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

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If the average (arithmetic mean) of the three different positive integ  [#permalink]

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New post 11 Oct 2018, 03:13
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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

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Re: If the average (arithmetic mean) of the three different positive integ  [#permalink]

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New post 11 Oct 2018, 03:56
T1101 wrote:
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


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
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If the average (arithmetic mean) of the three different positive integ  [#permalink]

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New post 15 Oct 2018, 01:16
T1101 wrote:
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


(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
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If the average (arithmetic mean) of the three different positive integ   [#permalink] 15 Oct 2018, 01:16
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