SOLUTION: a small corporation borrowed $500,000 to expand its product line. Some of the money was borrowed at 9%, some at 10% and some at 12%. How much was borrowed at each rate if the ann

Algebra ->  Test -> SOLUTION: a small corporation borrowed $500,000 to expand its product line. Some of the money was borrowed at 9%, some at 10% and some at 12%. How much was borrowed at each rate if the ann      Log On


   



Question 44917: a small corporation borrowed $500,000 to expand its product line. Some of
the money was borrowed at 9%, some at 10% and some at 12%. How much was
borrowed at each rate if the annual interest was $52, 000 and the amount
borrowed at 10% was two and a half times the amount borrowed at 9%?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
x = amount borrowed at 9%
y = amount borrowed at 10%
z = amount borrowed at 12%
x+%2B+y+%2B+z+=+500000
Annual interest charged at 9% = .09%2Ax
Annual interest charged at 10% = .10%2Ax
Annual interest charged at 12% = .12%2Ax
Total annual interest = 52000
.09%2Ax+%2B+.10%2Ay+%2B+.12%2Az+=+5.2+%2A+10%5E4
according to problem, y+=+2.5%2Ax
.09%2Ax+%2B+.10%2A2.5%2Ax+%2B+.12%2Az+=+5.2+%2A+10%5E4
.34%2Ax+%2B+.12%2Az+=+5.2+%2A+10%5E4
also,
x+%2B+y+%2B+z+=+500000
x+%2B+2.5%2Ax+%2B+z+=+50+%2A+10%5E4
3.5%2Ax+%2B+z+=+50+%2A+10%5E4
multiply both sides of this equation by (-.12)
-.42%2Ax+%2B+%28-.12%29%2Az+=+%28-.12%29%2A50+%2A+10%5E4
-.42%2Ax+-.12%2Az+=+-6+%2A+10%5E4
and the other equation was
.34%2Ax+%2B+.12%2Az+=+5.2+%2A+10%5E4
add these equations
-.08%2Ax+=+-.8+%2A+10%5E4
divide both sides by (-.08)
x+=+10%2A10%5E4
x+=+100000 amount borrowed at 9%
y+=+2.5%2A100000
y+=+250000 amount borrowed at 10%
z+=+500000+-+350000+
z+=+150000 amount borrowed at 12%
check
3.5%2Ax+%2B+z+=+50+%2A+10%5E4
35%2A10%5E4+%2B+15%2A10%5E4+=+50+%2A+10%5E4
50%2A10%5E4+=+50%2A10%5E4
OK