Online Word Problems #1

[Carla]

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.

[StaceyKoprince]

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!

[Carla]

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

[Guest]

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?

[StaceyKoprince]

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.

[Abdulla]

If we already told that set T consists of Y integers and 0<y<7.. Why when we are picking numbers we picked negative numbers ??
As I understood, we can pick up any number between 0-7 so all of them should be positive numbers ..
In choice A) you picked {-1,0,4} why ?

[StaceyKoprince]

0 to 7 does not apply to the value of the integers, but rather to the number of integers in the set. The set could have 1 integer, or 4 integers, or 6 integers - basically, anything between 0 and 7.

The actual value of those different integers could be anything, as long as the average of the numbers is positive (because this is given in the problem). A set of numbers could include a negative number and still return a positive average.

For example, the set could be: {-4, 14}
That set consists of y = 2 integers, so that fits (because y has to be between 0 and 7).
The average of the two numbers is positive, so that fits.

[jmacavoy]

2 Questions...

1) How much time should be spent on a problem like this? I imagine even when you are very efficient, it's going to take a long time.

2) Is there an alternative method that can be applied to reduce the burden of testing all of those numbers?

[jmacavoy]

Follow Up...

Is the quick answer that in order for (2/7)y to have a positive integer x as the average, y must be a multiple of 7? Since y is 1, 2, 3, 4, 5, or 6, it is impossible for (E) to be the answer.

[tim]

exactly. that should answer both of your questions, right? :)