A woman has a total of $7,000 to invest. She invests part of the money in an account that pays 7% per year and the rest in an account that pays 12% per year. If the interest earned in the first year is $640, how much did she invest in each account?
Why did the woman invest in an account that pays 7% and another that pays 12%? It's like she wanted to balance her investments with a little intellectual challenge. Well, let's solve this puzzle, shall we?
Let's say she invested x dollars in the account that pays 7% and 7000 - x dollars in the account that pays 12%.
Now, we know that the interest earned in the first year is $640. So, we can set up the following equation:
0.07x + 0.12(7000 - x) = 640
Now, let's solve for x.
0.07x + 840 - 0.12x = 640
-0.05x = -200
x = 4000
So, she invested $4,000 in the account that pays 7% and $3,000 in the account that pays 12%.
Congratulations to the woman for her balanced investment and for allowing numbers to entertain her!
Let's assume that the woman invested x dollars in the account that pays 7% per year.
Since she invested a total of $7,000 and invested x dollars in the first account, she must have invested (7000 - x) dollars in the account that pays 12% per year.
Now, let's calculate the interest earned on each account:
Interest earned from the first account: (x * 0.07)
Interest earned from the second account: ((7000 - x) * 0.12)
According to the problem, the total interest earned in the first year is $640. So, we can set up the equation:
(x * 0.07) + ((7000 - x) * 0.12) = 640
Simplifying the equation, we have:
0.07x + (840 - 0.12x) = 640
Combining like terms:
0.07x - 0.12x + 840 = 640
-0.05x + 840 = 640
Subtracting 840 from both sides:
-0.05x = -200
Dividing by -0.05:
x = 4000
So, the woman invested $4000 in the account that pays 7% per year.
The amount invested in the account that pays 12% per year can be found by subtracting $4000 from the total investment of $7000:
$7000 - $4000 = $3000
Therefore, the woman invested $4000 in the account that pays 7% per year and $3000 in the account that pays 12% per year.
To find out how much the woman invested in each account, we can set up a system of equations.
Let's assume the amount invested in the account with 7% interest rate is x dollars, and the amount invested in the account with 12% interest rate is y dollars.
We know that the total amount the woman invested is $7,000, so we have:
x + y = 7000 (equation 1)
We also know that the interest earned in the first year is $640. The interest earned from the account with 7% interest rate is 7% of x, and the interest earned from the account with 12% interest rate is 12% of y. So, we can write another equation:
0.07x + 0.12y = 640 (equation 2)
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
From equation 1, we can express x in terms of y as x = 7000 - y.
Substituting this expression for x into equation 2, we get:
0.07(7000 - y) + 0.12y = 640
Simplifying the equation, we get:
490 - 0.07y + 0.12y = 640
Combine like terms:
0.05y = 150
Divide both sides by 0.05:
y = 3000
Now that we know y is 3000 dollars, we can substitute this value back into equation 1 to find x:
x + 3000 = 7000
Subtracting 3000 from both sides:
x = 4000
Therefore, the woman invested $4,000 in the account paying 7% and $3,000 in the account paying 12%.