9 Green C Math Class

EXAMPLE 4 Hari Bahadur purchased 500 shares for Rs. 100 each of ADB Bank
which issued 5000 shares. If the bank distributed 50% of net profit of
Rs. 3,50,000. How much money does Hari Bahadur get as dividend?

Solution : Here,
The total number of share = 500
Price of each share = Rs. 100
Total net profit = Rs. 3,50,000
Profit amount for dividend = 50% of Rs. 3,50,000
= 15000 × Rs. 3,50,000
= Rs. 1,75,000
∴ The amount of dividend for Hari = Net dividend × No. of Shares

Total no. of share
= Rs. 175000 × 500

5000
= Rs. 17500

EXERCISE: 3.2

A. Very Short Questions

1. Fill in the gaps.
a. Bonus Amount = Bonus rate .............................
b. Define bonus.
c. Rate of dividend = q ............................... ×..................r%
......................
d. Dividend amount = ...................... × ............................

B. Short Questions

2. The net profit of a factory is Rs. 6,60,000 in a year. The management of the factory
decided to distribute 40% as the bonus for 500 employees. How much money each
one will get?

3. Net proft of Focus Computer is Rs. 12,50,000. The management decided to distribute
35% of net profit as a bonus to employees. If each employee gets Rs. 4375 as a bonus,
how many employees are there?

4. A shopping mall distributed 3,000 shares for Rs. 100 per share. The net profit of this
shopping mall is Rs. 45,00,000. The management of mall decided to distribute 10%
dividend. How much amount of dividend did by each share receive?

GREEN Mathematics Book-9 45

C. Long Questions

5. Padhma Prakash high school made a profit of Rs. 150000 in the fiscal year 071/72. If
0.5% of the profit is divided equally among 32 teachers as bonus; how much will each
teacher get?

6. If a company provides 8% bonus from profit to its 25 staff equally from the profit of
Rs. 13,50,000 a year, how much bonus will each staff receive?

7. Each of 40 staff of a bank having Rs. 5000000 as yearly profit receives Rs. 15,000 as
bonus. If the bonus is divided among all staff equally, what percentage of profit is
divided as bonus.

8. Mr. Ram purchased 500 shares at Rs. 100 for each share of Axis Bank. The bank issued
1000 shares in total. If the bank distributed 25% of net profit of Rs. 25,00,000. How
much does Ram get as dividend?

9. Mr. Khaniya purchased 2000 shares at Rs. 100 for each share of Standared Charterd
Bank. The bank issued 20000 shares in total. If the bank distributed 15% of net profit
of Rs. 30,00,000. How much money does Khaniya get as dividend?

10. A hydropower company distributed 25,000 shares. The value of each share is Rs.
500. Muna purchased some share from the hydropower, if the company distributed
10% of net profit of Rs. 20,00,000 as dividend, how many shares did she purchase for
getting Rs. 3200 as dividend?

3.3 Tax

Tax

Generally, tax means the yearly payment made by an individual or an organization
to the local or central government or the related institutions for the services of special
work.

There are various kinds of taxes. b. Transportation tax
a. Income tax d. Road tax
c. VAT (Value added tax) f. Education tax, etc.
e. Municipality tax

Income Tax

The tax imposed by the Government on the income of a person or an organisation
is called the income tax. On yearly income up to certain limit as per the minimum
needs of a person is the tax free amount. The income above the given limit is called the
taxable amount.

Note :
i. Income tax = tax rate × taxable amount
ii. Taxable in-come = yearly income – tax free amount

46 GREEN Mathematics Book-9

For resident person

There has been a change in the Tax rate applicable to the resident person from the
budget this year. Slab rate has been increased by 1,00,000 for tax payers assessed as
individual and couple. Details of the same has been provided below:

Particulars FY 2072/73 FY 2073/74

Assessed as individuals

First tax slab 2,50,000 1% 3,50,000 1%

Next 1,00,000 15% 1,00,000 15%

Balance exceeding 3,50,000 25% 4,50,000 25%

Assessed as couple

First tax slab 3,00,000 1% 4,00,000 1%

Next 1,00,000 15% 1,00,000 15%

Balance exceeding 4,00,000 25% 5,00,000 25%

In case the taxable salary of an individual is more than Rs. 25 lakhs additional tax
at the rate of 40% shall be levied on the tax applicable on the income above Rs.
25,00,000 (i.e. above Rs. 25,00,000 tax rate of 35% is applicable.

Note :

Government of Nepal allows tax free amount on the following heading:

i. Providend fund ii. Civil service fund

iii. Life insurance iv. Religion purpose

v. Donation vi. Remote Allowance

vii. Travelling allowance of 75% for foregin country

viii. Medical allowance

Above expenditure should be deducted from the total income in year, the
remaining amount is taxable income for the person. Every year government may
change the tax rate.

Worked Out EXAMPLES

EXAMPLE 1 Find the amount of sale tax on the purchase of Rs. 4,000 at 8%.

Solution : Here,
Original price = Rs. 4,000
Sale tax rate = 8%
Now,
We know,
Sales tax amount = 8% of original price

= Rs. 8 × 4000 = Rs. 320
100

GREEN Mathematics Book-9 47

EXAMPLE 2 Calculate the tax paid by Sabin Dangol who earned Rs. 3,66,000 in a
year if his tax free amount was Rs. 3,20,000 and tax rate was 15%.

Solution : Here,
Yearly income of Sabin = Rs. 3,66,000
Tax free amount = Rs. 3,20,000

Now,

Taxable income = total yearly income – tax free amount

= Rs. 3,66,000 – Rs. 3,20,000

= Rs. 46,000

∴ Tax Amount = 15% of Rs. 46000
= Rs. 15 × Rs. 46000

100
= Rs. 15 × 460

= Rs. 6900

EXAMPLE 3 Mr. Binesh earns Rs. 42,000 in each month. If he takes yearly allowance
Solution : for providend fund 10%, social service 5%, annual insurance premium
3%, donation 4% and medical allowance 2%. Find the total income tax
in a year if he is married. (Tax rate given in the above table).

Here,

Binesh's monthly income = Rs. 42,000

Binesh's yearly income = Rs. 42000 × 12

= Rs. 5,04,000

His allowance for providend fund = 10% of Rs. 5,04,000
= 10 × Rs. 5,04,000

100
= Rs. 50,400

His allowance for social service = 5% of Rs. 5,04,000
= 5 × Rs. 5,04,000

100
= Rs. 25,200

His allowance for insurance = 3% of Rs. 5,04,000
= 3 × Rs. 5,04,000

100
= Rs. 15,120

48 GREEN Mathematics Book-9

His allowance for donation = 4% of Rs. 5,04,000
= 4 × Rs. 5,04,000

100
= Rs. 20,160
His medical allowance for = 2% of Rs. 5,04,000
= 2 × Rs. 5,04,000

100
= Rs. 10,080
The total tax allowance =R s. (50,400+25,200+20,160+15,120+10,080)
= Rs. 1,20,960
His taxable income in a year = Rs. 5,04,000 – Rs. 1,20,960
= Rs. 3,83,040
His tax amount upto Rs. 3,83,040 = 1% of Rs. 3,83,040
= 1 × Rs. 3,83,040

100
= Rs. 3830.40

EXERCISE: 3.3

A. Very Short Questions

1. Fill in the gaps.
a. Income tax = ................. × ....................
b. Taxable income = ................. – ....................
c. Tax rate = ....... × ...........................
.......

B. Short Questions
2. From the table given below, find the income tax required to be paid by Mukesh.

Heading of taxable income Rate of tax
Income tax upto Rs. 4,00,000 1%

Income tax upto Rs. 4,00,000 – Rs. 5,00,000 15%
Income tax upto Rs. 5,00,000 – Rs. 25,00,000 15%
Income tax upto Rs. 25,00,000 – Rs. 30,00,000 25%
Income tax Rs. 30,00,000 and above per Rs. 1,25,000 40%

a. In 2071 B.S. annual earning Rs. 12,00,000, total allowance Rs. 5,00,000
b. In 2072 B.S. annual earning Rs. 15,00,000, total allowance Rs. 6,50,000
c. In 2073 B.S. annual earning Rs. 15,50,000, total allowance Rs. 7,50,000
d. In 2074 B.S. annual earning Rs. 16,20,000, total allowance Rs. 7,80,000

GREEN Mathematics Book-9 49

3. In a school, there were 1200 students. Average tuition fees of each student is Rs. 1600,
how much education tax should be paid by the school to the government at the rate
of 1% tax?

C. Long Questions
4. Ramu earns Rs. 8,50,000 in a year. He is unmarried and his total allowances are

Rs. 3,50,000. If he has to pay 15% annual income tax, find his tax amount.
5. Out of the annual earning of Rs. 6,50,000, the taxable income is Rs. 4,00,000. If a man

has to pay 20% income tax, find his income tax.
6. Bhawana earns Rs. 8,20,000 in a year. Her allowances are Rs. 4,00,000 and she pays

the annual income tax of Rs. 63,000. Find the rate of income tax.
7. Gore earns Rs. 18,00,000 in a year. If he pays Rs. 1,30,000 income tax every year at the

rate of 25%, find the total allowances.
8. The monthly salary of a project manager is Rs. 55,000. He has annual allowances of

Rs. 4,00,000. Find his annual income tax, if he pays the income tax at the rate of 15%
on the first Rs. 2,00,000 and 25% on the remaining sum of the taxable income.
9. In a school, there were 500 students. Average tuition fees of each student is Rs. 1600,
how much education tax should be paid by the school to the government at the rate
of 1% tax?
10. The basic monthly salary of a married civil servent is Rs. 75,000. He/She gets 67% tax
allowance and pays the tax for the rest amount. What amount does he/she pay tax in
a year? (Use the above rate of tax).

Project Work:
a. Discuss the different types of tax in a class.
b. Collect the balance sheet of your school and find dividend, bonus and tax amount

paid by the school.
c. Collect the education tax of previous year of your school.

50 GREEN Mathematics Book-9

4

Household Arithmatics

Estimated Teaching Periods : 6

Albert Einstein was a German-born theoretical physicist. He developed
the theory of relativity, one of the two pillars of modern physics.
Einstein's work is also known for its influence on the philosophy of
science. “Compound interest is the eighth wonder of the world. He who
understands it, earns it ... he who doesn't ... pays it.”

Contents

4.1 Domestic and day to day life activities
4.2 Telephone bill
4.3 Water Bill
4.4 Taxi meter

Objectives

At the end of this unit, students will be able to:
solve the problems of domestic arithmetic
calculate the bills of electricity, water, telephone and taxi used

Materials

Bills: telephone, water supply, company, electricity and rate of taxi fare.

GREEN Mathematics Book-9 51

4.1 Domestic and Day to Day Life Activities

Electricity

Let see the given electricity bill. The given electricity bill
shows the meter reading of electricity meter in Bhadra 2073.
The meter reading for Bhadra 2073 is 24478
Similarly,
The meter reading for Shrawan 2073 is 24345
Finally,
Total units comsumed in Bhadra = 24478 – 24345 = 133 units
According to NEA, minimum charge up to 50 units is Rs. 399
and the rate per unit more than 50 units is Rs. 7.50 so, the cost
of electricity charge for the bill of Bhadra.
= minimum charge + extra charge
= Rs. 399 + (133 – 50) × Rs. 8.80
= Rs. 399 + 83 × Rs. 8.80
= Rs. 399 + Rs. 730.40
= Rs. 1129.4 [∵ Total energy charge Rs. 1129.4]
∴ Total bill amount including service charge is Rs. 1230.

The minimum charge of the electricity according to the capacity of meter are different
which are as follows:

Capacity of meter Minimum charge Minimum unit

5 Ampere Rs. 80 20

6 to 15 Ampere Rs. 399 50

16 to 30 Ampere Rs. 795 100

31 to 60 Ampere Rs. 1765 200

52 GREEN Mathematics Book-9

5 Ampere 15 Ampere 30 Ampere 60 Ampere

K.w/hr. Service Energy Service Energy Service Energy Service Energy
charge charge charge charge charge charge charge charge

0 - 20 30 3 50 4 75 5 125 6

21 - 30 50 7 75 7 100 7 150 7

31 - 50 75 8.80 100 8.80 125 8.80 175 8.80

51 - 150 100 10 125 11 150 11 200 11

151 - 250 125 11 150 12 175 12 225 12

251 - 400 150 12 175 13 200 13 250 13

Above 400 175 13 200 14 225 14 275 14

Note :

i. Fixed charge is known as standing or minimum charge [In given electricity bill
minimum charge is Rs. 399 and the extra for 50 units]

ii. 1 unit = 1 kilowatt = 1000 watt × 1 hour.

iii. The electricity charge = minimum charge + number of kw × per unit cost.

∴ Electricity consumed= final reading – initial reading .

iv. Rule of electricity charge for NEA are as follows:

Units Charge
up to 20 units Rs. 4 per units (minimum Rs 80)
21 to 50 units Rs. 7.30 per unit
51 to 150 units Rs. 8.80 per unit
151 to 250 units Rs. 9.50 per unit
251 units above Rs. 11.50 per unit

v. Rebate and fine:
– Up to 7 days from the meter reading day 3% rebate.
– From 8th day to 22nd day as per the bill
– From 23rd day to 30th day 5% fine
– From 31st day to 40th day 10% fine
– From 41st day to 60th day, 25% fine.

GREEN Mathematics Book-9 53

Worked Out EXAMPLES

EXAMPLE 1 If a house owner uses 10 bulbs of 60 watts for 7 hours 30 minutes per
day and an electric heater of 1500 watts 2 hours per day. How much
would the house owner pay as a bill in a particular month of 30 days at
the rate of Rs. 7.30 per unit.

Solution : Electricity consumed by bulb = 10 × 60 watt × 7 hours 30 minutes × 30 days
= 10 × 60 × 7.5 × 30 watt hr.

= 135,000 watt hr.

= 135 kwhr.

Electricity consumed by electric heater

= 1500 × 2 × 30

= 9000 watt hr.

= 9 kw hr.

Total power consumed = (135,000 + 9000) watt

= 144000 watt hr.

= 144 kw hr.

We know, 1 kw hr. = 1 unit

144 kw hr = 144 units

∴ Total payment at the rate of Rs. 7.30.

= 144 × Rs. 7.30

= Rs. 1051.20

EXAMPLE 2 The meter reading for the consumption of electricity of certain house-
hold was 3270 units on 10th Falgun and 3295 units on 10th Chaitra. Cal-
culate the amount of money to be paid if the charges are indicated as
below. (minimum charge up to 20 units = Rs 80, charge for extra unit =
Rs. 7.50 per unit)

Solution : Initial reading = 3270 units, final reading of Chaitra = 3295

Now, electric energy used = Final reading (Chaitra)–Initial reading (Falgun)

= 3295 – 3270

= 25 units

So, the cost of first 20 units (minimum charge) = Rs 80.

The charge of next (extra) units

= (25 – 20) × Rs. 7.50

= Rs. 5 × 7.50

= Rs. 37.50

Finally, total bill amount = minimum charge + charge of extra units

= Rs. 80 + 37.50

= Rs. 117.50.

54 GREEN Mathematics Book-9

EXAMPLE 3 The electricity charge for the first 50 units is Rs 399 and Rs. 7.50 is the
Solution : extra charge after minimum charge. If the electricity charge of a month
is 752.66, find the electricity consumption of the month.

The charge of a month = Rs. 752.66

Minimum charge = Rs.399.

Now,

Charge for extra units = Rs. (752.66 – 399)

= Rs. 353.66

Again, extra units consumed = Rs. 353.66 = 47
Finally, the total units Rs. 7.50

= Minimum charge units + extra charge units

= 50 + 47

= 97

EXAMPLE 4 The detail of electricity bill of Anusuya Ghimire, for Falgun 2073 is
given as follows :

Capacity Domestic 15A

Previous reading 4965
Current reading 5097

Give the answers for the following questions, according to the above
informations:

a. How much is the consumed units of electricity at Anusuya's house
on Falgun month?

b. How much energy is charged for her?

c. If she paid bill on 6th days from meter reading, how much rebate
was given to her?

d. If she paid bill on 21st days from meter reading, find amount for her.

e. If she paid bill on 28th days then find total bill amount for her.

Solution : Here, previous reading = 4965 units

Current reading = 5097 units

a. Consumed units = Current reading – previous reading

= 5097 – 4965

= 132 units

Therefore, consumed units of electricities at Anusuya's house on
Falgun 2073 is 132 units.

GREEN Mathematics Book-9 55

b. Here, her meter capacity is 15A, so, minimum charge of 15A is Rs.
365 up to 50 units.

NEA charges Rs. 8.60 rate/unit on range 51 - 150 units,

So, energy charge for 132 units = Minimum charge + extra
charge

= Rs. 365 + (132 – 50) × Rs. 8.60

= Rs. 365 + Rs. 705.20

= Rs. 1070.20

c. If she paid bill on 6th days from meter reading.

rebate for her is 3% of bill amount

= 1300 × Rs. 1070.20

= Rs. 32.10

d. The consumer should pay according to bill within 8th to 22nd day
of billing.

As per rule, if she paid on 21st day

Total bill amount is Rs. 1070.20

e. According to NEW, 5% fine is imposed if the bill is paid within
23rd to 30th day of billing.

As per rule, if she paid on 28th day

Total bill amount = Energy charge + 5% of bill amount

= Rs. 1070.20 + 5 × Rs. 1070.20
100

= Rs. 1070.20 + Rs. 53.51

= Rs. 1123.71

EXERCISE: 4.1

A. Very Short Questions
1. Define fixed charge or (Standing charge or minimum charge).
2. Fill in the gaps.
a. 1 unit = .................... kw hr
b. Fixed charge is also known as ....................

56 GREEN Mathematics Book-9

c. Electric energy = Final reading ....................

d. .................... watts = 1 kilowatt ....................

h. The meter reading of electricity in Sabin's house on 25th Jestha 072 is 02489 and
on 25th Asar 072 is 02505. So, total .................. units of electricity is used in
Ashad 2072.

i. The electricity charge = minimum charge + number of Kwh × .................. cost.

3. How many units of electricity energy is consumed in each of the following?

a. 1000 Watt b. 2500 WH

c. 700 WH d. 1350 WH

e. 76521 WH

4. Find the total unit of electricity when,

a. 10 bulbs of 100 watt are used for 5 hours.

b. 3 TV sets of 500 W of each watched for 5 hours.

c. 5 electric heaters each of 1500 W used for 2 hours.

B. Short Questions
5. The electricity charge up to 20 units is Rs. 4.00 per unit. and Rs. 7.50 per units from 21

to 200 units. How much should be paid for 72 units of electricity?

6. According to the rate of electricity charge the minimum charge of electricity up to 20
units is Rs. 78 and Rs. 7.30 per unit after 20 units. How much should be paid for the
consumption of 120 units?

7. According to the rate of NEA the cost upto 20 units is Rs. 31.25 and cost per unit from
21 to 150 units is Rs. 8.80 per units. How much should be paid for the consumption
of 140 units?

8. Find the total cost of consumption of energy, if one unit costs Rs. 8.60.
a. 5 bulbs each of 60W are used for 10 hours.

b. 2 electric heater of 500 W are used for 3 hours.

c. 3 electric irons each of 750 W are used 2 hours daily for 30 days.

d. 4 tubelights of 40W each and 10 bulbs of 60W used 3 hours for 60 days.

e. 2 refrigerators each of 500 W, 3 electric heaters each of 500W and 15 tubelights
each of 40 W are used 5 hours daily for 30 days.

GREEN Mathematics Book-9 57

C. Long Questions

9. According to rate of NEA a standing cost up to 20 units = Rs. 80 and cost per unit
from 21 to 250 is Rs. 7.50.
a. If the monthly charge of electricity is Rs. 437. 70, how many units were
consumed?

b. If the monthly charge of electricity is Rs. 700, how many units were consumed?

10. According to rate of NEA a standing cost up to 50 units is Rs. 399 and cost per unit
from 51 units to 250 is Rs. 7.50.
a. How much will 110 units cost?

b. How much will be charged for the consumption of 200 units?

11. Ramesh Maharjan joined electricity of 15A. Find the total amount he paid for NEA
according to the detail of previous unit and current unit given as follows:
a. Current unit = 5493, previous unit = 5313, if he paid bill on 5th day after billing.

b. Current unit = 6211, previous unit = 5854, if he paid bill on 54th day after billing.

c. Current unit = 7325, previous unit = 7077, if he paid bill on 31st day after billing.

Note : See NEA rule

12. In a house of 5A electricity transmission line, how much should the bill be paid for 85
units be, if the cost for first 20 units is Rs. 80 and the rest at the rate of Rs. 7.30 upto
21 - 50 units and Rs. 8.60 upto 51-150 units.

13. Sanulal joined 30A electricity line, the electricity meter reading at his house on 16th of
Magh was 29870 units and on 16th of Falgun was 30580 units. If the cost of electricity
for the first 250 units at the rate Rs. 9.50 and rest at Rs. 11 per unit, find the total cost,
if he paid on the 5th day after billing.

14. Umesh uses 5 tubelights of 40 W for 4 hours a day. An electric heater of 500 watt for
6 hours a day and electric water pump of 1500 watt for 2 hours a day. How much he
should pay to NEA at the rate Rs. 9.50 per unit in a month of 30 days?

15. A house owner pays an average of Rs. 1759 for the cost of 250 units of electricity. He
charges Rs. 9 per unit to the tenants person. If his normal consumption of electricity
is 50 units, then find.
a. How much does he charge to the tenants?

b. How much amount does he make profit by the electricity charge to tenants?

c. If he does not want to make profit apart from the free use of 50 units, how much
should he charge per unit to the tenants?

58 GREEN Mathematics Book-9

4.2 Telephone Bill

Look at the telephone bill. The given telephone bill shows the total of rental cost calls and
charges of extra calls used in Baishakh 2065.

Now, the bill reading of current rent calls is 234 . Similarly, The bill reading of previous
rent calls = 87. Finally, Total rental calls units
= current units – previous units
= 234 – 87
= 147 calls [ whcih is not more than least calls]
According to NT, minimum charge upto 175 calls is Rs. 200 and Re. 1 for 1 extra call.
So, there is not extra calls for the Baishakh 2065 in the bill.
= minimum charge + charge for extra calls
= Rs. 200 + 0
= Rs. 200
= Rs. 200

Now, The bill reading of current rent calls 234 . Similarly, The bill reading of previous
rent calls = 87. Finally, Total rental calls units
= current units – previous units
= 239 – 87
= 147 calls [∴ Whcih is not more than least calls]
According to NT, minimum charge upto 175 calls is Rs. 200 and Re. 1 for 1 extra call.
So for the Baishakh 2065 in the bill, the cost of total calls .
= minimum charge + charge for extra calls
= Rs. 200 + (147 + 0) × Re. 1

GREEN Mathematics Book-9 59

= Rs. 200 × Rs. 1
= Rs. 200
Note:
1. Telephone calls used = Final calls – Initial call

2. The telephone calls charge = Minimum charge + number of total excess calls × unit
cost per call.

3. Fixed call charge is known as standing free calls or minimum calls charge where Rs.
200 for first 175 calls and Rs. 1 per excess call.

4. The bill includes 10% TSC and 13% VAT on the amount including 10% telephone
service charge (TSC)

Worked Out EXAMPLES

EXAMPLE 1 A telephone line is such that only local calls can be made from its set.
The number of total calls recorded in the bill is 420. If the first 175 calls
cost Rs. 200 and the extra calls are charged at Re. 1 per call, find the total
bill amount including 10% TSC and 13% value added Tax (VAT).

Solution : Total calls = 420

Extra calls = Total calls – minimum calls

= 420 – 175

= 245 calls

Now, R ate of extra call is Re. 1.

So, Cost of extra calls = Re. 1 × 245

∴ Total bill amount = minimum calls charge + extra calls charge

= Rs. 200 + Rs. 245

= Rs. 445.

I ncluding 10% TSC charge

= Rs 445 + 10 × Rs. 445
100

= Rs. (445 + 44.5)

= Rs. 489.50

Finally, Including 13% VAT.

13
= Rs. 489. 5 + 100 × Rs. 489.5
= Rs. 489. 5 + Rs. 63.6

= Rs. 553.11

∴T he total bill amount is Rs. 553.11

60 GREEN Mathematics Book-9

EXAMPLE 2 Bikash uses 500 calls from his PSTN line. Find how much he should
pay to NTC for the bill of a month, if minimum charge for 175 calls Rs.
200; after that Re. 1 is charged for each exra calls with 10% telephone
service charge and 13% VAT.

Solution : Here,

Total no. of calls = 500 calls

Minimum charge for 175 calls = Rs. 200

Extra consumed calls = 500 – 175

= 325 calls

Charge for 325 calls = Re. 1 × 325

= Rs. 325

Total charge for 500 calls = Rs. 200 + Rs. 325

= Rs. 525

TSC charge = 10% of Rs. 525

= Rs. 52.50

Total charge with TSC = Rs. 525 + Rs. 52.50

= Rs. 577.50

VAT = 13% of charge with TSC

= 13% of Rs. 577.50

= 13 × Rs. 577.50
100

= Rs. 75.07

∴ The total charge with TSC and VAT for 500 calls

= Rs. 577.50 + Rs. 75.07

= Rs. 652.57

EXAMPLE 3 The minimum charge of NTC is Rs. 200 for 175 calls Re. 1 for each extra
call. How many extra calls did a customer make if he paid Rs. 651.50
with 10% TSC and 13% VAT.

Solution : Here,
Charge for minimum 175 calls = Rs. 200
Let the number of additional calls be 'x'
Charge for additional calls = Re. 1 × x

GREEN Mathematics Book-9 61

Total calls = x + 175

Total call charge = Rs. 200 + Rs. x

Charge with 10% TSC = Rs. (200 + x) + 10% of Rs. (200 + x)

= Rs. (200 + x + 20 + 0.1x)

= Rs. (220 + 1.1x)

Charge with 13% VAT = (Rs. 220 + 1.1x) + 13% of Rs. (220 + 1.1x)

= Rs. (220 + 1.1x + 28.6 + 0.143x)

= 248.6 + 1.243x

Now,

Total amount paid = Rs. 651.50

or, Rs. (248.60 + 1.243x) = Rs. 651.50

or, 1.243x = Rs. (651.50 – 248.60)

or, x = 402.9
1.243

= 324.13

= 325 calls

EXERCISE: 4.2

A. Very Short Questions

1. Define fixed charge or (Standing charge)
2. Fill in the gaps.

a. Fixed charge is also known as ....................
b. Total calls = Final reading ....................
c. What is the total bill amount of 275 calls of telephone charge including 10%

VAT?

B. Short Questions

3. The minimum charge of a telephone for the first 175 calls is Rs. 200. If the charge for
each additional call is Re.1, find the charge for 260 calls.

4. The standing charge of telephone calls for 100 calls is Rs. 250. If Rs. 3 is charged for
each excess calls, then find the charge for 286 calls.

5. The minimum charge of telephone for the first 100 calls is Rs. 250. If the charge for
each additional call is Re. 1, find the charge for 280 calls including 10% TSC tax.

62 GREEN Mathematics Book-9

6. Find total charge for 574 calls, where the minimum charge for 175 calls is Rs. 200 and
Re. 1 is the charge for each calls extra. If 10% TCS and 13% VAT are levied.

7. If minimum charge for 100 calls costs Rs. 250 and extra call costs Rs. 1.50. How much
will 540 calls cost including 10% TSC and 13% VAT?

C. Long Questions
8. The standing charge of telephone for 175 calls is Rs. 200 and the charge for each

additional call Re. 1.
a. How much will be charge for 654 calls?
b. Find the charge for 560 calls including 10% TSC and 13% VAT.
9. The minimum charge of telephone for the first 175 calls is Rs. 200, the cost for each
extra call is Rs. 1. If the service charge is 10% and VAT is 13%.
a. how much should 420 calls cost?
b. how much should 310 calls cost?
10. Sona paid Rs. 546.92 including 10% TSC and 13% VAT, find how many extra calls did
he make, if NTC charges Rs. 200 for 175 calls and Re. 1 per call for additional calls.
11. Tika Ram uses his land line telephone 2 hours a day for a month of 30 days. Find how
much he should pay if NTC charges Rs. 200 as a minimum charge for 175 calls, then
Re. 1 is imposed for each extra call for a period of 2 minutes including 10% TSC and
13% VAT.

Project Work:
Collect an electricity bill of a particular month of your own house. Check the detail
of that bill. Keep the record of the current meter reading and find out your daily
consumption of your house-hold for two days. Show the detail to your teacher.

GREEN Mathematics Book-9 63

4.3 Water Bill

Look at the water bill given below. The given water bill shows the total consumption of
water indicated by water meter.

The above bill contains Rs. 785, which includes the minimum charge Rs. 100 and the
extra charge for using more than 10 units of water. It also contains 50% of sewage
charge.

Note :
• According KUKL rule standing or minimum charge upto 10 units (10,000 liters) is

Rs. 100 and Rs. 10 per excess units.
• 50% water bill is included as sewerage charge.
• The customer will get 3% rebate if the bill is paid with in 15th day of meter reading.
• No additional fine will be charged by the end of the month from 15th day.
• 10% fine will be charged during the second month.
• 20% fine will be charged during the third month.
• 50% fine will be charged for more than 3 months.

Worked Out EXAMPLES

EXAMPLE 1 From the following meter reading calculate the total water bill with
50% of its sewerage charge.

Solution : Month 2nd Baishak 2nd Jestha
Units 01510 01530

Water consumption in Jesth = (01530 – 01510)

= 20 units

= (10 + 10) units

64 GREEN Mathematics Book-9

Charge for the first 10 units is Rs. 100 as a standing charge

and extra charge = Rs. 10 × 10

= Rs. 100

∴ Total charge = Rs. 100 + Rs. 100

= Rs. 200

Again, S ewerage charge = 50% of bill amount (Rs. 200)

= 50 × Rs. 200
100

= Rs. 100
Finally, the total bill amount = Rs. 200 + Rs. 100 = Rs. 300

EXAMPLE 2 Water supply office allows 3% rebate for a water bill till 15 days of bill-
ing. After that no rebate is allowed for next 15 days. If payment is made
for next month, a fine of 10% is charged. In a house, a bill of Rs. 1050
was given on 1st of Falgun 2073. How much amount is paid on:

i. 10th of Falgun, 2073 ii. 22nd of Falgun, 2073

iii. 15th of Chaitra, 2073

Solution : i. Here,
Bill amount on 1st Falgun, 2073 = Rs. 1050
If the bill paid on 10th of Falgun, 2073, 3% of the amount of the bill
is given as a rebate.
∴ The amount to be paid = Bill amount – 3% of the bill amount
= Rs. 1050 – 3% of Rs. 1050
= Rs. 1050 – Rs. 31.50
= Rs. 1018.50
ii. If the bill is paid on 22nd Falgun, 2073, there will be no rebate
∴ Paid amount is according to bill i.e. Rs. 1050
iii. If the bill is paid on 15th of Chaitra 2073, then 10% extra charges
should be paid as a fine.
∴ The amount to be paid = Bill amount + 10% of bill amount
= Rs. 1050 + 10% of Rs. 1050
= Rs. 1050 + Rs. 105
= Rs. 1155

GREEN Mathematics Book-9 65

EXAMPLE 3 If the amount of water bill for a household is Rs. 760, how much wa-
ter consumption is made by a household, where KUKL charges Rs. 100
upto 10 units and Rs. 15 for each extra units.

Solution : Let, additional unit be 'x'

Total consumption unit = (x + 10) unit

Now,

C harge for 10 unit + charge for x unit = Rs. 870

or, 100 + 15x = 870

or, 15x = 760

or, x = 71650

= 51.33

∴ Total consumption = 51.33
= 61.33 units

EXERCISE: 4.3

A. Very Short Questions

1. .................... water bill is included as sewerage charge.
2. What is the bill amount after 3% rebate is given to the bill of Rs. 580.
3. What is the amount after 10% fine is levied to the bill of Rs. 475.

B. Short Questions

4. The water meter reading on 5th Jestha 073 was 245 units and on the 5th Asar o73 was
259. Calculate the amount to be paid according to charge as given below.
i. Standing charge (up to 8 units) = Rs. 45
ii. Excess charge = Rs. 8 per unit

5. The minimum charge for the first 10,000 liters of water is Rs. 100. If the additional
charge is Rs. 15 per unit, find the charge of 28 units of water consumption.

6. The water meter reading on 01st Baishak 2072 was 450 units and that on 01st Jestha
072 was 475 units. Calculate the amount of money to be paid including rule of KUKL
[standing units (10 units) = Rs. 100. Excess cost = Rs. 15 per unit] with sewerage
service charge = 50% of total bill]

7. The amount of water bill of Ram Chandra's house for the month of Kartik, 2073 is
Rs. 320. Calculate the amount to be paid with a miscellaneous charge of Rs. 25, if the
payment made on the first week of Mangshir receiving 3% rebate.

66 GREEN Mathematics Book-9

C. Long Questions

8. According to the rate of KUKL, the standing charge of water consumption upto 10
units (10,000 liters) is Rs. 100 and Rs. 15 per unit for each extra units.
a. How much amount should be paid for the water consumption of 45 units?

b. If the total charge is Rs. 850, how much water consumption have been made?

9. The amount of water bills for a household for a month is Rs. 210. Calculate the amount
to be paid with 50% sewerage charge if the payment is made on the first week of next
month receiving a rebate of 3%.

10. A rebate of 3% is allowed for a water bill till 15 days of billing. After that no rebate
for next 15 days. If consumer pays the amount of bill next month, a fine of 10% is
charged. A bill of Rs. 760.20 was given on 32nd Shrawan. How much will the amount
of bill be paid on:

a. 3rd Bhadra b. 25 Bhadra c. Ashwin

Project Work:

Collect a water bill. Find out minimum charge, sewerage charge, fine etc from that bill
and discuss it in your class.

4.4 Taxi Meter

In taxi meter there are some rules and conditions
1. A standing or minimum charge is Rs. 7.
2. After minimum charge the fare goes on at the rate of Rs. 9 per km.
3. Total waiting charge is Rs. 3/min.

Worked Out EXAMPLES

EXAMPLE 1 The minimum charge for hiring a taxi is Rs. 7. After that Rs. 9 will be
charged for every km. How much money should be paid to travel 25km?

Solution : Here,
The minimum charge for hiring a taxi = Rs. 7
Excess per km = Rs. 9.
Distance travelled = 25km
∴Charge for 25km = 25 × Rs. 9
= Rs. 225
Total fare = Rs. 7 + Rs. 225
= Rs. 232
∴ Rs. 232 should be paid to travel 25km.

GREEN Mathematics Book-9 67

EXAMPLE 2 Manoj Gupta hired a taxi to travel from Birgunj to Simara a distance
Solution : of 18km. During the journey he made the taxi to wait for 15 minutes.
The taxi charges Rs. 14 at initial period and Rs. 7.20 for every 200m. If
waiting charge is Rs. 30 per minute, find total amount paid by Manoj
for the taxi.

Initial charge = Rs. 14

Distance travelled = 18km

= 18000m

Here,

Fare for every 200m = Rs. 7.20

Fare for every 1m = Rs. 7.20
200

Fare for every 18000m = Rs. 7.20 × 18000
200

= Rs. 648

Waiting charge for 1 minute = Rs. 3
Waiting charge for 15 minutes = Rs. 3 × 15
= Rs. 45
∴ Payment made by Manoj = Rs. 14 + Rs. 648 + Rs. 45
= Rs. 707

EXAMPLE 3 Swostika paid Rs. 137.60 as taxi fare. If the taxi charges initial charge
Rs. 14 and then Rs. 7.20 for every 200m. In the journey she made the taxi
to wait 10 minutes as well. If the waiting charge is Rs. 3/min, find how
long had she travelled by taxi?

Solution : Total payment = Rs. 137.60

Let, she travel by taxi for x km = (1000 × x m)

Fare for every 200m = Rs. 7.20

Fare for every 1000x m = Rs. 7.20 × 1000x
200

= Rs. 36x

Waiting charge = Rs. 3 × 10

= Rs. 30

Now, Total charge = Rs. 14 + Rs. 36x + Rs. 30

or, 44 + 36x = Rs. 137.60

36x = 93.60

x = 2.6 km
∴ Swostika travelled 2.6 km by taxi.

68 GREEN Mathematics Book-9

EXERCISE: 4.4

A. Very Short Questions

1. Fill in the gaps:
a. The total waiting charge of taxi is .................. per minute.

b. The charge for every 200m is .................

c. Initial charge ........................

B. Short Questions

2. The initial charge for tax hiring was Rs. 6.60 and Re. 1.50 was charged for every 200m.
How much amount one has to pay for a journey of 20 km?

3. The minimum fare is Rs. 7 when a taxi is hired. Then Rs. 9 will be charged for every
km. How much amount should be paid to travel 30km?

4. The minimum fare is Rs. 7 when a taxi is hired. Then the fare goes on at the rate of
Rs. 9 per km. Find the total fare to be paid to travel from Langankhel to Banepa if the
distance between them is 28 km.?

5. Binod hired a taxi from Koteshwor to Kalimati which is 7km in distance. Taxi shows
Rs. 14 at first and then changes Rs. 7.20 for every 200m. Find the total fare for taxi that
Binod paid.

6. Urmila took a taxi and travelled 8km and during the journey she made taxi to wait
for 12 minutes. Find total amount that she paid, if initial charge is Rs. 14, and Rs. 7.20
for every 200m and waiting charge is Rs. 10 for every 2 minutes.

C. Long Questions
7. The minimum charge for hiring a taxi is Rs. 7. Thereafter Rs. 9 will be charged for

every km.
a. If a passenger pays Rs. 171, what distance was travelled by him?

b. How much should he pay to travel 19km?

8. The minimum cost is Rs. 7 for the first km. After that Rs. 9 will be charged for every
km. for hiring a taxi.
a. If a passenger pays Rs. 245, how far does he travel?

b. To travel 32 km, how much should be paid?

9. Rachana paid Rs. 230 travelling by taxi. The initial charge for taxi is Rs. 14 and then
Rs. 7.20 for every 200m. She made the taxi to wait for 10 minutes at the rate of Rs. 7.20
for every 2 minutes, find how long she had travelled by taxi?

Project Work:

Collect a bill of taximeter. Find minimum amount (initial charge), standing or waiting
charge and other charge imposed by taxi to the user and demonstrate about it in your
next day's class.

GREEN Mathematics Book-9 69

5

Mensuration

Estimated Teaching Periods : 24

Archimedes (287-212 BC) of Syracuse was a Greek mathematician,
physicist, engineer, inventor, and astronomer. Although few details of his
life are known, he is regarded as one of the leading scientists in classical
antiquity.

Contents

5.1 Review
5.2 Area of pathways
5.3 Area, quantities and cost
5.4 Area of four walls, floor and ceiling
5.5 Area and volume of solids
5.6 Estimation of number of bricks and cost required for building wall

Objectives

At the end of this unit, students will be able to:
solve the problems related to area, cost estimation of carpeting, painting, plastering, etc.
find the area of inner paths, outer paths and cross paths in the rectangular or square
plots or grounds or objects.
find the area of cross section, curved surface, area, total surface area and volume of
prism.
estimate the number of bricks required for making a wall and estimation of cost.

Materials

Modes of outer paths, inner path, cross paths in rectangular and square grounds;
models of prism, nets of prism, etc.

Different shapes, Relations on chart paper.

70 GREEN Mathematics Book-9

5.1 Review

Review of formulae which are used to calculate the perimeter and area of the following
plane figures..

(i) Perimeter (P) = sum of the all sides = AB + BC + A
AC
M
Area (A) = 1 × base × height Base (b) height
2
B C
= 1 × b × h
2

(ii) Perimeter (P) = a + b + c A

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

Area (A) = s(s – a) (s – b) (s – c) Heron's formula

B a

(iii) Right angled triangle A
Perimeter (P) = p + b + h
Area (A) = 1 × p × b h
2 p

B C
b

(iv) Equilateral triangle A

Perimeter (P) = a + a + a = 3a aa

Area (A) = 3 × a² Ba C

4

(v) Isosceles triangle A

Perimeter (P) = a + 2b b b
Area (A) = a 4b² – a² Ba C

4

GREEN Mathematics Book-9 71

Name Figure
vi. Quadrilateral
D

Perimeter (P) = sum of four sides = A B + BC + CD + AP
AD M2

Area (A) = 1 . d (P1 + P2) = 1 AC (BM + DN) PN d
2 2 1

B C

vii. Rectangle A D
Perimeter (P) = 2(l + b) bd C
Area (A) = l × b
Length of diagonal (d) = b² + l² B
viii. Square l
Perimeter (P) = 4l­
Area (A) = l² = 1 d2 AD

2 d
Length of diagonal (d) = 2 × l
B C
l
ix. Kite
D

Perimeter (P) = AB + BC + CD + AD = 2(AB + BC) d2
A d1 C
Area (A) = 1 × d1 × d2
2

x. Arrow-head B
A

Perimeter (P) = AD + BD + BC + AC d2
D
1 d1 C
2
Area (A) = × d1 × d2

B

xi. Parallelogram A D
Perimeter (P) = 2(sum of adjacent sides) = 2( AB + C
h
BC) D
Area (A) = b × h B C
Mb
xii. Rhombus
Perimeter (P) = 4a A d1
B d2
1 a
Area (A) = 2 × d1 × d2

72 GREEN Mathematics Book-9

xiii. Trapezium Aa D
C
Perimeter (P) = AB + BC + CD + AD hd
Mb
Area (A) = 1 h(a + b) = 1 × h(Sum of ||Sides) B
2 2

xiv. Circle

Circumference (C) = 2πr or πd r
Area (A) = πr² or 1πd²

4
xv. Semi-circle

Perimeter (P) = πr + 2r AB
Area (A) = 1πr² or 1πd² r Or

28

xvi. One - forth of circle A

Perimeter (P) = 1πr + 2r r B
2 O

Area (A) = 1πr² r
4

5.2 Area of pathways

Area of path running just outside a rectangular ground/garden.

Let ABCD be a rectangular ground in which shaded E d H
D
region is the path of uniform width d and EFGH is the A l d
d
ground with path. d l + 2d C b + 2d
If BC = l and AB = b db G

B

Now, area of path = Area of shaded part F

= Area of EFGH – Area of ABCD

= (l + 2d) (b + 2d) – l b

= lb + 2l d + 2bd + 4d² – lb

= 2d (l + b + 2d)

∴ Area of path running just outside a rectangular ground/garden etc. (A) = 2d(l + b + 2d)

Note:
If the ground/garden is square then area of outer path (A)
= 2d (l + l + 2d) = 4d (l + d)

GREEN Mathematics Book-9 73

Area of path running just inside a rectangular ground/garden.

Let ABCD be the rectangular ground in which shaded A l D
region is the path of uniform width d and EFGH is the E d
H

ground excluding the path. d d b

If BC = AD = l and AB = CD = b then EH = FG = l – 2d F

and EF = GH = b – 2d B d G
C

Area of path = Area of shaded part

= Area of ABCD – Area of EFGH

= l b – [(l – 2d) ( b – 2d)]

= l b – l b + 2ld + 2bd – 4d²

= 2d (l + b – 2d)

∴ Area of path running just inside a rectangular ground/garden etc. (A) = 2d (l + b – 2d)

Note :

If the ground/garden is square then area of inner path (A) = 2d (l + l – 2d)

= 4d (l – d)

Area of paths crossing each other perpendicularly

Let two paths (shaded regions) of uniform width A Kl D
'd' runs from the middle of every side of the rectan- d

gular ground or field crossing each other at right P d wx d Qb
R
angles. S zy
d
If BC = AD = l, AB = CD = b and wide of path = d, then C
MN
B l

Then area of crossing path = Area shaded region

= Area of PQRS + Area of KLNM – Area of square WXYZ

= l d + bd – d²

= d(l + b – d)

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

Note :
If the ground / field is square then, area of crossing paths (A) = d (l + l – d)
= d (2l – d)

74 GREEN Mathematics Book-9

Area of circular path A B
Or w
Let OA = r be the radius of a circular ground and shaded region be
the path of uniform width w. A
Then, radius of ground with path (R) = r + w. Rw
Now,
Area of path (A) = Area of bigger circle – Area of smaller circle O
= πR² – πr²
= π (R² – r²) or, π(2rw+ w²)
If the radius of the ground OA = r and shaded part is the path of
uniform width w inside the ground then radius of inner circle/
ground excluding path (R) = r – w.
Area of path (A) = πr² – π(r – w)2
= πr² – π(r² – 2rw + w2)
= 2rw – w2
= w(2r – w)

5.3 Area, quantities and cost

Carpeting a room

Let l and b be the length and breadth of a room and l1 and b1 respectively be the
length and breadth of a carpet.

We know that

area of room = area of carpet

l × b = l1 × b1

Thus, l1 = l×b ∴ b1 = l×b
b1 l1

Cost of carpeting

Let C be the rate of carpeting per meter and l1 be the length of carpet required, then
The total cost of carpeting a room (T) = length of carpet × rate of cost of carpeting
per meter.
= l1 × C

GREEN Mathematics Book-9 75

Number of bricks, tiles etc.

Let 'A' be the area of path/ground etc. and a be the area of tile/brick required to pave
and N be the number of bricks or tiles then N = A

a

Cost of paving stones/ tiles

Let N be the number of stones/tiles required to pave path/ ground and C be the cost
of a stone/ tile then
Total cost of paving stones/tiles (T) = number of stones/tiles × cost of a stone/tile
= N × C

Total cost of constructing path/carpeting a room

Let 'A' be the area of path/room and C be the cost of constructing/carpeting per m²
then
Total cost (T) = Area × Rate of cost
=A×C

Worked Out EXAMPLES

EXAMPLE 1 A
Solution :
Find the area of the shaded part and perimeter of the

sector AOB. O B

The given figure is 1 of a circle having radius 7cm. C
4 B
Now, Area of shaded part (A) = Area of 1 πr² – 1 . AO × OB
42
= 1 × 22 × 7² – 1 × 7 × 7
47 2

= (38.5 – 24.5)cm²

= 14cm²

Then, A
Perimeter of the sector AOBC(P) = AO + OB + arc AB

= 7 + 7 + 1 × 2πr O

1 4 22
4 7
= 14 + × 2 × × 7

= (14 + 11)cm

= 25cm

76 GREEN Mathematics Book-9

EXAMPLE 2 The length and breadth of a hall are 10m and 8m respectively. How
many pieces of carpet of length 1m and breadth 80cm are required for
carpeting the hall? Find the cost of carpeting the hall if each piece of
carpet costs Rs. 300.

Solution : Here, length of the hall (l) = 10m
breadth of the hall (b) = 8m
∴
Area of the hall (A) = l × b = 10 m × 8 m = 80m²

length of the carpet (l1) = 1m
breadth of the carpet (b1) = 80cm = 0.8 m
∴
Now, Area of the carpet (a) = l1 × b1 = 1m × 0.8cm = 0.8m²

Number of pieces of carpet (N) = A = 80 = 100 pieces
a 0.8
T otal cost of carpeting the hall (T) = N × C

= 100 × Rs. 300

= Rs. 30000

EXAMPLE 3 In the given figure, the unshaded part 30m D
is the path of uniform width 2m in a A 2d 2
rectangular ground of length 30m and
breadth 20m. Find the cost of making 20m C
the path at Rs. 100 per m². Also find the
cost of planting the grass on the remain- B
ing part of the ground at Rs. 150 per m².

Solution : Length of the ground (l) = 30m

breadth of the ground (b) = 20m

width of path (d) = 2m

Now, Area of path (A) = d(l + b – d)

= 2(30 + 20 – 2)m2

= 96m²

Rate of constructing the path (C) = Rs. 100/m²

T otal cost of constructing the path (T) = A × C

= 96 × Rs. 100

= Rs. 9600

GREEN Mathematics Book-9 77

Then,
Area of remaining part (A) = Area of ground – area of path
= 30 × 20 – 96
= (600 – 96)m2
= 504m²
Rate of planting grass (C) = Rs. 150 per m²
Total cost of planting grass (T) = A × C
= 504 × 150
= Rs. 75600.

EXAMPLE 4 A path of uniform width is surrounded outside a rectangular garden
of length 15m and breadth 10m. If the total cost of constructing the
path at Rs. 100 per m² is Rs. 2600, find the width of the path.

Solution : Length of garden (l) = 15m

Breadth of garden (b) = 10m

R ate of constructing the path (C) = Rs. 100 per m²

T otal cost of constructing the path (T) = Rs. 2600

Now, Area of outer path (A) = T
or, C
or,
2d(l + b + 2d) = 2600
100

2d (15 + 10 + 2d) = 26

or, d(25 + 2d) = 13

or, 2d² + 25 d – 13 = 0

or, 2d² + (26 – 1 )d – 13 = 0

or, 2d2 + 26d – d – 13 = 0

or, (d + 13) (2d – 1) = 0

either d + 13 = 0 or, d = – 13

or, 2d – 1 = 0 or, d = 0.5

The width of path can not be negative. So we neglect – 13.

∴ The width of the path (d) = 0.5 m.

78 GREEN Mathematics Book-9

EXAMPLE 5 A rectangular court is twice as long as its breadth and its perimeter
Solution : is 540 m. Find the number of bricks of size 20cm × 12cm to pave the
court. If the rate of cost of bricks is Rs. 950 per 1000, find also the cost
of paving the court.

Let the breadth of the court (b) = x m

∴ Length of the court (l) = 2xm

Perimeter (P) = 540m

or, 2(l + b) = 540

or, (2x + x) = 270

or, 3x = 270

∴ x = 90

Thus, b = 90m

l = 2 × 90 = 180 m

Now,

Area of the court (A) = l × b

= 180m × 90m

= 16200m² [∴1m = 100cm]

= 162000000cm²

Area of a brick (a) = 20cm × 12cm

= 240cm²

No. of bricks (N) = A
a

= 162200400000

= 675000 pieces

Cost of 1000 bricks = Rs. 950

∴ Cost of 1 bricks (R) = Rs. 0.95

Then,

Total cost of paving the court (C) = N × R

= 675000 × 0.95

= Rs. 641250

GREEN Mathematics Book-9 79

EXERCISE 5.1

A. Very Short Questions
1. i. If a, b, and c are sides of a triangles, then its
a. Perimeter (P) = ........................... b. Area (A) = .....................
ii. If l, d1 and d2 are length of a sides and diagonals of a rhombus, then its
a. Perimeter (P) = ......................... b. Area (A) = ......................
iii. If l and b be the length and breadth of a ground respectively and d is width

of a path, find
a. area of a path inside the ground (A) = ..........................
b. area of a path outside the ground (A) = ....................
iv. If h and b are height and base of a parallelogram, then area of the parallelo-

gram is .............................
v. If l and b are length and breadth respectively of a rectangle, then
a. Perimeter (P) = ......................... b. Length of a diagonal (d) = .............
vi. If R and r respectively are external and internal radii of gold ring then area of

a gold ring is .............................
vii. If l and b be the length and breadth of a play ground respectively and d is width

of a crossing path respectively, then area of a path (A) = .......................................
viii. If x is the length of a square football ground and d the width of a track around it

then the area of track = .......................................
ix. If y is the length of a square cricket ground and x the width of a track inside it

then the area of track = .......................................
x. a is the length of a diagonal of quadrilateral. If h1 and h2are height drawn from op-

posite vertices on the diagonal a then the area of quadrilateral is = .........................
xi. x and y are parallel sides of a trapezium. If h be the height of trapezium then

area of trapezium is = .....................................

80 GREEN Mathematics Book-9

B. Short Questions 14cm c A

2. Find the area of the given figures. Q

A

a bP

7cm
13cm
5cm
15cm 5cm
BC 3.5cm
M C
R B 8cm
12cm

A D A 6.5cm D f A 6cm D
B C
d 4cm e 5cm C 8cm
B
B 3cm 12cm C

g 10cm D
A

5cm

BC

14cm

3. a. The perimeter of a square field is 900m. Find its side.

b. Find the perimeter of a square garden if its area is 1369 m².

4. a. The area of the rectangular field is 320 m² and its breadth is 16m, find its perim-
eter.

b. The sum of the length and breadth of a rectangular field is 27m. What will be
the perimeter of the field?

5. a. If the sum of three sides of an equilateral triangle is 18cm, find the area of the
triangle.

b. If the area of an equilateral triangle is 16 3 cm², find its perimeter.

6. a. Find the area of the path 3m wide which runs just out side a rectangular garden
of length 40m and breadth 25m.

b. Find the area of the path of width 2m which is surrounded inside a square
garden of length 50m.

7. a. A square garden of length 600m has a cross path of width 15m at the middle of
the garden. Find the area of the path.

b. The radius of a circular pond is 30m. Calculate the area of path of width 3m
outside the pond.

GREEN Mathematics Book-9 81

8. Find the area of the shaded regions.

a 3m b 20m 12m 2m
e
3m 3m 3m c
20m 16m 2m
24m M P 12cm20m
S
4m 0.7m f 14cm
18cm 16m
d O
4m 4m
3.5m QR
24m 14cm

P S A PS
R Q 15cm R
g 19.8cm h 4cm i

Q B C
3cm
14m
C. Long Questions

9. The lower part of the window is square shape of side 2.1 m and upper 2.1m
part is a semi circular with the side of square as diameter. Find the total
surface area covered by the window and the perimeter of the window. 2.1m

10. A rectangular room is 8m long and 5m broad. Find the cost of carpeting
the room at Rs. 100 per meter, if the width of the carpet is 2m.

11. The length and breadth of a hall are in the ratio 3:2 and its perimeter is 40m. Find the
cost of carpeting the floor of the room at Rs. 100 per 2 sq meter.

12. A path of width 2m round outside a square field is of length 8m. Find the total cost
for constructing the path at the rate of Rs. 25 per square meter.

13. The length of a rectangular field is three times its breadth. If the area of the field is
75m², find the cost of fencing the ground 5 times at the rate of Rs. 80 per meter.

14. The breadth of a rectangular garden having length 120m is one third less than its
length. Find the cost of paving the garden with square stones of length 40cm at Rs.
20 for each stone.

15. The cost of gravelling a path 3m wide inside the boundary of a square park at Rs. 35
per m² is Rs. 50,400. Find the cost of covering the empty space with turfs at Rs. 20 per
sq.m.

16. The circumference of the circular field is 264m. A path of uniform width 1.4m is run-
ning around the field. Find the cost of gravelling the path at the rate of Rs. 80 per sq.
meter.

17. A circular meadow contains a uniform path surrounding outside of it. If the differ-
ence of the circumference of the outer and inner path is 132 meters, find the width of
the path.

18. A handkerchief is 20m long and 18m broad. How much breadth must be decreased
to cover a surface of 324m²?

82 GREEN Mathematics Book-9

5.4 Area of four walls, floor and ceiling

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

In the figure, A D
length = FG = BC = EH = AD = l E H
breadth = BF = CG = AE = DH = b
height = AB = CD = EF = GH = h B h C
b Fl G

Area of floor BFGC = l × b

Area of ceilling AEHD = l × b

Area of four walls = Area of (ABCD + EFGH + ABEF + CDHG)

= l × h + l × h + b × h + b × h

= 2l h + 2bh

= 2h (l + b)

Thus, Area of four walls (A) = 2h (l + b)

= Perimeter of room × height of room

Area of four walls and floor (A) = 2h (l + b) + l b

Area of four walls and ceiling (A) = 2h(­l + b) + lb

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

Area of four walls excluding doors and windows

Let the length and height of the door and window of a room l2 l2
be l1 and h1 and l2 and h2 respectively. Suppose m and n be the h2 h2
number of doors and windows respectively then, area of four
walls excluding doors and windows (A) h1
l1
= Area of four walls – area of doors and window

= 2h(l + b) – m (l1 × h1) – n (l2 × h2)

GREEN Mathematics Book-9 83

Worked Out EXAMPLES

EXAMPLE 1 The perimeter and height of a room are 22m and 4m respectively. Find
the total cost of painting its four walls at Rs. 50 per sq.m.

Solution : Perimeter of the room (P) = 22m
height of the room (h) = 4m
Area of four walls of the room (A) = P × h
= 22m × 4m
= 88 m²
Rate of painting the wall (C) = Rs. 50 per sq. m
∴ Total cost of painting the four walls (T) = A × C
= 88 × Rs. 50
= Rs. 4400

EXAMPLE 2 A room of 12m long, 8m broad and 10m high has two doors each mea-
suring 4m by 2m and four windows each of area 3m². Find the cost of

Solution : painting its walls at Rs. 30 per sq.m.

Length of room(l ) = 12m

breadth of room (b) = 8cm

height of room (h) = 10m

Area of four walls excluding door and windows (A) = Area of four
walls – Area of two doors and four windows

= 2h(l + b) – 2(l1 × h1) – 4 × 3
= 2 × 10 (12 + 8) – 2 (4 × 2) – 12

= 400 – 16 – 12

= 372m²

Cost of painting the walls (R) = Rs. 30 per sq.m.

∴ Total cost of painting the wall (T) = A × R

= 372 × Rs. 30

= Rs. 11160.

84 GREEN Mathematics Book-9

EXAMPLE 3 The conference hall 8m × 6m × 4m has 4 windows each of the area
1.5m × 1m and 2 doors each of area 2.5m × 2m. Find the rate of painting
its four walls and ceiling if the total cost is Rs. 21600.

Solution : Area of four walls and ceiling (A1) = 2h (l + b) + l b
= 2 × 4 (8 + 6) + (8 × 6)
= 160m²
Area of 2 doors and 4 windows (A2) = 2(2.5 × 2) + 4(1.5m × 1m)
= 10 + 6
= 16m²
Area of four walls and ceiling excluding doors
windows (A) = A1 – A2

= 160m² – 16m²
= 144m²

∴ Total cost of painting four walls and ceiling (T) = Rs. 21600

Now, rate of cost for painting (A) = T
A

= Rs. 21600
144

= Rs. 150 per m²

EXAMPLE 4 The cost of plastering 4 walls of a room of 8m long and 4.5 m high at
Rs. 45 per m² is Rs. 5670, find the cost of carpeting the floor at Rs. 150
per. m².

Solution : Rate of plastering (C) = Rs. 45 per m²

Total cost of plastering (T) = Rs. 5670

Now, A rea of four walls of the room (A) = T
C

or, 2h(l + b) = 5670
45

or, 2 × 4.5 (8 + b) = 126

or, b + 8 = 14

∴ b = 6m

Then, area of room (A) = l × b

= 8m × 6m = 48m²

Rate of carpeting (C) = Rs. 150 per sq.m.

∴ Total cost of carpeting (T) = A × C

= 48 × Rs. 150

= Rs. 7200

GREEN Mathematics Book-9 85

EXAMPLE 5 The cost of plastering the floor of a room at Rs. 18.50 per sq. meter is
Rs. 1480 and the cost of painting its walls at Rs. 12.50 per sq. meter is
Rs. 2700. If the length of the room is 10m, find the height of the room.

Solution : The cost of plastering the floor (T) = Rs. 1480

Rate of plastering the floor (C) = Rs. 18.50
Thus, Area of the room (A) = T

C
or, l × b = 1480

18.50

or, l × b = 80
10 × b = 80
∴ b = 8 m

Again, Area of four walls (A) = Cost of painting the walls
Rate of cost for painting

or, 2h(l + b) = 2700
12.50
or,
∴ 2h(10 + 8) = 216
So,
h = 6 m

the height of the room (h) = 6m

EXERCISE 5.2

A. Very Short Questions

1. a. The length, breadth and height of a room are 8m, 6m and 4m respectively. Find
the area of four walls of the room.

b. The area of 4 walls and ceiling of a room is 107 m². If the room is 6m long and
5m wide, how high is the room?

2. a. The perimeter and height of a room are 22m and 4m respectively. Find the area
of four walls of the room.

b. The area of four walls of a room is 88m² and its height is 5m. Find the sum of
the length and breadth of the room.

B. Short Questions

3. a. The area of four walls of a room is 192 m². It has a door of area 6m² and two
windows each of area 3m². Find the area of four walls excluding door and
windows.

86 GREEN Mathematics Book-9

b. The area of four walls and ceiling of a room is 96 sq.m. It has 3 windows and
a door. If the area of four walls and ceiling excluding door and windows is
4. a. 88.5sq. m, find the area of 1 door and 3 windows.
b.
The area of four walls and ceiling of a room is 450 m². Find the cost of painting
its walls and ceiling at Rs. 40 per sq. meter.

The total cost of plastering four walls of a room at Rs. 30 per sq. m is Rs. 5760.
Find the area of four walls of the room.

5. a. The dimension of a rectangular room is 6m × 5m × 3m. Find the number of
pieces of wall paper required to cover the wall if the size of each wall paper is
1m × 60cm.

b. 100 pieces of wall paper is required to cover the wall of a rectangular room, If
the size of each wall paper is 110cm × 80cm, find the area of four walls of the
room.

C. Long Questions

6. a. Find the cost of colouring the four walls and ceiling of a room of size 7m × 6m
× 3m at the rate of Rs. 150 per sq. meter.

b. Height of a square room is 5m. The total cost of plastering the four walls of the
room at the rate of Rs. 20 per square meter is Rs. 2400. Find the perimeter of the
room.

7. a. The length and breadth of a rectangular room are 10m and 8m respectively. If
the cost of plastering the four walls at the rate of Rs. 40 per square meter is Rs.
5400, find the height of the room.

b. The height of a square room is 3m. If the cost of painting its four walls at the
rate of Rs. 25 per sqm is Rs. 1800, find the length of the room.

8. a. A room is 6m × 5m × 3m with three windows each of size 2.5m × 1.5m and
a door of size 1.5m × 2.5. Find the cost of plastering the walls excluding 3
window and a door at the rate Rs. 30 sq.m.

b. A hall of 18m × 12m × 4m consists of a door of size 2m × 1m and 4 windows
each of size 3m × 1m. Find the number of tiles of size 15cm × 20cm to pave
on four walls and the floor. If the cost price per tile is Rs. 45, find total cost of
paving the tile.

9. a. A square room is 6m high. The cost of carpeting the floor at the rate of Rs. 40
per square meter is Rs. 4000. Find the cost of painting its walls at the rate of Rs.
30 per square meter.

b. The cost of carpeting the floor of a room whose breadth is twice the height and
length is twice its breadth at Rs. 500 per sq. meter is Rs. 64,000. What will be
cost of plastering its 4 walls at the rate of Rs. 20 per sq. meter?

c. The cost of carpeting a room which is 5m high and length is twice its breadth
is Rs. 36,000 at the rate of Rs. 500/m². What will be cost of plastering its 4 walls
at the rate of Rs. 20/m² ?

GREEN Mathematics Book-9 87

10. a. The cost of carpeting the floor of a room at Rs. 60 per square meter is Rs. 4800.
b. The cost of painting its four walls at the rate of Rs. 45 per square meter is
Rs. 9720. If length of room is 10m, find its height.

The cost of painting its four walls, floor and ceiling of a room at Rs. 45 per sq.
meter, is Rs.31,500. If the cost of painting its floor and ceiliing at Rs. 25 per sq.
meter is Rs. 10,000 and its breadth is 10m, find its height.

5.5 Area and volume of solids

Cuboid, cube, sphere, hemisphere, cylinder, pyramid and cone are some of the examples
of solid objects. These solid have shape and surfaces and they occupy space. Here we
learn to find the surface area and volume of some of solid objects.

A D G lH
E H h
h
B h C l Dh H G
b Fl G A b
G b
bb
E l Ch F l E
h
hB
h

El F

Cuboid is a solid having six rectangular surfaces. Two opposite surfaces of a cuboid are

equal and parallel. When a cuboid is opened, we get the net as shown in the above figure.

The sum of the area of six rectangular surfaces is the total surface area of cuboid.

Total surface area of a cuboid (A) = Area of rectangles shown in the net of cuboid

= lb + lb + bh + bh + l h + l h

= 2lb + 2bh + 2lh

= 2(lb + bh + lh)

Lateral surface area (LSA) = curved surface area (CSA)
Lateral surface area (LSA) = Area of four lateral rectangular faces
= lh + l h + bh + bh
= 2lh + 2bh
= 2h(l + b)
= 2(l + b) × h = p × h
∴ LSA = Perimeter of base × height
Volume of cuboid (V) = Area of base × height of the cuboid
= l × b × h
= A × h

88 GREEN Mathematics Book-9

Area and volume of cube

Cube is a solid having six equal faces. The adjoining figures is the cube having its length
l. In the cube length, breadth and height are equal.

Total surface area of cube = Area of six square faces = 6l²

Volume of cube (V) = l × l × l

= l³ l
Lateral surface area of the cube (LSA) = 4l2

5.6 Prisms and their cross sections

h h h

Cuboid h Square based Circular prism h
(Rectangular prism) Triangular prism cylinder L-shaped prism

prism

(Triangular base prism) (Circular base prism)

In the above solids the shaded faces are equal and parallel. Such solids are prism. The
equal and parallel faces are called bases (ends) of the prism and distance between them is
called the height or length of the right prism.

The prism are named after the base figures. If the base is triangle, the prism is called a
triangular prism. If the base is circle, the prism is called circular prism. So, the prism is
a solid with two opposite faces equal and parallel and the lateral faces are all rectangles.

When a prism is cut then the new surface formed is parallel to bases, the new surface is
called, cross section of the prism. The area of cross section is equal to the area of base.

Cross section Cross section

Volume of prism = Area of cross section × height of the prism

V = A × h

L ateral surface area of prism (LSA) = Perimeter of cross section × height of the prism

= P × h

Total surface area of prism (TSA) = Lateral surface area + 2 × area of cross section

GREEN Mathematics Book-9 89

Worked Out EXAMPLES 8cm
12cm
EXAMPLE 1 Find the cross section area, lateral surface 4cm10cm
area, total surface area and the volume of the 4cm 15cm
given prism.
4cm
Solution : Area of cross section = (10 × 4) + (8 × 4)

= 72cm²

Lateral surface area = Perimeter of cross section × height

= (10 + 12 + 4 + 8 + 6 + 4) × 15 4cm

= 44 × 15 10cm 8cm
6cm

= 660cm²

Total surface area = Lateral surface area + 2 × area of cross section

= 660cm² + 2 × 72 cm²

= 804cm²

Volume = Area of cross section × height

= 72cm² × 15cm

= 1080 cm³

EXAMPLE 2 The volume of the prism shown in the adjoining 4cm 4cm
figure is 640cm³. Find the value of x and the total 4cm
surface area of the prism. 4cm 4cm
12cm xcm
Solution : Area of cross section = (12 × 4) + (4 × 4)cm² 4cm
4cm
= 64cm² 12cm 4cm
Volume Cross section

Height of prism (h) = Area of cross secession

640cm³
= 64cm²

∴ x = 10cm

Total surface area = (Perimeter of cross section × height) + 2(area cross section)

= (12 + 4 + 4 + 4 + 4 + 4 + 4 + 4) × 10 + (2 × 64)

= 400 + 128
= 528cm²

90 GREEN Mathematics Book-9

EXAMPLE 3 If the total surface area of the given prism is 246cm², 2cm
find the value of x.
5cm
Solution : Area of cross section = (10 × 3) + (2 × 2) + (2 × 2)cm² 2cm

6cm xcm
10cm

= 38cm² 2cm 2cm

P erimeter of cross section (P) = 10 + 5 + 2 + 2 +6 + 2 + 2 + 5 2cm
2cm

= 34cm 6cm (5 –2)
Height of prism (h) = xcm = 3cm
10cm
Cross section

Total surface area of prism = (p × h) + 2 (area of cross section)

or, 246 = (34 × x) + (2 × 38)

or, 246 – 76 = 34x

or, 170 = 34x

∴ x = 5cm

EXAMPLE 4 At the rate of 1l per 10 seconds a cubical water vessel can be completely
filled in 10,000 seconds;
a. find the volume of vessel.
b. find the internal total surface area if it is lidless.

Solution : a. Volume of water filled in 10 seconds = 1l
1

Volume of water filled in 1 second = 10l
1

∴ Volume of water filled in 10,000 seconds = 10 × 10,000l
= 1000l

Thus, the internal volume of the vessel = 1000l

b. We have, internal volume = 1000l = 1000000cm3

or, l³ = (100cm)3 [∴1l = 1000cm3]
∴ l = 100cm

I nternal total surface area of the vessel(A) = 5l² [It is hidless] 91
= 5 × (100cm)²
= 5 × 1m²
= 5m2

GREEN Mathematics Book-9

EXERCISE: 5.3

A. Very Short Questions EF
AG

1. a. In the figure, area of shaded part is 27cm². Find its volume. B C 20cm

b. If the perimeter of the shaded region is 27cm, find the lateral surface area of the

solid.

c. Find the total surface area of the solid.

2. a. The area of cross section and the lateral surface area of a prism are 36cm2 and
320cm² respectively. Find the total surface area of prism.

b. The perimeter and area of cross section of a prism are 34cm and 48cm²
respectively. If the length of the prism is 8cm find the total surface area of the
prism.

c. The total surface area and the lateral surface area of a prism are 936cm² and
672cm² respectively. Find the area of cross section of the prism.

3. a. The area of cross section and the height of the prism are 154cm² and 20cm re-
spectively. Find the volume of the prism.

b. The area of the shaded part is 29cm² and volume of the x
solid is 290cm³. Find the value of x.

B. Short Questions

4. a. The length, breadth and height of a cuboid are 8cm, 6cm
and 5cm respectively. Find the lateral surface area and the total surface area of
the cuboid.

b. The length and breadth of a cuboid are 10cm and 5cm respectively. If the total
surface area is 220cm², find its thickness.

5. a. Find the volume of a cuboid if its base area is 60cm² and height is 5cm. .

b. If the total inner surface area of a cubical water tank is 24m², how much water
does it hold?

C. Long Questions

6. Find the cross section area, the lateral surface area, the total surface area and the vol-
ume of the following prisms.

a. b. c. d.

4cm 2cm 2cm 8cm

2cm 5cm
2cm

8cm
4cm

4cm
4cm
6cm 7cm 4cm 4cm 5cm
10cm
6cm 10cm 4cm 10cm
10cm 10cm
8cm

92 GREEN Mathematics Book-9

7. a. A water tank of dimensions 1.2m × 1m × 50cm is made of zinc sheet. If the tank
is opened at the top, find:

i. the area of zinc sheet required to make the tank?

ii. how much does it cost to make the tank if the cost of 1m² of zinc sheet is
Rs. 700.

iii. how much water does the tank hold?

b. A class room is 24m long, 20m wide and 15m high. How many children can be
accommodated in it allowing 15 sq. m. of floor for each child? How many cubic
meters of space will there be for each child?

8. a. Three metal cubes whose edges are 3cm, 4cm and 5cm are melted and formed
a single cube. Find the edge of the single cube.

b. The dimensions of a metallic cuboid are 100cm × 80cm × 64cm. If it is melted
and recast into a cube, find the surface area of the cube.

9. a. A 4cm cube is cut into 1cm cubes. Find the ratio of the surface area of small
cubes to that of large cube.

b. The length of a cubical packet of biscuits is 6cm. What size of carton is needed
to keep 20 packets of biscuits?

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

If the number of bricks required to make a wall is 'N' Wall
and the volume of a brick is 'v' then volume of the wall
(V) is equal to the product of the number of bricks and Volume of wall → V
the volume of a brick. Number of bricks used = N
Brick
So, V = Nv Volume = v
Thus,
N = V and ∴N = Large (V)
v Small (v)

v = V
N

If the cost of a brick is C then, total cost of bricks (T) = N × C

Thus, N = T
C

and C = T
N

If the wall contains doors and windows then volume of wall excluding doors and windows
= Volume of wall – space occupied by doors and windows.

GREEN Mathematics Book-9 93

Worked Out EXAMPLES

EXAMPLE 1 A wall 25m long, 4m high and 30cm wide. If it contains 2 windows
each of 2m × 1.5m and a door size 2m × 3m, find:

Solution : i. the number of bricks each of 15cm × 10cm × 5cm
ii. the cost of bricks at the rate of Rs. 5000 per 1000 bricks.
i. Volume of wall (V1) = 25m × 4m × 30cm
= 25m × 4m × 0.3m

= 30000000cm³

Volume of a door and 2 windows (V2)
= 2m × 3m × 30cm + 2(2m × 1.5m × 30)cm³

= (200 × 300 × 30)cm³ + 2 (200 × 150 × 30)cm³

= 3600000cm³

Volume of wall excluding a door and two windows (V) = V1 – V2
= (30000000 – 3600000)cm3

= 26400000cm³

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

= 750 cm³

Now, Number of bricks (N) = V
v

= 26400000 = 35200
750

ii. Cost of 1000 bricks = Rs. 5000

Cost of 1 brick (C) = Rs. 5000 = Rs. 5
1000

∴ C = Rs. 5 per brick

Total cost of bricks (T) = N × C

= 35200 × Rs. 5

= Rs. 176000

94 GREEN Mathematics Book-9