DS MEDIAN

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Q: A SET A CONSIST OF 5 NO. OF THE 5 NUMBERS THE LARGEST IS 4 GREATER THAN THE MEDIAN.
IS THE MEAN GREATER THAN THE MEDIAN ?

A: THE LARGEST NO. PLUS THE MEDIAN IS = 34
B: THE MEDIAN MINUS THE SMALLEST NO. IS = 10

[puneet.goel]

IMO C.

1st tells us that median is 15 and the largest no is 19.
We dont know anything about the smallest numbers.

2nd tells us about the shortest number and median relationship but neither we know median nor we know the range.

1+2 tells us that the series is something like this :

5,a,15,b,15

where
5<= a<=15 and 15<=b<=19

let us maximize a and b to get the max mean :
then a can be 15 and b can be 19.

so mean = 5+15+15+19+19 /5 = 73/5 which is 14.6. and median is 15. so we can get the answer from C.

[ashutoshkumar7]

I think the answer is B

There are five number consider them as
A,B,C,D,E

In the premises it is said E= C+4
A = C-10

B has to <= C and >=A

consider B =C and D =C+4
then the sum will be 4C+8-10 = 4c-2 so them mean will be less than C.

You can see that in the above scenario i have tried to make them mean more than C.

so with B it can be said that the mean is < median always

[tpr-becky]

The formula for average is sum/#

Therefore we are being asked if sum/5> m (where m represents the median).

you can rewrite this to ask if sum >5m

Statement one fills in some information you know that the median is 15 but we don't know what the two lowest numbers are, could be that the sum is 1+1+15+15+19 = 51 which is <5(15) or it could be 15+15+15+ 19 + 19 = 82 which is greater than 5(15)- Eliminate AD.

Statement 2 tells you that the lowest number is m - 10. now we can know what all the numbers are in relation to m. -

m - 10, (some number between m - 10 and m), m , (some number between m and m+4) and m+4.

if we add the lowest, median and highest we get 3m - 7. Then if you make the other two numbers as high as possible (m and m+4 respectively)trying to get the sum greater than 5m you get 5m - 3 which will always be less than 5m - thus B is sufficient.

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[1947]

How much time should it take to solve this ?
And How tough would you rate this problem ?

[tpr-becky]

I didn't really time myself but it was under 2 minutes - the key to quick solving is to know how to set up based on the words and spend your time solving the particulars of the problem.

the word median meant draw 5 lines adn put the numbers in order. Since there were no numbers but I know the relationship between the median and he largest I filled in that relationship.

The word average means sum/# = avg - I filled in what I knew and left blanks. The question set up an equation for me sum/# <Median.

All of this happened almost as I read it - you need to have trigger/response for what words mean and have a consistent way of working with them on the GMAT - that is why it is so important to only study questions that are official or have been vetted by a good company to look as close to official as possible.

Kick off your GMAT prep with a free practice exam. After the test, we'll give you a detailed performance report with personalized tips on how to improve your score. https://www.princetonreview.com/business ... -test.aspx.

[1947]

Thanks I will try on these lines to solve this.

[mehrasa]

let's say largest No in the set= X
and the median= Y... we know from Q that x=4+y

stat1) x+y=34 also we have x-y=4 ==> x=19 the median is 15
but we do not have any idea about mean==> insufficient

stat2)median - smallest no= 10 we don't know none of them ==> insufficient

together)
median=15, largest no=19 and smallest is 15-z(smallest)= 10 ==> smallest =5
now the set is as follow
5, ,15, ,19
we have the fisr and last one in the set ==> mean= Average*number of members in the set [(largest+smallest)/2]*5 ==> mean=12*5=60 then mean>median ==>sufficient

OA: C

[apex231]

we have the fisr and last one in the set ==> mean= Average*number of members in the set [(largest+smallest)/2]*5 ==> mean=12*5=60 then mean>median ==>sufficient

This formula applies only when numbers are consecutive.