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If x<y<z and y-x>5, where x is an even integer

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by GMATGuruNY » Sat Jul 27, 2013 7:08 am
guerrero wrote:If x<y<z and y-x>5, where x is an even integer and Y and Z are odd integers, what is the least possible value of z-x?

A. 6
B. 7
C. 8
D. 9
E. 10

OA d
y < z
y-x < z-x.

Since y-x > 5, we get:
5 < y-x < z-x.

Since y and z are ODD and x is EVEN, y-x = odd-even = ODD and z-x = odd-even = ODD.
Thus, the least possible options are y-x = 7 and z-x = 9.

The correct answer is D.
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by Matt@VeritasPrep » Sat Jul 27, 2013 7:46 am
Maybe a more intuitive approach:

If x is even, then the minimum value of x is 2.

If y - x > 5, then y > 5 + x, so y > 5 + 2, so y > 7. Since y is odd, the minimum y is 9.

If the minimum y is 9, and z > y, then the minimum z is 11 (since z is odd).

So z - x = 11 - 2 = 9
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by GMATGuruNY » Sat Jul 27, 2013 9:17 am
Plugging in values is a great approach here, but we have to be careful.
The question stem asks for the least POSSIBLE value of z-x.
After we try one case, we should confirm -- either by using reason or by plugging in a second combination of values -- that there is no way to yield a smaller value for z-x.

Case 2: x=4.
Since y > x+5, y > 9.
Since y must be odd, the minimum value for y = 11.
Since z>y, and z also must be odd, the minimum value for z = 13.
The result:
The minimum value for z-x = 13-4 = 9.

As the second case shows, if we change the value of x, the minimum distances between y and x and between z and x stay the same.
Thus, the least possible value of z-x = 9.
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by [email protected] » Sat Jul 27, 2013 2:29 pm
Hi guerrero,

All of the other explanations here should provide you the necessary tactics to crush this question (and its variations), so there's only one thing that I'll add to it:

This question uses Number Property vocabulary, which makes it likely that TESTing values will be an easy way to answer the question. In these situations, you are allowed to use ANY values that fit the description. While I would normally NOT use the number 0 (except for DS questions), this questions allows for the value of 0 (because the question asks for the LEAST POSSIBLE VALUE, every option that's permissible is applicable).

So, I'd set x = 0 (it's the easiest value for x and requires less math than the number 2)

Now, we have y - 0 > 5 AND y must be odd...The least value would be....

y = 7

Now, we need z>y AND z must be odd....The least value would be....

z = 9

So, the correct answer is still D and you were able to same a little time. Finding these little shortcuts will help you to avoid the pacing problem that many Test Takers have in the Quant.

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by Matt@VeritasPrep » Sat Jul 27, 2013 2:54 pm
Yup, Mitch - well-cautioned! You and I have had a role reversal on this question: you usually recommend the sensible number-picking approach while I like to bother with proofs. :D

That said, given that we have an even number minus an odd number, once we've proved 9 works, we only need to test for 7. Warning: proof ahead!

Given that y > x + 5 and z ≥ y + 2 -- since z and y are both odd and z > y -- we can combine the inequalities and get z > (x + 7), meaning z - x > 7 and 9 is our answer.
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by shruti93 » Sun Jul 28, 2013 7:02 am
hi! i am confused. since 0 is an even integer and all we know is y-x>5. if x=0 we can get the value of y=7(least possible value) so minimum value of z= 9. and minimum value of z-x=7. what is wrong with this?


ps-0 is neither positive nor negative but it is even.
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by [email protected] » Sun Jul 28, 2013 10:47 am
Hi shruti93,

Using x = 0 DOES make the math easier (see my explanation above), but you made a math mistake in your explanation:

We're looking for z - x

x = 0
z = 9

z - x = 9 NOT 7

These are exactly the little mistakes that keep Test Takers from scoring what they want to be scoring.

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Re: If x<y<z and y-x>5, where x is an even integer

by Scott@TargetTestPrep » Tue Jan 21, 2020 7:52 am
guerrero wrote: ↑
Sat Jul 27, 2013 6:43 am
 If x<y<z and y-x>5, where x is an even integer and Y and Z are odd integers, what is the least possible value of z-x?

A. 6
B. 7
C. 8
D. 9
E. 10

OA d
Since y < z and y and z are odd integers, the smallest value that z can be, in terms of y, is z = y + 2.


We know that y - x > 5, so we might think that the minimum difference between y and x would be 6. However, this is impossible. Since x is even and y is odd, their difference cannot be an even number (because even – odd = odd). Thus, we see that the smallest difference between even number x and odd number y is not 6, but rather it is 7. So we have y - x = 7.

Therefore, the smallest difference between z and x is:

z - x = (y + 2) - x = (y - x) + 2 = 7 + 2 = 9.

Answer: D

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