Chaitanya_1986 wrote:The mean of four integers will not change if all the integers are multiplied by any constant. What is always true about this set of numbers?
I. The mean of the set is 0
II. The sum of the largest member and the smallest member of the set is 0
III. The set contains both positive and negative integers
I only
II only
III only
I and II only
I, II, and III
OA is A)
The "logic" here might be to start thinking about the problem's original constraint in general. Think of a simple set of numbers: {1,2,3}. If I multiply them all by
any constant (other than 1), the mean will change. Try it and see.
Ok, so it seems that a set that allows this multiplication by a constant must be fairly special. Let's see if the choices offer any guidance for our thinking.
I. The mean of the set is 0
Let's come up with a couple of sets that fit this. What about {-1,0,1} = I made it easy by putting 0 in the physical middle of a consecutive set. It has a mean of 0, and when I multiply all of the terms times a constant, I still get a mean of zero. Try again. What about something non-consecutive like {0,0,0}. That has a mean zero as well when I multiply all of the terms times any constant.
So let's add theory to see if this will be true of any set that can be multiplied by a constant and not have the mean change. The formula for average is the Sum/Number = Average. If each element in the sum is multiplied by the same constant (let's call it K), it would be the same as just multiplying the Sum by K. So:
(k*Sum)/Number = Zero
Divide both sides by K and I still get zero on the right hand side. So the average will remain 0.
Now make the average some non-zero number:
(k*sum)/Number = (Non-Zero#)
Divide both sides by K and the new average is
sum/number = (non-zero#)/k
This means that if the average is not-zero, unless K is 1, the new average, after multiplication, will be different. Statement I must be true all the time = Eliminate B and C.
II. The sum of the largest and smallest is 0.
So, we know that the mean has to be 0 (from statement I). This is a harder nut to crack because coming up with "test" sets is a bit more difficult (so we might want to skip this initially and test statement III to help eliminate more). But we will go ahead and proceed. I need to force a set to have a mean 0, but have the lowest and highest numbers a different distance from 0. So let me pick a small set and try to solve.
Remember, if I can get my set to SUM to zero, I will have a zero average. So what if I start picking in the negative.
{-3, -1,
can I pick a number that "cancels" those 2 out? Sure, -3 + -1 = 4, so if the 3rd term is 4, the average is going to be 0.
{-3, -1, 4}, and the sum of the largest and smallest is NOT 0. Here is my NO case, so eliminate anything with statement II in it. That just leaves A.
III. The set contains both positive and negative integers
We have actually already proved this with the most frequently overlooked sets - the set where all numbers are the same {0,0,0}, or the set that contains just one number {0}. Here are our NO results so eliminate any choices with E.
The answer is A
