An investment of $37,000 was made by a business club. The investment was split into three parts and lasted for one year. The first part of the investment earned 8% interest, the second 6%, and the the third 9%. Total interest from the investments was $3030. The interest from the first investment was 6 times the interest from the second. Find the amounts of the three parts of the investment.
amount invested at 8% --- x
amount invested at 6% --- y
amount invested at 9% ---- 37000-x-y
.08x + .06y + .09(37000-x-y) = 3030
times 100
8x + 6y + 9(37000-x-y) = 303000
-x - 3y = -30000
x + 3y = 30000 (#1)
also .08x = 6(.06)y
8x = 36y
2x = 9y
x = 9y/2 (#2)
sub into #1
9y/2 + 3y = 30000
times 2
9y + 6y = 60000
y = 4000
then x = 9(4000)/2 = 18000
and 37000-4000-18000 = 15000
$18000 at 8%
$4000 at 6%
$15000 at 9%
Got it before I saw this xD thanks
Let's assume the amount of money invested in the first part is 'x'.
Since the interest from the first investment was 6 times the interest from the second, the interest from the second investment is (1/6) * interest from the first investment.
Therefore, the interest from the second investment is (1/6) * (0.08 * x) = 0.0133 * x.
The interest from the third investment is the total interest minus the interest from the first and second investments:
3030 - (0.08 * x) - (0.0133 * x) = 0.09 * (37000 - x).
Simplifying this equation, we get:
3030 - 0.0933x = 3330 - 0.09x.
Rearranging terms, we have:
0.0033x = 300.
Dividing both sides by 0.0033, we find that:
x = 90909.09.
Therefore, the amount invested in the first part is approximately $90909.09.
The amount invested in the second part is (1 - 6) * interest from the first part:
37000 - 90909.09 = $46090.91.
The amount invested in the third part is the remaining amount:
37000 - 90909.09 - 46090.91 = $0.00.
Therefore, the amounts of the three parts of the investment are approximately $90909.09, $46090.91, and $0.00.
To solve this problem, we can set up a system of equations based on the given information.
Let's assume the first part of the investment is x dollars.
The second part of the investment will be y dollars.
And the third part will be z dollars.
Since the first part earned 8% interest, the interest from the first part would be (0.08x).
Similarly, the second part earned 6% interest, so the interest from the second part would be (0.06y).
The third part earned 9% interest, so the interest from the third part would be (0.09z).
According to the given information, the interest from the first part was 6 times the interest from the second part. This can be expressed as:
0.08x = 6 * (0.06y)
Simplifying this equation, we get:
0.08x = 0.36y
Now, we can use this equation along with the other given information to form a system of equations:
x + y + z = 37000 ----(1) (since the total investment was $37,000)
0.08x + 0.06y + 0.09z = 3030 ----(2) (since the total interest was $3,030)
0.08x = 0.36y ----(3) (given information)
We can solve this system of equations to find the values of x, y, and z, which represent the amounts of the three parts of the investment.
To solve the system of equations, we can use substitution or elimination method. Let's use the elimination method:
From equation (3), we can rewrite it as:
0.08x - 0.36y = 0 ----(4)
Now, we can eliminate the x variable from equations (2) and (4) by multiplying equation (4) by 0.06 and equation (2) by 100:
0.06 * (0.08x - 0.36y) = 0.06 * 0
0.48x - 2.16y = 0 ----(5)
100 * (0.08x + 0.06y + 0.09z) = 100 * 3030
8x + 6y + 9z = 303000 ----(6)
Multiply equation (1) by 8:
8x + 8y + 8z = 296000 ----(7)
Now, subtract equation (5) from equation (7) to eliminate the x variable:
(8x + 6y + 9z) - (0.48x - 2.16y) = 303000 - 0
7.52x + 8.16y + 9z = 303000
Now, we have two equations:
7.52x + 8.16y + 9z = 303000 ----(8)
x + y + z = 37000 ----(9)
We can solve these two equations using a calculator or advanced algebraic techniques to find the values of x, y, and z.