SOLUTION: jeff invests $5500 in two different accounts. the first account paid 11%, thr second account paid 9% in intrest. at the end of the first year he had earned $525 in intrest. how muc
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Question 1147775: jeff invests $5500 in two different accounts. the first account paid 11%, thr second account paid 9% in intrest. at the end of the first year he had earned $525 in intrest. how much was in each account? Answer by greenestamps(13209) (Show Source):
The two rates are 11% (0.11) and 9% (0.09)
The two amounts are, respectively, x and 5500-x
The two amounts of interest are 0.11(x) and 0.09(5500-x)
The total interest is $525:
Solve using basic algebra (I leave that to you)
A much easier and faster way to solve "mixture" problems like this:
(1) Determine the amounts of interest if the whole amount was invested at each rate.
(2) Determine where the actual amount of interest lies between those two amounts. That exactly determines how the total must be split between the two rates.
$5500 all at 11% yields $605 interest
$5500 all at 9% yields $495 interest
$525 is 3/11 of the way from $495 to $605 (525-495 = 30; 605-495 = 110; 30/110 = 3/11)
That means 3/11 of the total was at the higher rate.
ANSWER: 3/11 of the $5500, or $1500, at 11%; the other $4000 at 9%
