HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y|

Question

x, y and z are integers such that |x|, |y| and |z| are distinct numbers. If xyz = 36, what is the least possible value of the average (arithmetic mean) of x, y and z?

A. -12
B. -7
C. -19/3
D. 7
E. 38/3

Kinshook · Answer

Given: x, y and z are integers such that |x|, |y| and |z| are distinct numbers.

Asked:If xyz = 36, what is the least possible value of the average (arithmetic mean) of x, y and z?

Since product xyz is positive and he have to find least value of average, 2 large numbers need to be negative and 1 number positive.

Let us take x=-36; y=-1; z=1; but since |y| & |z| are distinct numbers; not feasible.
Let us take x = - 18; y = -2; z = 1; |x|, |y| and |z| are distinct numbers; Average =( - 18 - 2+ 1)/3 = - 19/3

IMO C

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urshi · Answer

Answer - C

Given - |x|, |y| and |z| are distinct numbers. If xyz = 36

In order to minimize the value of the average, we need to take 2 values as negative , so that xyz remains positive

36 = 2²*3²= -18*-2*1

Average = (x+y+z)/3
= (-18-2+1)/3
= -19/3

Answer Option C

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Nidzo · Answer

For us to have a positive number after multiplying 3 numbers, x, y and z have to either be all positive, or two have to be negative and one needs to be positive for.

As we want the smallest possible mean, we are looking two negatives and one positive.

In this case we want 1 to be our largest number to maintain the positive sign when x, y and z are multiplied together and for the other two numbers to yield 36.

As we cannot use -1, the best pair of will be -2 and -18.

(1)(-2)(-18)= 36

(1)+(-2)+(-18)= -19

Answer is C.

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EatMyDosa · Answer

Constraints :

1. x, y, z = Integers

2. |x|, |y|, and |z| are not equal (i.e. distinct integers)

Given: xyz = 36
 
 What can we infer from xyz = 36 ? Since the product of 3 integers is positive, then only two scenarios are possible:

Scenario 1 :  All three integers  =  POSITIVE

Scenario 2 :  Two integers  (out of 3) =  NEGATIVE

Since we have to find the least possible value of average (x,y,z),   scenario 1 can be rejected   since  it will always give a positive value.

Now, under  scenario 2 ,  multiple cases are possible  (shown below)

Case 1 : x = 1, y = -2, z =-18 --->   Average   = (1-2-18)/3 =   -19/3

Case 2 : x = 1, y = -3, z =-12 --->Average = (1-3-12)/3 = -14/3

Case 3 : x = 1, y = -4, z =-9 ---> Average = (1-4-9)/3 = -12/3

Case 4 : x = 2, y = -3, z =-6 ---> Average = (1-3-6)/3 = -7/3

Among four cases above,  case 1 gives the least value of average.

Option C  (correct)

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EgmatQuantExpert · Answer

Given

• x, y, and z are integers such that |x|, |y| and |z| are distinct numbers.
• x y z = 36

To Find

• The least possible value of the average (arithmetic mean) of x, y, and z.

Approach and Working Out

• As we need to minimize the number and need to take the different absolute values, we can take it as,
 o x = - 18,
o y = - 2,
o z = 1

• Average =  \frac{-19}{3}

Correct Answer: Option C

Adarsh_24 · Answer

Factoring 36=1*2*2*3*3 we can have -ve of the same factors as long as no. Of -ves even.

Least possible. So let’s check max negative.
Highest -ve sum we can get -19.
We can get 1+-2+-18=-19

So -19/3

HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y|

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HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y|

Prep Toolkit

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