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

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

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y-x is >5. y is odd and x is even then y-x will be odd. Lowest possible value of y-x is 7.
For lowest possible value of z-x, y and z should be close, it means y and z are consecutive odd integers or z = y + 2.
Hence z-x = y + 2 - x = 7 + 2 = 9.
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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eybrj2 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


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

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New post 02 Aug 2012, 04:48
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


Y is odd and X is even, so (y - x) has to be odd. ( Y - X ) > 5 means ( Y - X ) can have a value of 7.

Y - X = 7; Y = X + 7

Main equation will become,
X < X + 7 < Z

Subtract X from all,

0 < 7 < Z - X

Since Z is odd and X is even, (Z - X) has to be odd.

As (Z - X) > 7, the least possible value of Z-X will be 9. ANSWER.


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

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New post 04 Aug 2012, 21:52
eybrj2 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


Z-x will be odd. therefore, option A, C, E is eliminated.

y-x> 5, if x= 2 then Y> 7 (8, 9 etc). so minimum value from z-x is 9
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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New post 19 Dec 2012, 01:27
I did it like this and i am getting 7 as the answer. Kindly tell me where i went wrong.

Given
x<y<z
y-x>5

From the first equation i subtracted x so 0<y-x<z-x
From the second equation multiply by (-1) so -y+x<-5
adding the above 2 i got 0<z-x-5 ie z-x>5
We now that z-x is odd so the next odd number is 7.
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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New post 19 Dec 2012, 03:54
maddyboiler wrote:
I did it like this and i am getting 7 as the answer. Kindly tell me where i went wrong.

Given
x<y<z
y-x>5

From the first equation i subtracted x so 0<y-x<z-x
From the second equation multiply by (-1) so -y+x<-5
adding the above 2 i got 0<z-x-5 ie z-x>5
We now that z-x is odd so the next odd number is 7.


You got z-x>5 but we also have y-x>5, so the least value of y-x is 7 and since z>y then the least value of z-x is 9.
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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New post 20 Dec 2012, 12:30
We have:
1) x<y<z
2) y-x>5
3) x=2k (x is an even number)
4) y=2n+1 (y is an odd number)
5) z=2p+1 (z is an odd number)
6) z-x=? least value

z-x=2p+1-2k=2p-2k+1=2(p-k)+1 - that means that z-x must be an ODD number. We can eliminate answer choices A, C and E
we are asked to find the least value, so we have to pick the least numbers
since y is odd and x is even, y-x must be odd. since y-x>5, the least value for y-x must be 7, the least value for x must be 2, and, thus, the least possible value for y must be 9 (y-2=7, y=9)
2<9<z, since z is odd, the least possible value for z is 11
z-x=11-2=9

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

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New post 23 Apr 2013, 07:33
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 :(


Question asks least possible value, thus we can substitute by least possible numbers to get to answer

A, C and E are out since they are even

Left with B and E

y-x>5 which means least value of y-x=7 (since y is odd and x is even, result will be odd)
Work back: y= 5 (least value) x= -2 (least value). thus y-x = 5 - (-2) = 7 (least possible odd integer greater than 5)

Since z> y, thus least possible value of z = 7
Therefore, z-x = 7 - (-2) = 9

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

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New post 28 Mar 2014, 01:46
ACE are out as they are even. z = odd and x is even therefore Z-X is odd. out of B or D we need to see that we have to get the minimum value of z-x so we have to minimize z and maximize x. Hence z-x is 9
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If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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New post 25 Nov 2015, 01:38
eybrj2 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?


From where x is an even integer & least possible value of z - x?

we can have x < y as 2 < 7 , where y - x = 5

From x < y < z and y and z are odd integers

We have x < y < z = 2 < 7 < 9

So, least possible value of z - x? = 9 - 2 => 7
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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New post 08 Mar 2016, 07:35
eybrj2 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


here let y be 13 => then x will be 8 atleast and z will be 15 atleast => 9 is the difference
thus D .
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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eybrj2 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



Think about it this way: x < y < z
Difference between y and x is more than 5 so it is at least 6. But x is even and y is odd so their difference would be odd. Hence the diff between them will be at least 7.
Now z is greater than y by at least 2 (since z is odd too), hence diff between x and z is at least 9.

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

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New post 02 Jun 2017, 11:38
Since y-x>5; let us assume x to be -4 (even no); such that we get the least value of y.
Hence y has to be either 1 or 3, but since y=1 give y-x=5, this. Ant be right. Thus we can conclude y to be 3
Therefore next available odd integer is z=5
Z-x = 5-(-4)=9


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

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New post 04 Sep 2017, 02:40
x=0; y=5; z=7 - if we take this set of values, then all the conditions satisfy and the answer becomes 7.
Where did I go wrong?
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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New post 04 Sep 2017, 04:33
DAakash7 wrote:
x=0; y=5; z=7 - if we take this set of values, then all the conditions satisfy and the answer becomes 7.
Where did I go wrong?


y - x needs to be greater than 5.
If x = 0, y = 5, then y - x = 5

So this set of values does not satisfy the conditions.
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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]

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New post 23 Oct 2017, 05:11
Easy one. Not sure if it is a 700 level question.
Given x < y < z and y-x > 5 and x is even and y&z are odd.
y-x>5. Now y is odd and x is even, therefore, Y can be 7, 9, 11.... and x is even and can be 0,2,4...
let us plug in a set of values
7-0>5
9-0>5
9-2>5
One more condition mentioned is z>y and is odd therefore z can be 9, 11, 13....
we now have
0<7<9 and z-x=9-0=9
2<9<11 and z-x=11-2=9 Answer D
Re: If x < y < z and y-x > 5, where x is an even integer and y   [#permalink] 23 Oct 2017, 05:11
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