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# GMAT Diagnostic Test Question 42

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Re: GMAT Diagnostic Test Question 42 [#permalink]  22 May 2010, 09:23
Sorry, but the stem says that m is an integer. 3.6 is not an integer.

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

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]  05 Jan 2012, 08:15
dzyubam wrote:
Explanation:
 Rating:

Official Answer: C

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.

I still can not prove S1+S2 are always sufficient....

for example, m=4, as in the example above, but now a=1, b=2, c=3, d=5 and e=6. In this case D=square root of 19/5, which is less than 2, therefor s1+s2 is not sufficient now....
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Re: GMAT Diagnostic Test Question 42 [#permalink]  23 Aug 2012, 00:16
Here is my solution why C is always correct -

From S1, we have 5 different integers
From S2, m is an integer but different from the above 5 integers

Now we have to say Yes or No for D>2. This is equivalent to saying that the Numerator should be greater than 20 in expression of D.
(Since sqrt (20/5) = sqrt (4) = 2)

Therefore, we want sum of 5 separate squares to be equal to 20.
Since all are integers, squares of the differences will be integers. Also, since m is different, 0 as a difference is ruled out. So at minimum, we have
2*1^2 + 2*2^2 +1*3^2 = 19 [Expression 1]

However, since m is an integer, a+b+c+d+e = 5n (n is any integer)
Subtracting this value from 5m; 5m - (a+b+c+d+e) = 5m - 5n
Therefore, (m-a) + (m-b) + (m-c) + (m-d) + (m-e) = 5(m-n)

For values to satisfy Expression 1 ; 1+1+2+2+3 = 5(m-n)
Therefore, 9 = 5(m-n) which is false since m and n are integers.

Hence, D has to be greater than 2, since Expression 1 always will be greater than 20, to satisfy S1 and S2.
And since there is at least 1 example (1,2,3,6,8) proves that C is the correct answer
Re: GMAT Diagnostic Test Question 42   [#permalink] 23 Aug 2012, 00:16

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# GMAT Diagnostic Test Question 42

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