![Solution to Quiz #5<br />Payments of PhP2500 and PhP6000 are due two and four years from today. They are to be replaced by two payments due one year and three years from today. The first payment is to be twice the amount of the second payment. What is the size of the first replacement payment if money can earn 8% compounded quarterly?<br />Timeline:<br />FV2 = 6000FV1 = 2500<br />43210<br />FV3 = ??FV4 = ??PVtotal = ??<br />Conditions:-Replace FV1 with payment #1 (FV3) on Year 1-Replace FV2 with payment #2 (FV4) on Year 3Given:<br />FV1 = 2500<br />FV2 = 6000<br />1. Let PVtotal = PV1 + PV2a) Find PV1PV1 = FV1 (1 + i)= 2500 (1 + 0.02)= 2500 (1.02)= 2133.73b) Find PV2PV2 = FV2 (1 + i)= 6000 (1 + 0.02)= 6000 (1.02)= 4370.67c) PVtotal = 2133.73 + 4370.67= 6504.40d) FVnew = PVtotal (1 + i)= 6504.40 (1 + 0.02)= 7040.572. Let FVnew be PVnew Where: PVnew = PV3 + PV4PV3 = 2 FV4PV4 = FV4 (1 + i)Thus:PVnew = 2 FV4 + FV4 (1 + i)FV4 = PVnew / 2 + [(1 + i) ]m = 4 <br />n1 = 4 x 2 = 8<br />-nn2 = 4 x 4 = 16<br />-n-8-8j = 8%<br />-n-n-n<br />-16-16<br />-n-83. Solve for FV4 FV4 = PVnew / 2 + [(1 + i) ]= 7040.57 / 2 + [(1 + 0.02) ]= 7040.57 / 2 + 0.8534904= 7040.57 / 2.8534904= 2467.354. Solve for FV3FV3 = 2FV4= 2 x 2467.35= 4934.70<br />4n<br />Answer:Payment #1 on Year 1 = PhP 4934.70Payment #2 on Year 3 = PhP 2467.35<br />](https://image.slidesharecdn.com/solutiontoquiz04-101209212807-phpapp02/85/Solution-to-quiz-04-1-320.jpg)
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Solution to Quiz #5 on compound interest payments
- 1. Solution to Quiz #5<br />Payments of PhP2500 and PhP6000 are due two and four years from today. They are to be replaced by two payments due one year and three years from today. The first payment is to be twice the amount of the second payment. What is the size of the first replacement payment if money can earn 8% compounded quarterly?<br />Timeline:<br />FV2 = 6000FV1 = 2500<br />43210<br />FV3 = ??FV4 = ??PVtotal = ??<br />Conditions:-Replace FV1 with payment #1 (FV3) on Year 1-Replace FV2 with payment #2 (FV4) on Year 3Given:<br />FV1 = 2500<br />FV2 = 6000<br />1. Let PVtotal = PV1 + PV2a) Find PV1PV1 = FV1 (1 + i)= 2500 (1 + 0.02)= 2500 (1.02)= 2133.73b) Find PV2PV2 = FV2 (1 + i)= 6000 (1 + 0.02)= 6000 (1.02)= 4370.67c) PVtotal = 2133.73 + 4370.67= 6504.40d) FVnew = PVtotal (1 + i)= 6504.40 (1 + 0.02)= 7040.572. Let FVnew be PVnew Where: PVnew = PV3 + PV4PV3 = 2 FV4PV4 = FV4 (1 + i)Thus:PVnew = 2 FV4 + FV4 (1 + i)FV4 = PVnew / 2 + [(1 + i) ]m = 4 <br />n1 = 4 x 2 = 8<br />-nn2 = 4 x 4 = 16<br />-n-8-8j = 8%<br />-n-n-n<br />-16-16<br />-n-83. Solve for FV4 FV4 = PVnew / 2 + [(1 + i) ]= 7040.57 / 2 + [(1 + 0.02) ]= 7040.57 / 2 + 0.8534904= 7040.57 / 2.8534904= 2467.354. Solve for FV3FV3 = 2FV4= 2 x 2467.35= 4934.70<br />4n<br />Answer:Payment #1 on Year 1 = PhP 4934.70Payment #2 on Year 3 = PhP 2467.35<br />