Online Question Bank : Word Problems : #1

I went over the online solution and looked at the different cases for each for example {-1,0,4} can be used to eliminate (A) because it has a median of 0 etc..

For this question I wonder if you could offer some advice on an method to apply to problems like this one.

I had started with a sample set of {1,2,3,4,5,6} and it took me a long time to try out the different cases that I just ended up trying to make a guess and went with the wrong answer choice.

Hi, Carla

I'd be happy to help! Please post the entire question and answer choices for any questions you have. (I'm sorry that I can't do this for you, but I have to respond to everyone's posts and I wouldn't have time to get through everything if I also had to look up and post all of the questions and answers for each request!) Thanks!

Hi Stacey,
I am sorry I must have missed this one.. the actual question is as follows:
Thank you so much for all of your help!!
-Carla


T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?


0
x
–x
(1/3)y
(2/7)y

I think the answer is 2/7y.

Y is the number of integers in the set
X is the mean of the Y numbers
The median could be 0 with 2 numbers, 1, -1
The median could be x with 3 numbers, 1,2,3
The median could be -x with 3 numbers, -1,-2,-3
The median could be y/3 with 3 numbers 0,1,2
The median couldn't be 2y/7 with 2-6 numbers

Is that right?

Yes, that's the right answer! Just be careful, as some of the examples you used did not follow the constraint given in the question (that the average has to be a positive integer).

Carla, the best approach, since this is a negative question (which could NOT be the median) is to try to disprove each answer choice.

The problem tells you that set T consists of y integers with y between zero and 7 - you chose to use 1 through 6, but this did not mean that the integers in T are zero thru 7; it means that the # of integers in T is somewhere between zero and 7. That is, you could have 1 integer in set T - and that integer could be anything.

This makes it easier to test the answer choices b/c you do not need to use so many numbers.

The problem tells us the average is a positive integer, so we have to follow this constraint. Other than that, I want to make my life as easy as possible, so I'm going to use the smallest numbers (and the smallest number of numbers in set T) that I can to test the choices.

A) zero. Can I make zero the median? I can have 3 integers with zero in the middle (__,0,__). As long as the average of the three is a positive integer, I'm fine. Let's minimize the negative number (to the left of zero) and call that -1. (-1,0,__). Now I just need a positive integer that will make the average also a positive integer. The smallest possible number that works is 4. (-1,0,4) the median is zero and the average is (-1+0+4)/3 = 1.

B) x. Can I make the median x? I can do any three consecutive positive integers. The median of the set will equal the average and the average will be a positive integer, as required. (1,2,3) is the simplest example.

C) -x. (__,-x,__) So the two left numbers have to be negative, and the right number has to be bigger to make the average positive. (-2,-1,__) is the simplest to start. That adds to -3. If I choose +6 for the third number, the three numbers will add to +3 and average (x) to 1. -x = -1. (-2,-1,6)

D) (1/3)y. (__,y/3,__). y has to be an integer between 0 and 7. It also has to be divisible by 3 (because the numbers in set T are all integers). My two possibilities for y, then, are 3 and 6. Start with the smallest one, as always. If y=3, then y/3 = 1. (__,1,__) Simplest possibility is (0,1,2) which has a positive integer average and a median equal to y/3.

E) (2/7)y. (__,2y/7,__). As before, y has to be an integer between 0 and 7, and 2y/7 also has to be an integer. 1 doesn't work. Neither do 2, 3, 4, 5, or 6. None of those will create an integer for 2y/7. This is the correct answer.