Arithmetic and Geometric(2)

Tieria

2009-11-15 7:47 am

1. Carmen bought a stamp from a stamp shop, and the value of the stamp now is $200. It is given that the value of the stamp increases by 4% every month.
(a) Find the value of the stamp after one year.
(Give the answer correct to 3 significant figures,)
(b) If Carmen decides to sell the stamp when its value just exceeds $1000, after how many months from now will she sell the stamp?

The answers are (a) $320 and (b) 42 months.

2. The annual depreciation rates for car A and car B are 10% and 20% respectively. If the values of car A and car B after one year (from now) will be $120000 and $230000 respectively, find
(a) the values of car A and car B after n years and express the answers in terms if n,
(b) the smallest value of n such that the value of car A exceeds that of car B after n years.

The answers are (a) car A: $120000(0.9)^(n-1), car B: $230000(0.8)^(n-1) and (b) 7.

回答 (1)

用戶9112

2009-11-15 8:04 am

✔ 最佳答案

(a) The value of the stamp after one year = $200 * (1 + 4%)^12
= $200 * 1.04^12
= $320 (Correct to 3 significant figures,)
(b) Let the number of months required be n
$200 * (1.04)^n > $1000
1.04^n > 5
nlog(1.04) > log 5
n > log5/log(1.04) = 41.04
She needs to wait for 42 months
2. (a) Value of car A now = $120000 / 0.9
Value of car A n years later = $120000 / 0.9 * 0.9^n
= $120000(0.9)^(n-1)
Value of car B now = $230000 / 0.8
Value of car B n years later = $230000 / 0.8 * 0.8^n
= $230000(0.8)^(n-1)
(b) $120000(0.9)^(n-1) > $230000(0.8)^(n-1)
(0.9/0.8)^(n-1) > 230000/120000
(n-1)log(0.9/0.8) > log(23/12)
n - 1 > log(23/12)/log(0.9/0.8) = 5.52
n > 6.52
The minimum number of years is 7

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