Stuart pays back two student loans over a 4-yr period. One loan charges the equivalent of 3% simple interest and the other charges the equivalent of 5.5% simple interest. If the total amount borrowed was $24,000 and the total amount of interest paid after 4 yr is $3280, find the amount borrowed from each loan.

Answer: the amount of each loans are $4000 and $20000

Step-by-step explanation:

Let the amount borrowed for the first loan be $x

Let the amount borrowed for the second loan be $y

The formula for simple interest will be

I = PRT/100

Where

I is the interest

P = principal

R = rate

T = time in years

Considering the first loan x,

Principal = x

R = 3%

T = 4 years

I = (x * 3 * 4) / 100

Ix = interest on x

Ix = 12x/100

Considering the second loan y,

Principal = y

Rate = 5.5 %

T = 4 years

Iy = interest on y

Iy = (y * 5.5 * 4) / 100 = 22y/100

If the total amount borrowed was $24,000, it means that

x + y = 24000

the total amount of interest paid after 4 yr is $3280. This means that

Ix + Iy = 3280. Therefore

12x/100 + 22y/100 = 3280

Cross multiplying,

12x + 22y = 3280 * 100 = 328000

12x + 22y = 328000

Substituting x = 24000 - y into 12x + 22y = 328000, it becomes

12 (24000 - y) + 22y = 328000

288000 - 12y + 22y = 328000

- 12y + 22y = 328000 - 288000

10y = 40000

y = 40000/10 = $4000

x = 24000 - y = 24000 - 4000

x = $20000

The first loan L1 = $20,000

This is the loan with 3% simple interest

The second loan L2 = $4,000

This is the loan with 5.5% simple interest

Step-by-step explanation:

L1 + L2 = $24,000 ... (1)

4 (3% of L1) + 4 (5.5% of L2) = $3,280 ... (2)

Where the first term in equation (2) represents the total interest paid on loan 1 after 4 years

The second term represents total interest accruing to loan 2 after 4 years

From equation (1), we single out L1

L1 = 24,000 - L2

Substitute this value for L1 in equation (2)

288,000 + 10L2 = 328,000

L2 = $4,000

L1 = $24,000 - $4,000 = $20,000