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

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If x < y < z and y-x > 5, where x is an even integer and y a [#permalink]

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-z-and-y-x-5-where-x-is-an-even-integer-and-y-129754.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Oct 2013, 03:07, edited 1 time in total.
Renamed the topic and edited the question.
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Re: number properties [#permalink]

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New post 07 Apr 2011, 06:41
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Eliminating the choices since answer is odd. Only B and D left.

B Vs D
Let x is 0. The smallest value of y is 7. The next odd is 9. Hence z = 9.
z - x = 9

Answer D.

Acer86 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


The answer i am getting is 7..thought original answer is something else...can someone help me out :(

Last edited by gmat1220 on 07 Apr 2011, 07:39, edited 1 time in total.
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Re: number properties [#permalink]

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New post 07 Apr 2011, 06:44
Acer86 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


The answer i am getting is 7..thought original answer is something else...can someone help me out :(

x<y<z
To find the least possible value for z-x; we need to find the values for z and x that can be closest to each other. If x is some even number, then what could be minimum possible odd z.

If x is some even number
y-x>5; y>x+5; minimum value for y=x+5+2=x+7[Note: x+5 is as even+odd=odd and nearest odd greater than x+5 is x+5+2]
Minimum value for z=y+2=x+7+2=x+9 [Note: z=y+2 because both z and y are odd. Difference between two odd numbers is 2]

Thus,
z-x = x+9-x= 9

Ans: "D"
********************************

Or, just use substitution:

Pick x as any even number
x=10
y> x+5 and Y is ODD
y> 15 and is ODD
Closest odd from 15 which is greater than 15 is 15+2=17
z is odd but more than y
Thus, z=y+2=17+2=19

z-x=19-10=9

Ans: "D"
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Re: number properties [#permalink]

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New post 07 Apr 2011, 06:55
z - x is least when z is least and x is max

y > 5 + x

For y to be minimum, x = 0 (lowest even), and y > 5, so y = 7

hence z = 9, because z is odd too

so 9 - 0 = 9

Answer - D
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Re: number properties [#permalink]

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New post 07 Apr 2011, 07:22
z - x is least when z is least and x is max ---> I think this is not true. The least value of z is -Infinity and max value of x is +Infinity. So by this rule the min value is -Infinity

So the assumption should be z and x are closeby for z-x to be minimum.
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Re: number properties [#permalink]

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Re: If x < y < z and y-x > 5, where x is an even integer and y a [#permalink]

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New post 11 Oct 2013, 03:08
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


We want to minimize \(z-x\), so we need to maximize \(x\).

Say \(z=11=odd\), then max value of \(y\) will be 9 (as \(y\) is also odd). Now, since \(y-5>x\) --> \(9-5>x\) --> \(4>x\), then max value of \(x\) is 2 (as \(x\) is even).

Hence, the least possible value of \(z-x\) is 11-2=9.

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-y-z-and-y-x-5-where-x-is-an-even-integer-and-y-129754.html
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Re: If x < y < z and y-x > 5, where x is an even integer and y a   [#permalink] 11 Oct 2013, 03:08
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