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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 19 Oct 2010, 00:58
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 11 is equal to the median of five numbers, what is the value of X

(1) 7 < x < 11
(2) x is the median of the five numbers

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first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 20 Oct 2010, 08:46
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satishreddy
if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X

1) 7<X<11
2) X is the median of five numbers

first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem

I am not convinced with the above answers...

from 1 ) 7 < X< 11 === > X is like 7.1, or 7.2 or 7.5 or 8,8.1 or 9 0r 9.5 ...

there are so many possibilites so i can not take only 8,9,10 as X

so i can not decide from 1

from 2) X is the median of five numbers then === > X, 7,2,16,11 ....

==> how can i arrange if i dont know the value of X in either descending or ascending..

so from the both also it wont be possible to get the ans i believe...
So ans E ...

Do let me know if i am wrong....
Show more

There is one more thing we know from the stem: "the average arithmetic mean of the five numbers x, 7, 2, 16, 11 is equal to the median of five numbers" --> \(\frac{x+2+7+11+16}{5}=median\). As there are odd number of terms then the median is the middle term when arranged in ascending (or descending) order but we don't know which one (median could be x, 7 or 11).

(1) 7 < x < 11 --> when ordered we'll get 2, 7, x, 11, 16 --> \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

(2) x is the median of five numbers --> the same info as above: \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

Answer: D.
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 19 Oct 2010, 03:22
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satishreddy
if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X

1) 7<X<11
2) X is the median of five numbers

first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem
Show more


For finding out median you have to arrange the numbers in increasing or decresing order, then only you have to take the middle term as median.
So now the order will be 2,7,x,11,16
And in that way ,you will have the same median value i.e, 9.

So answer is D.


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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 20 Oct 2010, 06:24
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satishreddy
if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X

1) 7<X<11
2) X is the median of five numbers

first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem
Show more

I am not convinced with the above answers...

from 1 ) 7 < X< 11 === > X is like 7.1, or 7.2 or 7.5 or 8,8.1 or 9 0r 9.5 ...

there are so many possibilites so i can not take only 8,9,10 as X

so i can not decide from 1

from 2) X is the median of five numbers then === > X, 7,2,16,11 ....

==> how can i arrange if i dont know the value of X in either descending or ascending..

so from the both also it wont be possible to get the ans i believe...
So ans E ...

Do let me know if i am wrong....
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 02 Dec 2012, 08:11
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The average of the five numbers x,7,2,16,11 is equal to the median of the five numbers. What is x?

1) 7<x<11

2)x is the median of the 5 numbers


STAT1
7<x<11
we can write the numbers in increasing order as
2,7,x,11,16
Avg = Median (and x will be the median as its the center number)

(2+7+x+11+16)/5 = x
=> x= 9
So, SUFFICIENT

STAT2
x is the median of the 5 numbers
so, if we write the numbers in increasing order then again x will come in middle.
Using the logic above x will again be 9
so, SUFFICIENT

Hence , answer will be D
Quote:

Hi,
I have a doubt regarding this. The answer says D. I can understand, why??

My question is - why only integers are allowed. In this question, we can also have decimal numbers like 8.8, 9.2. In that case, the answer should be E.
Show more

In the solution above it was NOT ASSUMED that x is an integer. (we got x as integer only by solving the equations)
And, given the conditions in the questions we will get only integer values of x and that too 9.

Hope it helps!
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 25 Sep 2013, 03:52
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1) allow us to say that the first and last numbers of the series are 2 and 16 giving an average = 9 and, therefore, a median = 9.
giving that x must be a value between 8, 9, 10 we came to the conclusion that x=9 -> sufficient
2) the information that x=median therefore x=average is sufficient to allow to conclude that x=9. This information, in fact allow to exclude x<2 and x>16 and at the same time provide indications that the value is somewhere between the second and 4th value of the series.
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 09 Oct 2013, 11:17
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The above set is not evenly spaced. Is it possible to have a non-evenly spaced set where \(mean=median\)?

Thank you for your help.
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The above set is not evenly spaced. Is it possible to have a non-evenly spaced set where \(mean=median\)?

Thank you for your help.
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Yes, it is. Consider: {0, 1, 1, 2}.
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 29 Mar 2016, 12:07
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Dear Bunuel , I have a doubt on statement B

Statement B says - x is the median of the 5 no's. So x can be 7,8,9,10,11 . Then how can we determine the value of x using B alone ?
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 29 Mar 2016, 12:19
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Dear Bunuel , I have a doubt on statement B

Statement B says - x is the median of the 5 no's. So x can be 7,8,9,10,11 . Then how can we determine the value of x using B alone ?
Show more

This is explained above: if-the-average-arithmetic-mean-of-the-five-numbers-x-103134.html#p803613

There is one more thing we know from the stem: "the average arithmetic mean of the five numbers x, 7, 2, 16, 11 is equal to the median of five numbers" --> \(\frac{x+2+7+11+16}{5}=median\). As there are odd number of terms then the median is the middle term when arranged in ascending (or descending) order but we don't know which one (median could be x, 7 or 11).
(1) 7 < x < 11 --> when ordered we'll get 2, 7, x, 11, 16 --> \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

(2) x is the median of five numbers --> the same info as above: \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

Answer: D.
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 18 Oct 2017, 01:37
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Bunuel
Hi Bunuel,
I understood the solution however,
Elementary doubt here. Since Mean=Median, Shouldn't the numbers in the set be consecutive, However I cannot seem understand the consecutive nature of the above mentioned set.
Thanks
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 18 Oct 2017, 01:48
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neilphilip10

Bunuel
Hi Bunuel,
I understood the solution however,
Elementary doubt here. Since Mean=Median, Shouldn't the numbers in the set be consecutive, However I cannot seem understand the consecutive nature of the above mentioned set.
Thanks
Show more

For an evenly spaced set (arithmetic progression), the median equals to the mean. Though the reverse is not necessarily true. Consider {0, 1, 1, 2} --> median = mean = 1 but the set is not evenly spaced.
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 22 May 2018, 12:05
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Hi, I just found a mistake in a question and would like to let you know

You can answer the question even without looking at the statements 1 and 2 - the question stem itself gives sufficient information to find the value of x, which is 9

My reasoning:

if we order the numbers in an increasing order we get 2, 7, 11, 16. we also have x. Regardless of the value of x, the median will be a number that is 7≤X ≤11

The question also says that median equals the mean. (36+x)/5 is the mean, which is also median. (36+x)/5 this tells us that x cannot be 7, because 36/5 is already greater than 7. If x were 7 the median would not be equal to the mean.

If x were 8, the median would be 8, but the mean would be greater than mean. You can check all numbers that satisfies 7≤X ≤11. The only number that satisfies 7≤X ≤11 and also makes median equal to the mean is 9


You can answer the question without looking at the statements
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 23 May 2018, 05:58
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kablayi
Hi, I just found a mistake in a question and would like to let you know

You can answer the question even without looking at the statements 1 and 2 - the question stem itself gives sufficient information to find the value of x, which is 9

My reasoning:

if we order the numbers in an increasing order we get 2, 7, 11, 16. we also have x. Regardless of the value of x, the median will be a number that is 7≤X ≤11

The question also says that median equals the mean. (36+x)/5 is the mean, which is also median. (36+x)/5 this tells us that x cannot be 7, because 36/5 is already greater than 7. If x were 7 the median would not be equal to the mean.

If x were 8, the median would be 8, but the mean would be greater than mean. You can check all numbers that satisfies 7≤X ≤11. The only number that satisfies 7≤X ≤11 and also makes median equal to the mean is 9


You can answer the question without looking at the statements
Show more

That's not correct.

From the stem, x can be -1, 9 or 19.

If x = -1, then the set is {-1, 2, 7, 11, 16} --> median = mean = 7.
If x = 9, then the set is {-2, 7, 9, 11, 16} --> median = mean = 9.
If x = 19, then the set is {2, 7, 11, 16, 19} --> median = mean = 11.
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 16 Feb 2020, 02:18
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satishreddy
if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X

1) 7<X<11
2) X is the median of five numbers


There is one more thing we know from the stem: "the average arithmetic mean of the five numbers x, 7, 2, 16, 11 is equal to the median of five numbers" --> \(\frac{x+2+7+11+16}{5}=median\). As there are odd number of terms then the median is the middle term when arranged in ascending (or descending) order but we don't know which one (median could be x, 7 or 11).

(1) 7 < x < 11 --> when ordered we'll get 2, 7, x, 11, 16 --> \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

(2) x is the median of five numbers --> the same info as above: \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

Answer: D.
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Hi Bunuel

I usually get confused in this type of sum knowing that value has to be 9 but by combination (C) instead of Either (D)

Because it's not mentioned in both conditions that the number has to be integer or divisible by 5. So how do we decide in (x+36)/5 will be without combination.

Please help!
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 16 Feb 2020, 02:40
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Bunuel
if the average arithmetic mean of the five numbers X, 7,2,16.11 is equal to the median of five numbers, what is the value of X

1) 7<X<11
2) X is the median of five numbers


There is one more thing we know from the stem: "the average arithmetic mean of the five numbers x, 7, 2, 16, 11 is equal to the median of five numbers" --> \(\frac{x+2+7+11+16}{5}=median\). As there are odd number of terms then the median is the middle term when arranged in ascending (or descending) order but we don't know which one (median could be x, 7 or 11).

(1) 7 < x < 11 --> when ordered we'll get 2, 7, x, 11, 16 --> \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

(2) x is the median of five numbers --> the same info as above: \(\frac{x+2+7+11+16}{5}=median=x\) --> \(\frac{x+36}{5}=x\) --> \(x=9\). Sufficient.

Answer: D.

Hi Bunuel

I usually get confused in this type of sum knowing that value has to be 9 but by combination (C) instead of Either (D)

Because it's not mentioned in both conditions that the number has to be integer or divisible by 5. So how do we decide in (x+36)/5 will be without combination.

Please help!
Show more


From the stem we got that \(\frac{x+2+7+11+16}{5}=median\).

From (1) we get that the set IN ASCENDING ORDER is 2, 7, x, 11, 16. So, the median is the middle term, x. Substitute: \(\frac{x+2+7+11+16}{5}=x\).

From (2) we are directly told that the median = x, so again \(\frac{x+2+7+11+16}{5}=x\).

Does this make sense?
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 14 Jun 2021, 11:20
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kind of confused as to why we are assuming that only integers are included?
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kind of confused as to why we are assuming that only integers are included?
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You can consider anything for x. Lets say x is not an integer. It is fraction or decimal. But you need to find x using the statements.
St1 says x is in between 7 and 11. Let it be anything-Integer, decimal or fraction but it is in between 7 and 11. So, x is the median because it is the middle term. Now as per question Mean = Median and Median = x => Mean = x

Sum of all terms /5 = mean
=> 2+7+11+16+x = mean * 5 = median * 5 = 5x ( mean = median = x)
=> 36+x = 5x
=> 4x = 36
=> x = 9

The equation gave x = 9 which is an integer. We started with assuming x= anything among integer / decimal / fraction but based on Statement 1, x came out to be an integer.

Hope it helps !!! Hit Kudos
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 21 Jul 2021, 05:13
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 11 is equal to the median of five numbers, what is the value of X

(1) 7 < x < 11
(2) x is the median of the five numbers

Show Spoiler
first statement is definitely sufficient, but in the second statement, if X is the median, it can take three numbers, 10. 9, 8, so ...how is it sufficient, i might be missing some basic point here,,,,so need help,,,,as the second statement is also sufficient for this problem
Show more

As with many problems on the GMAT, this problem is made much easier by doing some pre-work in the stem.

Stem analysis:
First let's organize the numbers provided: 2,7,11,16 and x, BUT we don't know exactly where x should go relative to the other numbers. But we do know that three scenarios are possible:
(1) X IS the median, which would result in the following equation: (2+7+11+16+x)/5 = x
(2) X is below the median, which would result in the following equation: (2+7+11+16+x)/5 = 7
(3) X is above the median, which would result in the following equation: (2+7+11+16+x)/5 = 11

If any of the statements provide us with enough information to determine where x is relative to the other numbers in the list then we have sufficient information to determine the value of x.

Statement 1:
Given this condition x MUST be equal to the median itself.

Statement 1:
Explicitly stated that x is the median so sufficient itself.

The answer is D.
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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and 24 Aug 2021, 11:47
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St(1):-7 < x < 11

For median, lets arrange the terms in ascending order.We have 2, 7, x, 11, 16.

Median would be x = Mean = 2 + 7+ x+ 11 + 16 /5 = (x + 36) / 5

Its a solvable linear equation. (Sufficient)

St(2):-x is the median of five numbers

It provides us the same information as in St(1).
(Sufficient)

(option d)

D.S
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