Prepared by Dr. Richard Ng

1. BBMP 1103
Mathematic Management
Exam Preparation Workshop Sept 2011
Part 1 - Compound Interest
Presented By: Dr Richard Ng
26 Nov 2011
2ptg – 4ptg
Prepared by Dr Richard Ng (2011) Page 1

2. Exam Format:
 Assignment – 30%
 Final Exam – 70%
Final Exam:
 Part A – 5 short questions (Answer All)
 Part B – 3 questions (Answer 2 only)
 Part C – 2 questions (Answer 1 only)
Exam Date:
 22 December 2011 (Tuesday) – 12.00pm to 2.30pm
Prepared by Dr Richard Ng (2011) Page 2

3. 1. Focus on Compound Interest
Question: 1 (September 2008)
Prepared by Dr Richard Ng (2011) Page 3

4. Suggested Answers:
nk
a) Given: r
S P1
k
P = 20,000
( 20 )( 2)
r = 10% = 0.1 0.1
S 20000 1
2
n = 20
k=2 200001.05 40
200007.03999
140 ,799 .77
Prepared by Dr Richard Ng (2011) Page 4

5. b) Given: nk
r
S P1
S = 80,000 k
( 25)( 4)
r = 10% = 0.1 0.1
80000 P 1
n = 25 4
k=4 80000 P 1.025100
80000 P 11.81372
80000
P 6,771.79
11.81372
Prepared by Dr Richard Ng (2011) Page 5

6. c) Given: nk
r
S P1
P = 40,000 k
n ( 4)
S = 100,000 0.08
100000 40000 1
r = 8% = 0.08 4
k=4 2.5 1.02 4n
ln 2.5 (4n) ln 1.02
ln 2.5 0.9162907
4n 46.27
ln1.02 0.0198026
n 11.57 or 11 years and 7 mths
Prepared by Dr Richard Ng (2011) Page 6

7. Question:2 (May 2010)
Prepared by Dr Richard Ng (2011) Page 7

8. Suggested Answers:
nk
a) Given: r
S P1
k
P = 480,000
(10 )(12 )
r = 12% = 0.12 0.12
S 4800001
12
n = 10
k = 12 480000 .01120
1
4800003.300387
1584185
.71
Prepared by Dr Richard Ng (2011) Page 8

9. b) Given:
nk
r
S P1
S = 465,308.35 k
r = 9% = 0.09 ( 20 ( 2)
0.09
n = 20 46530835 P 1
.
2
k=2
46530835 P 1.045 40
.
46530835 P 5.81636
.
46530835
.
P 7999992
.
5.81636
Prepared by Dr Richard Ng (2011) Page 9

10. nk
c) Given: r
S P1
k
P = 1,500,000
n (1)
0.08
S = 4,500,000 4500000 15000001
1
r = 8% = 0.08
3 1.08 n
k=1
ln 3 ln 1.08 n
ln 3 (n) ln1.08
ln 3 1.0986
n 14.27
ln1.08 0.0769
n = 14 years and 3 months
Prepared by Dr Richard Ng (2011) Page 10

11. Question: 3 (January 2011)

12. nk
r
a) Given: P = 180,000 S P1
k
r = 12% = 0.12 ( 7 )( 4)
0.12
1800001
k=4 4
n=7 180000 .03 28
1
1800002.287928
[ ]
411,826 .98
Hence, the compounded value is => RM411,826.98

13. nk
b) Given: S = 285,000 r
S P1
k
r = 8% = 0.08 (30 ( 2)
0.08
285000 P 1
k=2 2
n = 30
285000 P 1.04 60
285000 P[10.519627
]
285000
P
10.519627
P 27 ,092 .22
Hence, the amount of investment is => RM27,092.22

14. nk
c) Given: P = 200,000 r
S P1
k
S = 700,000 n (12 )
0.12
700000 2000001
12
r = 12% = 0.12
3.5 1.01 n(12)
k = 12
ln 3.5 ln 1.01 12n
ln 3.5 12 n ln 1.01
ln 3.5 1.25276
12n 125.91
ln1.01 0.00995
Hence, the number of years is => 125.91 or 125 years and
11 months

15. End of
Part 1