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# GMAT Diagnostic Test Question 42

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GMAT Diagnostic Test Question 42 [#permalink]

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07 Jun 2009, 01:07
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GMAT Diagnostic Test Question 42
Field: combinations
Difficulty: 750
 Rating:

Set $$X$$ has 5 integers: $$a$$, $$b$$, $$c$$, $$d$$, and $$e$$. If $$m$$ is the mean and $$D$$, where $$D = \sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$$, is the standard deviation of the set $$X$$, is $$D \gt 2$$?

(1) $$a$$, $$b$$, $$c$$, $$d$$, and $$e$$ are different integers
(2) $$m$$ is an integer not equal to any elements of the set $$X$$

A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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08 Jul 2009, 04:32
Explanation:
 Rating:

Statement 1 is not sufficient. The answer to the question might be YES or NO depending on the numbers in the set. Any set with significant range will have $$D \gt 2$$ (e.g. 1, 2, 3, 4, 100). On the other hand, $$D$$ can be as low as $$\sqrt{2}$$ for a set consisting of 1, 2, 3, 4, and 5 (mean of this set equals 3):

$$D = \sqrt{\frac{(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-3)^2}{5}} = \sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$$

The answer to the question can be either YES or NO. Not sufficient.

Statement 2 is not sufficient by itself either. Again, $$D$$ can be very big if the range is great (see an example frm S1). Consider this set for $$D$$ to be as low as 2: 1, 1, 1, 1, 6. The mean is 2, $$D$$ is calculated as follows:

$$D = \sqrt{\frac{(1-2)^2+(1-2)^2+(1-2)^2+(1-2)^2+(6-2)^2}{5}} = \sqrt{\frac{1 + 1 + 1 + 1 + 16}{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$$

The answer to the question can be either YES or NO. Not sufficient.

Combining both statements, we have enough information to answer the question. The answer is YES. To prove that we have to think of a set with minimum possible range under restrictions of S1 and S2. We will use this set: 1, 2, 3, 6, 8. Its mean is 4. The standard deviation is calculated as follows:

$$D = \sqrt{\frac{(1-4)^2+(2-4)^2+(3-4)^2+(6-4)^2+(8-4)^2}{5}} = \sqrt{\frac{9 + 4 + 1 + 4 + 16}{5}} = \sqrt{\frac{34}{5}} = \sqrt{6.8} > 2$$

S2+S1 is sufficient.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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28 Jul 2009, 17:59
The different elements of the set are a,b,c,d,e.
The standard deviation is d.
So, I actually got confused if the question asks to find if - both the standard deviation and the fourth element - are greater than 2.

Harry
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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29 Jul 2009, 00:51
The question was meant to ask if the standard deviation was greater than 2. I'll put a "D" to stand for standard deviation to avoid confusion. Thanks!
charimaalu wrote:
The different elements of the set are a,b,c,d,e.
The standard deviation is d.
So, I actually got confused if the question asks to find if - both the standard deviation and the fourth element - are greater than 2.

Harry

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Re: GMAT Diagnostic Test Question 42 [#permalink]

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30 Jul 2009, 00:04
I've updated the PDF and the question above to avoid confusion.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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23 Aug 2009, 09:58
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dzyubam wrote:
Explanation:
 Rating:

Statement 1 is not sufficient. The answer to the question might be YES or NO depending on the numbers in the set. Any set with significant range will have $$d \gt 2$$ (e.g. 1, 2, 3, 4, 100). On the other hand, $$d$$ can be as low as $$\sqrt{2}$$ for a set consisting of 1, 2, 3, 4, and 5 (mean of this set equals 3):

$$d = \sqrt{\frac{(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-3)^2}{5}} = \sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$$

The answer to the question can be either YES or NO. Not sufficient.

Statement 2 is not sufficient by itself either. Again, we $$d$$ can be very big if the range is great (see an example frm S1). Consider this set for $$d$$ to be as low as 2: 1, 1, 1, 1, 6. The mean is 2, $$d$$ is calculated as follows:

$$d = \sqrt{\frac{(1-2)^2+(1-2)^2+(1-2)^2+(1-2)^2+(6-2)^2}{5}} = \sqrt{\frac{1 + 1 + 1 + 1 + 16}{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$$

The answer to the question can be either YES or NO. Not sufficient.

Combining both statements, we have enough information to answer the question. The answer is YES. To prove that we have to think of a set with minimum possible range under restrictions of S1 and S2. We will use this set: 1, 2, 3, 6, 8. Its mean is 4. The standard deviation is calculated as follows:

S2+S1 is sufficient.

$$d = \sqrt{\frac{(1-4)^2+(2-4)^2+(3-4)^2+(6-4)^2+(8-4)^2}{5}} = \sqrt{\frac{9 + 4 + 1 + 4 + 16}{5}} = \sqrt{\frac{34}{5}} = \sqrt{6.8} > 2$$
can't prove "S2+S1 is sufficient", it is only one example.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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23 Aug 2009, 15:44
Could somebody prove that the answer is C?

An example is not enough to prove it.

I can't find a counterexample, most likely because the answer is indeed C, but would be good if it could be proved.

BTW, shouldn't this question be on the "statistics" field?
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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23 Aug 2009, 23:16
I guess the answer could be B, but can anyone give a prove? Thanks.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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24 Aug 2009, 00:12
defeatgmat wrote:
I guess the answer could be B, but can anyone give a prove? Thanks.

Its already been proved. Look at the official explanation above; dzyubam provided a counterexample that makes statement 2 not sufficient.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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24 Aug 2009, 01:11
powerka wrote:
defeatgmat wrote:
I guess the answer could be B, but can anyone give a prove? Thanks.

Its already been proved. Look at the official explanation above; dzyubam provided a counterexample that makes statement 2 not sufficient.

The example only proves 2) individually is insufficient;

The example after it gave only one instance, can't prove S1+S2 are always correct.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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24 Aug 2009, 05:27
In my proof that C is the correct option I've used the numbers that produce the least standard deviation given the conditions in S1 and S2. I can't think of a better way to prove it. You're welcome to come up with one if you can. Anybody?
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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25 Aug 2009, 17:18
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If D=2 is the minimum, we can say D>2.
can anyone prove this?
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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25 Aug 2009, 18:54
flyingbunny wrote:
If D=2 is the minimum, we can say D>2.
can anyone prove this?

Yep, I believe this would be the key.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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07 Sep 2009, 12:46
Think it should be E. Please let me know where my logic is incorrect.

What I did first to breakdown this problem is in order for std D> 2, have to find out the mininum the numerator should be
D = $$\sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$$

next step, I squared both sides for D > 2:
2^2 > $$\frac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}$$

2^2 * 5 > $$(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2$$

20 > sum of numerator

Combining information from the two statements together, this is my counterexample:
I used this for my set: 1, 2, 4, 5, 6. Its mean is 3.6. The standard deviation is calculated as follows:

is numerator greater than 20?
= $$\sqrt{\frac{(1-3.6)^2+(2-3.6)^2+(4-3.6)^2+(5-3.6)^2+(6-3.6)^2}{5}}$$
= $$\sqrt{\frac{(2.6)^2+(1.6)^2+(0.4)^2+(1.4)^2+(2.4)^2}{5}}$$
= 6.76 + 2.56 + 0.16 + 1.96 + 5.76
= 17.xx

in this case, it is a NO, D < 2
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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07 Sep 2009, 14:35
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There's one problem with your counter example. $$m$$ should be an integer according to S2.
beckee529 wrote:
Think it should be E. Please let me know where my logic is incorrect.

What I did first to breakdown this problem is in order for std D> 2, have to find out the mininum the numerator should be
D = $$\sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$$
...

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Re: GMAT Diagnostic Test Question 42 [#permalink]

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09 Sep 2009, 08:58
bb wrote:
GMAT Diagnostic Test Question 42
Field: combinations
Difficulty: 750
 Rating:

Set $$X$$ has 5 integers: $$a$$, $$b$$, $$c$$, $$d$$, and $$e$$. If $$m$$ is the mean and $$D$$, where $$D = \sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$$, is the standard deviation of the set $$X$$, is $$D \gt 2$$?

(1) $$a$$, $$b$$, $$c$$, $$d$$, and $$e$$ are different integers
(2) $$m$$ is an integer not equal to any elements of the set $$X$$

Sticking to the question and the conditions above suffices to say that D>2.

Things to be noted:

1. a, b, c, d, and e, Elements of the set X, are different integers.
2. m is different from the above 5 elements of the set X.
3. Calcualation of D is only using the given formula i.e.

$$D = \sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$$

Statements 1 and 2 are sufficient and therefore C is correct.

dzyubam wrote:
Explanation:
 Rating:

Statement 1 is not sufficient. The answer to the question might be YES or NO depending on the numbers in the set. Any set with significant range will have $$D \gt 2$$ (e.g. 1, 2, 3, 4, 100). On the other hand, $$D$$ can be as low as $$\sqrt{2}$$ for a set consisting of 1, 2, 3, 4, and 5 (mean of this set equals 3):

$$D = \sqrt{\frac{(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-3)^2}{5}} = \sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}} = \sqrt{\frac{10}{5}} = \sqrt{2}$$

The answer to the question can be either YES or NO. Not sufficient.

Statement 2 is not sufficient by itself either. Again, $$D$$ can be very big if the range is great (see an example frm S1). Consider this set for $$D$$ to be as low as 2: 1, 1, 1, 1, 6. The mean is 2, $$D$$ is calculated as follows:

$$D = \sqrt{\frac{(1-2)^2+(1-2)^2+(1-2)^2+(1-2)^2+(6-2)^2}{5}} = \sqrt{\frac{1 + 1 + 1 + 1 + 16}{5}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$$

The answer to the question can be either YES or NO. Not sufficient.

Combining both statements, we have enough information to answer the question. The answer is YES. To prove that we have to think of a set with minimum possible range under restrictions of S1 and S2. We will use this set: 1, 2, 3, 6, 8. Its mean is 4. The standard deviation is calculated as follows:

$$D = \sqrt{\frac{(1-4)^2+(2-4)^2+(3-4)^2+(6-4)^2+(8-4)^2}{5}} = \sqrt{\frac{9 + 4 + 1 + 4 + 16}{5}} = \sqrt{\frac{34}{5}} = \sqrt{6.8} > 2$$

S2+S1 is sufficient.

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Re: GMAT Diagnostic Test Question 42 [#permalink]

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15 Sep 2009, 16:05
edit: nvm, got it. the restriction that m is an integer is key.
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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25 Nov 2009, 17:06
I have a question about the OE for this problem. You have assumed that the integers in set X to be 1,1,1,1,6 which gives us m=2. however, this satisfies only S2. S1 says that all the elements of set X must be "Different Integers" or as I assumed to be "Distinct Integers".
The smallest set I came up with was x = {1,2,3,4,15} gives us m = 5 and SD = 26^1/2
Ofcourse, after the test, I realized that I never considered negative integers.
I came up with choice C as a calculated guess since thats all I could manage in 2 mins
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Re: GMAT Diagnostic Test Question 42 [#permalink]

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26 Nov 2009, 02:59
I assumed that when I was explaining why S2 ONLY can't be sufficient. We have to be sure to regard information presented in S1 and S2 separately as long as we consider A or B as possible answers.

I hope this helps .
amittilak wrote:
I have a question about the OE for this problem. You have assumed that the integers in set X to be 1,1,1,1,6 which gives us m=2. however, this satisfies only S2. S1 says that all the elements of set X must be "Different Integers" or as I assumed to be "Distinct Integers".
The smallest set I came up with was x = {1,2,3,4,15} gives us m = 5 and SD = 26^1/2
Ofcourse, after the test, I realized that I never considered negative integers.
I came up with choice C as a calculated guess since thats all I could manage in 2 mins

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Re: GMAT Diagnostic Test Question 42 [#permalink]

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05 Jan 2010, 18:06
Oh, that's what i missed too. Thanks.

dzyubam wrote:
There's one problem with your counter example. $$m$$ should be an integer according to S2.
beckee529 wrote:
Think it should be E. Please let me know where my logic is incorrect.

What I did first to breakdown this problem is in order for std D> 2, have to find out the mininum the numerator should be
D = $$\sqrt{\frac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$$
...
Re: GMAT Diagnostic Test Question 42   [#permalink] 05 Jan 2010, 18:06

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