A total of $7000 is invested: part at 6% and the remainder at 10% How much is invested at each rate if the annual interest is $680?
Well, let's call the amount invested at 6% "x" and the amount invested at 10% "y". According to the problem, we know that the total amount invested is $7000 and the total interest earned is $680.
Now, we can set up a system of two equations based on the given information.
Equation 1: x + y = 7000 (since the total amount invested is $7000)
Equation 2: 0.06x + 0.10y = 680 (since the total interest earned is $680)
To find the answer, let's solve these equations using a bit of math and a pinch of humor!
Let's solve Equation 1 for x:
x = 7000 - y
Now, we can substitute this value of x into Equation 2:
0.06(7000 - y) + 0.10y = 680
Now, let's solve it step-by-step and have some fun along the way:
420 - 0.06y + 0.10y = 680
Combine like terms:
0.04y = 260
Now, let's divide both sides by 0.04 and unravel this riddle:
y = 6500
Fantastic! We found that $6500 is invested at 10%.
Now, let's substitute this value back into Equation 1 to determine the amount invested at 6%:
x + 6500 = 7000
Subtract 6500 from both sides:
x = 500
Ta-da! We've found that $500 is invested at 6%.
To wrap it up, $6500 is invested at 10% and $500 is invested at 6%.
Let's assume that the amount invested at a 6% interest rate is x dollars.
According to the problem, the amount invested at the remaining 10% interest rate is the remaining amount, which is $7000 - x.
Now, let's calculate the interest at each rate:
Interest at 6% = x * (6/100)
Interest at 10% = (7000 - x) * (10/100)
According to the problem, the total annual interest is $680.
So, we can set up the equation:
Interest at 6% + Interest at 10% = 680
x * (6/100) + (7000 - x) * (10/100) = 680
Simplifying the equation:
0.06x + 0.1(7000 - x) = 680
0.06x + 700 - 0.1x = 680
Combining like terms:
-0.04x + 700 = 680
Subtracting 700 from both sides:
-0.04x = -20
Dividing both sides by -0.04:
x = -20 / -0.04
x = 500
Therefore, $500 is invested at a 6% interest rate and $6500 ($7000 - $500) is invested at a 10% interest rate.
To find out how much is invested at each interest rate, you can solve the problem using a system of equations.
Let's assume that the amount invested at 6% is x dollars. Since the total amount invested is $7000, the amount invested at 10% would be (7000 - x) dollars.
The interest earned from the amount invested at 6% is 0.06x dollars, and the interest earned from the amount invested at 10% is 0.10(7000 - x) dollars.
According to the problem, the total interest earned is $680. So we can set up the following equation:
0.06x + 0.10(7000 - x) = 680
Let's solve this equation to find the value of x:
0.06x + 700 - 0.10x = 680
-0.04x + 700 = 680
-0.04x = -20
x = -20 / -0.04
x = 500
Therefore, $500 is invested at 6%, and the remaining amount, $7000 - $500 = $6500, is invested at 10%.