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Yep, answer is D. Statement 2 is the easy one, cross multiply ant that's it.
The other way of looking at 1 is to see that if all your guys are equal to, say, 4 each, then they add up to one, but we are told that z is less than all the other guys, so the other fractions will be less than 1/z, so to allow the other fractions to be more than 1/4, z must be less than 4.
Mike, Per question: w > x > y > z , take the reciprocal(reverse inequality ) 1/w < 1/x < 1/y < 1/z Add all: 1/w + 1/x + 1/y + 1/z now since 1/z is the largest of the 4 terms the sum has to be less than 4/z or 1/w + 1/x + 1/y + 1/z < 4/z But Lft side =1 per 1 1 < 4/z z < 4 2: is straight forward, just cross multiply w < 4 and z < w , hence z < 4. D
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(2) 4 > w > z > 0, so z < 4, sufficient. We are left with B and D
ReplyDelete(1) 1/w < 1/x < 1/y < 1/z (from the question stem)
1/w + 1/x + 1/y + 1/z < 4/z
1 < 4/z
z < 4
So, (D) is the answer
Yep, answer is D. Statement 2 is the easy one, cross multiply ant that's it.
ReplyDeleteThe other way of looking at 1 is to see that if all your guys are equal to, say, 4 each, then they add up to one, but we are told that z is less than all the other guys, so the other fractions will be less than 1/z, so to allow the other fractions to be more than 1/4, z must be less than 4.
I keep getting B as the correct answer.
ReplyDelete1/w > 1/4, so w < 4.
from original we know w > z > 0, so 3.999 > z > 0, thus Z must be less than 4. What am I missing? Thanks! -Mike
Mike,
ReplyDeletePer question:
w > x > y > z , take the reciprocal(reverse inequality )
1/w < 1/x < 1/y < 1/z
Add all:
1/w + 1/x + 1/y + 1/z now since 1/z is the largest of the 4 terms the sum has to be less than 4/z
or 1/w + 1/x + 1/y + 1/z < 4/z
But Lft side =1 per 1
1 < 4/z
z < 4
2: is straight forward, just cross multiply
w < 4 and z < w , hence z < 4.
D