Paula invested into two accounts; one pays 5% simple interest and the other 6% simple interest. She invested $3000 more in the account paying 6% than in the other account 5%. At the the end of the first year, Paula total return was 730. How much did she invest in each account?
To solve this problem, let's go step by step:
Let's assume that Paula initially invested an amount "x" in the account paying 5% simple interest.
According to the problem, she invested $3000 more in the account paying 6% interest. Therefore, she invested an amount of "x + $3000" in that account.
Now, let's calculate the interest earned in each account:
Interest earned in the account paying 5% = x * 5% = 0.05x
Interest earned in the account paying 6% = (x + $3000) * 6% = 0.06(x + $3000)
According to the problem, Paula's total return after one year was $730. Therefore, we can set up the equation:
Total return = Interest earned in the account paying 5% + Interest earned in the account paying 6%
$730 = 0.05x + 0.06(x + $3000)
Simplifying the equation:
$730 = 0.05x + 0.06x + 0.06($3000)
$730 = 0.11x + $180
Subtracting $180 from both sides:
$730 - $180 = 0.11x
$550 = 0.11x
Dividing both sides by 0.11:
x = $550 / 0.11
x = $5000
Therefore, Paula initially invested $5000 in the account paying 5% interest.
To find out how much she invested in the account paying 6% interest, we can substitute the value of x into the expression we found earlier:
Amount invested in the account paying 6% = x + $3000 = $5000 + $3000 = $8000
So, Paula invested $5000 in the account paying 5% interest and $8000 in the account paying 6% interest.
$PO @ 5%.
Po+3000) @ 6%.
I = Po*r*t + (Po+3000)*r*t = $730
Po*0.05*1 + (Po+3000)*0.06*1 = 730
0.05Po + 0.06Po+180 = 730
0.11Po = 730-180 = 550
Po = $5,000 @ 5%.
Po+3000 = $8,000 @ 6%.