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Mean median and mode

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by leonswati » Mon Apr 23, 2012 11:03 am
The mean median and mode for a set of data are equal: 30,40,80,50,x. What is X?

My doubt is that the Mode cannot be more than 2 and how can the median be 2?Am I reading the question correctly?
Swati
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Source: — Problem Solving |

by neelgandham » Mon Apr 23, 2012 11:41 am
Swati,

The mode is the number that is repeated more often than any other number in the list and not the number of times that number is repeated.So, in this case mode can't be 2. Let me try and solve the question.
The mean,median and mode for a set of data are equal: 30,40,80,50,x. What is x?
Since we know that we have only one mode, we are sure than x is one among 30,40,50,80.

If x = 30, Mode = 30 and Mean = (30+30+40+50+80)/5 = 46. Since Mode != Mean, x !=30
If x = 40, Mode = 40 and Mean = (30+40+40+50+80)/5 = 48. Since Mode != Mean, x !=40
If x = 50, Mode = 50 and Mean = (30+40+50+50+80)/5 = 50, Median = 5-0 Since Mode = Mean = Median, x =50
Anil Gandham
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by leonswati » Mon Apr 23, 2012 12:39 pm
Ohh my God... Thank you sooooo much. How could I think mode to be the number of times a number is repaeated. But thanks for clearing my doubt..
Swati
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by LalaB » Tue Apr 24, 2012 1:29 am
our set is

30,40,50,80,x

lets assume that x>50
then median =50

mean =median
(30+40+50+80+x)/5= 50
x=50

then mode=50
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by shantanu86 » Tue Apr 24, 2012 3:36 am
Intuitive approach-

Arrange serially- 30, 40, 50 & 80
Current mean= 50
Going by the options adding 50 to set will not displace mean and will become new median :)

Mathematical approach-

Put X in set- 30, 40, x, 50 & 80 (as x is new meadian)
Since x is mean also, we have-

(30 + 40 + x + 50 + 80)/5 = x => x =50.

Hope it helps!!
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by [email protected] » Tue Apr 24, 2012 4:03 am
Well What neelgandham said was perfectly alright... The hint or the clue to break this question was the 'mode' was a part of this question. That is what the clue is. If mode is present so far in the data i.e 30,40,80,50,x.

Therefore the x has to repeat itself and it cannot be any different term. And then the trial and error would help you get to the answer 50...

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