Peter has money in two savings accounts. One rate is 9% and the other is 11%. If he has $200 more in the 11% account and the total interest is $68, how much is invested in each savings account?
Let x represent mount at 9%
Let x + 200 represent amount at 11%
Total interest earned is 68$
0.09x + 0.11 (x + 200) = 68
0.09x + 0.11x + 22 = 68
0.2x + 22 = 68
0.2x + 22 - 22 = 68 - 22
0.2x = 46
0.2x / 0.2 = 46 / 0.2
x = $230 is amount at 9%
Amount at 11%= 230 + 200 = $430
Well, Peter must be quite the financial acrobat to balance two savings accounts! Let's solve this money riddle together.
Let's assume Peter invested $x in the 9% account (because, as we all know, numbers are the building blocks of wealth!). Now, since he has $200 more in the 11% account, he must have invested $x + $200 in that account.
To calculate the interest, we can multiply the amount invested in each account by their respective interest rates and add them together, which should equal $68 in total. So our equation looks like this:
0.09x + 0.11(x + 200) = 68
Simplifying the equation a bit, we get:
0.09x + 0.11x + 22 = 68
Now let's do some mathematical tricks to solve it.
0.20x + 22 = 68
Subtracting 22 from both sides, we have:
0.20x = 46
Now comes the grand finale! Divide both sides by 0.20:
x = 230
So, Peter invested $230 in the 9% account. Since he had $200 more in the 11% account, that means he invested $230 + $200 = $430 in the 11% account.
Voila! Peter is now the ringmaster of his savings accounts, with $230 invested in the 9% account and $430 invested in the 11% account.
Let's assume the amount invested in the 9% account is x.
Accordingly, the amount invested in the 11% account is x + $200.
The interest earned from the 9% account is (x * 9%) = 0.09x.
The interest earned from the 11% account is ((x + $200) * 11%) = 0.11(x + $200).
Given that the total interest earned is $68, we can set up the following equation:
0.09x + 0.11(x + $200) = $68
Simplify the equation:
0.09x + 0.11x + $22 = $68
Combine like terms:
0.20x + $22 = $68
Subtract $22 from both sides of the equation:
0.20x = $68 - $22
0.20x = $46
Divide both sides by 0.20:
x = $46 / 0.20
x = $230
Therefore, Peter has $230 invested in the 9% account and $430 ($230 + $200) invested in the 11% account.
To solve this problem, let's assign variables to represent the amount of money Peter has in each savings account.
Let's say the amount in the 9% account is x dollars. Therefore, the amount in the 11% account is x + $200.
Now, let's calculate the interest earned from each account.
The interest earned from the 9% account = (9/100) * x = 0.09x dollars.
The interest earned from the 11% account = (11/100) * (x + $200) = 0.11(x + $200) dollars.
According to the problem, the total interest earned from both accounts is $68. So, the equation can be written as:
0.09x + 0.11(x + $200) = $68
Now, let's solve this equation to find the value of x, which represents the amount of money in the 9% account.
0.09x + 0.11x + 0.11($200) = $68
0.09x + 0.11x + $22 = $68
0.20x + $22 = $68
0.20x = $68 - $22
0.20x = $46
Dividing both sides of the equation by 0.20, we get:
x = $46 / 0.20
x = $230
Therefore, the amount of money invested in the 9% account is $230.
The amount of money invested in the 11% account is x + $200 = $230 + $200 = $430.
So, Peter has $230 in the 9% account and $430 in the 11% account.