INTRODUCTORY MATHEMATICS AND STATISTICS 6TH EDITION SOLUTIONS MANUAL John S. Croucher Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd2 Solutions Manual, Chapter 5 Chapter 5 Simple interest Quick Qui z 1. (d) 2. (a) 3. (b) 4. (c) 5. (c) 6. (c) 7. (d) 8. (c) 9. (a) 10. (b) Practice Questions 1.Total interest earned =$1125 2.Rate =3.75% per annum 3.Rate =120% per annum 4.Flat rate of interest =16.67% 5.Effective rate of interest =32% 6.Equivalent rate of interest =7.56% 7.Flat rate of interest =4.1% 8.Annual rate of simple interest =3.2% 9.Effective rate/flat rate =2 ( 1) ( 1) 111 1 1N N N NN N N+ + = = ++ + + 1 since number of repayments is at least 1 thus effective rate flat rate 10. Reserve Bank of Australiahttp://www.rba.gov.au/ Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd3 Solutions Manual, Chapter 5 Exercises 5.1(a) Simple interest =$4000 0.08 4 =$1280 (b) Simple interest =$12 500 0.06 5 =$3750 (c) Simple interest =$25 000 0.07 3 =$5250 5.2(a) Rate of simple interest =$100 =0.02 or 2% $1000 5 (b) Rate of simple interest = $1600 0.1066 or 10 23% $5000 3 (c) Rate of simple interest =$312.50=0.025 or 2.5% $2500 x 5 5.3(a) r =3.25%(b) r =7.6%(c) r =9.5% 5.4(a) Period = $300 =2 years $3000 0.05 (b) Period =$1023.75 =3.5 years $6500 0.045 or 3 years 6 months (c) Period =$12 397.50=7.25 years $18 000 0.095 or 7 years 3 months 5.5(a) T =3 years 4 months (b) T =5 years 10 months (c) T =6 years 5.6(a) Total simple interest =$600 0.075 1.25 =$56.25 (b) Maturity value of principal =$600 +$56.25 =$656.25 5.7(a) Total interest =$845.25 (b) Maturity value =$3145.25 5.8Amount to invest = $3000 $1804.51 1 +0.1325 5 5.9P = $100001 0.1325 8 + =$4854.37 Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd4 Solutions Manual, Chapter 5 5.10 MonthInterest Due ($) March 20130 April 20130.61875 May 20130.61875 J une 20130.87083 Total2.11 5.11 MonthInterest Due ($) March 20130 April 20131.04 May 20131.04 J une 20131.46 Total$3.54 5.12(a) Total simple interest =$3500 0.0625 3 =$656.25 (b) Equivalent effective rate of interest = 2 36 0.0625 0.1216 or 12.16% 36 +1 5.13Equivalent flat rate of interest =181 0.14 =0.070 or 7.0% 360 5.14Equivalent flat rate of interest =181 0.185 =0.093 or 9.30% 360 5.15R =31 0.14 =0.072 or 7.2% 60 5.16(a) Total cost of sewing machine =($38.50 24) +(0.16 $1000) =$924 +$160 =$1084 (b) Total interest paid =$1084 - $1000 =$84 (c) Flat rate of interest =$84 =0.050 or 5% $840 2 (d) Effective rate of interest = 2 24 0.050=0.096 or 9.6% 24 +1 5.17(a) Total cost =($12.25 48) =$588 (b) Total interest paid =$588 $500 =$88 (c) Flat rate of interest =88=0.044 or 4.4% 500 4 (d) Effective rate of interest =2 48 0.044=0.0862 or 8.62% 48 +1 Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd5 Solutions Manual, Chapter 5 5.18Minimum annual rate of interest = $1250=0.0893 or 8.93% $3500 4 5.19Time = $15 000=6.32 years $25 000 0.095 5.20(a) Equivalent flat rate of interest =33 0.155 =0.08 or 8.0% 64 (b) Total simple interest =$20 000 0.08 8 =$ 12 800 5.21Time = $ 5312. 50=4.25 years =4 years 3 months $20 000 0.0625 5.22Time = 1 =13.33 years or 13 years 4 months 0.075 5.23Rate of interest =$2250=0.0375 or 3.75% per annum $5000 (12) 5.24(a) Total simple interest =$64 000 0.085 6 =$32 640 (b) Effective rate of interest = 2 72 0.085=0.168 or 16.8% 73 (c) Total cost =$16 000 +$64 000 +$32 640 =$112 640 5.25Annual interest rate =73% 5.26(a) Total interest earned=$4 0000.05 120365 =$65.75 (b) Total amount repaid =$4 065.75 5.27(a)23 Sep 2013 (b) Total interest =$149.44 (c) Repayment =$91.61 per week 5.28(a) Total repayment =3240 (1 +0.05 259365 ) =$3354.95 Repayment per week =$3354.95 =$90.67 37 (b) Total payment =$3354.95 +$360 =$3714.95 Under the deal in Exercise 5.27 the customer would pay a total of $3749.44. Hence this new deal is slightly better since it would save $34.49 in interest. 5.29(a) Amount required =$8268.73 Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd6 Solutions Manual, Chapter 5 (b) Repayment =$1378.12 per month 5.30(a) Let total interest to borrow =$I. Then I =(25 000 +I) 0.065 1.5 which yields I =2 700.83. The consultant should borrow $27 700.83. (b) Monthly repayments =$27 700.83 =$1538.94 18 5.31(a) Interest earned =$50 000 (0.0725)(4.5) =$16 312.50 (b) Maturity value =$50 000 (1 +(0.0725)(4.5)) =$66 312.50 5.32Equivalent flat rate =7.04% 5.33Flat rate of interest = 301 0.1758.78%600= 5.34(a) Total interest =$108 000 (b) Equivalent effective rate =13.09% 5.35(a) Total simple interest =$75 000 10 0.055 =$41 250 (b) Effective rate of interest =40 0.055 =0.1048 or 10.48% 21 5.36(a) Flat rate interest =$14 250 Effective rate of interest =$13 837.50 (b) Take the effective rate, since the total interest payable is lower. 5.37(a) Flat rate of interest =6.5% Total simple interest =$15 000 0.065 5 =$4875 Effective rate of interest =12% Equivalent flat rate of interest =21 0.12 =0.063 40 Total simple interest =$15 000 0.063 5 =$4725 (b) Therefore the hairdresser should take the effective rate of interest of 12% as the total simple interest due is lower. Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd7 Solutions Manual, Chapter 5 5.38(a) (b) Total interest =$7.47 5.39(a) MonthMinimum monthly balanceInterest ($) September 201200 October 20125002.29 November 20125002.29 December 20123201.47 J anuary 20133201.47 February 20133201.47 March 20135202.38 (b) Total interest =$11.37 5.40No. The total interest will be $2700, so she will have only a total of $11 700, still $300 short of her goal. 5.41Simple interest =$28 500 0.006 7.5 =$1282.50 5.42Amount Emily should invest now =$25 000=$12 658.23 1 +0.065 15 5.43(a) Total simple interest =$20 000 0.08 10 =$16 000 (b) Equivalent effective rate of interest =2 120 0.08 =0.159 or 15.90% 120 +1 5.44In this case, E =0.11, R =0.0645 and N is the unknown. N = E =0.11=5.79 2R E2(0.0645) 0.11 Hence, 5.79 repayments would have to be made. This would be adjusted to6 payments. MonthMinimum balance ($) Interest ($) J an00 Feb1500.41 Mar1500.41 Apr1500.41 May6501.76 J une3250.88 J uly3250.88 Aug3250.88 Sept3250.88 Oct3550.96 Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd8 Solutions Manual, Chapter 5 5.45Same as the answer to Exercise 5.44. The loan amount is irrelevant. 5.46Amount Angela should invest now =$20 000= $15 625.00 1 +0.035 8 5.47In this case, S =$25 000, R =0.048 and T =8. P =$25 000= $25 000 =$18 063.58 1 +0.048 81.384 Hence she should invest $18 063.58 now in order to have $25 000 at the end of8 years. 5.48Annual interest payable =4 $400 =$1600 Annual interest rate =100 $1600 =160% $1000 5.49(a)Rate of simple interest =$600=1.8 or 180% $1000 412 (b) No. The rate of interest in Exercise 5.48 is a better deal for the client because it is lower than the rate in Exercise 5.48. 5.50Answer depends on the model of financial calculator selected. 5.51P =$1900, R =2.5100 =0.025, T =8 I =P R T=$1900 13thly balan =$380 5.52Rate of simple interest =$299.20 (1360 8) =0.0275 or 2.75% 5.53R 2 6061=6% R =3.05% 5.54Simple interest earned =$30 000 0.0565 4 =$6780 5.55Check the answer is reasonable. 5.56Simple interest earned =$160 000 5.25% 144 =$35 700 5.57 Annual simple interest charged =$1953.13 ($25 000 1.25) =0.0625 or 6.25% Introductory Mathematics and Statistics 6e, John S Croucher, 2013 McGraw-Hill Education (Australia) Pty Ltd9 Solutions Manual, Chapter 5 5.58P =$400001 5.5% 16 + =$21276.6 5.59P =$18 000, T =0.50, I =$1012.50 R =

=1012.5018 000 0.50 =0.1125 or 11.25% 5.60(a) $31.80 (b) $31.80 6.75% 112 =$0.18