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Bunuel
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A veterinarian calculates the median weight of ten dogs. Later, she ad 16 Dec 2019, 01:32
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95% (hard)

Question Stats:

31% (02:18) correct 69% (01:44) wrong based on 158 sessions

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A veterinarian calculates the median weight of ten dogs. Later, she adds the weights of two additional dogs to her calculations. Is the median of the new list of twelve dogs' weights greater than the median of the original ten?

(1) The weight of one of the added dogs is greater than the average of the original ten, and the weight of the other added dog is less than the average of the original ten.

(2) The weights of the two added dogs are each greater than the median weight of the original ten.


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A veterinarian calculates the median weight of ten dogs. Later, she ad 16 Dec 2019, 02:24
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Median of 12 > Median of 10 ??

Statement 1: Insufficient. We do not know the average/median of weights. The average can be greater than median and vice versa. Both situations possible

A D / B C E

Statement 1: Sufficient. The weight of 2 dogs is more than the median of other 10. Surely, there will be a increasing shift of Median.

'B' is the winner.

Bunuel
A veterinarian calculates the median weight of ten dogs. Later, she adds the weights of two additional dogs to her calculations. Is the median of the new list of twelve dogs' weights greater than the median of the original ten?

(1) The weight of one of the added dogs is greater than the average of the original ten, and the weight of the other added dog is less than the average of the original ten.

(2) The weights of the two added dogs are each greater than the median weight of the original ten.


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A veterinarian calculates the median weight of ten dogs. Later, she ad 27 Dec 2019, 12:03
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Answer is E definitely.
Old median could be mode, thus even the shift cause median to remain same. This may or may not be the case. Hence insufficient.
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A veterinarian calculates the median weight of ten dogs. Later, she ad 03 Oct 2020, 13:45
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Bunuel can you please provide the explanation for this question?
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A veterinarian calculates the median weight of ten dogs. Later, she ad 04 Oct 2020, 08:49
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Bunuel
A veterinarian calculates the median weight of ten dogs. Later, she adds the weights of two additional dogs to her calculations. Is the median of the new list of twelve dogs' weights greater than the median of the original ten?

(1) The weight of one of the added dogs is greater than the average of the original ten, and the weight of the other added dog is less than the average of the original ten.

(2) The weights of the two added dogs are each greater than the median weight of the original ten.
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The issue is whether the two added dogs are both greater than the original median (median increases), or if one is greater and the other is less (median stays the same). You should also think about what happens if the all the 10 or 12 dogs have the same weight.

Stat (1)
Although this seems like it would be sufficient because both would be greater than the original mean, there are subtle differences between median and average (mean).

For instance, if the weights are {10, 10, 10, 10, 10, 10, 11, 11, 11, 10000}, then the average will be over 1000. In this contrived example, the two added weights could be 999 and 1100, which would move the median from 10 to 10.5 because the new set is {10, 10, 10, 10, 10, 10, 11, 11, 11, 999, 1100, 10000}.
Insufficient

Stat (2)
Clearly, the median could change. But it could also stay the same: for instance, if the original 10 dogs weigh 10 lbs each, then the median won't change no matter what the 2 added dogs weigh.
Insufficient

Stat (1+2)
The examples described above are both still possible.
Insufficient

(e) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
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