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If Set Z has a median of 19, what is the range of Set Z? (1) [#permalink]
14 Jan 2008, 05:23

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C

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E

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Question Stats:

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If Set Z has a median of 19, what is the range of Set Z?

(1) Z = {18, 28, 11, x, 15, y}

(2) The average (arithmetic mean) of Set Z is 20. A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient D. EITHER statement BY ITSELF is sufficient to answer the question E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem

[spoiler]OA is C[/spoiler]

Last edited by AVakil on 14 Jan 2008, 05:53, edited 1 time in total.

Re: DS: Median and Range [#permalink]
14 Jan 2008, 06:45

AVakil wrote:

If Set Z has a median of 19, what is the range of Set Z?

(1) Z = {18, 28, 11, x, 15, y}

(2) The average (arithmetic mean) of Set Z is 20. A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient D. EITHER statement BY ITSELF is sufficient to answer the question E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem

[spoiler]OA is C[/spoiler]

Z = {11, 15, 18, x, 28, y}

We no nothing about x and y.

So it can be : x, 11,15,y,18,28..... in this case y=23 ( if median is 19) but range depends on x, which we can't determine. Further there can be several permutations for x and y. Thus (i) is insufficient.

From (ii)

We know only that x+y= 58. And nothing else, so insufficient.

Re: DS: Median and Range [#permalink]
14 Jan 2008, 07:11

Expert's post

LM wrote:

But if we combine, (i) and (ii)

11, 15,18,20,28,38 Range = 38-11 = 27

10, 11,15,18,28,48 range = Range = 48-10 = 38

Thus insufficient.

The answer should be "E".

median of {10, 11,15,18,28,48} is (15+18)/2=16.5<19 that contradicts Q. the average of {11,15,18,20,28,38} is 21.7>20 that contradicts ii _________________

Re: DS: Median and Range [#permalink]
14 Jan 2008, 07:25

I would vote for C. From S1 and S2 it can be determined what either x or y is (20) since median is 19. The other value can be calculated from the mean definition. Once both x and y are known the value of the range is known.

Re: DS: Median and Range [#permalink]
14 Jan 2008, 10:35

AVakil wrote:

If Set Z has a median of 19, what is the range of Set Z?

(1) Z = {18, 28, 11, x, 15, y}

(2) The average (arithmetic mean) of Set Z is 20.

1. we have 6 numbers. the median of a set of an even number of elements is the average b/w the 3rd and the 4th. so: 11 15 18 20 28 y, in order for the median to be 19. alone is not suff

2. alone is not suff. we don't even know the number of elements in the set.

c. 20*6=120, thus we can calculate y and then the range.

Re: DS: Median and Range [#permalink]
14 Jan 2008, 10:58

Answer is C Looking at statement II (mean=20) we can rearrange first three terms of set as { 11,15,18._,_,_) Also,using mean value we can get 11+15+18+28+x+y/6=20 x+y=48 ...........(a) From statement I (median=19) 18+Z/2=19.......(I used Z because this term can be either x or y) Z=20 using equation (a) other term would be 48-20=28

now, we can complete our set as {11,15,18,20,28,28} so the range would be 28-11=17

Re: DS: Median and Range [#permalink]
15 Jan 2008, 04:47

Okay, this is what Im not getting. Just because you know the average, how do you know the order of the terms in the set that allow you to calculate the median ?

gmatclubot

Re: DS: Median and Range
[#permalink]
15 Jan 2008, 04:47

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...