To find if the median of the new list of twelve dogs' weights greater than the median of the original tenStatement 1The 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 1Lets 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 2Lets 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 sufficientStatement 2The 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 sufficientCombining 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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