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1999-2000

by goyalsau » Tue Oct 05, 2010 6:37 pm
The price of a house decreased by x% from 1998 to 1999 and increased by y% from 1999 to 2000. If the house cost M dollars in 2000, how much did it cost in 1998?

100*((M/1+X)/1-y)

(M/(1 + (s/100) ))/(1 - (y/100))

M/(1 + (x/100) - (y/100))

10,000M/((100+y)/(100-x))

10,000M/((100+x)/100-y))

OA is D
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by Rahul@gurome » Tue Oct 05, 2010 6:49 pm
Let the price of house in 1998 = $P
Then price in 1999 = P - (0.0x)P
Price in 2000 = [P - (0.0x)P] + 0.0y[P - (0.0x)P] = P(1 + 0.0y)[1 - (0.0x)]
So, P(1 + 0.0y)[1 - (0.0x)] = M
P = M/(1 + 0.0y)[1 - (0.0x)] =10,000M / [(100 + y)(100 - x)]

[spoiler]The correct answer is (D).[/spoiler]
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by diebeatsthegmat » Tue Oct 05, 2010 7:02 pm
goyalsau wrote:The price of a house decreased by x% from 1998 to 1999 and increased by y% from 1999 to 2000. If the house cost M dollars in 2000, how much did it cost in 1998?

100*((M/1+X)/1-y)

(M/(1 + (s/100) ))/(1 - (y/100))

M/(1 + (x/100) - (y/100))

10,000M/((100+y)/(100-x))

10,000M/((100+x)/100-y))

OA is D
are you sure about the answer??? because there is something not right... it should be 10000M/((100-x)*(100+y))

this is how i solved it....
if 1998 is 100% so 1999=(100-x)%
suppose 100-x : 100%
so 2000: A% : 100+y
from this i will have A% of 2000 ( the percent increased y% in 2000)=(100-x)(100+y)/100=M ( M is known)
price of 2000 :(100-x)(100+y)/100=M
price of 1998 100......................=x so x=10000M/((100-x)*(100+y))

this is a bit time comsuming because i dont understand why answer is
10,000M/((100+y)/(100-x))
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by goyalsau » Tue Oct 05, 2010 7:06 pm
I am not able to do it by plugging in value for x and y and M

Can't we do it like that..
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by goyalsau » Tue Oct 05, 2010 7:37 pm
price of 1998 100......................=x so x=10000M/((100-x)*(100+y))

Yes you are right, Now i able to do it by plugging in values.
Even i was very confused why i am not able to do it by plugging in values in any option.

I would say no body should join Wisdom Mart, Worst education Center in INDIA.
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