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If x, y and z are different integers, what is the value of y

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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Tue Jul 31, 2018 4:53 am
BTGmoderatorDC wrote:If x, y and z are different integers, what is the value of y?

1) The average (arithmetic mean) of x, y and z is 2
2) x<y<z
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by Jay@ManhattanReview » Tue Jul 31, 2018 4:54 am
BTGmoderatorDC wrote:If x, y and z are different integers, what is the value of y?

1) The average (arithmetic mean) of x, y and z is 2
2) x<y<z
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by Jay@ManhattanReview » Tue Jul 31, 2018 5:01 am
BTGmoderatorDC wrote:If x, y and z are different integers, what is the value of y?

1) The average (arithmetic mean) of x, y and z is 2
2) x<y<z
Given: x, y and z are different integers
Question: What is the value of y?

Let's take each statement one by one.

1) The average (arithmetic mean) of x, y and z is 2.

=> x + y + z = 6. Insufficient.

2) x < y < z

Insufficient.

(1) and (2) together

There are can be many combinations of x, y, and z keeping the three conditions: 1). x + y + z = 6 and 2). x < y < z 3). x, y and z are different integers

For example: 1). x = 0, y = - 4 and z = 10; 2). x = 1, y = -5 and z = 10. No unique value of y.

The correct answer: E

Hope this helps!

-Jay
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by deloitte247 » Sat Aug 04, 2018 7:09 am
What is the value of x
Statement 1 = The average (arithmetic mean) of x, y and z is 2.
This means that the sum of the 3 integers is 6
$$\frac{Efx}{f}=\ \frac{\left(x+y+2\right)}{3}$$
With this we have multiple solutions because,
$$\frac{\left(0+1+5\right)}{3}=\ \frac{6}{3}=2$$
$$\frac{\left(2+2+2\right)}{3}=\ \frac{6}{3}=2$$ $$\frac{\left(1+2+3\right)}{3}=\ \frac{6}{3}=2$$
Hence, statement 1 is NOT SUFFICIENT.

Statement 2 = x
From my own view, there is no sufficient information on about y and z,
Hence, statement 2 is NOT SUFFICIENT.

Therefore, statement 1 and 2 are both not sufficient, hence leaving us with option E as the correct answer.
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