Name:
MATHEMATICS Compulsory Part 1
Class:
(
)
F5 Final Exam 2017
PAPER 1
Question-Answer Book
Marker’s Use Only
Question No.
Time allowed: 2 hours 15 mins
This paper must be answered in English.
INSTRUCTIONS
1.
2.
3.
4.
5.
6.
7.
Write your Name, Class and Class No.
in the space provided on Page 1.
This paper consists of THREE sections,
A(1), A(2) and B. A(1) and A(2)
carries 33 marks and Section B carries
34 marks.
Attempt ALL questions in this paper.
Write your answers in the spaces
provided in this Question-Answer Book.
Do not write in the margins.
Unless otherwise specified, all working
must be clearly shown.
Unless otherwise specified, numerical
answers should be either exact or
correct to 3 significant figures.
The diagrams in this paper are not
necessarily drawn to scale.
Graph paper and supplementary answer
sheets will be supplied on request.
Write your name, class and class no.
and put them INSIDE this book.
1
/3
2
/3
3
/3
4
/4
5
/4
6
/4
7
/4
8
/4
9
/4
10
/6
11
/6
12
/6
13
/7
14
/8
15
/6
16
/6
17
/4
18
/8
19
/10
A1(33)
Total
F5 Final Exam 2017
–1–
Marks
A2(33)
B(34)
SECTION A(1) (33 marks)
Page total
Answer ALL questions in this section and write your answers in the spaces provided.
1. Simplify
(x 6 y 1) 2
(x 3 )5
and express the answer with positive indices.
2. Make x the subject of the formula
F5 Final Exam 2017
Ax  (5 x  B )C .
–2–
(3 marks)
(3 marks)
Page total
3.
Consider a quadratic equation
x 2  7 x  3  0.
(a)
Find the value of discriminant of the equation.
(b)
State the nature of the roots of the equation.
(3 marks)
.
4. Factorize
F5 Final Exam 2017
(a)
7m  14n ,
(b)
m 2  2mn  8n 2 ,
(c)
m2  2mn  8n 2  7m  14 n .
–3–
(4 marks)
Page total
5.
Given that log 2 = a and log 3 = b, express each of the following in terms of
(a)
log 12,
(b)
log 15,
(c)
log 30 .
F5 Final Exam 2017
a
and
b.
(4 marks)
–4–
Page total
6.
It is given that y varies inversely as x .
When x = 169, y = 400.
(a)
Express y in terms of x .
(b)
If the value of x is increased from 169 to 256, find the change in the value of y .
(4 marks)
F5 Final Exam 2017
–5–
Page total
7. Eric is going to arrange
8
different magazines on a bookshelf.
(a)
How many ways are there to arrange the magazines?
(b)
If two computer magazines should be put together, how many ways are there to arrange the magazines?
(4 marks)
8. Simplify
17i
and express the answer in standard form .
5  3i
(4 marks)
F5 Final Exam 2017
–6–
Page total
9. The following stem-and-leaf diagram shows the weights of 20 boys.
Stem (10 kg)
Leaf (1 kg)
3
5
5
6
8
4
0
0
0
1
2
5
2
4
6
7
7
6
2
7
2
4
5 7
(a)
Find the median weight.
(1 mark)
(b)
Find the inter-quartile range of the weight of the boys.
(1 mark)
(c)
Construct a box-and-whisker diagram to represent the data in the graph paper.
(2
marks)
F5 Final Exam 2017
–7–
Page total
Section A(2) (33 marks)
Answer ALL questions in this section and write your answers in the spaces provided.
2
10. Consider a quadratic equation ax  2bx  c  0 .
(a)
Find the discriminant of the equation.
(b)
If a, b, c form a geometric sequence,
(c)
(1 mark)
(i)
show that the equation has one double root;
(ii)
express the root of the equation in terms of
a
and
b.
Using the result of (b)(ii) or otherwise, solve 5 x 2  2 10 x  2 5  0.
F5 Final Exam 2017
–8–
(3 marks)
(2 marks)
Page total
F5 Final Exam 2017
–9–
Page total
11.
2
The Figure 1 shows the graph of L1 : y   x  4 . L1 cuts the x-axis at P.
3
Figure 1
(a)
Find the coordinates of P.
(2 marks)
(b)
Suppose a straight line L2 intersects L1 at P and L 2  L1 . Find the equation of L2 .
(4 marks)
F5 Final Exam 2017
– 10 –
Page total
12.
Consider
(a)
(b)
2(1  sin )
 tan( 90   ) for 0    180 .
cos
Rewrite the equation in terms of sin only.
2(1  sin )
Hence solve the equation
 tan( 90   ) .
cos
(3 marks)
(Give the answers correct to 1 decimal place.)
(3 marks)
F5 Final Exam 2017
– 11 –
Page total
13.
In Figure 2, BD is an angle bisector of  ADC.
AD is a diameter of the circle. If DAC  42,
(a) find ADC ,
(3 marks)
(b) find CAB .
(4 marks)
Figure 2
F5 Final Exam 2017
– 12 –
Page total
14. There are
n
rows of seats in a theatre.
It is known that the first row has 15 seats and
each row has 2 seats
more than the preceding row, as shown in the Figure 3.
Figure 3
(a)
Find the number of seats in the nth row in terms of n, where n > 1.
(b)
Find the total number of seats in the first
(c)
If the number of seats in the theatre is 792, find the number of rows of seats in the
n
rows.
theatre.
F5 Final Exam 2017
(3 marks)
(1 mark)
(4 marks)
– 13 –
Page total
Section B (34 marks)
Answer ALL questions in this section and write your answers in the spaces provided.
15.
Mary wants to buy 7 boxes of chocolate from a supermarket.
She chooses randomly from 9 boxes of dark chocolate, 5 boxes of milk chocolate & 8 boxes of white chocolate.
(a) Find the probability that only dark chocolate are chosen.
(2 marks)
(b) Find the probability that 2 boxes of dark chocolate, 3 boxes of milk chocolate and 2 boxes of white
chocolate are chosen.
(2 marks)
(c) Find the probability that she chooses exactly 4 boxes of milk chocolate.
F5 Final Exam 2017
– 14 –
(2 marks)
Page total
16.
The table below shows the means and standard deviations of the time that a large group of students used to
finish a 100m run in 2 fitness tests:
Test
Mean
Standard Deviation
I
20s
2s
II
18s
1s
The standard score of Billy in Test I is 1.5 .
(a) Find the time that Billy used to finish the run in Test I.
(3 marks)
(b) Assume that the time distributions in each of the above tests are normally distributed.
Billy uses 20s to finish the race in Test II.
He claims that comparing to other students, he performs better in Test II than that in Test I.
Is his claim correct?
F5 Final Exam 2017
Explain your answer.
(3 marks)
– 15 –
Page total
17.
The Figure 4 shows the graph of log y against x for the relationship log y  mx  c , where m and c are
constants. The graph passes through (0, 5) and (6, 2).
(a)
Find the values of
m
and
c.
(b)
Hence express y in terms of x .
(2 marks)
(2 marks)
Figure 4
F5 Final Exam 2017
– 16 –
Page total
18.
In Figure 5, ABC is a triangular paper card. D is a point lying on AB such that CD is perpendicular to AB.
It is given that AC = 15cm, CAD  40 and CBD  25 .
Figure 5
(a)
Find AD and BD.
(b)
The triangular paper card in Figure 5 is folded along CD such that ABD lies on the horizontal
plane as shown in Figure 6.
(3 marks)
Let N be a point on AB such that DN is the shortest distance between
D and AB.
Figure 6
(i) If the distance between A and B is 15cm, find DAB and DN.
(ii) Let P be a movable point on AB. Describe how CPD varies as P moves
from A to B. Explain your answer.
(5 marks)
F5 Final Exam 2017
– 17 –
Page total
F5 Final Exam 2017
– 18 –
Page total
19.
In a city, the controller of an incinerator P predicts that P can handle waste of weight W(n) tonnes
in the nth year since the start of its operation, where n is a positive integer.
It is given that W (n)  k  1000 a n , where k and a are positive constants.
It is found that the weights of waste handled by P in the 1st year and the 2nd year since the start of its
operation are 363 900 tonnes and 364 650 tonnes respectively.
(a)
(i)
Find
a
and
k.
(ii)
Starting from which year will the weight of waste that P needs to handle increase by
more than 40% annually?
(b)
(i)
(4 marks)
Express, in terms of n, the total weight of waste handled by P in the first n years since the
start of its operation.
(ii)
Find the total weight of waste handled by P in the first 18 years since the start of its operation,
correct to the nearest tonne.
(c)
(3 marks)
It is given that the upper limit of the weight of waste that P can handle in the 1st year is
700 000 tonnes, and this limit rises by 100 000 tonnes each year afterward.
Suppose the limit can rise continually. The controller claims that in the 19th year and
thereafter, the weight of waste exceeds the limit that P can handle each year.
Do you agree? Explain your answer.
F5 Final Exam 2017
(3 marks)
– 19 –
Page total
F5 Final Exam 2017
– 20 –
Page total
END OF PAPER
F5 Final Exam 2017
– 21 –
Page total
F5 Final Exam 2017
– 22 –
Page total
Solution
SECTION A(1) (33 marks)
(x 6 y 1) 2
1. Simplify
and express the answer with positive indices.
(x 3 )5
(3 marks)
(x 6 y 1) 2
(x 3 )5
=
x12 y 2
1M
x 15
= x12( 15) y 2
=
1M
x 27
1A
y2
2. Make x the subject of the formula
1M
x( A  5C )  BC
1M
BC
( A  5C )
1A
Consider a quadratic equation
x 2  7 x  3  0.
(a)
Find the value of discriminant of the equation.
(b)
State the nature of the roots of the equation.
(a)
(3 marks)
Ax  5Cx  BC
Ax  5Cx  BC
x
3.
Ax  (5 x  B )C .
2
Discriminant = b – 4ac
(3 marks)
1M
= 72 – 4(1)(3)
1A
= 37
(b) Since  > 0,
1A
the equation has two distinct real roots.
4. Factorize
(a)
(b)
(c)
(a)
7m  14n ,
(b)
m 2  2mn  8n 2 ,
(c)
7m  14n
= 7(m  2n)
m2  2mn  8n 2  7m  14 n .
(4 marks)
1A
m 2  2mn  8n 2
= (m  4n)( m  2n)
1A
m2  2mn  8n 2  7m  14 n
= m2  2mn  8n 2  (7 m  14 n)
= (m  4n)( m  2n)  7(m  2n)
1M
= (m  4n  7)( m  2n)
1A
F5 Final Exam 2017
– 23 –
Page total
5.
Given that log 2 = a and log 3 = b, express each of the following in terms of
(a)
log 12,
(b)
log 15,
(c)
log 30 .
a
and
b.
(4 marks)
2
(a) log 12  log( 2  3)
 log 22  log 3
1M
 2 log 2  log 3
 2a  b
(b)
1A
30
2
3  10
 log
2
 log 3  log 10  log 2
log15  log
 b  a 1
(c) log 30 
1A
1
log 30
2
1
log(3  10)
2
1
 (log3  log10)
2
1
 (b  1)
2


6.
1 1
 b
2 2
1A
It is given that y varies inversely as x .
When x = 169, y = 400.
(a) Express y in terms of x .
(b)
If the value of x is increased from 169 to 256, find the change in the value of y .
(4 marks)
6. (a)Let y 
k
x
, where
k is a non-zero constant.
k
So, we have 400 
1A
.
169
Solving, we have
Thus, we have y 
(b)
k =5200
5200
1A
x
Note that the initial value of y is 400.
The final value of y
5200

(by (a))
256
 325
The decrease in the value of y
1M
 400  325 =75
1A
(or change = -75)
F5 Final Exam 2017
– 24 –
Page total
7. Eric is going to arrange
8
different magazines on a bookshelf.
(a)
How many ways are there to arrange the magazines?
(b)
If two computer magazines should be put together, how many ways are there to arrange the magazines?
(4 marks)
1M
1A
(a)
Number of ways  8!
 40320
(b)
Number of ways  (6 + 1)!  2!
 10080
17i
5  3i
8. Simplify
1M
1A
and express the answer in standard form .
(4 marks)
17i 5  3i
17i

=
5  3i 5  3i 5  3i


1M
17i (5  3i )
52  (3i )2
85i  51i 2
25  9i 2
51  85i

34
3 5
  i
2 2
1A
2A
9. The following stem-and-leaf diagram shows the weights of 20 boys.
Stem (10 kg)
Leaf (1 kg)
3
5
5
6
8
4
0
0
0
1
2
5
2
4
6
7
7
6
2
7
2
4
5 7
(a)
Find the median weight.
(1 mark)
(b)
Find the inter-quartile range of the weight of the boys.
(1 mark)
(c)
Construct a box-and-whisker diagram to represent the data in the graph paper.
(2 marks)
(a)
(b)
Median
42  44

kg
2
 43 kg
1A
40  40
kg  40 kg
2
54  56
Q3 
kg  55 kg
2
Q1 
Inter-quartile range  Q3  Q1
 (55  40) kg
 15 kg
F5 Final Exam 2017
1A
– 25 –
Page total
(c)
2A
Section A(2) (33 marks)
Answer ALL questions in this section and write your answers in the spaces provided.
2
10. Consider a quadratic equation ax  2bx  c  0 .
(a)
Find the discriminant of the equation.
(b)
If a, b, c form a geometric sequence,
(c)
(a)
(b)
(1 mark)
(i)
show that the equation has one double root;
(ii)
express the root of the equation in terms of
a
and
b.
Using the result of (b)(ii) or otherwise, solve 5 x 2  2 10 x  2 5  0.
  (2b) 2  4(a )(c)
 4b 2  4ac
(i)
(3 marks)
(2 marks)
1A
Let r be the common ratio.

b  ar
c  ar 2
1M
By the result of (a),
  4(ar ) 2  4a ( ar 2 )
 4a 2 r 2  4a 2 r 2
1M
0

The equation has one double root.
1A
(ii)
The root of the equation
2b  0
2a
b

a

(c)
1A
Let a  5 , b  10 and c  2 5 .
b
10

 2
a
5
c 2 5

 2
a
10

1M
5 , 10 , 2 5 form a geometric
sequence.
By (b)(ii),
x
10
 2
1A
5
F5 Final Exam 2017
– 26 –
Page total
11.
2
The Figure 1 shows the graph of L1 : y   x  4 . L1 cuts the x-axis at P.
3
Figure 1
11.
(a)
Find the coordinates of P.
(2 marks)
(b)
Suppose a straight line L2 intersects L1 at P and L 2  L1 . Find the equation of L2 .
(4 marks)
(a)
When y  0,
0
2
x4
3
1M
2
x 4
3
x6

(b)
1A
The coordinates of P are (6, 0).
Slope of L1  
2
3
Since L2  L1 ,
slope of L2  slope of L1  1
1M
 2
slope of L2      1
 3
3
slope of L 2 
2
The equation of L 2 :
1A
3
y 0

2
x6
1M
2y  3x  18
3x  2y  18  0
F5 Final Exam 2017
1A
– 27 –
Page total
12.
Consider
(a)
(b)
2(1  sin )
 tan( 90   ) for 0    180 .
cos
Rewrite the equation in terms of sin only.
2(1  sin )
Hence solve the equation
 tan( 90   ) .
cos
(3 marks)
(Give the answers correct to 1 decimal place.)
(3 marks)
2(1  sin  )
 tan( 90   )
cos 
2  2 sin 
1

cos
tan 
2  2 sin
1

sin
cos
cos
(a)
1M
cos2   2 sin  2 sin2 
1M
1  sin2   2 sin  2 sin2 
3 sin 2   2 sin   1  0
(b)
1A
2(1  sin  )
 tan(90   )
cos
3 sin 2   2 sin   1  0
(3 sin   1)(sin   1)  0
sin  
1
3
1M
or sin  1 (rejected)
  19.47122 or 180  19.47122

F5 Final Exam 2017
  19.5 or 160.5 (cor. to 1 d. p.)
1A+1A
– 28 –
Page total
13.
In Figure 2, BD is an angle bisector of  ADC.
AD is a diameter of the circle. If DAC  42,
(a) find ADC ,
(3 marks)
(b) find CAB .
(4 marks)
Figure 2
(a)
ACD  90
( in semicircle)
1A
In ACD,
DAC  ADC  ACD  180
42  ADC  90  180
ADC  48
(b)

( sum of )
1A
1A
BD is an angle bisector of ADC.
1
BDC  ADC
2
 24
CAB  BDC
1A
1A+1A
1A
(s in the same segment)
 24 
14. There are
n
rows of seats in a theatre.
It is known that the first row has 15 seats and each row has 2 seats
more than the preceding row, as shown in the Figure 3.
Figure 3
(a)
Find the number of seats in the nth row in terms of n, where n > 1.
(b)
Find the total number of seats in the first
(c)
If the number of seats in the theatre is 792, find the number of rows of seats in the
n
rows.
theatre.
(a)
(3 marks)
(1 mark)
(4 marks)
First term  15
1M
Common difference  2
T(n)  15 + (n  1)(2)
1M
 13 + 2n

F5 Final Exam 2017
1A
There are 13 + 2n seats in the nth row.
– 29 –
Page total
(b)
(b)
The total number of seats in first n rows
n
 (15  13  2n)
2
 n 2  14n
1A
1M
n 2 + 14n = 792
n 2 + 14n  792 = 0
(n + 36)(n  22) =0

n= 22 or 36 (rejected)

The number of rows of seats
1M + 1A
1A
in the theatre is 22.
Section B (34 marks)
Answer ALL questions in this section and write your answers in the spaces provided.
15.
Mary wants to buy 7 boxes of chocolate from a supermarket.
She chooses randomly from 9 boxes of dark chocolate, 5 boxes of milk chocolate & 8 boxes of white chocolate.
(a) Find the probability that only dark chocolate are chosen.
(2 marks)
(b) Find the probability that 2 boxes of dark chocolate, 3 boxes of white chocolate and 2 boxes of white
chocolate are chosen.
(2 marks)
(c) Find the probability that she chooses exactly 4 boxes of milk chocolate.
(a)
P(only dark chocolate are chosen)
=
(b)
𝐶79
𝐶722
=
3
14212
= 0.000211089
= 0.000211(cor. to 3 sig. fig.)
1M+1A
P(2 boxes of dark chocolate, 3 boxes of milk chocolate and 2 boxes of white chocolate are chosen)
=
𝐶29 ×𝐶35 ×𝐶28
𝐶722
=
10080
210
= 3553
170544
=0.059104981
=0.0591(cor. to 3 sig. fig.)
(c)
1M+1A
P(exactly 4 boxes of milk chocolate)
=
=
𝐶45 ×𝐶317
𝐶722
3400
170544
25
= 1254
=0.019936204
=0.0199(cor. to 3 sig. fig.)
16.
(2 marks)
1M+1A
The table below shows the means and standard deviations of the time that a large group of students used to
finish a 100m run in 2 fitness tests:
Test
Mean
Standard Deviation
I
20s
2s
II
18s
1s
The standard score of Billy in Test I is 1.5 .
F5 Final Exam 2017
– 30 –
Page total
(a) Find the time that Billy used to finish the run in Test I.
(3 marks)
(b) Assume that the time distributions in each of the above tests are normally distributed.
Billy uses 20s to finish the race in Test II.
He claims that comparing to other students, he performs better in Test II than that in Test I.
Is his claim correct?
16.
(a)
Explain your answer.
(3 marks)
Let Ts be the time that Billy used to finish the run in test 1.
𝑇−20
2
= 1.5
1M
T = 23
2A
Thus Billy used 23 s to finish the run in test 1.
(b) Standard score of Billy in test II =
20−18
1
= 2 > 1.5
1M
No, I disagree with his claim.
(Because the shorter the time, the better the performance)
2A
17. The Figure 4 shows the graph of log y against for the relationship log y  mx  c , where m and c are constants.
The graph passes through (0, 5) and (6, 2).
(a)
Find the values of
m
and
c.
(b)
Hence express y in terms of x .
(2 marks)
(2 marks)
Figure 4
F5 Final Exam 2017
– 31 –
Page total
(a)
1M
When x = 0, log y = 5.
5  m(0)  c
c5
When x = 6, log y = 2.
2  6m  5
m
1
2
(b) ∴ log y  
1A
1
x5
2
1M
from (a)
1A
x
 5
2
y  10
F5 Final Exam 2017
– 32 –
Page total
18.
In Figure 5, ABC is a triangular paper card. D is a point lying on AB such that CD is perpendicular to AB.
It is given that AC = 15cm, CAD  40 and CBD  25 .
Figure 5
(a)
Find AD and BD.
(b)
The triangular paper card in Figure 5 is folded along CD such that ABD lies on the horizontal
plane as shown in Figure 6.
(3 marks)
Let N be a point on AB such that DN is the shortest distance between
D and AB.
Figure 6
(i)
If the distance between A and B is 15cm, find DAB and DN.
(ii)
Let P be a movable point on AB. Describe how CPD varies as P moves
from A to B. Explain your answer.
(5 marks)
18.
(a)
AD
15
= cos40°
1M
AD = 15cos40°
= 11.5 cm
1A
CD = 15sin40°
CD
BD
= tan25°
BD =
15sin40°
tan25°
= 20.676937
= 20.7 cm
1A
(b)
BD2 = AD2 + AB2 − 2(AD)(AB)cos∠DAB
F5 Final Exam 2017
– 33 –
Page total
20.6769372 = (15cos40°)2 + (15)2
(i)
− 2(15cos40°)(15)cos∠DAB
∠DAB = 102°
(b)
(ii)
19.
1M
1A
∵ ∠DAB is an obtuse angle. ∴ D is at A.
The required distance = AD = 11.5 cm
CD
Note that tan∠CPD = PD,
1M
When P moves from A to B, PD increases from
AD(11.5cm) to BD(20.7cm).
Thus, ∠CPD decreases as P moves from A to B.
1A
1A
or consider
CD
sin∠CPD = PC
In a city, the controller of an incinerator P predicts that P can handle waste of weight W(n) tonnes
in the nth year since the start of its operation, where n is a positive integer.
It is given that W (n)  k  1000 a n , where k and a are positive constants.
It is found that the weights of waste handled by P in the 1st year and the 2nd year since the start of its
operation are 363 900 tonnes and 364 650 tonnes respectively.
(a)
(i)
Find
(ii)
Starting from which year will the weight of waste that P needs to handle increase by
a
and
k.
more than 40% annually?
(b)
(i)
(4 marks)
Express, in terms of n, the total weight of waste handled by P in the first n years since the
start of its operation.
(ii)
Find the total weight of waste handled by P in the first 18 years since the start of its operation,
correct to the nearest tonne.
(c)
(3 marks)
It is given that the upper limit of the weight of waste that P can handle in the 1st year is
700 000 tonnes, and this limit rises by 100 000 tonnes each year afterward.
Suppose the limit can rise continually. The controller claims that in the 19th year and
thereafter, the weight of waste exceeds the limit that P can handle each year.
Do you agree? Explain your answer.
F5 Final Exam 2017
(3 marks)
– 34 –
Page total
19. (a)(i)
When n = 1,
k + 1 000a = 363 900 ………. (1)
When n = 2,
k + 1 000a2 = 364 650 ……… (2)
(2)  (1):
1 000a2  1 000a = 750
4a2  4a  3
=0
(2a  3)(2a + 1)
=0
a
= 1.5
or
1M
0.5 (rejected)
Substitute a = 1.5 into (1).
k + 1 000(1.5) = 363 900
1A
k = 362 400
(ii)
W(n) = 362 400 + 1 000(1.5)n
W (n)  W (n  1)
 100%
W (n  1)
> 40%
362 400  1 000(1.5) n  [362 400  1 000(1.5) n  1 ]
362 400  1 000(1.5) n 1
1 000(1.5)  1 000(1.5)
n
n1
1 000(1.5)n  1(1.5  1.4)
1.5
n 1
100%
> 40%
> 144 960 + 400(1.5)
n1
1M
> 144 960
> 1 449.6
(n  1)log 1.5
> log 1 449.6
n  1 > 17.952 328 79
n
∴
> 18.952 328 79
Starting from the 19th year, the weight of waste that P needs to handle
1A
will increase by more than 40% annually.
(b)(i)
Total weight of waste handled by P in the first n years
= [(k + 1 000a) + (k + 1 000a2) + … + (k + 1 000an)] tonnes
= [nk + 1 000(a + a2 + … + an)] tonnes

1 000a(a n  1) 
nk



a 1
 tonnes
= 

1 000(1.5)(1.5n  1) 
362
400
n



1.5  1


=
tonnes
1M
= [362 400n + 3 000(1.5  1)] tonnes
n
1A
F5 Final Exam 2017
– 35 –
Page total
(ii)
Total weight of waste handled by P in the first 18 years
= [362 400(18) + 3 000(1.518  1)] tonnes
 10 953 875.64 tonnes
= 10 953 876 tonnes, cor. to the nearest tonne
1A
F5 Final Exam 2017
– 36 –
Page total
(c)
W(19)
= 362 400 + 1 000(1.5)19
 2 579 237.82
Upper limit of the weight of waste in the 19th year
1M
= [700 000 + (19  1)(100 000)] tonnes
= 2 500 000 tonnes
< 2 579 237.82 tonnes
∴ The weight of waste exceeds the limit that P can handle in the 19th
∵
year.
2 579 237.82  40%
1M
= 1 031 695.128
> 100 000
∴ By (a)(ii), the weight of waste exceeds the limit that P can handle in
∴
F5 Final Exam 2017
every subsequent year after the 19th year.
The claim is agreed.
– 37 –
1A