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x, y, and z are positive integers and average of x, y, and z is 10.

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x, y, and z are positive integers and average of x, y, and z is 10.  [#permalink]

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New post 19 Jan 2019, 08:50
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x, y, and z are positive integers and average of x, y, and z is 10. If x<=y<=z and z-x=3, which of the following could be the median of x, y, and z?

I. 9

II. 10

III. 11

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I and III only

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Location: India
Re: x, y, and z are positive integers and average of x, y, and z is 10.  [#permalink]

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New post 19 Jan 2019, 09:16
rohan2345 wrote:
x, y, and z are positive integers and average of x, y, and z is 10. If x<=y<=z and z-x=3, which of the following could be the median of x, y, and z?

I. 9

II. 10

III. 11

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) I and III only


Given
Average of x, y, and z is 10-------------(a)
x<=y<=z => x=y=z and x<y<z --------------(b)
z-x=3 -----------------------(c)

x+ y + z = 30

Will start from II= 10
median will be y in all the cases because of (b)
Now you can say x=y=z=10(will give median as y), but when you consider z-x = 3, you will never find (c) to be true
9<10<11, 11-9 = 2
10=10=10, this will just prove (a), but (c) will never be true.
You can remove B and D now.

Case I y = 9
9 <= 9 < 12, this will satisfy all the conditions, 12-9 =3 & 9+9+12 = 30

Case III, y = 11
8 < 11 <= 11, this will satisfy all the conditions, 11 - 8 =3 & 8+11+11 = 30

Answer E.
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Re: x, y, and z are positive integers and average of x, y, and z is 10.   [#permalink] 19 Jan 2019, 09:16
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x, y, and z are positive integers and average of x, y, and z is 10.

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