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I think median of a combined set cannot be less than median of each.

If this right, I) would suffice. The information provided in A is required to calculate the average (which will be around 77) and since the median is at least 78, we know the answer.

II) We can't decide on average based on this info, as the average could come close to B's average if there are lost of students in B and less in A.

The median of a combined set can't be smaller than the smallest median of the individual sets, and can't be larger than the largest median of the individual sets.

Hong Hu you seem to have a good collection of questions. Sad its in Chinese

Not so sad Hong Hu are you living in China ? Where did you buy this book ? I am a french guy lost in Shenzhen but I can speak chinese so i am interested...

No actually I'm now in US, I bought the book from a Chinese friend that came from Beijing about a year ago, I think. I'm sure you can find GMAT preparation books in China if you ask the right people.

The median of a combined set can't be smaller than the smallest median of the individual sets, and can't be larger than the largest median of the individual sets.

OK, this is the first time I hear this. Can you prove it?

Where can I learn more about mean, median, mode, S.D? I just want to know all the properties/theories behind. My math knowledge is not from American education system. Back then I didn't get to know about those and probability until I moved to the US. So there is my weakness.

I don't have any algebric formula to explain this but I definitely agree with HongHu. The global median has to be inside the set of the 2 others median, I put some drawing attached with this post, it doesn't explain anything but it just help me to figure it out...

Let U(A)=the smallest x so that at least half the values are <x, and
V(A)=the largest x so that at least half the values are >x, then that
makes M(A)=(U(A)+V(A))/2: i.e. no need to distinguish between even or
odd number of elements.

We can assume without loss of generality that U(A)<=U(B): otherwise we
just swap A and B. This gives three different possibilities:

a) U(A)<=V(A)<U(B)<=V(B)
b) U(A)<=U(B)<=V(A)<=V(B)
c) U(A)<U(B)<=V(B)<V(A)

In case a, for x<V(A), no more than half the values of A are <x; since
x<U(B), less than half the values are <x. Hence, less than half the
values of A+B are <x, and we must have U(A+B)>x. As this holds for all
x<V(A), this makes V(A+B)>=V(A). Similarly, U(A+B)<=U(B). Hence,
M(A)<=V(A)<=M(A+B)<=U(B)<=M(B).

If U(A)<=U(B)<=V(A), for x<U(B) less than half the values in B are <x,
whereas x<V(A) makes at most half the values of A <x; thus less than
half the values of A+B are <x, which makes x<U(A+B). However, for
x=U(B), at least half than values of both A and B are <x, so this is
also true for A+B. This proves that U(A+B)=U(B).

Similarly, if U(B)<=V(A)<=V(B), then V(A+B)=V(A). Case b now follows
as 2*M(A)=U(A)+V(A)<=U(B)+V(A)=U(A+B)+V(A+B)=2*M(A+B) etc.

In case c, the above arguments gives U(A+B)=U(B) and V(A+B)=V(B),
which makes M(A+B)=M(B).

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...