The mean of four numbers

[BMAD]

Assume that a < b < c < d.  The mean of four whole numbers is 78.  The median of a and b is 65.  When c is increased by one, d has to decrease by two to keep the same median. What is the possible range of d values?

[araver]

Agreeing with Rob. Or kinda disagreeing with both

Median for two numbers equals mean. So a +b = 130. A+b+c+d = 4*78= 312 so c+d = 182.

Median for 4 numbers (a,b,c,d already sorted) equals initially the mean of b and c.

So, if c increases by 1 and d has to decrease by 2 to preserve a median, the median in that sentence is ambiguous.

- it cannot be the median of c and d since that is the mean and that is constant only if d decreases the same amount

- it cannot be the median of a set that does not contain d since that would not change, hence it can be the median of b,c,d or a,b,c,d ( biggest three numbers or all numbers)

Assuming the latter (median of all numbers denoted M=(b+c)/2) b/c of symmetry and Occam's razor.

Then as c increases with 1 and d decreases with 2 and since a <b<c<d, then d-2 cannot remain the biggest number since the median would be the mean of b and c+1 which clearly cannot be equal to M. So c+1 > d-2 (if equal, then again c+1 is part of the mean, hence it cannit be constant)

We have 3 cases:

1) a<b<=d-2< c+1. Equality of medians leads to b+c = b + d-2. But c+d = 182 so d= 92 and c= 90. Any 65 <=b<90 works.

2) a<=d-2<b<c+1. Equality of medians leads to the same c=d-2 so d=92, c=90, b however cannot be grater than d-2=c by intial hypothesis, so this case is impossible.

3) d-2 < a<b<c+1. Equality of medians means b+c = b +d which is impossible as well.

So d = 92.

Edited August 19, 2015 by araver
mobile and spoilers don't mix well

[CaptainEd]

If we interpret the clause as "...d has to decrease by two to keep the same median of (a,b,c,d)", I get an unequivocal answer (edited from a moment ago)

There appear to be 4 possible orders of (a, b, c+1, d-2)

( a < b < c+1 < d-2) median (b+c+1)/2
( a < b < c+1=d-2) median (b+c+1)/2
( a < b < d-2 < c+1) median (b+d-2)/2
( a < b=d-2 < c+1) median b

The only case that can have a median = (b+c)/2 is
( a < b < d-2 < c+1) median (b+d-2)/2
in which case c = d-2.

Since avg (a,b) = 65 and avg (a,b,c,d) = 78, a+b=130, a+b+c+d=312, so c+d=182.

That makes c=90, d=92

Edited August 19, 2015 by CaptainEd

[Rob_G]

Is that supposed to be median or mean? Because d has no effect on the median.

Unless you meant the median of c and d.

[BMAD]

It is meant to be median,  the condition a <b <c <d only holds true in the initial conditon.  If c is increased and d is decreased but the median is maintained... the implication should be clear.

[CaptainEd]

BMAD, I'm also having difficulty interpreting the words "...d has to decrease by two to keep the same median". The only median mentioned so far is the median of a and b. As Rob and Araver have said, the value of d has no effect on the median of a and b. So, should we interpret the fragment above as: "...d has to decrease by two to keep the same median of (a,b,c,d)"?

[BMAD]

Yes. I see now that i left out a bit in the OP.  It was meant to be read as  "when c is increased by one, d has to decrease by two to keep the same OVERALL median.

Edited August 20, 2015 by BMAD