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Re: If x < y < z and y-x > 5, where x is an even integer and y
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28 Mar 2012, 01:24
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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.
Re: If x < y < z and y-x > 5, where x is an even integer and y
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28 Mar 2012, 01:26
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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.
Re: If x < y < z and y-x > 5, where x is an even integer and y
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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
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19 Dec 2012, 02:54
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
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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
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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
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Re: If x < y < z and y-x > 5, where x is an even integer and y
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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
Re: If x < y < z and y-x > 5, where x is an even integer and y
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02 Jun 2017, 08:07
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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
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02 Jun 2017, 10:38
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
Re: If x < y < z and y-x > 5, where x is an even integer and y
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23 Oct 2017, 04:11
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
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