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Re: If X is a set of numbers {1,3,5} and Y is a set of number [#permalink]
D for me too.

Yes even I went the long way......
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Re: If X is a set of numbers {1,3,5} and Y is a set of number [#permalink]
can you guys explain your answers....


I get 6...and I dont see that option

I get the |X-Y| as 0, 1, 2, 3, 4, 5....

3-3=0
6-5=1
5-3=2
6-3=3
6-2=4
6-1=5

Sine we have | | we can neglect the other results that give +ve and -ve
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Re: If X is a set of numbers {1,3,5} and Y is a set of number [#permalink]
some of your numbers are subtracting y-x..

for example your
Quote:
6-5 = 1
is Y-X

the questions is asking for X-Y
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Re: If X is a set of numbers {1,3,5} and Y is a set of number [#permalink]
81rabbit wrote:
some of your numbers are subtracting y-x..

for example your
Quote:
6-5 = 1
is Y-X

the questions is asking for X-Y

|X-Y|=|Y-X|
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Re: If X is a set of numbers {1,3,5} and Y is a set of number [#permalink]
thenine wrote:

I also got D, but just by working it out. I don't know a short cut to solve it, but working it out took far less than 2 minutes


OA is D.

the possible values of lx-yl are 9, but some of the values are repetitive, therefore there are only 5 diff values. i think this is the only possible short cut way to solve this problem.
l1-2l=1
l1-3l=2
l1-6l=5

l3-2l=1
l3-3l=0
l3-6l=3

l5-2l=3
l5-3l=2
l5-6l=1

the values are 0,1,2,3 and 5.
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Re: If X is a set of numbers {1,3,5} and Y is a set of number [#permalink]
fresinha12 wrote:
can you guys explain your answers....


I get 6...and I dont see that option

I get the |X-Y| as 0, 1, 2, 3, 4, 5....

3-3=0
6-5=1
5-3=2
6-3=3
6-2=4
6-1=5

Sine we have | | we can neglect the other results that give +ve and -ve

fresinha,

2 is not in set x. therefore l6-2l =4 is not a value of lx-yl.
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Re: If X is a set of numbers {1,3,5} and Y is a set of number [#permalink]
aye aye aye.....damn, how could I have screwed this one up...your right, 2 is in the same set...ma bad...if i can only stop making these stupid mistakes....ahhhhhhhhh!


MA wrote:
fresinha12 wrote:
can you guys explain your answers....


I get 6...and I dont see that option

I get the |X-Y| as 0, 1, 2, 3, 4, 5....

3-3=0
6-5=1
5-3=2
6-3=3
6-2=4
6-1=5

Sine we have | | we can neglect the other results that give +ve and -ve

fresinha,

2 is not in set x. therefore l6-2l =4 is not a value of lx-yl.




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