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arithmatic mean

by pappueshwar » Tue Mar 13, 2012 8:29 am
If there are different positive 3 integers and the average (arithmetic mean) is 8, what are the value of the 3 integers?

1) The largest integer is 2times the smallest integer.
2) There is 9 among the 3 integers.


OA IS A

IMO C. if we take 3 integers as x,y,z, if z is the largest integer and x is the smallest then z= 2x

there fore X+ Y+2X / 3 = 8, we still can find the values since Y is unknown.

let me know. thanks
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by GMATGuruNY » Tue Mar 13, 2012 9:43 am
pappueshwar wrote:If there are different positive 3 integers and the average (arithmetic mean) is 8, what are the value of the 3 integers?

1) The largest integer is 2times the smallest integer.
2) There is 9 among the 3 integers.


OA IS A

IMO C. if we take 3 integers as x,y,z, if z is the largest integer and x is the smallest then z= 2x

there fore X+ Y+2X / 3 = 8, we still can find the values since Y is unknown.

let me know. thanks
Sum = (number of integers)(average) = 3*8 = 24.
In ascending order, let the 3 integers be x, y and z, so that x<y<z.

Statement 1: z = 2x.
Thus:
x + y + 2x = 24.
3x + y = 24.
y = 24-3x.

To determine the range of x, rephrase the inequality x<y<z in terms of x.

Substituting y=24-3x and z=2x into x<y<z, we get:
x < 24-3x < 2x
4x < 24 < 5x.
The only integer value that satisfies this inequality is x=5.
Thus, x=5, z=10, and y=9.
SUFFICIENT.

Statement 2: 9 is among the 3 integers.
Thus, the sum of the other two integers = 24-9 = 15.
It's possible that the 3 integers are 7, 8, and 9.
It's possible that the 3 integers are 1, 9, and 14.
INSUFFICIENT.

The correct answer is A.
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by pappueshwar » Tue Mar 13, 2012 9:55 am
thanks mitch ...wonderful expln
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