SOLUTION: At 9:00am on the super sale day at Clothing city, Amy saw the coat she wants priced at $65.00. Amy only has $65.00. Every hour the price on the coats will be reduced 10% from the p

Algebra ->  Test -> SOLUTION: At 9:00am on the super sale day at Clothing city, Amy saw the coat she wants priced at $65.00. Amy only has $65.00. Every hour the price on the coats will be reduced 10% from the p      Log On


   

Question 42303: At 9:00am on the super sale day at Clothing city, Amy saw the coat she wants priced at $65.00. Amy only has $65.00. Every hour the price on the coats will be reduced 10% from the previous hour's price.
1. At what time will Amy be able to buy the coat for $46.00 or less, provided the coat is still avaliable?
2. Explain in detail how you found your anwser.
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
At 9 am the price is $65.00.

Each hour the price decreases by 10% of its value in the previous hour.

After red%281%29 hr the price will be $(65.00 - 0.1x65.00) = $(0.9%5E%28red%281%29%29x65.00).

So after 2 hours the price will be 10% less than the price after 1 hour.
After red%282%29 hrs the price will be $(0.9x65.00 - 0.1x0.9x65.00) = $(0.9%5E%28red%282%29%29x65.00).

Thus, after red%283%29 hrs the price will be $(0.9%5E2x65.00 - 0.1x0.9x0%5E2.1x65.00) = $(0.9%5E%28red%283%29%29x65.00).

Similarly, after red%28n%29 hrs the price will be $(0.9%5E%28red%28n%29%29x65.00).

[Note the similarity between the numbers marked in red ink]

Let us assume that the price becomes $46 or less after 'n' hours from 9 am.

Then, 0.9%5Enx65+%3C+46
or 0.9%5En+%3C+46%2F65
or 0.9%5En+%3C+0.708 ______(1)

By trial,
when n = 2, LHS = 0.81 > RHS
when n = 3, LHS = 0.729 > RHS
when n = 4, LHS = 0.6561 < RHS

These value of 'n' satisfies the inequation (1) and so Amy will be able to purchase the dress at price below $46 after 4 hrs from 9 am.
Then the price of the dress will be $(0.9%5E4x65.00) = $42.65 (approx).