Janet invested $26,000, part at 6% and part at 3%. If the total interest at the end of the year is $1,080, how much did she invest at 6%?
That is a very strange way to spell "arithmetic"
X*.06+ Y*.03=1080
X+Y=26000
solve these for X
To find out how much Janet invested at 6%, we need to set up an equation using the information given.
Let's assume Janet invested an amount, x, at 6% interest. The amount she invested at 3% interest can be determined as the difference between the total investment and the amount invested at 6%. Since the total investment is $26,000, the amount invested at 3% would be $(26,000 - x).
The interest earned from the investment at 6% is calculated by multiplying the amount invested at 6% by the interest rate and dividing by 100. Similarly, the interest earned from the investment at 3% is calculated by multiplying the amount invested at 3% by the interest rate and dividing by 100.
So, the equation can be set up as follows:
(6/100) * x + (3/100) * (26,000 - x) = 1,080
To solve this equation, we can simplify it step by step:
0.06x + 0.03(26,000 - x) = 1,080
0.06x + 780 - 0.03x = 1,080
0.03x = 1,080 - 780
0.03x = 300
x = 300 / 0.03
x = 10,000
Therefore, Janet invested $10,000 at 6%.