Percentage Rate

Question

A company invests a total of ​$30,000 of surplus funds at two annual rates of​ interest:

4​% and 7.5%.

It wishes an annual yield of no less than 6.5​%.

What is the least amount of money that the company must invest at the 7.5% ​rate

Answer 1

The interest rate range is (7.5-4) = 3.5. The difference between the desired interest rate and the bottom interest rate is (6.5-4) = 2.5. The ratio of these two numbers is 2.5/3.5 = 0.7142. Therefore, the amount of money that must be invested is $30,000 x 0.7142 = $21,429.

Answer 2

To the nearest cent at 7.5% , at least $21,428.57 is needed

To the nearest dollar at 7.5%, at least $21,429 is needed

You want to earn

30000*.065= 1950

You can use of system of equations and substitution to solve with the rates above

Equation 1 for the total amount invested

x + y = 30000

Equation 2 for the total interest you want to earn

0.075x + 0.04y =

Equation one allows for an easy substitution

y = 30000 -x

Substitute the quantity above into Equation 2

0.075x + 0.04(3000 -x) = 1950

Multiply to remove the parentheses

0.075x + 1200 - 0.04x = 1950

Combine like terms

0.035x + 1200 = 1950

Subtract 1200 from both sides of the equation

0.035x = 750

Divide both sides by 0.035

x = 21428.57

y = 30000 - x = 30000 - 21428.57 = 8571.43

y = 8571.43

Checking in both equations

0.075(21428.57) + 0.04(8571.43) = 1950

1607.14 + 342.86 = 1950

1950 = 1950

And

x + y = 30000

21428.57 + 8571.43 = 30000