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17 Feb 2022, 05:45
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The median is larger than the average for which one of the following sets of integers?
(A) {8, 9, 10, 11, 12}
(B) {8, 9, 10, 11, 13}
(C) {8, 10, 10, 10, 12}
(D) {10, 10, 10, 10, 10}
(E) {7, 9, 10, 11, 12}
(A) {8, 9, 10, 11, 12}
(B) {8, 9, 10, 11, 13}
(C) {8, 10, 10, 10, 12}
(D) {10, 10, 10, 10, 10}
(E) {7, 9, 10, 11, 12}
17 Feb 2022, 07:33
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Bunuel
The median is larger than the average for which one of the following sets of integers?
(A) {8, 9, 10, 11, 12}
(B) {8, 9, 10, 11, 13}
(C) {8, 10, 10, 10, 12}
(D) {10, 10, 10, 10, 10}
(E) {7, 9, 10, 11, 12}
(A) {8, 9, 10, 11, 12}
(B) {8, 9, 10, 11, 13}
(C) {8, 10, 10, 10, 12}
(D) {10, 10, 10, 10, 10}
(E) {7, 9, 10, 11, 12}
Useful property: If the values in a certain set are equally spaced, then the mean of the set = the median of the set.
Scan the answer choices.....
Notice that, if answer choice E were {8, 9, 10, 11, 12}, then the five values would be equally spaced, in which case the mean = median = 10
However, since the actual set of values is {7, 9, 10, 11, 12}, the median is still 10, but the mean is now slightly less than 10.
Answer: E
gmatchile
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17 Feb 2022, 09:19
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Do you remember what median and mean are?
Let's remember:
What is median?
Median is the central term of a sample, once the data has been sorted (increasing or decreasing).
What is the average?
The average, or arithmetic mean, is the average of all values in a sample.
ie: Average=Sum of all terms divided by the number of terms.
Observing the alternatives, we see that in all of them the median is 10.
So now we would have to see in which of the alternatives the average is less than 10?
To go faster, we will apply the concept of arithmetic progression.
What is an arithmetic progression?
If in a sample (once ordered), the difference of two consecutive terms is a value that does not change (a constant value), then said ordering is an arithmetic progression.
And why is it important to identify arithmetic progressions?
Why in arithmetic progressions is there more than one way to find average?
Try to see how else, besides adding the terms and dividing by the number of terms, you can get the average, in an arithmetic progression.
Give yourself your time.
Very well!
Since the median's predecessor and successor data are the same distance apart (equally spread out), we can by simple inspection add the median predecessor and successor data and divide by two to get the average.
One question, does this continue to hold until the first and last term of the sample? Why?
You can, give an answer and explain why.
Excellent!
Indeed, if we add the first and last term of an arithmetic progression, and then divide it by 2, we obtain the average.
Very interesting, right?
Let's see which of the alternatives gives arithmetic progression or very similar to arithmetic progression?
That's how it is!
Alternatives A) B) and E) deliver orders very similar to arithmetic progression.
In A) The median is equal to the mean.
In B) the only difference with A) is that the last term exceeds by 1 (12 and 13 respectively) the last term of A). Then in B) the mean will be greater than the median of B).
In D) the only difference with A), is that the first term of D) is less than one unit, than the first term of A), for that reason the average of D) will be less than the median of D).
So the correct answer is D.
Data Sufficiency Challenge in the following link
https://gmatchile.cl/spip.php?article14029
Let's remember:
What is median?
Median is the central term of a sample, once the data has been sorted (increasing or decreasing).
What is the average?
The average, or arithmetic mean, is the average of all values in a sample.
ie: Average=Sum of all terms divided by the number of terms.
Observing the alternatives, we see that in all of them the median is 10.
So now we would have to see in which of the alternatives the average is less than 10?
To go faster, we will apply the concept of arithmetic progression.
What is an arithmetic progression?
If in a sample (once ordered), the difference of two consecutive terms is a value that does not change (a constant value), then said ordering is an arithmetic progression.
And why is it important to identify arithmetic progressions?
Why in arithmetic progressions is there more than one way to find average?
Try to see how else, besides adding the terms and dividing by the number of terms, you can get the average, in an arithmetic progression.
Give yourself your time.
Very well!
Since the median's predecessor and successor data are the same distance apart (equally spread out), we can by simple inspection add the median predecessor and successor data and divide by two to get the average.
One question, does this continue to hold until the first and last term of the sample? Why?
You can, give an answer and explain why.
Excellent!
Indeed, if we add the first and last term of an arithmetic progression, and then divide it by 2, we obtain the average.
Very interesting, right?
Let's see which of the alternatives gives arithmetic progression or very similar to arithmetic progression?
That's how it is!
Alternatives A) B) and E) deliver orders very similar to arithmetic progression.
In A) The median is equal to the mean.
In B) the only difference with A) is that the last term exceeds by 1 (12 and 13 respectively) the last term of A). Then in B) the mean will be greater than the median of B).
In D) the only difference with A), is that the first term of D) is less than one unit, than the first term of A), for that reason the average of D) will be less than the median of D).
So the correct answer is D.
Data Sufficiency Challenge in the following link
https://gmatchile.cl/spip.php?article14029
