An investment of $63,000 was made by a business club. The investment was split into three parts and lasted one year. The first part of the investment earned 8% interest, the second 6% and the third 9%. Total interest from the investments was $4950. The interest from the investment was 4 times the interest from the second. Find totals of the three parts of the investment. What is the amount of the first part of the investment?
the initial investment of $3000 earns 7% interest compounded continuously. what will the investment be worth in 20 years?
Let's assume the amounts invested in the first, second, and third parts of the investment are x, y, and z respectively.
According to the information given, we can set up the following equations:
x + y + z = $63,000 --- Equation 1
0.08x + 0.06y + 0.09z = $4,950 --- Equation 2 (representing the total interest earned)
0.08x = 4(0.06y) --- Equation 3 (interest from the investment is 4 times the interest from the second)
Let's solve these equations step-by-step:
From Equation 3, we can rewrite it as:
0.08x = 0.24y --- Equation 4
Now, let's solve the system of equations by substitution:
We can substitute the value of y from Equation 4 into Equation 2:
0.08x + 0.06y + 0.09z = $4,950
0.08x + 0.06(0.24y) + 0.09z = $4,950 (substituting 0.24y for 0.08x based on Equation 4)
0.08x + 0.0144y + 0.09z = $4,950
Now, let's substitute the value of y from Equation 4 into Equation 1:
x + y + z = $63,000
x + 0.24y + z = $63,000 (substituting 0.24y for x based on Equation 4)
Now, we have a system of equations:
x + 0.24y + z = $63,000 --- Equation 5
0.08x + 0.0144y + 0.09z = $4,950 --- Equation 6
To solve this system, we can use the substitution method. Let's solve Equation 5 for x:
x = $63,000 - 0.24y - z
Now, substitute the value of x into Equation 6:
0.08($63,000 - 0.24y - z) + 0.0144y + 0.09z = $4,950
$5,040 - 0.192y - 0.08z + 0.0144y + 0.09z = $4,950
Combine like terms:
0.2024y + 0.01z = $90
Now, let's solve Equation 5 for z:
z = $63,000 - 0.24y - x
Substitute the value of z into the equation above:
z = $63,000 - 0.24y - ($63,000 - 0.24y - z)
z = $63,000 - 0.24y - $63,000 + 0.24y + z
Combine like terms:
2z = 0
This indicates that z can be any value since it cancels out. Therefore, we cannot determine the value of z.
However, we can still find the values of x and y. Let's substitute z = 0 into Equation 5:
x + 0.24y + 0 = $63,000
x + 0.24y = $63,000
Now, rewrite Equation 6 with z = 0:
0.08x + 0.0144y = $4,950
To solve this system of equations, we can use the method of elimination. Let's multiply Equation 1 by 0.08 and Equation 2 by 0.24:
0.08(x + 0.24y) = 0.08($63,000)
0.08x + 0.0192y = $5,040 --- Equation 7
0.24(0.08x + 0.0144y) = 0.24($4,950)
0.0192x + 0.003456y = $1,188 --- Equation 8
Now, subtract Equation 8 from Equation 7:
(0.08x + 0.0192y) - (0.0192x + 0.003456y) = $5,040 - $1,188
0.0608y = $3,852
Divide both sides by 0.0608:
y = $3,852 / 0.0608
y ≈ $63,447.37
Now, substitute the value of y into Equation 5 to find x:
x + 0.24($63,447.37) = $63,000
x + $15,227.37 = $63,000
x ≈ $63,000 - $15,227.37
x ≈ $47,772.63
Therefore, the amount of the first part of the investment is approximately $47,772.63.
To find the amounts of the three parts of the investment, let's assume the first part is x dollars, the second part is y dollars, and the third part is z dollars.
We are given that the total investment is $63,000, so we can set up the equation:
x + y + z = 63000 (equation 1)
We are also given that the interest from the investment is $4950, so we can set up the equation:
0.08x + 0.06y + 0.09z = 4950 (equation 2)
Furthermore, we know that the interest from the investment is 4 times the interest from the second part. Therefore, we can set up the equation:
0.08x = 4 * 0.06y (equation 3)
Now, we have a system of three equations (equations 1, 2, and 3) that we can solve to find x, y, and z.
Let's solve the system of equations:
From equation 3, we can rewrite it as:
0.08x = 0.24y
Simplifying the equation, we get:
x = 3y (equation 4)
Now we can substitute equation 4 into equations 1 and 2:
Substituting x = 3y into equation 1, we get:
3y + y + z = 63000
4y + z = 63000 (equation 5)
Substituting x = 3y into equation 2, we get:
0.08(3y) + 0.06y + 0.09z = 4950
0.24y + 0.06y + 0.09z = 4950
0.30y + 0.09z = 4950 (equation 6)
Now we have two equations (equations 5 and 6) with two unknowns (y and z) that we can solve.
From equation 5, we can solve for z:
z = 63000 - 4y (equation 7)
Substituting equation 7 into equation 6, we get:
0.30y + 0.09(63000 - 4y) = 4950
0.30y + 5670 - 0.36y = 4950
-0.06y = -720
y = -720 / -0.06
y = 12000
Now we can substitute the value of y back into equation 7 to solve for z:
z = 63000 - 4(12000)
z = 63000 - 48000
z = 15000
Finally, we can substitute the values of y and z into equation 4 to solve for x:
x = 3(12000)
x = 36000
Therefore, the amounts of the three parts of the investment are $36,000 for the first part, $12,000 for the second part, and $15,000 for the third part.