Last week the average (arithmetic mean) of the daily costs of John's

Question

Last week the average (arithmetic mean) of the daily costs of John's lunch on Monday, Tuesday, and Wednesday was $6. What was the median of the daily costs of John's lunch for the 3 days? (1) The cost of John's lunch on Monday of last week was $5. (2) The cost of John's lunch on Tuesday of last week was $6. 2024-01-24_12-29-46.png

Bunuel · Accepted Answer

Last week the average (arithmetic mean) of the daily costs of John's lunch on Monday, Tuesday, and Wednesday was $6. What was the median of the daily costs of John's lunch for the 3 days?

Essentially, we need to calculate the median of three values, knowing that their average is 6, so the sum is 18. The median of an odd count of values is the middle term when the values are in order.

(1) The cost of John's lunch on Monday of last week was $5.

This implies that one of the values is 5. If one of the remaining values is less than 5 and another is more than 5, say 1, and 12, respectively, then the median will be 5. However, if both of the remaining values are more than 5, say 6 and 7, then the median would be 6. Not sufficient.

(2) The cost of John's lunch on Tuesday of last week was $6.

Since one of the values equals the average, the remaining two values cannot be both greater or both less than 6 (in one case the sum will be more than 18 and in another less than 18). Thus, it must be that (value 1) ≤ (value 2 = 6) ≤ (value 3). Therefore, the median value must be 6. Sufficient.

Answer: B.

sayan640 · Answer

Whenever you find mean equal to one of values in the set , that has to be the median. Solved it under 50 secs. 
6=6  only present in option B.

Else ,

Using statement 1 , two possibilities are there :-  5,6,7  ==> 6 is the median  OR 5,5,8  ===> 5 is the median

Using statement 2 , only one median is possible no matter any combinations of numbers in the set are considered , 
for example , 6, 6,6  ==> 6 is the median

5,6,7  ===> 6 is the median

3,6,9 ===> 6 is the median

Hence B is the ans.

GMATinsight

av722 · Answer

(M+T+W)/3=6

(1) The cost of John's lunch on Monday of last week was $5.
From this, (5+T+W)/3=6
5+T+W=18
T+W=13

There many different combinations of T and W to get 13. We could have 5 and 8, 6 and 7, 10 and 3, the list goes on. Each scenario would have a different median. A and D are out.

(2) The cost of John's lunch on Tuesday of last week was $6.
Although at first glance it seems like this would be ruled out for the same reason (1) is ruled out, this is sufficient. 
If the cost of John's lunch is $6 on Tuesday, we could enter two possible scenarios:

a) M and W's cost is also $6. This is one way we'd have an average of $6. 
b) M and W's costs are different. In order to get an average of 6, though, one value has to be less than $6 and the other has to be greater than $6.

For example, 
if M = $5 and W = $7, we'd get an average of 6. 
However, if M = $3 and W = $3, our average would be (3+3+6)/3 = 4. 
Notice how one term has to be lesser and one term has to be greater than 6 in order for the average to be equal to 6. As a result, we know the median would have to be 6.

B is sufficient .

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Last week the average (arithmetic mean) of the daily costs of John's

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