Lesson 4 Household Arithmetic

4.1 Household expenses for use of electricity

Exercise 4.1

1. A meter of capacity 5 ampere is installed in Dambar Kumari Tamang’s house. The meter reading of her house from the month Kartik to Fagun of the year 2076 BS is given in the following table. Study the table and answer the questions given below:

Meter Reading Unit | 3528 | 3593 | 3700| 3904 | 3980

(a) What is the total unit of electricity consumed on the month of Manshir?

(b) In which months did Dambar Kumari’s house consume the most and the least amount of electricity?

Solution,

(a) Consumed unit in the month of Manshir = Meter reading in Manshir - Meter reading in Kartik

= 3593 - 3528

= 65 units

(b)

Kartik to Manshir (Manshir’s consumption) = 3593 - 3528 = 65 units

Manshir to Push (Push’s consumption) = 3700 - 3593 = 107 units

Push to Magh (Magh’s consumption) = 3904 - 3700 = 204 units

Magh to Fagun (Fagun’s consumption) = 3980 - 3904 = 76 units

Most consumption = 204 units, which occurred in Magh

Least consumption = 65 units, which occurred in Manshir

Hence, Dambar Kumari’s house consumed the most electricity in Magh and the least in Manshir.

2. According to the current tariff rate, calculate the tariff to be paid to Nepal Electricity Authority in the following cases:

(a) Present reading: 2575, Previous reading: 2472, Capacity of meter: 5 Ampere, The bill is paid on the 4th day of meter reading.

(b) Present reading: 3036, Previous reading: 2831, Capacity of meter: 15 Ampere, The bill is paid on the 41st day of meter reading.

(c) Present reading: 3603, Previous reading: 3294, Capacity of meter: 30 Ampere, The bill is paid on the 39th day of meter reading.

(d) Present reading: 4311, Previous reading: 3944, Capacity of meter: 60 Ampere, The bill is paid on the 14th day of meter reading.

(e) Present reading: 5555, Previous reading: 5107, Capacity of meter: 30 Ampere, The bill is paid on the 17th day of meter reading.

(f) Present reading: 6452, Previous reading: 6443, Capacity of meter: 5 Ampere, The bill is paid on the 57th day of meter reading.

Solution:

(a)

Here, Present Reading = 2575, Previous Reading = 2472

Units Consumed = 2575 - 2472 = 103 units

Capacity of Meter = 5 Ampere

Service Charge = Rs 30

We know that,

Energy Charge: First 20 units at Rs 4/unit = 20 × 4 = 80

Next 83 units at Rs 7.5/unit = 83 × 7.5 = 622.50

Total Energy Charge = 80 + 622.50 = 702.50

Now, Total Bill (before penalty) = Service Charge + Energy Charge

= 30 + 702.50 = 732.50

Then, Payment on 4th day (within 7 days), so no penalty applies.

Thus, the tariff to be paid is: Rs 732.50.

(b)

Here, Present Reading = 3036, Previous Reading = 2831

Units Consumed = 3036 - 2831 = 205 units

Capacity of Meter = 15 Ampere

Service Charge = Rs 75

We know that,

Energy Charge: First 20 units at Rs 5/unit = 20 × 5 = 100

Next 185 units at Rs 8.5/unit = 185 × 8.5 = 1,572.50

Total Energy Charge = 100 + 1,572.50 = 1,672.50

Now, Total Bill (before penalty) = Service Charge + Energy Charge

= 75 + 1,672.50 = 1,747.50

Then, Payment on 41st day (after 7 days), so 5% penalty applies.

Again, Penalty = (5/100) × 1,747.50 = 87.375

Next, Final Bill = 1,747.50 + 87.375 = 1,834.88

Thus, the tariff to be paid is: Rs 1,834.88.

(c)

Here, Present Reading = 3603, Previous Reading = 3294

Units Consumed = 3603 - 3294 = 309 units

Capacity of Meter = 30 Ampere

Service Charge = Rs 125

We know that,

Energy Charge: First 20 units at Rs 6/unit = 20 × 6 = 120

Next 230 units at Rs 9.5/unit = 230 × 9.5 = 2,185

Remaining 59 units at Rs 9.5/unit = 59 × 9.5 = 560.50

Total Energy Charge = 120 + 2,185 + 560.50 = 2,865.50

Now, Total Bill (before penalty) = Service Charge + Energy Charge

= 125 + 2,865.50 = 2,990.50

Then, Payment on 39th day (after 7 days), so 5% penalty applies.

Again, Penalty = (5/100) × 2,990.50 = 149.525

Next, Final Bill = 2,990.50 + 149.525 = 3,140.03

Thus, the tariff to be paid is: Rs 3,140.03.

(d)

Here, Present Reading = 4311, Previous Reading = 3944

Units Consumed = 4311 - 3944 = 367 units

Capacity of Meter = 60 Ampere

Service Charge = Rs 175

We know that,

Energy Charge: First 20 units at Rs 7/unit = 20 × 7 = 140

Next 230 units at Rs 11/unit = 230 × 11 = 2,530

Remaining 117 units at Rs 11/unit = 117 × 11 = 1,287

Total Energy Charge = 140 + 2,530 + 1,287 = 3,957

Now, Total Bill (before penalty) = Service Charge + Energy Charge

= 175 + 3,957 = 4,132

Then, Payment on 14th day (after 7 days), so 5% penalty applies.

Again, Penalty = (5/100) × 4,132 = 206.60

Next, Final Bill = 4,132 + 206.60 = 4,338.60

Thus, the tariff to be paid is: Rs 4,338.60.

(e)

Here, Present Reading = 5555, Previous Reading = 5107

Units Consumed = 5555 - 5107 = 448 units

Capacity of Meter = 30 Ampere

Service Charge = Rs 125

We know that,

Energy Charge: First 20 units at Rs 6/unit = 20 × 6 = 120

Next 230 units at Rs 9.5/unit = 230 × 9.5 = 2,185

Remaining 198 units at Rs 9.5/unit = 198 × 9.5 = 1,881

Total Energy Charge = 120 + 2,185 + 1,881 = 4,186

Now, Total Bill (before penalty) = Service Charge + Energy Charge

= 125 + 4,186 = 4,311

Then, Payment on 17th day (after 7 days), so 5% penalty applies.

Again, Penalty = (5/100) × 4,311 = 215.55

Next, Final Bill = 4,311 + 215.55 = 4,526.55

Thus, the tariff to be paid is: Rs 4,526.55.

(f)

Here, Present Reading = 6452, Previous Reading = 6443

Units Consumed = 6452 - 6443 = 9 units

Capacity of Meter = 5 Ampere

Service Charge = Rs 30

We know that,

Energy Charge: First 9 units at Rs 4/unit = 9 × 4 = 36

Total Energy Charge = 36

Now, Total Bill (before penalty) = Service Charge + Energy Charge

= 30 + 36 = 66

Then, Payment on 57th day (after 7 days), so 5% penalty applies.

Again, Penalty = (5/100) × 66 = 3.30

Next, Final Bill = 66 + 3.30 = 69.30

Thus, the tariff to be paid is: Rs 69.30.

4.2 Household expenses for use of water

Exercise 4.2

1. There is 1½ inch drinking water supply pipe connected in Kopila’s house. The current and previous reading in the month of Shrawan are 4225 and 4197 respectively. Solve the following questions on the basis of tariff table given on the previous page:

(a) How much does Kopila have to pay for 50% of that month’s sewage service?

(b) What is the total bill amount?

(c) If the bill is paid within the first and second month of distribution, how much tariff will Kopila have to pay?

(d) If the bill is paid within the third month of distribution, how much tariff will she have to pay?

Solution:

Here, Present Reading = 4225

Previous Reading = 4197

Units Consumed = 4225 - 4197 = 28 units

We know that,

Charge of minimum 10 units (10000 litre) = Rs.100

Here, 28 units = 10 units + 18 units

Minimum charge Rs.100 at the rate of Rs.32 per unit

Now, Total tariff = 100 + 18 × 32

= 100 + 576

= Rs.676

(a)

Then, Sewerage Charge = 50% of Rs.676

= (50/100) × 676

= Rs.338

Thus, Kopila has to pay Rs.338 for 50% of that month’s sewage service.

(b)

Again, Total Bill = Total Tariff + Sewerage Charge

= 676 + 338

= Rs.1,014

Thus, the total bill amount is Rs.1,014.

(c)

Next, If the bill is paid within the first or second month, a 3% discount applies.

Discount = 3% of Rs.1,014

= (3/100) × 1,014

= Rs.30.42

Then, Amount after Discount = 1,014 - 30.42

= Rs.983.58

Thus, if the bill is paid within the first or second month, Kopila has to pay Rs.983.58.

(d)

Next, if the bill is paid within the third month, a 10% fine applies.

Fine = 10% of Rs.1,014

= (10/100) × 1,014

= Rs.101.40

Then, Amount with Fine = 1,014 + 101.40

= Rs.1,115.40

Thus, if the bill is paid within the third month, Kopila has to pay Rs.1,115.40.

2. 423 units of water was consumed in a month at a hotel from a 1.5" sized water supply pipe. Solve the following questions on the basis of tariff table given on the previous page:

(a) What will be the total tariff including sewerage in that month?

(b) If the bill is paid within five months of the distribution of the bill, how much penalty has to be paid?

Solution:

Here, Total consumption of water = 423 units

We know that,

Charge of minimum 10 units (10000 litre) = Rs.100

Here, 423 units = 10 units + 413 units

Minimum charge Rs.100 at the rate of Rs.32 per unit

Now, Total tariff = 100 + 413 × 32

= 100 + 13,216

= Rs.13,316

Then, Sewerage Charge = 50% of Rs.13,316

= (50/100) × 13,316

= Rs.6,658

(a)

Again, Total Bill = Total Tariff + Sewerage Charge

= 13,316 + 6,658

= Rs.19,974

Thus, the total tariff including sewerage in that month is Rs.19,974.

(b)

Next, if the bill is paid within five months, a 10% fine applies.

Fine = 10% of Rs.19,974

= (10/100) × 19,974

= Rs.1,997.40

Thus, the penalty to be paid is Rs.1,997.40.

Household expenses for use of telephone

Exercise 4.3

1. The current reading telephone calls of Ramlal’s house in the month of Baishakh is 4444 and previous reading is 3992, calculate:

Solution:

a) How many total calls have been made?

Total calls = Current reading – Previous reading

= 4444 – 3992

= 452 calls

Hence, the total calls made are 452 calls.

b) What is the total tariff (TC) of the calls if Rs 200 is charged for the first 175 calls and after that Re 1 per call?

Now, calculate the additional calls:

452 calls = 175 calls + (452 – 175) calls

= 175 calls + 277 calls

Minimum charge Rs 200 at the rate of Re 1 per call:

Total tariff (TC) = Rs 200 + 277 × Re 1

= Rs 200 + Rs 277

= Rs 477

Hence, the total tariff (TC) is Rs 477.

c) Calculate the service charge and value-added tax amount.

Service charge (TSC) = 13% of Rs 477

= (13/100) × 477

= Rs 62.01

Total tariff including service charge (C + TSC) = Rs 477 + Rs 62.01

= Rs 539.01

Then, value-added tax (VAT) = 13% of Rs 539.01

= (13/100) × 539.01

= Rs 70.0713

Hence, the service charge (TSC) is Rs 62.01, and the value-added tax (VAT) is Rs 70.0713.

d) Calculate the total tariff after adding value-added tax.

Total tariff = C + TSC + VAT amount

= 477 + 62.01 + 70.0713

= Rs 609.0813

Total tariff = Rs 609.08

Hence, the total tariff after adding value-added tax is Rs 609.08.

2. Calculate the total tariff including 13% service charge (TSC) and 13% value added tax for the following telephone calls: (Where Rs.200 is charged for first 175 calls and after that Re.1 per call is added)

Solution:

a) 550 calls

Here, calculating the total calls:

Total calls = 550 calls

Extra calls = 550 – 175 = 375 calls

Total tariff (TC) = minimum charge + extra charge

= Rs 200 + Rs 375 × 1

= Rs 200 + Rs 375

= Rs 575

Now, using the alternative method:

Tariff including TSC and VAT = (113/100) × (113/100) × TC

= (113/100) × (113/100) × 575

= (113 × 113 × 575) / (100 × 100)

= (12769 × 575) / 10000

= 7342175 / 10000

= Rs 734.2175

Hence, the total tariff for 550 calls is Rs 734.22.

b) 695 calls

Here, calculating the total calls:

Total calls = 695 calls

Extra calls = 695 – 175 = 520 calls

Total tariff (TC) = minimum charge + extra charge

= Rs 200 + Rs 520 × 1

= Rs 200 + Rs 520

= Rs 720

Now, using the alternative method:

Tariff including TSC and VAT = (113/100) × (113/100) × TC

= (113/100) × (113/100) × 720

= (113 × 113 × 720) / (100 × 100)

= (12769 × 720) / 10000

= 9193680 / 10000

= Rs 919.368

Hence, the total tariff for 695 calls is Rs 919.37.

c) 793 calls

Here, calculating the total calls:

Total calls = 793 calls

Extra calls = 793 – 175 = 618 calls

Total tariff (TC) = minimum charge + extra charge

= Rs 200 + Rs 618 × 1

= Rs 200 + Rs 618

= Rs 818

Now, using the alternative method:

Tariff including TSC and VAT = (113/100) × (113/100) × TC

= (113/100) × (113/100) × 818

= (113 × 113 × 818) / (100 × 100)

= (12769 × 818) / 10000

= 10447042 / 10000

= Rs 1044.7042

Hence, the total tariff for 793 calls is Rs 1044.70.

3. Minimum tariff for the first 175 calls is Rs 200 and after that Re 1 is charged for each extra call. How many telephone calls can be made from the following amount excluding service charge (TSC) and value added tax?

Solution:

a) Rs 275

Total tariff (TC) = Rs 275

Extra charge = TC – minimum charge

= 275 – 200

= Rs 75

Extra calls = Extra charge / Rate per call

= 75 / 1

= 75 calls

Total calls = 175 + extra calls

= 175 + 75

= 250 calls

Hence, 250 calls can be made from Rs 275.

b) Rs 695

Total tariff (TC) = Rs 695

Extra charge = TC – minimum charge

= 695 – 200

= Rs 495

Extra calls = Extra charge / Rate per call

= 495 / 1

= 495 calls

Total calls = 175 + extra calls

= 175 + 495

= 670 calls

Hence, 670 calls can be made from Rs 695.

c) Rs 890

Total tariff (TC) = Rs 890

Extra charge = TC – minimum charge

= 890 – 200

= Rs 690

Extra calls = Extra charge / Rate per call

= 690 / 1

= 690 calls

Total calls = 175 + extra calls

= 175 + 690

= 865 calls

Hence, 865 calls can be made from Rs 890.

4. Minimum tariff for the first 175 calls is Rs 200 and after that Re 1 is charged for each extra call. If the total tariff including 13% service charge and 13% value added tax is Rs 696.08, how many telephone calls had been made?

Solution:

Here, the total tariff including TSC and VAT is Rs 696.08.

Using the alternative method:

Tariff including TSC and VAT = (113/100) × (113/100) × TC

696.08 = (113/100) × (113/100) × TC

696.08 = (12769/10000) × TC

Now, solve for TC:

TC = 696.08 / (12769/10000)

= 696.08 × (10000/12769)

= (696.08 × 10000) / 12769

= 6960800 / 12769

= 545

Total tariff (TC) = Rs 545

Extra charge = TC – minimum charge

= 545 – 200

= Rs 345

Extra calls = Extra charge / Rate per call

= 345 / 1

= 345 calls

Total calls = 175 + extra calls

= 175 + 345

= 520 calls

Hence, 520 telephone calls had been made.

5. Minimum service charge (Rental charge) of GSM postpaid mobile service of Nepal Telecom is Rs 300. After applying 13% service charge on this amount, 13% value added tax is added. If after that again 2% ownership tax is added on the total tariff, how much minimum monthly tariff has to be paid by a postpaid SIM user?

Solution:

Here, the minimum service charge (rental charge) is Rs 300.

Service charge (TSC) = 13% of Rs 300

= (13/100) × 300

= Rs 39

Total tariff including service charge (C + TSC) = Rs 300 + Rs 39

= Rs 339

Now, value-added tax (VAT) = 13% of Rs 339

= (13/100) × 339

= Rs 44.07

Total tariff including TSC and VAT = Rs 339 + Rs 44.07

= Rs 383.07

Then, ownership tax = 2% of Rs 383.07

= (2/100) × 383.07

= Rs 7.6614

Total minimum monthly tariff = Rs 383.07 + Rs 7.6614

= Rs 390.7314

Total tariff = Rs 390.73

Hence, the minimum monthly tariff to be paid by a postpaid SIM user is Rs 390.73.

4.4 Calculation of amount of taxi meter

Exercise 4.4

1. Shreenath took a taxi to travel 9 km from Baneshwar to Bhaktapur. Initially the taxi meter charged a minimum of Rs.14 and then Rs. 7.80 per 200 meters. During the journey, if 6 minutes waiting charge at the rate of Rs. 7.80 per 2 minute was also charged, how much taxi fare did he pay? Find it.

Solution

Minimum (initial) charge = Rs 14

The distance travelled by the taxi = 9 km

Fare per km = Rs 39

Waiting time = 6 minutes

The fare of waiting charge per 2 minutes = Rs 7.80

The amount for waiting charge for 6 minutes = (7.80 / 2) × 6 = Rs 23.40

Total fare = Rs 14 + (39 × 9) + 23.40

= Rs 14 + 351 + 23.40

= Rs 388.40

2. Shristi took a taxi to travel 18 km from Jagati, Bhaktapur to Sanepa, Lalitpur. At first, the taxi meter charged a minimum of Rs.14 and then Rs. 7.80 per 200 meters. During the journey, if 20 minutes waiting charge at the rate of Rs. 7.80 per 2 minute was also charged, how much taxi fare did he pay? Find it.

Solution:

Minimum (initial) charge = Rs 14

The distance travelled by the taxi = 18 km

Fare per km = Rs 39

Waiting time = 20 minutes

The fare of waiting charge per 2 minutes = Rs 7.80

The amount for waiting charge for 20 minutes = (7.80 / 2) × 20 = Rs 78

Total fare = Rs 14 + (39 × 18) + 78

= Rs 14 + 702 + 78

= Rs 794

3. Calculate the taxi fare in the following cases based on the rate mentioned in this lesson:

Distance (in km) Time Waiting time (in minute)

(a) 6 8 am 15

(b) 15 3:30 pm 10

(d) 8.75 4 am 7

Solution:

Minimum (initial) charge = Rs 14

The distance travelled by the taxi = 6 km

Fare per km = Rs 39

Waiting time = 15 minutes

The fare of waiting charge per 2 minutes = Rs 7.80

The amount for waiting charge for 15 minutes = (7.80 / 2) × 15 = Rs 58.50

Total fare = Rs 14 + (39 × 6) + 58.50

= Rs 14 + 234 + 58.50

= Rs 306.50

Solution:

Minimum (initial) charge = Rs 14

The distance travelled by the taxi = 15 km

Fare per km = Rs 39

Waiting time = 10 minutes

The fare of waiting charge per 2 minutes = Rs 7.80

The amount for waiting charge for 10 minutes = (7.80 / 2) × 10 = Rs 39

Total fare = Rs 14 + (39 × 15) + 39

= Rs 14 + 585 + 39

= Rs 638

Solution:

Minimum (initial) charge = Rs 14

The distance travelled by the taxi = 7.5 km

Fare per km = Rs 39

Waiting time = 0 minutes

The amount for waiting charge = Rs 0

Total fare = Rs 14 + (39 × 7.5) + 0

= Rs 14 + 292.50

= Rs 306.50

Solution:

Minimum (initial) charge = Rs 14

The distance travelled by the taxi = 8.75 km

Fare per km = Rs 39

Waiting time = 7 minutes

The fare of waiting charge per 2 minutes = Rs 7.80

The amount for waiting charge for 7 minutes = (7.80 / 2) × 7 = Rs 27.30

Total fare = Rs 14 + (39 × 8.75) + 27.30

= Rs 14 + 341.25 + 27.30

= Rs 382.55

4. Samyog pays Rs.228.50 for taking a taxi with 10 minutes of waiting charge. Taxi meter shows the fare at first, a minimum of Rs.14 and then Rs.7.80 per 200 meters. If the waiting charge is at the rate of Rs. 7.80 per 2 minutes, what distance does he travel?

Solution:

Total fare = Rs 228.50

Minimum (initial) charge = Rs 14

Waiting time = 10 minutes

The fare of waiting charge per 2 minutes = Rs 7.80

The amount for waiting charge for 10 minutes = (7.80 / 2) × 10 = Rs 39

We know that,

Fare for distance = Total fare - Minimum charge - Waiting charge

= 228.50 - 14 - 39

= Rs 175.50

Fare per km = Rs 39

Again, Distance travelled = Fare for distance / Fare per km

= 175.50 / 39

= 4.5 km

5. Phulmaya paid Rs.1027.20 for taking a taxi including 12 minutes of waiting charge. She had to pay Rs 14 at first, and then Rs. 11.70 per 200 meters. If the waiting charge was at the rate of Rs. 11.70 per 2 minutes, what distance did she travel?

Solution:

Here, total fare = Rs 1027.20

Minimum (initial) charge = Rs 14

Waiting time = 12 minutes

The fare of waiting charge per 2 minutes = Rs 11.70

The amount for waiting charge for 12 minutes = (11.70 / 2) × 12 = Rs 70.20

We know that,

Fare for distance = Total fare - Minimum charge - Waiting charge

= 1027.20 - 14 - 70.20

= Rs 943

Fare per km = Rs 58.50

Again, Distance travelled = Fare for distance / Fare per km

= 943 / 58.50

= 16.1196 km

= 16.12 km

Vedanta Excel in Mathematics - Book 9

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Electricity Bill

EXERCISE 4.1

General section

1. a) The meter reading of previous month was 3450 units and the meter reading of present month is 3467 units, what is the consumed units of electricity in 1 month?

b) The meter readings for the consumption of electricity of a household is 1140 units on 1 Bhadra and 25 units of electricity is found to be consumed in Bhadra, what is the meter reading of 1 Aswin?

1. a)

Solution:

The number of units of electricity consumed is calculated by subtracting the meter reading of the previous month from the meter reading of the present month.

Consumed electricity = Meter reading of the present month – Meter reading of the previous month

= 3467 units – 3450 units

= 17 units

Hence, 17 units of electricity were consumed in 1 month.

1. b)

Solution:

The meter reading on 1 Aswin can be found by adding the consumed units in Bhadra to the meter reading on 1 Bhadra.

Meter reading on 1 Aswin = Meter reading on 1 Bhadra + Units consumed in Bhadra

= 1140 units + 25 units

= 1165 units

Hence, the meter reading on 1 Aswin is 1165 units.

2. a) What do you mean by 1 unit of electricity?

b) How many units of electricity is consumed when a refrigerator of 100 watts is used for 20 hours?

c) The power of a water pump installed at a house is 4 kW. It is used 15 minutes everyday. How many units of electricity is consumed if it is used for 16 days?

2. a)

1 unit of electricity is defined as the amount of electricity consumed by an electric appliance with a power rating of 1000 watts (1 kilowatt) operating for 1 hour. It is also known as 1 kilowatt-hour (kWh). Mathematically:

1 unit of electricity = 1 kilowatt-hour (kWh) = 1000 watt-hours

This means if a 1000-watt appliance runs for 1 hour, it consumes 1 unit of electricity.

2. b) How many units of electricity is consumed when a refrigerator of 100 watts is used for 20 hours?

Solution:

Here, Power of the refrigerator = 100 watts

Time of usage = 20 hours

We know that,

Energy consumed (in watt-hours) = Power × Time

= 100 watts × 20 hours

= 2000 watt-hours

Since 1 unit = 1000 watt-hours, we divide the energy consumed by 1000 to get the units.

Units of electricity consumed = 2000 watt-hours ÷ 1000

= 2 units

Hence, the refrigerator consumes 2 units of electricity when used for 20 hours.

2. c)

Solution:

Here, Power of the water pump = 4 kW

Time of usage per day = 15 minutes = 0.25 hours

We know that,

Energy consumed per day (in kWh) = Power × Time

= 4 kW × 0.25 hours

= 1 kWh

Since 1 kWh = 1 unit, the water pump consumes 1 unit of electricity per day.

Total units consumed for 16 days = Units per day × Number of days

= 1 unit/day × 16 days

= 16 units

Hence, the water pump consumes 16 units of electricity when used for 16 days.

3. a) According to the recent electricity tariff rate, the service charge up to 20 units is Rs 30 but energy charge is free; find the charge of consumption of 10 units.

b) There is no energy charge but the service charge is Rs 30 up to 20 units; how much bill should be charged by Nepal Electricity Authority (NEA) for the consumption of 18 units?

3. a)

Solution:

Here, the consumption of electricity = 10 units

According to the recent tariff rate, for consumption up to 20 units:

Service charge = Rs 30

Energy charge = Free (Rs 0 per unit)

We know that,

Total charge = Service charge + Energy charge

= Rs 30 + (10 units × Rs 0)

= Rs 30 + Rs 0

= Rs 30

Hence, the charge for the consumption of 10 units is Rs 30.

3. b)

Solution:

Here, the consumption of electricity = 18 units

According to the tariff rate, for consumption up to 20 units:

Service charge = Rs 30

Energy charge = Free (Rs 0 per unit)

We know that,

Total bill = Service charge + Energy charge

= Rs 30 + (18 units × Rs 0)

= Rs 30 + Rs 0

= Rs 30

Hence, the bill charged by Nepal Electricity Authority (NEA) for the consumption of 18 units is Rs 30.

4. a) The electricity tariff rate for the consumer using 5 A meter box are given alongside. Answer the following questions.

(i) What is the electricity charge for the consumption of 26 units?

(ii) What is the electricity charge for the consumption of 28 units?

4. a) (i)

Solution:

Here, the consumption of electricity = 26 units

From the tariff table for a 5 A meter box:

- For 0–20 units: Service charge = Rs 30, Energy charge = Rs 0 per unit

- For 21–30 units: Service charge = Rs 50, Energy charge = Rs 6.50 per unit

We know that,

The consumption of 26 units falls in the 21–30 units slab, but the tariff is applied progressively. So, we calculate the charge for the first 20 units and then for the remaining units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 30 + (20 units × Rs 0)

= Rs 30

Remaining units = 26 – 20 = 6 units (falls in the 21–30 units slab)

Charge for the remaining 6 units (21–30 units slab):

= Energy charge for 6 units

= 6 units × Rs 6.50 per unit

= Rs 39

Total charge for 26 units = Service charge for 21–30 units slab + Energy charge for the first 20 units + Energy charge for the remaining 6 units

= Rs 50 + (20 units × Rs 0) + (6 units × Rs 6.50)

= Rs 50 + Rs 0 + Rs 39

= Rs 89

Hence, the electricity charge for the consumption of 26 units is Rs 89.

4. a) (ii)

Solution:

Here, the consumption of electricity = 28 units

From the tariff table for a 5 A meter box:

- For 0–20 units: Service charge = Rs 30, Energy charge = Rs 0 per unit

- For 21–30 units: Service charge = Rs 50, Energy charge = Rs 6.50 per unit

We know that,

The consumption of 28 units falls in the 21–30 units slab, but the tariff is applied progressively. So, we calculate the charge for the first 20 units and then for the remaining units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 30 + (20 units × Rs 0)

= Rs 30

Remaining units = 28 – 20 = 8 units (falls in the 21–30 units slab)

Charge for the remaining 8 units (21–30 units slab):

= Energy charge for 8 units

= 8 units × Rs 6.50 per unit

= Rs 52

Total charge for 28 units = Service charge for 21–30 units slab + Energy charge for the first 20 units + Energy charge for the remaining 8 units

= Rs 50 + (20 units × Rs 0) + (8 units × Rs 6.50)

= Rs 50 + Rs 0 + Rs 52

= Rs 102

Hence, the electricity charge for the consumption of 28 units is Rs 102.

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

Solution:

Here, meter reading on 1 Baisakh = 01250 units

Meter reading on 1 Jestha = 01280 units

Consumed electricity in Baisakh = Meter reading on 1 Jestha - Meter reading on 1 Baisakh

= 01280 units - 01250 units

= 30 units

Now, remaining units = 30 - 20 = 10 units

Charge for remaining 10 units,

Energy charge = 10 units x Rs 6.50 per unit = Rs 65

Total charge for 30 units = Charge for the first 20 units + Charge for the remaining 10 units

= Rs 30 + Rs 65

= Rs 95

Hence, the electricity charge for the month of Baisakh for Sandhya's house is Rs 95.

Creative section

5. a) The electricity tariff rate fixed by NEA for the consumer using 15 A meter are given alongside.

(i) The meter readings of electricity of a household was 3015 units on 1 Magh and 3040 units on 1 Falgun. Find the electricity charge for the month of Magh.

(ii) The meter readings of electricity of a household was 6780 units on 1 Chaitra and 6830 units on 1 Baishakh. Find the electricity charge for the month of Chaitra.

5. a) (i)

Solution:

Here, meter reading on 1 Magh = 3015 units

Meter reading on 1 Falgun = 3040 units

We know that,

Consumed electricity in Magh = Meter reading on 1 Falgun – Meter reading on 1 Magh

= 3040 units – 3015 units

= 25 units

From the tariff table for a 15 A meter box:

- For 0–20 units: Service charge = Rs 50, Energy charge = Rs 4.00 per unit

- For 21–30 units: Service charge = Rs 75, Energy charge = Rs 6.50 per unit

The consumption of 25 units falls in the 21–30 units slab, but the tariff is applied progressively. So, we calculate the charge for the first 20 units and then for the remaining units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 50 + (20 units × Rs 4.00)

= Rs 50 + Rs 80

= Rs 130

Remaining units = 25 – 20 = 5 units (falls in the 21–30 units slab)

Charge for the remaining 5 units (21–30 units slab):

= Energy charge for 5 units

= 5 units × Rs 6.50 per unit

= Rs 32.50

Total charge for 25 units = Service charge for 21–30 units slab + Energy charge for the first 20 units + Energy charge for the remaining 5 units

= Rs 75 + (20 units × Rs 4.00) + (5 units × Rs 6.50)

= Rs 75 + Rs 80 + Rs 32.50

= Rs 187.50

Hence, the electricity charge for the month of Magh is Rs 187.50.

5. a) (ii)

Solution:

Here, meter reading on 1 Chaitra = 6780 units

Meter reading on 1 Baishakh = 6830 units

We know that,

Consumed electricity in Chaitra = Meter reading on 1 Baishakh – Meter reading on 1 Chaitra

= 6830 units – 6780 units

= 50 units

From the tariff table for a 15 A meter box:

- For 0–20 units: Service charge = Rs 50, Energy charge = Rs 4.00 per unit

- For 21–30 units: Service charge = Rs 75, Energy charge = Rs 6.50 per unit

- For 31–50 units: Service charge = Rs 75, Energy charge = Rs 8.00 per unit

The consumption of 50 units falls in the 31–50 units slab, but the tariff is applied progressively. So, we calculate the charge for each slab up to 50 units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 50 + (20 units × Rs 4.00)

= Rs 50 + Rs 80

= Rs 130

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 units × Rs 6.50 per unit

= Rs 65

Remaining units = 50 – 30 = 20 units (falls in the 31–50 units slab)

Charge for the remaining 20 units (31–50 units slab):

= Energy charge for 20 units

= 20 units × Rs 8.00 per unit

= Rs 160

Total charge for 50 units = Service charge for 31–50 units slab + Energy charge for the first 20 units + Energy charge for the next 10 units (21–30 slab) + Energy charge for the remaining 20 units (31–50 slab)

= Rs 75 + (20 units × Rs 4.00) + (10 units × Rs 6.50) + (20 units × Rs 8.00)

= Rs 75 + Rs 80 + Rs 65 + Rs 160

= Rs 380

Hence, the electricity charge for the month of Chaitra is Rs 380.

5. b) Nepal Electricity Authority has fixed the electricity tariff rates for the consumer using 30 A meter are given alongside.

(i) The meter readings of electricity of a household was 1250 units on 1 Kartik and 1300 units on 1 Mansir. Find the electricity charge for the month of Kartik.

(ii) The meter readings of electricity of a household was 4040 units on 1 Asar and 4114 units on 1 Saun. Find the electricity charge for the month of Asar.

5. b) (i)

Solution:

Here, meter reading on 1 Kartik = 1250 units

Meter reading on 1 Mansir = 1300 units

We know that,

Consumed electricity in Kartik = Meter reading on 1 Mansir – Meter reading on 1 Kartik

= 1300 units – 1250 units

= 50 units

From the tariff table for a 30 A meter box:

- For 0–20 units: Service charge = Rs 75, Energy charge = Rs 5.00 per unit

- For 21–30 units: Service charge = Rs 100, Energy charge = Rs 6.50 per unit

- For 31–50 units: Service charge = Rs 100, Energy charge = Rs 8.00 per unit

The consumption of 50 units falls in the 31–50 units slab, but the tariff is applied progressively. So, we calculate the charge for each slab up to 50 units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 75 + (20 units × Rs 5.00)

= Rs 75 + Rs 100

= Rs 175

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 units × Rs 6.50 per unit

= Rs 65

Remaining units = 50 – 30 = 20 units (falls in the 31–50 units slab)

Charge for the remaining 20 units (31–50 units slab):

= Energy charge for 20 units

= 20 units × Rs 8.00 per unit

= Rs 160

= Rs 100 + (20 units × Rs 5.00) + (10 units × Rs 6.50) + (20 units × Rs 8.00)

= Rs 100 + Rs 100 + Rs 65 + Rs 160

= Rs 425

Hence, the electricity charge for the month of Kartik is Rs 425.

5. b) (ii)

Solution:

Here, meter reading on 1 Asar = 4040 units

Meter reading on 1 Saun = 4114 units

We know that,

Consumed electricity in Asar = Meter reading on 1 Saun – Meter reading on 1 Asar

= 4114 units – 4040 units

= 74 units

From the tariff table for a 30 A meter box:

- For 0–20 units: Service charge = Rs 75, Energy charge = Rs 5.00 per unit

- For 21–30 units: Service charge = Rs 100, Energy charge = Rs 6.50 per unit

- For 31–50 units: Service charge = Rs 100, Energy charge = Rs 8.00 per unit

- For 51–100 units: Service charge = Rs 125, Energy charge = Rs 9.50 per unit

The consumption of 74 units falls in the 51–100 units slab, but the tariff is applied progressively. So, we calculate the charge for each slab up to 74 units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 75 + (20 units × Rs 5.00)

= Rs 75 + Rs 100

= Rs 175

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 units × Rs 6.50 per unit

= Rs 65

Charge for the next 20 units (31–50 units slab):

= Energy charge for 20 units

= 20 units × Rs 8.00 per unit

= Rs 160

Remaining units = 74 – 50 = 24 units (falls in the 51–100 units slab)

Charge for the remaining 24 units (51–100 units slab):

= Energy charge for 24 units

= 24 units × Rs 9.50 per unit

= Rs 228

Total charge for 74 units = Service charge for 51–100 units slab + Energy charge for the first 20 units + Energy charge for the next 10 units (21–30 slab) + Energy charge for the next 20 units (31–50 slab) + Energy charge for the remaining 24 units (51–100 slab)

= Rs 125 + (20 units × Rs 5.00) + (10 units × Rs 6.50) + (20 units × Rs 8.00) + (24 units × Rs 9.50)

= Rs 125 + Rs 100 + Rs 65 + Rs 160 + Rs 228

= Rs 678

Hence, the electricity charge for the month of Asar is Rs 678.

5. c) The table represents the electricity tariff rates for the consumer using 60 A meter.

(i) The meter readings of electricity of a household was 2575 units on 1 Ashoj and 2620 units on 1 Kartik. Find the electricity charge for the month of Ashoj.

(ii) The meter readings of electricity of a household was 2002 units on 1 Baishakh and 2080 units on 1 Jeth. Find the electricity charge for the month of Baishakh.

Baisakh Jestha

011981 012234

Month Falgun Chaitra

Meter reading 2400 2425

Units (kWh) Service charge Energy charge

0-20 Rs 50 Rs 4

21-30 Rs 75 Rs 6.50

31-50 Rs 75 Rs 8

Units (kWh) Service charge Energy charge

0-20 Rs 75 Rs 5

21-30 Rs 100 Rs 6.50

31-50 Rs 100 Rs 8.00

51-100 Rs 125 Rs 9.50

Units (kWh) Service charge Energy charge

0-20 Rs 125 Rs 6

21-30 Rs 125 Rs 6.50

31-50 Rs 125 Rs 8.00

51-100 Rs 150 Rs 9.50

Units (kWh) Service charge Energy charge

0-20 Rs 30 Rs 3

21-30 Rs 50 Rs 6.50

5. c) (i)

Solution:

Here, meter reading on 1 Ashoj = 2575 units

Meter reading on 1 Kartik = 2620 units

We know that,

Consumed electricity in Ashoj = Meter reading on 1 Kartik – Meter reading on 1 Ashoj

= 2620 units – 2575 units

= 45 units

From the tariff table for a 60 A meter box:

- For 0–20 units: Service charge = Rs 125, Energy charge = Rs 6.00 per unit

- For 21–30 units: Service charge = Rs 125, Energy charge = Rs 6.50 per unit

- For 31–50 units: Service charge = Rs 125, Energy charge = Rs 8.00 per unit

The consumption of 45 units falls in the 31–50 units slab, but the tariff is applied progressively. So, we calculate the charge for each slab up to 45 units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 125 + (20 units × Rs 6.00)

= Rs 125 + Rs 120

= Rs 245

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 units × Rs 6.50 per unit

= Rs 65

Remaining units = 45 – 30 = 15 units (falls in the 31–50 units slab)

Charge for the remaining 15 units (31–50 units slab):

= Energy charge for 15 units

= 15 units × Rs 8.00 per unit

= Rs 120

Total charge for 45 units = Service charge for 31–50 units slab + Energy charge for the first 20 units + Energy charge for the next 10 units (21–30 slab) + Energy charge for the remaining 15 units (31–50 slab)

= Rs 125 + (20 units × Rs 6.00) + (10 units × Rs 6.50) + (15 units × Rs 8.00)

= Rs 125 + Rs 120 + Rs 65 + Rs 120

= Rs 430

Hence, the electricity charge for the month of Ashoj is Rs 430.

5. c) (ii)

Solution:

Here, meter reading on 1 Baishakh = 2002 units

Meter reading on 1 Jeth = 2080 units

We know that,

Consumed electricity in Baishakh = Meter reading on 1 Jeth – Meter reading on 1 Baishakh

= 2080 units – 2002 units

= 78 units

From the tariff table for a 60 A meter box:

- For 0–20 units: Service charge = Rs 125, Energy charge = Rs 6.00 per unit

- For 21–30 units: Service charge = Rs 125, Energy charge = Rs 6.50 per unit

- For 31–50 units: Service charge = Rs 125, Energy charge = Rs 8.00 per unit

- For 51–100 units: Service charge = Rs 150, Energy charge = Rs 9.50 per unit

The consumption of 78 units falls in the 51–100 units slab, but the tariff is applied progressively. So, we calculate the charge for each slab up to 78 units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 125 + (20 units × Rs 6.00)

= Rs 125 + Rs 120

= Rs 245

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 units × Rs 6.50 per unit

= Rs 65

Charge for the next 20 units (31–50 units slab):

= Energy charge for 20 units

= 20 units × Rs 8.00 per unit

= Rs 160

Remaining units = 78 – 50 = 28 units (falls in the 51–100 units slab)

Charge for the remaining 28 units (51–100 units slab):

= Energy charge for 28 units

= 28 units × Rs 9.50 per unit

= Rs 266

Total charge for 78 units = Service charge for 51–100 units slab + Energy charge for the first 20 units + Energy charge for the next 10 units (21–30 slab) + Energy charge for the next 20 units (31–50 slab) + Energy charge for the remaining 28 units (51–100 slab)

= Rs 150 + (20 units × Rs 6.00) + (10 units × Rs 6.50) + (20 units × Rs 8.00) + (28 units × Rs 9.50)

= Rs 150 + Rs 120 + Rs 65 + Rs 160 + Rs 266

= Rs 761

Hence, the electricity charge for the month of Baishakh is Rs 761.

6. In a household of 5A electricity transmission line, the meter reading of 1 Bhadra was 05724 units and that of 1 Aswin was 05844 units. Answer the following questions under the given rates and billing system.

Units 0-20 21-30 31-50 51-100 101-250

Rate per unit Rs 3 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50

Service charge Rs 30 Rs 50 Rs 50 Rs 75 Rs 100

a) Calculate the charge amount required to pay the electricity bill on the following days.

(i) 6th days of meter reading

(ii) 11th day of meter reading

(iii) 22nd day of meter reading

(iv) 40th day of meter reading

b) How much more amount should be required if the payment is made on 60th day rather than it is within 7 days after meter reading?

Solution:

Here, meter reading on 1 Bhadra = 05724 units

Meter reading on 1 Aswin = 05844 units

We know that,

Consumed electricity = Meter reading on 1 Aswin – Meter reading on 1 Bhadra

= 05844 units – 05724 units

= 120 units

The consumption of 120 units falls in the 101–250 units slab, but the tariff is applied progressively. So, we calculate the charge for each slab up to 120 units.

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 30 + (20 units × Rs 3.00)

= Rs 30 + Rs 60

= Rs 90

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 units × Rs 6.50 per unit

= Rs 65

Charge for the next 20 units (31–50 units slab):

= Energy charge for 20 units

= 20 units × Rs 8.00 per unit

= Rs 160

Charge for the next 50 units (51–100 units slab):

= Energy charge for 50 units

= 50 units × Rs 9.50 per unit

= Rs 475

Remaining units = 120 – 100 = 20 units (falls in the 101–250 units slab):

Charge for the remaining 20 units (101–250 units slab):

= Energy charge for 20 units

= 20 units × Rs 9.50 per unit

= Rs 190

Total charge for 120 units = Service charge for 101–250 units slab + Energy charge for the first 20 units + Energy charge for the next 10 units (21–30 slab) + Energy charge for the next 20 units (31–50 slab) + Energy charge for the next 50 units (51–100 slab) + Energy charge for the remaining 20 units (101–250 slab)

= Rs 100 + (20 units × Rs 3.00) + (10 units × Rs 6.50) + (20 units × Rs 8.00) + (50 units × Rs 9.50) + (20 units × Rs 9.50)

= Rs 100 + Rs 60 + Rs 65 + Rs 160 + Rs 475 + Rs 190

= Rs 1050

(i) 6th day of meter reading:

The 6th day falls within 7 days of the meter reading. No penalty applies.

Total charge = Base charge

= Rs 1050

Hence, the charge on the 6th day is Rs 1050.

(ii) 11th day of meter reading:

The 11th day falls between the 8th and 30th day. A 5% penalty applies.

Penalty = 5% of Rs 1050

= (5/100) × Rs 1050

= Rs 52.50

Total charge = Base charge + Penalty

= Rs 1050 + Rs 52.50

= Rs 1102.50

Hence, the charge on the 11th day is Rs 1102.50.

(iii) 22nd day of meter reading:

The 22nd day falls between the 8th and 30th day. A 5% penalty applies.

Penalty = 5% of Rs 1050

= (5/100) × Rs 1050

= Rs 52.50

Total charge = Base charge + Penalty

= Rs 1050 + Rs 52.50

= Rs 1102.50

Hence, the charge on the 22nd day is Rs 1102.50.

(iv) 40th day of meter reading:

The 40th day falls after the 30th day. A 10% penalty applies.

Penalty = 10% of Rs 1050

= (10/100) × Rs 1050

= Rs 105

Total charge = Base charge + Penalty

= Rs 1050 + Rs 105

= Rs 1155

Hence, the charge on the 40th day is Rs 1155.

Payment on the 60th day:

The 60th day falls after the 30th day. A 10% penalty applies.

Penalty = 10% of Rs 1050

= (10/100) × Rs 1050

= Rs 105

Total charge on the 60th day = Base charge + Penalty

= Rs 1050 + Rs 105

= Rs 1155

Payment within 7 days:

No penalty applies.

Total charge within 7 days = Rs 1050

Additional amount required = Charge on the 60th day – Charge within 7 days

= Rs 1155 – Rs 1050

= Rs 105

Hence, the additional amount required if the payment is made on the 60th day rather than within 7 days is Rs 105.

7. a) Jasmin’s house has a 5A electricity meter box. She made the payment of Rs 235.20 with service charge on the 3rd day of meter reading. How many units of electricity was consumed in the month? Calculate it by using the following rates.

Units 0-20 21-30 31-50

Rate per unit Rs 3 Rs 6.50 Rs 8.00

Service charge Rs 30 Rs 50 Rs 75

Payment within 7 days of meter reading: 2% rebate

Solution:

Let the total bill before the rebate be Rs. x.

After a 2% rebate, the amount paid is 98% of the total bill:

0.98 × x = 235.20

or, x = 235.20 / 0.98

or, x = 240

So, the total bill before the rebate is Rs 240.

Now, we calculate the number of units consumed (u) such that the total bill (including service charge) equals Rs 240.

Service charge for 31–50 units slab = Rs 75

Energy charge for the first 20 units (0–20 units slab) = 20 × Rs 3.00 = Rs 60

Energy charge for the next 10 units (21–30 units slab) = 10 × Rs 6.50 = Rs 65

Let the remaining units be (u – 30).

Energy charge for the remaining units (31–50 units slab) = (u – 30) × Rs 8.00

Total bill = Service charge + Energy charge for the first 20 units + Energy charge for the next 10 units + Energy charge for the remaining units

75 + (20 × 3) + (10 × 6.50) + (u – 30) × 8 = 240

or, 75 + 60 + 65 + (u – 30) × 8 = 240

or, 200 + (u – 30) × 8 = 240

or, (u – 30) × 8 = 240 – 200

or, (u – 30) × 8 = 40

or, u – 30 = 40 / 8

or, u – 30 = 5

or, u = 35

Hence, the number of units of electricity consumed in the month is 35 units.

7. b) A company has fixed a 15A electricity meter box. It made the payment of Rs 1100 with service charge on the 50th day of meter reading. Apply the rates given in the following table to find the of units of electricity consumed in a month.

Units 0-20 21-30 31-50 51-100 101-250

Rate per unit Rs 4 Rs 6.50 Rs 8.00 Rs 9.50 Rs 9.50

Service charge Rs 50 Rs 75 Rs 75 Rs 100 Rs 125

Payment from 40th to 60th day of meter reading: 25% extra fine

Solution:

Since the payment is made on the 50th day, which falls between the 40th and 60th day of meter reading, a 25% extra fine applies.

Let the total bill before the fine be Rs. x.

After a 25% fine, the amount paid is 125% of the total bill:

1.25 × x = 1100

or, x = 1100 / 1.25

or, x = 880

So, the total bill before the fine is Rs 880.

Now, we calculate the number of units consumed (u) such that the total bill (including service charge) equals Rs 880.

The consumption falls in the 51–100 units slab (as determined by trial):

Service charge for 51–100 units slab = Rs 100

Energy charge for the first 20 units (0–20 units slab) = 20 × Rs 4.00 = Rs 80

Energy charge for the next 10 units (21–30 units slab) = 10 × Rs 6.50 = Rs 65

Energy charge for the next 20 units (31–50 units slab) = 20 × Rs 8.00 = Rs 160

Let the remaining units be (u – 50).

Energy charge for the remaining units (51–100 units slab) = (u – 50) × Rs 9.50

Total bill = Service charge + Energy charge for the first 20 units + Energy charge for the next 10 units + Energy charge for the next 20 units + Energy charge for the remaining units

100 + (20 × 4) + (10 × 6.50) + (20 × 8) + (u – 50) × 9.50 = 880

or, 100 + 80 + 65 + 160 + (u – 50) × 9.50 = 880

or, 405 + (u – 50) × 9.50 = 880

or, (u – 50) × 9.50 = 880 – 405

or, (u – 50) × 9.50 = 475

or, u – 50 = 475 / 9.50

or, u – 50 = 50

or, u = 100

Hence, the number of units of electricity consumed in the month is 100 units.

8. Dinesh, who is using a 5A meter box, recorded the meter reading of first 6 months of last year. Study the following table and answer the questions according to the latest tariff rates of electricity.

Month 1 Baishakh 1 Jestha 1 Asar 1 Shrawan 1 Bhadra 1 Aswin

Meter reading 5465 5500 5540 5570 5616 5660

a) How many units of electricity was consumed in the month of Asar?

b) Which month has maximum consumption?

c) How much electricity charge did he pay for the month of

(i) Baishakh if the payment was made on 5 Jestha?

(ii) Jestha if the payment was made on 20 Asar?

Solution:

First, calculate the consumption for each month:

Baishakh (1 Baishakh to 1 Jestha): 5500 – 5465 = 35 units

Jestha (1 Jestha to 1 Asar): 5540 – 5500 = 40 units

Asar (1 Asar to 1 Shrawan): 5570 – 5540 = 30 units

Shrawan (1 Shrawan to 1 Bhadra): 5616 – 5570 = 46 units

Bhadra (1 Bhadra to 1 Aswin): 5660 – 5616 = 44 units

Consumption in Asar = 5570 – 5540 = 30 units

Hence, 30 units of electricity were consumed in the month of Asar.

Baishakh: 35 units

Jestha: 40 units

Asar: 30 units

Shrawan: 46 units

Bhadra: 44 units

The maximum consumption is in Shrawan with 46 units.

Hence, the month of Shrawan has the maximum consumption.

(i)

Consumption in Baishakh = 35 units (falls in the 31–50 units slab):

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 30 + (20 × Rs 3.00)

= Rs 30 + Rs 60

= Rs 90

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 × Rs 6.50

= Rs 65

Remaining units = 35 – 30 = 5 units (falls in the 31–50 units slab):

Charge for the remaining 5 units (31–50 units slab):

= Energy charge for 5 units

= 5 × Rs 8.00

= Rs 40

Total charge for 35 units = Service charge for 31–50 units slab + Energy charge for the first 20 units + Energy charge for the next 10 units + Energy charge for the remaining 5 units

= Rs 50 + (20 × Rs 3.00) + (10 × Rs 6.50) + (5 × Rs 8.00)

= Rs 50 + Rs 60 + Rs 65 + Rs 40

= Rs 215

Payment on 5 Jestha (4 days after 1 Jestha, within 7 days): No penalty applies.

Total charge = Rs 215

Hence, the electricity charge for Baishakh, paid on 5 Jestha, is Rs 215.

(ii)

Consumption in Jestha = 40 units (falls in the 31–50 units slab):

Charge for the first 20 units (0–20 units slab):

= Service charge + Energy charge

= Rs 30 + (20 × Rs 3.00)

= Rs 30 + Rs 60

= Rs 90

Charge for the next 10 units (21–30 units slab):

= Energy charge for 10 units

= 10 × Rs 6.50

= Rs 65

Remaining units = 40 – 30 = 10 units (falls in the 31–50 units slab):

Charge for the remaining 10 units (31–50 units slab):

= Energy charge for 10 units

= 10 × Rs 8.00

= Rs 80

Total charge for 40 units = Service charge for 31–50 units slab + Energy charge for the first 20 units + Energy charge for the next 10 units + Energy charge for the remaining 10 units

= Rs 50 + (20 × Rs 3.00) + (10 × Rs 6.50) + (10 × Rs 8.00)

= Rs 50 + Rs 60 + Rs 65 + Rs 80

= Rs 255

Payment on 20 Asar (19 days after 1 Asar, between 8th and 30th day): A 5% penalty applies.

Penalty = 5% of Rs 255

= (5/100) × 255

= Rs 12.75

Total charge = Base charge + Penalty

= Rs 255 + Rs 12.75

= Rs 267.75

Hence, the electricity charge for Jestha, paid on 20 Asar, is Rs 267.75.

d) If he paid the electricity charge of Rs 223 on 10 Kartik for the bill of Aswin, what was the meter reading of 1 Kartik?

Payment Rebate/fine

Within 7 days 2% rebate

8th – 15th days –

16th – 30th days 5% fine

31st – 40th days 10% fine

41st – 60th days 25% fine

Solution:

Payment on 10 Kartik is 9 days after 1 Kartik (the meter reading date for Aswin bill), which falls in the 8th – 15th days range. Therefore, no rebate or fine applies.

Total bill = Rs 223

Now, we calculate the number of units consumed (u) in Aswin such that the total bill (including service charge) equals Rs 223.

The consumption falls in the 31–50 units slab (as determined by trial):

Service charge for 31–50 units slab = Rs 50

Energy charge for the first 20 units (0–20 units slab) = 20 × Rs 3.00 = Rs 60

Energy charge for the next 10 units (21–30 units slab) = 10 × Rs 6.50 = Rs 65

Let the remaining units be (u – 30).

Energy charge for the remaining units (31–50 units slab) = (u – 30) × Rs 8.00

Total bill = Service charge + Energy charge for the first 20 units + Energy charge for the next 10 units + Energy charge for the remaining units

50 + (20 × 3) + (10 × 6.50) + (u – 30) × 8 = 223

or, 50 + 60 + 65 + (u – 30) × 8 = 223

or, 175 + (u – 30) × 8 = 223

or, (u – 30) × 8 = 223 – 175

or, (u – 30) × 8 = 48

or, u – 30 = 48 ÷ 8

or, u – 30 = 6

or, u = 36

So, the number of units consumed in Aswin is 36 units.

Meter reading on 1 Aswin = 5660 units

Meter reading on 1 Kartik = Meter reading on 1 Aswin + Units consumed in Aswin

= 5660 + 36

= 5696 units

Hence, the meter reading on 1 Kartik is 5696 units.

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

Solution:

Here, Total power = 5 × 15 watts = 75 watts

Energy consumed by CFL bulbs per day = 75 watts × 4 hours = 300 watt-hours

Now, the electric heater of 1200 watts:

Energy consumed by heater per day = 1200 watts × 1 hour = 1200 watt-hours

Then, the total daily energy consumption = 300 + 1200 = 1500 watt-hours

Daily consumption in kWh = 1500 / 1000 = 1.5 kWh = 1.5 units

Calculating the monthly consumption (assuming 30 days):

Monthly consumption = 1.5 units/day × 30 days = 45 units

The consumption of 45 units falls in the 31–50 units slab:

Charge for the first 20 units (0–20 units slab):

= 20 units × Rs 3.00 = Rs 60

Charge for the next 10 units (21–30 units slab):

= 10 units × Rs 6.50 = Rs 65

Remaining units = 45 – 30 = 15 units (falls in the 31–50 units slab):

Charge for the remaining 15 units (31–50 units slab):

= 15 units × Rs 8.00 = Rs 120

Total energy charge = Rs 60 + Rs 65 + Rs 120 = Rs 245

Total charge = Energy charge + Service charge

= Rs 245 + Rs 75

= Rs 320

Payment on the 10th day falls in the 8th – 15th days range (assuming a typical structure). No rebate or fine applies.

Total charge = Rs 320

Hence, the cost of payment of the bill for the month is Rs 320.

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

Solution:

Here, calculating the daily electricity consumption:

5 LED bulbs of 10 watts each:

Total power = 5 × 10 watts = 50 watts

Energy consumed by LED bulbs per day = 50 watts × 4 hours = 200 watt-hours

Now, for 2 televisions of 60 watts each:

Total power = 2 × 60 watts = 120 watts

Energy consumed by televisions per day = 120 watts × 5 hours = 600 watt-hours

For the refrigerator of 250 watts:

Energy consumed by refrigerator per day = 250 watts × 2 hours = 500 watt-hours

Total daily energy consumption = 200 + 600 + 500 = 1300 watt-hours

Daily consumption in kWh = 1300 ÷ 1000 = 1.3 kWh = 1.3 units

Then, calculating the monthly consumption (assuming 30 days):

Monthly consumption = 1.3 units/day × 30 days = 39 units

Again, the consumption of 39 units falls in the 31–50 units slab:

Charge for the first 20 units (0–20 units slab):

= 20 units × Rs 4.00 = Rs 80

Charge for the next 10 units (21–30 units slab):

= 10 units × Rs 6.50 = Rs 65

Remaining units = 39 – 30 = 9 units (falls in the 31–50 units slab):

Charge for the remaining 9 units (31–50 units slab):

= 9 units × Rs 8.00 = Rs 72

Total energy charge = Rs 80 + Rs 65 + Rs 72 = Rs 217

Total charge = Energy charge + Service charge

= Rs 217 + Rs 75

= Rs 292

Payment on the 35th day falls in the 31st – 40th days range (assuming a typical structure). A 10% fine applies:

Fine = 10% of Rs 292

= (10/100) × 292

= Rs 29.20

Total charge with fine = Rs 292 + Rs 29.20 = Rs 321.20

Hence, the cost of payment of the bill for the month is Rs 321.20.

Water bill

EXERCISE 4.2

General section

a) How many litres of water is to be consumed to make 1 unit?

Solution:

We know, 1 unit = 1,000 litres of water (as per the facts provided).

Hence, 1,000 litres of water is consumed to make 1 unit.

b) According to water tariff rates of Nepal Water Supply Corporation, what is the minimum charge up to 10 units for the consumption of water by using a half-inch pipe?

Solution:

From the NWSC tariff rates for a half-inch (1/2") pipe:

Minimum consumption = 10,000 litres (10 units)

Minimum charge = Rs 110

Hence, the minimum charge up to 10 units is Rs 110.

c) What is the minimum charge for the consumption of water up to 56,000 litres by using a 1" pipe according to NWSC?

Solution:

From the NWSC tariff rates for a 1" pipe:

Minimum consumption = 56,000 litres (56 units)

Minimum charge = Rs 3,420

Hence, the minimum charge for up to 56,000 litres is Rs 3,420.

a) The meter reading for the consumption of water in the previous month was 1380 units and the meter reading of present month is 1400 units, find the consumed units of water in 1 month.

Solution:

Here, the meter reading of the previous month = 1380 units

The meter reading of the present month = 1400 units

∴ Consumed units of water in 1 month = (1400 – 1380) units = 20 units

Hence, 20 units of water were consumed in 1 month.

b) The meter reading of consumption of water in 1 Baishakh was 6699 units and that of 1 Jestha was 7070 units, how many units of water was consumed in the month of Baishakh?

Solution:

Here, the meter reading on 1 Baishakh = 6699 units

The meter reading on 1 Jestha = 7070 units

∴ Consumed units of water in the month of Baishakh = (7070 – 6699) units = 371 units

Hence, 371 units of water were consumed in the month of Baishakh.

c) On 1 Poush, the meter readings of water of a household is 1140 units and the household consumed 15 units of water in Mansir. What was the meter reading of 1 Mansir?

Solution:

Here, the meter reading on 1 Poush = 1140 units

Consumed units of water in Mansir = 15 units

We know, Consumed units = Present reading – Former reading

∴ 15 = 1140 – Meter reading on 1 Mansir

Meter reading on 1 Mansir = 1140 – 15 = 1125 units

Hence, the meter reading on 1 Mansir was 1125 units.

a) The water tariff rates fixed by Nepal Water Supply Corporation for the consumers using half-inch pipe are given alongside. Answer the following questions.

Minimum consumption Minimum charge (Rs) Additional charge (Rs)

10 units Rs 110 Rs 25 per unit

Sewerage service charge: 50%

(i) How much should a household pay for the consumption of 15 units of water in a month?

Solution:

Minimum charge up to 10 units = Rs 110

The additional units of water consumption = (15 – 10) units = 5 units

Now, charge of 5 additional units of water = 5 × Rs 25 = Rs 125

∴ Total charge = Rs 110 + Rs 125 = Rs 235

Also, sewerage service charge = 50% of Rs 235 = 50/100 × 235 = Rs 117.50

Total bill including sewerage service charge = Rs 235 + Rs 117.50 = Rs 352.50

Hence, the household should pay Rs 352.50 for 15 units.

(ii) How much will be charged for a household for the consumption of 24 units of water in a month?

Solution:

Minimum charge up to 10 units = Rs 110

The additional units of water consumption = (24 – 10) units = 14 units

Now, charge of 14 additional units of water = 14 × Rs 25 = Rs 350

∴ Total charge = Rs 110 + Rs 350 = Rs 460

Also, sewerage service charge = 50% of Rs 460 = 50/100 × 460 = Rs 230

Total bill including sewerage service charge = Rs 460 + Rs 230 = Rs 690

Hence, the household will be charged Rs 690 for 24 units.

b) Kathmandu Upatyaka Khanepani Limited (KUKL) has implemented the water tariff rates for the consumers using half-inch pipe are given in the table. Answer the following questions.

From KUKL tariff: Minimum consumption = 10,000 litres (10 units), Minimum charge = Rs 100, Additional charge = Rs 32 per unit

Sewerage service charge: 50% (assumed as per NWSC pattern, since not specified differently)

(i) How much should a household of Kathmandu pay for the consumption of 18 units of water in a month?

Solution:

Minimum charge up to 10 units = Rs 100

The additional units of water consumption = (18 – 10) units = 8 units

Now, charge of 8 additional units of water = 8 × Rs 32 = Rs 256

∴ Total charge = Rs 100 + Rs 256 = Rs 356

Also, sewerage service charge = 50% of Rs 356 = 50/100 × 356 = Rs 178

Total bill including sewerage service charge = Rs 356 + Rs 178 = Rs 534

Hence, the household should pay Rs 534 for 18 units.

(ii) How much will be charged for a household of Madhyapur, Thimi for the consumption of 40 units of water in a month?

Solution:

Minimum charge up to 10 units = Rs 100

The additional units of water consumption = (40 – 10) units = 30 units

Now, charge of 30 additional units of water = 30 × Rs 32 = Rs 960

∴ Total charge = Rs 100 + Rs 960 = Rs 1,060

Also, sewerage service charge = 50% of Rs 1,060 = 50/100 × 1,060 = Rs 530

Total bill including sewerage service charge = Rs 1,060 + Rs 530 = Rs 1,590

Hence, the household will be charged Rs 1,590 for 40 units.

Creative section

a) Mr. Upreti has a metered tap using half-inch pipe in his house in Banepa. The meter reading of his house on 1 Baishakh was 1050 units and on 1 Jestha was 1075 units. According to NWSC, the minimum charge up to 10 units is Rs 110 and Rs 25 per unit for additional consumption of water, and 50% of the total charge is to be paid as sewerage service charge. Answer the following questions.

(i) How many litres of water is consumed in Baishakh month?

Solution:

Here, the meter reading on 1 Baishakh = 1050 units

The meter reading on 1 Jestha = 1075 units

∴ Consumed units of water in Baishakh = (1075 – 1050) units = 25 units

We know, 1 unit = 1,000 litres of water

∴ 25 units = 25 × 1,000 litres = 25,000 litres

Hence, 25,000 litres of water were consumed in Baishakh.

(ii) How much sewerage service charge should he pay?

Solution:

Minimum charge up to 10 units = Rs 110

The additional units of water consumption = (25 – 10) units = 15 units

Now, charge of 15 additional units of water = 15 × Rs 25 = Rs 375

∴ Total charge = Rs 110 + Rs 375 = Rs 485

Also, sewerage service charge = 50% of Rs 485 = 50/100 × 485 = Rs 242.50

Hence, he should pay Rs 242.50 as sewerage service charge.

(iii) Find the total bill including 50% sewerage service charge.

Solution:

Total charge = Rs 485 (from part ii)

Sewerage service charge = Rs 242.50 (from part ii)

Total bill including sewerage service charge = Rs 485 + Rs 242.50 = Rs 727.50

Hence, the total bill is Rs 727.50.

(iv) If he paid only Rs 465 for the consumption of water in Asar, how many units of water was consumed in the month of Asar?

Solution:

Let the total charge of water excluding sewerage service charge be Rs x.

Then, x + 50% of x = Rs 465

or, x + 50/100 × x = Rs 465

or, 3x/2 = 465

or, x = 465 × 2 / 3 = 310

Also, the charge for additional units of water = total charge of water – minimum charge = Rs 310 – 110 = Rs 200

∴ The additional units of water = Rs 200 / Rs 25 = 8 units

Total consumed units of water = minimum units + additional units = 10 + 8 = 18 units

Hence, 18 units of water were consumed in Asar.

b) Mr. Maharjan uses a 3/4" pipe in his house in Kathmandu. The meter reading of his house on 1 Aswin was 2020 units and on 1 Kartik was 2050 units. According to KUKL, the minimum charge up to 27 units is Rs 1,910 and Rs 71 per unit for additional consumption of water, and 50% of the total charge is to be paid as sewerage service charge. Answer the following questions.

(i) How many units of water is consumed in Aswin month?

Solution:

Here, the meter reading on 1 Aswin = 2020 units

The meter reading on 1 Kartik = 2050 units

∴ Consumed units of water in Aswin = (2050 – 2020) units = 30 units

Hence, 30 units of water were consumed in Aswin.

(ii) How much sewerage service charge should he pay?

Solution:

Minimum charge up to 27 units = Rs 1,910

The additional units of water consumption = (30 – 27) units = 3 units

Now, charge of 3 additional units of water = 3 × Rs 71 = Rs 213

∴ Total charge = Rs 1,910 + Rs 213 = Rs 2,123

Also, sewerage service charge = 50% of Rs 2,123 = 50/100 × 2,123 = Rs 1,061.50

Hence, he should pay Rs 1,061.50 as sewerage service charge.

(iii) Find the total bill including 50% sewerage service charge.

Solution:

Total charge = Rs 2,123 (from part ii)

Sewerage service charge = Rs 1,061.50 (from part ii)

Total bill including sewerage service charge = Rs 2,123 + Rs 1,061.50 = Rs 3,184.50

Hence, the total bill is Rs 3,184.50.

(iv) If he paid only Rs 3,717 for the consumption of water in Falgun, how many units of water was consumed in the month of Falgun?

Solution:

Let the total charge of water excluding sewerage service charge be Rs x.

Then, x + 50% of x = Rs 3,717

or, x + 50/100 × x = Rs 3,717

or, 3x/2 = 3,717

or, x = 3,717 × 2 / 3 = 2,478

Also, the charge for additional units of water = total charge of water – minimum charge = Rs 2,478 – 1,910 = Rs 568

∴ The additional units of water = Rs 568 / Rs 71 = 8 units (since 568 ÷ 71 = 8 exactly)

Total consumed units of water = minimum units + additional units = 27 + 8 = 35 units

Hence, 35 units of water were consumed in Falgun.

a) A household of Lahan consumed 32 units of water in a month by using half-inch pipe. Calculate the payment of the bill as per the rules implemented by NWSC including 50% sewerage service charge, if the payment is made within the first month of the bill issued.

Solution:

From NWSC tariff for half-inch pipe: Minimum charge up to 10 units = Rs 110, Additional charge = Rs 25 per unit

Minimum charge up to 10 units = Rs 110

The additional units of water consumption = (32 – 10) units = 22 units

Now, charge of 22 additional units of water = 22 × Rs 25 = Rs 550

∴ Total charge = Rs 110 + Rs 550 = Rs 660

Also, sewerage service charge = 50% of Rs 660 = 50/100 × 660 = Rs 330

Total bill including sewerage service charge = Rs 660 + Rs 330 = Rs 990

Since payment is made within the first month, a 3% rebate is allowed:

Required payment = Rs 990 – 3% of Rs 990 = Rs 990 – (3/100 × 990) = Rs 990 – 29.70 = Rs 960.30

Hence, the payment of the bill is Rs 960.30.

b) 127 units of water is consumed by using 3/4" pipe in a hotel in Pokhara. If the payment of the bill is made within the fifth month after the bill issued, how much money is required to clear the bill with 50% sewerage charge? Calculate it as per the tariff rates provisioned by NWSC.

Solution:

From NWSC tariff for 3/4" pipe: Minimum consumption = 27,000 litres (27 units), Minimum charge = Rs 1,490, Additional charge = Rs 40 per unit

Minimum charge up to 27 units = Rs 1,490

The additional units of water consumption = (127 – 27) units = 100 units

Now, charge of 100 additional units of water = 100 × Rs 40 = Rs 4,000

∴ Total charge = Rs 1,490 + Rs 4,000 = Rs 5,490

Also, sewerage service charge = 50% of Rs 5,490 = 50/100 × 5,490 = Rs 2,745

Total bill including sewerage service charge = Rs 5,490 + Rs 2,745 = Rs 8,235

Since payment is made within the fifth month, a 20% fine is charged:

Required payment = Rs 8,235 + 20% of Rs 8,235 = Rs 8,235 + (20/100 × 8,235) = Rs 8,235 + 1,647 = Rs 9,882

Hence, Rs 9,882 is required to clear the bill.

a) A hospital at Dhulikhel is using a 3/4" of water pipe with meter. The meter reading for the consumption of water of the hotel was 1540 units on 1 Kartik and 1600 units on 1 Mansir. The tariff rates implementing by the NWSC is given below.

Size of pipe Minimum consumption Minimum charge Additional charge

3/4" 27,000 litres Rs 1,490 Rs 40 per 1,000 litres

Calculate the charge to be paid including 50% sewerage service charge if the payment of the bill is made in the following schedule.

Solution:

Here, the meter reading on 1 Kartik = 1540 units

The meter reading on 1 Mansir = 1600 units

∴ Consumed units of water in the month = (1600 – 1540) units = 60 units

Minimum charge up to 27 units = Rs 1,490

The additional units of water consumption = (60 – 27) units = 33 units

Now, charge of 33 additional units of water = 33 × Rs 40 = Rs 1,320

∴ Total charge = Rs 1,490 + Rs 1,320 = Rs 2,810

Also, sewerage service charge = 50% of Rs 2,810 = 50/100 × 2,810 = Rs 1,405

Total bill including sewerage service charge = Rs 2,810 + Rs 1,405 = Rs 4,215

(i) within the first month after the issue of bill

3% rebate is allowed:

Required payment = Rs 4,215 – 3% of Rs 4,215 = Rs 4,215 – (3/100 × 4,215) = Rs 4,215 – 126.45 = Rs 4,088.55

Hence, the charge to be paid is Rs 4,088.55.

(ii) within the third month after the issue of bill

No rebate and no fine:

Required payment = Rs 4,215

Hence, the charge to be paid is Rs 4,215.

(iii) within the fifth month after the issue of bill

20% fine is charged:

Required payment = Rs 4,215 + 20% of Rs 4,215 = Rs 4,215 + (20/100 × 4,215) = Rs 4,215 + 843 = Rs 5,058

Hence, the charge to be paid is Rs 5,058.

(iv) within the seventh month after the issue of bill

50% fine is charged:

Required payment = Rs 4,215 + 50% of Rs 4,215 = Rs 4,215 + (50/100 × 4,215) = Rs 4,215 + 2,107.50 = Rs 6,322.50

Hence, the charge to be paid is Rs 6,322.50.

b) A factory run in Lalitpur is using a 1" of water pipe. The meter reading for the consumption of water in the factory was 3500 units on 1 Falgun and 3580 units on 1 Chaitra. The tariff rates implementing by the KUKL is given below.

Size of pipe Minimum consumption Minimum charge Additional charge

1" 56,000 litres Rs 3,960 Rs 71 per 1,000 litres

Calculate the charge to be paid including 50% sewerage service charge if the payment of the bill is made in the following schedule.

Solution:

Here, the meter reading on 1 Falgun = 3500 units

The meter reading on 1 Chaitra = 3580 units

∴ Consumed units of water in the month = (3580 – 3500) units = 80 units

Minimum charge up to 56 units = Rs 3,960

The additional units of water consumption = (80 – 56) units = 24 units

Now, charge of 24 additional units of water = 24 × Rs 71 = Rs 1,704

∴ Total charge = Rs 3,960 + Rs 1,704 = Rs 5,664

Also, sewerage service charge = 50% of Rs 5,664 = 50/100 × 5,664 = Rs 2,832

Total bill including sewerage service charge = Rs 5,664 + Rs 2,832 = Rs 8,496

(i) within the second month after the bill issued

3% rebate is allowed:

Required payment = Rs 8,496 – 3% of Rs 8,496 = Rs 8,496 – (3/100 × 8,496) = Rs 8,496 – 254.88 = Rs 8,241.12

Hence, the charge to be paid is Rs 8,241.12.

(ii) within the fourth month after the bill issued

10% fine is charged:

Required payment = Rs 8,496 + 10% of Rs 8,496 = Rs 8,496 + (10/100 × 8,496) = Rs 8,496 + 849.60 = Rs 9,345.60

Hence, the charge to be paid is Rs 9,345.60.

(iii) within the sixth month after the bill issued

50% fine is charged:

Required payment = Rs 8,496 + 50% of Rs 8,496 = Rs 8,496 + (50/100 × 8,496) = Rs 8,496 + 4,248 = Rs 12,744

Hence, the charge to be paid is Rs 12,744.

Telephone bill

EXERCISE 4.3

General section

a) The telephone reading of the former month was 5500 calls and the call reading of current month is 6000 calls, how many telephone calls is made in 1 month?

Solution:

Here, the call reading of the former month = 5500 calls

The call reading of the current month = 6000 calls

∴ Number of telephone calls made in 1 month = 6000 – 5500 = 500 calls

Hence, 500 telephone calls were made in 1 month.

b) The call reading of Baishakh was 8800 calls and that of Jestha was 9900. If the minimum charge up to 175 calls is Rs 200 and Re 1 for each additional call, calculate the subtotal amount of telephone bill.

Solution:

Here, the call reading of Baishakh = 8800 calls

The call reading of Jestha = 9900 calls

∴ Number of calls made in the month = 9900 – 8800 = 1100 calls

Minimum charge up to 175 calls = Rs 200

Additional number of calls = 1100 – 175 = 925 calls

Charge for additional number of calls = 925 × Re 1 = Rs 925

∴ Sub-total charge (C) = Minimum charge + Additional charge = Rs 200 + Rs 925 = Rs 1,125

Hence, the subtotal amount of the telephone bill is Rs 1,125.

a) The subtotal amount of a telephone bill is Rs 1,000. Find the grand total amount of the bill with 13% TSC and 13% VAT.

Solution:

Here, sub-total charge (C) = Rs 1,000

TSC = 13% of sub-total charge (C) = 13/100 × 1,000 = Rs 130

Total charge (TC) = Sub-total charge (C) + TSC = Rs 1,000 + Rs 130 = Rs 1,130

VAT = 13% of total charge (TC) = 13/100 × 1,130 = Rs 146.90

Grand total charge = C + TSC + VAT = Rs 1,130 + Rs 146.90 = Rs 1,276.90

Hence, the grand total amount of the bill is Rs 1,276.90.

b) What will be the grand total amount of a telephone bill with 13% TSC and 13% VAT if the subtotal amount is Rs 2,000?

Solution:

Here, sub-total charge (C) = Rs 2,000

TSC = 13% of sub-total charge (C) = 13/100 × 2,000 = Rs 260

Total charge (TC) = Sub-total charge (C) + TSC = Rs 2,000 + Rs 260 = Rs 2,260

VAT = 13% of total charge (TC) = 13/100 × 2,260 = Rs 293.80

Grand total charge = C + TSC + VAT = Rs 2,260 + Rs 293.80 = Rs 2,553.80

Hence, the grand total amount of the bill is Rs 2,553.80.

Creative section

a) A school made a total of 2175 calls in the month of Baishakh. According to Nepal Telecommunication Authority (NTA), the minimum charge up to 175 calls is Rs 200 and Re 1 for each additional call. 13% TSC and 13% VAT are applied as per the Government rule.

(i) How much monthly rental charge is fixed?

Solution:

Here, the minimum charge up to 175 calls = Rs 200

Hence, the monthly rental charge is Rs 200.

(ii) What is the sub total amount of telephone bill?

Solution:

Total number of calls = 2175

Minimum charge up to 175 calls = Rs 200

Additional number of calls = 2175 – 175 = 2000 calls

Charge for additional number of calls = 2000 × Re 1 = Rs 2,000

∴ Sub-total charge (C) = Minimum charge + Additional charge = Rs 200 + Rs 2,000 = Rs 2,200

Hence, the sub total amount of the telephone bill is Rs 2,200.

(iii) Find the TSC and VAT amounts.

Solution:

Sub-total charge (C) = Rs 2,200 (from part ii)

TSC = 13% of sub-total charge (C) = 13/100 × 2,200 = Rs 286

Total charge (TC) = Sub-total charge (C) + TSC = Rs 2,200 + Rs 286 = Rs 2,486

VAT = 13% of total charge (TC) = 13/100 × 2,486 = Rs 323.18

Hence, the TSC amount is Rs 286 and the VAT amount is Rs 323.18.

(iv) How much total amount should the school pay to clear the bill?

Solution:

Sub-total charge (C) = Rs 2,200 (from part ii)

TSC = Rs 286 (from part iii)

VAT = Rs 323.18 (from part iii)

Grand total charge = C + TSC + VAT = Rs 2,200 + Rs 286 + Rs 323.18 = Rs 2,809.18

Hence, the school should pay Rs 2,809.18 to clear the bill.

b) Nepal Telecommunication Authority (NTA) has determined the rental charge of Rs 200 with 175 free calls, Re 1 for each extra call and, 13% TSC and 13% VAT. In a hospital, the telephone reading of the months of 1 Aswin and 1 Kartik was 7438 calls and 9601 calls respectively.

(i) How many calls were made in Aswin?

Solution:

Here, the call reading on 1 Aswin = 7438 calls

The call reading on 1 Kartik = 9601 calls

∴ Number of calls made in Aswin = 9601 – 7438 = 2163 calls

Hence, 2163 calls were made in Aswin.

(ii) What is the sub total amount of telephone bill?

Solution:

Total number of calls = 2163

Minimum charge up to 175 calls = Rs 200

Additional number of calls = 2163 – 175 = 1988 calls

Charge for additional number of calls = 1988 × Re 1 = Rs 1,988

∴ Sub-total charge (C) = Minimum charge + Additional charge = Rs 200 + Rs 1,988 = Rs 2,188

Hence, the sub total amount of the telephone bill is Rs 2,188.

(iii) Find the TSC and VAT amounts.

Solution:

Sub-total charge (C) = Rs 2,188 (from part ii)

TSC = 13% of sub-total charge (C) = 13/100 × 2,188 = Rs 284.44

Total charge (TC) = Sub-total charge (C) + TSC = Rs 2,188 + Rs 284.44 = Rs 2,472.44

VAT = 13% of total charge (TC) = 13/100 × 2,472.44 = Rs 321.42

Hence, the TSC amount is Rs 284.44 and the VAT amount is Rs 321.42.

(iv) How much amount should the hospital pay to clear the bill?

Solution:

Sub-total charge (C) = Rs 2,188 (from part ii)

TSC = Rs 284.44 (from part iii)

VAT = Rs 321.42 (from part iii)

Grand total charge = C + TSC + VAT = Rs 2,188 + Rs 284.44 + Rs 321.42 = Rs 2,793.86

Hence, the hospital should pay Rs 2,793.86 to clear the bill.

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

Solution:

Total number of calls = 475

Minimum charge up to 175 calls = Rs 200

Additional number of calls = 475 – 175 = 300 calls

Charge for additional number of calls = 300 × Re 1 = Rs 300

∴ Sub-total charge (C) = Minimum charge + Additional charge = Rs 200 + Rs 300 = Rs 500

TSC = 13% of sub-total charge (C) = 13/100 × 500 = Rs 65

Total charge (TC) = Sub-total charge (C) + TSC = Rs 500 + Rs 65 = Rs 565

VAT = 13% of total charge (TC) = 13/100 × 565 = Rs 73.45

Grand total charge = C + TSC + VAT = Rs 565 + Rs 73.45 = Rs 638.45

Hence, the charge for 475 calls is Rs 638.45.

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

Solution:

Charge for 1 minute STD call = Rs 1.25

Total call duration = 10 minutes

Sub-total charge (C) = 10 × Rs 1.25 = Rs 12.50

TSC = 13% of sub-total charge (C) = 13/100 × 12.50 = Rs 1.625

Total charge (TC) = Sub-total charge (C) + TSC = Rs 12.50 + Rs 1.625 = Rs 14.125

VAT = 13% of total charge (TC) = 13/100 × 14.125 = Rs 1.83625

Grand total charge = C + TSC + VAT = Rs 14.125 + Rs 1.83625 = Rs 15.96125

Hence, Mr. Rajbanshi should pay Rs 15.96 (rounded to two decimal places).

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

Solution:

Charge for 1 minute ISD call = Rs 20

Total call duration = 5 minutes

Sub-total charge (C) = 5 × Rs 20 = Rs 100

TSC = 13% of sub-total charge (C) = 13/100 × 100 = Rs 13

Total charge (TC) = Sub-total charge (C) + TSC = Rs 100 + Rs 13 = Rs 113

VAT = 13% of total charge (TC) = 13/100 × 113 = Rs 14.69

Grand total charge = C + TSC + VAT = Rs 113 + Rs 14.69 = Rs 127.69

Hence, Nirmal should pay Rs 127.69.

a) The minimum charge of telephone calls up to 175 calls is Rs 200. The charge for each extra call of 2 minutes duration is Re 1. If Mr. Malla paid Rs 510.76 with 13% TSC and 13% VAT for his telephone bill, answer the following questions.

(i) Find the charge of telephone calls without VAT.

Solution:

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

Then, x + 13% of x = Rs 510.76

or, 113/100 × x = Rs 510.76

or, 1.13x = Rs 510.76

or, x = 510.76 / 1.13 = Rs 452

Hence, the charge of telephone calls without VAT is Rs 452.

(ii) Find the charge of telephone calls without TSC.

Solution:

Let the charge of telephone calls without TSC be Rs y.

Then, y + 13% of y = Rs 452 (from part i)

or, 113/100 × y = Rs 452

or, 1.13y = Rs 452

or, y = 452 / 1.13 = Rs 400

Hence, the charge of telephone calls without TSC is Rs 400.

(iii) Find the total number of calls that he made.

Solution:

Sub-total charge (C) = Rs 400 (from part ii)

Minimum charge up to 175 calls = Rs 200

Charge for extra calls = Rs 400 – Rs 200 = Rs 200

Number of extra calls = Rs 200 / Re 1 = 200 calls

Total number of calls = 175 + 200 = 375 calls

Hence, the total number of calls made is 375.

b) Mrs. Gurung has a shop of cultural dresses. She uses a landline phone. She paid Rs 1,276.90 with 13% TSC and 13% VAT for her telephone bill for the month of Poush. Given that the minimum charge of telephone calls up to 175 calls is Rs 200 and the charge for each extra call is Re 1.

(i) Find the charge of telephone calls without VAT.

Solution:

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

Then, x + 13% of x = Rs 1,276.90

or, 113/100 × x = Rs 1,276.90

or, 1.13x = Rs 1,276.90

or, x = 1,276.90 / 1.13 = Rs 1,130

Hence, the charge of telephone calls without VAT is Rs 1,130.

(ii) Find the charge of telephone calls without TSC.

Solution:

Let the charge of telephone calls without TSC be Rs y.

Then, y + 13% of y = Rs 1,130 (from part i)

or, 113/100 × y = Rs 1,130

or, 1.13y = Rs 1,130

or, y = 1,130 / 1.13 = Rs 1,000

Hence, the charge of telephone calls without TSC is Rs 1,000.

(iii) Find the number of total calls made in Poush.

Solution:

Sub-total charge (C) = Rs 1,000 (from part ii)

Minimum charge up to 175 calls = Rs 200

Charge for extra calls = Rs 1,000 – Rs 200 = Rs 800

Number of extra calls = Rs 800 / Re 1 = 800 calls

Total number of calls = 175 + 800 = 975 calls

Hence, the total number of calls made in Poush is 975.

a) Mr. Regmi uses a postpaid NTC mobile. The monthly rental charge for Nepal Telecom GSM postpaid mobile service is Rs 300. If on-net call charge is Rs 1.50 per minute, 13% TSC is added and 13% VAT is levied on it, then 2% ownership tax (OT) is charged.

(i) How much minimum charge should he pay in a month?

Solution:

Here, monthly rental charge = Rs 300

Charge with 13% TSC = 113% of Rs 300 = 113/100 × 300 = Rs 339

Charge with 13% VAT = 113% of Rs 339 = 113/100 × 339 = Rs 383.07

Charge with 2% OT = 102% of Rs 383.07 = 102/100 × 383.07 = Rs 390.73

Hence, he should pay a minimum charge of Rs 390.73 in a month.

(ii) How long can he make calls on a mobile with recharge of Rs 100?

Solution:

Let the call charge without OT be Rs x.

Then, x + 2% of x = Rs 100

or, 102/100 × x = Rs 100

or, 1.02x = Rs 100

or, x = 100 / 1.02 = Rs 98.04

Let the call charge without VAT be Rs y.

Then, y + 13% of y = Rs 98.04

or, 113/100 × y = Rs 98.04

or, 1.13y = Rs 98.04

or, y = 98.04 / 1.13 = Rs 86.76

Let the call charge without TSC be Rs z.

Then, z + 13% of z = Rs 86.76

or, 113/100 × z = Rs 86.76

or, 1.13z = Rs 86.76

or, z = 86.76 / 1.13 = Rs 76.78

On-net call charge = Rs 1.50 per minute

Talk time for Rs 76.78 = 76.78 / 1.50 = 51.19 minutes

51.19 minutes = 51 minutes + 0.19 × 60 seconds = 51 minutes + 11.4 seconds

Hence, he can make calls for 51 minutes and 11.4 seconds.

b) Mrs. Magar uses a postpaid NTC mobile. The monthly rental charge for Nepal Telecom GSM postpaid mobile service is Rs 300. If on-net call charge is Rs 1.50 per minute, 13% TSC is added and 13% VAT is levied on it, then 2% ownership tax (OT) is charged.

(i) How much minimum charge should she pay in a month?

Solution:

Here, monthly rental charge = Rs 300

Charge with 13% TSC = 113% of Rs 300 = 113/100 × 300 = Rs 339

Charge with 13% VAT = 113% of Rs 339 = 113/100 × 339 = Rs 383.07

Charge with 2% OT = 102% of Rs 383.07 = 102/100 × 383.07 = Rs 390.73

Hence, she should pay a minimum charge of Rs 390.73 in a month.

(ii) How long can she make calls on a mobile recharge of Rs 500?

Solution:

Let the call charge without OT be Rs x.

Then, x + 2% of x = Rs 500

or, 102/100 × x = Rs 500

or, 1.02x = Rs 500

or, x = 500 / 1.02 = Rs 490.20

Let the call charge without VAT be Rs y.

Then, y + 13% of y = Rs 490.20

or, 113/100 × y = Rs 490.20

or, 1.13y = Rs 490.20

or, y = 490.20 / 1.13 = Rs 433.81

Let the call charge without TSC be Rs z.

Then, z + 13% of z = Rs 433.81

or, 113/100 × z = Rs 433.81

or, 1.13z = Rs 433.81

or, z = 433.81 / 1.13 = Rs 383.90

On-net call charge = Rs 1.50 per minute

Talk time for Rs 383.90 = 383.90 / 1.50 = 255.93 minutes

255.93 minutes = 255 minutes + 0.93 × 60 seconds = 255 minutes + 55.8 seconds

255 minutes = 4 hours (240 minutes) + 15 minutes

Hence, she can make calls for 4 hours, 15 minutes, and 55.8 seconds.

Calculation of taxi fare in a taximeter

EXERCISE 4.4

General section

a) Answer the following questions.

(i) Which department is responsible for implementation of the rates of taxi fare?

Solution:

The Nepal Bureau of Standards and Metrology (NBSM) is responsible for implementing and monitoring the rates of taxi fare.

Hence, the department responsible is the Nepal Bureau of Standards and Metrology (NBSM).

(ii) What is the minimum fare of taxi from 6:00 a.m. to 9:00 p.m.?

Solution:

From the NBSM rates, the minimum fare from 6:00 a.m. to 9:00 p.m. is Rs 50.

Hence, the minimum fare is Rs 50.

(iii) What is the taxi fare for each 200 m from 6:00 a.m. to 9:00 p.m.?

Solution:

From the NBSM rates, the fare for each 200 meters from 6:00 a.m. to 9:00 p.m. is Rs 10.

Hence, the taxi fare for each 200 meters is Rs 10.

(iv) What is the minimum fare of taxi from 9:00 p.m. to 6:00 a.m.?

Solution:

From the NBSM rates, the minimum fare from 9:00 p.m. to 6:00 a.m. is Rs 75.

Hence, the minimum fare is Rs 75.

(v) What is the taxi fare for each 200 m from 9:00 p.m. to 6:00 a.m.?

Solution:

From the NBSM rates, the fare for each 200 meters from 9:00 p.m. to 6:00 a.m. is Rs 15.

Hence, the taxi fare for each 200 meters is Rs 15.

b) According to rates of taxi fare implemented by Nepal Bureau of Standards and Metrology (NBSM), answer the following questions.

(i) What is the waiting charge for 2 minutes from 6:00 a.m. to 9:00 p.m.?

Solution:

From the NBSM rates, the waiting charge from 6:00 a.m. to 9:00 p.m. is Rs 10 per 2 minutes.

Hence, the waiting charge for 2 minutes is Rs 10.

(ii) What is the waiting charge for 2 minutes from 9:00 p.m. to 6:00 a.m.?

Solution:

From the NBSM rates, the waiting charge from 9:00 p.m. to 6:00 a.m. is Rs 15 per 2 minutes.

Hence, the waiting charge for 2 minutes is Rs 15.

a) Suntali hired a taxi and travelled 6 km at 3:30 p.m. If the minimum fare is Rs 50 and the fare goes on at the rate of Rs 10 per 200 metres, calculate the total fare paid by her.

Solution:

Here, 3:30 p.m. lies between 6:00 a.m. and 9:00 p.m.

Minimum fare = Rs 50

Distance travelled = 6 km = 6000 m

Fare per 200 m = Rs 10

Number of 200 m segments = 6000 / 200 = 30

Fare for distance = 30 × Rs 10 = Rs 300

Total fare = Minimum fare + Fare for distance = Rs 50 + Rs 300 = Rs 350

Hence, the total fare paid by Suntali is Rs 350.

b) Madan Bahadur hired a taxi and travelled 9 km at 8:30 a.m. If the minimum fare is Rs 50 and the fare goes on at the rate of Rs 10 per 200 metres, calculate the total fare paid by him.

Solution:

Here, 8:30 a.m. lies between 6:00 a.m. and 9:00 p.m.

Minimum fare = Rs 50

Distance travelled = 9 km = 9000 m

Fare per 200 m = Rs 10

Number of 200 m segments = 9000 / 200 = 45

Fare for distance = 45 × Rs 10 = Rs 450

Total fare = Minimum fare + Fare for distance = Rs 50 + Rs 450 = Rs 500

Hence, the total fare paid by Madan Bahadur is Rs 500.

c) Mr. Sampang hired a taxi and travelled 15 km at 10:00 p.m. If the minimum fare is Rs 75 and the fare goes on at the rate of Rs 15 per 200 metres, calculate the total fare paid by him.

Solution:

Here, 10:00 p.m. lies between 9:00 p.m. and 6:00 a.m.

Minimum fare = Rs 75

Distance travelled = 15 km = 15000 m

Fare per 200 m = Rs 15

Number of 200 m segments = 15000 / 200 = 75

Fare for distance = 75 × Rs 15 = Rs 1,125

Total fare = Minimum fare + Fare for distance = Rs 75 + Rs 1,125 = Rs 1,200

Hence, the total fare paid by Mr. Sampang is Rs 1,200.

Creative section

a) Mrs. Kandel and her family wished to visit zoo at Jawalakhel which is at a distance of 4.6 km from her home located in Kalanki. For this, she hired a metered taxi and asked it to wait for 6 minutes, then they travelled to zoo at 10:00 a.m.

(i) What was the minimum fare appeared immediately after the meter was flagged down?

Solution:

Here, 10:00 a.m. lies between 6:00 a.m. and 9:00 p.m.

From the NBSM rates, the minimum fare is Rs 50.

Hence, the minimum fare appeared immediately after the meter was flagged down is Rs 50.

(ii) What was the waiting charge?

Solution:

Waiting charge for 2 minutes from 6:00 a.m. to 9:00 p.m. = Rs 10

Waiting charge for 1 minute = Rs 10 / 2 = Rs 5

Waiting time = 6 minutes

Waiting charge for 6 minutes = Rs 5 × 6 = Rs 30

Hence, the waiting charge is Rs 30.

(iii) Calculate the total fare paid by her.

Solution:

Minimum fare = Rs 50 (from part i)

Distance travelled = 4.6 km = 4600 m

Fare per 200 m = Rs 10

Number of 200 m segments = 4600 / 200 = 23

Fare for distance = 23 × Rs 10 = Rs 230

Waiting charge = Rs 30 (from part ii)

Total fare = Minimum fare + Fare for distance + Waiting charge = Rs 50 + Rs 230 + Rs 30 = Rs 310

Hence, the total fare paid by Mrs. Kandel is Rs 310.

(iv) If she hired another taxi and returned back to her home through the same route, how much more or less fare did she pay?

Solution:

For the return trip, no waiting is mentioned.

Minimum fare = Rs 50

Distance travelled = 4.6 km = 4600 m

Fare per 200 m = Rs 10

Number of 200 m segments = 4600 / 200 = 23

Fare for distance = 23 × Rs 10 = Rs 230

Total fare for return = Rs 50 + Rs 230 = Rs 280

Fare paid to zoo = Rs 310 (from part iii)

Difference = Rs 310 – Rs 280 = Rs 30

Hence, she paid Rs 30 more while travelling to the zoo.

b) At 2:45 a.m., Mr. Poudel had to go hospital which was at a distance of 15 km from his home. He hired a metered taxi and asked it to wait for 4 minutes.

(i) What was the minimum fare appeared immediately after the meter was flagged down?

Solution:

Here, 2:45 a.m. lies between 9:00 p.m. and 6:00 a.m.

From the NBSM rates, the minimum fare is Rs 75.

Hence, the minimum fare appeared immediately after the meter was flagged down is Rs 75.

(ii) What was the waiting charge?

Solution:

Waiting charge for 2 minutes from 9:00 p.m. to 6:00 a.m. = Rs 15

Waiting charge for 1 minute = Rs 15 / 2 = Rs 7.50

Waiting time = 4 minutes

Waiting charge for 4 minutes = Rs 7.50 × 4 = Rs 30

Hence, the waiting charge is Rs 30.

(iii) Find the taxi fare paid by him.

Solution:

Minimum fare = Rs 75 (from part i)

Distance travelled = 15 km = 15000 m

Fare per 200 m = Rs 15

Number of 200 m segments = 15000 / 200 = 75

Fare for distance = 75 × Rs 15 = Rs 1,125

Waiting charge = Rs 30 (from part ii)

Total fare = Minimum fare + Fare for distance + Waiting charge = Rs 75 + Rs 1,125 + Rs 30 = Rs 1,230

Hence, the taxi fare paid by Mr. Poudel is Rs 1,230.

(iv) If he hired another taxi and returned back to his home at 7:00 a.m. through the same route, how much more or less fare would he pay than travelling to hospital?

Solution:

For the return trip at 7:00 a.m., it lies between 6:00 a.m. and 9:00 p.m., and no waiting is mentioned.

Minimum fare = Rs 50

Distance travelled = 15 km = 15000 m

Fare per 200 m = Rs 10

Number of 200 m segments = 15000 / 200 = 75

Fare for distance = 75 × Rs 10 = Rs 750

Total fare for return = Rs 50 + Rs 750 = Rs 800

Fare paid to hospital = Rs 1,230 (from part iii)

Difference = Rs 1,230 – Rs 800 = Rs 430

Hence, he would pay Rs 430 more while travelling to the hospital.

a) Bishwant hired a taxi and travelled a certain distance at 6:15 a.m. He paid the total fare of Rs 450 including the additional waiting charge for 20 minutes. If the minimum fare is Rs 50, the fare per 200 m is Rs 10 and the waiting charge is Rs 10 per 2 minutes, answer the following questions.

(i) How much was the waiting charge?

Solution:

Here, 6:15 a.m. lies between 6:00 a.m. and 9:00 p.m.

Waiting charge for 2 minutes = Rs 10

Waiting charge for 1 minute = Rs 10 / 2 = Rs 5

Waiting time = 20 minutes

Waiting charge for 20 minutes = Rs 5 × 20 = Rs 100

Hence, the waiting charge was Rs 100.

(ii) Calculate the fare excluding the waiting charge and minimum fare.

Solution:

Total fare = Rs 450

Minimum fare = Rs 50

Waiting charge = Rs 100 (from part i)

Fare excluding waiting charge and minimum fare = Rs 450 – Rs 50 – Rs 100 = Rs 300

Hence, the fare excluding the waiting charge and minimum fare is Rs 300.

(iii) Find the distance travelled by him.

Solution:

Fare for distance = Rs 300 (from part ii)

Fare per 200 m = Rs 10

Number of 200 m segments = Rs 300 / Rs 10 = 30

Distance travelled = 30 × 200 m = 6000 m = 6 km

Hence, the distance travelled by Bishwant is 6 km.

b) Dipika hired a taxi and travelled a certain distance at 9:40 p.m. She paid the total fare of Rs 675 including the additional waiting charge for 10 minutes. If the minimum fare is Rs 75, the fare per 200 m is Rs 15 and the waiting charge is Rs 15 per 2 minutes, answer the following questions.

(i) How much was the waiting charge?

Solution:

Here, 9:40 p.m. lies between 9:00 p.m. and 6:00 a.m.

Waiting charge for 2 minutes = Rs 15

Waiting charge for 1 minute = Rs 15 / 2 = Rs 7.50

Waiting time = 10 minutes

Waiting charge for 10 minutes = Rs 7.50 × 10 = Rs 75

Hence, the waiting charge was Rs 75.

(ii) Find the distance travelled by her.

Solution:

Total fare = Rs 675

Minimum fare = Rs 75

Waiting charge = Rs 75 (from part i)

Fare for distance = Rs 675 – Rs 75 – Rs 75 = Rs 525

Fare per 200 m = Rs 15

Number of 200 m segments = Rs 525 / Rs 15 = 35

Distance travelled = 35 × 200 m = 7000 m = 7 km

Hence, the distance travelled by Dipika is 7 km.

Wednesday, March 26, 2025

Class 9 Maths New Course Solution | Lesson 4 Household Arithmetic | CDC, Vedanta Excel

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