Measurments in increasing order

[datslc]

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?

16
15
14
13
12

How do I approach this?

[ridhamshah]

Answer is (A). By trial and error

Mean = (33+x)/6
Median = (33+x)/7

[sandydiwan]

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?

16
15
14
13
12

How do I approach this?

Median = (6+8)/2 = 7
Mean = (4+5+6+8+10+x)/6
Median = 6/7 Mean
7 = 6/7 ( (33+x)/6 ) => 7 = 1/7(33+x) => 49 = 33+ x => x = 16

[akhp77]

Since, all the values of x, given is question, is greater than 10

median = (6+8)/2 = 7

mean = (33+x)/6

median = mean * 6/7

x = 16

[StaceyKoprince]

You could use trial and error, you could set up the actual math - whatever you think is easiest for you. In this case, they give us enough to do a couple of relatively straightforward calculations, so I think that would be easier than trial and error.

Median = the number in the middle of the set (in the case of a set containing an even number of numbers, it's the average of the middle two).

We know that the middle two numbers are 6 and 8 (because the tell us x is larger than 10). So the median of the set is the average of 6 and 8, or 7.

then we're told:
median = 6/7(average)
7 = (6/7)(average)
49/6 = average

Yuck, that's kind of a messy average, huh?

average = sum / #
49/6 = (4+5+6+8+10+x)/6
49 = (4+5+6+8+10+x)
hey! that worked out nicely. The 6's cancelled out! The problem just got a whole lot easier.

I'm just pointing out here that, when you see something like 49/6, don't freak out. Chances are it'll get a lot easier again before the problem is over. :)

49 = 33+x
x = 16.

[rachelhong2012]

I used a different approach, with the concept similar to that of "residuals", I would call it "difference"

To better illustrate, let's look at an example:

1 3 7 8 12
median: 7
what's the mean? 6.2
Besides by doing sum/total amount of numbers in the set, there's another way to get to the mean, using median
the distance/difference between 1 and 7, the median, is -6 (to get 1, median subtract by 6), the difference between 3 and 7 is -4. On the left hand side, the difference between 7 and 8 is +1 (to get 8, you add 1 to 7), and difference between 7 and 12 is +5.

Adding all these "differences" up, you get -4. Distribute this sum of differences into 5 numbers, each number in the set, including the median, gets an additional -4/5=-0.8.
the median, which is 7, + (-0.8)=6.2, hence you get the mean, using the median and the differences between other numbers and the median.

going back to this question, what I did was:
mean=median + (sum of all differences, both positive and negative)/total amount of #'s in the set
mean=7/6 median = median + (sum of all differences, both positive and negative)/6
mean=7/6median = 6/6 median + (sum of all differences, both positive and negative)/6
since they have common denominators, we essentially have:
1/6 median = (sum of all differences, both positive and negative)/6

or median/6 = (sum of all differences, both positive and negative)/6

so 7=sum of all differences.

Now let's look at the difference between each number and the median:

to get from 7 to 4: -3
7 to 5: -2
7 to 6: -1

on the left hand side of 7, we have:
to get from 7 to 8: +1
7 to 10: +3

now you see +3 and -3 cancel out, so does -1 and +1, we're left with a difference of -2, and since we just found that the overall difference should be +7, let's mean that we have a +9 on the left hand side:

7 to x: +9,
x=16

This might seem LONG and unnecessary in this problem, but I found this approach comes in handy when you see a situation where you're being asked:

median < mean, what happens?
Basically, as you know to get from median to mean, you have to add the distributed "difference" to the median, that means that since we have a bigger mean than median, we have just added positive "difference" to the median to arrive at a mean that's bigger than median, where does that positive "difference" come from? It must be that there are more positive difference than the negative difference.

I hope I explain myself as clear as possible, this approach is great when it comes to a problem like this:

if-x-is-an-integer-is-the-median-of-the-5-numbers-shown-t3669.html

where you can use this concept to manipulate the average and the mean to make up situations with mean<median as well as median<mean.

[tim]

looks pretty good; thanks for sharing!