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Re: Last year Leo bought two paintings. This year he sold them for $2,000 [#permalink]
Would someone mind explaining this one more thoroughly and explain what I am doing wrong?

Profit margin = (Sales - Cost) / Sales


painting with profit:
1/4 = (2000 - x) / 2000

x = 1500

profit = 500


painting with loss:
-1/4 = (2000 - x)/2000

x = -2500

loss = -500

net = 0
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Re: Last year Leo bought two paintings. This year he sold them for $2,000 [#permalink]
bdsco337 wrote:
Would someone mind explaining this one more thoroughly and explain what I am doing wrong?

Profit margin = (Sales - Cost) / Sales


painting with profit:
1/4 = (2000 - x) / 2000

x = 1500

profit = 500


painting with loss:
-1/4 = (2000 - x)/2000

x = -2500

loss = -500

net = 0


bdsco337

we need to find the actual value of price for which it was sold for a profit @25% and loss 25 %:


Case 1 :Price of painting with 25% profit
x*1.25= 2000
=> 2000/1.25= 1600$

Price of painting with loss 25%=2000/.75= 2666$
total invested amount= $ 4266 ( 2666+1600) ; this will be our actual cost at which two paintings were purchased at or say amount invested

total SP= $4000 ( given )
so net gain or loss= SP-CP= 4000-4266= (-) $266, option C

Hope this helps..

Alternatively ; coming to your solution:


Solving using your formula
the formula which you have used :

for profit relation would be
.25= (2000-x)/x
x=1600 $

for loss
.25= (x-2000)/x
x= 2666$

Hope this helps
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Re: Last year Leo bought two paintings. This year he sold them for $2,000 [#permalink]
LT2018 wrote:
Is there any other way?



LT2018 :


for profit relation would be
.25= (2000-x)/x
x=1600 $

for loss
.25= (x-2000)/x
x= 2666$

total amount invested : 2666+1600 = $4266

net selling price : 2000+2000 = 4000

so net profit or loss : SP - CP = 4000-4266 = (-) $266; option C
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Re: Last year Leo bought two paintings. This year he sold them for $2,000 [#permalink]
Given : Selling price of both the items is 4000$ ( 2000 + 2000 ) and one was sold at 25% profit, other 25% loss

Assume cost price of one object was x $ and other y $. Say x was sold at 25% profit and y at loss =>

x + (x/4) = 2000 and y - (y/4) = 2000 [Since 25% of x is 25x/100 = x/4]

Solving we get x = 1600, y = 2666.66

Cost price of both items = 4266 $
Selling Price = 4000 $

So he sold at a loss > 100$
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Re: Last year Leo bought two paintings. This year he sold them for $2,000 [#permalink]
https://gmatclub.com/forum/percentages- ... l#p1596564

According to the method explained above net loss percentage = \((p^2)/100\)
25^2/100 = 625/100% = 6.25 % percentage loss
total cost = 2000*2 = 4000
total loss = 6.25*4000/100=~240
Hence answer C
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Re: Last year Leo bought two paintings. This year he sold them for $2,000 [#permalink]
Archit3110 wrote:
Bunuel wrote:
Last year Leo bought two paintings. This year he sold them for $2,000 each. On one, he made a 25% profit, and on the other he had a 25% loss. What was his net loss or profit?

A. He broke even.
B. He lost less than $100.
C. He lost more than $100.
D. He earned less than $100.
E. He earned more than $100.



Price of painting with 25% profit= 2000/1.25= 1600$
Price of painting with loss 25%=2000/.75= 2666$
total invested amount= $ 4266

total SP= $4000
so net gain or loss= SP-CP= 4000-4266= (-) $266, option C


I have a finance background so the concept of profit for me entails net income/revenue where net income = Revenue - COGS(Cost). I know your solution is correct but I am having a hard time wrapping my head around this. If the revenue in the first example is $2000 and the cost was $1600, that leaves a profit of $400. Profit Margin = (Net Income/Revenue) = ($400/$2000) = 20%. How can I visualize this problem differently?
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Re: Last year Leo bought two paintings. This year he sold them for $2,000 [#permalink]
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