1. The surface area of the ceiling can be calculated as follows:
SA = length x width = 20m x 15m = 300m2
Therefore, the cost of painting the ceiling would be:
300m2 x N$6.50/m2 = N$1,950
2. The surface area of the walls (excluding the windows and door) can be calculated as follows:
SA = (height x length) x 2 + (height x width) x 2
SA = (5m x 20m) x 2 + (5m x 15m) x 2
SA = 200m2 + 150m2
SA = 350m2
Therefore, the cost of painting the walls would be:
350m2 x N$8.50/m2 = N$2,975
3. The area of the floor can be calculated as follows:
Area = length x width = 20m x 15m = 300m2
The number of tiles required can be calculated by dividing the area by the area of each tile:
Number of tiles = Area of floor / Area of each tile
Number of tiles = 300m2 / (0.2m x 0.2m) = 7,500 tiles
Since each box contains 50 tiles, the number of boxes required would be:
Number of boxes = Number of tiles / Tiles per box
Number of boxes = 7,500 tiles / 50 tiles per box = 150 boxes
Therefore, the cost of tiling the floor would be:
150 boxes x N$50/box = N$7,500
4. The volume of the classroom can be calculated as follows:
Volume = length x width x height = 20m x 15m x 5m = 1,500m3
Therefore, the cost of providing air conditioning would be:
1,500m3 x N$60/m3 = N$90,000
William has been contracted to paint a school classroom. The classroom is 20m long, l5m
wide and 5m high. There are four windows (2m by 3m) and a door (2m by lm).
1.What is the cost of painting the ceiling at N$6.50 /m'z?
2.What is the cost of painting the walls (excluding the windows and door) at N$8.50
lm2?
3.What is the cost of tiling the floor using tiles (20cm by 20cm) that are sold in boxes
of 50 tiles each at N$50 per box?
4What is the cost of providing air conditioning for this classroom at N$60/m3?
3 answers
(a) George is buying a house for N$850,000. He has to pay a
from his bank repayable over 25 years at l2%o p.a. interest.
(i) What is his monthly instalment?
(l) 30 year payback period
(2) Reduced interest rate
l0% deposit and can secure a bond
(ii) Another bank is now giving him two other options i.e. to increase the payback period to 30
years or to reduce the rate by 15% p.a. Find the monthly instalment for each option.
(iii) How much money will he save per month using the cheapest option? (4 marks)
from his bank repayable over 25 years at l2%o p.a. interest.
(i) What is his monthly instalment?
(l) 30 year payback period
(2) Reduced interest rate
l0% deposit and can secure a bond
(ii) Another bank is now giving him two other options i.e. to increase the payback period to 30
years or to reduce the rate by 15% p.a. Find the monthly instalment for each option.
(iii) How much money will he save per month using the cheapest option? (4 marks)
(a)(i) Using the formula for calculating monthly payments on a loan, we have:
P = L[c(1 + c)^n]/[(1 + c)^n - 1]
where:
P = monthly payment
L = loan amount
c = monthly interest rate = annual interest rate/12
n = total number of monthly payments
For George's loan, L = N$850,000 and n = 25 years x 12 months/year = 300 months. The monthly interest rate, c, can be calculated as follows:
c = 12%o p.a. / 12 months = 1%o per month
Substituting these values into the formula, we get:
P = 850000[0.01(1 + 0.01)^300]/[(1 + 0.01)^300 - 1] ≈ N$8,919.31
Therefore, George's monthly instalment is approximately N$8,919.31 per month.
(a)(ii) Increasing the payback period to 30 years would result in a total of 30 years x 12 months/year = 360 monthly payments. Using the same formula as above, but with n = 360, we get:
P = 850000[0.01(1 + 0.01)^360]/[(1 + 0.01)^360 - 1] ≈ N$7,517.36 per month
Reducing the interest rate by 15% p.a. would result in a new monthly interest rate of:
c = (12%o p.a. - 15%o p.a.)/12 months = -0.25%o per month
Using the original loan amount and payback period, but with this new monthly interest rate, we get:
P = 850000[(-0.0025)(1 - 0.0025)^300]/[(1 - 0.0025)^300 - 1] ≈ N$6,638.77 per month
Therefore, George's monthly instalments for the increased payback period and reduced interest rate options are N$7,517.36 per month and N$6,638.77 per month, respectively.
(a)(iii) To calculate how much George saves per month using the cheapest option, we need to compare the monthly instalment for each option to the original monthly instalment of N$8,919.31.
The increased payback period option results in a monthly saving of:
N$8,919.31 - N$7,517.36 = N$1,401.95 per month
The reduced interest rate option results in a monthly saving of:
N$8,919.31 - N$6,638.77 = N$2,280.54 per month
Therefore, George would save more money per month by choosing the reduced interest rate option.
P = L[c(1 + c)^n]/[(1 + c)^n - 1]
where:
P = monthly payment
L = loan amount
c = monthly interest rate = annual interest rate/12
n = total number of monthly payments
For George's loan, L = N$850,000 and n = 25 years x 12 months/year = 300 months. The monthly interest rate, c, can be calculated as follows:
c = 12%o p.a. / 12 months = 1%o per month
Substituting these values into the formula, we get:
P = 850000[0.01(1 + 0.01)^300]/[(1 + 0.01)^300 - 1] ≈ N$8,919.31
Therefore, George's monthly instalment is approximately N$8,919.31 per month.
(a)(ii) Increasing the payback period to 30 years would result in a total of 30 years x 12 months/year = 360 monthly payments. Using the same formula as above, but with n = 360, we get:
P = 850000[0.01(1 + 0.01)^360]/[(1 + 0.01)^360 - 1] ≈ N$7,517.36 per month
Reducing the interest rate by 15% p.a. would result in a new monthly interest rate of:
c = (12%o p.a. - 15%o p.a.)/12 months = -0.25%o per month
Using the original loan amount and payback period, but with this new monthly interest rate, we get:
P = 850000[(-0.0025)(1 - 0.0025)^300]/[(1 - 0.0025)^300 - 1] ≈ N$6,638.77 per month
Therefore, George's monthly instalments for the increased payback period and reduced interest rate options are N$7,517.36 per month and N$6,638.77 per month, respectively.
(a)(iii) To calculate how much George saves per month using the cheapest option, we need to compare the monthly instalment for each option to the original monthly instalment of N$8,919.31.
The increased payback period option results in a monthly saving of:
N$8,919.31 - N$7,517.36 = N$1,401.95 per month
The reduced interest rate option results in a monthly saving of:
N$8,919.31 - N$6,638.77 = N$2,280.54 per month
Therefore, George would save more money per month by choosing the reduced interest rate option.
4 answers