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# Answer to Question #6021 in Economics of Enterprise for LaMarcus Streeter

Question #6021
John and Daphne are saving for their daughter Ellen's college education. Ellen just turned 10 at (t = 0), and she will be entering college 8 years from now (at t = 8). College tuition and expenses at State U. are currently $14,500 a year, but they are expected to increase at a rate of 3.5% a year. Ellen should graduate in 4 years--if she takes longer or wants to go to graduate school, she will be on her own. Tuition and other costs will be due at the beginning of each school year (at t = 8, 9, 10, and 11). So far, John and Daphne have accumulated$15,000 in their college savings account (at t = 0). Their long-run financial plan is to add an additional $5,000 in each of the next 4 years (at t = 1, 2, 3, and 4). Then they plan to make 3 equal annual contributions in each of the following years, t = 5, 6, and 7. They expect their investment account to earn 9%. How large must the annual payments at t = 5, 6, and 7 be to cover Ellen's anticipated college costs? a.$1,965.21
b. $2,068.64 c.$2,177.51
d. \$2,292.12

Explanation:

they need for studying:
14 500 * ((1.035)^8) + 14 500 * ((1.035)^9) + 14 500 * ((1.035)^10) + 14 500 * ((1.035)^11) = 80 479
Tuition the beginning of t=8: 14 500 * ((1.035)^7) = 18 448.05
Tuition the beginning of t=9: 14 500 * ((1.035)^8) = 19 093.73
Tuition the beginning of t=10: 14 500 * ((1.035)^9) = 19 762.01
Tuition the beginning of t=11: 14 500 * ((1.035)^10) = 20 453.68

Before the studying begins:
from accumulations they have: 1 500 * 1.09^7 = 27 420.59
from regular additions: 5 000 * 1.09^7 + 5 000 * 1.09^6 + 5 000 * 1.09^5 + 5 000 * 1.09^4 = 32 276.7

Total = 59 697.3

Cash flows during studying:
[table] they have tuition left (year&#039;s beginning)result ((year&#039;s end)*1.09)859697,3118448,0541249,2644961,6934944961,6919093,7325867,96328196,079671027196,0819762,017434,078103,1363118103,13620453,68-12350,5437 [/table]
So we have to save 9 536.8858 additionally before the study begins (12 350.5437/(1,09^3)) by 3 equal contributions: X*(1,09^3) + X*(1,09^2) +
X*1,09 = 9 536.8858, X = 2 292.12

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