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PKN
Bunuel
Set A = {a, b, c} and has a median of b. Is the average of a and c greater than b?

(1) The average of Set A is b
.

(2) c = 2a

Given A = {a, b, c} and has a median of b, implies that the elements of set A are either in the ascending order or in the descending order.
And 'b' is the median.

Question stem:- Is the average of a and c greater than b
Or, Is \(\frac{a+c}{2}>b\) ?
Or, Is \((a+c)>2b\) ?

St1:- The average of Set A is b
Note:-
1)The average value of a series of terms is equal to the median value when the terms are in A.P.(Arithmetic progression).
2) When all the terms in a series are equal, they are also in A.P. with common difference zero.
So, a, b, and c are in A.P.
Or, c+a=2b.

Answer to the question stem is No.
Sufficient.

St2:- c = 2a

1) If A={4,5,8}, then (a+c)>2b. Answer to question stem is Yes.
2) If A={4,7,8}, then (a+c)<2b. Answer to question stem is No.

Insufficient.

Ans. (A)

Hi PKN

I do not agree with the highlighted part above. Take the example below:

7,7,10,12,16..............Median= Mean=10..........it is not AP.
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Bunuel
Set A = {a, b, c} and has a median of b. Is the average of a and c greater than b?

(1) The average of Set A is b
.

(2) c = 2a

Given A = {a, b, c} and has a median of b, implies that the elements of set A are either in the ascending order or in the descending order.
And 'b' is the median.

Question stem:- Is the average of a and c greater than b
Or, Is \(\frac{a+c}{2}>b\) ?
Or, Is \((a+c)>2b\) ?

St1:- The average of Set A is b
Note:-
1)The average value of a series of terms is equal to the median value when the terms are in A.P.(Arithmetic progression).
2) When all the terms in a series are equal, they are also in A.P. with common difference zero.
So, a, b, and c are in A.P.
Or, c+a=2b.

Answer to the question stem is No.
Sufficient.

St2:- c = 2a

1) If A={4,5,8}, then (a+c)>2b. Answer to question stem is Yes.
2) If A={4,7,8}, then (a+c)<2b. Answer to question stem is No.

Insufficient.

Ans. (A)

Hi PKN

I do not agree with the highlighted part above. Take the example below:

7,7,10,12,16..............Median= Mean=10..........it is not AP.

My point is when the series is in AP , then mean=median. But vice-versa is not true.
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Hi PKN

I do not agree with the highlighted part above. Take the example below:

7,7,10,12,16..............Median= Mean=10..........it is not AP.

My point is when the series is in AP , then mean=median. But vice-versa is not true.

Actually, there is one exception to the highlighted part. When you have a set of 3 numbers only (as in the question) and mean equal to Median, then you conclude directly that it is AP. If the set has more than 3 numbers, then it true that it is not vice verse.
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Mo2men


Hi PKN

I do not agree with the highlighted part above. Take the example below:

7,7,10,12,16..............Median= Mean=10..........it is not AP.

My point is when the series is in AP , then mean=median. But vice-versa is not true.

Actually, there is one exception to the highlighted part. When you have a set of 3 numbers only (as in the question) and mean equal to Median, then you conclude directly that it is AP. If the set has more than 3 numbers, then it true that it is not vice verse.

Thank you. I am aware of the exception. ☺️
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Mo2men


Hi PKN

I do not agree with the highlighted part above. Take the example below:

7,7,10,12,16..............Median= Mean=10..........it is not AP.

My point is when the series is in AP , then mean=median. But vice-versa is not true.

Actually, there is one exception to the highlighted part. When you have a set of 3 numbers only (as in the question) and mean equal to Median, then you conclude directly that it is AP. If the set has more than 3 numbers, then it true that it is not vice verse.

Hello Mo2men

Can you explain below ?

Why--> When you have a set of 3 numbers only (as in the question) and mean equal to Median, then you conclude directly that it is AP.
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Given a 3-term set, one of the terms must always be the median. In this case, the median is b
this means that a
and c
are on either side of b
and that either a<b<c
or that c<b<a
. You are then asked whether the average of a
and c
is greater than b
.

Statement (1) gives that b
is the average of Set A as well as its median. Sets with the same median as average are always going to equally spaced sets around the median. And for any equally spaced set, the average of the two extremes (in this case a
and c
) will be the same as the average of the set as a whole. This means that the average of the set is b
and the answer to the question posed is a definitive "no".

If the rule described above wasn't immediately apparent, consider a set of three numbers 1, 2, and 3. The average of the three numbers will be 2, as will the median. Similarly, the average of the first and third terms will be 2.

Statement (2) gives you that c=2a
. While this gives you information about the relative values of c
and a
and would allow you to calculate the average of the terms in terms of a
, it does not give you any information about how these two numbers would relate to b
, so it is impossible to answer the question posed with the information given in statement (2). Statement (2) is therefore insufficient.

If you wanted to prove that with numbers, note that you can just place the median immediately adjacent to either a
or c
to move the average in either direction. For example, given that c=2a
, you could use 4 and 8. If you make the median 5, then the average of a
and c
, 6, is greater than the median (giving the answer "yes"); if you make the median 7, it's less than the median ("no"). Since you can answer the question either way, this proves that statement (2) is insufficient.

The correct answer is (A).
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