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Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number

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Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number  [#permalink]

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New post 14 Aug 2011, 14:07
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Question Stats:

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Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the numbers greater than 35 are greater than 40. Which of the following could be the median number?

I. 34
II. 35
III. 36

(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

I get till this part...200*0.165=x*0.333 =>x=99

Can someone help me with next steps??
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Re: Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number  [#permalink]

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New post 14 Aug 2011, 16:53
DeeptiM wrote:
Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the numbers greater than 35 are greater than 40. Which of the following could be the median number?
I. 34
II. 35
III. 36
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

I get till this part...200*0.165=x*0.333 =>x=99

Can someone help me with next steps??



yes I think you have deducted correctly till now. If 99 numbers are > 35 , it means rest 101 numbers are equal to or less than 35.

median in the list of a set of 200 members will be the average of 100th and 101st member of the set when we arrange the members of the set in ascending order. From deduction we know that 102nd member to 200th member of the set are greater than 35 , and 1st member to 101st member are less than or equal to 35. 100th and 101st member could be 34 or 35 . so 34 and 35 could be the median.
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Re: Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number  [#permalink]

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New post 14 Aug 2011, 17:05
The median occurs at term\(\frac{(n + 1)}{2}=\frac{201}{2}=100.5\), and for a set with an even count (200 in this case), it will be the average of terms \(100\) and \(101\).

With \(200*16.5% = \frac{1}{3}x\), you have established that the \(x = 200*16.5%*3=99\) highest observations are each greater than 35 and that the \(200*16.5%=33\) highest observations are each greater than 40. All other terms must be less than or equal to 35.

The average of the 101st and 100th terms is the median. The 99th term in descending order is the \(200-99+1=102\)nd term in ascending order. The average of the 101st term and the 100th term, which are each less than or equal to 35, can only be 34 or 35.
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Re: Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number  [#permalink]

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New post 01 Mar 2018, 20:09
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In my opinion, the fact that there are 200 numbers is irrelevant. We can deduce the median just by working with the percentages.

We're given that:
- 16.5% of the 200 numbers are greater than 40
- 33.3% of the numbers greater than 35 are greater than 40

Combining the 2 statements:
- 33.3% of the numbers greater than 35 are 16.5% of the 200 numbers

So, 100% of the numbers greater than 35 are 49.5% of the 200 numbers

Since the median is at the 50% mark when the numbers are arranged in order, and since 49.5% are greater than 35, then the median must be less than or equal to 35.

Answer: D
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Re: Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number  [#permalink]

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New post 29 May 2019, 01:29
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Re: Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number   [#permalink] 29 May 2019, 01:29
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