Taxation
Creative section - B
15. a) After allowing 15% discount on the marked price of a camera, 13% VAT was levied
and sold it. If the selling price of the camera with VAT is Rs 4,420 more than its price
after discount, find the marked price of the camera.
[Hint: SP. with VAT – S.P. = Rs 4,420]
b) A projector was sold after allowing 10% discount on the marked price and levying
13% VAT. If the selling price of the projector after discount is Rs 5,850, less than its
selling price with VAT, find the marked price of the projector.
16. a) The marked price of a digital watch is Rs 6,000. After allowing 10% discount and
including same percentage of value added tax, the watch is sold. By how much
percent is the VAT amount less than discount amount?
b) The marked price of a guitar is Rs 5,500. After allowing 10% discount and levying
same percentage of VAT, the guitar is sold. By how much percent is the VAT amount
less than discount amount?
17. a) A wholesaler sold a photocopy machine for Rs 48,000 to a retailer. The retailer spent
Rs 2,000 for transportation and Rs 1,500 for the local tax. If the retailer sold it at a
profit of Rs 4,500 to a customer, how much did the customer pay for it with 13% VAT?
b) The Buddha supplier sold a digital T-shirt printer for Rs 3,00,000 to Everest supplier.
The Everest supplier spent Rs 5,500 for transportation and Rs 2,500 for the local tax
and sold at a profit of 10% to a customer. How much did the customer pay for the
printers with 13% VAT?
c) A wholesaler purchased a washing machine for Rs 60,000 and sold it to a retailer at
10% profit. The retailer spent Rs 2,400 for transportation and Rs 1,600 for local tax.
Then she sold it to a customer at 12% profit. How much did the customer pay for it
with 13% VAT?
18. a) A retailer allowed 4% discount on his goods to make 20% profit and sold a refrigerator
for Rs 10,848 with 13% VAT. By how much is the discount to be increased so that he
can gain only 15%?
b) A supplier sold a scanner machine for Rs 41,400 with 15% VAT after allowing 10%
discount on it's marked price and gained 20%. By how much is the discount percent
to be reduced to increase the profit by 4%?
c) A retailer hired a room in a shopping mall at Rs 45,000 rent per month and started
a business of garments. He spent Rs 20,00,000 to purchase different garment items
in the first phase and marked the price of each item 30% above the cost price.
Then, he allowed 10% discount on each item and sold to customers. His monthly
miscellaneous expenditure was Rs 15,000 and the items of worth 10% of the
investment remained as stocks after two months. Find his net profit or loss percent.
Project work and activity section
19. a) Make groups of your friends. Collect different types of goods purchasing bill. Study
about the marked price, rate of discount, rate of VAT or other rates of taxes mentioned
in the bills. Prepare the reports and present in your class.
b) Visit the nearby department store or your local shops. Search and collect the marked
price, rate of discount and rate of VAT on different daily using goods. Prepare a report
and compare your report with that of your friends.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 47 Vedanta Excel in Mathematics - Book 9
Taxation
c) Let's become a problem maker and problem solver yourself. Write the values of the
variable of your own and find the unknown variables.
(i) (ii) (iii)
Given: Given: Given:
M. P. = .................. M. P. = .................. M. P. = ..................
Discount= ............. % Discount= ............. % S.P. = ..................
Discount amount = ? Discount amount = ? Discount amount = ?
S.P. = ? S.P. = ? Discount Percent = ?
(iv) (v) (vi)
Given:
Given: Given:
S. P. = ..................
VAT = .................. S. P. = .................. M. P. = ..................
S.P. with VAT = ? VAT = .................. % Discount = .............%
S.P. with VAT = ? VAT = ..................%
S.P. with VAT = ?
d) Using the above variables make a word problem of each of (i) to (vi) of your own,
related to the real life situations. Then, solve your problems and get the unknown
variables.
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. Which of the following is the taxable income?
(A) Provident fund (B) Festival expense (C) Allowance (D) Insurance premium
2. The rate of social security tax is
(A) 1% (B) 10% (C) 20% (D) 30%
3. Which of the following is the tax-exempt income?
(A) Salary (B) Festival expense (C) Donation (D) Interest
4. The annual income of an individual is Rs 4,50,000. How much income tax does s/he
pay in a year?
(A) Rs 4,000 (B) Rs 4,500 (C) Rs 9,000 (D) Rs 9,500
5. A couple is the sole proprietor of a firm, income tax is exempted up to
(A) Rs 4,00,000 (B) Rs 4,50,000 (C) Rs 5,00,000 (D) Rs 5,50,000
6. An indirect tax which is charged at the time of consumption of goods and services is
(A) VAT (B) Income tax (C) Customs duty (D) Excise duty
7. The current rate of Value Added Tax (VAT) in Nepal is
(A) 10% (B) 13% (C) 15% (D) 7%
8. Which of the following formulae is correct?
(A) S.P. with VAT = S.P. + VAT% of S.P. (B) S.P. with VAT = S.P. – VAT% of S.P.
(C) S.P. with VAT = C.P. + VAT% of C.P. (D) S.P. with VAT = M.P. – VAT% of M.P.
9. The Value Added Tax (VAT) is imposed on
(A) M.P. (B) S.P. (C) C.P. (D) Discount
10. If M.P. = Rs x, discount = Rs y and VAT = Rs z, then S.P. with VAT is
(A) Rs (x – y + z) (B) Rs (x + y – z) (C) Rs (x + y + z) (D) Rs (x – y – z)
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Unit 3 Commission, Bonus and Dividend
3.1 Commission – Introduction
Let's study the following examples and make the sense of commission.
a) Mr. Lama had a piece of land and he wanted to sell it. He announced to provide
2% of selling price to the agent for selling it. Mr. Sharma, a salesperson, came to
know about it and searched for the buyers. He sold the land for Rs 50,00,000 and
got the amount of 2% of Rs 50,00,000 i.e., Rs 1,00,000 for his service.
b) Mrs. Ghale wished to buy a house. She went through an agent. The agent found
a house on sale. Mrs. Ghale liked it and bought for Rs 1,20,00,000. She gave 1.5%
of the price of the house to the agent.
c) Mr. Chaudhary works in a hardware shop. His income of the month, Push is given
in the table. Answer the following questions.
(i) What is his monthly salary? Month: Push
(ii) What is the total sale of the Salary Rs 25,000
month? Total sales Rs 8,50,000
Commission rate 1.5%
(iii) What is the commission rate? Commission amount Rs 12,750
(iv) What amount of money did he get Total income Rs 37,750
as commission?
(v) What is his total income in Push?
Commission is the amount of money paid to an agent (or an employee) for
performing a business services such as buying and selling goods, property, etc.
The commission is usually a percent of the selling price. The percent is called the
commission rate.
For example,
A Publication House pays 5% commission of the total sales of books to its dealer
in a year.
A real estate company pays 2.5% commission to its agent for selling lands and
property.
In this way, we calculate the amount of commission as the given percent of the
selling price.
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Commission, Bonus and Dividend
Facts to remember
1. Amount of commission = percent of commission × total selling price.
2. Commission percent = Commission amount × 100%
S.P.
Worked-out examples
Example 1: A real estate company gives 1.5 % commission to its agent on selling a
Solution: piece of land for Rs 20,00,000 and 2 % commission for additional amount
of selling price above the fixed price. If the agent sold the land for
Rs 28,50,000, how much commission did the agent receive?
Here, the fixed selling price of the land = Rs 20,00,000
The selling price of the land = Rs 28,50,000
Now, the commission received by the agent
= 1.5 % of Rs 20,00,000 + 2 % of (Rs 28,50,000 – Rs 20,00,000)
= 1.5 × Rs 20,00,000 + 2 × Rs 8,50,000
100 100
= Rs 30,000 + Rs 17,000 = Rs 47,000
Hence, the agent received the commission of Rs 47,000.
Example 2: A plywood factory provides commission to its dealers on the basis of the
following monthly transactions.
Sales up to Rs 5,00,000, 4 % commission.
Sales from Rs 5,00,000 to Rs 7,50,000, 5 % commission.
Sales more than Rs 7,50,000, 6 % commission.
Calculate the commissions of the following monthly transaction of
different dealers.
Solution: (i) Rs 4,80,000 (ii) Rs 7,20,000 (iii) Rs 8,60,000
(i) Commission of the sale of Rs 4,80,000 = 4 % of Rs 4,80,000 = Rs 19,200
(ii) Commission of the sale of Rs 7,20,000
= 4 % of Rs 5,00,000 + 5 % of (Rs 7,20,000 – Rs 5,00,000)
= Rs 20,000 + Rs 11,000 = Rs 31,000
(iii) Commission of the sale of Rs 8,60,000
= 4% of Rs 5,00,000 + 5% of (Rs 7,50,000 – Rs 5,00,000) + 6% of (Rs 8,60,000 – Rs 7,50,000)
= Rs 20,000 + Rs 12,500 + Rs 6,600 = Rs 39,100
Example 3: The monthly salary of an employee in a publication house is Rs 30,000
and 2 % commission is provided when the monthly sales is more than
Rs 5,00,000. If the sales of the publication house in a month is Rs 6,75,000,
find the income of the employee in the month.
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Commission, Bonus and Dividend
Solution:
Here, the monthly salary of the employee = Rs 30,000
Commission percent = 2%
Sales of the month = Rs 6,75,000
Now, sales for the commission = Rs 6,75,000 – Rs 5,00,000 = Rs 1,75,000
Amount of commission = 2 % of Rs 1,75,000
= 2 × Rs 1,75,000 = Rs 3,500
100
Again, the income of the employee of the month = Rs 30,000 + Rs 3,500 = Rs 33,500
Hence, the income of the employee in the month is Rs 33,500.
Example 4: The monthly salary of an employee in a hardware shop is
Rs 27,000 and a certain commission is given as per the monthly sales. If
the sales of a month is Rs 5,00,000 and the total income of the employee
of the month including commission is Rs 32,500, calculate the rate of
commission.
Solution:
Here, the monthly salary of the employee = Rs 27,000
Sales of the month = Rs 5,00,000
Total income of the employee of the month = Rs 32,500
Now, the amount of commission of the month = Rs 32,500 – Rs 27,000 = Rs 5,500
Again, the rate of commission = Amount of commission × 100%
Sales of the month
= 5,500 × 100% = 1.1%
Hence, the required rate of commission is 1.1%. 5,00,000
EXERCISE 3.1
General section
1. Find the commission amount received by agent from the table given below.
S.N. Property Total sale Rate of commission Commission
a) Land Rs 60,50,000 1% ……….
b) House Rs 2,40,00,000 1.5% ……….
c) Car Rs 60,00,000 2% ……….
2. a) An agent gets 1.5% commission while selling a house for Rs 2 crores 50 lakhs.
(i) How much commission did the agent get?
(ii) How much amount did the house owner get after paying the commission?
b) A real estate company gives 3% commission to it’s agents. An agent sold a piece of
land for Rs 12,50,000.
(i) How much commission did the agent get?
(ii) How much amount did the company receive after paying the commission?
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Commission, Bonus and Dividend
3. a) An automobile company gives 10% commission to it’s agents for selling second-hand
motorbike. If an agent received Rs 16,290 by selling a scooter, at what price did the
agent sell the scooter?
b) A farmer sold his seasonal vegetables through an agent by giving 2% commission. If
the agent got Rs 2,200 commission, at what price did the agent sell the vegetables?
c) Mr. Kulman sold a piece of land through a real estate agent. After paying commission
at 3.5% to the real estate agent, he received Rs 1,78,52,500 for selling his land.
Calculate the selling price of the land.
d) A noodles factory gives 10% commission to a wholesaler for selling its products.
After giving the commission to the wholesaler, the company receives Rs 3,15,000 in
Baishakh. Find the total sales of noodles by the wholesaler in Baishakh.
4. a) If a sales agent received Rs 30 thousands commission when he sold a second hand
taxi for Rs 12 lakhs, what rate of commission did he get?
b) A building owner fixed the cost of his building as Rs 27,50,400 and the price above
the fixed cost goes to an agent as his commission. If the agent sold it for Rs 29,01,672,
find his commission percent.
Creative section
5. a) A real estate company gives 5 % commission on selling a piece of land for
Rs 10,00,000 and 7 % commission for the additional amount of selling price above
the fixed price. If the agent sold the land for Rs 12,99,000, how much commission did
he/she receive from the company?
b) An insurance company offered 2 % commission for the first 10 lakh and 2.5 % for
the rest sum of money collected from new clients by its agents. If an agent is able to
collect a sum of Rs 15,30,000 from his new clients, find his total commission.
c) A noodle factory provides commission to its sales agents on the basis of the following
weekly transactions.
Sales Rate of commissions
Upto Rs 50,000 2.5 %
From 50,000 - Rs 1,00,000 5%
Above Rs 1,00,000 6%
Now, calculate the commission of the following weekly sales.
(i) Rs 48,600 (ii) Rs 72,500 (iii) Rs 1,10,700
6. a) The monthly salary of a salesperson of a subway restaurant is Rs 21,600, and an
additional incentive of 1.5 % on the total monthly sale is provided as commission.
(i) Calculate his/her total income in a month if he/she makes a total sale of
Rs 5,80,000 in that month.
(ii) What should be his/her total sale in the next month so that he/she can receive a
total income of Rs 31,350 in the month?
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Commission, Bonus and Dividend
b) Mr. Bibek is an online salesperson in an online shopping store. His monthly salary is
Rs 18,700 and 2% commission is given to him when the monthly sales is more than
5 lakh rupees. If the sales of the store in a month is Rs 7,20,000, calculate his total
income of the month.
c) Mrs. Nepali draws Rs 19,800 as her monthly salary in a wholesale cosmetic shop
and a certain commission is given as per the monthly sales. If the sales of a month is
Rs 12,00,000 and her total income of the month including commission is Rs 31,800,
find the rate of commission.
7. a) Mr. Ghising decides to sell a plot of 10 Aanas of land he owns at the rate of
Rs 5,00,000 per Aana. He promises a commission of 3% on the sale of the land to a
broker. If he gives a discount of 2% on the original price to the buyer, find his income
from the sale of the land, after paying the commission to the broker.
b) Mr. Chauhan is a car dealer who purchases a car for Rs 50,00,000. He intends selling
the car for Rs 55,00,000. However, he gives a discount of 3% on this price to the buyer
and a commission of 2% to a broker. How much profit does he make?
3.2 Bonus
Let's study the following examples and get the idea about bonus.
a) Mr. Bajracharya is an officer of a corporation. His monthly salary is Rs 40,000.
This year, the corporation decided to distribute a bonus amount that is equivalent
to his two month’s salary. So, he received Rs 80,000 bonus.
b) The board of management of an enterprise had decided to distribute equal
amount of bonus to its 50 employees from 10% of its net profit.
Bonus is an extra amount of money that is given to an employee as a reward for his/
her good performance. It is an incentive given to an employee besides his/her fixed
salary.
Bonus is calculated as a certain percent of profit and it is decided by the board of
management of any business organization.
Facts to remember
1. Total bonus amount = Rate of bonus × Net profit
2. Bonus rate = Total bonus amount × 100%
Net profit
Total bonus amount
3. Bonus amount received by each employee = Number of employees
Example 1: The annual salary of an employee in a business company is Rs 3,02,400.
Besides, the company provides 15% bonus from its net profit at the
end of each fiscal year. If the company made a profit of Rs 60,50,000
in a fiscal year and decided to distribute bonus to its 30 employees
equally, how much income did the employee make in the year?
Solution:
Here, the annual salary of the employee = Rs 3,02,400
Profit of the company = Rs 60,50,000
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Commission, Bonus and Dividend
The total amount of bonus = 15 % of Rs 60,50,000 = Rs 9,07,500
Now, the amount of bonus received by each employee = Rs 907500 = Rs 30,250
30
Again the annual income of the employee = Rs 3,02,400 + Rs 30,250 = Rs 3,32,650
Hence, the employee made an income of Rs 3,32,650 in the year.
Example 2: A cement factory made a net profit of Rs 2,60,00,000 in the last year.
The management of the factory decided to distribute 10% bonus from
the profit to its 130 employees equally.
(i) By what percent should the bonus be increased so that each
employee can receive Rs 30,000?
(ii) What should be the profit of the factory so that it can provide Rs
44,500 to each employee at 20% bonus?
Solution:
Here, net profit of the factory = Rs 2,60,00,000,
bonus percent = 10% and number of employees = 130
(i) Bonus amount received by each employee = Rs 30,000, bonus percent = ?
Bonus amount =130× Rs 30,000 = Rs 39,00,000
Bonus percent = Total bonus amount × 100% = Rs 39,00,000 × 100% = 15%
Net profit Rs 2,60,000
Hence, the bonus should be increased by 15% - 10% = 5%
(ii) Bonus percent= 20%, number of employees = 130,
bonus amount received by each worker = Rs 44,500
Now, bonus amount =130× Rs 44,500 = Rs 57,85,000
Let, net profit of the factory be Rs x
Then, bonus amount = bonus % of net profit
or, Rs 57,85,000 = 20% of x
20x
or, 57,85,000 = 100
∴ x = Rs 2,89,25,000
Hence, the profit of the factory should be Rs 2,89,25,000.
Example 3: When a commercial bank increased its profit from 20% to 25%, the
amount of profit increased to Rs 1,25,00,000. If the company decided
to distribute 80% bonus to its 50 employees equally from the increased
Solution: amount of profit, how much bonus will each employee receive?
Let, the annual income of the bank be Rs x.
According to question,
25% profit of the income = Rs 1,25,00,000
25
or, 100 × x = Rs 1,25,00,000
∴ x = Rs 5,00,00,000
When the profit was 20%, profit amount = 20% of Rs Rs 5,00,00,000 = Rs 1,00,00,000
Increased amount of profit = Rs 1,25,00,000 – Rs 1,00,00,000 = Rs 25,00,000
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Commission, Bonus and Dividend
Now, bonus amount = 80% of Rs 25,00,000 = Rs 20,00,000
Bonus amount
∴ Bonus amount received by each employee = Number of employees
= Rs 20,00,000 = Rs 40,000
50
Hence, each employee received Rs 40,000 bonus.
EXERCISE 3.2
General section
1. Calculate the bonus amount distributed by each of the following companies from the
table given below.
S.N. Company Net profit Bonus rate Total bonus amount
a) Restaurant Rs 50,00,000 10% ……….
b) Bike showroom Rs 65,00,000 15% ……….
c) Insurance company Rs 90,80,500 20% ……….
2. a) A publication house announced to distribute 10 % bonus equally to its 20 employees
from the net profit of Rs 18,36,000 at the end of a fiscal year, find the bonus received
by each employee.
b) The net profit of a hotel in a fiscal year was Rs 80 lakhs. The board of management of
the hotel announced to distribute 25% bonus equally to its 50 employees from the net
profit. How much bonus did each employee receive?
3. a) A garment factory announced 20 % bonus to its 25 workers from the net profit at the
end of last fiscal year. If every worker received Rs 18,500, how much was the profit of
the factory?
b) A trading corporation distributed 8% of its net profit equally among its 120 employees
last year. If each employee received Rs 24,000 bonus, how much net profit did the
corporation make in the last year?
4. a) A business company distributed bonus to its 24 employees from the net profit of Rs
16,48,000. If every employee received Rs 8,240, what was the bonus percent?
b) The board of management of an LP gas company announced to distribute Rs 15,500
bonus equally to its 60 employees from its net profit of Rs 46,50,000 in a fiscal year.
What percent of its net profit was distributed by the company for the employee bonus?
Creative section
5. a) A garment factory made a net profit of Rs 48,00,000 in the last year. The management
of the factory decided to distribute 18 % bonus from the profit to its 25 employees.
(i) Find the bonus amount received by each employee.
(ii) By what percent should the bonus be increased so that each employee can receive
Rs 38,400?
(iii) What should be the profit of the company so that it can provide Rs 40,000 to each
employee at 20 % bonus?
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Commission, Bonus and Dividend
b) A hydro power company made a net profit of Rs 2,50,00,000 in the last year. The
board of management of the company decided to distribute 10% bonus from the profit
to its 125 employees.
(i) Find the bonus amount received by each employee.
(ii) By what percent should the bonus be increased so that each employee can receive
Rs 24,000?
(iii) What should be the profit of the company so that it can provide Rs 25,000 to each
employee at 8% bonus?
6. a) When a publication house increased its profit from 20% to 25%, the amount of profit
increased to Rs 52,08,000. If the company decided to distribute 60% bonus to its 30
employees equally from the increased amount of profit, how much bonus does each
employee receive?
b) When cement factory increased its profit from 20% to 25% the amount of profit
increased to Rs 87,50,000. If the company decided to distribute 80% bonus to its 50
employees equally from the increased amount of profit, how much bonus will each
employee receive?
Project work and activity section
7. a) Is your any family member involved in any organisation or corporation. If so, discuss
with him/her and learn more about bonus. Prepare a report on bonus and present in
the class.
b) Make the groups of 5 of your friends. Take the information regarding bonus distribution
of any 5 public or private organisations. Present your report in the class.
3.3 Dividend
Let's study the following examples and get the idea about dividend.
a) At the end of a fiscal year, a bank announced Rs 25 dividend per share to its shareholders.
Mr. Wagle had 250 shares of the company and received Rs 6,250 cash dividend.
b) A person has bought 1,500 shares out 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
dividend received by the person as per his/her number of shares is called Rs 1,05,084.
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.
Facts to remember
1. Total dividend amount = Rate of dividend × Net profit
2. Dividend rate = Total dividend amount × 100%
Net profit
Total dividend amount
3. Dividend per share = Total number of shares
4. Dividend for a shareholder with 'n' shares = n × dividend per shares
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Commission, Bonus and Dividend
Example 1: Mrs. Bhatta bought 750 shares out of 20,000 shares from a Business
Solution: 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?
Here, the net profit of the company = Rs 90,00,000
Now, 10% of Rs 90,00,000 = 10% × Rs 90,00,000 = Rs 9,00,000
100
Again, the profit of 20,000 shares = Rs 9,00,000
The profit of 1 share = Rs 9,00,000 = Rs 45
20,000
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% = 15%
Rs 2,50,00,000
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 = Rs 125
400
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
Now, the percent of profit as dividend = Dividend for 1 share × 100%
Profit for 1 share
= Rs 125 × 100% = 15%
Rs 2500/3
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Commission, Bonus and Dividend
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) Calculate the dividend amount from the table given below.
S.N. Company Net profit Dividend rate Dividend
(i) Finance Rs 25,00,000 50% ……….
(ii) Hydropower Rs 80,00,000 60% ……….
(iii) Commercial bank Rs 1,75,40,000 45% ……….
b) Find the dividend per share from the table given below.
Company Net profit Dividend rate sNhuamrebser of Dividend per share
2,000
(i) A Rs 60,00,000 40% 9,000 ……….
(ii) B Rs 98,00,000 45% 30,700 ……….
(iii) C Rs 2,45,60,000 70% ……….
2. 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.
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Commission, Bonus and Dividend
3. 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?
4. 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
5. 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.
6. 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?
7. 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 and activity section
8. 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.
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Commission, Bonus and Dividend
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. Commission is
(A) a payment given to the employees based on the sales they made.
(B) a payment to the shareholders by the company from the net profit.
(C) a payment to the employees by the company from the net profit.
(D) a payment to an employee as a reward for good performance.
2. An extra amount of money that is given to an employee as a reward for good
performance is called
(A) commission (B) bonus (C) dividend (D) tax
3. The payment made by a corporation to its shareholders from certain portion of net
profit is called
(A) bonus (B) commission (C) allowance (D) dividend
4. Commission is calculated on the basis of
(A) cost price (B) marked price (C) selling price (D) net profit
5. Bonus is calculated as a certain percent of
(A) profit (B) loss (C) selling price (D) cost price
6. The commission in selling a dozer for Rs 60,00,000 at 0.30% is
(A) Rs 18,000 (B) Rs 15,000 (C) Rs 1,800 (D) Rs 1,80,000
7. Mr. Gopal paid a commission to the agent while buying a house for Rs 1,35,00,000 at
0.75% is
(A) Rs 11,250 (B) Rs 1,01,250 (C) Rs 1,10,250 (D) Rs 1,12,500
8. A farmer sells the fruits through an agent. If the agent charges 5% commission for
selling the fruits worth Rs 45,000; the amount received by the farmer is
(A) Rs 2,250 (B) Rs 42,750 (C) Rs 47,750 (D) Rs 40,500
9. The monthly salary of a salesman in a departmental store is Rs 15,000 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 6,60,000 in that month?
(A) Rs 15,000 (B) Rs 6,600 (C) Rs 8,400 (D) Rs 2,1600
10. A doctor working in a clinic has monthly salary Rs 40,000. 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 60,000?
(A) 8 lakhs (B) 10 lakhs (C) 15 lakhs (D) 20 lakhs
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Unit 4 Household Arithmetic
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
0-20
21-30 (Rs.) per unit) (Rs.) (Rs.per (Rs.) (Rs.per (Rs.) (Rs.per
31-50
51-100 unit) unit) unit)
101-250
Above 250 30.00 0.00 50.00 4.00 75.00 5.00 125.00 6.00
50.00 6.50
50.00 8.00 75.00 6.50 100.00 6.50 125.00 6.50
75.00 9.50 75.00 8.00 100.00 8.00 125.00 8.00
100.00 9.50 100.00 9.50 125.00 9.50 150.00 9.50
150.00 11.00
125.00 9.50 150.00 9.50 200.00 9.50
175.00 11.00 200.00 11.00 250.00 11.00
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Household Arithmetic
Facts to remember
1. For the consumers with 5 Ampere meter box, the energy charge is Rs 3 per unit
upto 20 units incase of consuming more than 20 units of electricity.
2. The payment schedule after the meter reading are as follows.
(i) Within 7 days of meter reading 2 % rebate is allowed.
(ii) From the 8th day to the 15th day, the payment will be according to the bill
(iii) From the 16th day to the 30th day, 5 % extra fine
(iv) From the 31st day to the 40th day, 10 % extra fine
(v) From the 41st day to the 60th day, 25 % extra fine
(vi) 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: 20 units is Rs 30 but energy charge is free; find the charge of consumption
of 15 units of electricity.
Here, consumption of electricity = 15 units.
Service charge = Rs 30
Rate of energy charge up to 20 units = Rs 0 per unit
Now, total charge of electricity with service charge = Rs 30 + 15×Rs 0 = Rs 30
Hence, the required charge of consumption of 15 units of electricity is Rs 30.
Example 2: 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
∴ 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 3: In a household of 5A electricity transmission line, the meter reading of
1 Aswin was 1492 units and that of 1 Kartik was 1572 units. Answer the
following questions under the given rates and billing system.
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Household Arithmetic
Units 0-20 21 – 30 31 – 50 51 - 100
Rs 3 Rs 6.50 Rs 8.00 Rs 9.50
Rate of charge per units
Service charge = Rs 75
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 = 1572 units – 1492 units = 80 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
51-100 80 – 50 = 30 Rs 9.50 30 ×Rs 9. 50 = Rs 285
∴ Total charge = service charge + total electricity charge
= Rs 75 + (Rs 60 + Rs 65 + Rs 160 + Rs 285) = Rs 645
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 645 – 2% of Rs 645 = Rs 632.10
(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 645
(iii) If the payment was made on the 22nd day of meter reading, there would be 5% fine
∴ Total charge to be paid = Rs 645 + 5% of Rs 645 = Rs 677.25
(iv) If the payment was made on the 35th day of meter reading, there would be 10% fine
∴ Total charge to be paid = Rs 645 + 10% of Rs 645 = Rs 709.50
(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 645 + 25% of Rs 645 = Rs 806.25
Example 4: 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 - 100 101-250 Above 250
Rate of charge Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50 Rs 11
per units
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Household Arithmetic
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
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
51-100 100 – 50 = 50 Rs 9.50 50 ×Rs 9. 50 = Rs 475
101-250 250 – 100 = 150 Rs 9.50 150 ×Rs 9.50 = Rs 1425
Above 250 260 – 250 = 10 Rs 11 10 ×Rs 11= Rs 110
∴ Total charge = service charge + total electricity charge
= Rs 175 + (Rs 80 + Rs 65 + Rs 160 + Rs 475 + Rs 1425 + Rs 110) = Rs 2,490
Since, the payment was made on the 28th day of meter reading, there was 5% extra fine
∴ Total charge with fine = Rs 2,490 + 5% of Rs 2,490 = Rs 2,614.50
Again, if the customer paid the bill within 7 days of meter reading, there would be 2%
rebate.
∴ Total charge after rebate = Rs 2,490 – 2% of Rs 2,490 = Rs 2,440.20
Difference between two payments = Rs 2,614.50 – Rs 2,440.20 = Rs 174.30
Rs 174.30
∴ Less % in payment = Rs 2614.50 × 100% = 6.67%
Hence, the payment of the bill would be 6.67% less if it was paid within 7 days of meter
reading.
Example 5: 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 - 100 101-250
Rate of charge per units Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50
Payment within the 31st to the 40th day of meter reading:10% extra fine.
Solution:
Let the number of consumed unit of electricity of the month be x units.
Now,
Consumption block No. of units Rate of charge Electricity charge
Rs 4 20 ×Rs 4 = Rs 80
0-20 20 – 0=20
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
51-100 100 – 50 = 50 Rs 9.50 50 × Rs 9.50 = Rs 475
101-250 x – 100 Rs 9.50 (x – 100) ×Rs 9. 50 = Rs (9.5x – 950)
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Household Arithmetic
∴ Total charge = service charge + total electricity charge
= Rs 125 + (Rs 80 + Rs 65 + Rs 160 + Rs 475 + Rs 9.5x – Rs 950)
= Rs (9.5x – 45)
According to question,
Payment of the bill with fine = Rs 1100 Vedanta ICT Corner
or, R s (9.5x – 45) + 10% of Rs (9.5x – 45) = Rs 1100 Please! Scan this QR code or
or, (9.5x – 45) + 0.1 (9.5x – 45) = 1100 browse the link given below:
or, (9.5x – 45) + (0.95x– 4.5) = 1100 https://www.geogebra.org/m/yph4p67w
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 20 units is
Rs 30 but energy charge is free; find the charge of consumption of 10 units.
b) There is no energy charge but the service charge is Rs 30 up to 20 units; how much
bill should be charged by Nepal Electricity Authority (NEA) for the consumption of
18 units?
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?
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. 01198 1 01223 4
a) The meter reading of Sandhya's house on 1 Baisakh Baisakh Jestha
and 1 Jestha are shown in the meter box alongside.
Find the electricity charge for the month of Baisakh.
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Household Arithmetic
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
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 14 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.
Units 0-20 21 – 30 31 – 50 51 - 100 101-250
Rate of charge per units Rs 3 Rs 6.50 Rs 8.00 Rs 9.50 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.
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Household Arithmetic
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, Rs 9.50 per unit from 51 to 100 units and Rs 9.50 per
unit from 101 to 250 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 – 100 101-250 Above 250
Rate of charge Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50 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.
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, Rs 9.50 per unit
from 51 to 100 units and Rs 9.50 per unit from 101 to 250 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.
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Household Arithmetic
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.
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
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Household Arithmetic
\ 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.
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.
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Household Arithmetic
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
3 1" 56,000 3420 40 9200 2700
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 21" 10,000 100 32 785
2 34" 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
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
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Household Arithmetic
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) -g]kfn u0' f:t/ tyf gfktf}n 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
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.
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Household Arithmetic
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.
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?
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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?
AAsharouwsaesho1l2d60usuensits12"anodf
c) water pipe. The meter reading of the household on 1st of
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.
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 and activity section
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.
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Household Arithmetic
c) 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
Let’s tick (√) the correct alternative.
1. 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.
2. The consumption of electricity of recent month is calculated by
(A) Meter reading of recent month – Meter reading of previous month
(B) Meter reading of following month – Meter reading of recent month
(C) Meter reading of previous month – Meter reading of recent month
(D) Meter reading of following month – Meter reading of previous month
3. The meter readings of the month Baishakh, Jesth and Asar are 1127, 1155 and 1180
respectively. The consumption of the electricity in the month Jestha is
(A) 28 units (B) 25 units (C) 53 unts (D) 38 units
4. According to payment schedule of consumption of electricity, if the payment is made
within 7 days of meter reading then there is provision of
(A) 3% of rebate is allowed. (B) 2% of rebate is allowed.
(C) 5% extra fine (D) No rebate no fine
5. What is the grand total in the telephone bill if the sub-total is Rs 400?
(A) Rs 440 (B) Rs 452 (C) Rs 497.20 (D) Rs 406.80
6. According to Nepal Telecommunications Authority, the minimum charge for 175 calls
is
(A) Rs 100 (B) Rs 200 (C) Rs 175 (D) Rs 350
7. The compulsory provision of sewerage service charge in water bill is
(A) 20% of water consumption charge (B) 40% of water consumption charge
(C) 50% of water consumption charge (D) 100% of water consumption charge
8. When the water bill is paid within the third month of bill issued, there will be
(A) 3% rebate in bill (B) no rebate no fine (C) 10% fine (D) 20% fine
9. The consumer gets 3% rebate in water bill if the payment is made within
(A) 1st month after the bill issued (B) 2nd month after the bill issued
(C) 1st or 2nd month after the bill issued (D) 3rd month after the bill issued
10. 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 7.20 (B) Rs 10.80 (C) Rs 14 (D) Rs 21
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Unit 5 Mensuration (I): Area
5.1 Area – Looking back
Classwork-Exercise
1. Fill in the blanks with correct answers as quickly as possible.
a) The formula to calculate the area of rectangle is …………………
b) The formula to find the area of square is …………………
c) The formula to find the area of parallelogram is …………………
d) The formula to calculate the area of triangle is …………………
2. Calculate the areas mentally and write in the blank space as quickly as possible.
a) The length of a rectangular carpet is 3 m and its breadth is 2 m. Its area is ……..
b) A squared garden is 30 feet long. Its area is ….….
c) The base of a parallelogram is 15 cm and its height is 8 cm. Its area is ….….
d) The area of triangle whose base is 20 inch and height is 9 inch is ….….
e) A right angled triangle has base 26 cm and perpendicular 10 cm. Its area is ..…...
5.2 Mensuration
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.
Mensuration is used in the field of architecture, medicine, construction, etc. It
is necessary for everyone to learn formulae used to find the perimeter and area of
2-dimensional figures as well as the surface area and volume of 3-dimensional solids
in day to day life.
5.3 Perimeter and area of plane figures - review
Perimeter of a plane figure is the total of the length of its boundary. The unit of perimeter
is the same as the unit of length, for example centimeter (cm), metre (m), etc. are the
units of perimeter.
Perimeter of a triangle = a + b + c, where, a, b and c are the length of sides of a triangle.
Perimeter of a rectangle = 2(l + b), where, l is the length and b is the breadth of a
rectangle.
Perimeter of a square = 4l, where, l is the length of each side of a square.
Perimeter of a circle = 2πr, where, r is the radius of a circle.
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Mensuration (I): Area
Area of a plane figure is the measure of the plane surface enclosed by its boundary
line. The unit of area is square centimetre (cm2 ) square metre (m2 ), etc. It is noted that
‘square metre’ and ‘metre square’ have different meanings. 9 square metre means the
area is 9 m2 . However, 9 metre square means a square each of whose side is 9 m long.
5.4 Area of triangle
We can find the area of different types of triangles by using special formulae which are
generalised on the basis of the given measurements of various parts of the triangles.
(i) Area of a triangle in terms of its base and height (altitude)
In the adjoining D ABC, AD ⊥ BC. So, AD is the height and BC is the base of the
triangle ABC. AA
Here, area of D ABC = 1 base × height
2
1
= 2 BC × AD h
= 1 b × h B D CD Bb C
2
(ii) Area of a right-angled triangle
In the given right-angled D ABC right angled at B, the perpendicular AB is the
height of the triangle and BC is its base.
Area of right-angled D ABC = 1 base × height A
2 p
B
= 1 base × perpendicular
2
= 1 BC × AB Cb
2
(iii) Area of an equilateral triangle
In the given equilateral triangle ABC, AB = BC = CA = a (suppose). AD
perpendicular to BC is drawn. A
In an equilateral triangle, perpendicular drawn from a
vertex to its opposite side bisects the side.
∴ BD = DC = a aa
2 h
By using Pythagoras Theorem in right-angled D ABD,
AD = AB2 – BD2 = a2 – a2 = 32 a B a D a C
4 2 a 2
Now, area of D ABC = 1 base × height
2
= 1 BC × AD = 1 a × 23 a = 43 a2
2 2
Thus, area of equilateral triangle = 43 (side)2
(iv) Area of an isosceles triangle
In the given isosceles triangle ABC, AB = AC = a and BC = b (suppose).
AD perpendicular to BC is drawn.
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Mensuration (I): Area
We know that, in an isosceles triangle, perpendicular A
drawn from the vertical angle to the base bisects the base.
Therefore, BD = DC = b a a
2
By using Pythagoras Theorem in right-angled D ABD,
AD = AB2 – BD2 = a2 – b2 = 4a22– b2
4
Bb D bC
Now, area of D ABC = 1 base × height 22
2
1
= 2 BC × AD
= 1 b × 4a2 – b2 = 1 b 4a2 – b2
2 2 4
Thus, area of an isosceles triangle = 14ofbits4ath2 –rebe2 sides (Heron's formula)
(v) Area of scalene triangle in terms
Heron of Alexandria [A.D. 10 – A.D. 70 (CE)] was a Greek
mathematician and engineer. He lived in the city of Alexandria in
Egypt, and is one of the greatest “experimenter” of antiquity. He wrote
books on mathematics, mechanics and physics. In mathematics, he
is mostly remembered for Heron’s formula that allows us to calculate
the area of a triangle using only the lengths of its sides.
In the adjoining scalene triangle ABC, suppose the A
sides BC = a, CA = b and AB = c. Let's draw AD⊥BC. ch b
Therefore, AD = h is the height of DABC. x
Let the length of DC be x cm.
∴ The length of BD = (a – x) cm B a–x D a C
Here, the perimeter of D ABC = 2s = BC + CA + AB
=a+b+c
∴ 2s = a + b + c
oNro, ws ,=inari+ghb2t-+ancglewdhDichACisDc,abllyedustihneg semi-perimeter of the triangle.
Pythagoras Theorem,
AD2 = AC2 – DC2
or, h2 = b2 – x2 .................(i)
Also, in right-angled D ABD, Vedanta ICT Corner
AD2 = AB2 – BD2 Please! Scan this QR code or
or, h2 = c2 – (a – x)2 .................(ii) browse the link given below:
From equations (i) and (ii), we get,
c2 – (a – x)2 = b2 – x2 https://www.geogebra.org/m/ejsvzhgy
or, c2 – a2 + 2ax – x2 = b2 – x2
or, x = a2 + b2 – c2 ................. (iii)
2a
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Mensuration (I): Area
Again, from the equations (i) and (iii), we get,
h2 = b2 – a2 + b2 – c2 2
2a
= b + a2 + b2 – c2 b – a2 + b2 – c2
2a 2a
= 2ab + a2 + b2 – c2 2ab – a2 – b2 + c2
2a 2a
= (a + b)2 – c2 c2 – (a – b)2
2a 2a
(a + b + c) (a + b – c) (c + a – b) (c – a + b)
= 2a 2a
= (a + b + c) (a + b + c – 2c) (a + b + c – 2b) (a + b + c – 2a)
4a2
Now, substituting, a + b + c = 2s, we get,
h2 = 2s (2s – 2c) (2s – 2b) (2s – 2a)
4a2
16s (s – a) (s – b) (s – c)
= 4a2
h = 2 s(s – a) (s – b) (s – c)
a D ABC =
Then, area of 1 base × height
2
1
= 2 BC × h
= 1 a × 2 s(s – a) (s – b) (s – c)
2 a
= s(s – a) (s – b) (s – c)
Thus, when three sides of a triangle are given, its area A = s(s – a) (s – b) (s – c)
Worked-out examples
Example 1: Find the area of the triangle and the quadrilateral given below.
a) A b) 13 cm C
B
14 cm 13 cm 12 cm
4 cm
Solution: B 15 cm C
14 + 15 + 13 A 3 cm D
2
a) In ∆ABC, s = cm = 21 cm
∴ Area of ∆ABC = s (s – a) (s – b) (s – c)
= 21 (21 – 15) (21 – 13) (21 – 14)
= 21 × 6 × 8 × 7 = 7 × 3 × 3 × 2 × 2 × 4 × 7
= 7 × 3 × 2 × 2 = 84 cm2.
b) In ∆ABD, by Pythagoras theorem, we have,
BD = AB2 + AD2 = 42 + 32 = 5 cm
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Mensuration (I): Area
Now, area of ∆ABD = 1 × AB × AD = 1 × 4 × 3 = 6 cm2
2 2
In ∆BDC, s = 13 + 12 + 5 cm = 15 cm
2
∴ Area of ∆BDC = s (s – a) (s – b) (s – c)
= 15 (15 – 13) (15 – 12) (15 – 5) = 15 × 2 × 3 × 10 = 30 sq. cm.
Hence, area of quad. ABCD = Area of (∆ABD + ∆BDC) = 6 sq. cm. + 30 sq. cm = 36 sq. cm.
Example 2: Find the area of the given parallelogram, in which AB = DC = 41 cm and
AD = BC = 14 cm and diagonal BD = 28 cm.
Solution: DC
The semi-perimeter of ∆ABD is 28 cm
41 cm + 28 cm + 15 cm 15 cm
2
s = = 42 cm
Now, area of ∆ABD = s (s – a) (s – b) (s – c) A 41 cm B
= 42 (42 – 41) (42 – 28) (42 – 15)
= 42 × 1 × 14 × 27 = 126 cm2
∴ Area of parm. ABCD = 2 × area of D ABD = 2 × 126 cm2 = 252 cm2
Example 3: The traffic signal board is an equilateral triangle with
perimeter 180 cm, find its area.
Solution:
Here, the perimeter of equilateral traffic signal board = 180 cm
or, 3a = 180 cm
or, a = 60 cm
∴Semi-perimeter (s) = 180 cm = 90 cm Alternatively,
2
Area of the equilateral triangular
Now, area (A) = s (s – a) (s – b) (s – c) board (A) = 43 a2
= 90 (90 – 60) (90 – 60) (90 – 60) = 43 × (60 cm)2
=1558.85 cm2
= 90 × 30 × 30 × 30
= 2430000 = 1558.85 cm2
Hence, the area of the signal board is 1,558.85 cm2
Example 4: Find the area of an isosceles triangle whose base is 10 cm and length of each
of the equal side is 13 cm
Solution: 13 cm 13 cm
Here, in an isosceles triangle, base (b) = 10 cm and equal sides (a) = 13 cm
∴Semi-perimeter (s) = 13 cm + 13 cm + 10 cm = 18 cm
2
Now, area of the triangle (A) = s (s – a) (s – b) (s – c) 10 cm
= 18 (18 – 13) (18 – 13) (18 – 10)
= 18 × 5 × 5 × 8 = 3600 = 60 cm2
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Mensuration (I): Area
Alternatively,
Area of isosceles triangle (A) = b 4a2 – b2
4
= 140 4(13)2 – 102 5 676 – 100 = 25×24= 60 cm2
= 2
Hence, the area of isosceles triangle is 60 cm2.
Example 5: The perimeter of a triangle is 48 cm. If the ratios of its sides are 4 : 5 : 3, find
its area.
Solution:
Here, the perimeter of the triangle = 48 cm
∴ The semi-perimeter of the triangle = 48 cm = 24 cm
2
Again, let the sides of the triangle are 4x cm, 5x cm and 3x cm.
∴ 4x + 5x + 3x = 48 cm
or, 12x = 48 cm
or, x = 48 cm = 4 cm.
12
Now, 4x = 4 × 4 cm = 16 cm, 5x = 5 × 4 cm = 20 cm and 3x = 3 × 4 cm = 12 cm.
∴ a = 16 cm, b = 20 cm and c = 12 cm.
∴ Area of the triangle = s (s – a) (s – b) (s – c)
= 24 (24 – 16) (24 – 20) (24 – 12) = 24 × 8 × 4 × 12= 96 cm2.
Hence, the required area of the triangle is 96 cm2.
Example 6: The sides of a triangular garden are in the ratio 13:14:15. If the area of the
garden is 336 m2, find its perimeter.
Solution:
Let, the sides of the garden be a = 13x m, b =14x m and c = 15x m.
Now, semi-perimeter (s) = 13x + 14x + 15x = 42x = 21x
2 2
Also, Area of the garden = s (s – a) (s – b) (s – c)
or, 336 m2 = 21x(21x – 13x)(21x – 14x)(21x – 15x)
or, 336 = 21x × 8x × 7x × 6x
or, 336 = 7056x4
or, 336 = 84x2
or, x2 = 4
or, x = 2
Thus, the sides of the garden are;
a = 13x = 13 × 2 m = 26 m, b= 14x = 14 × 2m = 28m and c = 15x = 15 × 2m = 30m
Hence, the perimeter of the garden is a+ b + c = 26 m + 28m + 30m = 84 m.
Example 7: Mr. Pattel is a farmer in Janakpur. He has a field in the shape of a rhombus.
The perimeter of the field is 800m and one of its diagonal is 240m. Find the
area of the field in hectare. (1 hectare = 10,000 m2)
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Mensuration (I): Area
Solution: A D
Here, perimeter of the field in the shape of rhombus ABCD = 800m 240 m
or, 4× length of each side (l) = 800m
or, l = 200m
∴AB = BC = CD = AD = 200m B C
In ∆BCD, semi-perimeter (s) = 200m + 200m + 240m = 640m = 320m
2 2
∴ Area of ∆BCD = s (s – a) (s – b) (s – c)
= 320(320 – 200)(320 – 200)(320 – 240)
= 320 × 120 × 120 × 80 = 368640000 = 19,200 m2
Thus, the area of the field ABCD = 2 × Area of ∆BCD = 2 × 19,200 m2 =38,400 m2
Again, 10,000 m2 = 1 hectare
∴ 38,400m2 = 38400 hectare = 3.84 hectares
10000
Hence, the area of his field is 3.84 hectares.
Example 8: The umbrella given alongside is made by stitching
8 triangular pieces of cloth, each piece measuring 61
cm, 61 cm and 22 cm. how much cloth is required for
the umbrella? If the cost of 1 cm2 is Rs 0.25, find the
total cost of cloth.
Solution:
Here, the sides of a triangular piece of cloth are
a = 61 cm, b = 61 cm and c = 22 cm
Now, semi-perimeter (s) = 61 cm + 61 cm + 22 cm = 144 cm = 72m
2 2
∴ Area of a triangular piece = s (s – a) (s – b) (s – c)
= 72(72 – 61)(72 – 61)(72 – 22)
= 72 × 11 × 11 × 50 = 435600 = 660 cm2
Also, the area of cloths required to make the umbrella (A) = 8×660 cm2 = 5,280cm2
Again, the total cost of cloth (T) = Area (A) ×Rate (R)
= 5280 ×Rs 0.25
= Rs 1,320
Hence, the total cost of cloth required to make the umbrella is Rs 1,320.
Example 9: The area of right-angled triangle is 30 cm2 and its perimeter is 30 cm. Find
the sides of the triangle.
Solution:
Let, p, b and h are the perpendicular, base and hypotenuse of the right- p h
angled triangle.
Now, area of right angled triangle (A) = 1 base (b)×perpendicular (p)
2
b
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Mensuration (I): Area
or, 30 = 1 b×p
2
∴ pb = 60 … (i)
Also, perimeter = 30 cm
or, p + b + h = 30
or, p + b + p2 + b2 = 30 [ h = p2 + b2 ]
or, p2 + b2 = 30 – (p + b)
Squaring on both sides, we get
p2 + b2 ={30 – (p + b)}2
or, p2 + b2 = 302 – 2 × 30 × (p + b) + (p + b)2
or, p2 + b2 = 900 – 60(p + b) + p2 + 2pb + b2
or, 0 = 900 – 60(p + b) + 2 × 60 [ pb = 60]
or, 60 (p + b) =900 + 120
or, 60 (p + b) = 1020
∴ p + b = 17 … (ii)
Again, we have, (p – b)2 = (p + b)2 – 4pb
= 172 – 4 × 60 = 289 – 240 = 49 = 72
∴ p – b = 7 ... (iii)
Now, adding equations (ii) and (iii), we get,
p + b + p – b = 17 + 7
or, 2p = 24
or, p = 12
Putting the value of p in equation (ii), we get,
12 + b = 17
or, b = 17 – 12 =5
Also, h = p2 + b2 = 122 + 52 = 144 + 25 = 169 = 13
So, the required sides of the right angled triangle are 12 cm, 5 cm and 13 cm.
EXERCISE 5.1
General section
1. a) What is the area of triangle whose base is x cm and height is y cm?
b) Write down the area of right angled triangle in which base is m cm and perpendicular
is n cm.
c) What is the area of equilateral triangle whose side is ‘x’ cm?
d) If the length of each equal sides of an isosceles triangle is p cm and base is q cm,
what is its area?
e) The sides of a triangle are x, y and z units respectively, what is the semi-perimeter
of the triangle?
f) The semi-perimeter of triangle having side lengths a cm, b cm and c cm is s cm.
Find its area.
g) What is the area of triangle whose sides are p cm, q cm and r cm?
Vedanta Excel in Mathematics - Book 9 82 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (I): Area
2. a) Find the area of the following triangles.
(i) A (ii) P (iii) X (iv) L 41 cm N
8 cm 10 cm 13 cm 15 cm 25 cm 12 cm
B Z
41 cm 18 cm
6 cm C Q 14 cm R Y 17 cm
M
b) The edges of a triangular kitchen garden are 20 ft, 21 ft and 29 ft long. Calculate the
area of the garden.
c) The sides of the triangular park are 193 m, 194 m and 195 m. Find its area.
d) Find the area an isosceles triangle in which each of equal sides is 25 cm and the
base is 14 cm.
e) Find the area of an equilateral triangle whose perimeter is 12 cm.
Creative section - A
3. Find the area of the following quadrilaterals. D
a) D C b) P S c)
12 cm
15 cm 14 cm 48 cm A 25 cm
A 13 cm B Q 25 cm R 9 cm 17 cm
d) A 13 cm D e) Z f) B 10 cm C
C
16 cm D 21 cm
20 cm 11 cm 30 cm
Y 12 cm 13 cm
B 21 cm W 17 cm X A 16 cm B
C
4. a) The shape of a piece of land is a parallelogram whose adjacent sides are 12 m and
b) 17 m and the corresponding diagonal is 25 m. Find the area of the land.
c) A park is in the shape of parallelogram. The lengths of two adjacent edges are 51m
d) and 52 m and the corresponding diagonal is 53 m, what is the area of the park?
A garden is in the shape of a rhombus whose each side is 15 m and its one of two
5. a) diagonals is 24 m. Find the area of the garden.
b) A farmer has a field in the shape of a rhombus. The perimeter of the field is 400m
c) and one of its diagonals is 120m. Find the area of the field in hectare. (1 hectare =
d) 10,000 m2)
If the ratio of the sides of a triangle is 3: 4: 5 and its perimeter is 36 cm, find the
area of the triangle.
The perimeter of a triangular park is 84 m and the ratio of its sides of a triangle is
13: 14: 15, find the area of the triangle.
The ratio of the sides of a triangular field is 12:5:13 and its area is 120 m2, find
the perimeter of the field.
The area of triangular kitchen garden is 324 sq. ft and its edges are in the ratio
9:10:17. Find the perimeter of the garden.
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Mensuration (I): Area
Creative section - B
6. a) The area of a right angled triangle is 24 cm2 and perimeter is 24 cm. Find the
b) sides of the triangle.
The perimeter of a garden which is in the shape of right angled triangle is 40 ft
and its area is 60 ft2. Find the length of sides of the garden.
c) The perimeter of a right angled triangular park is 120 m and its area is 600 m2.
Find the length of sides of the park.
7. a) An umbrella is made by stitching 8 triangular pieces of cloth of two different colors
as shown in the figure. Each piece measures 50 cm, 50 cm and 28 cm. How much
cloth of each color is required for the umbrella? If the rate of cost of the cloth is Rs
0.15 per sq. cm, find the cost of the cloths required to make the umbrella.
b) Mrs. Limbu made a rangoli which has 8 identical triangles. 13cm
Each triangle measures 20 cm, 13 cm and 11 cm. How much
surface of the floor is covered with the design? If the rate of 20cm
cost of the colour is Rs 2.50 per sq. cm, find the cost of the 11cm
colours required to make the design.
c) A floral design on a floor of a room is made up of 16 triangular
pieces of tiles. The sides of each piece of tile are 28 cm, 15 cm
and 41 cm. Find the area of the design. Also, estimate the cost of
polishing the tiles at Rs 1.50 per sq. cm.
5.5 Local Land Measurement Units in Nepal
In Nepal, people from different parts of the country have different culture, custom,
life style and languages. Similarly, the people use different customary units to
measure the area of land in different regions though the metric system has been the
official standard since 1968 AD. In Terai region, Bigha-Kattha-Dhur measurements
are common to measure the area of land while Ropani-Aana measurements are
common in hilly and mountainous regions.
Facts to remember Conversion of Hilly customary units into
Conversion of Terai customary units into other local and international units
other local and international units
(i) 1 Ropani = 16 Aana
(i) 1 Bigha = 20 Kattha (ii) 1 Aana = 4 Paisa
(ii) 1 Kattha = 20 Dhur (iii) 1 Paisa = 4 Daam
(iii) 1 Bigha = 400 Dhur (iv) 1 Ropani = 508.72 m² = 5476 sq.ft.
(iv) 1 Bigha = 6772.63 m² = 72900 sq.ft. (v) 1 Aana = 31.80 m² = 342.25 sq. ft.
(v) 1 Kattha = 338.63 m² = 3645 sq.ft. (vi) 1 Paisa = 7.95 m² = 85.56 sq. ft.
(vi) 1 Dhur = 16.93 m² = 182.25 sq.ft. (vii) 1 Daam = 1.99 m² = 21.39 sq. ft.
(vii) 1 Bigha = 13.31 Ropani
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Mensuration (I): Area
Worked-out examples
Example 1: The lengths of sides of a triangular park are 51 m, 25 m and 74 m. Find the
area of the park in (i) Aana (ii) Kattha
(1 Aana = 31.80 m2, 1 Kattha = 338.63 m²)
Solution:
Here, the sides of triangular park are; a = 51 m,
b = 25 m
and c = 74 m
Now, Semi-perimeter (s) = 51 m + 25 m + 74 m = 75 m
2
Again, area of the park (A) = s (s – a) (s – b) (s – c)
= 75 (75 – 51) (75 – 25) (75 – 74)
= 75 × 24 × 50 × 1 = 90000 = 300 m2
(i) 1 Aana = 31.80 m2. So, the area of the park = 300 Aana = 9.43 Aana
31.80
300
(ii) 1 Kattha = 338.63 m2. So, the area of the park = 338.63 Kattha = 0.89 Kattha
Example 2: Mr. Sherpa has an apple farm in Mustang which is in the shape of a rhombus.
The diagonals of the field are 240 m and 320 m respectively. Find the area
of his farm in Ropani. (1 Ropani = 508.72 m2)
Solution: AD
In a rhombus ABCD, the diagonals AC (d1) =240 m and BD (d2) = 320 m
Now, the area of the rhombus ABCD = 1 × AC (d1) × BD (d2)
2
BC
= 1 × 240 m × 320 m
2
= 38,400 m2
Again, 508.72 m2 = 1 Ropani
∴ 38,400 m2 = 38,400 Ropani = 75.48 Ropani
508.72
Hence, the area of the apple farm is 75.48 Ropani.
Example 3: Mrs. Chaudhary has a piece of land in the shape of quadrilateral with
the longer diagonal 36 m. The lengths of the perpendiculars from the
opposite vertices on this diagonal are 13.5 m and 26.5 m. If she sold the
land to Mr. Rai at Rs 15,00,000 per Kattha, how much did Mr. Rai pay to
Mrs. Chaudhary? (1 Kattha = 338.72 m2)
Solution: D
Here, in the quadrilateral ABCD; AC is the longer diagonal. C
BP ⊥ AC, DQ ⊥ AC, AC = 36 m, BP = 13.5 m and DQ =26.5 m Q
P
Now, area of the field ABCD = Area of (∆ABC + ∆ACD)
AB
= 1 AC × BP + 1 AC × DQ
2 2
1
= 2 AC (BP + DQ)
= 1 × 36 (13.5 + 26.5) = 720 m2
2
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Mensuration (I): Area
We know, 338.72 m2 = 1 Kattha. So, the area of the land = 720 Kattha =2.13 Kattha
338.72
Again, the value of the land = 2.13 ×Rs 15,00,000 = Rs 31,95,000.
Hence, Mr. Rai paid Rs 31,95,000 to Mrs. Chaudhary.
EXERCISE 5.2
General section
1. a) How many Aana are there in 1 Ropani?
b) How many Paisa are there in (i) 1 Aana? (ii) 1 Ropani?
c) How many Daam are there in (i) 1 Paisa? (ii) 1 Aana? (iii) 1 Ropani?
d) How many Kattha are there in 1 Bigha?
e) How many Dhur are there in (i) 1 Kattha? (ii) 1 Bigha?
f) How many Ropani are there in 1 Bigha?
2. a) A rectangular field is 5 m long and 4.77 m wide. Convert the area of the field in
Paisa. (1 Paisa = 7.95 m2)
b) The side of a squared garden is 10 m. Find its area in Dhur. (1 Dhur = 16.93m2).
c) The edges of a triangular kitchen garden are 13 m, 14 m and 15 m respectively, find
the area of the garden in Aana. (1 Aana = 31.80 sq. m.)
d) How many Bigha are there in a triangular land having sides 723 ft., 724 ft. and
725 ft.? (1 Bigha =72900 sq.ft.)
Creative Section
3. a) Mr. Dhital bought a plot of land of 5 Aanas last year. Now, he is building a house on
the plot covering its space of 40 ft. by 30 ft. Calculate the area of remaining part of
the plot in square feet. (1 Aana = 342.25 sq. ft.)
b) Mr. Lama has a piece of land of 10 Kattha. He constructed a square pond of length
55 m for fish farming within it. How much space of his land is left? Find it in
sq. m? (1 Kattha = 338.63 m2)
4. a) A hospital is going to be built in the land of the shape of a rhomus having its
diagonals 250 m and 480 m respectively in a mountainous district of Nepal. Find
the area covered by the hospital in Ropani. (1 Ropani = 508.72 m2).
b) An international airport is built in the land of the shape of a quadrilateral in Terai
region. The longer diagonal of the land is 800 m and the length of perpendiculrs
from opposite vertices on the diagonal are 150 m and 200 m, calculate the area of
the land covered by the airport in Bigha? (1 Bigha = 6772.63 m²)
5. a) Mr. Gosain has a piece of land in Bhaktapur. The shape of the land is a rhombus.
The diagonals of the land are 16 m and 15.9 m respectively. He sold the land to
Mrs. Ojha at the rate of Rs 25,00,000 per Aana. How much did Mrs. Ojha pay to
Mr. Gosain?(1 Aana = 31.8 m2)
b) A real estate agent bought two triangular ‘Kittas’ having a 20 m
common side as shown in the figure from two people for 21 m
Rs 20,00,000 altogether. If he immediately sold the land in the 29 m
shape of a quadrilateral at Rs 15,00,000 per Kattha, find his 36 m
profit or loss.(1 Kattha= 338.63 sq. m)
25 m
Vedanta Excel in Mathematics - Book 9 86 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (I): Area
Project work and activity section
6. a) Cut chart papers into a few number of triangles and quadrilateral shapes. Measure
b) the dimensions of each triangular and quadrilateral piece. Then, find the area of
each shape.
The table given below shows the different types of triangles having same perimeter.
Find the remaining sides of triangles and their areas. Compare the areas and draw
the conclusion on what type of triangle possesses the greatest area though they
have equal perimeter.
Types of triangle Perimeter Possible side length Area
Equilateral triangle 60 cm (i) …., (ii) …., (iii)…. .....
Isosceles triangle 60 cm (i) 10 cm, (ii) …. , (iii) …. …..
Scalene triangle 60 cm (i) 18 cm, (ii) 22 cm, (iii) …. …..
c) Prepare a report on the units of measuring the land in your locality. Then, present
it in the class.
5.6 Area of 4 walls, floor and ceiling
A rectangular room has four walls, a floor and a ceiling. The floor and ceiling are
congruent. The walls along the length are congruent and the walls along the breadth
are congruent.
If l, b, and h be the length, breadth and the height of a room, Ceiling
(i) Area of floor = l × b h
(ii) Area of ceiling = Area of floor = l × b Floor
l
b
(iii) Area of opposite walls along the length = l × h + l × h = 2lh
(iv) Area of opposite walls along the breadth = b × h + b × h = 2bh
(v) Area of 4 walls of the room = 2lh + 2bh = 2h (l + b)
Facts to remember
1) The perimeter of the floor of the room (P) = 2 (l + b)
The area of four walls = 2h (l +b) = 2(l +b) × h = P × h
2) For a square room, length = breadth = l and height = h
a) Area of floor = Area of ceiling = l × l = l2
b) Area of four walls = 2h (l + l) = 2h (2l) = 4lh
c) Area of room including floor and ceiling = 2l2 + 4lh
3. For a cubical room, length = breadth = height = l
a) Area of floor = Area of ceiling = l × l = l2
b) Area of four walls = 2l (l + l) = 2l (2l) = 4l2
c) Area of room including floor and ceiling = 2l2 + 4l2 = 6l2
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Mensuration (I): Area
If la1 adnodorh, 1 be the length and height of a window, l2 and h2 be the length and height
of
(i) Area of 4 walls excluding a window
= 2h (l + b) – l 1 h1 h h1 l1 h2 l2
(ii) Area of 4 walls excluding a door b l
= 2h (l + b) – l 2 h2
(iii) Area of 4 walls excluding a window and a door
= 2h (l + b) – l1 h1 – l2 h2
Facts to remember
Let A = Area of floor, four walls, ceiling etc.
a = Area of each piece of carpet, wallpaper etc.
N = Number of pieces of carpet required for the floor, wallpaper for the walls etc.
n = Number of designs made in 1 sq. units in the floor, walls etc.
Then, (i) N = Aa (ii) a = A (iii) A = N × a (iv) N = n × A
N
Worked-out examples
Example 1: The length, breadth, and height of a 6.5 m
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. 20 m 15 m
(iii) Find the area of opposite walls
along the length.
(iv) Find the area of opposite walls along the breadth.
(v) Find the area of its four walls.
(vi) Find the area of the room including floor and ceiling.
Solution:
Here, the length of the hall (l) = 20 m
The breadth of the hall (b) = 15 m
The height of the hall (h) = 6.5 m
(i) The area of the floor (A1) = l × b =20 m×15 m = 300 m2
(ii) The area of the ceiling (A2) = l × b =20 m×15 m = 300 m2
(iii) The area of the opposite walls along the length = 2 (l × h)
=2 × 20 m × 6.5 m = 260 m2
(iv) The area of the opposite walls along the breadth = 2 (b × h)
=2 × 15 m × 6.5 m = 195 m2
(v) The area of the four walls (A3) = 2h (l + b)
= 2 × 6.5 m (20 m+15 m)
= 13 m × 35 m = 455 m2
Vedanta Excel in Mathematics - Book 9 88 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Mensuration (I): Area
(vi) The area of the room including floor and ceiling (A) = A1+A2+A3
= 300 m2 + 300 m2 + 455 m2
= 1,055 m2
Example 2: The guest room in Sandhya’s house is 10 m long, 8 m wide and 4.5 m high. It
contains two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. If
her parents wish to paste the walls with 3D paper, find the area of the walls
covered with paper. If the size of each piece of paper is 8 sq. m, how many
pieces of wallpaper are required for the room?
Solution:
Here, the length of the room (l) = 10 m
The breadth of the room (b) = 8 m
The height of the room (h) = 4.5 m
Now, the area of its 4 walls = 2h (l + b) = 2×4.5 m (10 m + 8 m) = 162 m2
Also, the area of 2 windows =2 (l1×h1) = 2 (2 m × 1.5 m) = 6 m2
The area of a door = l2×h2= 1 m × 4 m = 4 m2
∴The area of 4 walls excluding windows and door = (162 – 6 – 4) m2 = 152 m2
Hence, the area of the walls covered with paper is 152 m2.
Again, the size of each piece of wallpaper (a) = 8 m2
∴Number of paper required for the room (N)= Area of walls excluding windows and door (A)
Area of each piece of wall paper (a)
= 152 m2 = 19
8 m2
Hence, 19 pieces of 3D paper are required for pasting on the walls of the room.
Example 3: The dining room of a restaurant is 25 m long, 18 m wide and 6 m high.
It contains six windows of size 2.5 m × 2 m each and two doors of size
1.2 m × 5 m each.
(i) How many dining table sets are required to arrange on its floor at 1
table set per 15 sq. meters?
(ii) How many lamps are required to hang on its ceiling at 3 lamps per
30 sq. meters?
(iii) How many paintings are required to hang on the walls at 5 painting per
79 sq. meters?
Solution:
Here, the length of the dining room of the restaurant (l) = 25 m
The breadth of the dining room of the restaurant (b) = 18 m
The height of the dining room of the restaurant (h) = 6 m
(i) Area of floor = l×b = 25 m×18 m = 450 m2
Now, in 15 sq. meters, 1 table sets can be arranged.
or, In 1 sq. meters, 1 table sets can be arranged.
15
1
or, In 450 sq. meters, 15 × 450 = 30 table sets can be arranged.
Hence, 30 table sets are required to arrange on its floor.
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Mensuration (I): Area
(ii) Area of ceiling = l×b = 25 m×18 m = 450 m2
Also, in 30 sq. meters, 3 lamps can be hanged.
ooHrre,, nIInnce41,54s0q5.slqma.memtpeesrtsae,rr3es3,0r1e1l0qaum×ipre4sd5c0taon=hba4en5hglaaonmngpethds e.cacneibliengh.anged.
(iii) The area of its 4 walls = 2h (l + b) = 2×6 m (25 m + 18 m) = 516 m2
Also, the area of 6 windows = 6 (l1×h1) = 6 (2.5 m × 2 m) = 30 m2
The area of 2 doors = 2(l2×h2) =2 ×1.2 m × 5 m = 12 m2
∴The area of 4 walls excluding windows and door = (516 – 30 – 12) m2 = 474 m2
Also, in 79 sq. meters, 5 paintings can be hanged.
417s4q.sqm.emteertse,r7s5,975p9ai×nt4in7g4s=ca3n0
or, In be hanged. be hanged.
or, In paintings can
Hence, 30 paintings can be hanged on the walls.
Example 4: The rectangular room is 24 ft long. It contains h
three windows of size 4 ft. × 3 ft. each and a
door of size 3 ft × 8 ft. If 4 pieces of carpets each b 24 ft.
of size 12 ft × 8 ft are required for carpeting its
floor and 8 pieces of paper each of size 82.5
sq. ft. are required for pasting on its walls,
determine the height of the room.
Solution:
Here, the length of the room (l) = 24 ft.
Area of 4 pieces of carpet = 4 × (12 ft. × 8 ft.) = 384 ft2.
Now, area of floor = area of 4 pieces of carpet
or, l × b = 384 ft2.
or, 24 × b = 384
∴ b = 16 ft.
Also, the area of its 4 walls = 2h (24 +16) = 80h ft2.
Again, the area of 3 windows =3 (l1×h1) = 3 (4 ft. × 3 ft.) = 36 ft2.
The area of a door = l2×h2 = 3 ft. × 8 ft. = 24 ft2.
∴The area of 4 walls excluding windows and door = (80h– 36–28) ft2= (80h – 60) ft2.
But, area of walls excluding windows and door = area of 8 pieces of wallpaper
or, 80h – 60 = 8 × 82.5 ft2 Vedanta ICT Corner
or, 80h – 60 = 660 ft2 Please! Scan this QR code or
or, 80h = 720 browse the link given below:
∴ h = 9 ft.
Hence, the required height of the room is 9 ft. https://www.geogebra.org/m/dbjdzpej
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Mensuration (I): Area
EXERCISE 5.3
General section
1. a) What is the relation between the areas of floor and ceiling in a rectangular room?
b) Write the relation between the areas of opposite walls of a rectangular room.
2. a) From the adjoining room, write down the area of its zm
following parts.
(i) Floor (ii) Ceiling
(iii) Floor and ceiling (iv) Opposite walls along ym
the length xm
(v) Opposite walls along the breadth (vi) Four walls
(vii) Walls and floor (viii) Walls and ceiling
b) The perimeter of the floor of a rectangular room is P m and the height of the room
is Q m, what is the area of its four walls?
c) The length of a square room is x m and its height is y m, what is the area of its four
walls?
d) If the length of a cubical room is a ft., write down the area of its four walls and
ceiling.
e) In a rectangular room, if the area of the floor is A m2 and the area of its four walls
is B m2, what is the total surface area of the room?
3. a) The length of a square room is 10 m, find the area of its floor.
b) A square room is 25 ft. long. Find the area of its ceiling.
c) A cubical room is 12 m long. What is the area of its four walls?
d) If the perimeter of a rectangular room having height 4 m is 36 m, calculate the area
of its walls.
e) If 5 pieces of carpets having dimensions 10 ft. × 6 ft. of each piece are required for
carpeting the floor of a rectangular room, find the area of the floor.
f) If 60 tiles each of having dimensions 2 ft. × 2 ft. are required for paving the floor
of a rectangular room, find the area of the ceiling of the room.
g) 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.
Creative section-A
4. a) The dimensions of a room are given in the figure aside. 5m
(i) Find the area of its floor.
(ii) Find the area of its ceiling.
(iii) Find the area of opposite walls along the length. 15m 10m
(iv) Find the area of opposite walls along the breadth.
b)
(v) Find the area of its 4 walls.
(vi) Find the area of the rooms including floor and ceiling
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.
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Mensuration (I): Area
c) The length of cubical room is 15 ft.
(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.
5. a)
The area of four walls of a room is 210 m2. If the length and breadth of the room
b) are 12 m and 9 m respectively, find the height.
c) The area of four walls of a room is 639 ft2. If the breadth and height of the room
are 15.5 ft and 9 ft respectively, find the length.
A room is 24 ft. long and 16 ft. wide. If the area of its floor and ceiling is equal to
the area of four walls, find the height of the room.
6. a) A rectangular room is twice as long as it is broad and height is 4.5 m. If the area of
its 4 walls is 216 m2, find the area of the floor.
b) The length of a rectangular hall is two times of its breadth and the breadth is two
times of its height. If the area of its ceiling is 200 ft2, find the area of the four walls.
7. 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 area of walls
excluding windows and door.
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. If its walls are white washing, find the
area of walls covered with white washing.
c) Shambhu is painting the walls and ceiling of a recreation hall with length, breadth
and height 40 ft., 25 ft. and 18 ft respectively. It contains three windows of size 6
ft. × 5 ft. m each and two doors of size 4 ft. × 10 ft. How much part of the walls
and ceiling will be painted by him?
Creative section-B
8. a) The length, breadth and height of Shashwat’s classroom are 9 m, 6 m and 4.5 m
respectively. It contains two windows of size 1.7 m × 2 m each and a door of size
1.2 m × 3.5 m.
(i) Find the area of four walls excluding windows and door.
(ii) How many decorative chart papers are required to cover the walls at 2 chart
paper per 8 sq. meters?
b) A Taekwondo hall is 30 m long, 20 m wide and 5 m high.It contains four windows
of size 2.5 m × 2 m each and two doors of size 1.5 m × 4 m each.
(i) How many player's-posters are required to hang on the walls at 3 posters per
108 sq. meters?
(ii) How many taekwondo mats are required to pave on the floor at 5 mats per
15 sq. meters?
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Mensuration (I): Area
9. a) The study room of Sunayana is 8 m long, 6 m wide and 4.5 m high. It contains
two windows of size 2 m × 1.5 m each and a door of size 1 m × 4 m. If she is
papering the walls with 3D paper of size 4 m2 each, how many pieces of wallpaper
are required for her room?
b) There is an auditorium hall of length 50 m, breadth 30 m and height 10 m in
Chhiring’s school. It contains six windows of size 3 m × 2 m each and three doors
of size 2 m × 6.5 m each. If its walls and ceiling are covered with compressed
fiberboard of size 5 m2 each, find the number of fiberboard used in the hall.
10. a) The rectangular room is 20 ft long. It contains two windows of size 4 ft. × 3.5 ft.
each and a door of size 3 ft × 8 ft. If 5 pieces of carpets each of size 10 ft × 6 ft
are required for carpeting its floor and 8 pieces of paper each of size 81sq. ft. are
required for pasting on its walls, find the height of the room.
b) A rectangular room is twice its breadth. It contains four windows of size
2 m × 1.75 m each and two doors of size 1 m × 3.5 m. If 24 pieces of tiles of size
2 m × 1.5 m each are required for tiling its floor and 41 pieces of 3D paper of size
3 m2 each, are required for pasting on its 4 walls, find the height of the room.
Project work and activity section
11. a) Measure the length, breadth, and height of your bedroom. Also measure the
length and height of windows and doors using measuring tapes. Then, solve the
following problems and preset in the classroom.
(i) Find the area of the floor.
(ii) Find the area of the ceiling.
(iii) Find the area of the four walls including windows and doors.
(iv) Find the area of the windows and doors.
(v) Find the area of the four walls excluding windows and doors.
b) Make a group of 10 friends and measure the length, breadth, and height of your
classroom, its windows, doors and whiteboard and notice-board using measuring
tapes. Then, prepare a report about the following problems and present in the
classroom.
(i) The area of the floor. (ii) The area of the ceiling.
(iii) The area of the four walls including windows and doors.
(iv) The area of windows, doors and white board.
(v) The area of the walls and ceiling excluding windows and doors where the
white washing is done.
(vi) The estimated number of chart paper for papering on the walls excluding
the windows, doors and whiteboard.
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 93 Vedanta Excel in Mathematics - Book 9
Mensuration (I): Area
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. The semi-perimeter of a triangle having sides 25 cm, 15 cm and 20 cm is
(A) 60 cm (B) 30 cm (C) 45 cm (D) 15 cm
2. The amount of surface enclosed by the boundary line of a plane closed figure is its
(A) Area (B) Perimeter (C) Volume (D) Semi-perimeter
3. What is the area of triangle whose sides are a, b and c cm and semi-perimeter is s cm?
(A) s (s – a) (s – b) (s – c) cm2 (B) s(s + a)(s + b)(s + c) cm2
(C) (s – a) (s – b) (s – c) cm2 (D) s (s – a) (s – b) (s – c) cm2
4. The Heron’s formula to calculate the area of triangle is
(A) 43 a2 (B) b 4a2 – b2
4
(C) s (s – a) (s – b) (s – c) (D) 1 × b× h
2
5. The area of triangular park is 900 m2. What is the cost of covering the park with turfs
at the rate of Rs 150 per square meter?
(A) Rs 1,35,000 (B) Rs 5,13,000 (C) Rs 1,53,000 (D) Rs 3,15,000
6. If the area of an equilateral triangle is 9 3 cm2, what is its perimeter?
(A) 6 cm (B) 18 cm (C) 6 3 cm (D) 27 3 cm
7. The semi-perimeter of a triangle is 20 cm and two of its sides are 10 cm and 16 cm,
what is the third side of the triangle?
(A) 6 cm (B) 10 cm (C) 14 cm (D) 16 cm
8. In a rectangular room, which of the parts don’t have equal area?
(A) The floor and ceiling (B) The opposite walls along length
(C) The opposite walls along breadth (D) The four walls and the floor
9. A rectangular room is x m long, y m wide and z m high. What is the area of its four
walls?
(A) x (y + z) m2 (B) 2y (x + z) m2 (C) 2x (y + z) m2 (D) 2z (x + y) m2
10. The perimeter of the floor of a rectangular hall is 120 ft. and height is 12 ft. what is
area of its four walls?
(A) 132 ft2. (B) 108 ft2 (C) 1,440 ft. (D) 1,440 ft2.
Vedanta Excel in Mathematics - Book 9 94 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 6 Mensuration (II): Prism
6.1 Prism – Looking back
Classwork-Exercise
Let's study the figures given below and answer the questions as quickly as
possible.
a) (i) What is the perimeter of the rectangle? A D
(ii) What is the area of the rectangle? 9 cm
B 20 cm C
b) (i) What is the perimeter of the triangle? A
(ii) What is the area of the triangle? 3 cm 5 cm
B 4 cm C
c) (i) What is the perimeter of base of the cuboid? 10 cm
(ii) What is the area of base of the cuboid?
(ii) What is the total surface area of the cuboid? 15 cm 5 cm
(iii) What is the volume of the cuboid?
6.2 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.
(i) Area and volume of a cuboid
A cuboid has six rectangular faces. In the adjoining E F
cuboid, ABCD, BGFC, EFGH, CDEF, ADEH and ABGH D Ch
are the six rectangular faces of the cuboid.
Here, the total surface area of the cuboid is the sum H G
b
of the areas of its six faces. B
A = lb + lb + bh + bh + lh + lh Al
= 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
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 95 Vedanta Excel in Mathematics - Book 9
Mensuration (II): Prism
(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
6.3 Prisms
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 few examples of prism are rectangular prism (cuboid),
triangular prism, pentagonal prism etc.
Triangular prism Square prism Pentagonal prism
a) Cross-section of prism
(i) If a piece of biscuit is taken out from a packet of biscuits, it is the shape of
the piece of biscuit exactly the same to the shape of the biscuits remained
in the same packet?
(ii) If a 1000 rupee note is taken out from a bundle of 1000 rupees notes, is the
note congruent to the notes remained in the bundle?
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.
Vedanta Excel in Mathematics - Book 9 96 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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