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If r, s and w are positive numbers such that w=60r+80s and r [#permalink]
 17 Nov 2005, 19:21
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If r, s and w are positive numbers such that w=60r+80s and r+s=1, is w < 70? 1) r > 0.5 2) r > s -------------------------------------------------------------------------- Sorry for the confusion. I guess this question is messed-up. Would it be ok if it were "numbers" instead of "integers" ?
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Last edited by gamjatang on 17 Nov 2005, 21:19, edited 1 time in total.
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w = 60r + 80s
r+s = 1, then w = 60r + 80(1-r), w = 60r + 80 - 80r = w = 80-20r. In order for w to be greater than 70, it must be 71. So plug this in: 71 = 80 - 20r and r = 9/20. This value is less than 1/2. So if r is always greater than 1/2, w will always be less than 70. (1) is sufficient.
Statment (2) is a little contradicting. It says r>s, but we also know r and s are positive integers. Yet, r+s must be equal to 1. If so, s must be 0 but 0 is neither positive or negative. So i'm not too clear about (2).
A is my answer.
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how can r+s=1? the stem say both are positive integers?
this question is invalid in ma opinion..
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gamjatang wrote: If r,s and w are positive integers such that w=60r+80s and r+s=1, is w < 70?
1) r > 0.5
2) r > s
r+s=1 and r,s are positive integers ...it seems weird to me  . Remember 0 is not a positive integer.
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Yup, this question is mindboggling 
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fresinha12 wrote: how can r+s=1? the stem say both are positive integers?
this question is invalid in ma opinion..
good catch. r and s are only +ves to satisfy r+s=1.
good job freshina12. 
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After the edit, my answer is D. Both are sufficient.
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After edit...I would say D
(2) r>s
suppose r=0.99999999999999999999999 and s=0.000000000000000001
then in that case w~60...it is less than 70
suppose r=.50005 s=0.490005 then w~30+40~70...but we know it slightly less than 70...always...except when r and s both = 0.5 which is not the case
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gamjatang wrote: Edited: intergers -> numbers
--------------------------------------------------------------------------
If r, s and w are positive numbers such that w=60r+80s and r+s=1, is w < 70?
1) r > 0.5
2) r > s
--------------------------------------------------------------------------
Sorry for the confusion. I guess this question is messed-up.
Would it be ok if it were "numbers" instead of "integers" ?
you're correct, Matt ^_^
1) we have s=1-r --> w= 60r + 80 ( 1-r) = 80 - 20r
r>0.5 ---> 80-20r < 80 -20*0.5= 70
--->suff
2) r>s ---> r+r> r+s= 1 ---> 2r>1 ---> r>0.5 ---> similar to stmt 1 --->suff
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ywilfred wrote: w = 60r + 80s
r+s = 1, then w = 60r + 80(1-r), w = 60r + 80 - 80r = w = 80-20r. In order for w to be greater than 70, it must be 71. So plug this in: 71 = 80 - 20r and r = 9/20. This value is less than 1/2. So if r is always greater than 1/2, w will always be less than 70. (1) is sufficient.
Statment (2) is a little contradicting. It says r>s, but we also know r and s are positive integers. Yet, r+s must be equal to 1. If so, s must be 0 but 0 is neither positive or negative. So i'm not too clear about (2).
A is my answer.
r > s
r+s = 1, so r and s could be (.9,.1) (.8,.2), (.7,.3), (.6,.4)
For each of these sets, w = 60r+80s gives 62, 64, 66, 68 all of which is less than 70.
So after the edit, answer should be D.
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