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A small software corporation borrowed $ 250,(000) to expand its software line. The corporation borrowed some of the money at 9%, some at 10%, and some at 12%. Use a system of equations to determine how much was borrowed at each rate if the annual interest rate was $26,000 and the amount borrowed at 10% was 2.5 times the amount borrowed at 9%.

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Answer:

$50,000 was borrowed at 9%.

$125,000 was borrowed at 10%.

$75,000 was borrowed at 12%.

Explanation:

Let amount borrowed at 9% be x

Let amount borrowed at 10% be y

Let amount borrowed at 12% be z

Now equations form:


x+y+z=250000 ....(1)


0.09x+0.10y+0.12z=26000 ....(2)


y=2.5x ..... (3)

Substituting the value of y from (3) in (1) and (2)


x+2.5x+z=250000

=>
3.5x+z=250000 ......(4)


0.09x+0.10(2.5x)+0.12z=26000

=>
0.34x+0.12z=26000 ....(5)

Multiplying (4) by 0.12 and subtracting (5) from it, we get


0.42x+0.12z=30000 -
0.34x+0.12z=26000


0.42x-0.34x=30000-26000

0.08x=4000

x = $50000

As y = 2.5x

So,
y=2.5*50000=125000

y = $125000

And as x+y+z=250000

=>
z=250000-50000-125000

z = $75000

Therefore, $50,000 was borrowed at 9%.

$125,000 was borrowed at 10%.

$75,000 was borrowed at 12%.

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