The average (arithmetic mean) of the three integers x, y, and z is 10

Question

The average (arithmetic mean) of the three integers x, y, and z is 10 and \(x \leq y \leq z\). If \(z - x = 3\), which of the following is a possible value of the median x, y, and z ?

I. 9
II. 10
III. 11

A. I only

B. II only

C. III only

D. I and III only

E. I, II, and III

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Bunuel · Accepted Answer

Firstly, note that the median is the middle value, so essentially the question asks which of the following is a possible value of y. Next, the average being 10 implies the sum is 3*10 = 30.

Let's check the options:

I. 9

y = 9 would mean that x + z = 21. Since x = z - 3, then we'd get x = 9 and z = 12, which is possible since those are integers and x ≤ y ≤ z.

II. 10

y = 10 would mean that x + z = 20. Since x = z - 3, then we'd get x = 8.5 and z = 11.5, which is NOT possible since all integers have to be integers.

III. 11

y = 11 would mean that x + z = 19. Since x = z - 3, then we'd get x = 8 and z = 11, which is possible since those are integers and x ≤ y ≤ z.

Answer: D.

houston1980 · Answer

We can use odd and even properties value to know the answer.

z-x= 3 means one one of z and x is odd and the other is even.

z+x= odd, therefore the median should be an odd number.

Try 9 9 12 works
Try 8 11 11 works.

10 cannot be the median with the conditions given. Therefore the correct option is  D .

MartyMurray · Answer

The average (arithmetic mean) of the three integers  x ,  y , and  z  is  10  and  x \leq y \leq z . If  z - x = 3 , which of the following is a possible value of the median  x ,  y , and  z  ?

Since the average is  10 , the sum of  x ,  y , and  z  is  30 .

Since  x \leq y \leq z ,  y  is the median.

z = x + 3

So,  2x + 3 + y = 30 .

I.  9

2x + 3 + 9 = 30

2x = 18

x = 9   y = 9   z = 12

x \leq y \leq z

So,  9  works.

II.  10

2x + 3 + 10 = 30

2x = 17

\frac{17}{2}  is not an integer.

So,  10  does not work.

III.  11

2x + 3 + 11 = 30

2x = 16

x = 8   y = 11   z = 11

x \leq y \leq z

So,  11  works.

A. I only

B. II only

C. III only

D. I and III only

E. I, II, and III

Correct answer:  D

Purnank · Answer

The average (arithmetic mean) of the three integers x, y, and z is 10
x + y + z = 30 and x≤y≤z, y is the median, z - x = 3.
z = x + 3 and x+z = 2x + 3 = even + odd = odd
odd + odd = even = 30
only odd value possible for y 
now lets check for remainig value 9 and 11.
see if satisfies x≤y≤z after putting y=9 and y=11
for y=9, x=9 and z=12 and for y=11, x = 8 and z = 11
Answer D.

unraveled · Answer

The average (arithmetic mean) of the three integers x, y, and z is 10 and  x \leq y \leq z . If  z - x = 3 , which of the following is a possible value of the median x, y, and z ?

I. 9
II. 10
III. 11

A. I only

B. II only

C. III only

D. I and III only

E. I, II, and III

Here x + y + z = 30
Since x = y = z is a possibility 10=10=10 is the case here. 
But that falisfies the condition z - x  = 3
Hence y cannot be 10, eliminate II from all choices that have it.

Let's check for 9
that means x + z = 30 - y = 30 - 9 = 21 
Adding z - x = 3 to above we have 
2z = 24
z = 12 thus x = 12 - 3 = 9

Similarly, for y = 11
z = 11 and x = 8

I annd III, both possible.

Answer D.

SergejK · Answer

Important side note: 11 is the max here for the median. 13, even if odd wouldn't work as then y>z, which would be against our initial constraint.

KarishmaB · Answer

Use deviations for quick analysis. x \leq y \leq z so y is the middle number and the median in any case. Since mean = 10, let's keep all 3 numbers around 10 and the sum of their deviations from 10 should be 0. Greatest - Least = 3 I. 9 9, 9, 12 If mean = 10, Deficit = 2 and Excess = 2. Correct. 12 - 9 = 3. Correct Median = 9 is possible. II. 10 If deficit = excess, and y = 10, x should have deficit of 1.5 and z should have excess of 1.5 so that difference between them is 3. But then we don't get integers. Not possible. III. 11 8, 11, 11 If mean = 10, Deficit = 2 and Excess = 2. Correct. 11 - 8 = 3. Correct Median = 11 is possible. Answer (D) Here is a post on Deviations: https://anaprep.com/arithmetic-usefulne ... eviations/ Check out this question on mean median and range in DS: https://youtu.be/T_sPj1EKmn0

DanTheGMATMan · Answer

If you set up the sum, you'll notice that the median needs to be an odd number- you can double check the answers from there:
 https://www.youtube.com/watch?v=rNoJGlHZC50

Pr4n · Answer

I think a simple way to look at it is using integer properties that x,y and z are integers and as

x+y+z = 30

and z= x+3

hence,

2x+y =27

2x= 27-y

Now we already know that y is our median, and for x to be an integer y needs to be odd else x will become a decimal number.

And just put both the values of y in it to check whether it satisfies the constraints, as this eqn itself will make sure that the mean is 30.

If we put y=9, then x=9 and z= 12.
If we put y= 11, then x=8 and z=11

Satisfied. 10 is not a possible option this we know as soon as we see the integer constraint and the equation 2x= 27- y

As 2x has to be even and 27 - y will only be even if y is odd.

The average (arithmetic mean) of the three integers x, y, and z is 10

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The average (arithmetic mean) of the three integers x, y, and z is 10