Bill has d animals, 12 of which were sold at a profit of 10%

nice elimination technique, it's realy worth Kudos

Hi All,

Since this question asks for the TOTAL number of animals, and the answers are numbers, this is a perfect opportunity to TEST THE ANSWERS.

We're told a few Facts about a group of animals:
1) There are D total animals
2) 12 of them are sold at a 10% profit
3) The rest are sold at a 20% loss
4) The profits and losses "cancel out"

We're asked to solve for D.

Since 12 animals were sold for a profit, while the rest were sold for a loss, the total number of animals MUST be MORE than 12. Eliminate A and B.

Since we need the profits and losses to cancel out, we're likely looking for a situation in which there were FEWER animals sold for a loss than were sold for a profit (since each individual 'loss' is so much bigger than each individual 'profit').

Let's TEST Answer C....
IF there were 18 animals, 12 were sold for a profit and 6 were sold for a loss.

IF....each animal costs $100
The 12 that were sold for a 10% profit --> $110 each ---> 12($10 profit each) = $120 profit
The 6 that were sold for a 20% loss ---> $80 each --->6($20 loss each) = $120 loss

Here, the profits and losses cancel out, so C MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

20 % loss is twice 10 % profit thus we need twice the amount of sells to make up for the loss 1:2 or 6:12.

How can Someone Assume that All the animals had the same initial price EMPOWERgmatRichC ?????????????//?????///??

Just in tune with EMPOWERgmatRichC s reasoning , but using alligation theory



Further austin bro , we are told to find the



If you do not take the base ( Cost Price same ) it will be difficult to reach the overall profit/loss situation....


PS : However I accept your arguement that the price of the animals may be different , but for simplicity's sake ( coz we are only interested in the overall profit / loss) we can safely assume Cost Price as 100

PPS : If you consider the price of the items different ,

Say Price of 12 animals is 100 and price of D - 12 animals be 200 you can reach the conclusion but it will be very very lengthy and complicated process.

Hi stonecold,

You bring up a good point. Whoever wrote this question did NOT do as good a job with defining the 'restrictions' of the prompt as the GMAT question writers do. The 'intent' of the question is to assume that all of the animals were purchased at the same price - you cannot answer it otherwise (and if a question such as this one were to appear on your Official GMAT, the prompt would include that information).

GMAT assassins aren't born, they're made,
Rich

Answer:C
Lets consider the original cost to be C=100.
1- 10% profit ⇒ Selling S1=110
2- 20% loss ⇒ Selling S2=80
12 were sold at a profit of 10% ⇒ 12x110=12x100 (cost) + 12x10(profit) ⇒ For bill to make neither a profit nor a loss, he can loose no more than 12x10=120. From --2-- bill is loosing 20 dollars per animal ⇒ he can loose 6 animals for a total loss of 6x20=120 ⇒ total number of animals=12+6=18

-20----0----10
----1--:--2---

12=2x <-> x=6 <-> 3*6=18

Bill has d animals, 12 of which were sold at a profit of 10% each while the rest were sold at a loss of 20% each. Overall, Bill made neither a profit nor a loss. Which of the following is equal to d?

(A) 6
(B) 12
(C) 18
(D) 24
(E) 36

Using Weighted Averages
12 = w1
? = w2
a1 = 10
a2 = -20
aavg = 0 ( neither profit nor loss)

\(A2 - Aavg/Aavg - A1\) = W1/W2
\(-20 - 0/0 - 10\) = \(12/W2\)
\(2/1\) = \(12/W2\)
2W2 = 12
W2 = 6
Total = W1 + W2 = 12 + 6 = 18
C

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