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23 Sep 2010, 11:45
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A
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A list of measurements in increasing order is 4, 5, 6, 8, 10 and x. If the median of these measurements is 6/7 times their arithmetic mean, what is the value of x?
A. 16
B. 15
C. 14
D. 13
E. 12
Image1.JPG [ 8.42 KiB | Viewed 44003 times ]
A. 16
B. 15
C. 14
D. 13
E. 12
Attachment:
Image1.JPG [ 8.42 KiB | Viewed 44003 times ]
23 Sep 2010, 11:56
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A list of measurements in increasing order is 4, 5, 6, 8, 10 and x. If the median of these measurements is 6/7 times their arithmetic mean, what is the value of x?
A. 16
B. 15
C. 14
D. 13
E. 12
{4, 5, 6, 8, 10, x}
Median would be the average of two middle terms --> \(median=\frac{6+8}{2}=7\);
\(mean=\frac{4+5+6+8+10+x}{6}=\frac{33+x}{6}\);
Given: \(median=\frac{6}{7}*mean\) --> \(7=\frac{6}{7}*\frac{33+x}{6}\) --> \(x=16\).
Answer: A.
A. 16
B. 15
C. 14
D. 13
E. 12
{4, 5, 6, 8, 10, x}
Median would be the average of two middle terms --> \(median=\frac{6+8}{2}=7\);
\(mean=\frac{4+5+6+8+10+x}{6}=\frac{33+x}{6}\);
Given: \(median=\frac{6}{7}*mean\) --> \(7=\frac{6}{7}*\frac{33+x}{6}\) --> \(x=16\).
Answer: A.
23 Jul 2016, 00:53
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udaymathapati
Median of a set with even # of terms = Average of the middle two terms = \(\frac{(6+8)}{2} --->\frac{14}{2} ---> 7\)
Median is 7 ; Mean = \(\frac{(4+5+6+8+10+x)}{6}\)
Given that Median = \(\frac{6}{7}\) of Mean
Median =\(\frac{6}{7}* \frac{(4+5+6+8+10+x)}{6}\)
\(7 = \frac{6}{7}* \frac{(4+5+6+8+10+x)}{6}\)
\(7=\frac{6}{7} * \frac{(33+x)}{6}\)
\(49=33+x\)
\(x=49-33\)\(x=16\)
ANSWER IS A
