If the average (arithmetic mean) of the three different positive integ

Question

If the average (arithmetic mean) of the three different positive integers x,y and z is 3, what is the median of the three integers?

(1) x=5
(2) y=3

bashdeep · Answer

Answer:

Given that the average of three numbers is 3.
1) since x= 5, this means other two numbers must be less than 5 (because mean of xyz is 3, hence there sum must be 9, so if x =5, =>. y+z = 4, so neither y nor z can be greater than 5, also 4)

Lets try to find out what will be our second largest number (lets say y).

Can Y be 4?, No since y+z must be equal to 4 and both y and z must be positive

Can Y be 3? Yes, since for y=3, we can have z=1 and y+z will be equal to 4. ALSO WITH THIS WE HAVE OUR MEDIAN AS 3.

Can Y be 2? No, because then for having y+z = 4, z must be 2 which can not be true as all three numbers must be different.

so we can only have cases with either Y=3 and Z=1 (as shown) or vice-versa. In any of the two case, MEDIAN WILL BE 3. So option 1 is sufficient.

2) For y=3 and mean of X,Y,Z = 3, Y must be at the center. Why?
because if X and Z will be greater than Y, the 3 CANT BE THE MEAN ( as y=3 will be smallest). Sim, will be the case wit X,Z lower than Y.

Hence Y must be in the middle . This is also sufficient.

so both conditions are sufficient. Hence choice D is right

CounterSniper · Answer

(1) x=5

x+y+z=9
x=5
y+z=4
x,y,z can be arranged in 
x,y,z
or
x,z,y ( as x is the greatest)
the y,z pairs can be (3,1)
median when (3,1) = 3
in both the cases we can determine the median 
sufficient

(2) y=3
x+z=6
the x,z pairs can be (2,4)(3,3)(5,1)
median when (2,4) = 3
median when (5,1) = 3
in each of the case , we can determine the median 
sufficient

D

rrrkkk · Answer

i think this question is a bit confused, the questions doesn't explicitly state if we have to figure the median number or the elements. In other words to say if is asking about which element then option A is insufficient, vice versa. So yeah I wouldn't blame anyone who fault on this due to the confusion the author made.

varshas044 · Answer

Lets look at this:

(x + y + z) / 3 = 3 
X + y + z = 9

Statement 1: 
X = 5 --------- y + z = 4 
Case 1 ------ if x = 5 then y =1 and z = 3 ---- median 3 
Case 2 ---- if x = 5 then y = 3 and z = 1 -----median 3 
We can not have y = z = 2 since these are diff integers 
SUFFICIENT

Statement 2: 
Y = 3 ------ x + z = 6 
Case 1: if y = 3 then x = 1 and z = 5 -------- median = 3 
Case 2 : if y = 3 then x = 2 and z = 4 -------- median 3 
Case 3: if y = 3 then x = 4 and z = 2 ----- median 3 
And so on 
You can not have x = z = 3 since they are diff integers 
SIFFICIENT

Answer – D

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If the average (arithmetic mean) of the three different positive integ

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

If the average (arithmetic mean) of the three different positive integ

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards