Error in one of the Math Questions

[AndrewChan]

I found a mistake on one of the math questions:

x, y, x + y, x - 4y, xy, 2y

For the set of terms shown above, if y > 6 and the mean of the set equals y + 3, then the median must be

(x + y)/2
y + 3
y
3y/2
(x/3) +y


The explanation goes on:

The mean of a set is equal to the sum of terms divided by the number of terms in the set. Therefore,

(x + y + x + y + x - 4y + xy + 2y)/6

= y + 3




(3x + xy)/6

= y + 3


x (y + 3)/6

= y + 3




x (y + 3) = 6 (y + 3)

x = 6

Given that y > 6 and substituting x = 6, the terms of the set can now be ordered from least to greatest:
6 - 4y, 6, y, y + 6, 2y, 6y

The median of a set of six terms is the mean of the third and fourth terms (the two middle terms). The mean of the terms y and y + 6 is

(2y + 6) /2

= y + 3




The correct answer is B.

However, since y>6, then the correct order should be:
6 - 4y, 6, y, 2y, y + 6, 6y
not
6 - 4y, 6, y, y + 6, 2y, 6y
as it was in the answer.

The correct answer should be D (3y/2)

[RonPurewal]

However, since y>6, then the correct order should be:
6 - 4y, 6, y, 2y, y + 6, 6y
not
6 - 4y, 6, y, y + 6, 2y, 6y
as it was in the answer.

The correct answer should be D (3y/2)

nope, looks as though you're getting turned around.

y > 6. in other words, y is MORE than 6.
this means that 2y, which can be written as y + y, is y + (more than 6).
that's greater than y + 6.
therefore, the published answer is correct.

if you don't like the algebraic justification, plug any number greater than 6 into 2y and y + 6, and see for yourself which of them turns out to be greater.

[Guest]

I made a mistake when solving this problem by ignoring y>6. I simply proceeded as follows:

(x + y + x+y + x-4y + xy + 2y)/6 = y+3

(3x + xy)/6 = y+3

I did not factor out x and tried to solve for x and y instead:

3x + xy = 6y + 18
3x - 6y + xy -18 = 0
(x - 6)(y + 3) = 0

x = 6, y = -3

I plugged in these values into the set and arrived at the median = 0 = y + 3.

Clearly, there must be something wrong with the way I solved this because value y does not tie-up with the y>6 requirement, but nevertheless, I still arrived at the right answer. Why?

-Champion-

[JonathanSchneider]

Champion, your math is great but you missed one piece of logic. When we factor out a quadratic form that has been set equal to zero, why do we do that? We do it because we know that one of the factors must be equal to zero. In this case, you got two separate answers, one for x and one for y. However, you then plugged in both of these values. Be careful. Only one of these values needs to be correct. Notice that one does not fit the info given about y, so that is out. When you plug in the x value that remains, you will find the appropriate y value.

[umairmalick]

Hi guys,

I was a little confused here. What I did was, after even I got x=6, I was able to list all the number from least to greatest and was able to pinpoint that y+x+y/2 was the mean. I thought we were talking in terms of variables therefore I did not plug in 6 and therefore got confused and missed the question. It didn't explicitly ask to have a variable solution but it was confusing to plugin value for x whereas some answer choices have x in there.

Any thoughts are much appreciated!

Thanks,
Umair

[tim]

anytime you get a confirmed value for a variable, plug it in EVERYWHERE. this should fix the trouble you're having, but let me know if you still need further help with this one..

[divineacclivity]

However, since y>6, then the correct order should be:
6 - 4y, 6, y, 2y, y + 6, 6y
not
6 - 4y, 6, y, y + 6, 2y, 6y
as it was in the answer.

The correct answer should be D (3y/2)

nope, looks as though you're getting turned around.

y > 6. in other words, y is MORE than 6.
this means that 2y, which can be written as y + y, is y + (more than 6).
that's greater than y + 6.
therefore, the published answer is correct.

if you don't like the algebraic justification, plug any number greater than 6 into 2y and y + 6, and see for yourself which of them turns out to be greater.

Why's E not correct? Since x=6 given the condition in the question, y+3 = x/2+y since x is always 3 as long as y > 6. So, even E qualifies that way, no? What am I missing here? Thanks in advance.

[RonPurewal]

In the problem posted here, the last answer choice is x/3 + y, not x/2 + y.

6/3 is 2. Not 3.

[divineacclivity]

In the problem posted here, the last answer choice is x/3 + y, not x/2 + y.

6/3 is 2. Not 3.

Oh, sorry for that overlook. Thank you, Ron.

[RonPurewal]

no problem.