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If the price increased by x% from 2001 to 2002 and by y% fro

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If the price increased by x% from 2001 to 2002 and by y% fro  [#permalink]

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New post 18 Jan 2010, 00:29
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If the price increased by x% from 2001 to 2002 and by y% from 2002 to 2003, what is the percentage increase from 2001 to 2003?

(1) xy = 30
(2) 100x + 100y + xy = 800
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Re: Percentage problem  [#permalink]

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New post 18 Jan 2010, 04:54
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Price in 2002=(1+x)*Price-in-2001

Price in 2003=(1+y)*Price-in-2002=(1+x)(1+y)*Price-in-2001=(1+x+y+xy)*Price-in-2001

So basically, we need to determine (1+x+y+xy), where x=X/100, y=Y/100. 1+X/100+Y/100+XY/10,000


St1: XY=30 - > Clearly Insufficient

St2: 100X+100Y+XY=800; => X+Y+XY/100 = 8 => X/100+Y/100+XY/10,000 = 0.08 That's what we wanted. Sufficient


(B)
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Re: percent problem  [#permalink]

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New post 21 Mar 2011, 21:13
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This is a trap called "lone wolf". Wolves are loner. The question will imply that you will need two equations to solve since there are two variables. However we don't need two equations. Back to the question. If you consider the general formula for percentage change is
X+y+xy/100
S2 reduces to this form . Hence sufficient
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Re: percent problem  [#permalink]

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New post 21 Mar 2011, 21:36
Assume price in 2001 = 100
--> price in 2002 = 100(1+x/100) =100 + x
--> price in 2003 = (100+x)(1+y/100) = 100 + x + y + xy/100

from 2) x + y + xy/100 =8 --> sufficient.
1) is not sufficient.

it would of course be faster if one remembered the formula.
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Re: percent problem  [#permalink]

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New post 21 Mar 2011, 22:35
asmit123 wrote:
If the price increased by X% from 2001 to 2002 and by Y% from 2002 to 2003,what is the % increase from 2001 to 2003?
1. XY = 30
2. 100X + 100Y + XY = 800

Price 2001 =P
=>Price in 2002, P + x% of P = P + Px/100 = P(1 + x/100)
=> Price in 2003, P(1 + x/100)(1 + y/100)

Percentage Increase =>(P ((1 + x/100)(1 + y/100) -1)/P) *100
=>(((1+X/100)(1+Y/100))-1 ) *100
=>(100+X)(100+Y)-100
=>100X+100Y+XY+10000-100
=>800+10000-100

Hence B
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Re: percent problem  [#permalink]

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New post 22 Mar 2011, 00:54
%age increase = {P(1+x/100)(1+y/100) - P}/P

where P is the original price.

(10000 + 100y + 100x + xy)/100000 - 1

Clearly, the answer is B.
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Re: percent problem  [#permalink]

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New post 22 Mar 2011, 18:04
rephrasing question we have x+y+(xy/100) and thats what we need to find.

1. Not sufficient . there are different possible x and y that satisfy the given condition.

2. Sufficient . as we have the expression in similar form.

Answer is B.
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Re: If the price increased by x% from 2001 to 2002 and by y% fro  [#permalink]

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New post 20 Sep 2013, 10:58
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amitanand wrote:
If the price increased by x% from 2001 to 2002 and by y% from 2002 to 2003, what is the percentage increase from 2001 to 2003?

(1) xy = 30
(2) 100x + 100y + xy = 800


m12 q35

If the price increased by x% from 2001 to 2002 and by y% from 2002 to 2003, what is the percentage increase from 2001 to 2003?

The price in 2001 - \(p\);
The price in 2002 - \(p*(1+\frac{x}{100})\);
The price in 2003 - \(p*(1+\frac{x}{100})(1+\frac{y}{100})=p(1+\frac{y}{100}+\frac{x}{100}+\frac{xy}{10,000})\);

The percentage increase from 2001 to 2003 is \(\frac{2003-2001}{2001}*100=x+y+\frac{xy}{100}\).

(1) xy = 30. Not sufficient.

(2) 100x + 100y + xy = 800 --> divide both sides by 100: \(x+y+\frac{xy}{100}=8\), directly gives the answer. Sufficient.

Answer: B.
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Re: If the price increased by x% from 2001 to 2002 and by y% fro  [#permalink]

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New post 22 Sep 2013, 02:17
amitanand wrote:
If the price increased by x% from 2001 to 2002 and by y% from 2002 to 2003, what is the percentage increase from 2001 to 2003?

(1) xy = 30
(2) 100x + 100y + xy = 800


Let price, P be 100.

after x% increase, price P1 = 100+x

After y% further increase, price P2 = 100+x + (100+x)*(y/100)

% increase = Increase in price / original price = (p2-p)*100/p = (100 X +100Y+xy)/100.

B gives numerator value. so sufficient but A is not.
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Re: If the price increased by X% from 2001 to 2002 and by Y%  [#permalink]

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