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BBMP1103 - Sept 2011 exam workshop - part 1

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Prepared by Dr. Richard Ng

Prepared by Dr. Richard Ng

Published in: Education, Economy & Finance

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  • 1. BBMP 1103 Mathematic Management Exam Preparation Workshop Sept 2011 Part 1 - Compound Interest Presented By: Dr Richard Ng 26 Nov 2011 2ptg – 4ptgPrepared 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 2.30pmPrepared 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 .77Prepared 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 11.81372Prepared 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 mthsPrepared 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 .71Prepared 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 . 5.81636Prepared 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 monthsPrepared by Dr Richard Ng (2011) Page 10
  • 11. Question: 3 (January 2011)
  • 12. nk ra) 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. nkb) 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. nkc) 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 ofPart 1