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If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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28 Mar 2012, 02:00
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If x < y < z and yx > 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 yx > 5, where x is an even integer and y [#permalink]
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28 Mar 2012, 02:24
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yx is >5. y is odd and x is even then yx will be odd. Lowest possible value of yx is 7. For lowest possible value of zx, y and z should be close, it means y and z are consecutive odd integers or z = y + 2. Hence zx = y + 2  x = 7 + 2 = 9.



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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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28 Mar 2012, 02:26
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eybrj2 wrote: If x < y < z and yx > 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 \(zx\), 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 \(y5>x\) > \(95>x\) > \(4>x\), then max value of \(x\) is 2 (as \(x\) is even). Hence, the least possible value of \(zx\) is 112=9. Answer: D.
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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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02 Aug 2012, 04:48
If x < y < z and yx > 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 ZX will be 9. ANSWER.
 Ravender Singh



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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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04 Aug 2012, 21:52
eybrj2 wrote: If x < y < z and yx > 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 Zx will be odd. therefore, option A, C, E is eliminated. yx> 5, if x= 2 then Y> 7 (8, 9 etc). so minimum value from zx is 9
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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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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 yx>5
From the first equation i subtracted x so 0<yx<zx From the second equation multiply by (1) so y+x<5 adding the above 2 i got 0<zx5 ie zx>5 We now that zx is odd so the next odd number is 7.



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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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19 Dec 2012, 03:54



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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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20 Dec 2012, 12:30
We have: 1) x<y<z 2) yx>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) zx=? least value
zx=2p+12k=2p2k+1=2(pk)+1  that means that zx 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, yx must be odd. since yx>5, the least value for yx must be 7, the least value for x must be 2, and, thus, the least possible value for y must be 9 (y2=7, y=9) 2<9<z, since z is odd, the least possible value for z is 11 zx=112=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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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 yx>5 which means least value of yx=7 (since y is odd and x is even, result will be odd) Work back: y= 5 (least value) x= 2 (least value). thus yx = 5  (2) = 7 (least possible odd integer greater than 5) Since z> y, thus least possible value of z = 7 Therefore, zx = 7  (2) = 9 Correct choice D
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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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28 Mar 2014, 01:46
ACE are out as they are even. z = odd and x is even therefore ZX is odd. out of B or D we need to see that we have to get the minimum value of zx so we have to minimize z and maximize x. Hence zx is 9



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If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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25 Nov 2015, 01:38
eybrj2 wrote: If x < y < z and yx > 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 = 5From x < y < z and y and z are odd integersWe 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 yx > 5, where x is an even integer and y [#permalink]
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08 Mar 2016, 07:35
eybrj2 wrote: If x < y < z and yx > 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 yx > 5, where x is an even integer and y [#permalink]
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02 Jun 2017, 09:07
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eybrj2 wrote: If x < y < z and yx > 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 yx > 5, where x is an even integer and y [#permalink]
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02 Jun 2017, 11:38
Since yx>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 yx=5, this. Ant be right. Thus we can conclude y to be 3 Therefore next available odd integer is z=5 Zx = 5(4)=9
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Re: If x < y < z and yx > 5, where x is an even integer and y [#permalink]
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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 yx > 5, where x is an even integer and y [#permalink]
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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 yx > 5, where x is an even integer and y [#permalink]
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23 Oct 2017, 05:11
Easy one. Not sure if it is a 700 level question. Given x < y < z and yx > 5 and x is even and y&z are odd. yx>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 70>5 90>5 92>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 zx=90=9 2<9<11 and zx=112=9 Answer D




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