Miguel borrowed at 2%

Question

Rocco borrowed a total of $7000 from two student loans. One loan charged 2% simple interest and the other charged 1.5% simple interst, both payable after graduation. If the interest he owed after three yrs was $367.50, determine the amount of principal for each loan?

Answer 1

Hi Kassandra!

Let's start with what we know:

1. Rocco borrowed a total of $7000 from 2 loans
2. X = loan 1 principal amount (starting amount)
3. Y = loan 2 principal amount
4. X + y = 7000
5. The total interest from the 2 loans after 3 years is $367.50 (Interest 1 + Interest 2 = 367.50)
6. Formula for simple interest: I = Prt or (interest = principal x rate x time)
7. The rate of loan 1 is 1.5% or 0.015
8. The rate of loan 2 is 2% or 0.02
9. Time is 3 years

OK so now let's create a formula for the total interest

Interest 1 + Interest 2 = 367.50

Interest 1

I = Prt

1. as we see from above the principal is x, rate is 1.5% and time is 3 years. So let's plug it in

I = x(0.015)(3)

I = 0.045x

Interest 2

I = Prt

1. the principal is y, rate is 2% and time is 3 years. So let's plug it in

I = y(0.02)(3)

I = 0.06y

Now put it together for the total interest

0.045x + 0.06y = 367.50

1. Multiply both sides by 100 so that the coefficients are whole numbers. (It just makes it easier)

4.5x + 6y = 36,750

OK now remember x + y = 7000

So let's use these two equations to solve for x.

4.5x + 6y = 36,750

x + y = 7000

Multiply the second equation by -6 to eliminate the y.

-6(x + y = 7000) = -6x - 6y = -42,000

Combine the equations together

-6x - 6y = -42,000

4.5x + 6y = 36,750

-1.5x = -5250

Divide by -1.5 and solve for x

X = 3500 (principal for loan 1)

Y = 7000 - 3500 which is 3500

The starting amount for each loan is $3500

Answer 2

Hi there!

Here, it looks like we have a typical constraint problem with two variables and two equations. When we get problems like these, it's almost always a good idea to write out these equations and see where we can substitute a value from one into the other. To set this up, we should write down a couple of key equations:

1. I = Prt
2. P₁ + P₂ = 7000
3. P₁(.02)(3) + P₂(.015)(3) = 367.5 (using equation (1))

Where P₁ is the initial amount with 2% interest, and P₂ is the initial amount with 1.5% interest.

Now we can start to solve the problem using these equations. Multiplying out equation (3), we get

.06P₁ + .045P₂ = 367.5.

With this, we can use some simple substitution to solve for one of the initial values:

P₁ = 7000 – P₂

Substituting this new value of P₁ into equation (3), we get:

.06(7000 – P₂) + .045P₂ = 367.5

From here, we can simply multiply out the .06 and solve for P₂:

.06(7000) – .06(P₂) + .045P₂ = 367.5

420_nice – .06P₂ + .045P₂ = 367.5

420 – .015P₂ = 367.5

.015P₂ = 52.5

P₂ = 3500

Now that we have a value for P₂, we can refer back to equation (2) to figure out the value of P₁:

P₁ + P₂ = 7000

P₁ + 3500 = 7000

P₁ = 3500

So we get that P₁ and P₂ are both 3500. We can check this answer easily by making sure they satisfy both equation (2) and equation (3):

3500 + 3500 = 7000 (yup!)

.06(3500) + .045(3500) = 367.5 (yup!)

So we got the right answer. I hope this gave you a better idea of how to solve these kinds of problems, and let me know if you have any questions!

Answer 3

x=amt at 2% and y=amt at 1.5%

x+y=7000

interest is for 3 years

.06x+.045y=367.50

Multiply by 100 and divide by 15

4x+3y=24500

Multiply 1st equation by 3

3x+3y=21000

Subtract

x=3500 and then y=3500