Question

Gonzalez Manufacturing 30,000. Part of the money was borrowed at 6%, part at 8%, and part at 10%. The annual interest was $2440, and the total amount borrowed at 6% and 8% was twice the amount borrowed at 10%. Use Gaussian elimination or Gauss-Jordan elimination to find the amount borrowed at each rate.
How much money was borrowed at 6%?

How much money was borrowed at 8%?

How much money was borrowed at 10%?

Answer 1

Answer:

8,000 at 6%; 12,000 at 8%; 10,000 at 10%

Step-by-step explanation:

Let a, b, c represent the amounts borrowed at 6%, 8%, 10%, respectively. Then the problem statement gives rise to three equations:

... a + b + c = 30,000 . . . . . the total borrowed

... a + b - 2c = 0 . . . . . . . . . . the relationship between the quantities

... .06a +.08b + .10c = 2440 . . . . the total interest charge

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Subtracting the second equation from the first, it becomes

... 3c = 30,000

... c = 10,000 . . . . . divide by 3 . . . [equation 4]

Subtracting 0.06 times the first equation from the third, we have

... .02b +.04c = 640

Multiplying [equation 4] by 0.04 and subtracting from this, we get

... .02b = 240

... b = 12,000 . . . . divide by 0.02 .

Substituting for b and c into the first equation, we find ...

... a + 22000 = 30000

... a = 8000

Then the amounts are ...

• 8000 was borrowed at 6%
• 12000 was borrowed at 8%
• 10000 was borrowed at 10%

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Most graphing calculators are capable of solving matrix problems such as this one. The attachment shows a TI-84 solution.