Transcript May 2001: Paper 2 #5

May 2001: Paper 2 #5
The number (n) of bacteria in a colony after h hours is given
æ
ö. Initially, there are 1200
by the formula n= 1200ç30.25h ÷
÷
bacteria in the colony.
çè
÷
ø
(a) Copy and complete the table below, which gives values of n and h. Give
your answers to the nearest hundred.
Time in hours (h)
0
no. of bacteria (n) 1200
1
1600
2
3
2100 2700
4
3600
Be sure to copy the entire table in your answer booklet!
æ 0.25(1)ö
÷
When h = 1, we have
n= 1200ç
» 1600
÷
çç3
÷
÷
çè
ø
When h = 4, we have
æ 0.25(4)ö
÷
n= 1200ç
= 3600
÷
çç3
÷
÷
çè
ø
(b) On graph paper, draw the graph of the above function. Use a scale of 3
cm to represent 1 hour on the horizontal axis and 4 cm to represent 1000
bacteria on the vertical axis. Label the graph clearly.
Number of bacteria (n)
4000
(3,2700)
3000
2000
(4,3600)
(1,1600)
(2,2100)
1000
(0,1200)
1
2
3
Time in hours (h)
4
(c) Use your graph to answer each of the following showing your method clearly.
(i)
How many bacteria would there be after 2 hours and 40 minutes?
Give your answer to the nearest hundred.
(ii) After how long will there be approximately 3000 bacteria? Give
your answer to the nearest 10 minutes.
The key to this question is to using the graph that you have already made.
You need to use broken lines on the graph that clearly show the ordered
pairs.
For part (i), you do this by finding 2 hours 40 minutes on the horizontal axis
and showing which value on the vertical axis corresponds to this time.
For part (ii), you start on the horizontal axis and find 3000 bacteria. You are
then to show which value on the horizontal axis corresponds to this amount.
Number of bacteria (n)
4000
At 2 hrs 40 min, there are about 2500 bacteria.
3000 bacteria
3000
2500
2000
There will be 3000 bacteria at 3 hrs 20 min.
1000
2 hrs 40 min
1
2
3
3 hrs 20 min
Time in hours (h)
4
(ii) A picture is in the shape of a square of side 5 cm. It is surrounded
by a wooden frame width x cm, as shown in the diagram below.
5cm
x
x
l
The length of the wooden frame is l cm, and the area of the wooden frame is A cm2.
(a) Write an expression for the length of l in terms of x.
l = x + 5 + x = 2x + 5.
(b) Write an expression for the area A in terms of x.
The area we are looking for is only the area of the frame.
The area of the picture is given by
(5 cm)(5 cm) = 25 cm2.
The total area inside the frame is given by
2
l 2 = æççç 2x + 5ö÷÷÷ = 4x2 + 20x + 25
çè
÷
ø
5cm
x
x
A = (total area inside frame) – (area of picture)
Substituting these values, we see
æ
çç
çç
çè
ö
2
A= 4x + 20x + 25÷÷÷÷÷- æçççè 25öø÷÷÷÷= 4x2 + 20 x
÷
ø
l
(c) If the area of the frame is 24 cm2, find the value of x.
From part (b), we know
A= 4x2 + 20x
Substituting A = 24, we have
24 = 4x2 + 20x
Subtracting 24 from both sides gives us
0 = 4x2 + 20x- 24
Factoring completely yields
ö
÷
çç x- 1÷
÷
÷
0 = 4æçççè x + 6 öæ
÷
÷
ç
÷
÷
øè
ø
Thus, by the zero product property, we know
x = –6 or x = 1
Since x is a length, x  –6. Therefore, x = 1.
Note: You could just as easily
used the quadratic formula to
solve…
- 20± (20 )2- 4(4 )(- 24 )
x=
2(4 )
x= - 20±
400+ 384
8
784
x= - 20±
8
x= - 20± 28
8
x = 1 or x = - 6