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If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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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

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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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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7
5
vitamingmat wrote:
satishreddy wrote:
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....

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.

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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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1
satishreddy wrote:
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

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.

Please consider KUDOS if it is helpful to u .
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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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satishreddy wrote:
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....
Intern  Joined: 20 Sep 2013
Posts: 2
Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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1
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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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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3
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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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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2
2
emailmkarthik wrote:
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.

Yes, it is. Consider: {0, 1, 1, 2}.
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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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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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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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3
ThePlayer wrote:
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 ?

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.

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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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oh man..I just had this question in my GMATPrep test, and I panicked and picked A...
quick question to Bunuel, can we consider only INTEGER values, knowing that the question stem tells that we have NUMBERS (aka any numbers)?

1. x has values 8, 9, 10.
we know that (36+x)/5 = x
36+x=5x
x=9
so only value if x=9

2. x is the median of the five
36+x=5x
x=9
yes, 9 is the median.

D

I dismissed B because I considered non-integer values for x...but still..illogical..because if we apply what we are given..we always get x=9...
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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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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

Important to note: This is a purely CONCEPTUAL problem. Meaning, there's HARDLY ANY MATH REQUIRED/INVOLVED. You should be able to eye this and take 30-45 seconds MAX. I'll show you how.

1- set up #s in increasing order (common step whenever you see the word "median"): 2, 7, 11, 16, x --> quickly see 2+7=9...9+11=20, 20+16=36.
2- average = $$\frac{sum of #s in a set}{total numbers in a set}$$ --> $$\frac{36+x}{5}$$
3- set up question: $$\frac{36+x}{5}$$ = median.
> key here is to find out what x is (or what the median is)!

1) TELLS US X=MEDIAN! How do you know this? Well, there are 5 terms in the set.
> 2 terms are BELOW 7.
> 2 terms are ABOVE 11.
-- Since there are only 5 terms in a set, the 3rd term (middle) = median. Bingo. Eliminate B, C, E.

2) SAME AS 1!
-- Elim A.

Ans: D
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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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Bunuel wrote:
andrewwal wrote:
russ9

1) I believe that it can also be 11 since if x was 11 the median in fact would be 11 (but the median could never be 16 since if x=16, then you'd have 2, 7, 11, 16, 16 and the median would be 11 not 16).

2) Statement 1 and 2 are essentially saying the say thing, since the constraint "the mean is equal to the median" and that is the only value  that satisfies that constraint -- therefore it is D.

Please Kudos if this helped!

Correct. 15 was a typo there. Edited.

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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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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neilphilip10 wrote:
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

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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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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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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kablayi wrote:
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

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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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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Does this question even need any of the answer choices?

As in, this question can be answered without using either of the two options.

Is there any other question type where value of x needs to be calculated in series where the answer choices really are needed?

Bunuel wrote:
vitamingmat wrote:
satishreddy wrote:
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....

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.

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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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prekshita wrote:
Does this question even need any of the answer choices?

As in, this question can be answered without using either of the two options.

Is there any other question type where value of x needs to be calculated in series where the answer choices really are needed?

It is not possible. Without the statements, you CANNOT get a single solution. X can be anything! Statements are given either to find a solution or to prove something asked in the question.

P.S: By taking your statement into consideration, why would the test takers give such a question where the statements are useless? _________________

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Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and  [#permalink]

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Bunuel chetan2u I quickly recognized that only 9 will give an int answer (45/9=5) and satisfies the condition, and that there's only 1 number that gives us med = avg in that constraint.

But, is this a bad assumption since the prompt doesn't specify ints? What if in a different problem with a similar constraint the average and median were a non-int? We know that there's still only 1 correct value where med=avg, even if we don't calculate, right? Re: If the average (arithmetic mean) of the five numbers x, 7, 2, 16 and   [#permalink] 10 Jun 2019, 11:46
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