Suppose that Upper P 0 is invested in a savings account in which interest is compounded continuously at 6.4% per year. That is
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Answer:
Ans. The amount invested at 13% was $1,595.97 and $14,685.03 were invested at 15%
Explanation:
Hi, you can solve this by using 2 equations, so let X be the portion of the money invested at 15% and Y be the amount invested at 13%. So the equation for the whole amount is:
Now, the problem says that the money that you earn by investing at 15% exceeds the money received as interest in your investment of 13% by $1,995.27, this leads us to the second equation.
Now, to make it a little more friendly, we just have to go ahead and divide everything by 0.15, so we get.
Now, in our first equation, we substitute X fo 0.8867(Y)+13,301.8 and we will see this.
Now, we solve for Y
So the money invested at 13% was $1,595.97 therefore, the money invested at 15% was $16,281 - $1,595.97 = $14,685.03
And we can check this results like this. The money invested at 15% will return an amount of:
And the money invested at 13% will return
Substracting, we would found that the difference is:
Best of luck.