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Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]
28 Mar 2012, 01:24

1

This post received KUDOS

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.

Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]
28 Mar 2012, 01:26

2

This post received KUDOS

Expert's post

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.

Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]
19 Dec 2012, 00:27

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.

Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]
19 Dec 2012, 02:54

Expert's post

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. _________________

Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]
20 Dec 2012, 11:30

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

If x < y < z and y - x > 5, where x is an even integer and y and [#permalink]
23 Apr 2013, 06:33

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 _________________

"When the going gets tough, the tough gets going!"

Re: If x < y < z and y-x > 5, where x is an even integer and y [#permalink]
28 Mar 2014, 00:46

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

gmatclubot

Re: If x < y < z and y-x > 5, where x is an even integer and y
[#permalink]
28 Mar 2014, 00:46

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...