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15 Jul 2010, 13:45
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
81% (01:52) correct
19%
(01:48)
wrong
based on 111
sessions
History
Date
Time
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Not Attempted Yet
Is 20 the average (arithmetic mean) of u, v, w, and 20?
(1) u + v + w = 60
(2) u - v + w = 20, u + v - w =16
(1) u + v + w = 60
(2) u - v + w = 20, u + v - w =16
OA was A...
I believe it should be D - since adding the two equations provided in ST2 cancels out v + w, so we can solve for u, leaving us with 2 equations with 2 unknowns...
I believe it should be D - since adding the two equations provided in ST2 cancels out v + w, so we can solve for u, leaving us with 2 equations with 2 unknowns...
15 Jul 2010, 14:12
Kudos
Bookmarks
AndreG
Have you tried to solve these 2 equations?
When you add \(u-v+w=20\) and \(u+v-w=16\) you'll get \(u=18\) and if you substitute the value of \(u\) in these two equations you'll get: for 1. \(18-v+w=20\) --> \(w-v=2\) and for 2. \(18+v-w=16\) --> \(w-v=2\), so two identical equations, thus you won't be able to solve for \(w\), \(v\) or for \(w+v\).
Statement to be sufficient you need \(n\) distinct linear equations for \(n\) variables.
