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Problem Solving Territory: 700+ questions only..............

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Problem Solving Territory: 700+ questions only.............. [#permalink] New post 22 Sep 2006, 20:53
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Folks, I will be posting questions from Problem Solving section regularly in this thread.............

Here comes our first question

1. A shopkeeper increases the price of the article by X% and then decreases the resulting price by X%. As a result the price of the article is reduced by 180 $. Ater one more such cycle the price is further reduced by 153$. Find the oriiginal price of the article.
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)


Last edited by cicerone on 24 Sep 2008, 23:58, edited 1 time in total.
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 [#permalink] New post 22 Sep 2006, 23:10
CP (1+x)(1-x) = 180
180 (1+x)(1-x) = 153 => (1-x^2) = 153/180

=> CP/180 = 180/153
=> CP = 210.60 (approx)
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 [#permalink] New post 23 Sep 2006, 01:19
paddyboy wrote:
CP (1+x)(1-x) = 180
180 (1+x)(1-x) = 153 => (1-x^2) = 153/180

=> CP/180 = 180/153
=> CP = 210.60 (approx)


Hey paddyboy check out again..........
u r missing something........
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)


Last edited by cicerone on 24 Sep 2008, 23:57, edited 1 time in total.
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 [#permalink] New post 23 Sep 2006, 01:28
He definied x = X / 100 ;) It makes sens as x < X :)

Actually, he could set y=(1+X/100)(1-X/100) :)
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 [#permalink] New post 23 Sep 2006, 01:50
Fig wrote:
He definied x = X / 100 ;) It makes sens as x < X :)

Actually, he could set y=(1+X/100)(1-X/100) :)


Hey fig, that's fine.
But i said the price is decreased by 180 and not that the decreased price is 180.

So how can he write cp(1+x)(1-x) =180
CP(1+x)(1-x) is the reduced price where as 180 is the reduction in the prize.

The correct expression would be
CP(1+x)(1-x) = P-180, where P= initial price and x=X/100.
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)


Last edited by cicerone on 24 Sep 2008, 23:57, edited 1 time in total.
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 [#permalink] New post 23 Sep 2006, 01:55
Absolutly :)
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 [#permalink] New post 23 Sep 2006, 03:45
paddyboy wrote:
CP (1+x)(1-x) = 180
180 (1+x)(1-x) = 153 => (1-x^2) = 153/180

=> CP/180 = 180/153
=> CP = 210.60 (approx)


DAMN! 2 days to go and I'm crapping up!!! :evil:

Okay.. here goes.

CP - CP(1-x^2) = 180
=> CP*x^2 = 180 --(1)

CP(1-x^2) - CP(1-x^2)(1-x^2) = 153
=> CP(1-x^2)x^2 = 153 --(2)

Using (2)/(1), we get
(1-x^2) = 153/180
=> x^2 = 1 - 153/180 = 27/180 = 3/20 --(3)

From (1) and (3) we get,

CP(3/20) = 180 => CP = 1200.

Howzzat? Don't tell me I'm wrong again :(
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 [#permalink] New post 23 Sep 2006, 03:55
paddyboy wrote:
paddyboy wrote:
CP (1+x)(1-x) = 180
180 (1+x)(1-x) = 153 => (1-x^2) = 153/180

=> CP/180 = 180/153
=> CP = 210.60 (approx)


Oops! 2 days to go and I'm crapping up!!! :evil:

Okay.. here goes.

CP - CP(1-x^2) = 180
=> CP*x^2 = 180 --(1)

CP(1-x^2) - CP(1-x^2)(1-x^2) = 153
=> CP(1-x^2)x^2 = 153 --(2)

Using (2)/(1), we get
(1-x^2) = 153/180
=> x^2 = 1 - 153/180 = 27/180 = 3/20 --(3)

From (1) and (3) we get,

CP(3/20) = 180 => CP = 1200.

Howzzat? Don't tell me I'm wrong again :(


paddyboy, will u go on this monday for the GMAT? :)
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 [#permalink] New post 23 Sep 2006, 03:59
Tuesday actually, but I'm not counting it... :peek
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 [#permalink] New post 23 Sep 2006, 04:04
I c :) Best of luck for it :D U are on the right track for it :)

Next week seems to be the week for at least 3 of us ;)
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 [#permalink] New post 23 Sep 2006, 04:08
Who's the third?

And dude, :beer Best of luck! I'm sure we'll maul the GMAT :2gunfire:

Or as Futuristic put it, "Spread it out on the lawn and jump on it with hob-nailed boots!" :lol:
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 [#permalink] New post 23 Sep 2006, 04:31
I prefer to PM u it :)
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Re: Problem Solving Territory: 700+ questions only.......... [#permalink] New post 25 Sep 2006, 14:37
cicerone wrote:
Folks, I will be posting questions from Problem Solving section regularly in this thread.............

Here comes our first question

1. A shopkeeper increases the price of the article by X% and then decreases the resulting price by X%. As a result the price of the article is reduced by 180 $. Ater one more such cycle the price is further reduced by 153$. Find the oriiginal price of the article.


could you also give the answer choices. that way we can test the fastest way to go about this..
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 [#permalink] New post 27 Sep 2006, 19:58
Folks, this is again based on the basic funda of Percentages.

If a number is increased by x% and then decreased by x% there always will be a decrease.

It is given that the price (P)of the article is increased by x% and then decreased by x%. So the price will definitely decrease. Let the price decrease by y%

So clearly it is given that the price is decreased by 180 $.
Hence y%(P) = 180.............................(1)
Now the price of the artice is P-180. Since that same change is applied again on the present price this also will decrease by the same y%
Clearly it is given that the present price is decreased by 153$
Hence y%(P-180) = 153.....................(2)

From (1) and (2)
y%(180) = 27

ie y = 15.

Substituting the value of y in (1) we get
15%(P) = 180
Hence P=1200.

Now since u know the concept , u can use the options given in the exam to solve the questions like these.

Let's take 1200 as our first option
Initially the price is reduced by 180.

So the present price is 1020.

Now if 180 is the decrease on 1200, calculate the proportinate decrease on 1020.....

So i mean 1200------180
1020------??
ie (1020/1200)180 = (102/120)180 =153 which is exactly the second decrease in the given question..

Hence 1200 must be the answer........

I think i am lucid
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 [#permalink] New post 27 Sep 2006, 20:34
Ok folks, let's get into the next question.........

2.
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 [#permalink] New post 28 Sep 2006, 22:44
Hey, what happened?

Kevin, yezz, paddyboy,......... where r u?
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 [#permalink] New post 29 Sep 2006, 17:54
cicerone all of them are drinking beer !!!
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 [#permalink] New post 29 Sep 2006, 18:46
CE=1, not sure on how to calculate CD :roll:
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 [#permalink] New post 29 Sep 2006, 22:31
hey, I know this problem has to be solved based on similarity of triangles. But finding it difficult to identify which triangles need to be compared.

Cicerone, great initiative. Please keep posting questions.
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 [#permalink] New post 04 Oct 2006, 07:57
Answer pleaseeeeeeee
  [#permalink] 04 Oct 2006, 07:57
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