(c) An organization makes a profit of Rs. 1,50,000 during the year and decides
to distribute bonus at the rate of 2 % to each of its employees. If there are 10
employees, find the total amount of bonus.
(d) A firm makes a profit of Rs. 5,00,000 at the end of a year and distributes 16 % of
its profit to 40 employees. Find the bonus received by each.
3. (a) A company distributes bonus of Rs. 1000 to each of its 50 employees. If the annual
profit is Rs. 25,00,000, find the bonus rate.
(b) A restaurant management committee decides to pay Rs. 2500 as a bonus to each of
its 25 staff members. Find the rate of bonus if the total yearly profit is Rs. 12,50,000.
(c) A company makes a profit of Rs. 23,54,000. If the company decides to distribute
Rs. 5885 bonus each to its 64 employees, find what percent of profit is distributed
as bonus.
4. (a) Find the total profit made by a firm if at a rate of 2.5% bonus is distributed to 300
employees and each of them gets Rs. 1500.
(b) bAobnaunsk, wdhisattriibsuttheedto31toalf its profit to 140 employees. If each received Rs. 12000 as
profit of the bank?
5. The management of a soap industry decides to distribute bonus in percentage basis
of the monthly salary as follows.
Monthly salary Bonus
Rs. 5000 - Rs. 10,000 50% of salary
Rs. 10,000 - Rs. 12,000 40 % of salary
Rs. 12,000 - Rs. 15,000 25% of salary
If the total profit of the year is Rs. 5,00,000, find the bonus for the following monthly
salaries.
(a) Rs. 7,000 (b) Rs. 10,000 (c) Rs. 13,000.
Answer
1. Consult your teacher 2. (a) Rs. 6250 (b) Rs.9000 (c) Rs. 30,000 (d) Rs. 2,000
3. (a) 2% (b) 5% (c) 16% 4. (a) Rs. 1,80,00,000
(b) Rs. 50,40,000 5. (a) Rs. 3500 (b) Rs. 4,000 (c) Rs. 3250
3.4 Taxation
Income Tax
Ramesh is a government officer. His yearly income is Rs. 5,00,000. Nima is a
businessman. He earns more than Rs. 10,00,000 in a year. They have to pay some
Oasis School Mathematics-9 45
part of their income to the government. The amount paid by Ramesh and Nima to
the government is the income tax.
If the income of an individual or entity is more than a certain limit, then government
imposes certain part of extra income as tax.
Hence, an income tax is a government levy imposed on individuals or entities that
varies with income or profit of taxpayer.
Government manages its expense from the tax paid by the people.
The slogan of Nepal government on tax is "Tax payer friendly tax
administration, prosperity and good governance, our campaign".
æs/bftf dq} L s/ k|zf;g, ;d[l4 / ;z' f;g, xfdf| ] cleofgÆ
Tax rate in Nepal (FY 2073/74) upto 30 lakhs
Taxable income slab For Individual Tax amount
First Rs. 3,50,000 1% of 3,50,000
Tax rate
1% = Rs. 3500
15% of 1,00,000
Rs. 3,50,000 to Rs. 4,50,000 15%
25% = Rs. 15000
(For next 1,00,000) 35% 25% of 20,50,000
Rs. 4,50,000 to Rs. 25,00,000
= Rs. 5,12,500
(For next 20,50,000) 35% of Rs. 5,00,000
More than Rs. 25,00,000 (for next Rs.
5,00,000) = Rs. 1,75,000
Rs. 7,06,000
Total tax of yearly income Rs. 30,00,000
For couple
Taxable income slab Tax rate Tax amount
First Rs. 4,00,000 1% 1% of 4,00,000
Rs. 4,00,000 to Rs. 5,00,000 15% = Rs. 4,000
(For next Rs. 1,00,000) 15% of 1,00,000
Rs. 5,00,000 to Rs. 25,00,000 25%
(For next 20,00,000) = Rs. 15,000
More than Rs. 25,00,000 (for next 35% 25% of 20,00,000
Rs.5,00,000)
= Rs. 5,00,000
Total tax of yearly income Rs. 30,00,000 35% of Rs. 5,00,000
= Rs. 1,75,000
Rs. 6,94,000
46 Oasis School Mathematics-9
Note:
• Every employee has to pay 1% of his/her income for social security fund.
• An income is tax free on:
– the taxable income (upto 33%) kept in citizen investment fund
– the taxable income (10% of basic salary) kept on provident fund
– insurance premium (upto Rs. 20,000)
– the donation
– upto 75% of foreign allowance
• Medical expenditure
We can get the information related to the tax in the website http:www.ird.gov.np
Education Service Tax
Every private education institution has to pay 1% of their income from monthly fee and
admission fee as education service.
Let's be clear with an example:
A private Boarding School has 650 students and their average monthly fee per student is Rs.
2,200 and average admission fee per student is Rs. 5000. Find the amount of education service
tax.
Here, Total number of students = 650
Average monthly fee = Rs. 2200
Average yearly fee = Rs. 2200 × 12
= Rs. 26,400
Average admission fee = Rs. 5,000
Average yearly fee + Admission free = Rs. 26,400 + Rs. 5,000
= Rs. 31, 400
1% of Rs. 31,400 = 1 × 31,400
100
= Rs. 314
Total tax of 650 students = Rs. 314 × 650 = Rs. 2,04,100
Worked Out Examples
Example : 1
Monthly salary of Mr. Ramesh Bhattarai is Rs.38,000. He is a married man. If 1% tax is
imposed on his income upto Rs.4,00,000, find the amount of tax to be paid by him.
Solution:
Monthly salary of Ramesh Bhattarai = Rs.38,000
His yearly salary = Rs. 38,000 × 12
Oasis School Mathematics-9 47
= Rs.4,56,000
Tax on first Rs.4,00,000 = 1% of 4,00,000
1
= 100 × 4,00,000
= Rs. 4,000 ( Social security tax)
Remaining taxable income = Rs. 4,56,000 – Rs. 4,00,000
= Rs. 56,000
Tax on Rs. 5,6000 = 15% of 56,000
= 15 × 56,000 = Rs. 8400
100
∴ Total tax paid by him = Rs.4000 + Rs. 8400
= Rs. 12,400
Example : 2
Pemba Sherpa, an unmarried person, has a job in Tourism Board. His monthly salary is
Rs.32,000. Use tax rate of government of Nepal, find the amount of tax to be paid.
Solution:
Monthly salary of Pemba Sherpa = Rs. 32,000
His yearly salary = Rs. 32,000 × 12
= Rs. 3,84,000
For an unmarried person, tax for first Rs. 3,50,000 = 1% of Rs.3,50,000
= 1 × Rs. 3,50,000
100
= Rs.3500
Remaining income = Rs. 3,84,000 – Rs.3,50,000
= Rs. Rs. 34,000
Tax on Rs 34,000 = 15% of Rs. 34,000
= Rs. 15 × 34,000
100
= Rs. 5,100
∴Total tax = Rs.3,500 + Rs. 5,100
= Rs. 8,600.
Example : 3
Ram Bharosh, an unmarried person, works in a bank. His monthly income is Rs.60,000.
50% of his income is tax free. Using the above table of tax rate, calculate the amount of
48 Oasis School Mathematics-9
tax to be paid by him.
Solution: Monthly income of Ram Bharosh = Rs.60,000.
His yearly income = Rs. 60,000 × 12
= Rs. 7,20,000
Tax free income = 50% of 7,20,000
= 50 × 7,20,000
100
= Rs. 3,60,000
Taxable income = Rs.7,20,000 – Rs. 3,60,000
= Rs. 3,60,000
1
Now, 1% of Rs. 3,50,000 = Rs. 100 × 3,50,000
= Rs. 3,500
Remaining taxable income = Rs. (3,60,000 – 3,50,000)
= Rs. 10,000
Tax on Rs.10,000 = 15% of Rs. 10,000
15
= 100 × 10,000
= Rs. 1,500
∴His total tax = Rs. 3,500 + Rs. 1,500
= Rs. 5,000
Example: 4
Ashok is a married person earning Rs. 45,000 per month. He deposits 10% of the
income for C.I.F., 10% for P.F. and Rs. 1000 for premium of insurance every month. If
the rate of tax upto Rs. 4,00,000 is 15 % and above Rs. 4,00,000 is 25%, find the tax paid
by him per annum.
Solution:
Ashok's Monthly Income = Rs. 45,000
∴ Annual Income = Rs. 45,000 × 12
= Rs. 5, 40,000
Now, Exemption Limit (Tax free income) = Rs. 4,00,000
Amount deposited on C.I.F. = 10% of Rs. 5, 40,000
= Rs. 54,000
Amount deposited on P.F. = 10% of Rs. 5,40,000
= Rs. 54,000
Premimum of insurance = Rs. 1000 × 12
Oasis School Mathematics-9 49
= Rs. 12,000
∴ Total non–taxable income = Rs. (4,00,000 + 54,000 + 54,000 + 12,000)
= Rs. 5,20,000.
∴ Taxable Income = Total income – non–taxable income
= Rs. (Rs.5,40,000–5,20,000)
= Rs. 20,000
∴ Tax amount = 15 % of Rs. 20,000
15
= 100 × Rs. 20,000
= Rs. 3000.
Example : 5
The total number of students in a private boarding school is 408. Average monthly fee
of each student is Rs. 1250 and average admission fee is Rs.3500. Find the amount of
education service tax
Solution: Total number of students = 408
Average monthly fee = Rs. 1250
Average yearly fee = Rs. 1250×12=Rs. 15,000
Average admission fee = Rs.3500
Average yearly fee + Admission fee = Rs. 15000 + Rs. 3500
= Rs. 18500
Rate of education service tax = 1%
Education service tax = 1% of 18500
= 1 × 18500
100
= Rs. 185
Total amount = Rs. 408 × 185 = Rs. 75,480
Exercise 3.3
1. (a) What is the rate of Social Security Fund which every employee has to pay?
(b) Upto what percent of income kept in citizen investment fund is tax free?
(c) Upto what percent of income kept on provident fund is tax free?
(d) Upto what amout of insurance premium is tax free?
(e) Upto what percent of foreign allowance is tax free?
(f) What percent of a monthly fee of private education institute is to be paid as
Education Service Tax?
(g) For an unmarried person what amount of income is tax free?
(h) For a married person what amount of income is tax free?
50 Oasis School Mathematics-9
2. (a) Yearly income of Indra Bahadur Pariyar is Rs.3,50,000. Find the amount of tax to
be paid if 1% tax is imposed in the yearly income upto Rs.4,00,000.
(b) Moin Ali works in a finance company. He is an unmarried person. His yearly
income is Rs.3,40,000. If he has to pay 1% tax upto the income of Rs. 3,50,000, find
the amount of tax to be paid.
3. Yearly income of 4 friends are given below.
All of them are unmarried. 1% tax is imposed upto the income of Rs.3,50,000 and
15% on the income more than that. Find the amount of tax that each of them has to
pay in a year.
Person Income
Ram Saroj Yadhav Rs. 4,50,000
Gyaljen Tshring Sherpa Rs. 3,60,000
Bishnu Prasad Sharma Rs. 3,80,000
Namu Hang Rai Rs. 5,00,000
4. Yearly income of Puja, Dharana, Roshni and Namrata are given below. Using the
table of tax rate given above, find the amount of tax each of them has to pay. All of
them are married.
Person Yearly Income
Puja Rs. 4,40,000
Dharana Rs. 7,00,000
Roshni Rs. 5,60,000
Namrata Rs. 8,00,000
5. Monthly salary of Rahman, Aappa, Shankar and David is shown in the table. All of
them are married. Using the tax rate of government of Nepal, find the tax to be paid
by each of them in a year.
Person Monthly Income
Rahman Rs. 60,000
Aappa Rs. 55,000
Shankar Rs. 45,000
David Rs. 58,000
6. Monthly salary of four unmarried persons is given below. Using the tax rate of
Government of Nepal, find the amount of tax be paid by each of them.
Person Monthly Salary
A Rs. 47,000
B Rs. 55,000
C Rs. 37,000
D Rs. 41,000
Oasis School Mathematics-9 51
7. (a) Monthly salary of an army officer is Rs. 35,000. His 50% income is tax free.
How much tax does he have to pay, using the tax rate of Government of
Nepal.
(b) Monthly salary of a married finance officer is Rs. 55,000. The tax exemption limit
is Rs. 4,00,000. Using the government tax rate, find the amount of tax that he has
to pay.
8. (a) Annual income of a man is Rs. 4,50,000. If Rs. 4,00,000 is tax free income and
he pays Rs. 7,500 as annual tax, find the rate of tax.
(b) The yearly income of a Mr. Thapa is Rs. 4,80,000, in which tax is exempted upto
Rs. 4,00,000. If he pays Rs. 12,000 at the end of year as tax, find the rate of tax.
(c) Find the rate of tax paid by an officer if he earns Rs. 4,60,000 per year and the
exemption limit is Rs. 3,60,000, when he pays Rs. 1,000 as tax per month.
9. (a) Monthly income of Mr. Pitambar is Rs. 55,000. He deposits 15% of his income
for C.I.F., 10% for P.F. and he donates 5% of his income in a charity which is
also tax free. How much tax should he pay if,
(i) 1% tax is imposed upto the income of 4,00,000?
(ii) 15% tax is imposed above the income of Rs. 4,00,000?
(b) Monthly income of Hareram Ojha is Rs. 45,000. He deposits 10% of his income
for C.I.F., 10% for P.F., find the total tax to be paid if
(i) money deposited to C.I.F and P.F. are tax free.
(ii) 1% tax is imposed upto the income of Rs. 3,50,000
(iii) 15% tax is imposed on the income more than 3,50,000
(c) Monthly income of Raj Bhai Maharjan is Rs. 68,000. He is a married person. He
deposites 20% of his income for C.I.F., 10% of it for P.F. and he pays Rs.20,000 as
yearly premium of insurance. If the money deposited to C.I.F. and P.F. are tax
free and premium of insurance is also tax free, find the amount of tax to be paid
by him in a year.
10 (a) A private school earns Rs. 65,00,000 per year from the admission fee and monthly
fee. Find the amount of Education Service Tax to be paid by the school.
(b) Find the amount of Education Service Tax" to be paid by a school, if its total
income from monthly fee and admission fee is Rs. 85,67,000.
(c) A school earns Rs. 35,65,000 from admission fee and Rs. 85,68,000 from its
monthly fee. Find the amount of Education Service Tax to be paid by the school.
11. (a) Total number of students in a school is 545. Average monthly fee of the school is
Rs. 1,500 and average admission fee is Rs. 12,000 per student. Find
(i) total income of the school. (ii) amount of education service tax.
52 Oasis School Mathematics-9
(b) The total number of students in a school is 1240. Average monthly fee and
average admission fee are Rs. 1200 and Rs. 20,000 respectively. Find the amount
of Education Service Tax.
12. Given table shows the number of students, average monthly fee and average
admission fee of 5 different schools. Find the amount of Education Service Tax.
School Name Monthly Admission No. of
fee fee students
a. Sagarmatha Boarding School, Namche
b. Buddha Jyoti Boarding School, Taulihawa Rs. 1800 Rs. 15,000 370
c. Machhapuchhre Boarding School, Pokhara
d. Ram Janaki Boarding School, Janakpur Rs. 2500 Rs. 18,000 640
e. Old Capital Boarding School, Gorkha
Rs. 2800 Rs. 20,000 1250
Rs. 2200 Rs. 14,000 720
Rs. 3000 Rs. 15,000 840
Answer
1. Consult your teacher. 2. (a) Rs. 3500 (b) Rs. 3,400 3. Rs. 18,500, Rs. 5000, Rs. 8,000, Rs. 26,000
4. (a) Rs. 10000, (b) Rs. 69,000, (c) Rs. 34,000 (d) 94,000
5. Rs. 74,000, Rs. 59,000, Rs. 29,000, Rs. 68,000 6. Rs. 47,000, Rs. 71,000, Rs. 17,600, Rs. 29,000,
7. (a) Rs.2100 (b) Rs. 26,00 8. (a) 15% (b) 15% (c) 12% 9. (a) Rs.13,300
(b) Rs.15,800(c)Rs.31,800 10.(a)Rs.65000, (b)Rs.85,670 (c)Rs.1,21,330 1 1.(a)(i)Rs.1,63,50,000
(ii) Rs. 1,63,500 (b) Rs. 4,26,560 12. (a) Rs. 1,35,420, (b) Rs. 3,07,200, (c) Rs. 6,70,000,
(d) Rs, 2,90,880, (e) Rs. 4,28,400
Project work
I. Take the informations about the salary of bureaucrats of different level. Find the
yearly tax to be paid by an individual in both cases whether he/she is married or
unmarried. (Use the tax rate of Nepal).
Objective: To calculate the tax to be paid by an individual.
Material required: Salary sheet, table of tax rate, chart paper, A4 size paper, etc.
For married:
S. N. Level Monthly Yearly Tax Total tax
salary salary First slab Second Third Fourth
[Make the similar type of table for unmarried and calculate the tax.]
Oasis School Mathematics-9 53
II. Visit some private schools of your locality and take the informations about their
monthly salary, annual charges and number of students. Calculate the yearly income
of each school and the Education Service Tax to be paid by the school.
Objective: To calculate the education service tax.
Material Required: Chart paper, A4 size paper, colour pen, etc.
Observations
S.N. Name of school Number of Monthly fee Annual Yearly Education
students charge income service tax
Value Added Tax
Kriti, Smriti and Sristi went to a restaurant. They took coffee, Mo:Mo, tea, etc.
Altogether they took the food items of Rs. 700. But in bill they paid Rs. 791.
Why did they pay Rs. 91 more than the actual price?
Aadhya bought a mobile set of Rs. 18,000. But she paid Rs. 20340 for the mobile set.
Why did she pay Rs. 2340 more than the actual price?
The extra amount paid in both the cases is value added tax or VAT.
The tax imposed by the government on the selling price of goods is called "Value
Added Tax" in short 'VAT'. Besides the most necessary goods like rice, pulse, meat,
fish, milk, medicine, raw materials, etc. the buyer has to pay the VAT fixed by the
government on the sales price.
The current rate of VAT fixed by the government of Nepal is 13% i.e. if we buy the
goods we have to pay 13% of sales price extra as "VAT" .
The VAT so collected by the seller is deposited to the government fund and is used
for the development works and other expenses. VAT is always added to selling
price.
∴ VAT amount = VAT% of S.P.
Net S.P. = S.P. + VAT.
VAT% = VSA.PT. × 100%
54 Oasis School Mathematics-9
Note : • VAT was first applied on 10th April 1954 AD in France.
• VAT was introduced in Nepal in 1997 at the standard rate of 10%.
• Current VAT rate in Nepal is 13%.
Worked Out Examples
Example: 1
A restaurant charges Rs. 80 for one plate of "MO:MO". Find the bill to be paid by the
customer for it, including 13% VAT.
Solution:
The price of 1 plate "MO:MO" = Rs. 80
VAT amount = 13 % of Rs. 80
= 13 × Rs. 80
100
104
= Rs. 10
= Rs. 10.40
∴ Total bill amount = Rs. 80 + Rs. 10.40 = Rs. 90.40
Example: 2
The cost of an "Emergency Light" including 13% VAT is Rs. 565. Find the actual cost
without VAT.
Solution:
Let, Cost of light without VAT = Rs. x 13 13x
100 100
Here, VAT = 13% of Rs. x = × x = Rs.
By the question, x+ 13x = Rs. 565
100
or, 100x + 13x = Rs. 565
100
113x
or, 100 = Rs. 565
∴
x = 565 × 100 = Rs. 500
113
∴ Actual cost without VAT = Rs. 500
Example: 3
Marked price of an article is Rs. 7000. If 10% discount is allowed and 13% VAT is levied,
find (i) Discount (ii) S.P. (iii) VAT (iv) S.P. with VAT
Oasis School Mathematics-9 55
Solution:
Here, M.P. =Rs. 7,000 D% = 10%
We have, Discount = Discount% of M.P. = 10% of 7,000
= 11000 × 7,000 = Rs. 700
Again, S.P. = M.P.–Discount
= Rs. 6,300
Again, VAT = VAT% of S.P.
= 13% of 6,300
= 11030 × 6,300 = Rs. 819
S.P. with VAT
= S. P + VAT
= 6,300 + 819
= Rs. 7,119.
Exercise 3.4
1. (a) What is the current VAT rate in Nepal?
(b) What is the amount called which a customer has to pay while buying the goods?
(c) Write the formula to calculate VAT percent if S.P. is given.
(d) Write the formula to calculate VAT rate of net SP and spare given.
(e) Write the formula to calculate net S.P. if S.P. and VAT percent are given.
(f) If SP is x and VAT rate is 13% then what is the value of net S.P.?
2. (a) The price of a calculator excluding 13% VAT is Rs. 1,100. Find its price including
VAT.
(b) A radio costs Rs. 1,500 excluding the VAT. Find its price, if 13% VAT is added.
(c) Price of a pant is Rs 5,000. If 13% VAT is added to it, find the price of pant
including VAT.
3. (a) If the price of a TV set including 13 % VAT is Rs. 32,770, find its original price
without VAT.
(b) A gas heater costs Rs. 2,825 including 13% VAT. Find its original price without VAT.
(c) Cost of an article including 13% VAT is Rs. 6,554. Find the cost of the article
56 Oasis School Mathematics-9
excluding VAT.
4. (a) Find the rate of VAT if Rs. 5,876 is paid for an article which initially cost Rs.
5,200 without VAT.
(b) A table costs Rs. 650. If the customer pays Rs. 130 as the VAT over its costs price,
find the rate of VAT.
(c) Find the rate of VAT if a customer pays Rs. 1,469 for an article having cost
Rs.1,300.
5. (a) The price of a jacket in shop 'A' is Rs. 2,000 including VAT while the price of
the same jacket in shop 'B' is Rs. 1,800 excluding 13% VAT. Which is better
choice for a customer, A or B?
(b) There are two offers for customers to buy a laptop.
First offer: price Rs. 54000 including VAT.
Second offer : price Rs. 48000 excluding VAT.
If the VAT rate is 13%, identify which one is the better offer for customers.
6. (a) The marked price of a radio was Rs. 5,000. What will be the price of the radio
if 10% VAT was levied, after allowing 15% discount on it?
(b) The marked price of one set of computer was Rs. 65,500. What will
be the price of the computer if 15% VAT is levied after allowing 12%
discount on it?
(c) The marked price of a motorcycle was Rs. 85,000. What will be the price of the
motorcycle if 18% VAT was levied after allowing 15% discount on it?
Answer
1. Consult your teacher. 2. (a) Rs. 1243 (b) Rs. 1695 (c) Rs. 5650
3. (a) Rs. 29,000 (b) Rs. 2500 (c) Rs. 5800 4. (a) 13% (b) 20% (c) 13%
5. (a) A is better choice than B. (b) First offer 6. (a) Rs. 4,675 (b) Rs. 66,286 (c) Rs. 85,255
3.5 Dividend
Santosh has invested in a Bank. He has 450 shares. Value of each share is Rs. 100. Altogether
Bank has 80,000 shares. At the end of the year the Bank made the profit of Rs. 80,00,000.
Bank decided to distribute 40% of total profit to its shareholders.
This part of profit is the dividend.
Hence the dividend is a payment made by a corporation to its shareholders as a distribution
of profit. Dividend is a fraction of total profit.
Let's calculate the amount of dividend Santosh got.
Total profit = Rs. 80,00,000
Oasis School Mathematics-9 57
Dividend = 40% of 80,00,000
= 14000 × 80,00,000
= Rs. 32,00,000
Profit of 80,000 shares = Rs. 32,00,000
Profit of 1 share = Rs. 32,00,000
80,000
= Rs. 40
Profit of 450 shares = Rs. 450 × 40
= Rs. 18,000
∴ Dividend that Santosh got = Rs. 18,000
Note: Dividend is allocated as a fixed amount per share.
Shareholders receive the dividend in the proportion of their investment.
Worked Out Examples
Example: 1
There are altogether 4,00,000 shares in a hydropower company. Value of each share is Rs.
100. Company has decided to distribute Rs. 80,000 dividend. Find the dividend per share.
Solution:
Here,
Total number of shares = 4,00,000
Total dividend = Rs. 80,000
Dividend per share = Rs. 80,000
4,00,000
Example: 2
= Rs. 1 = Rs. 0.20
5
Anu has 800 shares in a hydropower company. Company has altogether 1,00,000 shares.
At the end of the year, company made a profit of Rs. 50,00,000. Company has decided to
distribute 5% of its profit to its shareholders find-
(i) total dividend.
(ii) dividend per share.
(iii) total dividend got by Anu.
58 Oasis School Mathematics-9
Solution:
Here,
Total profit made by a company = Rs 50,00,000
(i) Total dividend = 5% of 50,00,000
= 5 × 50,00,000
100
= Rs. 2,50,000
(ii) Total number of shares of company = 1,00,000
Dividend per share = 2,50,000 = Rs. 2.50
1,00,000
(iii) Total number of shares of Anu = 800
Her dividend = Rs.800 × 2.50
= Rs. 2,000
Example: 3
A finance company has altogether 50,000 shares of Rs. 100 each. At the end of a year
company made the profit of Rs. 5,00,000. Company decided to distribute 50% of its
profit as dividend. Find the dividend got by Sundar, Rehman, Pemba, Dayahang and
Ramsaroj if the number of shares they have is :
Shareholders Number of shares
Sundar 500
Rehman 600
Pemba 1500
Dayahang 1200
Ramsaroj 1000
Solution:
Total profit made by the finance company = Rs. 5,00,000
Total number of shares = 50,000
Total dividend = 50% of 5,00,000
= 50 × 5,00,000
100
= Rs. 2,50,000
Dividend per share = Rs. 2,50,000
50,000
= Rs. 5
Oasis School Mathematics-9 59
Dividend obtained by Sundar = Rs 5 × 500 = Rs 2,500
Dividend obtained by Rahman = Rs 5 × 600 = Rs 3,000
Dividend obtained by Pemba = Rs 5 × 1500 = Rs 7,500
Dividend obtained by Dayahang = Rs 5 × 1200 = Rs 6,000
Dividend obtained by Ramsaroj = Rs 5 × 1000 = Rs 5,000
Exercise 3.5
1. (a) If the total dividend of company is Rs. x and the number of shares is y then what
is the value of dividend per share?
(b) The dividend per share is Rs. ‘a’ the number of shares is ‘b’ then find the value of
total dividend.
(c) The total profit of a company is Rs. x and 50% of total profit is to be divided
among shareholders as dividend, find the value of total dividend.
2. (a) The yearly profit of a company is Rs. 5,80,000. Find the dividend if the company
has decided to distribute 20% of its profit as divided.
(b) The total profit made by a finance company is Rs. 15,60,000. If the company
decided to distribute 10% of its profit as dividend, find the amount of dividend.
3. (a) Find the total profit made by a company if its dividend is 3,50,000 which is 5% of
its total profit.
(b) A company distributed 10% of its total profit as the dividend. If its dividend is
Rs. 2,25,000, find the profit made by the company.
4. (a) A hydro-power company has 20,00,000 shares. Find the total dividend and
dividend per share if it makes a profit of Rs. 5,00,00,000 and 20% if total profit is
the dividend.
(b) The total number of shares in a company is 1,00,000. Value of each share is Rs.
100. Company made the total profit of Rs. 50,00,000 in a year. 20% of the total is
the dividend. Find
(i) total dividend (ii) dividend per share.
5. Given table shows the total profit, rate of dividend and number of share of 5 different
companies. Find the total dividend and dividend per share.
60 Oasis School Mathematics-9
Company Total number of Rate of Total profit
shares dividend
Bandana Finance Company 10% 55,00,000
Daraudi Hydro-power 1,10,000 90,00,000
Aruna Cement Factory 5% 1,20,00,000
Sadhana Publication Pvt. Ltd 80,000 8% 1,50,00,000
Tilicho Bikash Bank 1,20,000 1,00,00,000
12%
1,05,000 15%
50,000
6. (a) Rupsi Co-operative Pvt. (Ltd) has 50,000 shares. Value of each share is Rs. 100.
Tulasa has 2,000 shares in the co-operatives. Co-operative made a total profit
of Rs. 60,00,000 and it decided to distribute its 20% as dividend. Find
(i) total dividend
(ii) dividend per share
(iii) dividend got by Tulasa.
(b) Prathana Supermarket has 40,000 shares. Value of each share is Rs. 100.
Anugya has 3000 shares. Supermarket made a profit of Rs. 50,00,000 in a year.
It decided to distribute 40% of its profit as dividend, Find
(i) total dividend
(ii) dividend per share
(iii) dividend get by Anugya.
(c) Motherland cooperatives has 10,000 shares of Rs. 100 per share. Amrit has
1,500 shares in the co-operatives. It made a net profit of Rs. 8,00,000 in a year.
Cooperatives decided to distribute 50% of its profit as the cash dividend. Find
(i) total dividend
(ii) dividend per share
(iii) dividend get by Amrit.
Answer
1. Consult your teacher 2. (a) Rs. 1,16,000 (b) Rs. 15,600
3. (a) Rs. 70,00,000 (b) Rs. 22,50,000 4. (a) Rs. 1,00,00,000, Rs. 5 (b) Rs. 10,00,000, Rs. 10
5. a) Rs. 5,50,000, Rs. 5 (b) Rs. 4,50,000, Rs. 5.625 (c) Rs. 9,60,000, Rs. 8
(d) Rs. 18,00,000, Rs. 17.14 (e) Rs. 15,00,000, Rs. 30
6. (a) Rs. 12,00,000, Rs. 24, Rs. 48,000 (b) Rs. 20,00,000, Rs. 50, Rs. 1,50,000
(c) (i) Rs. 4,00,000 (ii) Rs. 40 (iii) Rs. 60,000
Oasis School Mathematics-9 61
Project work
Visit the managing director or CEO of some finance companies or industries to take
the informations about the number of shares, net profit, percentage of net profit to be
divided as the dividend. Calculate their dividend and dividend per share.
Objective: To calculate the dividend
Material required: Chart paper, colour pen, Newspaper, etc.
Observation:
S.N. Name of Total number Net profit Dividend Dividend
organisations of shares per share
62 Oasis School Mathematics-9
Unit
4 Household-Arithmetic
4.1 Warm-up Activities NEPAL
Discuss the following in your class and draw ELECTRICITY AUTHORITY
the conclusion.
Distribution & Consumer Services
• If the cost of 1 unit of electricity is Rs. 7.50,
what is the cost of 10 units of electricity? ELECTRICITY BILL
• What is the meaning of 1 unit electricity? BRANCH: KTM
BILL MTH & YEAR: PAU-2073
• How much consumption water is BL NO: 102035087109009457
equivalent to 1 unit?
• If the bill of telephone is Rs. 700, what is the SC NO: 038.08.00 KHA 3
VAT amount to be paid? CONSUMER ID: 35412
CATG:
• What is the minimum charge of NAME: Domestic (6-15A)
taxifare? ADDRESS: MR. BHATTARAI R.
METER NO:
4.2 Electricity Bills MF : MTR STS: KTM, NEPAL
APPROVED LOAD: DEF 006
Household arithmetic deals with the method PRESENT RDG:
of calculating the billing system of electricity, 21312
water, telephone and even taxi fare.
See the given meter reading for the two PREVIOUS RDG: 21562
successive months. UNITS: 250
The last digit shows the fraction of a unit of ENERGY CHARGES: 2250.00
the electricity, so ignoring that (highlight) SUBSIDY CHARGES: 0.00
OTHER CHARGES: 0.00
Bhadra 165 3 MTR RENT AMOUNT: 0.00
Asoj 198 7 CURRENT AMOUNT:
2250.00
N umber of units at the end of Bhadra = 165
Number of units at the end of Asoj = 198 ARREARS AMOUNT 0.00
BILL AMOUNT: 2250.00
∴ Number of units consumed during one INSTALLMENT:
month = 198 – 165 = 33 SBM ID: 0.00
READER:
Applying the same method for finding the
number of units consumed, we can find the CFL lrdsf] ko| f]u u/L ljBt' sf]
amount of bill for the particular month. dx;n' 36fcf}+ .
Oasis School Mathematics-9 63
Billing system
I. If the consumption is upto 20 units,
the rate of charge is Rs. 4 per unit.
II. If the consumption is upto 30 units,
• the rate for first 20 units is Rs. 4 per unit.
• the rate for remaining units is Rs 7.30 per unit.
III. If the consumption is upto 50 units,
• the rate for the whole consumption is Rs. 7.30 per unit.
IV. If the consumption is upto 150 units,
• the rate for first 50 units is Rs. 7.30 per unit
• the rate for remaining units is Rs. 8.60 per unit.
V. If the consumption is upto 250 units,
• the rate for first 150 units is Rs. 8.60 per unit
• the rate for remaining units is Rs. 9.50 per unit
IV. If the consumption is more than 250 units
• the rate for first 250 units is 9.50 per unit
• the rate for remaining units is Rs. 11.50 per unit
Extra charge and the rebate on the payment of bill
• If the payment is made within 7 days of mater reading, there is a rebate of 3%
on the total bill and if the bill is less than Rs. 80, there is a rebate of Rs. 4.
• If the payment is made between 8th and 22nd days of meter reading, there is
no extra charge and no rebate .
• If the payment is made between 23rd and 30th days of meter reading, there is
an extra charge of 5% on the bill.
• If the payment is made between 31st to 40th days of meter reading, there is an
extra charge of 10% on the total bill.
• If the payment is made between 41st to 60th days of meter reading, there is an
extra charge of 25% on the total bill.
Remember !
1 unit = 1 kw hr.
i.e. if an electric appliance of power 1kw is used for 1 hour, 1 unit electricity is
consumed.
64 Oasis School Mathematics-9
The minimum charge of the bill depends upon the meter capacity.
Meter capacity Minimum units Minimum changes (Rs.)
Upto 5 Ampere 20 80
6–15 Ampere 50 365
16–30 Ampere 100 795
31–60 Ampere 200 1765
Worked Out Examples
Example: 1
The electricity charge for first 20 units Rs. 4 per unit and from 20 units to 30 units the
rate is Rs. 7.30 per unit. Find the amount of bill to be paid if 27 units of electricity is
consumed.
Solution: Total energy consumed = 27 units
Charge for first 20 units = Rs. 20 × 4 = Rs. 80
Again, remaining units = (27 – 20) = 7 units.
Charge for remaining 7 units = Rs. 7 × 7.30
= Rs. 51.10
∴ Total amount of bill = Rs. 80 + Rs. 51.10
= Rs. 131.10
Example: 2
Given table shows the previous meter readings and current meter readings of Hirakaji's house.
Current reading Previous reading
3422 3195
(i) Find the total energy consumption in that month.
(ii) Using the rate of electricity given above, find the amount of bill.
(iii) If the payment is made on 25th of the next month, find the bill to be paid on that
day.
Solution:
Here, Present readings = 3422
Previous readings = 3195
Total energy consumed = 3422 – 3195
= 227 units
In this case, the rate for first 150 units.
Oasis School Mathematics-9 65
= Rs. 8.60 per unit
Charge of 150 units = Rs. 150 × 8.60
= Rs. 1290
Remaining units = (227 – 150) = 77 units.
Rate of 77 units = Rs 9.50 per unit
Charge for 77 units = 77 × 9.50
= Rs. 731.50.
∴ Total bill = Rs. (1290 + 731.50)
= Rs. 2021.50
Again, if the payment is made 25th of the next month,
Extra charge = 5% of 2021.50
= 5 × 2021.50
100
= Rs. 101.075
∴ Total bill = Rs. 2021.50 + Rs. 101.075 = Rs. 2122.575.
Example: 3
A household has a meter of capacity 6 – 15A. If the previous and current meter reading
are 21640 and 21900, find
(i) amount of bill.
(ii) amount of bill to be paid if the payment is made after 35 days of meter reading.
[Given that the rate of electricity for first 50 units is Rs. 9.50 per unit and the rate for
more than 250 units is Rs. 11.50 per unit.]
Solution:
Here,
Capacity of meter = 6 – 15A
Total units consumed = current reading – previous reading
= 21,900 – 21,640
= 260 units.
Since the capacity of meter is 6 – 15A,
Rate upto 250 units = Rs. 9.50 per unit
Charge upto 250 units = 250 × 9.50
= Rs. 2375
Charge for remaining 10 units = 10 × 11.50
= Rs. 115
66 Oasis School Mathematics-9
∴ Total charge = Rs. 2375 + Rs. 115
= Rs. 2490
If the payment is made after 35 days of meter reading, there is an extra charge of 25%
∴ Extra charge = 25% of 2,490
= 25 × 2,490
100
= Rs. 622.50
∴ Total bill = Rs. 2,490 + Rs. 622. 50
= Rs. 3112.50.
Exercise 4.1
1. (a) What is the meaning of 1 unit electric energy?
(b) If the meter capacity is upto 5 ampere, what minimum charge is to be paid?
(c) If the meter capacity is 6-15 ampere, what minimum charge is to be paid?
(d) If the meter capacity is 16-30 ampere, what minimum charge is to be paid?
(e) If the meter capacity is 31-60 ampere, what minimum charge is to be paid.
(f) If the meter capacity is 5 ampere, upto what unit of consumption, what minimum
charge is to be paid?
(g) If the meter capacity is 6-15 ampere, upto what unit of consumption, what
minimum charge is to be paid?
2. The given table shows the meter reading of a household in the beginning of
different months. Study the table and answer the questions give below.
Month Baishak Jestha Aashad Shrawan Bhadra
Meter reading 2374 2467 2560 2674 2772
Find the total energy consumed in different months.
3. (a) The minimum charge of consumption of electricity is Rs. 80 upto 20 units. If the
rate per unit is Rs. 7.30 for more than 20 units, find the amount of bill to be paid
by a person who uses 27 units in a month.
(b) The rate of electricity upto 150 units is Rs. 8.60 per unit. And the rate above
that is Rs. 9.50 per unit. Find the charge of electricity of the total electricity
consumed is 220 units.
(c) The rate of electricity upto 250 units is Rs. 9.50 per units. The rate above that is
11.50 per unit. Find the charge of electricity if the total energy consumed is 300
units.
Oasis School Mathematics-9 67
4 (a) Meter reading of a household on the 1st of Jestha is Rs. 565 and that on 1st of
Asadh is 720. If the rate of electricity is Rs. 8.60 for first 150 units and Rs 9.50
for the remaining units, find
(i) how much electricity is consumed on the month of Jestha?
(ii) calculate the amount of money to be paid for the month of Jestha.
(b) The present meter readings and previous meter readings in given.
(i) the energy consumed in the month
(ii) the amount to be paid if the rate of electricity Present readings 2765
for first 250 units is Rs. 9.50 per unit and the Previous readings 2502
rate more than that is Rs. 11.50 per unit.
5 The given table shows the current and previous meter readings of Ramlal's house.
Use the rules of billing system and find: Capacity 6 – 15A
(i) total energy consumed Previous reading 2472
(ii) amount of bill Current reading 2701
(ii) amount of bill to be paid if the payment is made on 5th day after meter reading.
6 Study the given table that shows previous readings and current readings of
electricity. Find the amount of bill to be paid in the given date:
(a) Capacity = 0 – 5A Previous reading : 2582 Current reading: 2685
Date of bill payment : 5th day after meter reading.
(b) Capacity : 0 – 5 A Previous reading : 378 Current reading = 456
Date of bill payment : 18th day after meter reading.
(c) Capacity : 6 – 15A Previous reading : 682 Current reading = 815
Date of bill payment : 27th day after meter reading.
(d) Capacity : 6 – 15A Previous reading : 7120 Current reading = 7385
Date of bill payment: 38th day after meter reading.
(e) Capacity : 6 – 15A Previous reading : 8295 Current reading = 8455
Date of bill payment : 56th day after meter reading.
7. The table given below shows the meter reading of electricity.
1st Baishak 1st Jestha 1st Ashad 1st Shrawan 1st Bhadra
1020 1080 1320 1665 1780
Meter is read on the first day of the next month. The rules of payment are as
follows.
68 Oasis School Mathematics-9
(i) Upto 7th day of meter reading, a rebate of 3 %.
(ii) From 8th day to 22nd day, according to the bill.
(iii) From 23rd to 30th day, additional charge of 5% .
(iv) From 30th to 45th day, additional charge of 10%.
(v) From 46th to 60th day, additional charge of 25%.
Find the
(a) bill of Baisakh is paid on 10th Jestha.
(b) bill of Jestha is paid on 5th Asadh.
(c) bill of Asad is paid on 28th Shrawan.
(d) bill of Shrawan is paid on 20th Aswin.
The rate is according to Nepal Electricity Authority.
Answer
1. Consult your teacher 2. Baishak-93 units, Jestha– 93 units, Asadh–114 units, Shrawan–98 units,
3. (a) Rs. 131.10 (b) Rs. 1955 (c) Rs. 2950 4. (a) i. 155 units ii. Rs. 1337.50
(b) i. 263 units ii. Rs. 2524.50 5. i. 229 units ii. Rs. 2040.50 iii. Rs. 1979.29
6. (a) Rs. 796.18 (b) Rs. 605.80 (c) Rs. 1132.74 (d) Rs. 2802.25 (e) Rs. 1731.25
7. (a) Rs. 451 (b) Rs. 2080.65 (c) Rs. 3640.87 (d) Rs. 1355
4.3 Water Bill
Everyone needs water. Water is supplied by Water Supply Corporation. Let's understand
the sample of bill, rate of bill and system of payment of bill.
Study the given water bill properly.
Size pipe Minimum Minimum (Rs.) Rate for extra
(Inch) Consumption of consumption (per
water (in litre) 100 (for first 10 units)
1/2 " 1,910 (for first 27 units) 1000 litre)
3/4" 10,000 3,960 (for first 56 units) 32
27,000 10,950 (for first 155 units) 71
1" 22,600 (for first 320 units)
1.5" 56,000 62,240 (for first 881 units) 71
2" 1,27,865 (for first 1810 units)
3" 1,55,000 71
4"
3,20,000 71
8,81,000 71
1,810,000 71
Oasis School Mathematics-9 69
NEPAL WATER SUPPLY CORPORATION
Branch Office Kathmandu
Name: Laxmi Shrestha Consumer No 6 - 32-2
Area : - 1-6-305
Address : Sinamangal, Kathmandu
Month : Falgun
Present Reading Former Reading Unit Consumed Minimum Cost Extra Cost Total
512 480 32 units Rs. 100 Rs. 704 Rs. 804
Meter reading of Meter reading of (512-480) Minimum charge i.e. 32–10 = 22 Units B. Chaudhary
last day of Falgun last day of Magh = 32 units of 10 units=Rs.100 22×32=Rs.704
Signature
Mini. cost+extra unit cost
i.e. Rs. 100+704=Rs.804
Note : Sewerage charge = 50% of the charge on water consumption
Rental charge = 4.5 per month
Billing system
According to current billing system, the rule of billing is as follows
• If the payment is made within first and second month from the date of billing,
there is a rabate of 3%.
• If the payment is made within third month from the date of billing, there is no
extra charge and no rabate.
• If the payment is made in the fourth month, there is an extra charge of 10% .
• If the payment is made in the fifth month, there is an extra charge of 20%.
• If the payment is made after fifth month, there is an extra charge of 50%
Worked Out Examples
Example: 1
A household consumed 40 units of water. If the minimum charge upto 10 units is Rs.
100 and charge for extra unit is Rs. 32 per unit, find the amount of bill.
Solution:
Total unit consumed = 40 units
Minimum charge for 10 units = Rs. 100
Remaining units = (40 – 10) units Use the rate
= 30 units. given above
Charge for extra 30 units = Rs. 30 × 32
70 Oasis School Mathematics-9
= Rs. 960.
Total bill = Rs. 100 + Rs. 960
= Rs. 1060.
Example: 2
Previous meter readings and current meter readings are 345 and 380. Find:
(i) total units consumed.
(ii) total charge if the charge for first 10 units is Rs. 10 per units and Rs. 32 per unit for
each extra unit.
(iii) total bill including 50% of total charge as sewerage charge.
Solution:
Previous meter readings = 345
Current meter readings = 380
Total units consumed = (380 – 345) = 35 units
Minimum charge upto 10 units = Rs. 100
Remaining units = (35 – 10) units = 25 units.
Charge for extra units = Rs. 25 × 32
= Rs. 800
Total charge = Rs. 800 + Rs. 100
= Rs. 900
Again, Sewerage charge = 50% of 900
= 50 × 900
100
= Rs. 450
∴ Total bill = Rs. (900+450) = Rs. 1350
Example: 3
A pipe of size 3" supplies water to Shanti Deep Public School. If the total consumption
4
of water is 50 units, the rate of first 27 units Rs. 1910 and the rate for each additional
unit is Rs. 71 per unit. Find the total bill to be paid including 50%, sewerage charge if
the payment is made in the fourth month after billing.
Solution: Total consumption = 50 units.
For the pipe of size 3" ,
4
Minimum charge for first 27 units = Rs. 1,910
Extra units = (50 – 27) units = 23 units.
Charge for extra 23 units = Rs. 23 × 71
= Rs. 1,633
Oasis School Mathematics-9 71
Total bill = Rs. 1910 + Rs. 1633
= Rs. 3543
Now, Sewerage charge = 50% of 3543
= 50 × 3543
100
= Rs. 1771.50
Now, Total bill = Rs. 3543 + Rs. 1771.50
= Rs. 5314.50
Again, i f the payment is made after fourth month of billing,
Extra charge = 10% of Rs. 5314.50 Use the rate
given above
= 10 × 5314.50
100
= Rs. 531.45
Total bill including extra charge = Rs. 5314.50 + 531.45 = Rs. 5845.95
Exercise 4.2
1. (a) What is the meaning of consumption of 1 unit water?
(b) If the size qoufatnhteitypiwpeatiesr?21 ", what is the minimum charge for consumption of
minimum
(c) If the size of pipe is 3 ", what is the minimum charge for minimum consumption
of water? 4
(d) If the size of pipe is 1", what is the minimum charge for minimum consumption
of water?
(e) If the size of pipe is 1.5", what is the minimum charge for minimum consumption
of water?
(f) According to the rule of Nepal Water Supply Corporation, what is the rate of
Sewerage Charge?
(g) What is the rule of payment of water bill if it is paid within the first and second
month from the date of billing?
(h) What is the rule of payment of water bill if it is paid on the fourth month from the
date of billing?
2. Find the amount of bill to be paid in each of the following cases if water is supplied
through the pipe of 12".
72 Oasis School Mathematics-9
[ Minimum charge for first 10 units is Rs. 100 and charge for each extra unit is Rs. 32 per unit]
(i) 25 units (ii) 30 units (iii) 50 units (iv) 60 unit.
3. Find the amount of bill to be paid in each of the following cases if water is supplied
through the pipe of 43".
[Minimum charge for first 27 units = Rs. 1910 and charge for each additional unit is Rs. 71]
(i) 60 units (ii) 75 units (iii) 80 units (iv) 56 units
4. (a) The previous and current meter reading of a household are 362 and 390. Water
is supplied through the pipe of 21". Using the billing system mentioned earlier, find
(i) total consumption of water
(ii) total charge
(iii) total charge including sewerage charge.
(b) The current reading and previous reading of water of a hotel is 3827 and 3918,
where water is supplied by the pipe of 1". Using the billing system mentioned
earlier, find
(i) total unit consumed
(ii) total charge
(iii) total charge including sewerage charge
5 The given table shows the previous readings, current readings, size of pipe and date
of payment of bill, find the total charge to be paid including 50% sewerage charge.
S.N. Previous reading Current reading Size of pipe Date of bill payment
i. 378 400 1" within 2nd month after
2 billing
ii. 6542 6580 in the 3rd month
1"
iii. 765 812 2 in the 4th month
iv. 2546 2618 3"
4 in the 5th month
1"
Answer
1. Consult your teacher
2. (i) Rs. 580 (ii) Rs. 740 (iii) Rs. 1380 (iv) Rs. 1700 3. (i) Rs. 4253
(iv) Rs. 3969 4. (a) (i) 28 units (ii) Rs. 676
(ii) Rs. 5318 (iii) Rs. 5673 (ii) Rs. 6445 (iii) Rs. 9667.50
(iii) Rs.5404.5 (iv) Rs. 9172.8
(iii) Rs. 1014 (b) (i) 91 units
5. (i) Rs. 704.22 (ii) Rs. 1494
Oasis School Mathematics-9 73
4.4 Telephone Bill
In modern time almost everyone is familiar with the use of telephone. Telephone
service is provided by Nepal Telecommunication Corporation (NTC) with certain
charge to its customers. Current charge is Rs. 200 for the minimum call cost upto
175 calls which is also known as rental charge. The charge per call for more than 175
calls is Re.1. In the bill, the bill amount adding the charge of rental and extra calls
is called sub total. Over sub total amount, 10% is added as telecom service charge
(TSC) which is called total bill amount. Finally 13 % VAT is added over the total bill
amount which is grand total bill amount. To clear the bill a customer is to pay the
grand total amount.
The model of telephone bill is printed below.
NEPAL TELECOM
Branch Office Naxal Kathmandu
No: 2598 Statement For the Month: Chaitra
Date : 2073-1-4
TELL NO : 01-225619 ACCOUNT CODE: A200817289 CHARGE SUMMARY Rs.
NAME : GOMA DEVI BASNET, KTM.
CLASS: PERMANENT LOCAL 2021
TYPE: ORDINARY LINE STATUS: RESIDENTAL STD: .......
ISD: .......
LOCAL CALL DETAILS NAT. OA: .......
INT OA: .......
PREVIOUS CURRENT TOT. CALLS RENTAL RS. EX. CALL TOTAL RS. ADJUSTMENT ........
24520 2021 SUB TOTAL 2021
26716 1996 200 1821 TSC 10% 202.10
TOTAL 2223.10
VAT 13% 289
GRAND TOTAL 2510.10
Reading of last day Reading of last day (26716-24520) Mini. charge of 1821 calls at
of Falgun=24520 of Chaitra=26716 = 1996 175 calls = 200 Re.1/call=1821
Rental+Extra 10% of Subtotal+TSC Total+13% Vat
calls cost=2021 sub-to- 2223.10 of total=2512.10
tal=202.10
Worked Out Examples
Example: 1
The telephone charge of first 100 calls is Rs. 180 and the charge for each additional call
is Rs. 1.50. How much will be charged for 400 calls including 10% TSC and 13% VAT?
Solution:
74 Oasis School Mathematics-9
Charge for first 100 calls = Rs. 180
Additional calls = 300 calls
Charge for 300 additional calls = Rs. (300 × 1.50) = Rs. 450
Total charge = Rs. 180 + Rs. 450
= Rs. 630
Charge including TSC = 630 + 10% of 630
= 630 + 10 × 630
100
= 630 + 63
= Rs. 693
Total bill including 13% VAT
= Rs. 693 + 13% of 693
= Rs. 693 + 13 × 693
100
= Rs. 693 + Rs. 90.09
= Rs. 783.09
Example: 2
Minimum charge of telephone upto 100 calls is Rs. 180. The charge for each extra call is
Rs. 1.50. If a man paid Rs. 726 with 10% TSC and 10% VAT, find the total number of
calls made in that month.
Solution:
Let the bill without adding 10% VAT be x
Now, x+10% of x = 726
x+ 10 × x = 726
100
or, 11x
10 = 726
x = 726 × 10
11
x = Rs. 660.
Again, let the bill without adding 10% TSC be y
Now, y + 10% of y = 660
or, y+ 10 × y = 660
or, 100
11y
or, 10 = 660
y = 660 × 10
11
y = Rs. 600.
Oasis School Mathematics-9 75
Now, m inimum charge + charge for additional calls = 600
180 + charge for additional calls = Rs. 600
∴ charge for additional calls = Rs. 600 – Rs. 180 = Rs. 420
Rate of additional call = Rs. 1.50 per call
Number of additional calls = 14.2500 = 280
∴ Total number of calls = 100 + 280 = 380.
Note: However NTC has offered the discounted rate for the telephone calls for
STD/ISD during certain days and hours as classified below. Business hour
(08:00 — 18:00), Normal hour (18:00 — 22:00), Off hour (6:00 — 8:00).
Exercise 4.3
1. (a) According to the rule of Nepal Telecom, what is the minimum charge for
minimum number of calls?
(b) What is the rate of Telephone service Charge?
2. (a) The minimum charge for 100 calls is Rs. 180. The charge for each additional call
is Re. 1. If a household makes total 450 calls during a month find the total bill
to be paid including 10% TSC and then 10% VAT.
(b) The minimum charge of telephone call upto 150 calls is Rs. 200. If the charge for
each additional call is Rs. 1.50, how much will be the charge of 500 telephone
calls with 15% TSC and 13% VAT.
(c) The minimum charge for 175 calls is Rs. 200. The charge for each additional call
is Re. 1. If Seema, a telephone user, makes a total of 530 calls during a month,
find her grand total bill to be paid including 10% TSC and 13% VAT.
3. (a) The minimum charge of telephone call upto 100 calls is Rs. 180 and the charge
of each additional call is Rs. 1.50. If Kavi Bhandari paid Rs. 330 with 10% VAT
of his one month telephone bill, find how many extra calls were made in that
month?
(b) The minimum charge of telephone call upto 100 calls is Rs. 180. The charge for
each extra call is Rs. 1.50. If a man paid Rs. 1320 with 10% VAT of his telephone
bill, find how much extra calls were made?
(c) The minimum charge of telephone call upto 100 calls is Rs. 180. The charge for
each extra call is Rs. 1.50. A man paid to Rs. 2178 including 10% TSC and 10%
VAT. Find the total number a calls he made.
76 Oasis School Mathematics-9
Answer
1. Consult your teacher 2. a) Rs. 641.30 (b) Rs. 942.14 (c) Rs. 689.87
3. (a) 80 calls (b) 680 calls (c) 1180 calls.
4.5 Taxi Fare
In some events we have to use taxi. We have to pay taxi fare according to the readings
of taxi meter. There is a meter reading attached in the taxi to indicate the minimum
charge and total charge to be paid.
Let's understand the given table.
Time Minimum Charge per additional 200m
6 A.M – 9. P.M. Rs. 14 Rs. 7.20
9 P.M. – 6 A.M. Rs. 21 Rs. 10.80
Waiting charge per 2 minutes = Rs. 7.20.
If we use taxi for the rent, we must have become familiar with the taxi fare. There
is meter reading attached in the taxi to indicate the minimum charge as well as
the charge for different distances in per km. basis. The concerned authority of
the government decides on all these rates to be paid by the taxi users. Currently,
it has been fixed that the minimum charge is Rs. 14 and charge per kilometer is
Rs. 22.80 as additional charge.
Worked Out Examples
Example: 1
The minimum charge of taxi is Rs. 14 and the rate for each additional 200m is Rs. 7.20,
find the taxi fare if the total distance covered is (i) 12 km (ii) 15 km 200m.
Solution:
(i) Minimum charge = Rs. 14
Charge for each 200m = Rs. 7.20
Charge for each km = Rs. 7.20 × 5 5 × 200m = 1000m
= 1 km.
= Rs. 36
Charge for 12 km = Rs. 36 × 12
= Rs. 432
∴ Total charge = Rs. 14 + Rs. 432
= Rs. 446
(ii) Minimum charge = Rs. 14
Charge for each 200m = Rs. 7.20
Oasis School Mathematics-9 77
Charge for each km = Rs. 7.20 × 5
= Rs. 36
Charge for 15 km = Rs. 36 × 15
= Rs. 540
∴ Total taxi fare = Rs. (14 + 540 + 7.20)
= Rs. 561.20.
Example: 2
A passenger travels 8km at night 10 P.M. The minimum charge is Rs. 21 and Rs. 10.80 for
each 200m. Find the total taxi fare if the taxi waits for 8 minutes and the rate of waiting
charge is Rs. 7.20 for each 2 minutes.
Solution: Minimum charge = Rs. 21
Rate for each 200m = Rs. 10.80
Rate for a km = Rs.10.80 × 5
= Rs. 54
Charge for 8 km = Rs. 8 × 54
∴ = Rs. 432
Total taxi fare = Rs. (432 + 21)
= Rs. 453
Again, waiting charge for 2 minutes = Rs. 7.20
waiting charge for 1 minute = Rs. 7.20
2
waiting charge for 8 minutes = Rs. 7.20 × 8
2
= Rs. 28.80
∴ Taxi fare including waiting charge = Rs. (453 + 28.80)
= Rs. 481.80
Example: 3
Using the above rate of taxi fare, calculate the total taxi fare in each of the following cases.
Distance Time Waiting time
10km 2:00 P.M. 4 minutes
6 km 11:00 P.M. 10 minutes
Solution:
Since, the time is 2:00P.M.
The minimum charge is Rs. 14.
and the rate for each 200m is Rs. 7.20
Then rate for each 1 km is Rs. 7.20 × 5 = Rs. 36.
78 Oasis School Mathematics-9
Taxi fare for 10 km = Rs 36 × 10
= Rs. 360
Waiting charge for 2 minutes = Rs. 7.20
RRss..77.22.2200× 4
Waiting charge for 1 minute =
Waiting charge for 4 minutes =
= Rs. 14.40
∴ Total taxi fare = Rs. (14 + 360 + 14.40)
= Rs. 388.40.
Again, Since the time is 11:00 P.M.,
the minimum charge is Rs. 21.
The rate for each 200m is Rs. 10.80
The rate for each km is Rs. 10.80 × 5
Rs. 54
The rate for 6 km = Rs. 54 × 6
= Rs. 324
Total taxi fare = Rs.(324 + 21)
= Rs. 345
Again, waiting charge for 2 minutes = Rs. 7.20
waiting charge for 1 minute = Rs. 7.20
2
waiting charge for 10 minutes = Rs. 7.20 × 10 = Rs. 36
2
T otal fare including waiting charge = Rs. (345 + 36) = Rs. 381.
Exercise 4.4
1. (a) What is the minimum taxi fare in the interval of time 6.A.M. to 9 P.M.?
(b) What is the minimum taxi fare in the interval of time 9 P.M. to 6 A.M.?
(c) During the time of 6 A.M. to 9 P.M., what is the additional fare for each exta 200m?
(d) During the time of 9 P.M. to 6 A.M., what is the additional fare for each extra
200 m?
(e) What is the rate of waiting charge in the day time?
(f) What is the additional taxi fare for 200 km the interval of time 6 A.M. to 9 P.M.?
2. The minimum taxi fare is Rs. 14 and the rate for each 200m is Rs. 7.20. Calculate the
taxi fare if the distance covered is
(a) 10 km (b) 12 km (c) 11km 400m (d) 18km 600m
Oasis School Mathematics-9 79
3. The minimum taxi fare is Rs. 21 and the rate for each 200m is Rs. 10.80. Calculate
the taxi fare if the distance covered is
(a) 12 km (b) 5 km (c) 6km 800m (d) 15km 200m
4. Use the rate mentioned above to calculate the total taxi fare in each case.
S.N. Distance Time Waiting time
(i) 6 km 11:00A.M. 6 minutes
(ii) 9 km 200m 8:00 A.M. 10 minutes
(iii) 15 km 11.00 P.M. 8 minutes
(iv) 20km 400m 4:00 A.M. 12 minutes.
5. (a) Sabhyata paid Rs. 194 taxi fare. The minimum charge is Rs. 14 and the rate for
each 200m is Rs. 7.20. Calculate the distance covered by her in taxi.
(b) Lochan paid Rs. 561 taxi fare. The minimum charge is Rs. 21 and the rate for
each 200m is Rs. 10.80 . Calculate the distance which he covered by taxi.
6. (a) Davi has to travel from Thamel to the Airport. Distance from Themel to
Airport is 6 km. How much he has to pay as taxi fare if he travels at (i) day
time with waiting time 6 min. (ii) night time with waiting time 10 min.
(b) Anil has to go to Bhaktapur from Balaju in taxi; distance between theere two
places is 18km. Calculate the amount of taxi fare to be paid it he travelled at
(i) day time with waiting time 12 minutes.
(ii) night time with waiting time 8 minutes.
Answer
1. Consult your teacher 2. (a) Rs. 374 (b) Rs. 446 (c) Rs. 424.40 (d) Rs. 683.60
3. (a) Rs. 669 (b) Rs. 291 (c) Rs. 388.20 (d) Rs. 841.80
4. (i) Rs. 251.60 (ii) Rs. 381.20 (iii) Rs. 859.80 (iv) Rs. 1165.80
5. (a) 5 km. (b) 10 km. 6. (a) (i) Rs. 251.60 (ii) Rs. 381 (b) (i) Rs. 705.20 (ii) Rs. 1021.80
Assessment Test Paper
Attempt all the questions. Full Marks: 50
Group 'A' [4 × 1 = 4]
1. (a) Find the profit percent when an article having cost Rs. 500 is sold at the profit
of Rs. 100?
(b) Find the amount of discount if an article having marked price Rs. 600 is sold at the
discount of 20%.
2. (a) Find the amount of commission at the rate of 5%, if the land costing Rs. 50,00,000
is sold.
80 Oasis School Mathematics-9
(b) What is the meaning of consumption of 1 unit electricity?
Group 'B' [8 × 2 = 16]
3. (a) An article is sold at Rs. 500 at a profit of 25%. Find its cost price.
(b) Find the profit percent if 5 pens are sold at the cost price of 6 pens.
4. (a) Marked price of an article is Rs. 600. Find the discount amount and selling price if
20% discount is given.
(b) A man paid Rs. 1170 for an article including 13% VAT. Find its selling price and
amount of VAT.
5. (a) A broker gets the commission of Rs. 50,000 from the land which was sold at Rs.
10,00,000. Find the rate of commission.
(b) Yearly income of an unmarried man is Rs. 3,00,000. Find the amount of tax to
be paid if 1% tax is imposed up to the income of Rs. 2,50,000 and 15% tax in the
income more than Rs. 2,50,000.
6. (a) The minimum charge of consumption of electricity is Rs. 80 for first 20 units and
Rs. 7.30 per unit for more than 20 units. Find the charge to be paid if the total
consumption of electricity is 28 units.
(b) A household made 250 calls. If the charge for first 175 calls is Rs. 200 and Re 1 for
each additional call. Find the amount of bill to be paid.
Group 'C' [5 × 4 = 20]
7. A shopkeeper sells an article at a discount of 20% and loses Rs. 600. If he allows 10%
discount, he gains Rs. 150, find the marked price and cost price of the article.
8. Monthly salary of a salesmen is Rs. 18,000. He gets a commission of 5% on the sale
more than 5,00,000. If he made the total sale of Rs. 6,00,000 in a month, find his
monthly income.
9. Monthly salary of a married man is Rs. 50,000. Using the tax rate given below, find
the tax to be paid by him in a year.
Yearly income Rate
Upto Rs. 4,00,000 1%
Rs. 4,00,000 to Rs. 5,00,000 2%
Rs. 5,00,000 to Rs. 25,00,000 25%
10. Samjhana has 450 shares in a hydropower company. Altogether the company has
2,00,000 shares. In a year, company made a profit of Rs. 40,00,000. Company decided
to distribute 10% of total profit as dividend. Find
Oasis School Mathematics-9 81
(i) the amount of dividend. (ii) dividend per share. (iii) dividend got by Samjhana.
11. Minimum charge of telephone upto 175 calls is Rs. 200. The charge for each extra call
is Re 1. If 350 calls were made find
(i) amount of bill. (ii) amount of bill including 10% TSC.
(iii) amount of bill including 13% VAT.
Group 'C' [2 × 5 = 10]
12. Monthly salary of Bir Bahadur Gurung is Rs. 67,000. He is an unmarried person. He
deposits 10% of his salary to P.F. 15% of his salary to C.I.F. He also pays Rs. 20,000
yearly as the premium of insurance.
(i) find the yearly amount deposited in P.F. and C.I.F.
(ii) find the tax free income if the amount deposited in P.F., C.I.F. and the premium
of insurance.
(iii) calculate the total tax to be paid using the given rate in the taxable income.
Slab Yearly income Rate
First upto Rs. 3,50,000 1%
Second from Rs. 3,50,000 to Rs. 4,50,000 15%
Third from Rs. 4,50,000 to Rs. 25,00,000 25%
13. Previous and current readings of water bill are 3712 and 3805. If the size of the pipe is
1" and the bill is paid in the 4th month. If the minimum charge upto 56 units is Rs. 3960
and Rs. 71 per extra unit; find the total water bill if 50% sewerage charge is added.
Project work
Take the meter reading of electricity in the first day of any two consecutive month.
Calculate the total bill to be paid if the payment is made on:
(i) 7 days of meter reading. (ii) 8th to 22nd days of meter reading.
(iii) 23rd to 30th days of meter reading. (iv) 31st to 40th days of meter reading.
(v) 41st to 60th days of meter reading.
Present all the reports in the classroom.
82 Oasis School Mathematics-9
Mensuration
20Estimated Teaching Hours
Contents
• Perimeter and area of plane figure
• Area of path ways
• Area of four walls, ceiling and floor
• Cost estimation of carpeting, fencing, plastering, etc.
• TSA and volume of prism
• Cost estimation.
Expected Learning Outcomes
At the end of this unit, students will be able to develop the
following competencies:
• To calculate the perimeter and area of plane figures like triangle,
rectangle, square, trapezium, etc.
• To estimate the cost of fencing, plastering, paving stones on the
plane surface
• To calculate the area of four wall, ceiling and floor
• To calculate the area of cross section, TSA and volume of prism
• To calculate the cost of making walls
Teaching Materials
• Models of plane shape, Net of cuboid, Model of prism, etc.
Oasis School Mathematics-9 83
Unit
5 Mensuration
5.1 Warm-up Activities
Discuss the following in your class and draw out the conclusion.
• What is the length of outer boundary of a closed figure called?
• What is the perimeter of given figure? 4cm 5cm
• How many altitudes can be drawn in a triangle?
• If all three sides of a triangle is 10cm, what is its
perimeter? 4.2cm 4.5cm
• Length of outer boundary of a field is 90m, find the total 4.6cm
cost of fencing if the cost per meter of fence is Rs. 50.
• What is the difference between cube and cuboid?
5.2 Perimeter and Area of Plane Figures
Perimeter is the total length of line segments or curves to enclose a plane figure,
while area is the measure of the space enclosed by a set of straight lines or curves in
a plane.
(i) Perimeter and area of a triangle:
Perimeter of a triangle is the sum of all three sides of the triangle. A
∴ Perimeter of a triangle = a + b + c cb
If three sides of a triangle are given, then
Area of a triangle = s (s – a) ( s – b) ( s – c) B a C
Where, a = side opposite to vertex A
b = side opposite to vertex B
c = side opposite to vertex C and
s = semi perimeter of a triangle
giv=e na, + 2b + c A
If altitude of triangle
the a is altitude (h)
area of a triangle = 1 base × altitude
2
= 1 b × h B D C
2 Base
84 Oasis School Mathematics-9
We can take any side of a triangle as base. Altitude is the length of perpendicular
from the opposite vertex to the given side. In a right angled triangle, out of two
perpendicular sides, one side is considered as a perpendicular and the other as a
base.
Therefore, A
Area of ∆ ABC = 1 base × altitude
2
= 1 BC. AB
2
Note: BC
'p', 'b' and 'h' represent perpendicular, base and hypotenuse of a right angled
1
triangle, then area of right angled triangle = 2 p × b.
In an equilateral triangle, all sides are equal.
Therefore,
A rea of an equilateral triangle = 3 4a2 [∵ a = b = c]
where, a = a side of an equilateral triangle.
(ii) Perimeter and area of a rectangle:
Perimeter of a rectangle = sum of all four sides
= 2 l + 2 b l
l
= 2 ( l + b) b b
= l × b
And, Area of a rectangle
(iii) Perimeter and area of a square:
All four sides of a square are equal.
Therefore, Perimeter of a square = 4l l
d
Area of a square = l² l l
= 1 d2 [∵ l2 + l2= d2] l
2
(iv) Circumference (perimeter) and area of a circle:
Circumference of a circle = 2πr r
O
Area of a circle = πr2
Where, r = radius of a circle
Oasis School Mathematics-9 85
(v) Circumference (perimeter) and area of a semicircle:
Circumference of a semicircle is the sum of circular part and diameter of a circle.
Therefore, πr
Circumference of a semi circle = πr + 2r
= r (π + 2) rO r
Points to remember !
1. Area of triangle = s(s–a) (s–b) (s–c) if three sides of a triangle are given.
= 1 base × height of base and height are given.
2
1
= 2 perpendicular × base in a right angled triangle.
2. Area of rectangle = length × breadth = l × b
3. Area of square = (length)2 = l2 = 1 d2 if diagonal is given.
2
• Area of circle = πr2
• Area of semicircle = 1 πr2
2
• Perimeter of a triangle = a+b+c
• Perimeter of rectangle = 2(l+b)
• Perimeter of a square = 4l
• Perimeter (circumference of a circle = 2πr
• Perimeter of semicircle = πr + 2r
Worked Out Examples
Example: 1
Find the area and perimeter of given figures.
(a) A (b) A D (c)
3cm
6cm
3cm 10cm O
B C 7cm
B 4cm C
(a) Solution:
Here, a = 4 cm, b = 3 cm, c = 3 cm
s =a + b + c = 4 + 3 + 3 cm
2 2
= 5 cm
∴ Perimeter of ∆ ABC = a + b + c = 10 cm
86 Oasis School Mathematics-9
Again, area of ∆ ABC = s(s − a)(s − b)(s − c)
= 5(5 − 4)(5 − 3)(5 − 3)
= 5.1.2.2
= 20
= 2 5 cm2
(b) Solution: BC = BD 2 − CD2 [Using Pythagoras theorem]
In ∆ BCD,
= (10)2 − (6)2
= 100 − 36
= 64
= 8 cm
In rectangle ABCD,
Length (l) = 8 cm
Breadth (b) = 6 cm
Perimeter of a rectangle = 2(l + b)
= 2(8 + 6)
= 28 cm
Again,
Area of rectangle ABCD = l × b
= 8 cm × 6 cm = 48 cm2
(c) Solution: Here,
Radius of the semicircle (r) = 7 cm
Its perimeter = πr + 2r
= (272 × 7 + 2 × 7) cm
= ( 22 + 14) cm
= 36 cm
Again, area of the semicircle = 1 × (πr2)
2
= 1 × 22 × (7)2
2 7
= 77 cm2
Example: 2
The length of a rectangular room is twice its breadth. If its area is 128 m², find its
perimeter.
Solution:
Let, the length of the room = 2x
Then, the breadth of the room = x
Oasis School Mathematics-9 87
Area of the room = l × b
= 2x × x = 2x2
It is given that,
Area of the room = 128m2
2x2 = 128
x2 = 64
x = 8
∴ Length (l) = 2x = 2 × 8m
= 16 m
Breadth (b) = x = 8 m
we have,
Perimeter of the room = 2 ( l + b)
= 2 (16 + 8) m
= 48 m
Example: 3
A square window of each side 2.1 m has a semi circular top. Calculate the perimeter and
area of window.
Solution: Here,
diameter of the semicircular part o f the window = 2.1m. 1.05m
∴ radius (r) = 2.1 m = 1.05 m 2.1 m
2
Length of each side of window = 2.1 m
From the figure,
Perimeter of the window = p erimeter of semi circular part + 3 (a side of window)
= πr + 3l
= (272 × 1.05 + 3 × 2.1)
= (3.3 + 6.3) m = 9.6 m
Again, area of the window = area of semicircular part + area of square
= 1 πr2 + l2
2
= 1 π(1.05)2 + (2.1)2
2
= 1 × 22 × (1.05)2 + (2.1)2
2 7
= 6.14m2
88 Oasis School Mathematics-9
Exercise 5.1
1. (a) Write the formula to calculate the area of a triangle if its all three sides are given.
(b) What is the semi perimeter of a triangle whose three sides are 'x' cm, 'y' cm and
'z' cm?
(c) What is the formula of area and perimeter of a rectangle?
(d) Write the formula for area and perimeter of square.
(e) If ‘d’ be the diagonal of a square write the formula to calculate the area of square
in terms of ‘d’.
(f) Write the formula to calculate the circumference and area of a circle.
(g) Write the formula to calculate the circumference and area of a circle in terms of
diameter ‘d’.
(h) Write the formula to calculate the circumference and area of semi circle in terms of r.
(i) Write the formula to calculating the circumference and area of a semicircle in
terms of d.
2. Find the area and perimeter of a triangle having :
(a) sides 5 cm, 6 cm, 8 cm. (b) base = 8 cm, hypotenuse = 10 cm
3. Find the area and perimeter of a rectangle having :
(a) length = 10cm, breadth = 8 cm (b) length = 12cm, diagonal = 13 cm
4. Find the area and perimeter of a square having:
(a) a side = 5 cm (b) diagonal (d) = 4 2 cm
5. Find the area and circumference of a circle having
(a) radius (r) = 7 cm (b) diameter (d) = 21 cm
6. (a) The area of a square plot is 2025m2. Find its length and perimeter.
(b) Area of a square is 625cm2, find its perimeter.
(c) Perimeter of a square is 200 cm, find its area.
(d) A diagonal of the square is 12 2 cm, find the area and perimeter of the square.
7. (a) Perimeter of a rectangular field is 130 m and the length of the field is 40 m, find
its breadth and its area.
(b) Perimeter of a rectangle is 60 cm. If the length is twice the breadth, find its
length, breadth and area.
(c) The perimeter of a rectangular field is 600 m. If the length of the field is twice
the breadth, find the area of the field.
Oasis School Mathematics-9 89
(d) Area of a rectangular field is 1200 cm2, if the ratio of length and breadth is 4:3,
find its perimeter.
(e) The length of a room is 4m more than its breadth. If its area is 140 m2., find its
perimeter.
8. (a) The area of a circular ground is 1386 m2. Find the circumference of the ground.
(b) The circumference of a circular field is 176 m. Find its area.
(c) Circumference of a semi circular field is 36m, find its area.
9. Find the perimeter and area of given figures.
(a) (b) (c)
14cm 14cm 30cm 16cm 7cm
7cm 7cm
(e) (f)
(d) 14cm 20 cm
20 cm 14cm
5cm
7cm
6cm 8cm 20cm
28 cm 20 cm
28 cm
Answer
1. Consult your teacher
2. (a) 19 cm, 14.98 cm² (b) 24 cm, 24 cm² 3. (a) 36 cm, 80 cm², (b) 34cm, 60cm²
4. (a) 20 cm, 25 cm² (b) 16 cm, 16 cm² 5. (a) 44 cm, 154 cm² (b) 66 cm, 346.5cm²,
6. (a) 45 m, 180 m, (b) 100cm (c) 2500cm2 (d) 144cm2, 48cm
7. (a) 25m, 1000m² (b) 20cm, 10cm, 200cm² (c) 20,000m² (d) 140cm (e) 48m
8. (a) 132m (b) 2464m2 (c) 77m2 9. (a) 64cm, 273cm2 (b) 86.4 cm, 460cm2 (c) 25cm , 40.47cm2
(d) 140cm, 1291.96cm2 (e) 33.72cm, 80.60 cm2 (f) 84 cm, 434 cm2
5.3 Cost Estimation
Fencing a ground:
While fencing a ground, the length of fencing material should be equal to the
perimeter of the ground.
If cost per unit length of fencing material is given,
Total cost (T) = P × C
where, P = Perimeter of ground
C = Cost per unit length
90 Oasis School Mathematics-9
Carpeting the room:
While carpeting the room, the area of a room must be equal to the area of carpet. If
the size of the room is not similar to the size of the carpet, the carpet must be cut into
suitable size to make the area equal.
If the cost of carpet per unit length is given,
The total cost (T) = l × C
Where, l = length of the carpet
C = cost per unit length
If the cost per unit area is given,
The total cost (T) = A × C
Where, A = area of a room (Carpet)
C = cost per unit area.
Paving stones, tiles or marbles:
While paving marbles (stones) in a room, first of all we need to find the number of
marbles required. If 'N' be the required number of marbles, 'A' be the area of room
and 'a' be the area of a marble, then
The area of a room = area of 'N' marbles
A = N×a
N = Aa
∴ Number of marbles (N) = Area of a room
Area of a marble
While calculating the total cost,
Total cost (T) = N × C
Where, N = Total number of marbles
C = Cost per piece of marble
Points to remember !
Total cost of fencing, T = P × C; where P = Perimeter of the field
Total cost of carpeting / plastering T = A × C
Total cost of paving stones T = N × C
• Number of stones (N) = Aa
• Cost of carpeting T = l × C; where l = length of carpet, C = Cost per meter
Oasis School Mathematics-9 91
Worked Out Examples
Example: 1
A rectangular court is 20 m by 16m. Find the cost of fencing the court at a rate of Rs. 125
per metre.
Solution:
Here, Length of the court (l) = 20 m
Breadth of the court (b) = 16 m
We have,
Perimeter of the court (P) = 2 (l + b)
= 2(20 + 16)m
= 72m.
Cost per metre of fencing material (C) = Rs. 125
We have, Total cost (T) = P × C
= 72 × Rs. 125
= Rs. 9,000.
Example: 2
The length of a rectangular hall is three times its breadth. If its perimeter is 160 m, find
the cost of carpeting its floor at the rate of Rs. 300 per m².
Solution:
Let, the breadth of the room (b) = x m
then, the length of the room (l) = 3x
We have,
Perimeter of the room = 2(l + b)
160 = 2(3x + x)
or, 8x = 160
or, x = 20 m
∴ Length (l) = 3x
= 3 × 20m
= 60m
Breadth (b) = x = 20 m
Again, Area of the floor (A) = l × b
= (60 × 20) m²
= 1200 m2
Cost per unit area (C) = Rs. 300 per sq.m.
92 Oasis School Mathematics-9
Total cost (T) = ?
we have, Total cost (T) = A × C
= 1200 × Rs. 300 = Rs. 3,60,000
Example: 3
A room is 7 m long and 6 m broad. Find the length of the carpet required to carpet the
room if the width of the carpet is 1.4 m .
Solution:
Here, Length of the room (l) = 7 m
Breadth of the room (b) = 6 m
Area of the room (A) = l × b
= 7m×6m
= 42 m2
Again, width of the carpet = 1.4 m
While carpeting the room,
Area of the room = Area of the carpet
42 = 1.4 × l
l = 42 m
1.4
∴ = 30 m
Length of the carpet = 30 m
Example: 4
The cost of carpeting a room at the rate of Rs. 215 per meter is Rs. 1720. Find the area of
the room if the breadth of the carpet is 150 cm.
Solution:
Here, cost per unit length of carpet (C) = Rs. 215 per meter.
Total cost (T) = Rs. 1720
We have, Total cost (T) = l × C
or, Rs. 1720 = l × 215
1720
or, l = 215 m
or, l = 8m
Breadth of the carpet (b) = 150 cm
= 1.5 m
Area of the carpet = l × b
= 8 × 1.5 m² = 12 m2
Since, area of the room = area of the carpet.
Area of the room = 12m2.
Oasis School Mathematics-9 93
Example: 5
A rectangular court is 30 m by 15 m. Find the number of stones of size 30cm × 10 cm
required to pave the court.
Solution:
Here, Length of the court (l) = 30 m
= 3000 cm
Breadth of the court (b) = 15 m
= 1500 cm
Area of the court (A) = l × b
= (3000 × 1500) cm2
= 45,00,000 cm2
Again, Area of a stone(a) = 30 cm × 10 cm
= 300 cm2
Number of stones (N) = A
a
= 45,00,000
300
= 15,000 stones.
Again, Cost per stone (C) = Rs. 20
Total cost = N × C
= 15,000 × Rs. 20
= Rs. 3,00,000
Exercise 5.2
1. (a) If 'c' be the cost per square meter of carpet, 'A' be the area and 'T' be the total
cost of carpeting, write the relation among T, A and C.
(b) If 'P' be the perimeter of the field and 'C' be the cost per meter and 'T' be the
total cost of fencing, write the relation among P, T and 'C'.
(c) If 'l' be the length of carpet, T be the total cost 'T' of carpeting and 'c' be the cost
per meter of carpeting, write the relation among 'T', 'l' and 'c'.
(d) If 'A' be the area of lawn, 'a' be the area of a stone and 'N' be the number of
stones required to pave on the ground, write the relation among 'A' 'a' and 'N'.
(e) If 'N' be the number of stones required to pave on the ground 'C' be the cost per
stone and 'T' be the total cost of paving, write the relation among C, T and N.
2. (a) Perimeter of a rectangular ground is 60m. Find the cost of fencing it at the rate
of Rs. 20 per meter.
94 Oasis School Mathematics-9
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Oasis Math 9
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