dferm wrote: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

Please explain....

We know X has to be the largest number, so we are safe with assuming that the numbers are in order. This is a key step (arranging the given numbers) before diving into Statistcis questions.

Given Condition:

Median = 6/7 ( Arithmetic Mean)

But, Median of an even number of number is the average of the middle 2 numbers. Therefore, Median = (6+8 ) / 2 = 14 / 2 = 7

Arithmetic mean is nothing but the average.
Average of these numbers = (4+5+6+8+10+x) / 6 = (33 + x) / 6

Going back to our condition:

Median = 6/7 ( Arithmetic Mean)
Or 7 = (6/7) (33+x) /6 ................The 6's get cancelled
Or, 7.7 = (33+x)........................Cross multiply
Or, 49 = 33 + X
Or, 49 - 33 = X

Or, X = 16.

QA = A

{ You could backsolve, but the returns are marginal when compared to those of Algebra in this case. In general, bcaksolving is a time- saver, but not when the equation setup is so straightforward}

dferm wrote: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

Please explain....

6/7Average = Median

Since there are 6 numbers, the median should be between the 3rd and 4th number = 7

Substitute x for Average such that 6/7x = 7
6x = 49
x=49/6

Average * Number of elements = Total

49/6 * 6 numbers = 49

Adding all the number but x : 4 + 5 + 6 + 8 + 10 = 33
Total - Remaining Total = x
49 - 33 = 16