SOLUTION: A person invested $4100, part of it in bank bonds bearing 3.10% interest and the remainder in Canada Savings Bonds bearing 4.85% interest and she received the same income fro

Question 962978: A person invested $4100, part of it in bank bonds bearing 3.10% interest and the remainder in Canada Savings Bonds bearing 4.85% interest

and she received the same income from each. How much, to the nearest dollar, was invested in each?
Do not round intermediate results. Round only final answers. Round to the nearest integer.
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A person invested $4100, part of it in bank bonds bearing 3.10% interest and the remainder in Canada Savings Bonds bearing 4.85% interest and she received the same income from each.
How much, to the nearest dollar, was invested in each?
Do not round intermediate results. Round only final answers. Round to the nearest integer.
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Equation:
Interest = Interest
0.031*x = 0.0485(4100-x)
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Multiply thru by 10^4 to get:
310x = 485*4100-485x
----
795x = 485*4100
x = $2501.26 (amt invested at 3.1%
4100-x = $1598.74 (amt invested at 4.85)
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Cheers,
Stan H.
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Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = amount invested in bank bonds
Interest from bank bonds is:

Interest from Canada savings bonds is:

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Interest from both of these is the same, so

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$2,501 was invested in bank bonds
$1,599 was invested in Canada savings bonds
----------------
check:

and

Looks pretty close, I think error
is due to rounding off

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