Stan invested $5,000, part at 8% and part at 17%. If the total interest at the end of the year is $490, how much did he invest at 8%?
.08 x + .17 (5,000 -x) = 490
assuming simple interest not compounded monthly or daily or anything.
If x was invested at 8%, 5000-x was invested at 17%.
Solve the equation
490 = 0.08x + 0.17 (5000 -x)
490 - 850 = -0.09 x
0.09x = 360
x = ?
It says that the total interest earned in two accounts, was $490.
Hi,
X is the amount invested not the interest so it could be greater than $490. If x was invested at 8%, 5000-x was invested at 17%.
Solve the equation
490 = 0.08x + 0.17 (5000 -x)
490 - 850 = -0.09 x
0.09x = 360
x = $4000
So the amount invested at 8% is $4000 and at 17% is $1000.
Hi,
X is the amount invested not the interest so it could be greater than $490. If x was invested at 8%, 5000-x was invested at 17%.
Solve the equation
490 = 0.08x + 0.17 (5000 -x)
490 - 850 = -0.09 x
0.09x = 360
x = $4000
So the amount invested at 8% is $4000 and at 17% is $1000.
To solve this problem, we can break it down into two equations based on the given information.
Let's assume Stan invested x dollars at 8% and y dollars at 17%.
The first equation is based on the total investment amount:
x + y = $5,000 --- Equation (1)
The second equation is based on the total interest earned:
0.08x + 0.17y = $490 --- Equation (2)
Now, we can solve this system of equations to find the values of x and y.
From Equation (1), we can express x in terms of y:
x = $5,000 - y
Substituting this value of x into Equation (2), we get:
0.08($5,000 - y) + 0.17y = $490
Now, we can solve this equation for y:
0.08($5,000 - y) + 0.17y = $490
400 - 0.08y + 0.17y = $490
0.09y = $490 - $400
0.09y = $90
y = $90 / 0.09
y = $1,000
Now, substitute the value of y back into Equation (1) to find x:
x + $1,000 = $5,000
x = $5,000 - $1,000
x = $4,000
Therefore, Stan invested $4,000 at 8%.