more than d?

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more than d?

by sanju09 » Fri Feb 03, 2012 4:15 am
Is the median of the four different positive integers a, b, c, and d, more than d?
(1) The average (arithmetic mean) of a, b, c, and d, is d.
(2) The average (arithmetic mean) of a, b, and c, is d.


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by Mike@Magoosh » Fri Feb 03, 2012 12:23 pm
Hi, there. I'm happy to help with this. :)

Prompt: Is the median of the four different positive integers a, b, c, and d, more than d?

Keep in mind, the median of a four-item list is going to be the average of the 2nd and 3rd numbers when the numbers are in numerical order. For example, the median of {2,3,5,8} is (3+5)/2 = 4. Also, incidentally, the median of {8,3,5,2} is also 4 --- changing the order doesn't change the median. To calculate the median you must put them all in numerical order.

Statement #1: The average (arithmetic mean) of a, b, c, and d, is d.

Well, in general, there's no guarantee what the size of the mean will be compared to the median. For sets in which all items are the same, or sets in which items are symmetrically distributed, the mean = median. If there are outliers one way or the other, outliers pull the mean away from the median.

So, in set {3, 3, 3, 3}, d = 3, the mean = 3, and the median = 3, so the median is not greater than d.

But, in set {4, 13, 13, 10}, d = 10, mean = 10, median = 11.5, median is greater than d.
(NOTE: The low outlier of 4 pulls the mean below the median)

So, consistent with the conditions, we can answer the question either way. Statement #1 is insufficient.

Statement #2: The average (arithmetic mean) of a, b, and c, is d.

So, now, the average of {a, b, c} is d, so when we add d to make a four-element set, that also will have an average of d. When you add the mean of a set as a new element in the set, the mean doesn't change. So, curiously, Statement #2 contains the information in Statement #1 --- it's not a "different" statement, mathematically.

Again, take set ={3,3,3}, mean = 3, add d = 3 to make {3,3,3,3}, median = 3 is not greater than d = 3

But take set = {4,13,13}, mean = 10, add d = 10 to make [4,13,13,10}, median = 11.5 is greater than d = 10.

Two answers possible, statement #2 is insufficient.

Since Statement #2 contains Statement #1, nothing new is added when we combine the statements. Combined statements are also insufficient.

Answer = E

I'm not familiar with this problem source, but I will say that I've never seen a real GMAT DS question in which one statement can be completely deduced from the other statement. For that reason, I am a bit suspicious of this problem source.

Did everything I said in the solution make sense? Please let me know if I can clarify anything.

Here's a more typical GMAT DS question concerning median:

https://gmat.magoosh.com/questions/938

The question at that link should be followed by a video explanation of the answer.

Mike :)
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by sanju09 » Sat Feb 04, 2012 4:14 am
Mike@Magoosh wrote:Hi, there. I'm happy to help with this. :)

Prompt: Is the median of the four different positive integers a, b, c, and d, more than d?

Keep in mind, the median of a four-item list is going to be the average of the 2nd and 3rd numbers when the numbers are in numerical order. For example, the median of {2,3,5,8} is (3+5)/2 = 4. Also, incidentally, the median of {8,3,5,2} is also 4 --- changing the order doesn't change the median. To calculate the median you must put them all in numerical order.

Statement #1: The average (arithmetic mean) of a, b, c, and d, is d.

Well, in general, there's no guarantee what the size of the mean will be compared to the median. For sets in which all items are the same, or sets in which items are symmetrically distributed, the mean = median. If there are outliers one way or the other, outliers pull the mean away from the median.

So, in set {3, 3, 3, 3}, d = 3, the mean = 3, and the median = 3, so the median is not greater than d.

But, in set {4, 13, 13, 10}, d = 10, mean = 10, median = 11.5, median is greater than d.
(NOTE: The low outlier of 4 pulls the mean below the median)

So, consistent with the conditions, we can answer the question either way. Statement #1 is insufficient.

Statement #2: The average (arithmetic mean) of a, b, and c, is d.

So, now, the average of {a, b, c} is d, so when we add d to make a four-element set, that also will have an average of d. When you add the mean of a set as a new element in the set, the mean doesn't change. So, curiously, Statement #2 contains the information in Statement #1 --- it's not a "different" statement, mathematically.

Again, take set ={3,3,3}, mean = 3, add d = 3 to make {3,3,3,3}, median = 3 is not greater than d = 3

But take set = {4,13,13}, mean = 10, add d = 10 to make [4,13,13,10}, median = 11.5 is greater than d = 10.

Two answers possible, statement #2 is insufficient.

Since Statement #2 contains Statement #1, nothing new is added when we combine the statements. Combined statements are also insufficient.

Answer = E

I'm not familiar with this problem source, but I will say that I've never seen a real GMAT DS question in which one statement can be completely deduced from the other statement. For that reason, I am a bit suspicious of this problem source.

Did everything I said in the solution make sense? Please let me know if I can clarify anything.

Here's a more typical GMAT DS question concerning median:

https://gmat.magoosh.com/questions/938

The question at that link should be followed by a video explanation of the answer.

Mike :)

Great effort Mike@Magoosh! I pledge a typical GMAT DS could have one statement be derived from the other, as this is one of the many favorite tricks of the ultimate group of authors. I'll produce one as example if I won't find an expert to commend this. Besides, [spoiler]I am Sanjeev K Saxena, author of this question; and I suggest the answer to this DS be D[/spoiler].

If four different positive integers a, b, c, and d are put in ascending order, then the average (arithmetic mean) of integers is more than the second integer and less than the fourth integer in that order read from left to right. If the average (arithmetic mean) of integers in such a case is one of the integers, d in present case, then d must be the third integer from left to right, when a, b, c, and d are put in ascending order. And then, the median will be the average (arithmetic mean) of d and another positive integer less than d, which in any case is less than d. So NO! The median of the four different positive integers a, b, c, and d, is NOT more than d.

Hence, [spoiler]each statement alone is sufficient.

My D
[/spoiler]
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by santhoshsram » Sat Feb 04, 2012 9:05 pm
If the average (arithmetic mean) of integers in such a case is one of the integers, d in present case, then d must be the third integer from left to right, when a, b, c, and d are put in ascending order.
From this explanation I can see that (1) is sufficient. However I'm not sure if (2) is sufficient.

(2) The average (arithmetic mean) of a, b, and c, is d

(a + b + c) = 3d

Let's plug in values.
Let 3d = 12 => d = 4, let us pick a = 1, b = 5, c = 6 (The question does not say anything about the order of a, b, c, and d so I think these values are valid). Now the median of 1(a) 4(d), 5(b), 6(c) is 4.5 > d.

Now, let 3d = 18 =>d = 6, let us pick a = 1, b = 2, c = 15. Now median of 1(a), 2(b), 6(d), 15(c) is 4 < d.

So I think (2) alone is insufficient. Did I miss something?
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by Mike@Magoosh » Mon Feb 06, 2012 8:36 am
sanju09 wrote:
If four different positive integers a, b, c, and d are put in ascending order, then the average (arithmetic mean) of integers is more than the second integer and less than the fourth integer in that order read from left to right. If the average (arithmetic mean) of integers in such a case is one of the integers, d in present case, then d must be the third integer from left to right, when a, b, c, and d are put in ascending order. And then, the median will be the average (arithmetic mean) of d and another positive integer less than d, which in any case is less than d. So NO! The median of the four different positive integers a, b, c, and d, is NOT more than d.
With all due respect, Mr. Saxena, I believe you are mistaken in your statement that: "If four different positive integers a, b, c, and d are put in ascending order, then the average (arithmetic mean) of integers is more than the second integer and less than the fourth integer in that order read from left to right."

For example, the set: {4, 99, 100, 101} has an average of 76, which is below the second number --- between the first and second number. If we have a set of four numbers, S = {a, b, c, d}, not all of which are equal to one another, then all we can say is that the min(S) < mean(S) < max(S). The mean could be between the first & second numbers, between the second & third, between the third & fourth, equal to the second, or equal to the third. All of those are possibilities.

As to the statement: "If the average (arithmetic mean) of integers in such a case is one of the integers, d in present case, then d must be the third integer from left to right, when a, b, c, and d are put in ascending order." Here, d could be either the second number or the third number.

In the set {1, 4, 5, 6), mean = 4, the second number (and hence, less than the median).

In the set {3, 4, 5, 8}, mean = 5, the third number (and hence, greater than the median).

So, knowing that the mean is an element of the set is not sufficient for determining whether the median is greater or less than the mean.

Given, this, I still believe the answer to the question is E.

Does what I've said make sense? Are we in agreement?

Mike :)

Means of sets are very tricky things.
Last edited by Mike@Magoosh on Tue Feb 07, 2012 8:25 am, edited 1 time in total.
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by sanju09 » Tue Feb 07, 2012 1:34 am
Mike@Magoosh wrote:
sanju09 wrote:
If four different positive integers a, b, c, and d are put in ascending order, then the average (arithmetic mean) of integers is more than the second integer and less than the fourth integer in that order read from left to right. If the average (arithmetic mean) of integers in such a case is one of the integers, d in present case, then d must be the third integer from left to right, when a, b, c, and d are put in ascending order. And then, the median will be the average (arithmetic mean) of d and another positive integer less than d, which in any case is less than d. So NO! The median of the four different positive integers a, b, c, and d, is NOT more than d.
With all due respect, Mr. Saxena, I believe you are mistaken in your statement that: "If four different positive integers a, b, c, and d are put in ascending order, then the average (arithmetic mean) of integers is more than the second integer and less than the fourth integer in that order read from left to right."

For example, the set: {4, 99, 100, 111} has an average of 76, which is below the second number --- between the first and second number. If we have a set of four numbers, S = {a, b, c, d}, not all of which are equal to one another, then all we can say is that the min(S) < mean(S) < max(S). The mean could be between the first & second numbers, between the second & third, between the third & fourth, equal to the second, or equal to the third. All of those are possibilities.

As to the statement: "If the average (arithmetic mean) of integers in such a case is one of the integers, d in present case, then d must be the third integer from left to right, when a, b, c, and d are put in ascending order." Here, d could be either the second number or the third number.

In the set {1, 4, 5, 6), mean = 4, the second number (and hence, less than the median).

In the set {3, 4, 5, 8}, mean = 5, the third number (and hence, greater than the median).

So, knowing that the mean is an element of the set is not sufficient for determining whether the median is greater or less than the mean.

Given, this, I still believe the answer to the question is E.

Does what I've said make sense? Are we in agreement?

Mike :)

Means of sets are very tricky things.

Oh yes, I am in complete agreement with you, Mike, with only one thing to disagree that "{4, 99, 100, 111} has an average of 76"; and so very happy to pull out [spoiler]my D[/spoiler] and place [spoiler]your E[/spoiler] there in the answer scheme prepared in some dim-witted alacrity. This pushes me push that very button that I push rather once in a blue moon. Thank you!!

How was this DS, when all's said and done, Mike?
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by Mike@Magoosh » Tue Feb 07, 2012 8:28 am
Mr. Saxena,

I'm sorry about the line: "the set: {4, 99, 100, 111} has an average of 76." I mistyped --- I meant to type 101, not 111, for the fourth number.

Overall, I think your question is a wonderfully challenging DS question. I like it quite a bit. If I may ask, for whom do you write questions?

With great respect,
Mike :)
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by kris610 » Tue Feb 07, 2012 10:28 am
Mike@Magoosh wrote:Hi, there. I'm happy to help with this. :)

Prompt: Is the median of the four different positive integers a, b, c, and d, more than d?

Keep in mind, the median of a four-item list is going to be the average of the 2nd and 3rd numbers when the numbers are in numerical order. For example, the median of {2,3,5,8} is (3+5)/2 = 4. Also, incidentally, the median of {8,3,5,2} is also 4 --- changing the order doesn't change the median. To calculate the median you must put them all in numerical order.

Statement #1: The average (arithmetic mean) of a, b, c, and d, is d.

Well, in general, there's no guarantee what the size of the mean will be compared to the median. For sets in which all items are the same, or sets in which items are symmetrically distributed, the mean = median. If there are outliers one way or the other, outliers pull the mean away from the median.

So, in set {3, 3, 3, 3}, d = 3, the mean = 3, and the median = 3, so the median is not greater than d.

But, in set {4, 13, 13, 10}, d = 10, mean = 10, median = 11.5, median is greater than d.
(NOTE: The low outlier of 4 pulls the mean below the median)

So, consistent with the conditions, we can answer the question either way. Statement #1 is insufficient.

Statement #2: The average (arithmetic mean) of a, b, and c, is d.

So, now, the average of {a, b, c} is d, so when we add d to make a four-element set, that also will have an average of d. When you add the mean of a set as a new element in the set, the mean doesn't change. So, curiously, Statement #2 contains the information in Statement #1 --- it's not a "different" statement, mathematically.

Again, take set ={3,3,3}, mean = 3, add d = 3 to make {3,3,3,3}, median = 3 is not greater than d = 3

But take set = {4,13,13}, mean = 10, add d = 10 to make [4,13,13,10}, median = 11.5 is greater than d = 10.

Two answers possible, statement #2 is insufficient.

Since Statement #2 contains Statement #1, nothing new is added when we combine the statements. Combined statements are also insufficient.

Answer = E

I'm not familiar with this problem source, but I will say that I've never seen a real GMAT DS question in which one statement can be completely deduced from the other statement. For that reason, I am a bit suspicious of this problem source.

Did everything I said in the solution make sense? Please let me know if I can clarify anything.

Here's a more typical GMAT DS question concerning median:

https://gmat.magoosh.com/questions/938

The question at that link should be followed by a video explanation of the answer.

Mike :)
Hello Mike,
I'm just wondering why you considered a set of identical elements when the question stem states that a b c and d are four *different* +ve integers.

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by Mike@Magoosh » Tue Feb 07, 2012 11:39 am
Dear kris610,

I suppose in my initial analysis I simply overlooked the stipulation that the four elements are all different, but whether that stipulation is in place or not does not alter the analysis. If you consider the two sets I cited in my response to Mr. Saxena above, {1, 4, 5, 6) and {3, 4, 5, 8}, these satisfy all constraints and give divergent answers to the prompt.

Does that make sense?

Mike :)
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by sanju09 » Wed Feb 08, 2012 12:10 am
Mike@Magoosh wrote:Mr. Saxena,

I'm sorry about the line: "the set: {4, 99, 100, 111} has an average of 76." I mistyped --- I meant to type 101, not 111, for the fourth number.

Overall, I think your question is a wonderfully challenging DS question. I like it quite a bit. If I may ask, for whom do you write questions?

With great respect,
Mike :)

Thanks for the good words, Mike. I was just kidding like a toddler's settling of scores by pin pointing that little error, which I knew is nothing more than typo, hence you don't really need to be sorry for that. Besides, I am writing questions and literature for [spoiler]Avenues Abroad, www.avenuesabroad.org (We are Overseas Education Consultants, website is under repair and subject to change)[/spoiler] a fast growing institution for the standardized tests like GMAT, in our city.

Regards
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