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Price Increase and Decrease Dillema!!

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Price Increase and Decrease Dillema!!

by KICKGMATASS123 » Wed May 20, 2009 3:03 pm
The price of a certain property increased by 10% in the first year, decreased by 20% in the second year, and increased by 25% in the third year. What was the amount of the dollar decrease in the property price during the second year?

(1) The price of the property at the end of the third year was $22,000.

(2) The decrease in the property price over the first two years was $2,000 less than the increase in the property price during the third year.

I understand the 1st statement.. But can someone explain the 2nd statement???? :( In detail please

oa is D
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by dmateer25 » Wed May 20, 2009 3:26 pm
(2) The decrease in the property price over the first two years was $2,000 less than the increase in the property price during the third year.


Let the property price = x

so the property price after 2 years will be:

(1.1)(.8)(x)

1.1 is for the 10% increase the first year. The .8 is for the 20% decrease the second year.

So the decrease over the 2 years will be equal to the original property price, which is x, minus (1.1)(.8)(x).

x - .88x = .12x

Now we are told the decrease is 2000 less than the increase in the 3rd year.

The price the third year is equal to (1.1)(.8)(1.25)(x), which equals 1.1x.

The increase in price from the 2nd to 3rd year will equal 1.1x - .88x = .22x.

So now we have the decrease over the first two years and the increase from year two to three.

The decrease is 2000 less than the increase, so

.12x + 2000 = .22x
2000 = .1x
x=20000.

Now you have the original amount and can find the amount of the decrease during the second year.

Note: you don't need to do ANY of this actual work other than to realize you can set up the equation for the original price of the house.
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by cramya » Wed May 20, 2009 3:53 pm
The explanation above is good.


If we can see that the price can be expressed as x and the subsequent increases or decreases and the yearly prices can be expressed in terms of x then u have 1 equation 1 variable which can be solved from the statements individdually/ Once we find x we can get to everything else. We dont even have to spend time setting it up(atleast for this question) since its a DS i.e. if something like this comes up on the real one . A problem like this can save us some time.

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by kyabe » Thu May 21, 2009 5:12 am
Hi dmateer25,

Just a query. isn't it that we need to take the price at the end of the year as the principal price for next year.. I mean as per question

Let original price be X

10% increase = 1.1X

20% decrease = (1.1 - .2)X2 = .88X

and 25 increase = (.88 + .22)X = 1.1X

Although the final equation produced by your approach and my approach is same. But still would you mind to clarify that??
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by siddhans » Sun Jun 19, 2011 5:29 pm
Can someone please tell me whats wrong with this approach=>

Let p denote the original price of the property

Price of property after 1st year = p + 10/100 p = 1.1p
Price of property after 2st year = 1.1p - 0.2p = 0.9 p <---- why is this 0.88 for others?
Price of property after 3rd year = 0.9p + .25p = 1.15 p

Statement1:
1.15 p = 22,000 <---- why cant be substitute 3rd year to get property price

Therefore, p = 22000/1.15 ???
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by phanideepak » Sun Jun 19, 2011 7:58 pm
siddhans wrote:Can someone please tell me whats wrong with this approach=>

Let p denote the original price of the property

Price of property after 1st year = p + 10/100 p = 1.1p
Price of property after 2st year = 1.1p - 0.2p = 0.9 p <---- why is this 0.88 for others?
Price of property after 3rd year = 0.9p + .25p = 1.15 p

Statement1:
1.15 p = 22,000 <---- why cant be substitute 3rd year to get property price

Therefore, p = 22000/1.15 ???
Siddhans

The mistake is in the part where you calculated the decrease for the second year.

The problem says that the price decreased 20% the second year. It means that the change in price of the property is WRT the price of the property at the end of first year.

What you have calculated is the price decrease WRT to the initial price(Start of first year). No where was it given that the 20% decrease is wrt to the initial price of the plot.



So your equations will look like


Price of property after 1st year = p + 10/100 p = 1.1p
Price of property after 2st year = 1.1p -(1.1p)(0.2) = 0.88 p
Price of property after 3rd year = 0.88p + (0.88p)(.25) = 1.1 p

Hope I made some sense.
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by amit2k9 » Sun Jun 19, 2011 10:51 pm
property price = p

third year it will be 1.25*0.8*1.1*p

a fits the bill. sufficient.

b difference = (1.1p-0.8*1.1p)
increase in 3 year price = (1.25p - 0.8*1.1p)

the equation is (1.1p-0.8*1.1p) = (1.25p - 0.8*1.1p) - 2000

thus p can be found out and hence the increase in 2nd year.

sufficient.

D it is.
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