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Source: — Problem Solving |
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by hwiya320 » Wed Dec 10, 2008 7:24 pm
hmmm, I thought I solved this on the forum before.

let's try this again

Machine A = $20,000
Machine B = $50,000

Downpayment
A = 20,000*20% = $4,000
B = 50,000*20% = $10,000

Remainders
A = 20,000-4000 = 16,000
B = 50,000-10,000 = 40,000

Interest = 40% of remaining cost
A = 16,000 * 40% = 6,400
B = 40,000 * 40% = 16,000

Total cost of
A = 20,000 + 6,400 = 26,400
B = 50,000 + 16,000 = 66,000

Question asks how much MORE is 1 of machine A than 2 of machine B?
2A = 52,800
1B = 66,000

$66,000 - $52,800 = $13,200!
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by canuckclint » Wed Dec 10, 2008 7:29 pm
********** Solve for As cost and then 2A cost: **********
K stands for thousands here, can replace with 000.

0.2 * 20K = 4K down
16K left
16K * 0.4 = 6.4K interest

4K + 16K + 6.4K = 26.4K total or 26,400

So 2A = 52,,800

********** Solve for B **********
0.2 * 50K = 10K down
40K left to pay
40K * 0.4 = 16K interest

10K + 40K + 16K = 66K

********** 2A - B **********

66,000 - 52,800 = 13,200
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