Hi,
I came across this question under GRE Quant Problem Solving (GregMat - GregMat) Tag - Mean, Median Mode, Type - Medium.
The correct answer to this question is marked as 17.6 and the process is explained in the video solution as well.
However, I think the answer to this question should be 20, as the question asks for mean greater than 16 and not equal to 16, for which the option 17.6 won’t work. Please correct me if I’m wrong.
Thanks
First arrange them in ascending order :
10.2 ,14.2, 15.6, 18.6, 19.8
Now, mean = \frac{78.4}{5}=15.68
and median = 15.6
Now, we need to add minimum in such a way that either of the following condition are satisfied :
Now, eliminate option A immediately as we need a value greater than our current mean(i.e. 15.68) to drag the value towards 16.
If we treat it like a weighted average problem we , can quickly solve it:
Now, tackle mean \geq 6 first ;
10.2 is 5.8 unit away from 16
14.2 is 1.8 unit away from 16
15.6 is .4 unit away from 16
18.6 is 2.6 ahead of 16
19.8 is 3.8 ahead of 16
or the above calculation can be re-written as:
-5.8 + -1.8 + -0.4 + 2.6 + 3.8
if we sum it , we’ll end up with -1.6 , thus we need our answer to be atleast 1.6 unit ahead of 16 (which is option C 17.6 ) as 17.6 -16 = 1.6
now checking on median condition :
10.2 ,14.2, (15.6 , 17.6) , 18.6, 19.8
median = \frac{15.6 + 17.6}{2}=16.6
Thus, option C is the min value which meets our condition
By choosing option C 17.6 only one of the two conditions is satisfied for the second condition, i.e Median >= 1.6 is satisfied. However, the mean remains 16, we have been asked for which the mean would be > 16 and not >= (greater than or equal to). Please correct me If I’m wrong.
its due the phrase approximate minimum in the question that we can select 17.6
I see, so 17.6 would be considered as 18? Ahh, so yeah that would make the mean greater than 16. That’s so clever, thanks for clearing this out.
