B
y>0, what is the value of x?
stmnt 1: |x-3|>y
or |x-3|>0 (y>0), also absolute value is always positive or 0.
x can be 3 or any other number. it gives no unique value. not sufficient
stment 2: |x-3|<-y
or |x-3|<0, absolute value can never be negative. the least it can be is 0, or x=3 suffient, hence B
x value
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Source: Beat The GMAT — Data Sufficiency |
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scoobydooby
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(i) Insufficient, x and y can take up many different values
(ii) Since the absolute value of a number is 0 or positve, it can never be negative.
So |x-3|> -y and can never be less than y.
So the only possibility is the |x-3|=-y (or 0), which is possible when x=3
Ans = B
Whats the OA
(ii) Since the absolute value of a number is 0 or positve, it can never be negative.
So |x-3|> -y and can never be less than y.
So the only possibility is the |x-3|=-y (or 0), which is possible when x=3
Ans = B
Whats the OA
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vittalgmat
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it is obvious stmt 1 is insufficient.
I used Ian's distance method.. it is so cool!!! I seem to have removed an "absolute" mindblock!!
Stmt 2 is tricky and at the outset looks like a wrong stmt.
but |x - 3| is less than or equal to negative value of a VARIABLE, not a constant. So it is ok.
As ppl have explained above, y can only take 0.
we can as well equation to 0
ie | x -3 | = 0. Ie the distance between x and 3 is 0 on the numberline.
This can happen only if x = 3.
B
Million thanks to Ian for pushing the distance idea for absolutes, in multiple posts.
HT helps
I used Ian's distance method.. it is so cool!!! I seem to have removed an "absolute" mindblock!!
Stmt 2 is tricky and at the outset looks like a wrong stmt.
but |x - 3| is less than or equal to negative value of a VARIABLE, not a constant. So it is ok.
As ppl have explained above, y can only take 0.
we can as well equation to 0
ie | x -3 | = 0. Ie the distance between x and 3 is 0 on the numberline.
This can happen only if x = 3.
B
Million thanks to Ian for pushing the distance idea for absolutes, in multiple posts.
HT helps
