$5400 is invested, part of it at 10%, and part of it at6%. For a certain year, the total yield is $444.00. How much was invested at each rate?
same type of question here
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just change the numbers.
To find out how much was invested at each rate, we can set up a system of equations based on the given information.
Let's denote the amount invested at 10% as x and the amount invested at 6% as y.
According to the problem, the total amount invested is $5400, so we have the equation:
x + y = 5400
We also know that the total yield (interest) after a year is $444. This means the interest from the amount invested at 10% plus the interest from the amount invested at 6% adds up to $444. The interest for an amount can be calculated by multiplying the amount by the interest rate.
The interest from x (amount invested at 10%) is given by: 0.1x
The interest from y (amount invested at 6%) is given by: 0.06y
So, we have the equation:
0.1x + 0.06y = 444
Now we have a system of equations:
x + y = 5400 (Equation 1)
0.1x + 0.06y = 444 (Equation 2)
To solve this system of equations, we can use the substitution method or the elimination method.
Let's start with the elimination method to solve this system:
First, let's multiply Equation 1 by 0.06: (this will help us eliminate y when we add the two equations)
0.06(x + y) = 0.06(5400)
0.06x + 0.06y = 324 (Equation 3)
Now, we can subtract Equation 2 from Equation 3 to eliminate the y term:
0.06x + 0.06y - 0.1x - 0.06y = 324 - 444
-0.04x = -120
Simplifying:
-0.04x = -120
Divide both sides by -0.04:
x = 3000
Now, substitute the value of x into Equation 1 to solve for y:
3000 + y = 5400
y = 2400
Therefore, $3000 was invested at 10%, and $2400 was invested at 6%.