Question 1124075: A person invested $8200 for 1 year, part at 8%, part at 9%, and the remainder at 15%. The total annual income from these investments was $987. The amount of money invested at 15% was $600 more than the amounts invested at 8% and 9% combined. Find the amount invested at each rate.
The person invested $ nothing at 8%,
$nothing at 9%,
and $ nothing at 15%.
Answer by ikleyn(52788)   (Show Source):
You can put this solution on YOUR website! .
A person invested $8200 for 1 year, part at 8%, part at 9%, and the remainder at 15%.
The total annual income from these investments was $987. The amount of money invested at 15% was $600 more
than the amounts invested at 8% and 9% combined. Find the amount invested at each rate.
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At the first glance, this problem is for 3 unknowns and 3 equations.
But it is really big deal to solve such a system.
Fortunately, there is a way to reduce the solution to two unknowns and two equations, which is much easier to solve.
Below I will show you how to do it.
Notice this phrase of the condition
"The amount of money invested at 15% was $600 more than the amounts invested at 8% and 9% combined".
It means that if you subtract $600 from the total sum of $8200, and then divide this difference by 2,
you will get how much was invested at 8% and 9% combined:  = 3800 dollars.
Thus you just know that the combined amount invested at 8% and 9% is $3800.
Hence, the rest $8200 - $3800 = $4400 was invested at 15%.
Also, now you can easily calculate interest from $4400 invested at 15%: it is 0.15*4400 = 660 dollars.
At this point you just, actually reduced the original problem to 2 unknowns and 2 equations.
Let x = the amount invested at 8% and y = the amount invested at 9%.
Then
x + y = 3800 dollars (1)
0.08x + 0.09y = 987 - 660 dollars (2)
Do you understand how I obtained the equation (2) and what does it mean ?
It simply means that combined interest from the 8% account and 9% account is equal to the total interest
of the three accounts minus the interest of the 15% account, which we calculated above.
OK, very good. Any way, we just reduced the original problem to the standard equations for two investments:
x + y = 3800 (1)
0.08x + 0.09y = 327 (2)
To solve it you can use the Substitution method. From eq(1) express x = 3800-y
and substitute it into eq(2). You will get
0.08*(3800-y) + 0.09y = 327
0.08*3800 - 0.08y + 0.09y = 327
0.01y = 327 - 0.08*3800
y =  = (at this point I use Excel spreadsheet in my computer and get the answer in one click) = 2300 dollars.
Thus you got that the amount at 9% is 2300 dollars.
Then the amount at 8% is the rest 3800 - 2300 = 1500 dollars.
Answer. $1500 at 8%; $2300 at 9% and $4400 at 15%.
Check. a) the combined amount invested at 8% and 9% is 1500 + 2300 = 3800 dollars;
It is $600 dollars less than $4400 invested at 15%. ! Correct !
b) Total investment is 1500 + 2300 + 4400 = 8200 dollars. ! Correct !
Solved.
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The lesson to learn from my solution is THIS :
At the first glance, this problem is for 3 unknowns and 3 equations.
But you learned, and I showed it to you, how to reduce the problem to 2 equation in 2 unknowns.
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To see many other similar problems solved by the same method, look into the lesson
- HOW TO algebreze and solve this problem on 2 equations in 2 unknowns
in this site.
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