Simple interest and simple discount
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Discussion Questions 1 Conceptual Framework
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- Exact Interest Ie = (4,500)(0.12)(255/365) Ie = P377 Ordinary Interest
- Answer: P15,000
| Name: Usi, Judielyn Gia F. Section: A-123 Time: MWF 1:30 – 2:30 pm SIMPLE INTEREST AND SIMPLE DISCOUNT Answer the following questions. Please show your solutions (3 points each number) Determine the principal that would have to be invested to provide P750 simple interest at the end of 21 months if the interest rate is 8%. P = ? I = P750 t = 21/12 r = 8% or 0.08 P = 750 (0.08) (21/12) P = 750 0.14 P = P5,357 Shan deposits P50,000 in a savings account that pays interest at the rate of 10% per year. How much is the interest and what will be the total amount after three years? I = ? P = P50,000 r = 10% or 0.10 t = 3 years I = (50,000) (0.10) (3) I = P15,000 Block Company deposited P200,000 in a bank account on Oct 12 and withdraws a total of P225,000 exactly on October 12 of the next year. What is the annual interest rate at which the company was paid? r = ? P = P200,000 I = P25,000 t = 1 Year r = 25,000 (200,000)(1) r = 25,000 200,000 r = 0.125 or 12.5% At what annual interest rate is P600 one year ago equivalent to P800 today? r = ? P = P600 t = 1 Year I = P200 r = 200 (600)(1) r = 200 600 r = 0.3333 or 33.33% How long will it take for an investment of P5,000 to grow to P9,500 if it earns 12% simple interest per year? t = ? P = P5,000 I = P4,500 r = 12% or 0.12 t = 4,500 (5,000)(0.12) t = 4,500 600
Determine the simple interest earned if P4,500 is invested at 12% interest rate in 255 days, a) using exact interest b) using ordinary interest I = ? P = P4,500 r = 12% or 0.12 t = 255 days Exact Interest Ie = (4,500)(0.12)(255/365) Ie = P377 Ordinary Interest Io = (4,500)(0.12)(255/360) Io = P382.5 Find the exact interest on a 150-day loan of P50,000 if the interest rate is 10%? I = ? t = 150/365 P = P50,000 r = 10% or 0.10 Ie = (50,000)(0.10)(150/365) Ie = P2,054.79 or P2,055 Determine the exact number of days from March 15, 2011 to December 20, 2012. 2011: March 31-15 = 16 April = 30 May = 31 June = 30 July = 31 Aug = 31
Sep = 30 Oct = 31
Nov = 30 Dec = 31 291
Feb = 29
Mar = 31 April = 30 May = 31 June = 30 July = 31 Aug = 31
Sep = 30 Oct = 31
Nov = 30 Dec = 20 646 + 291 = 646 days Find the ordinary interest on a 160-day loan of P500,000 if the interest rate is 8%? I = ? t = 160/360 P = P500,000 r = 8% or 0.08 Io = (500,000)(0.08)(160/360) Io = P17,777.77 or P17,778 Determine the approximate number of days from Dec. 15, 2018 to Oct. 10, 2019. Dec – Jan = 30 Jan – Feb = 30 Feb – Mar = 30 Mar – April = 30 April – May = 30 May – June = 30 June – July = 30 July – Aug = 30 Aug – Sep = 30
D = FdT
D = (7,500)(0.10)(6/12) D = P375
How much interest will be deducted from a loan worth P30,000 after 2 years with a discount rate of 5%? How much will the proceeds of the loan be? D = ? P = ? F = 30,000 T = 2 Years d = 5% or 0.05 D = (30,000)(0.05)(2) D = P3,000 P = 30,000 – 3,000 P = P27,000 Discount P15,500 at a 12% simple discount for 9 months. D = ? F = P15,500 d = 12% or 0.12 t = 9/12 D = (15,500)(0.12)(9/12) D = P1,395 How much interest will be deducted from a loan worth P50,000 after 3 years with a discount rate of 7.5%? How much will the proceeds of the loan be? D = ? P = ? F = P50,000 T = 3 years d = 7.5% or 0.075 D = (50,000)(0.075)(3) D = P11,250 P = 50,000 – 11,250 P = P38,750 Dominic wants to borrow P15,000 payable in two years at a 10% discount rate. How much will Dominic receive on the origin date? How much will he pay on the maturity date? Answer: P15,000 P = ? F = P15,000 t = 2 Years d = 10% or 0.10 P = F (1-dt) P = 15,000 [1 – (0.10)(2)] P = P12,000 Download 51.92 Kb. Share with your friends:
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