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What is the median value of the data displayed in the following frequ

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New post 14 Mar 2018, 00:23
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[GMAT math practice question]

What is the median value of the data displayed in the following frequency table?

Attachment:
a.png
a.png [ 2.35 KiB | Viewed 402 times ]


\(1) x=2\)
\(2) y=5\)
[Reveal] Spoiler: OA

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New post 14 Mar 2018, 09:05
MathRevolution wrote:
[GMAT math practice question]

What is the median value of the data displayed in the following frequency table?

Attachment:
a.png


\(1) x=2\)
\(2) y=5\)


As the data is arranged in ascending order, so \(y>5\). This implies that minimum numbers in the set will be \(= 1+2+3+4+5=15\).

Hence median will be a number that will in the \(8th\) or higher position depending upon the value of \(y\). Hence \(x\) cannot be median because last \(x\) will be at \(6th\) position if \(0<x<3\).

So essentially we need to know the value of \(y\)

Statement 1: nothing mentioned about \(y\). Insufficient

Statement 2: directly provides the value of \(y\). Sufficient

Option B
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Re: What is the median value of the data displayed in the following frequ [#permalink]

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New post 15 Mar 2018, 03:43
niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

What is the median value of the data displayed in the following frequency table?

Attachment:
a.png


\(1) x=2\)
\(2) y=5\)


As the data is arranged in ascending order, so \(y>5\). This implies that minimum numbers in the set will be \(= 1+2+3+4+5=15\).

Hence median will be a number that will in the \(8th\) or higher position depending upon the value of \(y\). Hence \(x\) cannot be median because last \(x\) will be at \(6th\) position.

So essentially we need to know the value of \(y\)

Statement 1: nothing mentioned about \(y\). Insufficient

Statement 2: directly provides the value of \(y\). Sufficient

Option B



we dont know the value of x. we cant assume that x will take value b/w 1 and 3
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What is the median value of the data displayed in the following frequ [#permalink]

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New post 15 Mar 2018, 04:02
varundixitmro2512 wrote:
niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

What is the median value of the data displayed in the following frequency table?

Attachment:
a.png


\(1) x=2\)
\(2) y=5\)


As the data is arranged in ascending order, so \(y>5\). This implies that minimum numbers in the set will be \(= 1+2+3+4+5=15\).

Hence median will be a number that will in the \(8th\) or higher position depending upon the value of \(y\). Hence \(x\) cannot be median because last \(x\) will be at \(6th\) position.

So essentially we need to know the value of \(y\)

Statement 1: nothing mentioned about \(y\). Insufficient

Statement 2: directly provides the value of \(y\). Sufficient

Option B



we dont know the value of x. we cant assume that x will take value b/w 1 and 3


hi varundixitmro2512

I am not saying that the value of x is between 1 & 3. The question states that the frequency of x is 3, hence it will not impact the median because frequency of y is more than 5

For eg. let x=100 & y=5,So if you arrange the numbers in ascending order, then it will be as follows -

0,1,1,3,3,3,4,4,4,4,4,100,100, you can calculate the median of this set.

you may take any value of x, as long as y>5, it will not impact the median.

For clarity I have edited the post
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New post 16 Mar 2018, 00:50
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

If \(y = 1\), the median is \(x\).
If \(y = 2\), the median is \(\frac{( x + 3 )}{2}\)
If \(3 ≤ y ≤ 9\), the median is \(3\).
If \(y = 10\), the median is \(\frac{( 3 + 4 )}{2} = 3.5\)
If \(y ≥ 10\), the median is \(4\).

Thus, the condition 2) is sufficient.

Therefore, the answer is B.

Answer: B
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New post 16 Mar 2018, 02:00
Statement 1 is anyways insufficient because value of y will make the median very from 2 to 4.

As per Statement 2, when y=5, 2 scenarios arise:

Situation 1: x=2, median =3
Situation 2: x=3.5, median =3.5

Thus, this statement also becomes insufficient. In my opinion, the answer should be C)

Am I missing something here?
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New post 16 Mar 2018, 05:12
mb2202 wrote:
Statement 1 is anyways insufficient because value of y will make the median very from 2 to 4.

As per Statement 2, when y=5, 2 scenarios arise:

Situation 1: x=2, median =3
Situation 2: x=3.5, median =3.5

Thus, this statement also becomes insufficient. In my opinion, the answer should be C)

Am I missing something here?


Hi mb2202

When frequency table is presented then generally data are arranged in ascending order. However it would have been great if the question stem had mentioned that the arrangement is in ascending order. The solution is based on the assumption that the data is in ascending order hence x<3, so x=3.5 is not possible here.
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New post 09 Apr 2018, 10:12
Quote:

I am not saying that the value of x is between 1 & 3. The question states that the frequency of x is 3, hence it will not impact the median because frequency of y is more than 5

For eg. let x=100 & y=5,So if you arrange the numbers in ascending order, then it will be as follows -

0,1,1,3,3,3,4,4,4,4,4,100,100, you can calculate the median of this set.

you may take any value of x, as long as y>5, it will not impact the median.

For clarity I have edited the post


Hi niks18,
Your answer is BASING on the assumption that x is not between 1&3. It actually MATTERS if x is between 1&3 and x is NOT between 1&3. If x =2 the Median is 3 and if x = 100 the median is 4. (x is repeated THREE TIMES NOT TWO TIMES)
What is the median value of the data displayed in the following frequ   [#permalink] 09 Apr 2018, 10:12
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