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Gmatprep PT1 question

Expert replies
by Reader » Wed Mar 18, 2009 5:33 pm
Image

My question is, how do we know? If 10 children have both or if 5 children developed both, then the answer will be different.
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Source: — Data Sufficiency |

by Goal to 800 » Thu Mar 19, 2009 6:14 am
(C)

1000 - 880 (who have nothing) = 120

120 - 20 (with fever) = 100 (with inflamm)

So the answer would be 100 children developed inflammation at the site...

The question is worded to say "some developed inflammation and some developed fever"

To me, that means that no children developed both at the same time, so (A) & (B) would be sufficient TOGETHER.

It seems trickier than is written because you are used to harder Venn Diagram problems like that...

Let me know if that helps you,

Jonathan.
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by Jose Ferreira » Thu Mar 19, 2009 8:10 pm
Hi,

I think the question allows for the possibility that some students get BOTH inflammation AND fever.

There are 1000 total students.

Statement 1: 880 have neither inflammation nor fever. This means that the number that either
1) Have fever only
2) Have inflammation only
OR
3) Have BOTH fever and inflammation
is 1000 - 880 = 120

Statement 2: 20 have fever.
This means that the sum of those who
1) Have fever only
OR
3) Have BOTH fever and inflammation
is 20

Together, the statements tell us that 1) + 2) + 3) = 120, and that 1) + 3) = 20. Therefore, 2), the number of people that get inflammation but not fever, is 120 - 20 = 100. The answer is C.

Note that we don't know how many fall in category 1), fever only, or in category 3), BOTH fever and inflammation. We just know that the sum of those groups is 100.
Jose Ferreira
Founder and CEO, Knewton, Inc.
https://www.knewton.com/gmat
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by Goal to 800 » Fri Mar 20, 2009 3:08 pm
Jose Ferreira wrote:Hi,

I think the question allows for the possibility that some students get BOTH inflammation AND fever.

There are 1000 total students.

Statement 1: 880 have neither inflammation nor fever. This means that the number that either
1) Have fever only
2) Have inflammation only
OR
3) Have BOTH fever and inflammation
is 1000 - 880 = 120

Statement 2: 20 have fever.
This means that the sum of those who
1) Have fever only
OR
3) Have BOTH fever and inflammation
is 20

Together, the statements tell us that 1) + 2) + 3) = 120, and that 1) + 3) = 20. Therefore, 2), the number of people that get inflammation but not fever, is 120 - 20 = 100. The answer is C.

Note that we don't know how many fall in category 1), fever only, or in category 3), BOTH fever and inflammation. We just know that the sum of those groups is 100.
I would normally agree with your strategy; however, the question does not give that much information. It states that the some got inflammation and some got fever. I cannot infer that you can get both from that and the information provided

880 out of 1000 have nothing

120 have either inflamm or fever

since 20 have fever, simple subtraction...

The question is too ambiguous to infer that you can get both...
(in reality, you can probably get both effects; however, we can only base our answers on information provided)

Let me know if I am thinking clearly...

Thanks,
Jonathan.
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by Reader » Sat Mar 21, 2009 11:58 pm
I sort figured it out , it doesn't matter how many children has both. We know 120 children that has to have fever or inflammation or both, of 20 has fever, the rest 100 will have only inflammation, we don't care if we have 100 people or 120 people have inflammation, we only care that we know 100 children has it only.
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