If the mean, median and mode of the list

Question

If the mean, median and mode of the list 2,5,2,10,4,2,x are arranged in increasing order, they form an non constant arithmetic progression. What is sum of all the possible values of x?

A. 3
B. 9
C. 16
D. 20
E. 23

aayushjoneja · Answer

The answer to this question should be 'B' (which is '9') not D (which is '20'). Can someone please care to explain?

Adamz · Answer

Mean = (25+x)/7
Mode = 2 (Most frequent number)
Median = 2, x and 4 (Only 3 possibilities if we arrange x in 2,2,2,4,5,10)

Since mean, median and mode are in AP (increasing order), the difference between consecutive terms will be same. Thus -

For Median: 2

2 - 2 = (25 +x)/7 - 2 |  x = -11

For Median: x

x - 2= (25 +x)/7 - x |  x = 3

For Median: 4

4 - 2 = (25 +x)/7 - 4 |  x = 17

Sum of possible values of x = -11 + 3 + 17 =  9

Answer:  Option B

lamthanhdat · Answer

Key word is "increasing order" so 2,2,2 is not considered?

modiassumenda · Answer

Hi  Adamz 
Nicely solved, I forgot to add non constant AP, meaning 2,2,2 will not be allowed.
Editing the que.

egmat · Answer

Hi lamthanhdat,

Yes, you've spotted exactly the flaw in that solution. Your instinct about "increasing order" is correct, and it's the whole reason the answer is  20 (D) , not  9 .

Look at what happens in Adamz's first case, where median =  2  gives  x = -11 :

- Mean = ( 25  + ( -11 ))/ 7  =  14/7 = 2 
- Median =  2 
- Mode =  2

So all three values come out to  2, 2, 2 . The problem says the three quantities are  arranged in increasing order  to form an arithmetic progression. "Increasing" means each term is genuinely larger than the one before it - the sequence has to go  up . A flat  2, 2, 2  doesn't increase at all, so it fails the condition and  x = -11  must be thrown out.

That's the step the earlier solution missed: it solved the equation but never checked that the resulting three numbers actually form an  increasing  list.

Dropping x =  -11  leaves the two valid cases:

-  x = 3  -  2, 3, 4  (increasing AP) ✓
-  x = 17  -  2, 4, 6  (increasing AP) ✓

Sum =  3 + 17 = 20 , which is  D .

Quick way to lock this in:  whenever a problem hands you a word like  increasing  (or  distinct , or  strictly ), treat it as a filter you apply  after  solving. Solve the equation, then ask: "Do my final values obey that word?" Any solution that makes the terms equal - or out of order - gets rejected. Here, that one filter is the difference between  9  and the correct  20 .

Answer: D

vanquisher · Answer

Any intutuive approach to this question?

egmat · Answer

Hi vanquisher,

The fastest way in is to notice that  one term is basically handed to you before you do any work.  Let me build the intuition on top of the thread you've been reading.

Anchor #1: the mode never moves

The value  2  already shows up  three  times (2, 2, 2), and no other value can catch up no matter what x is. So  mode = 2, always.  That means one of your  three  AP terms is  locked  at  2  - and since a mean and median of this list will sit at or above  2 , treat  2  as the  smallest  term of the increasing AP.

Anchor #2: the median has only 3 possible faces

Sort the fixed numbers:  2, 2, 2, 4, 5, 10 . The median is the  4th  value out of  7 , so wherever x slots in, the median can only be:

-  2  (if x ≤  2 )
-  x  (if  2  ≤ x ≤  4 )
-  4  (if x ≥  4 )

Median =  2  dies instantly: you'd have  2  and  2 , which can't be an  increasing  AP (this is exactly the filter e-GMAT pointed out). So the median is either  x  or  4 .

The one equation that finishes it

An increasing AP just means the middle term is the average of the ends. With  2  as the low end, the mean is the high end, so:

mean = 2 × (median) - 2 , and mean =  (25 + x)/7 .

Now plug in the two live medians:

-  Median = x:  (25 + x)/7 = 2x - 2 → 25 + x = 14x - 14 →  x = 3  → terms  2, 3, 4  ✓
-  Median = 4:  (25 + x)/7 = 6 →  x = 17  → terms  2, 4, 6  ✓

Sum =  3  +  17  =  20 , answer  D .

The whole shortcut is those two anchors:  mode is pinned at 2 , and  median has just three options  - so you're only ever solving one small equation at a time.

Answer: D

egmat · Answer

Hi aayushjoneja,

You're right that  B (9)  is the answer, and I can see exactly where the  20  comes from. If you sum only the two "clean" solutions  x = 3  and  x = 17 , you get  20  (choice D). The reason the real answer is smaller is that there's a  third  valid value hiding in the case most people skip.

The setup

The list has  7  numbers. The  mode is always 2  (it shows up three times). The  mean = (25 + x)/7 . The  median  is the  4th  value, and it depends on where x falls:

- If  x > 4 : median =  4  - gives the nice AP  2, 4, 6  -  x = 17 
- If  2 < x ≤ 4 : median =  x  - gives  2, 3, 4  -  x = 3

Those two are the ones you already found, and they add to  20 .

The case that's easy to miss

What if  x ≤ 2 ? Then the median is also  2  - same as the mode. So two of the three quantities are locked at  2 . For  2, 2, mean  to be an arithmetic progression in increasing order, the mean has to be  2  as well, giving the sequence  2, 2, 2 .

That's a perfectly legal AP - a constant list just has  common difference 0 . Solve it:

- (25 + x)/7 =  2  -  25  + x =  14  -  x = -11

Since  -11  is indeed ≤  2 , it's valid.

Putting it together

Possible values:  -11, 3, 17 . Sum =  -11  +  3  +  17  =  9  -  B .

Two things caused the  20 : the question never says x must be positive, so  x can be negative , and a run of equal numbers still counts as an AP. Whenever a problem says "sum of  all  possible values," it's worth checking the boundary case where two of your quantities collapse to the same number - that's usually where the extra solution lives.

Answer: B

PineToad · Answer

I think the most intuitive approach I could find is to spot the easiest of the three statistics first, which would be the mode. If doesn't matter what x is, the will be 2.

The mean looks a bit tricky, but the median follows straightforward logic:

If x < 2, then the median will be 2. If x > 4, then the median will be 4. If x is between 2 and 4 (so if x = 3), the median will be 3.

So that means we have the following combos for the mode and median:

2, 2
2, 4
2, 3

Notice that the questions says we want an arithmetic sequence that is non constant, meaning 2,2 gets eliminated from our list of potential combos.

Since these are arithmetic sequences, we just think about what the mean could be in these scenarios. 
For  2,4 , possible arithmetic sequences include. But remember x > 4 so our mean must be greater than our median!

2,4, 6 
 2, 3 ,4  mean will be less than median; invalid
  0 ,2,4  mean will be less than median; invalid

So now we know our mean could be 6, solve for x where    = 6, where x = 17

For  2,3 , we actually know that x = 3 here! So plug that into our mean formula, and we get the mean equals 4. We check to see if that's a valid sequence: 2, 3, 4!

I think this is the most intuitive explanation I could find. Sorry it was so long.

If the mean, median and mode of the list

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If the mean, median and mode of the list

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