A certain library assesses fines for overdue items as follows. On the first day the book is overdue, the total fine is $0.10. For each additional day the book is overdue, the total fine is either increased by $0.30 or doubled, whichever results in the lesser amount. What is the total fine for a book on the fourth day it is overdue?
A) $0.60
B) $0.70
C) $0.80
D) $0.90
E) $1.00
I know this seems like a very easy problem and I thought it would be but maybe I'm missing something very simple. This is how I worked it out:
D1 = $0.1
D2 = (0.1*2) or 0.30. Since $0.2 <$0.30 D2 - $0.2
D3 = (0.2*2) or 0.30. Since $0.30<0.40 D3 - $0.3
D4 = (0.3*2) or 0.30. Since 0.30<0.60. D4 - $0.3
Therefore, the total fine on the 4th day overdue is 0.1+0.2+0.3+0.3 = $1.00
This is obviously wrong but I'm not sure how to get to the right answer. Please help!
Thanks!
Simple Problem but tough to solve!
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sakhter wrote:A certain library assesses fines for overdue items as follows. On the first day the book is overdue, the total fine is $0.10. For each additional day the book is overdue, the total fine is either increased by $0.30 or doubled, whichever results in the lesser amount. What is the total fine for a book on the fourth day it is overdue?
A) $0.60
B) $0.70
C) $0.80
D) $0.90
E) $1.00
I know this seems like a very easy problem and I thought it would be but maybe I'm missing something very simple. This is how I worked it out:
D1 = $0.1
D2 = (0.1*2) or 0.30. Since $0.2 <$0.30 D2 - $0.2
D3 = (0.2*2) or 0.30. Since $0.30<0.40 D3 - $0.3
D4 = (0.3*2) or 0.30. Since 0.30<0.60. D4 - $0.3
Therefore, the total fine on the 4th day overdue is 0.1+0.2+0.3+0.3 = $1.00
This is obviously wrong but I'm not sure how to get to the right answer. Please help!
Thanks!
Solution:
On day 1, total fine is $0.1.
On day 2, total fine is min($0.2, $0.4) = $0.2.
On day 3, total fine is min($0.4, $0.5) = $0.4.
On day 4, total fine is min($0.8, $0.7) = $0.7.
The correct answer is B.
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