The range of a set is 56. Its median number is 50 more than

Question

The range of a set is 56. ITs median number is 50 more than the lowest number , and mean> median, at least how many numbers in this set?

A. 13
B. 15
C. 17
D. 19
E. 21

Paul · Answer

Wow, this one was tough. I got E but I used a calculator :cry:

. I approached this one intuitively: 0-50-50-50... 50 -56-56-56...56 Median has to be in center . The will be an even number on both right and left hand sides. However, to have mean>median, you have to max out right hand sides(56's) and also max out left hand side(50's). By backsolving, only 21 numbers works. You have 1*0, 10*50, 10*56 where average > median and where median = smallest number + 50. I wonder if there is a more practical approach though...

ruhi · Answer

Thats interesting Paul. But is it possible to work like this on a sum?? I would prefer not solving the sum then.

Dookie · Answer

I got 19 here. (D)

Let's say there are X times 50 and X times 56 in the set. 
First number is 0, last number is 56. Median is 50.

Total N= 2*X +1.

Mean = (50*X + 56*X)/N 

Solve Mean > Median and get that X>8.
So we pick X=9. 
N=2*9 + 1 = 19

venksune · Answer

Dookie,
Why 8?

is there any other data missing in the problem? whats the source of the problem. backsolving is critical here (both to assume the number of numbers and the potential spread) and both D and E satisfies.

Dookie · Answer

Venks, 
you just need to solve the equation i wrote:

N= 2*X +1.
Mean = (50*X + 56*X)/N 

106X/(2X+1) > 50
106X > 100X + 50
6X > 50
X> 50/6.
X is an Integer so, least X=9.

2*9+1 = 19.

bigpapi · Answer

The range of a set is 56. Its median number is 50 more than the lowest number, and mean > median, at least how many numbers in this set?


13, 15, 17, 19, 21

Dookie · Answer

https://www.gmatclub.com/phpbb/viewtopic.php?t=11555

MA · Answer

The range of a set is 56. Its median number is 50 more than the lowest number, and mean > median, at least how many numbers are in this set?

A. 13
B. 15
C. 17
D. 19
E. 21

Vijo · Answer

assuming x to be the smallest no in the set and the x+56 to be the greatest no in the set...

x-----------50---------x+56

now 50 is the median.. assuming the least possible case where x=0 and x+56=56, there are 6 nos between after 50(median) i.e. 51,52,53,54,55 and 56
thus there will be 6*2 +1=13 nos minimum..

(A).

christoph · Answer

median is the middle term. in your approach 50 is not the median, because there has to be an equal number of terms that are bigger and smaller than the median, i.e. => 123 4 567 => 4 is the median !

HongHu · Answer

I got 19 as well.

First since the real value is irrelevant, I'm assuming the first term is 0, last term is 56, median is 50 and mean is greater than 50. You can do x, x+56 and x+50 as well. But I'm just trying to make it simpler.

We want the mean to be greater than 50 with the least number of elements in the set, that means we need to have the other elements as big as possible. So I'm letting a elements equal to 56, and the other (a-1) elements equal to 50. 

So (0+50a+56a)/(2a+1)>50
106a>100a+50
6a>50
The smallest a would be 9
The total number of elements in the set is 2a+1=19.

Rupstar · Answer

Sorry but i dont quite understand what is being done here
if A elements are 56 and (A-1) are 50 ?? why is it that A-1 is 50?
in my opinion if A-1 will be 50 then #of terms that are 56 should be A-1 right? otherwise if the number of terms that are 56 are more than the number of terms that are 50 then 56 becomes the median .
I am still confused cause even with this we get

0+50A+56(A-1)/(2A+1)>50
and this doesnt really seem to work out
can someone help with the explanation

HongHu · Answer

Remember we've got a 0 in the first term. So I added (a-1) terms of 50 to the 50 that we already have, therefore making a 50s. And we have (a) terms of 56. This would make a total of 2a+1 elements, with the (a+1) term as the median, which is the last number 50.

Vijo · Answer

Hong I followed your approach and agree with it. But I really don't understand how to tackle these questions. Is there an easier approach for this????

MA · Answer

WELL GUYS, OA is D. Honghu's Approach is best.

lastochka · Answer

ditto that

I spent close to 5 minutes working on this problem....

HongHu · Answer

Hmmm that is a hard question. First thing about a Set question is that you need to make sure what info is given. Such as range(=xn-x1), mean(=(x1+x2+..+xn)/n), median(there are equal number greater than and smaller than it), mode(same elements appear the most of times), etc. Understanding these concept is the key, I believe. Then you'd have to try to write down these given info using letters and numbers, and go from there. For example: S is a collection of ten integers. The smallest number in S is 5. The mean of all the numbers in S is 8. What is the smallest range? You know that mean = (x1+x2+x10)/10=(5+x2+..+x10)/10=8 Therefore x2+...+x10=75 You want the biggest number x10 to be small, that means you want x2 to 8 to be as large as possible (but still not bigger than x10), because to get to the same mean if one number is smaller the other number would have to be bigger. So you figure that you would need three 9s and six 8s and that the range is 4. Not sure if this helps. Hopefully I didn't confuse you too much. :oops:

When you do more of this type of questions you may be able to get a feeling. Anybody has a better approach/explanation to this kind of questions?

lastochka · Answer

For example: S is a collection of ten integers. The smallest number in S is 5. The mean of all the numbers in S is 8. What is the smallest range?
You know that mean = (x1+x2+x10)/10=(5+x2+..+x10)/10=8
Therefore x2+...+x9=75
You want the biggest number x9 to be small, that means you want x2 to 8 to be as large as possible (but still not bigger than x9), because to get to the same mean if one number is smaller the other number would have to be bigger. So you figure that you would need three 9s and six 8s and that the range is 3.

Anybody has a better approach/explanation to this kind of questions? 

Pretty tough question too I thought, the part where you need to figure out the multiples so that you arrive at 75 with the least range possible is tough.

Regarding your example, you meant:
x2+...x 10 =75, and then you want x2 to x 9  to be as large as posible, right, otherwise you're only deealing with 8 integers.
Good practice

MA · Answer

Nice example.
 
the best way to find such a range is to keep in mind that the other numbers are as close as possible to the mean. 
 but i am confuesed with x9.

The range of a set is 56. Its median number is 50 more than

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