SOLUTION: A rollover IRA of $18,050 was invested in two mutual funds, one earning 12% interest and the other earning 10%. After 1 year, the combined interest income is $2,041. How much (in d

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A rollover IRA of $18,050 was invested in two mutual funds, one earning 12% interest and the other earning 10%. After 1 year, the combined interest income is $2,041. How much (in d      Log On

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Question 1204124: A rollover IRA of $18,050 was invested in two mutual funds, one earning 12% interest and the other earning 10%. After 1 year, the combined interest income is $2,041. How much (in dollars) was invested at each rate?

Found 3 solutions by ikleyn, MathLover1, greenestamps:
Answer by ikleyn(52765) About Me  (Show Source):
You can put this solution on YOUR website!
.
A rollover IRA of $18,050 was invested in two mutual funds, one earning 12% interest
and the other earning 10%. After 1 year, the combined interest income is $2,041.
How much (in dollars) was invested at each rate?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

x dollars invested at 12%;  (18050-x) dollars invested at 10%.


Write the total interest equation 

    0.12x + 0.1*(18050-x) = 2041  dollars.


To find x from this equation, simplify it step by step

    0.12x + 0.1*18050 - 0.1x = 2041

    0.12x - 0.1x = 2041 - 1805

        0.02x     =    236

           x     =    236/0.02 = 11800.


ANSWER.  $11800 invested at 12%;  the rest 18050 - 11800 = 6250 dollars invested at 10%.


CHECK.   0.12*11800 + 0.1*6250 = 2041 dollars, total interest.   ! correct !

Solved.



Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

invested: $18050
in two mutual funds: x and y
x%2By=18050....solve for x
x=18050-y....eq.1
one earning 12%=0.12 interest => let say mutual fund x
and the other earning 10%=0.10 interest => let say mutual fund y
if after 1 year, the combined interest income is $2041, we have
0.12%2Ax%2B0.10y=2041....eq.2
substitutex from eq.1
0.12%2818050-y%29%2B0.10y=2041
2166+-+0.12y%2B0.10y=2041
2166+-+0.02y=2041
2166+-+2041=0.02y
125=0.02y
y=125%2F0.02
y=12500%2F2
y=6250
go to eq.1
x=18050-y....eq.1, substitutey
x=18050-6250
x=11800

11800 was invested in mutual fund earning 12% interest
and
6250 was invested in mutual fund earning 10%. interest

check:
0.12%2A11800%2B0.10%2A6250=2041....eq.2
1416%2B625=2041
2041=2041





Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Here is an informal solution method that can be used for any 2-part "mixture" problem like this.

All $18,050 earning 10% would yield interest of $1805; all earning 12% would yield $2166.

Use a number line or any similar method to find that the actual interest of $2041 is 236/361 of the way from $1805 to $2166.

That means 236/361 of the total earned the higher interest rate.

(236/361)($18050) = 236($500) = $11800

ANSWERS: $11,800 was invested in the fund that earned 12% interest; the other $6250 was invested in the fund that earned 10% interest.

CHECK: .10($6250)+.12($11800) = $625+$1416 = $2041

Because of the "ugly" numbers in this problem, this informal method is not much faster than the standard formal algebraic solution method. But when the numbers are "nice", this method can be MUCH faster if your mental arithmetic is good.