math

Example 3: The profit of a company is Rs. 95,50,40,000. The bonus is distributed at
Solution: the rate of 9% equally among certain number of employees. If each
employee received Rs. 2,14,884, what is the number of employees?

Here,

Profit = Rs. 95,50,40,000

Bonus rate = 9%

Bonus received by each = Rs. 2,14,884

Number of employees = x (say)

Total Bonus = Rs. 214884x

Then,

Bonus = 9% of 95,50,40,000

or, 214884x = 9 × 95,50,40,000
100

 x = 400.

Hence, the number of employees is 400.

Example 4: The profit of a company is Rs 82,50,000. The bonus is distributed to the
employees of the company according to their salary as follows.

Monthly salary scale Bonus rate
Rs. 5000 – Rs 10,000 0.125%

Rs. 10,000 – Rs 20,000 0.25%

Rs. 20,000 – Rs 30,000 0.5%

(i) What is the bonus amount received by an employee whose salary is
Rs. 8500?

(ii) What is the bonus amount received by an employee whose salary is
Rs. 25000?

Solution: Here,

Profit = Rs. 82,50,000

(i) Bonus received by an employee whose salary is Rs 8500 = 0.125% of profit

= 0.125 × 82,50,000
100

= Rs. 10,312.50

(ii) Bonus received by an employee whose salary is Rs 25000 = 0.5% of 82,50,000

= 0.5 × 82,50,000
100

= Rs. 41,250

44 | Mathematics - 9 Arithmetic

Exercise 2.5

1. (a) A development bank makes a profit of Rs. 10,00,00,000. The management decided to
give 8% bonus of profit equally among 250 employees. Find the bonus amount that
each employee will receive.

(b) A business company made a profit of Rs. 20,50,00,000. It is decided to give a bonus at
the rate of 9% equally among 450 employees. What is each employee’s bonus amount?

(c) A finance company made a profit of Rs. 30,75,00,000. A bonus of 10% profit is
distributed equally among 150 shareholders. What is the bonus amount that each share
holder will receive?

(d) A bank made a profit of Rs. 90,00,00,000. The management decided to distribute 10% bonus
equally among its 250 employees. What is the bonus amount that each employee will receive?

2. (a) A business company made a profit of Rs. 80,00,00,000. There were 400 employees and
each received Rs 1,60,000 as a bonus. Find the bonus rate.

(b) There is a profit of Rs. 95,00,00,000 of a business firm. The firm distributed bonus
equally among 500 employees and each received Rs. 2,28,000. What is the bonus rate?

(c) A company decided to distribute bonus at the rate of 12% of the net profit
Rs. 94,50,00,000 equally among its employees. If each employee received Rs 2,52,000,
what is the number of employees?

(d) Each employee of a company received Rs. 3,20,125. If the company has distributed
bonus at the rate of 13% of profit Rs. 98,50,00,000, then, what is the number of
employees?

3. (a) A company made a profit of Rs. 2,50,00,000. Bonus is distributed among the employees
according to their salaries.

Monthly salary scale Bonus rate

Rs. 5000 – Rs 8,000 0.125%

Rs. 8,000 – Rs 12,000 0.25%

Rs. 12,000 – Rs 20,000 0.5%

(i) What is the bonus received by an employee whose salary is Rs. 6500?

(ii) What is the bonus received by an employee whose salary is Rs. 15000?

(b) A bank made a profit of Rs. 4,00,00,000. The employees received bonus according to
their salaries as follows.

Monthly salary scale Bonus rate

Rs. 4000 – Rs 8,000 0.25%

Rs. 8,000 – Rs 12,000 0.5%

Rs. 12,000 – Rs 16,000 1%

(i) Find the bonus received by an employee whose salary is between Rs. 8000 –
Rs 12000.

(ii) Find the bonus of an employee whose salary is in between Rs. 12000 –
Rs. 16000.

Arithmetic Mathematics - 9 |45

2.6 Share and Dividend

A firm: An association formed by few persons to carry a business investing small amount of capital is
called a firm.

A company: A firm with large amount of capital and numerous persons as partners is called a
company.

Shares: 1. Whenever more than one persons or firms invest time or money jointly in a profit making
service or business, the persons or firms are share holders and the profit made in divided among
themselves in proportion to their work or time or money invested. The amount received by each is
called dividend. (share over profit)

Example 1: As proprietors of a school, Himesh, Dipesh and pramesh have shares of
Solution: Rs 25,00,000, Rs 20,00,000 and Rs 15,00,000 respectively. In a certain
period the school made a net profit of Rs 30,00,000. Which they decided
to divide among themselves. Find each's share in the profit.

Himesh's share = Rs 25,00,000

Dipesh's share = Rs 20,00,000

Pramesh's share = Rs 15,00,000

 Their share in the ratio 25,00,000:20,00,000:15,00,000 = 5:4:3

Total profit made = Rs 30,00,000

Now Himesh's share in the profit = 5 × Rs 30,00,000
5+4+3

= 5 × Rs 30,00,000
12

= Rs 12,50,000

Dipesh's share in the profit = 4 × Rs 30,00,000
5+4+3

= 4 × Rs 30,00,000
12

And Pramesh's share in the profit = 3 × Rs 30,00,000
5+4+3

= Rs 7,50,000

Therefore, Himesh, Dipesh and Pramesh get Rs 12,20,000 Rs 10,00,000 and Rs
7,50,000 respectively.

2. When the capital of the business is beyond the financial capacity of the company/firm, a public
company of many partners are formed and required capital is divided into small units called shares of
nominal value. The nominal value of a share is not fluctuable. Obviously the company promoters
own a big portion of the shares and the remaining shares are divided among the public. A person who
purchases one or more shares is a share holder who is an owner of the company.

Dividend: At the end of a fiscal year, the company declares profit called dividend. The dividend is
expressed in terms of percentage of the nominal value. A share holder receives his/her dividend in
proportion to their investment.

46 | Mathematics - 9 Arithmetic

Market value: Once share holders buy shares or stocks of company/companies, they can keep the
shares always with them or sell them to other person who want to buy the shares. But the company
doesn't pay for the shares and take them back. Selling and buying of shares or transfer of shares can
take place only through a stock broker in stock exchange market.

Selling price of a share is called market value or quoted value, which fluctuates time to time. If the
marked value is equal to the nominal value (face value) it is said to be at par. If the market value is
greater than the nominal value, it is said to be at premium or above par and if the market value is less
than the nominal value, it is said to be at discount or below par. Shares own by a person or group
from one or more than one companies is called a stock or share market.

Example of shares in share market is shown below:

Nepal Stock Exchange

Transaction of 9 Kartik, 2073

Company Name Cost of Final Cost Transaction Difference
previous day Number -8
Agriculture 3856
Development Bank 653 645 -4
Api Power Co. Ltd. 1991 -51
Everest Bank Ltd. 590 586 14256
Total Transaction Cost 3601 3550
713,272,506 Transaction Number 573

Example 2: Find the dividend for 50 shares of Rs. 100 each at 8% rate of dividend.

Solution: Here,

Number of shares = 50

Nominal value of a share = Rs. 100.

Rate of dividend = 8%

 Nominal value of 50 shares = 50 x Rs. 100

= Rs. 5000

Now, dividend = 8% of Nominal value

= 8 × Rs. 5000
100

= Rs. 400

Note: Dividend is always taken as percentage of nominal value.

Example 3: The price of Rs. 100 shares is Rs. 150 and the company declares a
Solution: dividend of 15%. Find
(i) How many shares can be purchased for Rs. 18000?
(ii) Dividend.
(iii) Rate of interest on the investment.
Here,
Market value of a share = Rs. 150
Investment = Rs. 18000

Arithmetic Mathematics - 9 |47

i. Number of shares = Investment = Rs. 18000 = 120
Market Value Rs. 150

Nominal value of a share = Rs. 100

Nominal value of shares = 120 × Rs. 100 = Rs. 12000

ii.. Dividend = 15% nominal value

= 15 × Rs. 12000
100

= Rs. 1800

iii. Rate of interest (return on the investment)

= Dividend × 100%
Investment

= Rs. 1800 × 100% = 10%
Rs. 18000

Example 4: A company declares a dividend of 12% on the shares of face value
Solution: Rs. 100. A man buys some shares and gets 15% in his investment. Find
out for what price he bought the share.

Here,

Nominal value of a share = Rs. 100

Dividend in a share = 12% of nominal value= 12 × Rs. 100) = Rs. 12
100

Let the market price of a share = Rs. x.

Then, dividend = 15% of investment

Rs. 12 = 15 × x
100

 x = Rs. 80

Example 5: What is the interest percentage on a capital invested in 18% share when
Rs. 10 share cost Rs. 12?

Solution: Here,

Nominal value of a share = Rs. 10

Dividend = 18% of nominal value

= 18 × Rs. 10
100

= Rs. 1.8

Market value of a share = Rs. 12

 Interest = Dividend × 100%
Market value

= Rs. 1.8 × 100%
Rs. 12

= 15%

Note: Dividend distributed at the end of a year is equivalent to the interest on the investment i.e. market value.

48 | Mathematics - 9 Arithmetic

Exercise 2.6

1. (a) A village security committee appointed three persons A, B and C for night time security
of the village from 6 pm to 6 am. It A, B and C watch for 3 hours, 4 hours and 5 hours
respectively each day, and the committee provides Rs 24,000 per month for the security.
Find their salaries.

(b) A, B and C can do a work in 4 days, 5 days and 10 days respectively. Divide Wages of
Rs 33,000 of a certain week among them.

(c) Dinesh, Dipesh and Digdarshan invested Rs 10,00,000, Rs 15,00,000 and Rs 25,00,000
respectively to run a business and made a profit of Rs 12,00,000 in a certain year. Find
how much will the dividend be each's share.

(d) Investments of X, Y and Z are 45%, 25% and 30% respectively in a business form. Find
their dividends from the total profit of Rs 5,00,000.

2. (a) A development bank made a net profit of Rs 40,00,00,000. If the bank distributes 8% of
the profit to its share holders as dividend, find the total dividend.

(b) A company earned Rs 6,00,00,000 net profit in a year. It the company distributed Rs
34,80,000 to its share holders as dividend, what was the rate of dividend?

(c) A share holder of a hydro-electric power company owns certain number of shares of
nominal value Rs 100 each. If the company decided to distribute 20% dividend and the
share holder received Rs 1,00,000 as dividend, find the number of shares belonging to
the share holder.

(d) A share holder of a company owns 2,000 shares the company declares to distribute 12%
dividend for the year and the share holder received Rs 24,000 as dividend what is the
nominal value of a share.

3. (a) If the shares in a company stands at Rs. 118 each, how many shares can be bought for
Rs. 14160?

(b) A man invests Rs. 1,27,500 to purchase shares of a bank at Rs. 425 each. He sells them
when the share value becomes Rs. 450 each. What is his income?

(c) When a Rs. 100 shares costs Rs. 240 in the share market, Arnav invests Rs. 33600 to
buy shares of a company. At the end of the fiscal year, the company pays 13% dividend.
Find his income of that year from shares.

(d) Hari invests Rs. 1,20,000 in buying Rs. 100 shares. If the nominal value of the shares is
Rs. 75000. Find the market value of each share.

4. (a) A man invests Rs. 2880 to buy shares of a company at Rs. 90 each and sells them at
Rs. 120 each. Find his profit percentage.

(b) What is the percentage interest on the capital invested in 18% share, when Rs. 100 share
costs Rs. 120?

(c) Mrs. Gharti buys a Rs. 300 share in a company which pay dividend of 10% which she buys
at such a price that her profit is 20% of her investment. At what price is the share bought?

(d) A man buys some Rs. 100 shares when the market value is Rs. 250 per share for
Rs. 30,000 and makes a profit of 6.4% in his investment. What is the percentage
dividend allowed?

Arithmetic Mathematics - 9 |49

5. (a) A company declares a dividend of 15% and Rs. 100 share which is quoted at Rs. 240.

(i) Find how many shares can be purchased with Rs. 50,400.
(ii) What is the dividend?
(iii) What is the rate of interest on the investment?

(b) A company declares a dividend of 20% on a Rs. 100 share which is quoted at Rs. 150.

(i) What is the total cost of 100 shares?
(ii) What is the annual income of the shares?
(iii) What is the percentage return on the investment?

(c) A jobber buys Rs. 20 shares in a company which pays 12% dividend by investing
Rs. 12800. If his income is Rs. 960, find

(i) The market value of each share
(ii) The number of shares bought.
(iii) The nominal value of the shares.

(d) Imran invests RS. 3,00,000 in 15% Rs. 110 shares at Rs. 120. When the shares rise to
Rs. 130, he sells out enough shares to purchase a mobile set for Rs. 39000. Find

(i) The number of shares he still holds.
(ii) His loss in annual income.

2.7 Home Arithmetic

2.7.1 Electricity Billing Arithmetic

1 unit = 1kilowatt hour = 1000 watt hour. It means, if an electric
appliance of 1000 watt is used for 1 hour, it consumes 1 unit of
electricity. Thus, consumption of electricity is measured in number of
units.

How to calculate the units of electricity consumed? Say, the reading of
meter for the month of Baisakh is 2050 and that of Jestha is 2090
units. Units of electricity consumed in the month of Jestha = Reading
of Jestha (Final reading) - Reading of Baisakh (Initial reading)

= 2090 – 2050 = 40

Thus, units consumed = Final reading - Initial reading. Here, final
reading means units shown at the current month and initial reading
means units shown in previous month.

Charge rate of Electricity Authority is as follows.

(i) Charge for the 1st 20 units at the rate of Rs. 4 per unit is Rs. 80.

(ii) Charge for 21 units - 250 units is Rs. 7.30 per unit. It means
charge for next 230 units is at the rate of Rs. 7.30 per unit.

(iii) Charge for the units above 250 units is Rs. 9.90 per unit.

(iv) 3% rebate up to 7 days from the day of meter reading.

(v) From 8th day to 22nd day from the day of reading, no rebate.

50 | Mathematics - 9

(vi) From 23rd day to 30th day from the day of meter reading, 5% fine is charged.
(vii) From 31st day to 40th day from the day of meter reading, 10% fine is charged.
(viii) From 41st day to 60th day from the day of meter reading, 25% fine is charged.

Worked Out Examples

Example 1: The reading of the month, Baisakh is 2058 and that of the month Jestha
Solution: is 2208. What is the electricity charge for the month of Jestha?

Rules of Electricity Authority are as follows.

Minimum charge (upto 20 units) = Rs. 80

Cost per unit (21 - 250 units) = Rs. 7.30

Here,

Final reading = 2208 units

Initial reading = 2058 units

Units consumed = 2208 units – 2058 units = 150 units = 20 units + 130 units.

Cost of 1st 20 units = Rs. 80

Cost of next 130 units = Rs. (130 × 7.3) = Rs. 949

Total charge = Rs. (80 + 949) = Rs. 1029

Hence, the charge of electricity for the month of Jestha is Rs. 1029.

Arithmetic Mathematics - 9 |51

Example 2: The charge for the 1st 20 units is Rs. 80. The charge for the units from
Solution: 21 - 250 units is Rs. 7.50 per unit and the charge for the units above 250
Example 3: units is Rs. 9.90 per unit. What is the cost of 340 units of electricity?
Solution: Here,
Units consumed = 340 units = (20 + 230 + 90) units.
Example 4: Cost of 1st 20 units = Rs. 80
Solution: Cost of next 230 units = Rs. 230 × 7.50 = Rs. 1725
Cost of next 90 units = Rs. 90 × 9.9 = Rs. 891
Example 5: Hence, the total cost of 340 units of electricity = Rs. (80 + 1725 + 891) = Rs. 2696
Solution:
The charge of electricity for the first 20 units is Rs. 80 and the charge
from 21 to 250 units is calculated at the rate of Rs. 7.30 per unit. If the
total charge of electricity was Rs. 810, how many units of electricity was
consumed?

Here,
Cost of 1st 20 units = Rs. 80
Let, the extra units be x
Then, cost of x extra units = Rs. 7.30x
Total cost = 80 + 7.30x
or, 810 = 80 + 7.30x
or, 730 = 7.30x
or, x = 100units.

 Total units consumed = 100 + 20 = 120

The reading of Jestha is 4010 and that of Ashad is 4130. The cost of 1st
20 units is Rs. 80. The charge for the units from 21 - 250 units is Rs. 7.30
per unit. The meter reading is done on 30th Ashad. There is rebate of 3%
if payment is made on the 1st 7days from meter reading. How much is the
cost if payment is done on 2nd Shrawan?

Here,
Final reading = 4130 units
Initial reading = 4010 units
Units consumed = 4130 units – 4010 units = 120 units = 20 units + 100 units.
Cost of 1st 20 units = Rs. 80
Cost of next 120 units = Rs. (100 × 7.30) = Rs. 730
Total cost = Rs. (80 + 730) = Rs. 810
Rebate = 3% of 810 = Rs. 24.30

 Cost after rebate = Rs. (810 - 24.30) = Rs. 785.70

A house hold consume 71 units electricity in a math . Using the new
tariff given above of 5A : find the total bill of electricity of the household.

Here,

Total consumed units = 71

Service charge = Rs 100

Energy charge = 20 × 3 + 10 × 7 +20 ×8.50 + 21 × 10

52 | Mathematics - 9 Arithmetic

= 60 + 70 + 170 + 210

= Rs 510

Total bill = Service charge + energy charge

= 100 + 510

= Rs 610

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Example 6: The reading of 1st kartik is 9,345 units and reading on 1st mangsir is 9,660
Solution: units . As per rule of new tariff of NEA related to 5A, find the to electricity
bill of the household.

Here,
Consumed units 1st kartik = 9345 units
Consumed units 1st mangsir = 9660 units

Total consumed units = 9660 – 9345 = 315 units.

Total service charge = Rs 150

Energy charge = 20 × 3 + 10 × 7 + 20 × 8.5 + 100 × 10 + 100 × 11 + 65 × 12

= 60 + 70 + 170 + 1000 + 1100 + 780

= Rs 3180

Total charge = 150 + 3180 = Rs 3330

Exercise 2.7.1

1. (a) The reading of the month, Jestha is 4012 and that of the month, Ashad is 4142. The
electricity charge up to 20 units is Rs. 4 per unit and Rs. 7.30 per unit from 21 to 250
units. What is the charge for electricity?

(b) The minimum charge for the 1st 20 units is Rs. 80 and Rs. 7.50 per unit from 21 to 250
units. What is the charge for 180 units of electricity?

(c) The minimum charge for the 1st 20 units is Rs. 4 per unit. What is the charge for 18
units of electricity?

(d) The minimum charge for the 1st 20 units is Rs. 4 per unit. What is the charge for 10
units of electricity?

2. (a) The charge for the 1st 20 units is Rs. 80. The charge for the units from 21 - 250 is
Rs. 7.50 per unit and the charge for the units above 250 units is Rs. 9.90 per unit. What
is the charge for 300 units of electricity?

Arithmetic Mathematics - 9 |53

(b) The charge for the 1st 20 units is Rs. 80. The charge for the units from 21 - 250 is
Rs. 7.50 per unit and the charge for the units above 250 units is Rs. 9.90 per unit. What
is the charge for 350 units of electricity?

(c) The minimum charge for the 1st 20 units is Rs. 80. The charge for the units from 21 -
250 is Rs. 7.50 per unit. If a customer paid Rs. 1055 in a certain month, how many units
of electricity was consumed?

(d) The minimum charge for the 1st 20 units is Rs. 80. The charge for the units from 21 -
250 is Rs. 7.50 per unit. If a customer paid Rs. 1205 in a month, how many units of
electricity was consumed?

3. (a) The reading of Ashad is 4130 and that of Shrawan is 4330 units. The cost of 1st 20 units is
Rs. 80. The charge for the units from 21 - 250 is Rs. 7.50 per unit. The meter reading is done
on 30th Shrawan. There is a rebate of 3% if the payment is made on the 1st 7days from the day
of meter reading. How much does the customer have to pay if the payment is done on 4th
Bhadra?

(b) The reading of Shrawan was 4330 units and the reading of Bhadra was 4580 units. The
cost of 1st 20 units was Rs. 80 and the charge for the units from 21 - 250 units was
Rs. 7.30 per unit. The customer got 3% rebate as he paid on 1st 7 days from the day of
meter reading. How much was the cost for electricity?

(c) The electricity charge upto 20 units is Rs. 3.90 per unit and Rs. 6.50 per unit from 21 to
250 units. How much should the electricity charge be paid for 85 units?

(d) According to the rate of electricity charges the minimum cost upto 20 units is Rs. 78
and cost per unit from 21 units to 250 units is Rs. 6.50. How much does 110 units cost?

4. (a) The minimum charge for the first 20 Units is Rs. 3 per unit and service charge is Rs. 30.
What is the charge for 15 units of electricity?

(b) The reading of the Month, Ashwin 2073 is 3125 units and that of the month, Kartik
2073 is 3200 units. What is the electricity charge for the month of Ashwin 2073
according to the new billing system?

5. The charge for the first 20 units is Rs. 3 per unit. The charge for the units from 21-30 is Rs. 7
per unit, charge for the units 31-50 is Rs. 8.50 per unit, charge for the units 51-150 is Rs. 10
per unit and charge for the units 151-250 is Rs. 11 per unit. What is the cost of 230 units of
electricity if the service charge is Rs. 125?

6. The charge for the first 20 units is Rs. 3 per unit. The charge for the units from 21-30 is Rs. 7
per unit, charge for the units 31-50 is Rs. 8.50 per unit and the charge for the units 51-150 is
Rs. 10 per unit. The total charge of electricity was Rs. 1100, how many units of electricity was
consumed if the service charge was Rs. 100.

7. The reading of Mangshir 2073 is 5370 units. The cost of first 20 units is Rs. 3 per unit, charge
for the unit (21-30) units is Rs. 7 per unit, charge for the unit (31-50) is Rs. 8.50 per unit,
charge for the units (51-150) is Rs. 10 per unit and charge for the unit (151-250) is Rs. 11 per
unit and service charge is Rs. 125. The meter reading is done on Mangshir 30, 2073 is
Rs. 5545. There is rebate of 3% if payment is made on the first 7 days from the day of meter
reading. how much does the customer have to pay if the payment is done on 3rd Poush?

8. The reading of 1st Magh is 10,450 Units and reading on 30th Magh is 10,630 units. NEA charges 5%
fine if payment is made after 23 days from the day of meter reading. How much does the customer
have to pay if the payment is done on 25th Falgun according to new billing system?

54 | Mathematics - 9 Arithmetic

2.7.2 Water Billing

A house may have water facility with a tap with a meter or a tap without a meter. If there is no meter
connected with a tap, the customer has to pay fixed amount fixed by water supply corporation.

Water bill is prepared on the basis of quantity of water used by consumer as shown by the meter. 50%
of bill is charged as sewerage charge and Rs. 5 as miscellaneous charge.

Worked Out Examples

Example 1: Observe the meter readings of water supply of a certain house.

Months Baisakh Jestha

Units 4012 4020

Solution: The sewerage charge is 50% of water bill and Rs. 6 is the miscellaneous
charge. There is a rebate of 3% if the payment is made within the 2nd
week of next month. Find the charge that the consumer has to pay for
Jestha, if he pays on 3rd of Ashad. The cost upto 8 units is Rs. 26.
Here,
For the month of Jestha
Initial reading = 4012 units
Final reading = 4020 units
Units consumed = 4020 units – 4012 units = 8 units

Cost of 8 units = Rs. 26
Total cost with sewerage charge and miscellaneous charge = 26 + 50% of 26 + 6

= Rs. 45
But the payment is made on the 1st week of next month, so there is a rebate of 3%.

Charge after rebate = Rs. (45 - 3% of 45) = Rs. 43.65
Hence, the consumer has to pay Rs. 43.65

Example 2: Observe the meter readings of water supply of a certain house.

Months Bhadra Asoj

Units 4031 4040

Arithmetic Mathematics - 9 |55

Solution: The cost upto 9 units is Rs. 70. The sewerage charge is 50% of bill and
Rs. 10 is miscellaneous charge. If the payment of a bill is made after the
2nd week and within the next month, a customer doesn’t get rebate and
he/she is not fined. Find the charge for the month of Asoj, if the
customer pays on 25th Kartik.
Here, for Asoj

Initial reading = 4031 units

Final reading = 4040 units

Units consumed = Final reading - Initial reading = 4040 units – 4031 units = 9 units.

Cost of 9 units = Rs. 70

Cost with sewerage charge and miscellaneous charge = Rs. (70 + 50% of 70 + 10)

= Rs. 115
As payment is made on 25th Kartik, the customer neither gets rebate nor fined. So, the
customer has to pay Rs. 115.

Example 3: Observe the water meter reading of a certain household.

Months Shrawan Bhadra

Units 2012 2022

Solution: The charge upto 10 units is Rs. 50. The sewerage charge is 50% of bill
and Rs. 6 is miscellaneous charge. If the payment of a bill is made in the
third month, a customer is fined 10%. What is the charge if the customer
pays on 25th Mangsir?
Here, for Bhadra
Initial reading = 2012 units
Final reading = 2022 units
Units consumed = Final reading - Initial reading

= 2022 units – 2012 units = 10 units.
Charge with sewerage charge and miscellaneous charge = Rs. (50 + 50% of 50 + 6)

= Rs. 81
Fine = 10% of 81 = Rs. 8.1

Charge after fine = Rs. 81 + Rs. 8.10 = Rs. 89.10
Hence, the customer has to pay Rs. 89.10

Exercise 2.7.2

1. (a) Observe the water meter readings of a certain house.

Months Baisakh Jestha
Units 3000 3008

The cost upto 8 units is Rs. 40. 50% of water bill is charged as sewerage charge and

Rs. 10 as miscellaneous charge. There is a rebate of 3% if the payment is made within
the 2nd week of next month. Calculate the charge for the month of Jestha, if the payment
is made on the 2nd of Ashad.

56 | Mathematics - 9 Arithmetic

(b) Read the water meter readings given below:

Months Jestha Ashad
Units 3008 3017

The cost upto 9 units is Rs. 50. 50% of water bill is charged as sewerage charge and
Rs. 10 as miscellaneous charge. There is a rebate of 3% if the payment is made within
the 2nd week of next month. Calculate the charge for the month of Ashad if the payment
is made on the 3rd Shrawan.

(c) Observe the table given below, which shows water meter readings.

Months Ashad Shrawan
Units 3017 3027

The cost upto 10 units is Rs. 60. 50% of water bill is charged as sewerage charge and
Rs. 15 as miscellaneous charge. There is a rebate of 3% if the payment is made on the
2nd week of next month. Calculate the charge for the month of Shrawan, if the payment
is made on the 1st Bhadra.

(d) Observe the meter readings of water supply of a certain house hold.

Months Shrawan Bhadra
Units 3027 3037

The sewerage charge is 50% of water bill and Rs. 7 is the miscellaneous charge. The
cost upto 10 units is Rs. 60. There is a rebate of 3% if the payment is made on the 2nd
week of next month. Find the charge that the consumer has to pay for the month of
Bhadra, if he pays on the 4th of Ashoj.

2. (a) Observe the water readings.

Months Baisakh Jestha
Units 4012 4020

The cost upto 8 units is Rs. 80. The sewerage charge is 50% of water bill and Rs. 8 is
the miscellaneous charge. If the payment is made after the 2nd week and within the next
month there is neither rebate nor fine. Find the charge that the customer has to pay for
Jestha if the payment is made on the 2nd of Shrawan.

(b) Observe the table given below of water meter readings.

Months Jestha Ashad
Units 4020 4029

The charge upto 9 units is Rs. 90. The sewerage charge is 50% of water bill and Rs. 10
is the miscellaneous charge. If the payment is made after the 2nd week and within the
next month, there is neither rebate nor fine. Find the charge for the month of Ashad, if
the customer pays on the 28th Shrawan.

(c) Look at the meter readings of water supply of a certain house.

Months Ashad Shrawan
Units 4029 4039

The charge upto 10 units is Rs. 100. The sewerage charge is 50% of water bill and
Rs. 10 is the miscellaneous charge. If the payment is made after the 2nd week and within
the next month, there is neither rebate nor fine. Find the charge for the month of
Shrawan, if the customer pays on the 25th Bhadra.

Arithmetic Mathematics - 9 |57

(d) In the month of Bhadra, the water consumed by a family is 10 units. The charge upto 10
units is Rs. 60. The sewerage charge is 50% of water bill and Rs. 6 is the miscellaneous
charge. If the payment of a bill is made after the 2nd week and within the next month a
customer doesn’t get rebate and he/she is not fined. Find the charge for the month of
Bhadra which the customer pays on the 1st of Kartik.

3. (a) The reading of water meter for Shrawan was 2030 units and that of Bhadra was 2042
units. The charge upto 10 units is Rs. 50 and Rs. 5 for each extra unit. If the sewerage
charge is 50% of the bill and miscellaneous charge is Rs. 5, Calculate the charge if the
payment is made on time.

(b) The reading of water meter for Bhadra was 2042 units and that of Asoj was 2050 units.
The charge upto 8 units is Rs. 40. The sewerage charge is 50% of bill amount and Rs. 5
for miscellaneous charge. If the payment was made after the 4th month then 20% fine
was charged. What was the charge for Asoj, if the customer paid on the 4th Falgun?

(c) Observe the reading of water meter of a certain house given below:

Months Asoj Kartik
Units 5012 5025

The charge upto 10 units is Rs. 50 and Rs. 5 for each extra unit. The sewerage charge is
50% of the bill and miscellaneous charge is Rs. 6. If the payment is done after the 5th

month, 50% fine is charged. What is the charge for the month of Kartik if the payment
is made on the 2nd Baisakh of next year?

(d) The reading of a water meter in the month of Baisakh is 2560 units and that in the

month of Jestha is 2570 units. The charge upto to 10 units is Rs. 70. The sewerage
charge is 50% of bill and miscellaneous charge is Rs. 7. The customer paid after the 5th

month so he is charged 50% fine. What was the charge of water for the month of Jestha?

2.7.3 Telephone Billing

For the use of telephone, a person has to pay money. At present, on a land line 2 minutes call is
considered as a call and charges Re. 1. On it, telecom service charge (TSC) is added. After that, on the
total, VAT is added which gives the grand total which the customer has to pay.

58 | Mathematics - 9 Arithmetic

Worked Out Examples

Example 1: Raj made a call of 6 minutes in business hours. The time from 08:00 -
Solution: 18:00 hrs. is known as business hours. In business hours a call of 2
minutes is considered as 1 call. Find the charge if 10% TSC is imposed
on it and then 15% VAT. The charge for 1 call is Re. 1.

Here,

Time of call = 6minutes.

1 call is of 2 minutes.

No. of calls = 6 = 3
2

Cost of 3 calls = Rs. 3 × 1 = Rs. 3

TSC = 10% of Rs. 3 = Rs. 0.30

Total = Rs.(3+ 0.30) = Rs. 3.30

VAT = 15% of Rs. 3.30 = Rs. 0.50

Grand total = Rs. (3.30 + 0.50) = Rs. 3.80

Hence, the charge is Rs. 3.80.

Example 2: The rental call for a month is 200 calls. If Kabita made 198 calls, what is
Solution: the charge that Kabita has to pay after the imposition of 10% TSC and
13% VAT, where the charge per call is Re. 1?
Here,

Rental calls mean minimum calls that the customer has to pay.

Rental calls 200 calls mean, the customer has to pay the charge of 200 calls if he
makes calls less than 200.

Charge with TSC = Rs. 200 + 10% of Rs. 200 = Rs. 220

Then, Total charge with VAT = Rs. 220 + 13% of Rs. 220 = Rs. 248.60

Example 3: The time from 18:00 - 22:00 hours and from 06: 00 - 08:00 hours is known
Solution: as normal hours. In normal hours, a call of 4 minutes is considered as a
single call. Sangita talked with her daughter from 06:02 - 06:58 p.m. What
is the charge that Sangita has to pay after 10% TSC and 13% VAT, where
charge per call is Re.1?

Here,

Time of call = 06:58 - 06:02 = 56minutes

Number of calls = 56 = 14
4

Charge for 14 calls = Rs 14 × 1 = Rs. 14

Charge with TSC = Rs. 14 + 10% of Rs. 14 = Rs. 15.40

Charge with VAT = Rs. 15.40 + 13% of Rs. 15.40 = Rs. 17.40

Hence, the total charge is Rs. 17.40.

Arithmetic Mathematics - 9 |59

Example 4: The time from 22:00 - 06:00 hours is known as off hours. In off hours a
Solution: call of 8 minutes is known as 1 call. Dilasha made a call to her friend
from 11:00 - 11:56 p.m. what is the charge after 10% TSC and 13% VAT if
the charge per call is Re. 1?

Here,

Time of call = 11:56 - 11:00 = 56 minutes

No. of calls = 56 = 7
8

Charge for 7 calls = Rs. 7 × 1 = Rs. 7

Charge with TSC = Rs. (7 + 10% of 7) = Rs. 7.70

Charge with VAT = Rs. (7.70 + 13% of 7.70) = Rs. 8.70

Exercise 2.7.3

1. (a) In an office, the total calls of a day is 90 minutes in business hours. The time from
08:00 - 18:00 hours is known as business hours. In business hours a call of 2 minutes is
considered as a single call and charge is Re.1 per call. Find the charge that the office has
to pay after 10% TSC and 13% VAT.

(b) A travel agency made calls of 240 minutes in a certain day in business hours. The time from
08:00 - 18:00 hours is known as business hours and a call of 2 minutes is considered as a
single call. If the charge is Re. 1 per call, find the total charge after 10% TSC and 13% VAT.

(c) The rental call per month is 200 calls. If a man has made 190 calls in a certain month,
find the charge in a month after 10% TSC and 13% VAT where the charge is Re. 1/call.

(d) The rental call per month is 200 calls. If Raj has made 250 calls in a month, find the
charge in a month after 10% TSC and 13% VAT where the charge for rental call as well
as for extra call is Re. 1/call.

2. (a) The time from 18:00 - 22:00 hours and from 06:00 - 08:00 hours is known as normal
hours. In normal hours, a call of 4 minutes is considered as a single call. Mamata talked
with her friend from 07:20 - 08:00 p.m. what is the charge that Mamata has to pay after
10% TSC and 13% VAT where charge per call is Re. 1?

(b) In normal hours, Raj talked with his friend for 136 minutes. What is the charge after
10% TSC and 13% VAT where charge per call is Re. 1?

(c) The charge for certain number of calls with 10% TSC and 13% VAT is Rs. 79.552. Find
the number of calls if the charge per call is Re. 1.

(d) The charge for certain number of calls with 10% TSC and 13% VAT is Rs. 149.16. Find
the number of calls if the charge per call is Re. 1.

3. (a) The time from 22:00 - 06:00 hrs is known as off hours. In off hours, a call of 8 minutes
is considered as 1 call. Jay talked with his friend from 10:02 - 11:14p.m. What is the
charge after 10% TSC and 13% VAT if the charge per call is Re. 1?

60 | Mathematics - 9 Arithmetic

(b) In off hours, Jyoti made a call to her friend from 10:00 - 12:00 midnight. What is the
charge after 10% TSC and 13% VAT if the charge per call is Re. 1?

(c) In off hours, Nayan made a call of 3 hours. What is the charge after 10% TSC and 13%
VAT if the charge per call is Re. 1?

(d) In off hours, Suryaa made a call of 320 minutes. What is the charge after 10% TSC and
13% VAT if the charge per call is Re. 1?

2.7.4 Taxi Fare

Following are the rates of fare in taxi meter implemented by Nepal bureau of standards and
metrology.

Taxi Time Minimum fare Fare of 200m

Day time taxi 6:00Am to 9 : 00 Pm Rs14 Rs7.40

Night time taxi 9 : 00 Pm to 6: 00 Am Rs 21 Rs11.10

Waiting charge Rs 7.40 per 2 minutes
Day time taxi

Night time taxi Rs 11.10 per 2 minutes.

Worked Out Example

Example 1: Ram takes a taxi to travel 5 kilometers during day time. Find the fare he
Solution: has to pay with 3 minutes waiting.
Here,
Time of travel is day time.
Distance travelled = 5 km = 5 × 1000m. = 5000m.
Rate of fare = Rs. 7.40/200m
Waiting time = 3 minutes, which is equivalent to 4 minutes
Rate of wating charge = Rs. 7.40/2 minutes
Now,
Charge = Minimum charge + Fare of 5000 m + Waiting charge

= Rs. 14 + 5000 m × Rs. 7.40/200 m + 4 minutes × Rs. 7.40/2 minutes
= Rs. 14 + Rs. 25 × 7.40 + Rs. 2 × 7.40
= Rs. 213.80

 Total charge (fare) to be paid = Rs. 213.80.

Example 2 : Sailaja travelled a distance of 10km. before 9pm. and 8km. afterwards by
taxi. Find the total fare she has to pay with 5 minutes waiting time while
getting off.

Solution: Here,

Time before 9pm. is considered as day time and time after 9pm. is considered as night time.

Arithmetic Mathematics - 9 |61

Distance travelled during day time = 10km. = 10 × 1000 m.
= 10000 m.

Taxi Fare for day time journey at the rate of Rs. 7.40 per 200m.
= 10000 m × Rs. 7.40/200 m = Rs. 370

Distance travelled in night time = 8km. = 8 × 1000 m. = 8000 m.
Taxi Fare for night time journey at the rate of Rs. 11.10 per 200 m.

= 8000 m × Rs. 11.10/200 m
= Rs. 444
As she started travelling during day time, minimum charge = Rs. 14
Waiting time = 5 minutes, which is equivalent to 6 minutes
Waiting Charge = 6 minutes × Rs. 11.10/ 2 minutes (Night Time)
= Rs. 33.30
 Total Fare = minimum charge + fare of day time journey + fare of night time
journey + waiting charge = Rs. 14 + Rs. 370 + Rs. 444 + Rs. 33.30 = Rs. 861.30
Hence, she has to pay total fare of Rs. 861.30

Exercise 2.7.4

1. Find the total taxi fare between the two places given in table.

S.No. From To Time Distance Waiting time
a. Kalanki Chabahil 5 p.m 7 km. 3 minutes
b. Airport kalimati 2 A.m 10 km. -
c. Babazar Balain chook 11 A.m 4 km. 6 minutes

2. Sangam hired a taxi and travelled 15km. If the minimum rate is Rs. 14 and the fare goes on at
the rate of Rs. 7.20 per 200 meters. Calculate the total fare paid by him.

3. Mr. Ghising paid Rs. 122 for taxi fare. The minimum charge is Rs. 14 and additional charge
RS. 7.20 per 200meters. How many kilometers did he travel by the taxi?

4. Safal travelled the taxi from Singh Durbar to Sano Thimi of distance 7 km. during day time.
Minimum reservation charge is Rs. 14 and the additional charge Rs. 7.20 per 200 meters. There
was imposed waiting charge Rs. 7.20 per 2 minutes. Find what amount of fare Safal paid for
taxi.

5. Dakshyata hired a taxi to travel from Ratnapark to Godawari which is 20 km. away. The taxi
driver asked whether she wants to pay according to meter with usual rates and rules or a lump
sum Rs. 725. Which will she prefer? By how much is it cheaper?

6. Dikshant travelled 5km. before 9pm. and 4km. afterwards by taxi. Find the total fare he has to
pay with 3 minutes waiting time while getting off. (Use the rates given above)



62 | Mathematics - 9 Arithmetic

Unit Test

Time: 40 minutes F.M.- 24

Group-A (3×1=3)

1. What do you mean a unit of electricity?

2. When the selling price of a watch is Rs 440 and gain percentage is 10%, find the cost price.

3. The total sales of a company is Rs. 2,00,000 in a month and the company allows the
commission at the rate of 0.5% to the agent. Find the amount of commission.

Group-B (4×2=8)

4. The marked price of a pair of shoes Rs 4000. What is its selling price after discount of 10%.

5. The total number of shares of a company is 3,00,000. The company gets an annual profit of
Rs 3,50,00,000. The company board has decided to distribute the dividend 21.5% of total
profit. Find the total dividend amount.

6. Water is supplied through 1 inch pipe line in Hotel jungle lodge. If the Hotel consumed 580
units of water in a month, then find the charge of water with 50% additional service charge.
(minimum charge for 1" pipe upto 5600 liters of water is Rs 3960. Additional charge per
1000 liter is Rs71)

7. Hari travelled 25 km distance by a taxi at Rs 939. If the minimum fare of a taxi is Rs 14,
find the rate of taxi fare per km.

Group – C (2×4=8)

8. A secondary teacher's annual salary is Rs 3,78,000. If he/she pays an annual tax of Rs 14700
at the rate of 1% for the first 3,00,000 and 15% for the next Rs. 1,00,000 of taxable income,
what will be the monthly salary after deduction of tax?

9. Rupan chaudhary made a call of 8 min. in business hours. The time from 08:00-18:00hrs. is
known as business hours. In business hours, a call of 2 minutes is considered as 1 call. Find
the charge of 10% TSC is imposed on it and then 15% VAT. The charge for a call is Rs. 1.

Group – D (1×5 = 5)

10. The shopkeeper marks the price of an article 30% above its cost price and gives the
customers a discount of 10% so that he gains Rs. 34 on the article. Find the marked price of
the article.



Arithmetic Mathematics - 9 |63

Answers ____________________________________________________________

Exercise 2.1

1. (a) Rs 55 (b) Rs 800 (c) Rs 1200 (d) Rs 900
2. (a) 20% (b) 16.66% (c) 3331 % (d) 50%
3. (a) profit 28% (b) Rs 6 (c) Rs 690 (d) Rs 1060
4. (a) Rs 175 (b) Rs 16600 (c) Rs 220
5. (a) 25% (b) 9111 % (c) 50% (d) 25%

6. (a) 2.5% (b) Rs 1552.5 (c) 0.25% (d) Rs. 10000
3
7. (a) Rs 3000 (b) 1000 (c) 5
8. (a) Rs 24 (b) Rs Rs 52 (c) Rs 134 (d) 300 %
11

(d) 300

Exercise 2.2 (b) 700 (c) 34,500 (d) 11,000
(b) 64,00,000 (c) 6 (d) 5
1. (a) 3,20,000 (b) 72,80,000 (c) 1,12,500 (d) 1,00,000
2. (a) 45,00,000 (b) 90,000
3. (a) 61,75,000
4. (a) 55,000

Exercise 2.3

1. (a) 200,800 (b) 165 (c) 10
2. (a) 800 (b) 25%
3. (a) 1539 (b) 23408 (c) 8000
4. (a) 10,000 & 8000 (b) 9000 & 7500 (c) 25
5. (a) 25000 (b) 25 (c) Rs 2000
6. (a) 3750 (b) 6000 (c) 12%
7. (a) 2 (b) 15000
(c) 3500
Exercise 2.4 (c) 15
(c) 15
1. (a) 9,100 (b) 1000, 2,28,000 (d) 791.67, 2,40,500
2. (a) 30,000 (b) 40,000 (d) 12
3. (a)3,36,000 (b) 26,000 (d) 10
4. (a)1350 (b) 4250

Exercise 2.5

1. (a) 32,000 (b) 41,000 (c) 2,05,000 (d) 3,60,000
2. (a) 8 (b) 12 (c) 450 (d) 400
3. (a) 31,250 & 1,25,000 (b) 2,00,000 & 4,00,000

Exercise 2.6

1. (a) Rs 6,000, Rs 8,000, Rs 10,000 (b) Rs 15,000, Rs 12,000, Rs 6,000

(c) Rs 2,40,000, Rs 3,60,000, Rs 6,00,000 (d) Rs 2,25,000, Rs 1,25,000, Rs 1,50,000

2. (a) Rs 3,20,00,000 (b) 5.8% (c) 5000 (d) Rs 100

3. (a) 120 (b) 7500 (c) 1820 (d) 160

4. (a) 33.33% (b) 15% (c) 150 (d) 16%

64 | Mathematics - 9 Arithmetic

5. (a) (i) 180 (ii) 3600 (iii) 717 %
(iii) 1313 %
(b) (i) 15000 (ii) 2000 (iii) 8000

(c) (i) 32 (ii) 400 (c) 80
(d) (i) 2200 (ii) 4950 (c) 150
(c) 500.50
Exercise 2.7.1
7. Rs 1649
1. (a) 883 (b) 1280 (d) 80
2. (a) 2300 (b) 2795 (c)101.85 (d)170
3. (a) 1387.10 (b) 1706.23 (c) 160 (d) 663
4. (a) 90 (b) 650 (c) 155.25
5. 2305 6. 120 units 8. Rs 1365
(c) 248.60
Exercise 2.7.2 (c) 64 (d) 94.09
(c) 27.97 (d) 96
1. (a) 67.90 (b) 82.45 (d) 168
2. (a) 12.8 (b) 145 (c) Rs 184.20
3. (a) 95 (b) 78 (d) 310.75
(d) 120
Exercise 2.7.3 (d) 49.72

1. (a) 55.44 (b) 49.16
2. (a) 12.43 (b) 42.26
3. (a) 11.19 (b) 18.65

Exercise 2.7.4

1. (a) Rs 287.80 (b) Rs 576

2. Rs 554

3. 3km

4. Rs 280.40

5. The ump sum Rs 725, Cheaper by Rs 29

6. 443.2



Arithmetic Mathematics - 9 |65

g]kfnsf] gS;f

66 | Mathematics - 9 Arithmetic

3Chapter

Mensuration

Objectives:

At the end of this chapter, the
students will be able to:
 solve the problem related to

surface area and volume cubical
and cuboids soon.
 calculate the area of outside
path, inside path and cross path in
rectangular ground.
 calculate the cost of carpeting,
colouring, and plastering in a room
according as the cost per unit.

Teaching Materials:

Problem related objects which is
locally available i.e. brick, tiles, etc.
measuring instruments, like tape,
meter scale, locally available price list
of colouring materials, carpet, etc.
chart paper.

Historical fact

The Greek word “Peripheria”- meaning “ periphery” inspired the symbol  (pi) a
constant, now taken as the ratio of circumference to the diameter of a circle.

i.e.  = Circumference (C)
Diameter (D)

The accuracy by which the value of  was estimated in different ages by Willian Jones
different mathematicians differently.

 Egyptians (1650 B.C),  = 434

 Archimedes (240 B.C),  = 37101 <  < 37100

 Claudius Ptolemy (150A.D),  = 377
120

 Tsu Ching – Chih (480A.D),  = 355
113

 Aryabhatta (530 A.D),  = 62832
20000

 Francois Viete (1578 A.D) found  correct to nine decimal places.

 Ludolph Van Ceulen (1610A.D) of Netherland computed  to thirty-five decimal places. In his
will this number 3.1415926,5358979,3238462,6433832,7050288 was engraved on his tombstone
and this number is known as ‘ Ludolphian’ number.

In recent times, the approximate value of  has been calculated to more than 100,000 decimal places.

 The symbol  was used by English Mathematicians William Oughtred, Isaac Barrow and David
Gregory in 1737.

 The first person to use  as ratio of circumference to the diameter of a circle was an English
mathematician William Jones (1706).

 Johann Heinrich Lambert (1728-1777, Switzerland) first showed that  is irrational.

A mnemonic is often used to remember the value of  up to seven decimal places.

“May I Have A Large Container Of Coffee?”
π 3 1 4 1 5 9 2 6

68 | Mathematics - 9 Mensuration

Review Shapes Area Perimeter or
circumference
Name of the
figures A = 1 base x P = AB + BC + CA
2
1 Triangle
height
2 Square
A = l2 p = 4l
3 Rectangle
A = 1 d2
2

A=l×b P = 2(l + b)

4 Parallelogram A = base × P = AB + BC + CD
height + DA
5 Equilateral
Triangle A= 3 a2 p = 3a
4

Mensuration Mathematics - 9 |69

6 Isosceles A = a 4b2 – a2 p = a + 2b
Triangle 4

7 Rhombus A = 1 d1 x d2 p = 4a
8 Trapezium 2

9 Quadrilateral A = 1 (b1 + b2)xh P = sum of all sides.
10. Circle 2

11 Semi-circle A = 1 d(h1 + h2) P = sum of all sides.
2
70 | Mathematics - 9
A = r2 or C = 2r
A = (d+2/4) C = d

A = 1 r2 or C = r + 2r or
2 c = (d/2) + d

A = 1 d2
8

Mensuration

3.1 Area of Pathways

Area of Surrounding Paths

1. Area of surrounding path of fixed width running outside a rectangular field.

Roads or pathways sometimes run all around plots of land, gardens, squares, ponds etc and we need to

find the area covered by such pathways.

Let ABCD be a rectangular plot of a land of length l and breadth b. and a E F
pathway of uniform width w runs surrounding the plot outside. The length A B

and breadth of the plot including the path are HG = (l + 2w) and bw

FG = (b + 2w) respectively. l C
l + 2w G
 Area of plot, ABCD = A1 = l × b ... (i) D
H

Area of the plot including path, A2 = (l + 2w) (b + 2w)

= lb + 2lw + 2bw + 4w2 ... (ii)

 Area of surrounding path (outside) = A2 – A1
= lb + 2lw + 2bw + 4w2 – lb.

= 2w(l + b + 2w) square units.

2. Area of the pathway of fixed width running inside a rectangular field:

Let a pathway of uniform width w runs inside a rectangular plot ABCD of A B
length l and breadth b, then length and breadth of the plot EFGH excluding E F
l - 2w
the pathway are (l – 2w) and (b – 2w) respectively. b - 2w w

Area of the pathway = Area of rectangle ABCD – area of rectangle EFGH H G
= lb – (l – 2w) (b – 2w) D lC
= lb – (lb – 2lw – 2bw + 4w2)
= lb – lb + 2lw + 2bw - 4w2

= 2w (l + b - 2w) square units.
Note:
(i) Area of the path running outside the square plot of side a is A = 4w(a + w)
(ii) Area of the path running inside the square plot of side a is A = 4w(a – w).

3. Area of path of uniform width running outside a circular plot: w B
A
Let OA = r be the radius of the circular plot and OB = R be the radius of the
circular plot including path of width w i.e. r + w = R. r
O
 Area of the circular path = Area of outer circle – Area of inner circle.
= R2 - r2 = (R2 - r2) sq. units. R

Mensuration Mathematics - 9 |71

Area of Cross Paths A TU B
EF
Let two paths of width w run across a rectangular plot ABCD Q HG R
crossing each other. bw S
WlV
Area of the path PQRS parallel to length of the plot = lw. P C

Area of the path TUVW parallel to breath of the plot = bw. D
Area of the square EFGH, common to both the paths = w × w = w2

 Area of cross path = Area of PQRS + Area of TUVW – Area of EFGH.
= lw + bw - w2

= w(l + b – w) sq. units.

If the plot is a square, l = b

 Area of cross path = w(l + l – w) sq. units.

= w(2l - w) sq. units.

Worked Out Examples

Example 1: A rectangular garden of 32m by 22m is surrounded outside by a path 2m
Solution: wide. Find the cost of paving the path at Rs. 40 per m2.

Here,

Length of the rectangular garden (l) = 32m. 22m
Its breadth (b) = 22m
Width of the path (w) = 2m 32m
Now, area of surrounding path (outside) 2m

= 2w(l + b + 2w)

= 2 × 2m (32m + 22m + 2 × 2m)

= 232m2

Again, cost of paving the path = Area × Rate
= 232m2 × Rs. 40/m2 = Rs. 9280.

Therefore, the cost of paving the path is Rs. 9280.

Example 2: The length and breadth of a rectangular plot are in the ratio 3:2. A path
Solution: of width 1.5m runs around inside the plot. If the path has the area of
111m2, find the length and breadth of the path.
Here,
Area of the rectangular plot = 111m2

Width of the surrounding path (w) = 1.5m.

Let the length and breadth be 3x and 2x respectively. (Since length and breadth are in
the ratio 3:2)

72 | Mathematics - 9 Mensuration

We have,
Area of surrounding path (inside) = 2w(l + b – 2w)
or, 111m2 = 2 × 1.5m(3x + 2x – 2 × 1.5m)
or, 111m2 = 3m(5x – 3m)
or, 37m = 5x – 3m
or, 5x = 40m

x = 8m.

 Length (l) = 3x = 3 × 8m = 24m and breadth (b) = 2x = 2 × 8m = 16m.

Example 3: The circumference of a circular garden is 220m and a path 3.5m wide
Solution: runs around the garden. Find the cost of plastering the path at Rs. 60 per
sq. metre.

Here,

Circumference of the circular garden (C) = 220m.

If radius = r then

2r = 220m

or, 2 × 22 × r = 220m
7

r = 35m.
Width of the path(w) = 3.5m

Radius of the garden including path (R) = r + w

= 35m + 3.5m = 38.5m

Area of the path = (R2 - r2)

= 22 {(38.5m)2 – (35m)2}
7

= 22 {1482.25m2 – 1225m2}
7

= 22 × 257.5 = 808.5m2
7

The cost of plastering the path = Rs. 60 per sq.m.

Total cost of plastering the path = Area of path × Rate
= 808.5m2 × Rs. 60/m2

= Rs.48510.

Therefore, the cost of plastering the path is Rs. 48510.

Example 4: A rectangular plot of land is 64m long and 50m wide. Two paths, each of
width 2m one parallel to length and other parallel to width of the plot
cross each other inside the plot. Calculate,

(i) the cost of paving the path with bricks of size 16cm × 14 cm at Rs. 6
per brick.

(ii) The cost of turfing the remaining space at Rs. 4 per sq. m.

Mensuration Mathematics - 9 |73

Solution: Here, A TU B
Length of the rectangular plot (l) = 64m.

Breath of the plot (b) = 50m P 2m Q
Width of the cross path (w) = 2m. 50m 2m R
 Area of the cross path = w (l + b – w) W64mV
S C

= 2m(64m + 50m - 2m)

= 2m × 112m D

A = 224m2

 A = 2240000cm2

Area of a brick (a) = 16cm × 14cm = 224cm2

No of bricks needed = Area of paths = A = 222402400cm0c2m2= 10000.
Area of a brick a

(i) Total cost of bricks at Rs. 6 per brick = No. of bricks × Rate

= 10000 × Rs. 6 = Rs. 60000

(ii) Area of the plot including paths =l×b
= 64m × 50m = 3200m2

 Area of the remaining space = Area of plot - Area of path
= 3200m2 – 224m2 = 2976m2

 The cost of turfing the remaining space at Rs. 4 per sq.m. = 2976m2 × Rs. 4/m2

= Rs. 11904.

Therefore, the cost of turfing the remaining space is Rs. 11904 and the cost of
paving the paths is Rs. 60,000.

Example 5: Two pathways run across a field 80m × 40m crossing each other at a
Solution: right angle. The path parallel to the length is 2m wide and parallel to the
breadth is 3m wide. Find the area of the path inside the field. And
calculate the cost of the gravelling the path at Rs. 50 per sq.m.

Here, two crossing paths are of different widths. Length of the field (l) = 80m and

width of the path along length w1 = 2m. A TU B
Breadth of the field (b) = 40m.

Width of the path along breadth w2 = 3m P 2m Q
Now area of cross path 40m 3m R

= lw1 + bw2 - w1w2 S
= 80m × 2m + 40m×3m - 2m × 3m

= 160m2 + 120m2 - 6m2 D W80mV C
A = 274m2

 Area of the cross path is 274m2.

Now cost of gravelling sq.m (C) = Rs. 50

 Total cost (T.C) = A × C = 274 × Rs. 50 = Rs. 13700

74 | Mathematics - 9 Mensuration

Exercise 3.1

1. Calculate the area of the shaded regions of the following figures where width of the shaded
parts are uniform.
(a) (b)

4cm 18m
32cm 2m

(c) (d)

4m 24m
16m 3m

2. Find the area of the shaded parts in each of the following figures.
(a) (b)

0.7m 1.4cm

(c) 16m (d) 36cm
4m 4cm

4m 3cm

20m 42cm

Mensuration Mathematics - 9 |75

3. Find the area of the shaded portion with uniform width given in the figure below.
(a) (b)

30 m 60 m
6m
40 m

50 m 100 m

4. (a) If the rectangular ground has the length 80m, breadth 60m, uniform width of the path of
3m is running around outside of the ground, find the area of the path.

(b) A boarder 8cm wide runs around a towel 120 cm long and 80cm wide. Find the area of
the boarder of the towel.

(c) A path of uniform width of 1.5m runs all around a swimming pool 50m long and 25m
wide. Find the area of the path.

(d) Two paths each of width 4m run across the middle of a rectangular field 24m long and
16m wide. Calculate the area of the paths.

5. (a) A rectangular plot of land is 88m long and has an area 4840m2. If it has a path of
uniform width 2m within it, find the area of the path.

(b) A circular ground has a track of uniform width 2.1m. all around it. If the area of the
track is 568.26m2, find the area of the ground excluding the track.

(c) The cost of constructing a path running inside a rectangular field at Rs. 52 per square
metre is Rs. 7,280. Find the area of the path.

(d) 5000 bricks of size 20cm × 12cm each are required to pave the path running outside a
square field. Calculate the area of the path.

6. (a) A rectangular garden is surrounded by a path of uniform width 3m. If breadth of the
garden is 40m and area of the path is 588m2, find the length of the garden.

(b) A path of uniform width 4m runs outside a square flower bed and has an area of 448m2.
Find the area of the flower bed.

(c) A room measuring 6m × 4m is carpeted leaving space of 25cm all around. Find the area
of the carpet.

(d) The area of a square pond is 400 m2. A path of uniform width surrounds the pond and
its area is 176m2. Find the width of the path.

7. (a) Two cross paths run across a rectangular field of length 90m and breadth 60m. If the
area of the cross paths is 584m2, find the width of the path.

(b) Two cross paths each 3m wide run across a rectangular garden. If the area of the cross
path is 243m2 and the length of the garden is 48m, find the breadth of the garden.

(c) A circular plot of diameter 28m is surrounded by a path of uniform width. If the area of
the path is 346.5m2, find the width of the path.

(d) If the sum of the radii of two concentric circles is 28cm and difference is 12cm, find the
area of the annular part enclosed by the two circles.

8. (a) A rectangular field is 96m in length and 42m in breadth. A path of uniform width 3m runs
immediately inside its boundary. Find the cost of gravelling the path at Rs. 32 per square metre.

76 | Mathematics - 9 Mensuration

(b) A uniform path of width 2m surrounding a pool 20m × 16m is to be paved with marbles
of size 10cm × 8cm each. If a marble costs Rs. 5, find the cost of paving the path.

(c) A square lawn is surrounded by a path of uniform width 3m. The cost of paving the path
with marble at Rs. 200 per m2 is Rs. 36000. Find the area of the lawn and cost of
watering the lawn at Rs. 3 per m2.

(d) A square park 2025m2 in area is surrounded by a road 3m broad. Find the cost of paving
the road with concrete slabs 16cm × 12cm each at Rs. 10 per piece.

9. (a) A circular garden of diameter 56m has a path of uniform width 3.5m running
immediately outside it. Find the cost of plastering the path at Rs. 80 per m2.

(b) Fencing a circular field at Rs. 25 per meter costs Rs. 6600. A path 1.4 m wide runs
around the field. Find the cost of gravelling the path at Rs. 40 per sq. metre.

(c) Length and breadth of a rectangular field are in the ratio 2:1. The cost of gravelling two
cross paths of equal width across the middle of the field at Rs. 25 per sq.m is Rs. 11250
and if the cost of plastering the common part of the two cross paths at Rs. 40 is Rs. 360,
find the cost of turfing the empty space at Rs. 8 per m2.

(d) A park 36m × 20m is to be surrounded by a path of uniform width 2m, paving with
marble slabs 18cm × 12cm each at the rate of Rs. 20 per piece. If 10% of the space of
the path is occupied by cement, find the cost of paving the path.

3.2 Surface Area and Volume of Prisms

A prism is a solid having polygonal cross-section (base) and rectangular lateral surfaces.

Triangular based prism Square based prism Hexagonal based prism

Other prisms:

Mensuration Mathematics - 9 |77

Drawing three Dimensional View of a Solid

There are two basic views.
a. Isometric View: In this view none of the face (Top-bottom, front- back, side faces) are shown

rectangular.

(Use 30° - 60° set square)

Object line (dark)
Measurement line

3cm
Extension line

b. Oblique view: In this view, front face is shown rectangular.

(Use 45° - 45° set square)

Useful Relations E H
G
Area of a cross section (base) = a (depends upon shape of
the polygonal base). In the given rectangular based prism,

Area of the cross section ABCD (a) = l × b ... (i) AD F
Note:

i. Opposite faces which are congruent determine the l h

cross section. Here ABCD and EFGH are bases (cross

sections). Bb C

ii. Distance between these cross-sections is the length/height of the prism. Here CG = h is height.

78 | Mathematics - 9 Mensuration

iii. Rectangular faces between the cross sections or bases are lateral surfaces.

Lateral surface area of the prism = Area of (ABFE + BCGF + CDHG + ADHE)

= AB × AE + BC × CG + CD × CG + AD × AE

= l × h + b× h + l × h + b × h

= (2l + 2b)h

= 2(l + b)h

 L.S.A = P × h ... (ii) where p is the perimeter of the cross section.

 Total surface area (T.S.A) = L.S.A + 2 area of base

or, T.S.A = L.S.A + 2A ... (iii)

And volume of the prism = AB × BC × CG

=l×b×h

 V=A×h ... (iv)

Where A is the area of base.

Worked Out Examples

Example 1: Find the lateral surface area, total surface area and volume of the given
Solution: prism.

Here, D
the base is a right angled triangle with B = 90°,
A

BC = 7cm, AC = 25cm and height h = 30cm

 AB = AC2 - BC2 = (25cm)2 - (7cm)2 = 24cm E
C
1 B
2
 Base area/area of cross section (A) = .BC. AB F

= 1 × 7cm × 24cm = 84cm2.
2

Perimeter of the base (P) = AB + BC + CA

= 24cm + 7cm + 25cm = 56cm

Lateral surface area of the prism = P × h

= 56cm × 30cm

 L.S.A = 1680 cm2.

Total surface are area of the prism = L.S.A + 2A
= 1680cm2 + 2× 84cm2 = 1848cm2

And volume of the prism (V) =A×h
= 84cm2 × 30cm

V = 2520cm3.

Mensuration Mathematics - 9 |79

Example 2: Find the total surface area and volume of the prism given in the figure.
Solution:
Here,
Example 3:
Solution: ABCDEF is the cross-section. Dividing the cross A F
section in rectangles ABMF and EMCD, area of
cross-section = area of rect.(ABMF + EMCD)

= AB × BM + MC × CD B 5cm 4cm
= 12cm × 4cm + 5cm × 4cm ED
 A = 68cm2. M
Perimeter of the cross-section (p) = AB + BC + CD + 9cm C
DE + EF + FA

= 12cm + 9cm + 4cm + 5cm + 8cm + 4cm

= 42cm.

Lateral surface area of the prism (L.S.A) =P×h

= 42cm × 6cm [ h = 6cm]
= 252cm2

T.S.A = L.S.A + 2.A
= 252cm2 + 2 × 68cm2 = 388cm2

And volume (V) = A × h
= 68cm2 × 6cm

V = 408cm3

If the volume of a cube is 125cm3, find its total surface area.
Here,
Volume of cube = 125cm3.

If length of its side be l then,
l3 = 125cm3

 l = 5cm. = 6 × (5cm)2 = 150cm2
Now, total surface area of the cube (T.S.A) = 6l2
Therefore, T.S.A of the cube is 150cm2.

Exercise 3.2

1. Find the total surface area and volume of the following solids.
(a) (b)

18cm

80 | Mathematics - 9 Mensuration

(c) (d)

6cm 12cm

2. (a) The volume of a cube is 216cm3. Find its total surface area.

(b) If the total surface area of a cube is 384cm2, find its volume.

(c) The volume of a cubical block is 350cm3. If the area of its base is 70cm2, find its height.

(d) The total surface area of a rectangular parallelopipe with square cross-section is
6048cm2. If the area of the cross-section is 1296cm2, find its length.

3. Find the total surface area and volume of the given solids:

(a) 4cm (b) 3cm

1cm

5cm 3cm
1cm

9cm

2cm 2cm
(c) (d)

4cm

7cm (b)

4. Calculate the volume of the given solids:
(a) 3cm

4cm 4cm

Mensuration Mathematics - 9 |81

(c) (d)

6cm

3cm 6cm

3cm 50cm

6cm 50cm

5. (a) The adjoining figure shows a victory
stand. Find the volume and surface area of
the stand if the bottom of the stand is
open.

50cm

(b) The figure given alongside is a solid
metal with rectangular hole in it. Find the
volume of the metal.

5cm

(c) The figure given alongside is a window 13cm
frame. The wood used has cross-section 20cm
7.5cm × 8cm. Find the volume of the
wood used.

2m

(d) A swimming pool is 16m broad and 50m long. If its depth is increasing regularly from
1m to 2m, how much water will it contain in litres when it is fully filled?

6. A cubical wooden solid having side 16cm is cut down into 8 equal pieces of cube. Find the
length of side of a new cube.

7. Calculate the volume and total surface area of your class room.
8. Investigate the volume of the stage of program halls of your school.

82 | Mathematics - 9 Mensuration

3.3 Area of Four Walls, Floor and Ceiling

Given figure can be considered as the model of a room. Having H G
length l, breadth b and height h where ABCD is the floor, EFGH
is the ceiling and ABFE, BCGF, CDHG, DAEH are four walls. D h
l F
A comfortably constructed room is generally cuboidal in shape E
where, C
b
Area of floor = area of ceiling = l×b. ... (i)
B
Area of front wall = area of back wall = l×h. A
Area of right side wall = area of left side wall = b×h.

Area of 4 walls = 2lh + 2bh. ... (ii) Ceiling Walls
= 2h(l + b). ... (iii) Floor

Area of 4 walls, floor and ceiling
= 2lb + 2lh + 2bh.
= 2(lb + bh + bl).

A room also consists windows and doors. Area of 4 walls also
include doors and windows thus, on calculating area of 4 walls we must exclude the areas of doors,
windows, ventilators etc.

Diagonal of a Room H G
D
In a cuboidal room, line joining opposite corners of the room is h
l F
unknown as diagonal of the room. In the figure, DF is a diagonal. E
All the four diagonals are equal. C

Here, DAB = 90° b

 BD = AB2 + AD2 A B
= l2 + b2

Again in right DBF, where DBF = 90°

 DF = DB2 + BF2

= ( )l2 + b2 2 + h2

= l2 + b2 + h2

 Diagonal of a room = l2 + b2 + h2 .

Mensuration Mathematics - 9 |83

For Carpeting the Floor of a Room

Area of the carpet = area of the floor = A

If the cost of the carpet per square unit is C,

Then, total cost for the carpet = area × rate = A × C

Note:

i. Total cost = Area i. e. Total cost = A × C/m2
cost/sq.unit

ii. Total cost = Volume i.e. Total cost = V × C
cost/cu.unit m3

iii. Total cost = Number i.e. Total cost = N × C
cost/piece piece.

iv. Total cost = Length i.e. Total cost = l × C
cost/unit length m

Worked Out Examples

Example 1: Calculate area of 4 walls, area of 4 walls and floor and area of 4 walls
Solution: with floor and ceiling of the room 8m long, 6m wide and 4m high.
Here, Length of the room (l) = 8m.
Breadth (b) = 6m.
Height (h) = 4m.
We have,
Area of 4 walls = 2h(l + b)

= 2 × 4m (8m + 6m)
= 8m × 14m = 112m2
Therefore, area of 4 walls is 112m2.
Area of 4 walls and floor = 2h (l + b) + lb

= 2 × 4m (8m + 6m) + 8m × 6m
= 8m × 14m + 48m2
= 112m2 + 48m2 = 160m2
 Area of 4 walls and floor is 160m2.

And area of 4 walls, floor and ceiling = 2 (lb + bh + hl)
= 2 (8m × 6m + 6m × 4m + 4m × 8m)
= 2 (48m2 + 24m2 + 32m2)
= 2 × 104m2 = 208m2

 Area of 4 walls, floor and ceiling is 208m2.

84 | Mathematics - 9 Mensuration

Example 2: Perimeter of the floor of a room is 18m and its height 4m. Find the area
Solution: of 4 walls of the room.
Here, Perimeter of the floor (P) = 18cm
Height (h) = 4m
We have,
Area of 4 walls = 2h (l + b)

= 2(l + b) h
=P×h
= 18m × 4m = 72m2
 Area of 4 walls of the room is 72m2.

Example 3: A room with length 8m, breadth 6m and height 4m has a door 2m × 1m
Solution: and two windows 1m × 2m each. Find the area of 4 walls excluding door
and windows.
Here, For the room

Length (l) = 8m.

Breadth (b) = 6m.

Height (h) = 4m.

 Area of 4 walls (A) = 2h (l + b)

= 2 × 4m (8m + 6m)

= 8m × 14m
= 112m2
Area of a door (A1) = 2m × 1m. = 2m2
Area of 2 windows (A2) = 2(2m × 1m) = 4m2.

 Area of 4 walls excluding door and windows = A – A1 – A2
= 112m2 – 2m2 – 4m2 = 106m2

Therefore, area of 4 walls excluding door and windows is 106m2.

Example 4: A room is 10m long and 8m wide. If the area of 4 walls is 180m2, find the
Solution: height of the room.

Here, Length of the room (l) = 10m.

Breadth (b) = 8m.

Let height be h

Area of 4 walls of the room = 180m2

or, 2h(l + b) = 180m2

or, 2h(10m + 8m) = 180m2

or, h = 180m2
36m

 h = 5m.
Therefore, height of the room is 5m.

Mensuration Mathematics - 9 |85

Example 5: A room is 3.5m high. If the area of its 4 walls is 126m2, find the perimeter
Solution: of the floor.
Here, Height of the room (h) = 3.5m.
Area of 4 walls (A) = 126m2.
Perimeter of the room (P) = ?
We have,
Area of 4 walls = P × h
or, 126m2 = P × 3.5m

 P = 36m.
Therefore, perimeter of the floor is 36m.

Example 6: The area of the 4 walls of a square room is 84m2. If the height of the
Solution: room is 3.5 m, find the area of its floor.

Here, Height of the room (h) = 3.5m.

Let length of the square room be l

Area of 4 walls of the square room = 84m2.

 2h(l + b) = 84m2

or, 2 × 3.5m (l + l) = 84m2

or, 2l = 84m2
7m

 l = 6m.
Now,
Area of floor of the room = l2 = (6m)2 = 36m2.

Example 7: A room is 8m long and 6m wide. How long carpet is required to cover the
Solution: floor of the room if the width of the carpet is 2 m?
Here, Let length of the carpet be l
Its width (b) = 2m.
Length of the room (L) = 8m
Breadth of the room (B) = 6m
Since, Area of carpet = area of room.
l×b=L× B
or, l × 2m = 8m × 6m
or, l = 24m.
Therefore, 24m long carpet is required to cover the floor.

Example 8: A room is 10m long, 6m wide and 4m high. Find the cost of carpeting the
Solution: floor at the rate of Rs. 50 per square metre and painting the four walls at
the rate of Rs. 25 per square metre.

Here, Length of the room (l) = 10m

Breadth (b) = 6m

86 | Mathematics - 9 Mensuration

Height (h) = 4m

Area of floor =l×b
= 10m × 6m = 60m2

Cost of carpeting the floor at Rs. 50 per m2 (C1) = Area × Rate
= 60m2 × Rs. 50/ m2

= Rs. 3000

Area of 4 walls = 2h(l + b)

= 2 × 4m(10m + 6m)
= 8m × 16m = 128m2

Cost of painting 4 walls at Rs. 75 per m2 (C2) = Area of 4 walls × Rate
= 128m2 × Rs. 25/m2

= Rs. 3200.

Total cost = C1 + C2
= Rs. 3000 + Rs. 3200

= Rs. 6200.

Therefore, the total cost of carpeting the floor and painting the walls is Rs. 6200.

Example 9: A room is 12m long, 8m broad and 4m high. It has a door 2m × 1m and 2
Solution: windows each 2m × 1.5m. Calculate the total cost of papering on the
walls and the ceiling at the rate of Rs. 50 per square metre.

Here, Length of the room (l) = 12m.

Breadth (b) = 8m

Height (h) = 4m

Size of a door = 2m × 1m

Size of a window = 2m × 1.5m

Total area occupied by the door and windows

a = 2m × 1m + 2(2m × 1.5m)
= 2m2 + 6m2 = 8m2

Area of 4 walls and ceilings = 2h(l + b) + l × b

= 2 × 4m(12m + 8m) + 12m × 8m
= 8m × 20m2+ 96m2
= 160m2 + 96m2
= 256m2

Area of 4 walls and ceiling excluding door and windows = 256m2 – 8m2
= 248m2

Now cost of papering on the walls and ceiling at Rs. 50 per m2.

= Area × Rate

= 248m2 × Rs. 50/m2

= Rs. 12400.

Therefore, the cost of papering the walls and ceiling is Rs. 12400.

Mensuration Mathematics - 9 |87

Example 10: The cost of painting 4 walls of a room 4m high at Rs. 15 per square

metre is Rs. 2400. If breadth of the room is two third of its length, what
will be the cost of carpeting its floor at Rs. 200 per square metre?

Solution: Here, Height of the room (h) = 4m.

Rate of painting the walls = Rs. 15/m2.

Total cost of painting the walls = Rs. 2400.

Area of 4 walls = Total cost
Rate

= 2400 = 160m2
Rs15/m2

If length of the room is l, then

Breadth (b) = 2 l
3

2h(l + b) = 160m2

or, 2 × 4m (l + 2 l) = 160m2
3

or, 5l = 160 × 3 m
8

 l = 12m.

Breadth of the room (b) = 2 l = 2 × 12 m = 8m.
3 3

Now, area of floor = l × b
= 12m × 8m = 96m2

 Total cost of carpeting the floor at Rs. 200 per square metre = A × C
= 96m2 × Rs. 200/m2

= Rs. 19200.

Therefore, the cost of carpeting the floor of the room is Rs. 19200.

Example 11: The cost of plastering the walls of a square room at Rs. 10 per m2 is Rs.
3240 and the cost of carpeting its floor at Rs. 25 per m2 is Rs. 8100. Find

the height of the room.

Solution: Here, Cost of carpeting the room at Rs. 25 per m2 is Rs. 8100

Area of floor = T.C
C

or, l2 = Rs 8100
Rs 25

 l = 18m.

Again, cost of plastering the walls at Rs. 10 per m2 is Rs. 3240.

The area of 4 walls = T.C
C

2h (l + l) = Rs 3240
Rs 10/m2

or, 2 × h(18m + 18m) = 324m2

 h = 4.5m
Therefore, the height of the room is 4.5m.

88 | Mathematics - 9 Mensuration

Exercise 3.3

1. (a) Calculate the area of 4 walls of a hall of length 18m, breadth 12m and height 5m.

(b) Find the area of four walls and ceiling of a room 10 m long, 6m wide and 4m high.

(c) Find the area of 4 walls, floor and ceiling of a room 8m long, 6m width and 3.5m high.
(d) Find the area of 4 walls of a room 9m long, 6m wide and 4m high leaving 8m2 for doors

and windows.
2. (a) If the area of 4 walls of a room 6m wide and 3.5m high is 105m2, find the length of the

room.

(b) If the perimeter of a room 4m high is 44m, what is the area of 4 walls of the room?
(c) The area of 4 walls of a room is 56m2.Find the perimeter of the room if the height of the

room is 2 m.
(d) The area of 4 walls of a square room is 216m2. If the room is 4.5m high, find the length

of the room.

3. (a) A room is twice as long as it is broad and it is 3.5m high. If the area of its 4 walls is
105m2, find the area of its floor.

(b) The length and breadth of a chamber of a clock tower are 8m and 6m. If the diagonal of the
chamber is 26m, find the height of the chamber and hence find the area of its 4 walls.

(c) A room is 12m long, 10m broad and 4.5m high. It contains a door of size 1m × 3m and
two windows of size 2m × 1.5m each. Find the area of the four walls of the room
excluding the door and windows.

(d) Area of the ceiling of a room 4m high is 48m2. If the area of 4 walls of the room is
112m2, calculate the length and breadth of the room.

4. (a) If a room 8m long and 5m wide is laid with carpet 2m wide, find the total length of the
carpet required.

(b) A room 8m long and 4m broad is covered with carpet of a certain width. If the total
length of the carpet used is 12.8m, find the width of the carpet.

(c) How many pieces of paper each measuring 1.5m by 0.4m will be required to cover the four
walls of a hall 15m long, 9m wide and 5m high leaving 18m2 for doors and windows?

(d) How many pieces of paper each of area 1.25m2 are required to cover the 4 walls and
ceiling of a room 9m long, 7m wide and 3.5m high, if 7.5m2 of the wall is covered by
doors and windows?

5. (a) Find the cost of plastering 4 walls and ceiling of a room 9m long, 6.5m broad and 4m
high at a rate of Rs. 20 per square metre.

(b) Find the cost of laying a room 7m long and 6.5 m broad with parquet at Rs. 250 per
square metre.

(c) Find the cost of distempering the walls of a room 10m long, 8m wide and 5m high at
Rs.15 per square metre.

(d) The cost of plastering the walls of a room at Rs. 21.50 per m2 is Rs. 3870. Find the cost
of painting the walls at Rs. 16 per square metre.

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6. (a) The cost of painting the 4 walls of a room at the rate of Rs. 7 per m2 is Rs. 546. If the
room is 6m broad and 3m high, what is its length?

(b) Total cost of plastering 4 walls and ceiling of a room 7m long and 7m wide with
gypsum at Rs. 125 per square metre is Rs. 20125. Find the height of the room.

(c) If the perimeter of a square room is 54m and height 4m, find the total cost of carpeting
the floor at Rs. 48 per m2 and plastering its four walls at Rs. 12 per m2.

(d) Doors and windows of a room covers 559 % of the space of the walls. The cost of
distempering the remaining parts at Rs. 12 per m2 is Rs. 1428. If the sum of the length
and the breadth of the room is 14m, find the height of the room.

7. (a) The cost of plastering 4 walls of a room whose length is twice its breadth and thrice its
height at Rs. 7 per m2 is Rs. 1008. What will be the cost of carpeting the floor of the
room at Rs. 200 per m2?

(b) The height of a square room is one third of its length. If the cost of plastering its four walls
and ceiling at Rs. 9.5 per m2 is Rs. 3192, find the cost of carpeting its floor at Rs. 125 per m2.

(c) Length, breadth and height of a room are in the ratio 6:3:2. If the cost of carpeting the
floor at Rs. 120 per square metre is Rs. 13500, find the cost of plastering the walls and
ceiling at Rs. 30 per square metre.

(d) The cost of plastering four walls of a room 4m high and whose breadth is one third of
its length at Rs. 12 per m2 is Rs. 1536. What will be the cost of parqueting the floor at
the rate of Rs. 200 per square metre?

8. (a) The cost of carpeting the floor of a room at Rs. 75 per sq.m. is Rs. 11,250. If the room
contains 660 cu.m. of air, find the height of the room.

(b) The length of a room is twice its breath and it contains 396m3 of air. If the cost of
plastering its floor at Rs. 15 per sq.m. is Rs. 1080, find the cost of plastering its 4 walls
at Rs. 12 per sq.m.

3.4 Volume of Walls

We know,

Volume of a cuboid (V) = l × b × h

or, V= A × h

i.e. V = base area × height

Volume of a cube (V) = l3

Walls are to be built in different prismic and cylindrical shapes.

1. A single wall has as its dimension

Length × thickness × height

If length = l, thickness = t, and height = h, then

Volume (V) = l.t.h

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Walls also consist separate doors and windows as thick as the wall and of desired length and
height. For example, if width (length) and height of a door be l and h, then volume occupied
by the door.

V = l  t  h where t is thickness of wall.

2. Surrounding wall (inside)

DC

bh

b l

Al B

Let a wall of height h and thickness 't' is to be built surrounding inside a rectangular plot of
length l and breadth b. The base of the surrounding wall is just like a surrounding path of area.
A = 2t(l + b – 2t)

 Volume of the surrounding wall (inside),
V=A×h

 V = 2t(l + b – 2t)h
3. Surrounding wall (outside)

Let a wall of thickness t and height h is to be built surrounding outside a rectangular plot of
length l and breadth b, then the base of the wall is just like a surrounding path (outside) of area.
A = 2t(l + b + 2t)

 Volume of the surrounding wall (outside)
V=A×h

 V = 2t(l + b + 2t)h
4. Cross wall

t
tb

l

Let wall of height h and thickness 't' is to be built across a rectangular plot of length l and
breadth b.

Mensuration Mathematics - 9 |91

The base of the wall is just like a cross path of area.

A = t(l + b – t)

Volume of cross wall

V=A×h

V = t(l + b – t)h

Number of bricks and their cost required to build a wall

Let the volume of the wall = V

the Volume of a brick = v then

Number of bricks required (N) = Volume of wall = V
Volume of a brick v

If C is the cost of a brick, then total cost of bricks = Number of bricks × Rate

T.C = NC
Note: Units of all the dimensions of a wall and the bricks should be same.

Volume of material contained in a box

1. If external dimensions are l × b × h, external volume Vext = l × b × h. If thickness of the
material is t then internal dimensions are (l – 2t)(b – 2t)(h – 2t)

 Internal volume = Vint = (l – 2t)(b – 2t)(h – 2t)
 The volume of the material in the box = Vext - Vint.
2. If internal dimension l × b × h and thickness t are given,

Internal volume (Vint) = l × b × h
External volume (Vext) = (l + 2t)(b + 2t)(h + 2t)
 The volume of the material in the box = Vext - Vint.

Worked Out Examples

Example 1: Find the volume of a wall 25m long, 50cm wide and 4m high.
Solution: Here,
Length of the wall (l) = 25m
Thickness of the wall (b) = 50cm = 0.5m
Height (h) = 4m.

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Now,
We have, volume of the wall,
Volume of the wall (V) = l × b × h

= 25m × 0.5m × 4m = 50m3
Therefore, the volume of the wall is 50m3.

Example 2: There are 1200 bricks each of size 25cm × 10cm × 5cm in a wall. Find the
Solution: volume of the wall.
Here,
Size of a brick = 25cm × 10cm × 5cm

 Volume of a brick (v) = 0.25m × 0.10m × 0.05m.
v = 0.00125m3.

Number of bricks (N) = 1200

Volume of the wall (V) = N × v
= 1200 × 0.00125m3 = 1.5m3

Therefore, the volume of the wall (V) = 1.5m3

Example 3: A wall is 25m long, 4m high and 30cm thick. If it contains two windows
Solution: 2m × 1.5m each and a door 3m × 2m, find the cost of bricks of size
15cm ×10cm ×5cm at Rs. 4 per piece.

Here,

Length of wall (l) = 25m,

Height of the wall (h) = 4m,

Thickness of the wall (t) = 30cm = 0.30m

 Volume of wall = l × h × t = 25m × 4m × 0.30m = 30m3.

Volume of the wall occupied by doors and windows = 2(2m×1.5m×0.3m)+3m×2m×0.3m

= 1.8m3 + 1.8m3 = 3.6m3

Volume of the wall excluding the doors and windows (V)

= Volume of wall – volume of doors and windows.

= 30m3 – 3.6m3

= 26.4m3

Volume of a brick (v) = 15cm × 10cm × 5cm

= 0.15m × 0.10m × 0.05m

= 0.00075m3.

Now, number of bricks needed (N) = V
v

or, N = 26.4m3 = 35200
0.00075m3

Therefore, cost of bricks at Rs. 4 per piece = N × rate = 35200 × Rs. 4 = Rs. 140800

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