SOLUTION: Phyllis invested
44000
44000
dollars, a portion earning a simple interest rate of
5
5
percent per year and the rest earning a rate of
6
6
percent per year. After one
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-> SOLUTION: Phyllis invested
44000
44000
dollars, a portion earning a simple interest rate of
5
5
percent per year and the rest earning a rate of
6
6
percent per year. After one
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Question 1155825: Phyllis invested
44000
44000
dollars, a portion earning a simple interest rate of
5
5
percent per year and the rest earning a rate of
6
6
percent per year. After one year the total interest earned on these investments was
2430
2430
dollars. How much money did she invest at each rate? Found 2 solutions by MathLover1, greenestamps:Answer by MathLover1(20850) (Show Source):
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if Phyllis invested $ so that dollars earning a simple interest rate of percent, and dollars earning a simple interest rate of percent, we have
.....eq.1
if after one year the total interest earned on these investments was $, we have
......eq.2...substitute from eq.1
......solve for
go to
.....eq.1, substitute
so, Phyllis invested $ dollars earning a simple interest rate of percent, and $ dollars earning a simple interest rate of percent
Here is a quick and easy way to solve two-part mixture problems like this informally, with just a few mental calculations. If the numbers in the problem are "nice", you can solve the problem in 15-20 seconds if you are good with mental arithmetic.
The strategy:
(1) Find the amounts of interest if the whole had been invested at each rate.
(2) Consider those three interest amounts on a number line. Find what fraction of the way from the lowest amount to the largest amount it is from the lowest amount to the actual amount. (The words sound confusing, but the actual calculations below should make it clear.)
(3) That fraction is the fraction of the total that should be invested at the higher rate.
Note the basic concept is simple. If the actual interest is 3/4 of the way from the lower amount to the higher amount, then 3/4 of the money was invested at the higher amount. If the actual amount is 1/10 of the way from the lower amount to the higher amount, then 1/10 was invested at the higher rate. And so on....
The actual calculations for this example use "nice" numbers, so the answer is easily found.
All $44000 at 5% -- $2200 interest
All $44000 at 6% -- $2640 interest
$2200 to $2640 is $440; $2200 to the actual amount $2430 is $230. So the fraction of the total that was invested at the higher rate is 230/440 = 23/44.
Since the total invested was $44000, it is easy to see that $23000 was invested at 6% and the other $21000 at 5%
