SOLUTION: A data processing company has a computer which cost $1,000,000 when new and which depreciates linearly to a salvage value of $50,000 over 10 years. The company also has a computer

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Question 945558: A data processing company has a computer which cost $1,000,000 when new and which depreciates linearly to a salvage value of $50,000 over 10 years. The company also has a computer which cost $2,225,000 when new and depreciates linearly to a salvage value of $75,000 over a period of 10 years. If the computers were purchased at the same time when will the book value of the more expensive computer be exactly twice the book value of the less expensive computer?
Found 2 solutions by josgarithmetic, josmiceli:
Answer by josgarithmetic(39615) About Me  (Show Source):
You can put this solution on YOUR website!
Use functions, f(x) for the less expensive computer and g(x) for the more expensive computer. You have these data points as (x,y); x for time passage, y for current value; for each computer:

f, (0,1000000),(10,50000)
g, (0,2225000),(10,75000)

Find the formula for each function and formulate into "functional" form, or like slope-intercept form.

The meaning of the question is equivalent to for what value of x will be g%28x%29=2%2Af%28x%29 ?

The complete method to solve this problem should be clear. I did not supply all the details, but you should be able to.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The rate of depreciation for the 1st computer is
+%28+1000000+-+50000+%29+%2F+10+=+950000+%2F+10+
+950000+%2F+10+=+95000+ dollars / year
--------------------------------------
The rate of depreciation for the 2nd computer is:
+%28+2225000+-+75000+%29+%2F+10+=+2150000+%2F+10+
++2150000+%2F+10+=+215000+ dollars / year
----------------------------------------
Let +t+ = number of years since both computers
were purchased
Let +V+ = the current book value
--------------------------------
For 1st computer:
+V%5B1%5D+=+1000000+-+95000t+
For 2nd computer:
+V%5B2%5D+=+2225000+-+215000t++
-------------------------------
when is +V%5B2%5D+=+2%2AV%5B1%5D+ ?
+2225000+-+215000t+=+2%2A%28+1000000+-+95000t+%29+
+215000t+-+190000t+=+2225000+-+2000000+
+25000t+=+225000+
+t+=+9+
In 9 years the more expensive computer is
twice the value of the other one
------------
check:
+V%5B1%5D+=+1000000+-+95000%2A9+
+V%5B1%5D+=+1000000+-+855000+
+V%5B1%5D+=+145000+
and
+V%5B2%5D+=+2225000+-+215000%2A9++
+V%5B2%5D+=+2225000+-+1935000+
+V%5B2%5D+=+290000+
and
+2%2AV%5B1%5D+=+2%2A145000+
+2%2AV%5B1%5D+=+290000+
OK