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Help with Percentage Problem

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by cholloway » Mon Jun 01, 2009 6:39 am
Can someone please tell me understand where I am making a mistake?

In 1992 the value of a house was 3/4 of the original puchase price. In 1996, the price was 2/3 the original purchase price. By approximately what percent did the value of this house decrease from 1992 to 1996?

My solution:

Assume the original value of the house is $100. Therefore the value of the house was $75 in 1992, and $66 in $1996. 75-66/75 = 9/75 = 3/25 = 12/100. I get a 12 percent change.

The correct answer is supposed to be 11.1%.
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by tdadic84 » Mon Jun 01, 2009 7:15 am
hello...

dont assume 100...get a multiple of 3 and 4...since 100 will leave you with $66,666 and can throw your calc's off...

i choose 120...

so in 1992 = $90,000 and in 1996 = $80,000

therefore...$10,000/90,000 = 11.11%
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by dmateer25 » Mon Jun 01, 2009 7:19 am
You got 12 instead of 11.111 because you rounded 66.6666666 to 66

If you used a number such as 12 as the original price you wouldn't have to round numbers.

Assume original price = 12
1992 = 9

1996 = 8

New - Old/Old

(8 – 9)/9 = -1/9 = -.1111111

So the change is -11.1%
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by cholloway » Mon Jun 01, 2009 7:20 am
OK, that makes sense. Thanks!
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by blackarrow » Mon Jun 01, 2009 8:13 am
Or you can just leave it in fractions

Assuming the house value was 100 originally (its always comfotable to chose 100)

in 1992 - 75

in 1996 -200/3

thus decrease percentage = ((75-200/3)/75)*100

the answer is 100/9 == 11.11%
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Re: Help with Percentage Problem

by kvamsy » Mon Jun 01, 2009 7:49 pm
There is no need of assuming any original price because both are any way we can common factor is X both n numerator and denominator as given below.

(3/4-2/3)/3/4 - - - > 1/9

If u r calculating in percentage you can multiply by hundred that is 1/9*100 = 11.1%

As simple as that....
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