Unformatted text preview: Jaw EXERCISES 5.4 99 I  4 360
4 >330 P (1+H—1200) —1 150000 (1 + w 4/1200 :> P = 716.12. 1200 Monthly payments are $716.12. The amount of interest paid is 360(716.12) H 150000 = 107 803.20: If we let the remaining
principal after 15 years he B, then the accumulated value of B over the next 15 years (the
remaining time on the mortgage) must be equal to the value of the remaining 180 payments
of the mortgage at the time the mortgage is paid off. When we equate these two quantities, 4 130
B 1 4 180 716.12 [(1 + 120.0) — 1] .
. ( + %) _ F—MlW— => B m 00 813.80. (a) We equate the accumulated value of 30000 after 15 years to the value of an annuity of
180 payments of $P. At 8%, we obtain 8 130 P (1+ #1200) — 1]
30 000 (1 + 1200) =  8/1200 'Ibtal interest paid is 180(286.70) — 30 000 = 21600.00.
(b) We equate the accumulated value of 30 000 after 15 years to the value of an annuity of
180 payments of $P. At 7%, we obtain => P = 28670. 7 130
7 180 P [(1 + 1—200.)  I]
00 000 (1 + 1200.) = W— :> P = 260.65.  I'Ibtal interest paid is 180(260.05) — 30000 = 18537.00. 9. If P represents the amount of the deposit, then the value of P after 4 years must be equal
to the value of an annuity of 4 payments of $1500. At 4%, we obtain .(Hiygmt<l+nri = .4.
100 4/100 I? P 54448 . If P represents the amount of the deposit, then the value of P after 121 months must be equal to the value of an annuity of 120 payments of $500. At 5%, we obtain 5 130.
p (1 + ”1200) = 5,1200 =5 P = 46 5.07. (a) With 48 monthly payments of $284.58, the amount of interest paid is 48(28458) —
12 000 = 1050.84. ’ (b) With 60 monthly payments of $229.22, the amount of interest paid is 60(2292‘2) —
12 000 = 1753.20. (c) We equate the accumulated value of 12000 after 4 years to the value of an annuity of 4
payments of $P. At 6%, we obtain 6 4
6 4 P(1+m)—}
12000(1+—) =————— 100 _ 6/100 => P = 3463.10. . 11.. n._:..._u__ c\:_:..l. 04.4‘... 12...“... Mat4:”. :n Huif n. :n ﬁlll in Mail” nrnhihiterl ...
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 Fall '11
 Dr.Cornell
 Math

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