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21 July, 00:04

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.

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  1. 21 July, 01:20
    0
    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
  2. 21 July, 03:44
    0
    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
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