Sam invested $1900, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $32 more than twice the income from the 6% investment. How much did he invest at each rate?
work so far:
x + y = 1900
.06x + .08y = ?
Thank you.
So my two equations are...
x + y = 1900
.08y = 32 + .06x(2)
Yes?
You have tried to use the model where they give you the total interest earned. This one does not do that. Instead, if you read what was written, your second equation must be
.08y = 32 + 2*.06x
yes
you invested a total of $30,000 at 3% and 41/2% simple interest . during one year the two accounts earned $1200. how much did you invest in each account
Well, well, well, looks like Sam has become quite the savvy investor! Let's crunch those numbers and see what we can come up with.
From our work so far, we know that Sam invested a total of $1900. So, we can write the equation x + y = 1900, where x represents the amount invested at 6% and y represents the amount invested at 8%. Good job on that!
Now, let's move onto the second part of the puzzle. We're told that the yearly income on the 8% investment was $32 more than twice the income from the 6% investment. So, we can write another equation to express that relationship.
The income from the 6% investment is simply 0.06x (since it's given as a percentage). And the income from the 8% investment is 2(0.06x) + 32. Why? Well, twice the income from the 6% investment is 2(0.06x), and we add 32 to that to get $32 more. Make sense? Great!
So, our second equation is 0.06x + 32 = 0.08y.
Now, we've got ourselves a system of two equations. Time to plug 'em into the old calculating machine and see what comes out!
Solving the system of equations, x + y = 1900 and 0.06x + 32 = 0.08y, we find that x = 1000 and y = 900.
So, Sam invested $1000 at 6% and $900 at 8%.
That's it! Sam's investments are all figured out. Just remember, investing can be a rollercoaster ride, so make sure to hold on tight and enjoy the ups and downs. Happy investing!
To solve the problem, let's break it down step by step:
Step 1: Set up the equations using the given information.
Let's denote the amount invested at 6% as x, and the amount invested at 8% as y.
From the given information, we have:
Equation 1: x + y = 1900 (since the total amount invested is $1900)
Step 2: Convert the information about the interest income into an equation.
The yearly income on the 8% investment was $32 more than twice the income from the 6% investment.
This can be written as:
Equation 2: 0.08y = 2(0.06x) + 32
Step 3: Solve the system of equations.
We have two equations, so we can solve them simultaneously to find the values of x and y.
Let's solve Equation 1 for x:
x = 1900 - y
Substitute this value of x into Equation 2:
0.08y = 2(0.06(1900 - y)) + 32
Now simplify and solve for y:
0.08y = 2(0.06 * 1900 - 0.06y) + 32
0.08y = 2(114 - 0.06y) + 32
0.08y = 228 - 0.12y + 32
0.08y + 0.12y = 260
0.20y = 260
y = 260 / 0.20
y = 1300
Now substitute the value of y back into Equation 1 to find x:
x + 1300 = 1900
x = 1900 - 1300
x = 600
So, Sam invested $600 at 6% and $1300 at 8%.
To check if the solution is correct, substitute the values of x and y back into Equation 2 and see if it holds true:
0.08(1300) = 2(0.06(600)) + 32
104 = 2(36) + 32
104 = 72 + 32
104 = 104
Since the equation holds true, we can conclude that Sam invested $600 at 6% and $1300 at 8%.