After 6 months of use, the value of Pearl's computer dropped to $900. after 8 months, the value had gone down to $750. How much did the computer cost originally?

## How much did the computer cost originally?

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# 5 Answers

(6, 900)

(8, 750)

The slope is the rate of change.

900 - 750

----------

6 - 8

150

------

-2

-75

The value is decreasing by $75/month.

Original Value = 900 + 6*75

OV = 900 + 450 =

**$1350**after 6 months the computer is worth $900

after 8 months the computer is worth $750

$900-$750=$150

every two months the computer's value goes down $150

every two months the computer's value goes up $150

the value goes up $150 3 times(6/2)=3

the value goes up 3x$150=$450 to $900+$450=$1350

another approach: the rate of increase from $750 to $900 is (900-750)/750=150/750=1/5(leave this as a fraction)

let x=original price

the value increases 1/5 every two months four times

we are going from $750 to x and the rate of increase is 4/5

(x-750)/750=4/5

5x-3750=3000

5x=6750

x=$1350

more simply x=750+(4/5)(750)=750+600=$1350

even more simply x=1.8x750=$1350

Hi Ryan,

Since between 6 and 8 months, the value of the computer dropped from $900 to $750, this equates to a reduction of $150 every two months

*(8 mo. - 6 mo.= 2 months; $900 - $750 = $150)*.A decrease in value of $150 every two months is the same as saying $75 every one month. We now simply multiply

*$75 x 6 months*and add this amount to $900 to find out the original cost:$75 x 6=$450

$900+$450= $1350

So the original cost of the computer was $1350.

Hi Ryan;

[($900)-($750)]/(2 months)=rate of deterioration

$150/(2 month)=rate of deterioration

$75/month=rate of deterioration

x=original computer cost

[$750]+[($75/month)(8 months)]=x

Let's cancel units where appropriate...

In the second bracketed equation, months appears as both a numerator and denominator...

[$750]-[($75/month)(8
months)]=x

[$750]+[($75)(8)]=x

The unit of $ remains which is what we are solving for...

$750+$600=$1350

I APOLOGIZE FOR MY MISTAKE THIS MORNING. THAT IS THE LAST TIME I WYZANT IMMEDIATELY AFTER WAKING-UP!

One thing to consider in addition to the solution from Jason S. (from Katy, TX, above) is that the assumption made here is that the relationship is linear.

But with the help of a little technology, something interesting comes to be noted.

If we enter the given two points (6, 900) and (8,750) using the STAT - EDIT feature on the TI-83/84 graphing calculator (make sure the DiagnosticOn feature is enabled from the CATALOG menu), we can then use regression models.

First to find the LINEAR equation, use STAT-CALC-4:LinReg(ax+b), and you get the equation:

y = - 75 x + 1350 with r

^{2}=1 (ignore r^{2}- not analytically relevant) , but more importantly note thatr = - 1 (this means the equation has an EXACT negative correlation - a "perfect" fit to the given data)

Using this equation, the original amount (when x=0) is $1350, which is from Jason S.'s solution.

Next we look at the EXPONENTIAL equation. Use STAT-CALC-0:ExpReg, and you get the equation:

y = 1555.2 (.9128709292)^x with the SAME r

^{2}and r values as for the linear equation.This means that the exponential equation is ALSO an EXACT negative correlation - a "perfect" to the data.

So for this equation, the original amount (when x=0) is $1555.20, over $200 more than the linear equation yields.

So, it is important to solve this problem in context. If this problem is embedded among other problems involving linear relationships, then it's safe to assume it's linear.

Without any other context, you simply cannot assume the relationship is linear. The fact that BOTH the linear equation and the exponential equation are an exact fit means that technically, unless you have more information (like a third data point), then
the original value of the computer could be $1350 if you assume a linear decrease, or it's $1555.20, if you assume an exponential decrease.

If this were a multiple-choice question and only one of the answers were available, then obviously you choose that correct answer. If BOTH the linear and the exponential answers were available, then I would boldly challenge this problem with your teacher.