BrainDen.com - Brain Teasers

Question

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?

Recommended Posts

• 0

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 by araver
mobile and spoilers don't mix well
• 1
Share on other sites

• 1

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 by CaptainEd
Share on other sites

• 0

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.

Share on other sites

• 0

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.

Share on other sites

• 0

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)"?

Share on other sites

• 0

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.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account. ×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.