A business purchases a computer system for $3000. If the value of the system decreases at a rate of 15% per year, how much is the computer worth after 4 years?
you multiply by .85 repeatedly, so 3000*.85^4 .85 because 100% = 1.00 and 15% decrease = 1.00 - .15 = .85
Wait huh? :I
what part is confusing, I'll try to make it make sense...
3000 * .85 = 2550 (after one year) 2550 * .85 = 2167.5 (after two years) but look, 2550 = 3000*.85, so after the second year can be found by: (3000*.85)*.85 which equals 3000*.85^2 so after 4 years, the exponent becomes a 4
Each year its value decrease by 15% of its original value. In 4 years, it will decrease by 4*15%=60%. That means it will cost just 40% of its original value. So, its value after 4 years is: (40/100)*3000=$1,200
No, the answer is not 1200
because it is not losing 15% of 3000 each year. the second year, you lose 15% of a smaller number.
OMG, Stick with tonks :)
By AnwarA's calculation, you would have no money after 8 years, but you can lose 15% for ever
Aha, I'm still trying to figure the problem out myself
Monie, does using .85 make sense?
Yea it does
okay, then do it out slowly instead of any shortcuts
But why .85?
oh, if there were no change, you would have 100% the next year. Since there is a 15% decrease, you have 15% less than 100%, or 85% next year. Then move the decimal over to turn a percent into something you can multiply with for .85
I end up with 0.7225...Is that correct? .-.
hmmm, 3000*.85 = year 1 (3000*.85)*.85 = year 2 ((3000*.85)*.85)*.85 = year 3 (((3000*.85)*.85)*.85)*.85 = year 4 = 3000*.85*.85*.85*.85 = ...
do you see why each step works? What do you get?
I get 1556.01875.... But I'm guessing it to be 1556? Correct?
I think you should keep it to the penny, 1556.02
yes, do you get the .85 and the repeated multiplying?
Yes I do :). THank you so very much for helping me Tonks.
ur welcome
One more question if you don't mind? (:
If it is a math problem, post it as a new thread
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