to its shareholders, usually as a
distribution of profits. A dividend is allocated as a fixed amount per share with
shareholders receiving a dividend in proportion to their shareholding.

For example, a person has bought 1,500 shares out of total of 1,00,000 shares of
Rs 100 per share from a hydroelectricity corporation. If the corporation earned a net
profit of Rs 3,50,28,000 in a certain year and it decided to distribute 20% of the net
profit to its shareholders, the amount received by the person as per his/her number of
shares is called his/her dividend.

Worked-out examples

Example 1: Mrs. Bhatta bought 750 shares out of 20,000 shares from a Business
Company. If the company earned a net profit of Rs 90,00,000 and it
declared to distribute 10% dividend to its shareholders, how much money
did Mrs. Bhatta receive?

51Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Commission and Taxation

Solution: = Rs 90,00,000
Here, the net profit of the company
= 10% × Rs 90,00,000
Now, 10% of Rs 90,00,000 100

= Rs 9,00,000

Again,

The profit of 20,000 shares = Rs 9,00,000

The profit of 1 share = Rs 9,00,000
20,000

= Rs 45

The profit of 750 shares = 750 × Rs 45 = Rs 33,750
Hence, she received Rs 33,750 as her dividend.

Example 2: A commercial bank sold 30,000 shares. The bank earned a net profit of
Solution: Rs 2,50,00,000 in a year and distributed a certain percent of profit as
dividend. If a shareholder, who has bought 400 shares, received Rs 50,000
dividend, find what percent of profit was distributed as divided by the
bank?

Here, the dividend of 400 shares = Rs 50,000

the dividend of 1 share = Rs 50,000 = Rs 125
400

the dividend of 30,000 shares = Rs 30,000 × Rs 125 = Rs 37,50,000

∴ The total dividend = Rs 37,50,000

The net profit = Rs 2,50,00,000

Now, the percent of net profit as dividend = Dividend × 100%
Net profit

= Rs 37,50,000 × 100%
Rs 2,50,00,000

= 15%

Hence, 15% of the net profit was distributed as dividend by the bank.

Alternative process:

The dividend of 400 shares = Rs 50,000

The dividend of 1 share = Rs 50,000
400

= Rs 125

Also, the net profit for 30,000 shares = Rs 2,50,00,000

The net profit for 1 share = Rs 2,50,00,000 = Rs 2500
30,000 3

Vedanta Excel in Mathematics - Book 9 52 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Commission and Taxation

Now, the percent of profit as dividend = Dividend for 1 share × 100%
Profit for 1 share

= Rs 125 × 100%
Rs 2500/3

= 15%

Example 3: Harka Magar bought 500 shares out of 50,000 shares sold by a hydropower
Solution: company. When the company distributed 25% of its net profit, he received
Rs 43,850 as his dividend in a year. Calculate the net profit of the company.

Here, the profit of 500 shares = Rs 43,850

The profit of 1 share = Rs 43,850
500

= Rs 87.70

The profit of 40,000 shares= 50,000 × Rs 87.70

= Rs 43,85,000

Let, the net profit of the bank be Rs x.

Now, 25% of Rs x = Rs 43,85,000

or, 25x = Rs 43,85,000
100

∴ x = Rs 1,75,40,000

Hence, the net profit of the company is Rs 1,75,40,000.

EXERCISE 3.3
General section

1. a) A business company makes a net profit of Rs 80,00,000 in a year. The Board of
Directors declares 12% dividend from the net profit. If the company has sold 1000
shares, calculate the dividend for each share.

b) A development bank made a net profit of Rs 2,40,00,000 in a year and the management
announced to distribute 24% dividend from the net profit. If the bank has sold 2500
shares, find the divided for each share.

2. a) A finance company earned a net profit of Rs 45,20,000 in a year. The company declared
to distribute a certain percent of profit as dividend. If the total dividend amount to
Rs 9,04,000, what percent of net profit was distributed as dividend?

b) A man has 200 shares out of 1000 total shares of a business company. He received
Rs 96,000 as dividend in a year which was a certain percent of net profit of
Rs 24,00,000. At what percent of the net profit was the dividend distributed?

53Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Commission and Taxation

3. a) A publication house distributed 21% of dividend to its shareholders from the net
profit of a year. If the amount of distributed dividend was Rs 23,52,000, calculate the
net profit made by the publication house.

b) A small hydropower company distributed the total amount of dividend of
Rs 14,58,000 to its shareholders which was 30% of the net profit earned by the
company. Find the net profit earned by the company.

Creative section - A

4. a) Mrs. Rai bought 250 shares out of 10,000 shares from a finance company. The
company earned a net profit of Rs 85,20,000 and declared 17% dividend to its
shareholders. Calculate the amount of dividend received by Mrs. Rai.

b) A Business Company sold 2,500 shares at Rs 1,200 per share. Bishwant bought 450
shares. If the company earned a net profit of Rs 39,00,000 in a year and it announced
to distribute 18% dividend from the net profit to its shareholders, find the amount
of dividend received by Bishwant.

5. a) A Cable Car Company sold 3,000 shares to the local people. The company earned
a net profit of Rs 1,20,00,000 in a year and distributed a certain percent of profit as
dividend. If a shareholder who has bought 125 shares received Rs 1,10,000 dividend,
what percent of profit was distributed as dividend?

b) Suntali bought 200 shares out of 5,000 shares sold by a hydropower company to
the local people. The company earned a net profit of Rs 75,00,000 in a year and it
declared to distribute a certain percent dividend to its shareholders. If she received
Rs 81,000, what percent of the net profit was distributed as dividend?

6. a) Mr. Dhurmus bought 500 shares out of 10,000 shares sold by a commercial bank.
The bank earned some profit and it distributed 14% of the net profit as dividend in
a year. If Dhurmus received Rs 1,03,600 in the year, find the net profit of the bank.

b) A Life Insurance Company earned some profit and announced to distribute 40%
dividend from its net profit to its shareholders. If a shareholder who bought 300
shares out of 12,000 shares sold by the company received Rs 1,25,400, calculate the
net profit of the company.

Project work

7. a) Take the information from website or national level of daily news papers or from any
business papers and make a report about the share values of any five companies,
banks, or any business organisations. Present your report in the class.

b) Is your any family member involved in any business organisation as one of the
shareholders? If so, discuss with her/him and learn more about bonus, dividend, etc.
Then, prepare a report and present in the class.

Vedanta Excel in Mathematics - Book 9 54 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Unit Household Arithmetic

4

4.1 Introduction

Household arithmetic deals with the regular financial activities
based on the household purposes. Payments of electricity bill,
telephone bill, water bill, etc. are a few examples of household
financial activities. Money exchange, calculation of taxi fare are
also under the household arithmetic. In this unit, we shall discuss
about electricity bill, telephone bill, water bill, and taxi fare.

4.2 Electricity bill

We measure the consumption of electricity in the number

of units. Digital meter and dial meter are two types of electric

meters which are used to find the number of units of electricity

consumed.

1 unit of electricity = 1 kilowatt hour or 1000 watt hour
Thus, 1 unit of electricity means, it is the amount of consumption
of electricity by an electric appliance of 1000 watt power in 1 hour.
The number of units of electricity consumed is calculated as:
Reading of the recent month – Reading of the previous month

The meter reading to find the consumption of electricity of every
month by a household is performed at the end of each month.

The table given below shows the capacity of meter box of the
lower voltage limit 230-400 volt connected in our houses and
the minimum energy charges.

Domestic Consumers

Service and Energy Charges (Single Phase)

5 Ampere 15 Ampere 30Ampere 60 Ampere

kWh Service Energy Service Energy Service Energy Service Energy
(Monthly)
Charge Charge (Rs. Charge Charge Charge Charge Charge Charge
0-10
11-20 (Rs.) per unit) (Rs.) (Rs.per (Rs.) (Rs.per (Rs.) (Rs.per
21-30
31-50 30.00 0.00 50.00 unit) 75.00 unit) 125.00 unit)
51-100 30.00 3.00 4.00 5.00 6.00
101-150 50.00 6.50
151-250 50.00 8.00 50.00 4.00 75.00 5.00 125.00 6.00
251-400 75.00 9.50
Above 400 100.00 9.50 75.00 6.50 100.00 6.50 125.00 6.50
125.00 10.00
150.00 11.00 75.00 8.00 100.00 8.00 125.00 8.00
175.00 12.00 100.00 9.50 125.00 9.50 150.00 9.50
125.00 9.50 150.00 9.50 200.00 9.50

150.00 10.00 175.00 10.00 200.00 10.00

175.00 11.00 200.00 11.00 250.00 11.00

200.00 12.00 225.00 12.00 275.00 12.00

55Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Household Arithmetic

The payment schedule after the meter reading are as follows.
1. Within 7 days of meter reading 2 % rebate is allowed.
2. From the 8th day to the 15th day, the payment will be according to the bill
3. From the 16th day to the 30th day, 5 % extra fine
4. From the 31st day to the 40th day, 10 % extra fine
5. From the 41st day to the 60th day, 25 % extra fine
6. If the bill is not paid upto the 60th day, the electricity line will be disconnected.

Worked-out examples

Example 1: According to the rate of electricity charge, the service charge up to
Solution: 10 units is Rs 30 but energy charge is free; find the charge of consumption
of 7 units of electricity.

Here, consumption of electricity = 7 units.

Service charge = Rs 30

Rate of energy charge up to 10 units = Rs 0 per unit

Now, total charge of electricity with service charge = Rs 30 + 7×Rs 0 = Rs 30

Hence, the required charge of consumption of 7 units of electricity is Rs 30.

Example 2: The rate of energy charge up to 20 units is Rs 3 per unit. How much will be
Solution: charged for the consumption of 16 units of electricity with Rs 30 service
charge?

Here, consumption of electricity = 16 units.

Service charge = Rs 30

Rate of energy charge up to 20 units = Rs 3 per unit

Now, energy charge of 16 units = 16×Rs 3 = Rs 48

? Total charge of electricity with service charge = Rs 30 + Rs 48 = Rs 78

Hence, the required charge of consumption of 16 units of electricity is Rs 78.

Example 3: The rate of electricity charge up to 20 units is Rs 3 per unit and Rs 6.50
Solution: per unit from 21 to 30 units. If the meter reading of a household was
01045 units on 1 Baishakh and 01070 units on 1 Jestha, find the number of
units of electricity consumed in Baishakh and the charge of consumption
of electricity with Rs 50 service charge.

Here, consumption of electricity = meter reading of 1 Jestha – meter reading of 1 Baishakh

= 01070 – 01045 = 25 units.

Rate of charge up to 20 units = Rs 3 per unit

? Charge up to 20 units = 20×Rs 3= Rs 60

Also,

The excessive number of units = (25 – 20) units = 5 units

Rate of charge from 21 to 30 units = Rs 6.50 per unit

Vedanta Excel in Mathematics - Book 9 56 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Household Arithmetic

? The charge of 5 units = 5×Rs 6.50= Rs 32.50
? Total charge of electricity with service charge = Rs 50 + Rs 60 + Rs 32.50 = Rs 142.50
Hence, the required charge of consumption of 25 units is Rs 142.50.

Example 4: In a household of 5A electricity transmission line, the meter reading of
1 Aswin was 1492 units and that of 1 Kartik was 1634 units. Answer the
following questions under the given rates and billing system.

Units 0-20 21 – 30 31 – 50 51 - 150
Rate of charge per units Rs 3 Rs 6.50 Rs 8.00 Rs 9.50
Service charge = Rs 100

The rules of rebate/fine

Meter reading within 7 days within 8th – within 16th – within 31st – within 41st

Rebate/fine 2% rebate 15th days 30th days 40th days – 60th days
- 5% fine 10% fine 25% fine

a) Calculate the charge of the consumed units of electricity.

b) Calculate the electricity charge if the payment was made on the

following date.

(i) the 5th day of meter reading (ii) the 14th day of meter reading

(iii) the 22nd day of meter reading (iv) the 35th day of meter reading

Solution: (v) 20 Mansir

Here, consumption of electricity = 1634 units– 1492units = 142 units.

a) Now,

Consumption block No. of units Rate of charge Electricity charge
0-20 20 – 0=20 Rs 3 20 ×Rs 3 = Rs 60

21-30 30 – 20=10 Rs 6.50 10 ×Rs 6.50 = Rs 65

31-50 50 – 30=20 Rs 8 20 ×Rs 8 = Rs 160

51-150 142 – 50 = 92 Rs 9.50 92 ×Rs 9. 50 = Rs 874

? Total charge = service charge + total electricity charge

= Rs 100 + (Rs 60 + Rs 65 + Rs 160 + Rs 874) = Rs 1259

b) (i) If the payment was made on the 5th day of meter reading, there would be 2% rebate

? Total charge to be paid = Rs 1259 – 2% of Rs 1259 = Rs 1233.82

(ii) If the payment was made on the 14th day of meter reading, payment would be according
to the bill.

? Total charge to be paid = Rs 1259

(iii) If the payment was made on the 22nd day of meter reading, there would be 5% fine

? Total charge to be paid = Rs 1259 + 5% of Rs 1259 = Rs 1321.95

(iv) If the payment was made on the 35th day of meter reading, there would be 10% fine

? Total charge to be paid = Rs 1259 + 10% of Rs 1259 = Rs 1384.90

(v) If the payment was made on 20 Mansir i.e., within 40 -60 days of meter reading,
there would be 25% fine.

? Total charge to be paid = Rs 1259 + 25% of Rs 1259 = Rs 1573.75

57Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Household Arithmetic

Example 5: A household having 15A electricity meter consumed 260 units in the month
of Paush. If the bill was paid on the 28th day of meter reading, calculate
the total charge under the following condition.

Units 0-20 21-30 31 - 50 51 - 150 151-250 251-400
Rate of charge Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 10 Rs 11

per units

Service charge = Rs 175 Payment within 16th to 30th day 5% extra fine.

By what percent less would the bill be made if the customer paid the bill
within 7 days of meter reading?

Solution:

Here, consumption of electricity = 260 units

Now,

Consumption block No. of units Rate of charge Electricity charge
20 ×Rs 4 = Rs 80
0-20 20 – 0=20 Rs 4 10 ×Rs 6.50 = Rs 65
20 ×Rs 8 = Rs 160
21-30 30 – 20=10 Rs 6.50 100 ×Rs 9. 50 = Rs 950
100 ×Rs 10 = Rs 1000
31-50 50 – 30=20 Rs 8 10 ×Rs 11= Rs 110

51-150 150 – 50 = 100 Rs 9.50

151-250 250 – 150 = 100 Rs 10

251-400 260 – 250 = 10 Rs 11

? Total charge = service charge + total electricity charge
= Rs 175 + (Rs 80 + Rs 65 + Rs 160 + Rs 950 + Rs 1000 + Rs 110) = Rs 2,540

Since, the payment was made on the 28th day of meter reading, there was 5% extra fine
? Total charge with fine = Rs 2,540 + 5% of Rs 2,540 = Rs 2,667
Again, if the customer paid the bill within 7 days of meter reading, there would be 2%
rebate.
? Total charge after rebate = Rs 2,540 – 2% of Rs 2,540 = Rs 2,489.20

Difference between two payments = Rs 2,667 – Rs 2,489.20 = Rs 177.80

Rs 177.80
? Less % in payment = Rs 2667 × 100% = 6.67%

Hence, the payment of the bill would be 6.64% less if it was paid within 7 days of meter
reading.

Example 6: Sunayana’s house has a 15A electricity meter box. If she made the
payment of Rs 1100 with service charge Rs 125 on the 40th day of meter
reading, how many units of electricity was consumed in the month?
Calculate it by using the following rates:

Units 0-20 21 – 30 31 – 50 51 - 150
Rate of charge per units Rs 4 Rs 6.50 Rs 8.00 Rs 9.50

Solution: Payment within the 31st to the 40th day of meter reading:10% extra fine.

Let the number of consumed unit of electricity of the month be x units.

Now,

Vedanta Excel in Mathematics - Book 9 58 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Household Arithmetic

Consumption block No. of units Rate of charge Electricity charge
0-20 20 – 0=20 Rs 4 20 ×Rs 4 = Rs 80
21-30 30 – 20=10 Rs 6.50 10 ×Rs 6.50 = Rs 65
31-50 50 – 30=20 Rs 8 20 ×Rs 8 = Rs 160
x – 50 = 100 Rs 9.50 (x – 50) ×Rs 9. 50 = Rs (9.5x – 475)
51-150

? Total charge = service charge + total electricity charge
= Rs 125 + (Rs 80 + Rs 65 + Rs 160 + Rs 9.5x – Rs 475) = Rs (9.5x – 45)

According to question,
Payment of the bill with fine = Rs 1100
or, Rs (9.5x – 45) + 10% of Rs (9.5x – 45) = Rs 1100
or, (9.5x – 45) + 0.1 (9.5x – 45) = 1100
or, (9.5x – 45) + (0.95x– 4.5) = 1100
or, 10.45x = 1149.5
or, x = 110
Hence, 110 units of electricity was consumed in the month.

EXERCISE 4.1
General section

1. a) What do you mean by 1 unit of electricity?
b) The meter readings for the consumption of electricity of a household was 1140
units on 1 Bhadra and 1285 units on 1 Aswin. How many units of electricity was
consumed in the month of Bhadra?
c) How many units of electricity is consumed when an electric refrigerator of 2000
watts is used for 2 hours?

d) If the rate of electricity charge is Rs x per unit, consumed electricity is y units,
service charge is Rs z, and total bill is Rs t, write down the relation among t, x, y, and
z.

2. a) According to the recent electricity tariff rate, the service charge up to 5 units is Rs 30
but energy charge is free; find the charge of consumption of 8 units.

b) There is no energy charge but the service charge is Rs 30 up to 10 units; how much
bill should be charged by Nepal Electricity Authority (NEA) for the consumption of
8 units?

c) The energy rate of electricity up to 20 units is Rs 3 per units, how much should a
household pay for the consumption of 15 units with Rs 30 service charge?

d) The energy rate of electricity up to 20 units is Rs 3 per units. If a household consumes
18 units of electricity in a month, how much should s/he pay the electricity bill with
Rs 30 service charge?

3. a) The rate of electricity charge upto 20 units is Rs 3 per unit and Rs 6.50 per unit from
21 to 30 units. Find the charge of consumption of 28 units with Rs 50 service charge.

b) According to the recent rate of electricity charge the charge upto 20 units is
Rs 3 per unit and Rs 6.50 from 21 to 30 units. If 26 units of electricity is consumed
in Ritambhara's house in this month, how much charge should she pay with Rs 50
service charge?

59Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Household Arithmetic

4. The table given below shows the rate of electricity charge with service charge for a
5A meter.

KWh (units) Service charge Energy charge

0-20 Rs 30 Rs 3.00
21-30 Rs 50 Rs 6.50

Find the total of electricity charge in the following cases.

a) The meter reading of Sandhya's house on 1 Baisakh 01198 1 01223 4
and 1 Jestha are shown in the meter box alongside. Baisakh Jestha
Find the electricity charge for the month of Baisakh.

b) The meter reading of Kapil's house is shown Month Falgun Chaitra
in the table. Find the electricity charge of the
month. Meter reading 2400 2430

c) The rate of electricity charge up to 20 units is Rs 3 per unit and Rs 6.50 per unit from
21 to 30 units. If the meter reading of a household was 02021 units on 1 Baishakh
and 02045 units on 1 Jestha, find the number of units of electricity consumed in
Baishakh and the charge of consumption of electricity with Rs 50 service charge.

Creative section - A

5. Electricity tariff rates and rebate/fine rules are given below.

kWh 5 Ampere 15 Ampere 30 Ampere 60 Ampere
(Monthly)
Service Energy Service Energy Service Energy Service Energy
Units Charge Charge Charge Charge
Charge Charge Charge Charge
0-20 per unit per unit
per unit per unit Rs 75.00 Rs 5.00 Rs 125.00 Rs 6.00

Rs 30.00 Rs 3.00 Rs 50.00 Rs 4.00

The rules of rebate/fine

Meter reading within 7 days within 8-15 days within 16-30 days within 31-40 days within 41-60 days

Rebate/fine 2% rebate – 5% fine 10% fine 25% fine

From the tables given above, workout the following problems.
a) A household having 5 A electricity meter consumed 18 units of electricity in one

month. Find the amount of payment made by the household within 7 days.
b) A household having a 15 A meter consumed 16 units of electricity in one month.

Find the amount of payment made by the household on the 25th day of meter reading.
c) A house having a 30 A meter consumed 19 units of electricity in one month and if the

payment was made on the 36th day of meter reading, find the amount of payment.
d) A house having a 60 A meter consumed 20 units of electricity in one month and the

payment was made the 50th day of meter reading. Calculate the amount of payment
made by the house.

6. In a household of 5A electricity transmission line, the meter reading of 1 Bhadra was
050844 units and that of 1 Aswin was 050969 units. Answer the following questions
under the given rates and billing system.

Vedanta Excel in Mathematics - Book 9 60 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Household Arithmetic

Units 0-20 21 – 30 31 – 50 51 - 150

Rate of charge per units Rs 3 Rs 6.50 Rs 8.00 Rs 9.50

Service charge = Rs 100

The rules of rebate/fine are:

- If the payment is made within 7 days of meter reading, 2% rebate is allowed.

- If the payment is made within the 8th to the 15th days of meter reading, the

payment is according to the bill.

- If the payment is made within the 16th to the 30th days of meter reading, 5% extra fine.

- If the payment is made within the 31st to the 40th days of meter reading, 10% extra fine.

- If the payment is made within the 41st to the 60th days of meter reading, 25% extra fine.

a) Calculate the charge of the consumed units of electricity.

b) Calculate the electricity charge if the payment was made on the following date.

(i) the 6th day of meter reading (ii) the 11th day of meter reading

(iii) the 25nd day of meter reading (iv) the 32nd day of meter reading

(v) the 48th day of meter reading

7. a) Yankee’s house has a meter box of capacity of 5A. NEA charges the bill at the rate

of Rs 3 per unit for the first 20 units, Rs 6.50 per unit from 21 to 30 units, Rs 8 per

unit from 31 to 50 units and Rs 9.50 per unit from 51 to 150 units. The payment

within 7 days of meter reading allows 2% rebate and there is 5% fine for the

payment within the 16th to the 30th days of mater reading. If the consumption of

the electricity in the month of Falgun was 120 units and she paid the bill on the

20th days of the meter reading, how much did she pay to clear the bill with Rs 100

service charge? By what percentage less would the bill be made if she paid the bill

within 7 days?

b) A meter of capacity 15A is fixed in Dharmendra’s house. If the consumption of the

electricity was 280 units in the month of Magh and the bill was paid on 55th day

of meter reading, calculate the total charge under the following condition.

Units 0-20 21 – 30 31 – 50 51 – 150 151-250 251-400

Rate of charge Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 10 Rs 11

per units

Service charge = Rs 175

For the payment within 7 days of meter reading 2% rebate and the payment within

41st to 60th day of meter reading charges 10% extra fine

By what percentage less would the bill be made if the customer paid the bill within

7 days of meter reading?

8. a) Jasmin’s house has a 5A electricity meter box. If she made the payment of the

electricity Rs 777 with service charge Rs 75 on the 24th day of meter reading, how

many units of electricity was consumed in the month? Calculate it by using the

following rates

Units 0-20 21 – 30 31 – 50 51 - 100

Rate of charge per units Rs 3 Rs 6.50 Rs 8.00 Rs 9.50

Payment within the 16th to the 30th day of meter reading: 5% extra fine.

61Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Household Arithmetic

b) The rates of electricity up to 20 units is Rs 4 per unit, Rs 6.50 per unit from 21 to 30
units, Rs 8 per unit from 31 to 50 units and Rs 9.50 per unit from 51 to 150 units.
Tikaram’s house has a 15A electricity meter box. He paid the bill of Rs 1,166.20
with service charge Rs 125 on the 3rd day of meter reading getting 2% rebate; find
the consumed units of electricity.

9. a) Mr. Sharma has a 5A meter in his house. He uses 5 CFL bulbs of 15 watt each for
4 hours and an electric heater of 1200 watt for 1 hour everyday. Find the cost of
payment of the bill of a month at the rate of Rs 3 per unit up to 20 units, Rs 6.50 per
unit from 21 to 30 units and Rs 8 from 31-50 units with Rs 75 service charge, if the
payment is made on the 10th day of meter reading.

b) Mrs. Bajracharya's house has 15A electricity meter. She uses 5 LED bulbs of 10 watt
each for 4 hours, 2 televisions of 60 watt each for 5 hours and a refrigerator of 250
watt for 2 hours everyday. Find the cost of payment of the bill of a month at the rate
of Rs 4 per unit up to 20 units, Rs 6.50 per unit from 21 to 30 units and Rs 8 from 31
to 50 units with Rs 75 service charge, if the payment is made on the 35th day of meter
reading.

4.3 Telephone bill

The payment of telephone bill is also a regular financial household activity.
Here, we discuss about the telephone bill of PSTN/Land line
Study the following rules of payment of the telephone bills implemented by Nepal

Telecommunications Authority (gk] fn b/" ;~rf/ k|flws/0f).

– Minimum charge of 175 calls is Rs 200.
– Rate of charge above 175 calls is Re 1 per call
– Tariff (C) = Minimum charge (c) + charge of additional number of calls

C =c+n×r
Telecom service charge (TSC) = 10 % of C

VAT = 13 % of (C + TSC)
Total charge = C + TSC + VAT

No. : Nepal Telecom
Date :
Nepal Doorsanchar Company Ltd.

Statement For The Month:

Local call, STD call and ISD call are three types of telephone calls. The rates of charge
of these calls are different.

Vedanta Excel in Mathematics - Book 9 62 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Household Arithmetic

Worked-out examples

Example 1: The minimum charge up to 175 calls is Rs 200. If the charge for each
additional call is Re 1, how much is the charge for 280 calls with 10 %
TSC and 13 % VAT?

Solution:
Here, the minimum charge for 175 calls = Rs 200

The additional number of calls = 280 – 175 = 105

Charge for additional number of calls = 105 × Re 1 = Rs 105

? Charge for 350 calls (C) = Rs 200 + Rs 105 = Rs 305

Again, TSC = 10 % of Rs 305 = Rs 30.50

? Charge with TSC = Rs 305 + Rs 30.50 = Rs 335.50

Also, VAT = 13 % of Rs 335.50 = Rs 43.62

Now, total charge = C + TSC + VAT

= Rs 305 + Rs 30.50 + Rs 43.62 = Rs 379.12

Hence, the total charge of 350 calls is Rs 379.12.

Example 2: The minimum charge of telephone calls up to 175 calls is Rs 200. The
Solution: charge for each extra call of 2 minutes duration is Re 1. If a person paid
Rs 870.10 with 10 % TSC and 13 % VAT for his telephone bill, find the
number of extra calls made.

Let the charge of telephone calls without VAT be Rs x.

? x + 13 % of x = Rs 870.10

113x
or, 100 = Rs 870.10
or, x = Rs 770
Again, let the charge of telephone calls without TSC be Rs y.

y + 10 % of y = Rs 770

11y
or, 10 = Rs 770
or, y = Rs 700

The minimum charge up to 175 calls = Rs 200

The charge for the extra calls = Rs 700 – Rs 200 = Rs 500

Now, the number of extra calls = Rs 500 = 500
Re 1

Hence, the required number of extra calls is 500.

63Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Household Arithmetic

4.4 Water bill
Nepal Water Supply Corporation (gk] fn vfg]kfgL ;:+ yfg) is the concerned authority of

the Ministry of Water Supply and Sanitation, the Government of Nepal.

A household having water supply facility may have tap with a meter or without a
meter. If a meter is not connected in the tap, a customer should pay a lump sum
amount fixed by the Nepal Water Supply Corporation. For the payment of a water bill,
Nepal Water Supply Corporation has implemented the following rules.

S. Size of Tap with Meter Tap without Meter
No. Pipe

Minimum Minimum Additional Main Tap Branch
Consumption Charge Consumption Charge Tap Charge
(Rs) per thousand (Rs) (Rs)
(litre)
litre (Rs)

1 1" 10,000 110 25 560 200
2

2 3" 27,000 1490 40 3360 1600
4 56,000 3420 40 9200 2700

3 1"

There is also the compulsory provision of sewerage service charge which is 50% of
the water consumption charge.
On the other hand, Kathmandu Upatyaka Khanepani Limited (KUKL) has implemented
a slightly different water tariff rules.

S. No. Size of Pipe Tap with Meter

Minimum Minimum Additional Tap without
Consumption Charge Consumption Meter
(Rs) per thousand (Rs)
(litre)
litre
(Rs)

1 12" 10,000 100 32 785
2 43" 27,000 1910 71 4595
3 1" 56,000 3960 71 9540

Furthermore, Nepal Water Supply Corporation has implemented the following rules
and regulations about the schedule of the payment of the bills.

Payment is made after the bill issued Rebate/Fine
1. Within the first and second month 3% rebate
2. Within the third month No rebate and no fine
3. Within the fourth month 10% fine
4. Within the fifth month 20% fine
5. After fifth month 50% fine

Vedanta Excel in Mathematics - Book 9 64 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Household Arithmetic

Example 3: The meter reading for the consumption of water of a household was 1420
units on 1 Mangsir and 1470 units on 1 Poush. Calculate the charge to be
paid including 50% sewerage service charge if the payment of the bill is
made in the following schedule.
(i) within the second month after the issue of bill
(i) within the fourth month after the issue of bill
(iii) within the sixth month after the issue of bill

Solution:
Here, the meter reading on 1 Mangsir = 1420 units

The meter reading of 1 Poush = 1470 units

Consumed units of water = (1470 – 1420) units = 50 units

According to water tariff provisions of Nepal Water Supply corporation:

The minimum charge of 10,000 litres (i.e. 10 units) = Rs 110

The additional units of water consumption = (50 – 10) units = 40 units

Now, the charge of 35 units of water = 40 × Rs 25 = Rs 1,000

∴ The total charge = Rs 110 + Rs 1,000 = Rs 1,110

Again, the charge including sewerage service charge = Rs 1,110 + 50% of Rs 1,110

= Rs 1,665
(i) If the payment is made within the second month,

the required payment = Rs 1,665 – 3% of Rs 1,665 = Rs 1,615.05
(ii) If the payment is made within the fourth month,

the required payment = Rs 1,665 + 10% of Rs 1,665 = Rs 1,831.50
(iii) If the payment is made within the sixth month

the required payment = Rs 1,665 + 50% of Rs 1,665 = Rs 2,497.50

4.5 Calculation of taxi fare in a taximeter

Nowadays, with the increasing facilities of taxi services, specially in the urban area,
the payment of taxi fare is also becoming a regular financial activity. A device used in
taxis that automatically records the distance travelled and the fare payable is called
taximeter.

Nepal Bureau of Standards and Metrology (NBSM) -gk] fn u0' f:t/ tyf gfktfn} laefu_ is

responsible to implement the rules, regulations, and the rates of fare and monitoring
that it is carried out fairly or correctly.

Here, we shall discuss about the calculation of fare in a taximeter. Following are the
rates of fare in taximeter implemented by NBSM department:

Time Minimum fare Fare of per 200 meters
6:00 am to 9:00 pm Rs 14 Rs 7.20
9:00 pm to 6:00 am Rs 21 Rs 10.80

(or, 1.5 times of the fair (or, 1.5 times of the fair
of 6:00 am to 9:00 pm) of 6:00 am to 9:00 pm)

While travelling between 6:00 am to 9:00 pm, Rs 14 appears in the monitor of the
taximeter immediately when it is flagged down. Then, the fare goes on at the rate of
Rs 7.20 per 200 m. But, in the case of travelling between 9:00 pm to 6:00 am, when

65Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Household Arithmetic

the taximeter is flagged down Rs 21 appears in the monitor and the fare goes on at the
rate of Rs 10.80 per 200 m.

If the taxi is asked to wait by a passenger, an additional waiting charge of Rs 7.20 per

2 minutes is to be paid in the case of 6:00 am to 9:00 pm and this waiting charge is

Rs 10.80 per 2 minutes in the case of 9:00 pm to 6:00 am.

Example 4: Anil travelled 9 km by a metered taxi. The minimum fare of Rs 14 appeared
Solution: immediately after the meter was flagged down, then the fare went on
at the rate of Rs 7.20 per 200 meters. An additional waiting charge of
Rs 7.20 per 2 minutes was charged for the waiting of 6 minutes during the
journey. Calculate the total fare paid by Anil.

Here, the minimum fare = Rs 14

Distance travelled = 9 km = 9000 m

Now, the fare of 200 meters = Rs 7.20

The fare of 1 meter = Rs 7.20 × 9000 = Rs 324
The fare of 9000 meters = R2s0.072.0200

Also, the waiting charge of 2 minutes = Rs 7.20

The waiting charge of 1 minute = Rs 7.20
2
7.20
The waiting charge of 6 minutes = Rs 2 × 6 = Rs 21.60

∴ The total fare = Rs 14 + Rs 324 + Rs 21.60 = Rs 359.10

Hence, Anil paid the total fare of Rs 359.10.

Example 5: Sangita hired a taxi and travelled a certain distance at 10.30 pm. She
Solution: paid the total fare of Rs 345 including the additional waiting charge for
10 minutes. If the minimum fare is Rs 21, the fare per 200 m is Rs 10.80 and
the waiting charge is Rs 10.80 per 2 minutes, find the distance travelled
by her.

Here, the waiting charge for 2 minutes = Rs 10.80

The waiting charge for 1 minute = Rs 10.80 = Rs 5.40
2
The waiting charge for 10 minutes = 10 × Rs 5.40 = Rs 54

Now, the taxi fare excluding the minimum fare and waiting charge = Rs 345 – Rs 21 – Rs 54

= Rs 270

Again,

Rs 10.80 is the fare of 200 metres

Re 1 is the fare of 200 metres.
10.80

Rs 270 is the fare of 200 × 270 meters = 5000 meters = 5 km
10.80

Hence, she travelled 5 km.

Vedanta Excel in Mathematics - Book 9 66 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Household Arithmetic

EXERCISE 4.2

1. a) The minimum charge up to 175 calls is Rs 200. If the charge for each additional call
is Re 1, how much is the charge for 475 calls with 10 % TSC and 13 % VAT?

b) The charge for STD call from Bhadrapur to Surkhet for 1 minute is Rs 1.25. If
Mr. Rajbanshi makes a call for 10 minutes, how much should he pay with 10 % TSC
and 13 % VAT?

c) The charge of ISD call from Bharatpur to Melbourne, Australia is Rs 20 per minute. If
Nirmal made a call for 5 minutes, calculate the cost paid by him with 10 % TSC and
13 % VAT.

2. a) The minimum charge of telephone calls up to 175 calls is Rs 200. The charge for each
extra call is Re 1. If a person paid Rs 870.10 with 10 % TSC and 13 % VAT, find the
number of extra calls.

b) The minimum charge of telephone calls up to 175 calls is Rs 200. The charge for each
extra call of 2 minutes duration is Re 1. If a household paid Rs 633.93 with 10 % TSC
and 13 % VAT to clear the bill of a month, find the total number of calls made in the
month.

3. a) A household consumed 32 units of water in a month. Calculate the payment of

the bill including 50% sewerage service charge, if the payment is made within

the first month of the bill issued.

b) 127 units of water is consumed by using 34"pipe in a hotel. If the payment of the
bill is made within the fifth month after the bill issued, how much money is

required to clear the bill with 50% sewerage charge?

c) A household uses 1" of water pipe. The meter reading of the household on 1st of
2
Asar was 1260 units and on 1 Shrawan was 1330 units. Calculate the charge to

be paid including 50% sewerage service charge if the payment of the bill is made

in the following schedule.

(i) within the first month after the issue of bill
(ii) within the third month after the issue of bill
(iii) within the fifth month after the issue of bill
(iv) within the seventh month after the issue of bill

4. a) Mrs. Suntali hired a taxi and travelled 12 km at 3:30 p.m. If the minimum fare is
Rs 14 and the fare goes on at the rate of Rs 7.20 per 200 metres, calculate the
total fare paid by her.

b) Mr. Kattel travelled 15 km by a hired taxi at 5:00 a.m. The minimum fare of
Rs 21 appeared immediately after the meter was flagged down. Then, the fare
went on at the rate of Rs 10.80 per 200 metres. An additional waiting charge of
Rs 10.80 per 2 minutes was charged for waiting of 10 minutes. Calculate the
total of the taxi fare paid by him.

67Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Household Arithmetic

5. a) Rita hired a taxi and travelled a certain distance at 8:00 a.m. She paid the total
fare of Rs 194. If the minimum fare is Rs 14 and the fare per 200 meters is
Rs 7.20, find the distance travelled by her.

b) Bishwant hired a taxi and travelled a certain distance at 10:45 p.m. He paid the
total fare of Rs 615 including the additional waiting charge for 10 minutes. Find
the distance travelled by him.

Project work

6. a) Make a group of 10 friends, and prepare a report about the types of electricity
consumption meter-box fixed in each of 10 friends' house. Present your report in
the class.

b) Estimate the average number of units of electrical energy consumed in your house
in a month. Calculate the cost of energy paid by your house in a month.

c) Make a list of different electrical appliances being used in your house. How many
units of electricity do these appliances consume approximately in

(i) 1 hour (ii) 24 hours (ii) 1 month?

d) Mention a few ways to reduce the unnecessary use of electricity in your house.
Estimate the amount of reduction of cost in a month by applying these ways.

7. a) Collect the telephone bill and water bill of the recent months and bring to your
class. Discuss about the minimum cost, rate of cost, rebate, fine, etc. in the class.

b) Search in the available website and find the information about the rate of
cell-phone charge, internet charge, etc. mentioned by Nepal Telecom.

Objective Questions

1. If a goat is sold for Rs 2781 at 8% profit, its cost price is

(A) Rs 2575 (B) Rs 3003.48 (C) Rs 2502.90 (D) Rs 2755

2. A man expects a gain of 10 % on his cost price. On a day, if he sells a calculator for
Rs. 550, what is his profit?

(A) Rs 50 (B) Rs 55 (C) Rs 10 (D) Rs 65

3. If the selling price of a dozen of pens is equal to the cost price of 15 pens, find the
profit percent.

(A) 25% (B) 50% (C) 33.3% (D) 20%

4. A man bought some umbrellas for Rs 8000 and sold them for Rs 9000 by making Rs 50
profit in each, how many umbrellas did he buy and sell?

(A) 10 (B) 15 (C) 20 (D) 25

5. If an article is sold at three fourth of its marked price, what percent of discount is given?

(A) 5% (B) 10% (C) 20% (D) 25%

6. What does the difference between the price after value added tax (VAT) and price after
discount of any article give us?

(A) Market price (B) Discount (C) Selling price (D) VAT

Vedanta Excel in Mathematics - Book 9 68 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Household Arithmetic

7. If a bag was sold for Rs 2034 with 13% value added tax (VAT), what amount of VAT
was imposed?

(A) Rs 1800 (B) Rs 234 (C) Rs 264.42 (D) Rs 1769.58

8. The monthly salary of a salesman in a departmental store is Rs 15000 and additional
payment of 1% on the total monthly sale is provided as commission. What is his income
in a month if he makes a total sale of Rs 660000 in that month?

(A) Rs 15000 (B) Rs 6600 (C) Rs 8400 (D) Rs 21600

9. A doctor working in a clinic has monthly salary Rs 40000. He gets 2% commission on
the monthly sales of medicine above 5 lakhs in the clinic. What should be the monthly
sales in the clinic so that the monthly income of the doctor in that monthly is Rs
60000?

(A) 8 lakhs (B) 10 lakhs (C) 15 lakhs (D) 20 lakhs

10. A company distributed 15% bonus to its 50 employees equally from the net profit at
the end of last year. If every employee received Rs 16500, how much was the net profit
of the factory?

(A) Rs 55 lakhs (B) Rs 44 lakhs (C) Rs 45 lakhs (D) Rs 54 lakhs

11. The monthly salary of a professor is Rs 55000. If 1% social security tax is charged up to
the annual income of Rs. 450000, the rate of tax is 15% for the income of Rs. 450000 to
Rs. 550000 and the rate of tax is 25% for the rest, how much income tax should be paid
by the individual in the year?

(A) Rs 4500 (B) Rs 19500 (C) Rs 32000 (D) Rs 47000

12. The monthly salary of an unmarried individual is Rs 38500. If the tax free allowance is
upto annual income of Rs 400000 and he pays Rs 9300 as income tax in year, the rate of
income tax is:

(A) 10% (B) 12% (C) 15% (D) 20%

13. What does 1 unit of electricity mean?

(A) Consumed amount of electricity by electric appliance of 1 watt power in 1 hour.

(B) Consumed amount of electricity by electric appliance of 10 watt power in 1 hour.

(C) Consumed amount of electricity by electric appliance of 100 watt power in 1 hour

(D) Consumed amount of electricity by electric appliance of 1000 watt power in 1 hour.

14. Which one of the following is true for telephone bill?

(A) Sub Total = Minimum charge + Charge of additional number of calls

(B) Total = Sub Total + 10% telecom service charge (TSC) of Sub Total

(C) Grand Total = Total + 13% VAT of Total

(D) All of the above

15. According to Nepal Bureau of Standards and Metrology (NBSM), what is the minimum
taxi fare from 6:00 am to 9:00 pm?

(A) Rs 7.40 (B) Rs. 11.10 (C) Rs 14 (D) Rs 21

69Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Unit Mensuration

5

5.1 Mensuration – review

Mensuration is the branch of mathematics which studies the measurement of
the geometric figures. It deals with length, area, and volume of different types of
2 dimensional and 3 dimensional shapes.
Let's study the table given below to review the concept of measurements of perimeter
and area of different types of 2 dimensional plane figures.

Plane figures Measurements
Triangle 1

with the given base (b) Area of triangle = 12 perpendicular × base
and height (h) =2×b×h

Right-angled triangle Area of right-angled triangle
1
Scalene triangle
= 2 perpendicular × base
1

=2p×b

Perimeter of triangle (P) = a + b + c

Semi-perimeter (s) = a+b+c

2

Area of triangle = s(s – a) (s – b) (s – c)

Equilateral triangle Perimeter of equilateral triangle
= 3 × length of a side = 3a
2a
Height of equilateral triangle = a2 – a2 2
Rectangle
= 34a2 = 32 a
1
Area of equilateral triangle = 2 × b × h
1
= 2× 32 a = 34 a2

Perimeter of rectangle = 2 (l + b)
Area of rectangle = l × b

Vedanta Excel in Mathematics - Book 9 70 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Square Mensuration

d Perimeter of square = 4l
d Area of square = l2

Parallelogram Area of square = 12 d2

ha Perimeter of parallelogram = 2 (a + b)
Area of parallelogram = bh
d2 b
Rhombus Perimeter of rhombus = 4l
ll Area of rhombus = 12 d1 × d2
Wrhhomerbe,usd1 and d2 are the diagonals of the
d1 Perimeter of kite = 2(a + b)

ll 1
Area of the kite = 2 × d1 × d2, where d1 and d2
a Kite are the diagonals of the kite.
d2
b Perimeter of trapezium = sum of its four sides
a Area of trapezium = 12 h (a + b) , where a and
d1 b are the opposite parallel sides and h is the
b height of the trapezium.

Trapezium
b

h Perimeter of quadrilateral=sum of its four sides

a
Quadrilateral

h1 d Area of quadrilateral = 12 d (h1 + h2), where h1
h2
caonmd mh2onarebatshee heights of two triangles on the
Circle and semi circle quadrilateral. which is the diagonal (d) of the

Perimeter (or circumference) of a circle

1 = Sd or 2Sr
2
Perimeter of a semi circle= Sd + d = d(2S + 1)

dr or, Perimeter of semicircle = 21 × 2πr + 2r

r r = r (π + 2)

Area of a circle = 4S d2 or Sr2
Area of semicircle = 8S d2 or, 12 Sr2

71Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Mensuration

5.2 Area of pathways

(i) Area of a path running outside a rectangle

Let ABCD be a rectangular field. A path of uniform width d is running outside the field.
Here, area of rectangle ABCD = l × b
Area of rectangle PQRS = (l + 2d) × (b + 2d)
? Area of path = Area of PQRS – Area of ABCD

= (l + 2d) (b + 2d) – lb

= 2d (l + b + 2d)
? Area of path running outside a rectangle

= 2d (l + b + 2d)
In case of a square field, l = b
? Area of path running outside a square = 2d (l + l + 2d)

= 4d (l + d)

(ii) Area of a path running inside a rectangle

Let, ABCD be a rectangular field. A path of uniform width d is running inside the field.
Here, area of rectangle ABCD = l × b

Area of rectangle PQRS = (l – 2d) × (b – 2d)
? Area of path = Area of ABCD – Area of PQRS

= lb – (l – 2d) (b – 2d)
= 2d (l + b – 2d)
? Area of a path running inside a rectangle
= 2d (l + b – 2d)
In the case of a square field, l = b
? Area of a path running inside a square = 2d (l + l – 2d)

= 4d (l – d)

(iii) Area of paths crossing each other perpendicularly

Let, ABCD be a rectangular field.
Two paths PQRS and WXYZ of uniform width d are
running from the middle of every side of the field and
crossing each other at a right angle.
Here, the area of path PQRS = l × d
The area of path WXYZ = b × d
The area of square EFGH = d2

? The area of crossing paths= Area of (PQRS + WXYZ – EFGH)
= ld + bd – d2
= d (l + b – d)

? Area of crossing paths = d (l + b – d)

In case of a square field, l = b

? Area of crossing paths = d (l + l – d) = d (2l – d)

Vedanta Excel in Mathematics - Book 9 72 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Mensuration

(iv) Area of path running outside a circular field

Let r be the radius of a circular field.

A path of uniform width d is running outside the field.

Here, area of the circle XYZ = πr2

Area of the circle PQR = π (r + d)2

? Area of a path = Area of circle PQR – Area of circle XYZ
= π (r + d)2 – πr2

= πd (2r + d)

? Area of a path running outside a circular field = πd (2r + d)
Alternatively, if the radius of the circle PQR = r + d = R

Then, the area of the path = πR2 – πr2 = π (R + r) (R – r)

(v) Area of path running inside a circular field

Let, r be the radius of a circular field.

A path of uniform width d is running inside the field

Here, area of the circle XYZ = πr2

Area of the circle PQR = π(r – d)2

? Area of the path = Area of circle XYZ – Area of circle PQR
= πr2 – π (r – d)2

= πd (2r – d)

? Area of path running inside a circular field = πd (2r – d)
The table given below shows the summary of area of pathways.

Types of pathways Area of pathways

A path running d In the case of a rectangle:
outside a rectangle
and a square d d b+2d A = 2d (l + b + 2d)
d In the case of a square:
A = 4d (l + d)
l+2d

A path running inside d In the case of a rectangle:
a rectangle and
a square d b – 2d d b A = 2d (l + b – 2d)
l – 2d In the case of a square:
Two crossing paths A = 4d (l – d)
d
In the case of a rectangle:
l
A = d (l + b – d)
d b In the case of a square:
dd A = d (2l – d)

d
l

A path running A = Sd (2r + d)
outside a circle or

r+d A = S (R + r) (R – r)
r–d
A path running A = Sd (2r – d)
inside a circle

73Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Mensuration

5.3 Area, cost and quantities

Let A = Area of paths, floor, carpet, etc.
R = Rate of cost of paving, carpeting, etc.
C = Total cost of paving, carpeting, etc.
N = Number of bricks required to pave the path, number of pieces of carpet, etc.
a = Surface area of each brick, each piece of carpet, etc.

Now,

Total cost = Area × Rate C = A × R Then, A = C and R = C
R A

Number of bricks = Area of path N = Aa Then, A = N × a and a = A
Surface of a brick N
C C
Total cost = Number of bricks × Rate C=N×R Then, N = R and R = N

Worked-out examples

Example 1: A rectangular room is 9 m long and 6 m broad. Find the number of

pieces of carpet each of 4 m long and 1.5 m wide to cover the floor of

the room. If the rate of cost of carpet is Rs 110 per metre, find the cost

of carpeting the floor.

Solution:

Here, Length of the room (l) = 9 m

Breadth of the room (b) = 6 m

Width of the carpet (b1) = 1.5 m
Length of the carpet (l1) = 4 m
Rate of cost of carpet (R) = Rs 110 per metre

Now, the area of the floor of the room (A) = l × b

= 9 m × 6 m = 54 m2

Also, the area of carpet (a) = l1 × b1 = 4 m × 1.5 m = 6 m2

Now, the number of pieces of carpet = A = 54 m2 =9
a 6 m2
Hence, 9 pieces of carpet are required to cover the floor of the room.

Again, let x m be the total length of carpet required to cover the floor.

Here, area of each piece of carpet = Area of the floor

or, x × 1.5 m = 54 m2

or, x = 54 m2 = 36 m
1.5 m
Now, the total cost of carpeting the floor = length × rate = 36 × Rs 110 = Rs 3,960.

Example 2: The length of a rectangular room exceeds its breadth by 2 m and its
perimeter is 36 m. Find the cost of carpeting its floor at Rs 105 per sq.m.

Vedanta Excel in Mathematics - Book 9 74 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Mensuration

Solution:
Here, length of the room (l) = breadth + 2 m = (b + 2) m

Perimeter of the floor = 36 m
or, 2 (l + b) = 36 m
or, 2 (b + 2 + b) = 36 m
or, 4b + 4 = 36 m
or, b = 8 m
? l = (b + 2) m = (8 + 2) m = 10 m
Now, area of the floor (A) = l × b = 10 m × 8 m = 80 sq.m.
? Cost of carpeting the floor (C) = A × R

= 80 × Rs 105 = Rs 8,400
Hence, the required cost of carpeting the floor is Rs 8,400.

Example 3: The adjoining figure is a window with the lowermost part
a square of side 2.8 m and the uppermost part is the semi-
circular shape. Find the total length of the aluminium
frame of the window. Also, find the cost of covering the
window with curtains at Rs 540 per sq. m.

Solution:
Here, perimeter of the square shaped lower part= 3 l = 3 × 2.8 m = 8.4 m

Perimeter of the semi-circular upper part = 1 × 2πr = πr = 22 × 2.8 = 4.4 m
? The perimeter of the window 2 7 2
= 8.4 m + 4.4 m = 12.8 m

Total length of the aluminium frame of the window = perimeter of the window = 12.8 m

Again, area of the window = (2.8 m)2 + 1 πr2 = (2.8 m)2 + 1 × 22 × 2.8 × 2.8 m2
2 2 7 2 2

= 7.84 m2 + 3.08 m2 = 10.92 m2
? Area of the curtains = 10.92 m2
Now, the cost of curtains = Area × Rate = 10.92 × Rs 540 = Rs 5,896.80.

Example 4: A rectangular park measures 125 m by 78 m. It has a gravel path 3 m wide all

around on the outside. Find the area of the path and the cost of gravelling it

at Rs 120 per sq. m. 3m

Solution: 3m 78m

Here, the length of the park (l) = 125 m

The breadth of the park (b) = 78 m 125m
The width of the path = 3 m

Now, the area of the path (A) = 2d (l + b + 2d)

= 2 × 3 (125 + 78 + 2 × 3)

= 1,254 m2

Again, the cost of gravelling the path = Area × Rate

= 1,254 × Rs 120 = Rs 1,50,480

Hence, the area of the path is 1,254 m2 and the cost of gravelling it is Rs 1,50,480.

75Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Mensuration

Example 5: A rectangular garden is 64 m long and 46 m broad. Two crossing paths
Solution: each of 2 m wide running across the middle of the garden are at right
angle.
(i) Calculate the cost of paving the bricks of size 18 cm by 12 cm on the

paths at Rs 15,000 per 1000 bricks.
(ii) Find the cost of growing grass in the remaining space of the garden

at Rs 40 per sq. metre.

Here, length of the garden (l) = 64 m
Breadth of the garden (b) = 46 m
Width of the paths (d) =2m
Now,the area of the crossing paths (A) = d (l + b – d)
= 2 (64 + 46 – 2)
Also, the area of each brick (a) = 216 m2
= 18 cm × 12 cm = 216 cm2

(i) Number of bricks (N) = A = 216 × 100 × 100 = 10,000 bricks
a 216

Cost of 1000 bricks = Rs 15,000

Cost of 10000 bricks = Rs 15,000 × 10000 = Rs 1,50,000
1,000

Again, the area of garden= 64 m × 46 m = 2944 m2

The area of the remaining space = Area of garden – Area of paths

= 2944 m2 – 216 m2 = 2728 m2
(ii) ? The cost of growing grass in the remaining space = Area × Rate

= 2728 × Rs 40 = Rs 1,09,120
Hence, the cost of paving the paths is Rs 1,50,000 and the cost of growing grass in the
remaining space is Rs 1,09,120.

Example 6: The cost of gravelling a path 3 m broad inside the boundary of a square
park at Rs 110 per sq.m is Rs 1,54,440. Find the cost of covering the empty
Solution: space with turfs at Rs 50 per sq. m.
Here,
width of the path (d) = 3 m

Rate of cost of gravelling the path (R) = Rs 110

Total cost of gravelling the path (C) = Rs 1,54,440

Now, the area of the path (A) = C = 1,54,440 m2
or, R 110

4d (l – d) = 1404 m2

or, 4 × 3 (l – 3) = 1404 m2

or, l = 120 m

Again, the length of empty space = (120 – 3 – 3) m = 114 m

Then, the area of empty space = (114 m)2 = 12,996 m2

? The cost of covering the empty space with turfs = Area × Rate

= 12,996 × Rs 50 = Rs 6,49,800

Hence, the required cost of covering the empty space with turfs is Rs 6,49,800.

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Example 7: The circumference of a circular pond is 220 m. A path 3.5 m wide is
Solution: running around the pond. Find the cost of paving the path at Rs 150 per
sq. metre with stones.

Here, the circumference of the circular pond = 220 m

or, 2πr = 220 m

or, 2 × 22 × r = 220 m 3.5m
7
or, r = 35 m

Also, the width of the path (d) = 3.5 m

Now, the area of the path running around the pond = πd (2r + d)

= 22 × 3.5 (2 × 35 + 3.5) = 808.5 m2
7
Again, the cost of paving the path with stones = Area × Rate

= 808.5 × Rs 150 = Rs 1,21,275

Hence, the required cost of paving the path is Rs 1,21,275.

EXERCISE 5.1

General section
1. Answer the following questions.

a) a, b, and c are the length of sides of a triangle. Write the formula to find its area.

b) If the length of each side of an equilateral triangle is p cm, what is its area?

c) Length of the diagonals of a rhombus are x cm and y cm respectively. Write the formula
to find its area.

d) If the length of opposite parallel sides of a trapezium are p cm and q cm and height is
h cm, find its area.

e) Two paths of uniform width 'w' m are running from the middle of every side of a
rectangular field of length 'l' m and breadth 'b' m and crossing each other at right
angle. Write the formula to find the area of these crossing paths.

f) If 'A' be the area of a path 'a' be the surface area of each brick, write the formula to find
the number of bricks required to pave the path.

2. Find the area of the following plane shapes.
a) (i) A triangle of sides 15 cm, 14 cm, and 13 cm respectively.
(ii) A triangle of base 12 cm and height 8 cm
(iii) An equilateral triangle of each side 4 3 cm
(iv) A right angled triangle with perpendicular 15 cm and base 14 cm
b) A rectangle of length 8 cm and a diagonal 10 cm
c) A square of a diagonal 12 cm
d) A parallelogram of base 6.5 cm and height 4 cm
e) A rhombus with diagonals 5.4 cm and 7.5 cm

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f) A quadrilateral with diagonal 8 cm, and heights from opposite vertices to diagonal are
4 cm and 6 cm respectively.

g) A trapezium with opposite parallel sides 6.5 cm and 10 cm and height 7.2 cm.
h) A circle with diameter 14 cm
Creative section - A

3. Calculate the area of the shaded regions.

a) 6 cm b) 50 m c) d)

10 cm 3m 2m 14cm

3m 2m 7cm

54 m

50 m 75 m

e) 14 cm f) g) 14 m h)
45 m
8 cm 25 m

8 cm 10 cm

4. a) The given diagram is a piece of land in the shape of a 24 m 16 m
quadrilateral. Find the area of the land. 85 m

b) The adjoining figure is a park in the shape of a trapezium. 225 m
Calculate the cost of paving the park with stones at Rs 80 per 90m
sq. m.
75 m
c) The figure given alongside is a garden in the shape of a kite
of diagonals 120 m and 60 m respectively. Calculate the cost 60m
of growing grass in the garden at Rs 45 per sq.m.
120m

5. a) The frame of the given window is made of a rectangle with a
semicircular top. Find its perimeter and area.

b) Calculate the total area of the diagram given alongside.

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c) The adjoining circle with centre O has a radius of 14 cm. ABCD
is a square drawn inside the circle. Calculate the area of the
shaded region.

d) The shape alongside is one-quarter of a circle, with
radius of 14 cm. Find
(i) the length of arc AB
(ii) the perimeter of the figure
(iii) the area of the figure (iv) the area of ∆AOB
(v) the area of the shaded segment.

e) Find the perimeter and the area of the adjoining shape.

6. a) A rectangular hall is 12 m long and 10 m broad. Find the length of carpet 2 m wide
required for covering its floor. If the rate of cost of carpet is Rs 110 per metre, find the
cost of carpeting the floor.

b) The length and the breadth of a rectangular room are 10 m and 8 m respectively. How
many pieces of carpet 5 m long and 2 m wide are required to cover its floor? If the cost
of each piece of carpet is Rs 750, find the cost of carpeting the floor.

c) The length of a rectangular room is two times its breadth and the perimeter of its floor
is 45 m. Find the cost of carpeting its floor at Rs 110 per sq.m.

d) A rectangular court is twice as long as its breadth and its perimeter is 540 m. Find the
number of bricks of size 20 cm × 12 cm to pave the court. If the rate of cost of bricks
is Rs 950 per 1000, find the cost of paving the court.

Creative section - B

7. a) A rectangular land of length 64 m and breadth 45 m is surrounded by a 3 m wide path
from outside.

(i) Find the area of the path.

(ii) Calculate the cost of gravelling the path at Rs 130 per sq. metre.

b) A rectangular park is 250 m long and 140 m broad. A path 2 m wide is running around
inside the park.

(i) Calculate the cost of paving the path with stones at Rs 45 per sq. metre.

(ii) Calculate the cost of covering the empty space with turfs at Rs 25 per sq. metre.

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c) A rectangular park is 66 m long and 44 m broad. Two crossing paths at right angle
each of 3 m wide are running across the middle of the park. Calculate:

(i) the cost of paving the bricks of size 20 cm by 15 cm on the paths at Rs 15000 per
1000 bricks.

(ii) The cost of covering the empty space with turfs at Rs 20 per sq. metre.

8. a) The length of the side of a square land is 70 feet. For the purpose of the real estate
business, a path 10 feet wide inside the boundary of the land is made.

(i) Find the area of the path

(ii) How many stones each of 2 feet length and 1.5 feet breadth are required to pave
the path?

(iii) If a stone costs Rs 105, calculate the cost of paving the path.

b) The area of a square pond is 5625 m2 and a 2 m wide path is made around the pond.
(i) Find the area of the path
(ii) Calculate the number of tile each of 40 cm × 20 cm required to pave the path.
(iii) If the cost of a tile is Rs 35, find the cost of paving the path.

(c) A square garden is 113.5 m long. Two crossing paths each of 2 m wide running across
the middle of the garden are at right angle.

(i) Calculate the cost of paving the bricks of size 30 cm by 15 cm each on the paths
at Rs 17 per brick.

(ii) Find the cost of growing grass in the remaining space of the garden at Rs 10 per
square metre.

d) The cost of constructing a path 5 m broad inside the boundary of a square lawn at
Rs 36.25 per sq. metre is Rs 90625. What is the cost of covering the empty space with
turfs at the rate of Rs 20 per sq. metre?

9. a) A circular ground has radius 28 m. A path of uniform width 3.5 m is running outside
the ground. Find the cost of plastering the path at Rs 70 per sq. metre.

b) The circumference of a circular field is 264 m. A path 1.4 m wide runs around it from
inside. Find the cost of gravelling the path at Rs 30 per sq, metre.

c) A racetrack is in the form of a ring whose inner and outer circumferences are 437 m
and 503 m respectively. Find the width of track and its area.

d) The perimeter of a semicircular protractor is 36 cm. Find its diameter and its area.

7cm

10. a) The given window frame is made up of iron plate 7 cm wide. It is 7cm 7cm 2.1m
in the shape of a rectangle with a semi-circular top. Find the cost
of painting the frame at 60 paisa per sq. cm.

b) The diagram shows a running track with a straight of 7cm
100 m and two semicircular ends of diameter 56 m. 1.4m
(i) Find the cost of fencing the track at Rs 175 per meter
(ii) Find the cost of plastering the area covered by the
track at Rs 55 per sq. m.

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Project work
11. a) Measure the length and breadth of the floor of your classroom. Estimate the cost of

carpeting the floor at your local market rate.
b) Measure the length and breadth of your school compound by using measuring tape.

Estimate the cost of paving bricks at the rate of your local market.
c) Measure the length and breadth with frames of whiteboard inside your classroom.

Also, measure the length and breadth of the whiteboard without frames. Then
calculate the area of the plane surface of the frames.

12. a) Draw the following plane shapes. Draw boarder with uniform width around each
shape and shade the boarder. Then, calculate the area of the shaded regions.

(i) Rectangle (ii) Square (iii) Circle

b) Draw a square in a chart paper and cut it out. Fold the square along its one diagonal
and cut it out into two equal pieces. Fold one of the two halves through right angled
corner and make it half. Cut out this half parts also. Now, join these three pieces of the
square to form a rectangle of length equal to the diagonal.

d

d ddd d
2

c) Perform the similar activities with a rhombus and a kite.

5.4 Area of 4 walls, floor and ceiling

If l, b, and h be the length, breadth and the height of a room,

(i) Area of 4 walls of the room = 2lh + 2bh ceiling
l×b
= 2h (l + b)
floor
(ii) Area of floor =l×b h l×b

(iii) Area of ceiling = ×b l

(iv) Area of 4 walls, floor and ceiling = 2h (l + b) + lb + lb b
= 2h (l + b) + 2lb

If l1 and h1 be the length and height of a window, l2 and h2 be the length and height
of a door,

(v) Area of 4 walls excluding a window = 2h (l + b) – l1h1 h1 h2 h
(vi) Area of 4 walls excluding a door = 2h (l + b) – l2h2 l1 l2 b
(vii) Area of 4 walls excluding a window and a door

= 2h (l + b) – l1h1 – l2h2

l

Worked-out examples

Example 1: The length, breadth, and height of a rectangular room are 12 m, 10 m and
6 m respectively.
a) Find the cost of painting its wall and ceiling at Rs 75 per sq. m.
b) Find the cost of carpeting its floor at Rs 120 per sq. m.

81Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

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Solution:
Here, the length of the room (l) = 12 m

The breadth of the room (b) = 10 m

The height of the room (h) = 6 m

The rate of cost of painting (R) = Rs 75 per sq. m.

The rate of cost of carpeting the floor (R') = Rs 120 per sq. m.

a) Now, the area of 4 walls and ceiling (A) = 2h (l + b) + lb

= 2 × 6 (12 + 10) + 12 × 10 = 384 m2

Again, the cost of painting its wall and ceiling = A × R = 384 × Rs 75 = Rs 28,800

Also, the cost of carpeting the floor = Area floor × Rate = (12 × 10) × Rs 120 = Rs 14,400

Hence, the required cost of painting the walls and the ceiling of the room is Rs 28,800 and
the cost of carpeting the floor is Rs 14,400.

Example 2: A rectangular room is 15 m long, 10 m broad and 5 m high. If it contains
Solution: two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m, find
the cost of plastering its walls at Rs 50 per sq. metre.

Here, the length of the room (l) = 15 m

The breadth of the room (b)= 10 m

The height of the room (h) = 5 m

The rate of plastering walls (R) = Rs 50 per sq. m.

Now, the area of its 4 walls = 2h (l + b) = 2 × 5 m (15 m + 10 m) = 250 m2

Also, the area of 2 windows = 2 (2 m × 1.5 m) = 6 m2

The area of a door = 1 m × 4 m = 4 m2

? The area of 4 walls excluding windows and door = (250 – 6 – 4) m2 = 240 m2
Again, the cost of plastering its walls = A × R = 240 × Rs 50 = Rs 12,000

Hence, the required cost of plastering its walls is Rs 12,000.

Example 3: A room is 12 m long and 5.5 m high. If the cost of carpeting its floor at
Solution: Rs 80 per sq .metre is Rs 7,680, find the cost of colouring its walls at
Rs 35.50 per sq, m.

Here, the length of the room (l) = 12 m

the height of the room (h) = 5.5 m carpeting the floor
cost of Rate of cost
Now, area of the floor =

or, l×b = 7680 m2
80

or, 12 × b = 96 m2
96
or, b = 12 m =8 m

Again, area of 4 walls of the room (A) = 2h (l + b) = 2 × 5.5 m (12 m + 8 m) = 220 m2

? The cost of colouring the walls = Area × Rate = 220 × Rs 35.50 = Rs 7810
Hence, the required cost of colouring the walls is Rs 7810.

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Example 4: The cost of carpeting a square room at the rate of Rs 75 per sq. metre is

Rs 10,800. If the cost of plastering its walls at Rs 25 per sq. metre is Rs

Solution: 6000, find the height of the room.

Here, the area of the floor of the square room = cost of carpeting
Rate of cost
10800
or, l2 = 75 m2 = 144 m2

or, l = 12 m and, also b = 12 m

Again, area of 4 walls = cost of plastering the walls
Rate of cost
6000
or, 2h (l + b) = 25 m2

or, 2h (12 + 12) = 240 m2

or, h = 5 m

So, the required height of the room is 5 m.

Example 5: The cost of carpeting the floor of a room, whose breadth is twice the height
and the length is twice its breadth, at the rate of Rs 80 per sq. metre is
Rs 10,240. What is the cost of plastering its walls at Rs 30 per sq. metre?

Solution:
Let, the height of the room (h) = x m
Then, the breadth of the room (b) = 2x m
And, the length of the room (l) = 2b = 2 × 2x m = 4x m

Now, the area of the floor = cost of carpeting the floor
or, Rate of cost
10240
l×b = 80 m2

or, 4x × 2x = 128 m2
or, x2 = 16 m2
or, x = 4 m
? The height of the room (h) = x = 4 m

The breadth of the room (b) = 2x = 2 × 4 m = 8 m

The length of the room (l) = 4x = 4 × 4 m = 16 m

Again, area of 4 walls = 2h (l + b) = 2 × 4 m (16 m + 8 m) = 192 m2t

? The cost of plastering 4 walls = Area × Rate = 192 × Rs 30 = Rs 5760
So, the required cost of plastering its 4 walls is Rs 5760.

EXERCISE 5.2
General section

1. Find the area of floor, ceiling, 4 walls and area of room including floor and ceiling from
the given dimensions of the rooms.
a) l = 10 m, b = 8 m, h = 5 m b) l = 12 m, b = 10 m, h = 6 m

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2. a) The length, breadth, and height of a rectangular hall are 20 m, 15 m and 6.5 m
respectively.

(i) Find the area of its floor. (ii) Find the area of its ceiling.

(iii) Find the area of its 4 walls.

(iv) Find the area of the rooms including floor and ceiling.

b) A square room is 12 m long and 5 m high.
(i) Find the area of its floor and ceiling. (ii) Find the area of its 4 walls.

(iii) Find the area of the room including floor and ceiling.

c) The area of 4 walls with two windows and a door is 220 m2. If each window is 3 m2
and the door is 5 m2, find the area of the walls excluding windows and door.

3. a) The area of 4 walls and ceiling of a room is 242 m2. Find the cost of colouring its walls
and ceiling at Rs 60 per sq. metre.

b) If the cost of plastering 4 walls of a room at Rs 75 per sq. meter is Rs 12,150, find the
area of the 4 walls of the room.

Creative section - A

4. a) A rectangular room is 10 m long, 8 m wide, and 5 m high. Find the cost of colouring
its walls and ceiling at Rs 65 per sq. metre.

b) A square room is 15 feet long and 10 feet high. Find the cost of plastering its walls,
ceiling and floor at Rs 18 per sq. ft.

c) The cost of plastering the walls and the ceiling of a room at Rs 20 per sq. feet is
Rs 14,400. Find the cost of colouring the walls and ceiling at Rs 16 per sq. feet.

5. a) A rectangular room is 8 m long, 6 m broad and 4 m high. It contains two windows of size
2 m × 1.5 m each and a door of size 1 m × 4 m. Find the cost of painting its walls and
ceiling at Rs 54 per sq. metre.

b) A square hall is 15 m long and 5 m high. It contains three square windows each of
2 m long and two doors of size 1.5 m × 4 m. Find the total cost of plastering and
colouring its walls and ceiling at Rs 50 per sq. metre and Rs 45 per sq. metre respectively.

Creative section - B

6. a) A rectangular room is 10 m long and 8 m wide, It has two windows each of 2 m × 2 m
and a door of size 1.5 m × 4 m. If the cost of plastering its walls at Rs 55 per sq. metre
is Rs 9,130, find the height of the room.

b) A square room is 4 m high. The cost of carpeting its floor at Rs 95 per sq. metre is
Rs 6,080. Find the cost of painting its walls and ceiling at Rs 60 per sq. metre.

c) A rectangular room is 10 m long and 5 m high. If the cost of paving marbles at Rs 180
per sq. metre is Rs 10800, find the cost of papering its walls at Rs 45 per sq. metre.

d) The cost of carpeting a square room at Rs 110 per sq. metre is Rs 5,390. If the cost of
plastering its walls at Rs 56 per sq. metre is Rs 7,840, find the height of the room.

e) The cost of plastering the floor of a room, which is 10 m long, at Rs 54 pr sq. metre is
Rs 4,320. If the cost of painting its walls at Rs 48 per sq. metre is Rs 10,368, find the
height of the room.

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7. a) A rectangular room is twice as long as it is broad and its height is 4.5 m. If the cost of
papering its walls at Rs 40 per sq. metre is Rs 6,480, find the cost of paving marbles
on its floor at Rs 150 per sq. metre.

b) A rectangular room is three times as long as it is broad and its height is 4.6 m. If the
cost of carpeting its floor at Rs 85 per sq. metre is Rs 6,375, find the cost of colouring
the walls at Rs 50 per sq. metre.

c) The length of a rectangular hall is two times its breadth and the breadth is two times
its height. If the cost of papering its walls at Rs 36 per sq. metre is Rs 15,552, find the
cost of paving marbles on its floor at Rs 250 per sq. metre.

d) The length of a rectangular room is twice its breadth and thrice its height. If the cost
of carpeting the floor at Rs 112 per sq. metre is Rs 12,600, find the cost of plastering
its walls and ceiling at Rs 60 per sq. metre.

Project work
8. a) Make groups of friends and visit your local market to find the normal rate of cost of

the following things.
(i) Rate of cost of different qualities of carpets.
(ii) Rates of cost of colouring the walls with different qualities of colours.
(iii) Rate of cost of plastering walls, ceiling and floor.
b) Measure the length, breadth, and height of your classroom, windows and doors using
measuring tapes. Then, solve the following problems.
(i) Find the area of the floor and the cost of carpeting the floor at the local rate of

cost.
(ii) Find the area of 4 walls excluding the windows and doors. Then, find the total

area of 4 walls and ceiling.
(iii) Calculate the cost of plastering and colouring the walls and ceiling at the local

rate of cost.

5.5 Area and volume of solids

Solids are three dimensional objects. Therefore, they occupy space. Cube, cuboid,

cylinder, sphere, pyramid, cone, etc. are the examples of solid objects. Length, breadth

and height (or thickness) are three dimensions of cube and cuboid.
E
(i) Area and volume of a cuboid F

A cuboid has six rectangular faces. In the adjoining D Ch
cuboid, ABCD, BGFC, EFGH, CDEF, ADEH and ABGH

are the six rectangular faces of the cuboid. H G
b
Here, the total surface area of the cuboid is the sum A l B

of the areas of its six faces.

A = lb + lb + bh + bh + lh + lh

= 2 (lb + bh + lh)

So, the total surface area of cuboid (A) = 2 (lb + bh + lh)

Similarly, the volume of cuboid (V) = Area of the base × height

=l×b×h

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(ii) Area and volume of a cube

In case of a cube, its length, breadth and height are equal.

i.e., l = b = h

So, the total surface area of a cube = 2 (l.l + l.l + l.l)

= 2 × 3l2 = 6l2

Similarly, volume of a cube = l × l × l = l3

5.6 Prisms and their cross sections

In geometry, a prism is a flat solid which has two opposite congruent and parallel
faces called bases. All prisms have any kind of polygonal bases and all the other faces
are rectangular in shape.

A cross section of a prism is a plane surface which is congruent and parallel to the
bases of the prism. A cross section is obtained by cutting the prism into plane surface
(like a slice) perpendicular to the height or length of the prism.
A prism has uniform and infinite number of cross sections.

A triangular prism A rectangular prism A square prism with
with triangular bases with rectangular bases square bases

The cross section of a The cross section of a The cross section of
triangular prism is a triangle. cuboid is a rectangle. a cube is a square.

The cross section of The cross section of The cross section of this
this prism is -shaped.
this prism is L-shaped. prism is H -shaped.

We have already discussed that a cross section of a prism is congruent to the bases of
the prism. Therefore, area of cross section (or cross sectional area) is, of course, the
area of the base of a prism.

Then, the volume of a prism is the product of its cross sectional area and height.

i.e., Volume of a prism = Area of cross section × height

The lateral surfaces of a prism are the surfaces excluding the bases of the prism. The
lateral surface area of a prism is the product of the perimeter of its cross-section and
the length of the prism.

i.e., Lateral surface area of a prism = perimeter of cross section × length

Also, the total surface area of a prism is the sum of the lateral surface area and two
times the area of its cross section (or two times the area of a base).

i.e., Total surface area of a prism = lateral surface area + 2 × area of cross section

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Example 1: Worked-out examples 5cm 2cm5cm 2cm
Solution: 10 cm 8 cm
Calculate (i) the area of cross section
(ii) lateral surface area (iii) total surface area
(iv) volume of the adjoining prism.

D E 5 cm C
2 cm
(i) Here, the area of cross section = Area of (ABCD – FGCE) F
Alternative process 10 cm G
= 10 × (2 + 2) cm2 – 5 × 2 cm2 2 cm
B
= 30 cm2 A

Area of cross section = Area of trapeziums (ABCD + ADEF)

= 1 ×2 (10 + 5) cm2 + 1 × 5 (4 + 2) cm2 F 5cm E
2 2 D2cm5cm C

= 15 cm2 + 15 cm2 = 30 cm2 2cm

(ii) Lateral surface area = Perimeter of cross section × length A 10 cm B

= (10 + 2 + 5 + 2 + 5 + 4) cm × 8 cm

= 224 cm2

(iii) Total surface area = lateral surface area + 2 × cross sectional area

= 224 cm2 + 2 × 30 cm2

= 284 cm2

(iv) Volume of the prism = Area of cross section × height

= 30 cm2 × 10 cm

Example 2: = 300 cm3
Calculate (i) the area of cross section (ii) lateral
surface area (iii) total surface area (iv) volume
of the adjoining prism.

Solution:

(i) Area of cross section = (9 × 4) cm2 + (3 × 4) cm2 3 cm
= 48 cm2
4 cm 4 cm
(ii) Lateral surface area = Perimeter of cross section × length
3 cm 3 cm 4 cm

= (9 + 4 + 3 + 4 + 3 + 4 + 3 + 4) cm × 8 cm 9 cm

= 34 cm × 8 cm = 272 cm2

(iii) Total surface area = lateral surface area + 2 × cross sectional area

= 272 cm2 + 2 × 48 cm2

= 368 cm2

(iv) Volume of the prism = Area of cross section × height

= 48 cm2 × 8 cm

= 384 cm3

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Example 3: A rectangular metallic block is 30 cm long, 25 cm broad and 8 cm high.

How many pieces of rectangular slices each of 5 mm thick can be cast

lengthwise from the block? Alternative process:
Solution:
Here, length of the block = 30 cm 8cm

Breadth of the block = 25 cm 25cm
Height of the block = 8 cm
30 cm
? Volume of block = 30 cm×25 cm×8 cm
Also, volume of each slice = 0.5 cm×25 cm× 8cm Here,

thickness of each slice = 5 mm

? Number of slices = volume of block = 0.5 cm
volume of each slice Length of the block = 30 cm
30 × 25 × 8 When it is cut lengthwise,
= 0.5 × 25 × 8 number of slices
= 30 cm÷0.5 cm = 60
= 60

Hence, the required number of rectangular slices is 60.

Example 4: A cubical vessel of length 10 cm is completely filled with water. If the

water is poured into a rectangular

vessel 25 cm × 20 cm × 10 cm, 10 cm
10 cm
find the height of the water level

in the rectangular vessel. How 10 cm 10 cm 25 cm 20 cm
much more water is poured into
the vessel to fill it completely?

(1 l = 1000 cm3)

Solution:

Here,volume of water in cubical vessel = 10 cm × 10 cm × 10 cm

Volume of water in rectangular vessel = 25 cm × 20 cm × h cm

Now, 25 × 20 cm × h cm = 10 cm × 10 cm × 10 cm

or, h = 10 cm × 10 cm × 10 cm = 2 cm
25 cm × 20 cm

Again, remaining height of rectangular vessel to be filled = 10 cm - 2 cm = 8 cm
Then, volume of empty space of the vessel = 25 cm × 20 cm × 8 cm = 4000 cm3

Now, 1000 cm3 = 1 l ? 4000 cm3 = 1 × 4000 l = 4 l
1000

Hence, the required height of water level in the rectangular vessel is 2 cm and 4 l of water is
needed to pour into it to fill it completely.

EXERCISE 5.3

General section
1. a) What is a prism? Define with examples.

b) Define a cross section of a prism. What are the cross section of a cube and a cuboid?

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c) Define lateral surface of a prism. What are the lateral surfaces of a triangular prism?
d) If the area of a base of a triangular prism is 24 cm2, what is its cross sectional area?
2. a) The area of cross section of a prism is 36 cm2 and its height is 15 cm. Find its

volume.
b) The perimeter of cross section of a prism of length 10 cm is 48 cm. Find its lateral

surface area.
c) The lateral surface area and the cross sectional area of a prism are 175 cm2 and

40 cm2 respectively. Find total surface area of the prism.
d) The perimeter of cross section and the area of cross section of a prism are 36 cm

and 54 cm2 respectively. If the length of the prism is 15 cm, find its total surface
area and volume.
3. a) The volume of a cube is 125 cm3. Find its total surface area.
b) The total surface area of a cube is 96 cm2. Find its volume.
c) The length and the breadth of a cuboid are 15 cm and 8 cm respectively. If the
volume of the cuboid is 720 cm3, find its height and calculate the total surface
area.
d) The area of the rectangular base of a metallic block is 192 cm2 and its volume is
768 cm3. Find its thickness.
e) A rectangular metallic block is 40 cm long, 24 cm broad and 10 cm high. How
many pieces of rectangular slices each of 8 mm thick can be cast lengthwise from
the block?

Creative section - A

4. Calculate the cross sectional area, the lateral surface area, the total surface area and

volume of each of the following prisms.

a) 8 cm b) 2cm c)
4cm
2cm 4cm
4cm 9 cm
5 cm 6cm

8cm 6 cm 6 cm 3cm
16 cm 10 cm 4cm
5 cm 10cm

d) e) f) 10cm
2cm 10cm10cm 10cm
3cm 3m 3m 10cm

3cm
2cm
5m
2 cm 2 cm 5m 30 cm
3cm 5 cm
6m 30 cm
4m

5. a) A cubical water tank is filled in 1296 seconds at the rate of 1 litre per 6 seconds.

(i) Calculate the internal volume and the length of side of the tank.

(ii) Calculate the total internal surface area of the tank.

89Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

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b) A lidless rectangular water tank made of zinc plates is 2 m long, 1.5 m broad and
1 m high.
(i) How many square metres of zinc plates are used in the tank?
(ii) How many litres of water does it hold when it is full?
(iii) Find the cost of zinc plates at Rs 1200 per sq. m.

c) A rectangular carton is 80 cm × 60 cm × 40 cm.

(i) How many packets of soaps can each of 10 cm × 5 cm × 4 cm be kept
inside the carton?

(ii) By how many centimetres should the height of the carton be increased to keep

1200 packets of soaps?

6. a) A cubical wooden block of length 12 cm is cut into 8 equal cubical pieces. Find the
length of edge of each piece.

b) 8 metallic cubical blocks of equal size are melted and joined together to form a
bigger cubical block. If each smaller block is 10 cm thick, find the thickness of the
bigger block.

c) A rectangular metallic block is 50 cm × 20 cm × 8 cm. If it is melted and re-formed
into a cubical block, find the length of the edge of the cube.

d) Vegetable ghee is stored in a rectangular vessel of internal dimensions
60 cm × 10 cm × 45 cm. It is transferred into the identical cubical vessels. If the
internal length of each cubical vessel is 10 cm, how many vessels are required to
empty the rectangular vessel?

Creative section - B 20 cm
7. a) The adjoining figure is a rectangular glass vessel of
12 cm 30 cm
length 40 cm, breadth 30 cm, and height 20 cm. If it
contains some water upto the height of 12 cm, how 40 cm
many litres of water is to be poured into it to fill the
vessel completely? (1 l = 1000 cm3)

b) In the given figure, a cubical vessel 20 cm 15 cm
of length 20 cm is completely filled

with water. If the water is poured 20 cm

into a rectangular vessel of length 20 cm 32 cm 25 cm
32 cm, breadth 25 cm, and height
15 cm find the height of the water level

in the rectangular vessel. How much more water is required to fill the rectangular

vessel completely? (1l = 1000 cm3)

5.7 Estimation of number of bricks and cost required for building wall

The adjoining diagram is a part of a compound wall. Its shape is

like a rectangular block.

Let l, b, and h be the length, breadth and the height of a wall.

(i) Then, volume of the wall (V) = l × b × h h

The shape of a brick is also like a rectangular block. If l1, b1 and h1 l
be the length, breadth, and the height (thickness) of a brick,
b

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(ii) then, volume of a brick (v) = l1 × b1 × h1 NEPAL h1
b1
Again, if N be number of bricks required for building a wall, l1

(iii) then, number of bricks= volume of the wall
Volume of a brick
V
N = v

Let, C be the total cost of bricks and R be the rate of cost of one brick.

(iv) Then, total cost of bricks = Number of bricks × Rate C = N × R
C C
Also, N = R and R = N

Further more, if the wall contains a window of length l2 and height h2, and a door of
length l3 and height h3,
(v) then, the space occupied by a window = l2 × h2 × thickness of wall (t)
(vi) The space occupied by a door = l3 × h3 × thickness wall (t)
(vi) The volume of wall excluding a window and a door = V – (l2h2t + l3h3t)

Worked-out examples

Example 1: A part of a compound wall is 40 m long, 2 m high and 20 cm wide. How
many bricks each of 20 cm × 10 cm × 5 cm are required to build the wall?
Also find the cost of bricks at Rs 15,000 per 1000 blocks.

Solution:
Here, the length of the wall (l) = 40 m = 40 × 100 cm = 4000 cm

The breadth of the wall (b) = 20 cm
The height of the wall (h) = 2 m = 2 × 100 cm = 200 cm

Now, volume of the wall (V) = l × b × h
= (4000 × 20 × 200) cm3 = 16000000 cm3

Also, volume of each brick (v) = (20 × 10 × 5) cm3 = 1000 cm3

? Required number of bricks (N) = V = 16000000 = 16000
v 1000

Again, the rate of cost of each brick (R) = Rs 15000 = Rs 15
1000

? The required cost of bricks (C) = N × R = 16000 × Rs 15 = Rs 2,40,000

Hence, the required number of bricks is 16000 and the cost is Rs 2,40,000.

Example 2: A wall of a hall is 36 m long, 6 m high, and 30 cm wide. It contains three
windows each of 2 m × 1.5 m and two doors of size 1.5 m × 4 m.
(i) Find the number of concrete blocks each of 30 cm × 15 cm × 10 cm

required to construct the wall.
(ii) Find the cost of blocks at the rate of Rs 35,000 per 1000 blocks.

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Mensuration

Solution:

Here, the length of the wall (l) = 36 m

The width of the wall (b) = 30 cm = 0.3 m

The height of the wall (h) = 6 m

(i) Now, volume of the wall (V) = l × b × h

= 36 m × 0.3 m × 6 m = 64.8 m3

Also, volume of three windows = 3 (2 m × 0.3 m × 1.5 m) = 2.7 m3

And, volume of two doors = 2 × 1.5 m × 0.3 × 4 m = 3.6 m3

?Volume of the wall excluding windows and doors = 64.8 m3 – (2.7 + 3.6) m3

= 58.5 m3

= 58.5 × 100 × 100 × 100 cm3

Again, volume of each blocks (v) = 30 cm × 15 cm × 10 cm = 4500 cm3

?The required number of blocks (N) = V = 58.5 × 100 × 100 ×100 = 13,000
v 4500
(ii) Also, the cost of 1000 blocks = Rs 35,000

The cost of 1 block = Rs 35000 = Rs 35
The cost of 13,000 blocks 1000

= 13000 × Rs 35 = Rs 4,55,000

Hence, the required number of blocks is 13,000 and the cost is Rs 4,55,000.

Example 3: A square room contains 180 m3 of air. The cost of plastering its four walls
Solution: at Rs 60 per sq. metre is Rs 7,200. Find the height of the room.

Let, the length (l) = breadth (b) = x m

Here, volume of the room = Volume of the air

or, l × b × h = 180 m3

or, x × x × h = 180 m3

or, h = 180 m ..................... (i)
x2
Cost of plastering of 4 walls
Again, area of 4 walls = Rate of cost

or, 2h (l + b) = 7200 m2
60

or, 2h (x + x) = 120 m2

or, h = 30 m ..................... (ii)
x

From equation (i) and (ii) we get,

180 = 30
x2 x

or, x = 6 m

Now, putting the value of x in equation (ii), we get,

h = 30 = 5 m
6

Hence, the required height of the room is 5 m.

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EXERCISE 5.4

General section
1. a) A part of a compound wall is 30 m long, 4 m high and 20 cm wide. It has two

windows each of 2 m × 1.5 m. Find the volume of the wall excluding window.

b) The length, width and height of a wall of a courtyard are 50 m, 30 cm and 6 m
respectively. It contains four windows of size 2.5 m × 1.6 m each and two doors of
size 2 m × 3 m each. Find the volume of wall excluding the windows and doors.

2. a) The volume of a wall of a room is 120 m3. Find the number of bricks of size
20 cm × 12 cm × 10 cm required to build the wall.

b) How many concrete blocks of size 25 cm × 15 cm × 10 cm are required to construct
a wall of dimensions 60 m × 6.5 m × 20 cm?

Creative section
3. a) A part of a compound wall is 36 m long, 4 m high and 30 cm wide. It contains two

windows each of 2 m × 1.5 m and a door of size 2 m × 3 m.

(i) Find the number of bricks each of 18 cm × 10 cm × 8 cm required for the
construction of the wall.

(ii) Find the cost of bricks at Rs 17,000 per 1000 bricks.
b) The dimensions of a part of a compound wall is 30 m × 5 m × 20 cm. It contains

three windows each of 2 m × 1.5 m. How many bricks of size 20 cm × 10 cm × 5 cm
are required to build the wall leaving 10 % of the space for the cement work? Also
find the cost of bricks at the rate of Rs 18,000 per 1000 bricks.

4. a) One of the four walls of a hall is 20 m long and 20 cm thick. If 4,700 bricks each
of size 25 cm × 16 cm × 10 cm are required to construct the wall containing two
windows of size 2 m × 1.5 m, find the height of the wall.

b) A wall is 4 m high and 40 cm wide. It contains two windows of dimensions
2.5 m × 1 m and a gate way of dimensions 2 m × 3 m. If 2450 cubical slabs each of
length 20 cm are required to construct the wall, find the length of the wall.

5. a) A room is 12 m long and 8 m broad and it contains 480 cu. metre of air. Find the
cost of colouring its walls at Rs 54 per sq metre.

b) A room contains 600 cu. m. of air. If the cost of carpeting its floor at Rs 90 per
sq. metre is Rs 13,500, find the height of the room.

c) A square room contains 288 m3 of air. The cost of carpeting the room at Rs 105 per
sq. metre is Rs 6,720. Find the cost of painting its walls at Rs 45 per sq. metre.

d) A square room contains 256 cu. m. of air. If the cost of painting its 4 walls at Rs 50
per sq. metre is Rs 6,400, find the cost of carpeting its floor at Rs 99 per sq. metre.

e) A square room contains 220.5 cu. m. of air. If the cost of colouring its 4 walls at
Rs 55 per sq. metre is Rs 6,930, find the height of the room.

6. a) A room is two times longer than its breadth and it contains 360 m3 of air. If the cost
of plastering its floor at Rs 40 per sq. metre is Rs 2,880, find the cost of plastering
its walls at Rs 45 per sq. metre.

b) The length of a room is two times its breadth and it contains 210 m3 of air. If the
cost of painting its walls at Rs 54 per sq.m. is Rs 6,804, find the height of the room.

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Mensuration

Project work
7. a) Measure the length, breadth, height, and thickness of walls of your classroom. Also

measure the length and height of windows and doors of the classroom.
(i) Find the volume of wall excluding windows and doors.
(ii) Find the average volume of a brick available in your surroundings.
(iii) Find the number of bricks required to construct the wall excluding windows

and doors.
(iv) Estimate and cost of bricks required to construct the walls as per the local rate

of cost.
b) Measure all the dimensions of a part of the compound wall of your school or your

house. Calculate the volume of the part of the wall. Find the volume of a brick and
calculate the number of such bricks required to construct the wall. Also, calculate
the cost of bricks as per the local rate of cost.

Objective Questions

1. The area of a rectangle having length 9 cm and breadth 4 cm is equal to that of a
square, what is the perimeter of the square?

(A) 6cm (B) 12 cm (C) 24cm (D) 26cm

2. Four squares of different length of sides are 2cm 3cm 4cm 5cm
depicted in the given figure. Which of the (i) (ii) (iii) (iv)
following relation is true with respect to their
areas?

(A) (i) + (ii) = (iii) (B) (ii) + (iii) = (iv)

(C) (i) + (iii) = (iv) (D) (i) + (ii) + (iii) = (iv)

3. The area of path of width d units runs around outside the rectangular pond of length l
units and breadth b units then the area of path is:

(A) 2d (l + b – 2d) (B) 2d (l + b + 2d) (C) d (l + b – d) (D) d (l + b +d)

4. The area of path of width d units runs around inside the square garden of length l units.
Then the area of path is

(A) 2d (l – d) (B) 2d (l + d) (C) 4d (l – d) (D) 4d (l +d)

5. A wire is bent to form a square of side 22 cm. If the wire is re-bent to form a circle, its
radius is:
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm

6. Perimeter of a semi-circle with radius r is:

(A) πr (B) ) 2πr (C) ) r(π + 2) (D) r(π - 2)

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7. The area of a circular park is 7546 m2. A path 2.1 m wide is running inside around the
park. Then, the cost of gravelling the path at Rs 50 per sq. m is

(A) Rs 33033 (B) Rs 377300 (C) Rs 31647 (D) Rs 345653

8. A room is 8m long, 5 m broad and 4 m high. How many pieces of carpets each of 2.5 m
long and 2 m wide are required to carpeting the floor of a room?

(A) 4 (B) 5 (C) 8 (D) 16

9. The guest room of a house is 12 m long and 5.5 m high. If the total cost of plastering its
four walls at Rs 45 is Rs 9900, how broad is the room?
(A) 6 m (B) 8 m (C) 9 m (D) 10m

10. The cost of carpeting a square room at the rate of Rs 75 per square metre is Rs 10800.
If the cost of plastering its walls at Rs 25 per square metre is Rs 6000, the height of the
room is

(A) 4 m (B) 4.5 m (C) 5 m (D) 5.5 m

11. Three metallic cubes of edges 3cm, 4cm, and 5 cm are melted, and recast a single cube.
What is the edge of the new cube?

(A) 4.5 cm (B) 5.5 cm (C) 6.5 cm (D) 6 cm

12. The total cost of carpeting a bedroom of 6m long and 4.5m broad at the rate of Rs 470 is
Rs 6345, how long is the carpet?

(A) 9.6 m (B) 10.4 m (C) 13.5 m (D) 16.2 m

13. Three cubical boxes each of side 10 cm are joined end to end. What is the total surface
area of the resultant figure?

(A) 700 cm2 (B) 1400 cm2 (C) 1500 cm2 (D) 3000 cm2

14. How many brick each of 15cm × 10cm × 5 cm are required to build a wall of 20 m
long, 3 m high, and 20 cm wide?

(A) 16000 (B) 18000 (C) 20000 (D) 24000

15. 960 identical packets of soaps, each of 10 cm long and 5 cm wide, can be kept inside a
rectangular carton of dimensions 80 cm × 60 cm × 40 cm, the height of each packet of
soap is:

(A) 3 cm (B) 4 cm (C) 5 cm (D) 6 cm

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Unit Algebraic Expressions

6

6.1 Factors and Factorisation - Review x2 x

In the given figure, x unit is the length of a square.

So, the area of the square = x2 sq. unit. x
Let, the length of the square is increased by 2 units. x+2
Then, the length of the rectangle = (x + 2) units.

Now, the area of the rectangle = x × (x + 2) sq. units x2 2x x
= (x2 + 2x) sq. units

Here, x2 + 2x is the product of x and (x + 2).
Therefore, x and (x + 2) are the factors of the expression x2 + 2x. x 2

Thus, factorisation is the process of expressing a polynomial as the product of two or

more polynomials.

When we factorise an expression, we write it as the product of its factors. The process

of getting the factors becomes easier if we apply the selected method of factorisation

for the particular type of expression. So, it is important and very much useful to know

about the types of expressions which are to be factorised.

(i) Expression having a common factor in each of its term

Let's take an expression ax + ay. Here, both terms contain a common term, a.
In such an expression, the factor which is present in all terms of the expression
is taken out as common and each term of the expression should be divided by the
common factor to get another factor.

Worked-out examples

Example 1: Factorise (i) 4ab + 6ac (ii) 2x2y – 4xy2 + 6xy
Solution:
(iii) 2a (x – y) + 7b (y – x) (iv) 3p2q (a – b) – 6pq2 (b – a)

(i) 4ab + 6ac = 2a (2b + 3c)

(ii) 2x2y – 4xy2 + 6xy = 2xy (x – 2y + 3)

(iii) 2a (x – y) + 7b (y – x) = 2a (x – y) – 7b (x – y)

= (x – y) (2a – 7b)

(iii) 3p2q (a – b) – 6pq2 (b – a) = 3p2q(a – b) + 6pq2(a – b)

= 3pq (a – b) (p + 2q)

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Algebraic Expressions

(ii) Expression having common factors in the groups of terms

Let's take an expression ax + by – ay – bx. It can be regrouped as ax – ay – bx + by
(or ax – bx – ay + by). In ax – ay – bx + by, the group ax – ay has the common factor
a and the group – bx + by has the common factor b. Then,

ax – ay – bx + by = a(x – y) –b(x – y) = (x – y) (a – b)

Similarly, ax – bx – ay + by = x(a – b) –y (a – b) = (a– b) (x – y)

In this way, in such expressions, the terms are arranged in suitable groups such that each
group has a common factor.

Example 2: Factorise (i) x2 – ax + ab – bx (ii) ac (b2 + 1) + b (a2 + c2)

Solution:

(i) x2 – ax + ab – bx = x2 – ax – bx + ab = x (x – a) – b (x – a) = (x – a) (x – b)
(ii) ac (b2 + 1) + b (a2 + c2) = ab2c + ac + a2b + bc2

= ab2c + a2b + bc2 + ac

= ab (bc + a) + c (bc + a) = (bc + a) (ab + c)

Example 3: Simplify x2 2x2 – 6x – 6
Solution: + 2x – 3x

2x2 – 6x = 2x (x – 3) = (x 2x (x – 3) = x 2x
x2 + 2x – 3x – 6 x (x + 2) – 3 (x + 2) + 2) (x – 3) +2

EXERCISE 6.1

General section

1. In each of the following figures, write the polynomial as the product of its factors.

a) b) 2 c) 1 d) 3

xx xa

x1 x x2 a3

Creative section

2. Factorise: b) 4p2 – 6p c) 6a2b + 9ab2
a) 2ax + 4bx e) 6x3y2 + 9x2y3 – 3x2y2 f) 2x (a + b) + 3y (a + b)
d) 2px2 – 4px + 6p2x

g) 3a (x – y) – (x – y) h) x (x + 2) + 3x + 6 i) 2t (t – 1) – t + 1

3. Resolve into factors:

a) ax + by + ay + bx b) pm – qn + pn – qm c) a2 + ab + ca + bc
f) x2 + 4x + 3x + 12
d) mx2 + my2 – nx2 – ny2 e) xy – 2y + 3x – 6 i) a2 – a (b + c) + bc

g) p2 – 8p – p + 8 h) 16x2 – 4x – 4x + 1

j) x2 – (y – 3) x – 3y k) pq (r2 + 1) – r (p2 + q2) l) y (x + z) + z (x + y) + y2 + z2

4. Simplify:

a) a2 + a b) 3x2 – 6xy c) 3p2 6p2 – 2p – 1
2ab + 2b 2xy – 4y2 – p + 3p

d) 2x2 – xy + 2xy – y2 e) x2 + 2x + x + 2 f) x2 – 4x – 3x + 12
6xy – 3y2 x2 – 3x + 2x – 6 x2 – 4x + 2x – 8

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Algebraic Expressions

(iii) Expression having the difference of two squared terms
(Expression of the form a2 – b2)

Let's take a square sheet of paper of length a units. From a corner of the sheet, another

square of length b units is cut out.

a aa b a b a
D C
a–b a–b
ab a bb
a a2 b b2 A B

a+b
Area = (a + b) (a – b)
Area = a2 a–b
Area of shaded region is a2 – b2

Here, area of the rectangle ABCD is a2 – b2, which is the product of length (a + b)
and breadth (a – b)

? length × breadth = area of rectangle

i.e., (a + b) (a – b) = a2 – b2

Here, the expression a2 – b2 is the difference of two squared terms a2 and b2 and it is
the product of (a + b) and (a – b). So, (a + b) and (a – b) are the factors of a2 – b2.

Thus, to factorise an expression of the form a2 – b2, we should use the formula,

a2 – b2 = (a + b) (a – b)

Worked-out examples

Example 1: Factorise (i) 8x3y – 18xy3 (ii) 81ax5 – 16ax
Solution:

(i) 8x3y – 18xy3 = 2xy (4x2 – 9y2) = 2xy[(2x)2 – (3y)2]
= 2xy (2x + 3y) (2x – 3y)

(ii) 81ax5 – 16ax = ax (81x4 – 16)
= ax [(9x2)2 – (4)2]
= ax (9x2 + 4) (9x2 – 4)
= ax (9x2 + 4) [(3x2) – 22]
= ax (9x2 + 4) (3x + 2) (3x – 2)

Example 2: Resolve into factors. a) 1 – 9 (a – b)2 b) a2 – b2 + 2b – 1
Solution:

a) 1 – 9 (a – b)2 = 12 – [3 (a – b)]2
= [1 + 3 (a – b)] [1 – 3 (a – b)]
= (1 + 3a – 3b) (1 – 3a + 3b)

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Algebraic Expressions

b) a2 – b2 + 2b – 1 = a2 – (b2 – 2b + 1)
= a2 – (b – 1)2 Using (a – b)2 = a2 – 2ab + b2
= (a + b – 1) [a – (b – 1)]
= (a + b –1) (a – b + 1)

Example 3: Factorise x2 – 6x – 40 + 14b – b2
Solution:

x2 – 6x – 40 + 14b – b2 = (x2 – 6x) – 40 + 14b – b2
= (x2 – 2.x.3 + 32 – 32) – 40 + 14b – b2
= (x – 3)2 – 9 – 40 + 14b – b2
= (x – 3)2 – (49 – 14b + b2)
= (x – 3)2 – (72 – 2.7.b + b2)
= (x – 3)2 – (7 – b)2 Using a2 – 2ab + b2 = (a – b)2
= (x – 3 + 7 – b) (x – 3 – 7 + b)
= (x – b + 4) (x + b – 10)

Example 4: Factorise (w2 – x2) (y2 – z2) – 4wxyz

Solution:

(w2 – x2) (y2 – z2) – 4wxyz = w2y2 – w2z2 – x2y2 + x2z2 – 4wxyz

= (wy)2 – 2.wxyz + (xz)2 – (wz)2 – 2wxyz – (xy)2

= (wy)2 – 2.wy.xz + (xz)2 – [(wz)2 + 2.wz.xy + (xy)2]

= (wy – xz)2 – (wz + xy)2

= (wy – xz + wz + xy) (wy – xz – wz – xy)

= (wy + wz + xy – xz) (wy – wz – xy – xz)

(iv) Expression of the form a4 + a2b2 + b4

The expressions of the form a4 + a2b2 + b4 are also factorised by using the similar

method of factorisation of the expression of a2 – b2 form. Following formulae are

useful while factorising these types of expressions.

a2 + 2ab + b2 = (a + b)2 a2 + b2 = (a + b)2 – 2ab

a2 – 2ab + b2 = (a – b)2 a2 + b2 = (a – b)2 + 2ab

a2 – b2 = (a + b) (a – b)

Example 5: Factorise a) a4 + a2b2 + b4 b) x4 – 3x2y2 + y4

Solution:

a) a4 + a2b2 + b4 = (a2)2 + (b2)2 + a2b2

= (a2 + b2)2 – 2a2b2 + a2b2 Using a2 + b2 = (a + b)2 – 2ab

= (a2 + b2)2 – (ab)2

= (a2 + b2 + ab) (a2 + b2 – ab) Using a2 – b2 = (a + b) (a – b)

= (a2 + ab +b2) (a2 – ab + b2)

99Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Vedanta Excel in Mathematics - Book 9

Algebraic Expressions

b) x4 – 3x2y2 + y4 = (x2)2 + (y2)2 – 3x2y2
= (x2 – y2)2 + 2x2y2 – 3x2y2
= (x2 – y2)2 – (xy)2 Using a2 + b2 = (a – b)2 + 2ab
= (x2 – y2 + xy) (x2 – y2 – xy)
= (x2 + xy – y2) (x2 – xy – y2)

Example 6: Resolve into factors a) 9x4 + 14x2 + 25 b) x4 + x2 +1 c) a4 + 4b4
Solution: y4 y2

a) 9x4 + 14x2 + 25 = (3x2)2 + (5)2 + 14x2

= (3x2 + 5)2 – 2.3x2.5 + 14x2

= (3x2 + 5)2 – 16x2

= (3x2 + 5)2 – (4x)2

= (3x2 + 4x + 5) (3x2 – 4x + 5)

b) x4 + x2 +1 = x2 2 + (1)2 + x2 c) a4 + 4b4 = (a2)2 + (2b2)2
y4 y2 y2 y2

= x2 + 1 2 – 2. x2 .1 + x2 = (a2 + 2b2)2 – 2a2.2b2
y2 y2 y2 = (a2 + 2b2)2 – (2ab)2
=(a2 + 2ab + 2b2) (a2 – 2ab + 2b2)
= x2 + 1 2 x2
y2 y
–

= x2 + x +1 x2 – x +1
y2 y y2 y

EXERCISE 6.2
General section

1. Factorise:

a) 9x2 – 4 b) 25a2b2 – 1 c) 48ax2 – 75ay2 d) x4 – y4

e) 16x5y – 81xy5 f) 625a4 – 256b4 g) 4 – (m – n)2 h) 1 – (a – b)2

i) 16 – 25(p – q)2 j) (2a – b)2 – (a – 2b)2 k) a2 + 2ab + b2 – c2 l) p2 – q2 – r2 – 2qr

m) a2 – b2 + 4b – 4 n) 16a4 – 4a2 – 4a – 1 o) ax2 – ay2 – x – y p) a2 – (a – b)x – b2

2. Resolve into factors:

a) x2 + x2y2 + y4 b) a4 + a2 + 1 c) 9x4 + 2x2y2 + y4

d) m4 + 4m2n2 + 16n4 e) 4x4 – 8x2y2 + 49y4 f) 9p4 – 34p2q2 + 25q4

g) 36y4 – 25y2 + 4 h) a4 + 4 i) 64x4 + 1

j) 100a4 – 45a2 + 81 k) 169b4 – 35b2x2 + 961x4 l) a4 + a2 + 1
b4 b2

m) p4 + q4 + 1 n) x4 – 7x2 +1 o) b4 – 15db22 + 9
q4 p4 y4 y2 d4

Vedanta Excel in Mathematics - Book 9 100 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur