Power Prep II section 2

[nickiss20002000]

For the question below, I chose only answer b and c, but answer d was also included in the right answers.

if x y and z are positive integers, and 3x<2y<4z the which of the statements is true? Indicate all that are true.
a. x=y
b. y=z
c. y>z
d. x>z

I read a post related to this question on this blog, but it did not answer my question. Why is x>z a right choice.
Thanks

[n00bpron00bpron00b]

x,y,z any positive integers

which satisfies the condition : 3x < 2y < 4z

let's consider x = 20 => 3x = 60

We want a value of "y" such that the value of 2y is slightly greater than 3x

let's consider y = 31 => 2y = 62 (satsfying the condition 3x < 2y)

Similarly,

We want a value of "z" such that the value of 4z is slightly greater than 3x & 2y and the value of "z" is less than the value of "x"

let's consider z=19 (z<x) = 4 * 19 = 76

x = 20, y = 31, z = 19 (satisfying the condition x>z)

also satisfies the condition 3x (60) < 2y (62) < 4z (76)

[tommywallach]

Nick, are you certain the original question didn't say either:

a) "indicate all that MUST be true"

or

b) "indicate all that COULD be true"

?

It makes a huge difference, and I have literally NEVER seen a GRE question that didn't clarify this issue. Please always post from the actual text of the question, not from memory. The answer to MUST be true would be different, and my guess is you may have confused the two.

-t

[nickiss20002000]

Hey Tommy,
The question states which of the statement could be true.
Indicate all such statements.

I rephrased it in my mind as "indicate all that could be true", so I am aware that some of these answers may not be true all the time.

I think I know the difference between the MUST and the could be true.
My issue is I can't seem to find a reasoning as to why x>z could be true. I don't seem to find a way to quickly find the right numbers as I did with the others.
I guess my really issue is how to prove that x>z could be true, I can see clearly that it's not always true.
Thanks

Nick, are you certain the original question didn't say either:

a) "indicate all that MUST be true"

or

b) "indicate all that COULD be true"

?

It makes a huge difference, and I have literally NEVER seen a GRE question that didn't clarify this issue. Please always post from the actual text of the question, not from memory. The answer to MUST be true would be different, and my guess is you may have confused the two.

-t

[tommywallach]

Cool. Well Noob provided the numbers, so that's the proof there. In terms of how you find this in the future, remember you are trying to check extremes. With a question like this, that will mean making the constraint true in all available ways (can you make them all really close to each other? can you make them all really far away from each other? etc.) Notice Noob said we want something "slightly" larger. This is the kind of game you have to play on questions like this.

Hope it makes more sense now. Let me know if you have further questions.

-t

[nickiss20002000]

Thanks N00b! I went over your explanation and I got it!

x,y,z any positive integers

which satisfies the condition : 3x < 2y < 4z

let's consider x = 20 => 3x = 60

We want a value of "y" such that the value of 2y is slightly greater than 3x

let's consider y = 31 => 2y = 62 (satsfying the condition 3x < 2y)

Similarly,

We want a value of "z" such that the value of 4z is slightly greater than 3x & 2y and the value of "z" is less than the value of "x"

let's consider z=19 (z<x) = 4 * 19 = 76

x = 20, y = 31, z = 19 (satisfying the condition x>z)

also satisfies the condition 3x (60) < 2y (62) < 4z (76)

[nickiss20002000]

Thank you Tommy.

Cool. Well Noob provided the numbers, so that's the proof there. In terms of how you find this in the future, remember you are trying to check extremes. With a question like this, that will mean making the constraint true in all available ways (can you make them all really close to each other? can you make them all really far away from each other? etc.) Notice Noob said we want something "slightly" larger. This is the kind of game you have to play on questions like this.

Hope it makes more sense now. Let me know if you have further questions.

-t

[irs031]

I believe that the most efficient way to solve the problem is picking numbers and using inequalities.

3x<2y<4z

x=10
3(10) = 30

2y>30
y>15, or y>=16 (equal or greater than 16)

4z>2(16)
4z>32
z>8, so z>=9 (greater than or equal to 9)

So when x=10, y will be equal or greater than 16 and z will be equal or greater than 9.

x=y - can not be true because x=10 and y>=16 (no overlap)
y=z - could be true because y>=16 and z >=9 (there is overlap on the number line at 16 and above)
y>z - could be true because z>=9 and y>=16; true when z=9, 10, 11, 12, 13, 14 or 15
x>z - could be true because x=10 and z>=9; true when z=9

I believe the fastest way to solve this problem is to draw a number line and plot the relationships between all 3 variables.

[tommywallach]

Agreed, irs. Great example too.

-t