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14 Mar 2018, 00:23
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Difficulty:
35%
(medium)
Question Stats:
61% (01:13) correct
39%
(01:11)
wrong
based on 142
sessions
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Not Attempted Yet
[GMAT math practice question]
What is the median value of the data displayed in the following frequency table?
a.png [ 2.35 KiB | Viewed 2972 times ]
\(1) x=2\)
\(2) y=5\)
What is the median value of the data displayed in the following frequency table?
Attachment:
a.png [ 2.35 KiB | Viewed 2972 times ]
\(1) x=2\)
\(2) y=5\)
14 Mar 2018, 09:05
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MathRevolution
[GMAT math practice question]
What is the median value of the data displayed in the following frequency table?
\(1) x=2\)
\(2) y=5\)
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
15 Mar 2018, 03:43
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niks18
MathRevolution
[GMAT math practice question]
What is the median value of the data displayed in the following frequency table?
\(1) x=2\)
\(2) y=5\)
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
