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# A veterinarian calculates the median weight of ten dogs. Later, she ad

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A veterinarian calculates the median weight of ten dogs. Later, she ad  [#permalink]

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12 Jul 2018, 19:45
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Difficulty:

85% (hard)

Question Stats:

22% (01:41) correct 78% (01:54) wrong based on 23 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.

A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient.

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Re: A veterinarian calculates the median weight of ten dogs. Later, she ad  [#permalink]

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12 Jul 2018, 21:29
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To find if the median of the new list of twelve dogs' weights greater than the median of the original ten

Statement 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.

Case 1

Lets assume the weights of the dogs as 1,2,3,4,5,5,5,5,5,5

In this case the median is 5, mean is $$\frac{39}{6} = 6.5$$

Lets assume the weights of new dogs as 6 and 10, since one of the dog weighs less than the average of 10 and the other dog weights more than average of ten dogs.

median weight of 12 dogs is still 5

In this case the median of 12 dogs is same as median of 10 dogs

case 2

Lets assume the weights of the dogs as 1,2,3,4,5,6,7,8,9,10

In this case the median is $$\frac{5+6}{2} = 5.5$$, mean is $$5.5$$

Lets assume the weights of new dogs as 12 and 12, since both the new dogs weigh more than the median weight of 10 dogs

The new median is $$\frac{6+7}{2} = 6.5$$

new median is greater than old median

Since two different cases are possible, statement 1 is not sufficient

Statement 2

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

case 1

The weights of 10 dogs are 1,2,3,4,5,6,7,8,9,10

In this case the new median is greater than old median

case 2

The weights of 10 dogs are 5,5,5,5,5,5,5,5,5,5

In this case the new median is equal to old median

Statement 2 not sufficient

Combining statements 1 and 2 together

We still get two cases where the new median is greater than old median and where the new median is equal to old median.

Hence option E
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Re: A veterinarian calculates the median weight of ten dogs. Later, she ad &nbs [#permalink] 12 Jul 2018, 21:29
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