SOLUTION: Will someone please help me with this problem?A stockbroker has money in three accounts. The interest rates on the three accounts are 9%, 10%, and 11%. If she has twice as much mo

Question 76645: Will someone please help me with this problem?A stockbroker has money in three accounts. The interest rates on the three accounts are 9%, 10%, and 11%. If she has twice as much money invested at 10% as she has invested at 9%, three times as much at 11% as she has at 9%, and the total interest for the year is $930, how much money is invested at each rate? Enter your answer in the form without the units. Let be the amount of money invested at 9%, be the amount of money invested at 10%, and be the amount of money invested at 11%.
Thank You,
Sabrina

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
I'm not sure what the last sentence in your problem is supposed to mean. But let's work
the problem based on the preceding givens. First let's say that N is the amount that she has
invested at 9% (or in decimal form 0.09). Then let's say that T is the amount that she
has invested at 10% (or in decimal form 0.10). Finally, let's say that E is the amount that
she has invested at 11% (or in decimal form 0.11).
.
The problem says that the money invested at 10% is twice the amount invested at 9%. So we
can say that T = 2*N. And it also says that the money invested at 11% is three times the
amount invested at 9%. So we can say that E = 3*N.
.
The next thing we know is that the sum of all the interest earned for the year is $930.
And we know that interest is computed by multiplying the rate times the corresponding
amount that is invested at that rate. So we can write that the total interest is:
.

.
Then since we know that T = 2*N and E = 3*N, we can substitute 2*N for T and 3*N for E to
change the equation to:
.

.
Do the multiplications on the left side to get:
.

.
Add all the terms on the left side and the equation becomes:
.

.
Finally solve for N by dividing both sides by 0.62 to get:
.

.
The amount invested at 9% is $1,500. Twice as much, or $3,000, is invested at 10% and
three times as much, or $4,500, is invested at 11%.
.
Hope this gives you some insight into how to do this problem. Remember that to calculate
the amount of interest you multiply the amount invested by the rate of interest that it
earns.
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