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A charity seeks to raise money by telephoning people at random from a call centre and asking them to donate.
Over the years, this charity has found that the amount of money raised ($A) is related to the number of telephone calls made (n). A graph of this relationship is shown.
It costs the charity $2100 per week to run the call centre. It also costs an average of 50 cents per telephone call.
(i) Write an equation to represent the total cost, C, of running the call centre for a week in which n phone calls are made. (1 mark)
(ii) By graphing this equation on the axes above, determine the number of phone calls the charity needs to make in order to break even. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
i. C=$2100+$0.50n♦ Mean marks of 48% and 32% for parts (i) and (ii) respectively.
ii.
From the above graph, the charity needs to
make 700 calls to break even.
Sue and Mikey are planning a fund-raising dance. They can hire a hall for $400 and a band for $300. Refreshments will cost them $12 per person.
(i) Write a formula for the cost ($C) of running the dance for x people. (1 mark)
The graph shows planned income and costs when the ticket price is $20 .
(ii) Estimate the minimum number of people needed at the dance to cover the costs. (1 mark)
(iii) How much profit will be made if 150 people attend the dance? (1 mark)
(iv) Sue and Mikey plan to sell 200 tickets. They want to make a profit of $1500.
What should be the price of a ticket, assuming all 200 tickets will be sold? (3 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) $C =400+300+(12×x)
=700+12x
(ii) Using the graph intersection
Approximately 90 people are needed
to cover the costs.
(iii) If 150 people attend
Income =150×$20
=$3000
Costs =700+(12×150)
=$2500
∴ Profit =3000−2500
=$500
(iv) Costs when x=200:
C =700+(12×200)
=$3100
Income required to make $1500 profit
=3100+1500
=$4600
∴ Price per ticket =4600/200
=$23
Jack needs to find the number of years, t, it will take for a population of bats to first exceed 18 000.
He uses a ‘guess-and-check’ method to estimate t in the following equation
5 × 3^t = 18 000.
Here is his working:
(i) Jack’s next guess is t=6. Show Jack’s correct working for this guess, including the calculation and conclusion. (1 mark)
(ii) Continue using the ‘guess-and-check’ method to find the number of years, t, it will take for the population to first exceed 18 000, if t is a whole number. Include the calculations and conclusions. (2 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
i. When t=6,
5×3^6=3645
⇒ Too small
ii. When t=7,
5×3^7=10 935
⇒ Too small
When t=8,
5×3^8=32 805
⇒ exceeds 18 000
∴t=8
Wind turbines, such as those shown, are used to generate power.
In theory, the power that could be generated by a wind turbine is modelled using the equation
T = 20 000w^3
where T is the theoretical power generated, in watts
w is the speed of the wind, in metres per second.
(i) Using this equation, what is the theoretical power generated by a wind turbine if the wind speed is 7.3 m/s ? (1 mark)
(ii) In practice, the actual power generated by a wind turbine is only 40% of the theoretical power.
If A is the actual power generated, in watts, write an equation for A in terms of w. (1 mark)
The graph shows both the theoretical power generated and the actual power generated by a particular wind turbine.
(iii) Using the graph, or otherwise, find the difference between the theoretical power and the actual power generated when the wind speed is 9 m/s. (1 mark)
(iv) A particular farm requires at least 4.4 million watts of actual power in order to be self-sufficient.
What is the minimum wind speed required for the farm to be self-sufficient? (1 mark)
(v) A more accurate formula to calculate the power (P) generated by a wind turbine is
P = 0.61 × π × r^2 × w^3
where r is the length of each blade, in metres
w is the speed of the wind, in metres per second.
Each blade of a particular wind turbine has a length of 43 metres.
The turbine operates at a wind speed of 8 m/s.
Using the formula above, if the wind speed increased by 10%, what would be the percentage increase in the power generated by this wind turbine? (3 marks)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) T=20 000w^3
If w=7.3
T =20 000×(7.3)^3
=7 780 340 watts
(ii) We know A=40%×T
⇒A =0.4×20 000×w^3
=8000w^3
(iii) Solution 1
At w=9
A=5.8 million watts (from graph)
T=14.6 million watts (from graph)
Difference =14.6 million −5.8 million
=8.8 million watts
Alternative Solution
At w=9
T =20 000×9^3
=14 580 000 watts
A =8000×9^3
=5 832 000 watts
Difference =14 580 000 −5 832 000
=8 748 000 watts
(iv) Find w if A=4.4 million
8000w^3 =4 400 000
w^3 =4 400 000/8000
=550
∴w =550‾‾‾‾√3
=8.1932…
=8.2 m/s (1 d.p.)
∴ The minimum wind speed required is 8.2 m/s
(v) Find P when w=8 and r=43
P =0.61×π×r^2×w^3
=0.61×π×43^2×8^3
=1 814 205.92 watts
When speed of wind ↑⏐10%
w′=8×110%=8.8 m/s
Find P when w′=8.8
P =0.61×π×43^2×8.8^3
=2 414 708.08 watts
Increase in Power =2 414 708.08 −1 814 205.92
=600 502.16
∴ % Power increase =600 502.16/1 814 205.92
=0.331
=33% (nearest %)
(i) The number of bacteria in a culture grows from 100 to 114 in one hour.
What is the percentage increase in the number of bacteria? (1 mark)
(ii) The bacteria continue to grow according to the formula n=100(1.14)^t, where n is the number of bacteria after t hours. What is the number of bacteria after 15 hours? (1 mark)
(iii) Use the values of n from t=0 to t=15 to draw a graph of n=100(1.14)^t.
Use about half a page for your graph and mark a scale on each axis. (4 marks)
(iv) Using your graph or otherwise, estimate the time in hours for the number of bacteria to reach 300. (1 mark)
This response will be awarded full points automatically, but it can be reviewed and adjusted after submission.
(i) Percentage increase
=(114−100)/100×100
=14%
(ii) n=100(1.14)^t
When t=15,
n =100(1.14)^15
=713.793 …
=714 (nearest whole)
(iii)
(iv) Using the graph
The number of bacteria reaches 300 after
approximately 8.4 hours.