Source: Gmat Prep, mba.com, Test II
I am looking for the most efficient (fastest) way to solve this DS problem. Thanks!
Is |x| = y - z?
(1) x + y = z
(2) x < 0
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9 postsPage 1 of 1
- GMAT 5/18
- Guest
Is A the answer?
Here is how I arrived at it...
is the difference between y and z is absolute x? ie diff between 7 and 4 or 4 and 7 in either direction, arrives at 3?
1) clues either the difference is either positive or negative.. but since the stem accepts any as long as there is an abosute value. is sufficient A
2) x <0 relationship with y and z not defined...
I could be wrong.. but let me know
Here is how I arrived at it...
is the difference between y and z is absolute x? ie diff between 7 and 4 or 4 and 7 in either direction, arrives at 3?
1) clues either the difference is either positive or negative.. but since the stem accepts any as long as there is an abosute value. is sufficient A
2) x <0 relationship with y and z not defined...
I could be wrong.. but let me know
- GMAT 5/18
Guest,
The answer is C.
Here are my steps in solving.
1. Rephrase the question:
x = y - z or x = -y + z
2. Eliminate BD (clearly insufficient)
3. Make x the subject in Statement I:
x = z - y (or -y + z)
As this matched my rephrased question, I too selected A, although incorrectly.
Can someone please explain the flaw in my method? I think I know why C is the answer, as for the rephrase question to be x = -y + z, we need x to be < 0 (which is statement II). If someone can confirm, that would be awesome.
Cheers.
The answer is C.
Here are my steps in solving.
1. Rephrase the question:
x = y - z or x = -y + z
2. Eliminate BD (clearly insufficient)
3. Make x the subject in Statement I:
x = z - y (or -y + z)
As this matched my rephrased question, I too selected A, although incorrectly.
Can someone please explain the flaw in my method? I think I know why C is the answer, as for the rephrase question to be x = -y + z, we need x to be < 0 (which is statement II). If someone can confirm, that would be awesome.
Cheers.
- Guest
DS
GMAT 5/18 -
Here's how I approached it:
Fact (1) gives you x=-y+z
Is this sufficient by itself? No, it's not. If x>0 then |x|=x and x does not equal y-z. However is x<0 then |x|=-x and x does equal y-z.
Fact (2) tells you x<0. This is clearly insufficient by itself. However, combined with (1), it tells you that |x|=-x and therefore it is true that x=y-z. So the correct answer is C.
Cheers,
Jeff
Here's how I approached it:
Fact (1) gives you x=-y+z
Is this sufficient by itself? No, it's not. If x>0 then |x|=x and x does not equal y-z. However is x<0 then |x|=-x and x does equal y-z.
Fact (2) tells you x<0. This is clearly insufficient by itself. However, combined with (1), it tells you that |x|=-x and therefore it is true that x=y-z. So the correct answer is C.
Cheers,
Jeff
- christiancryan
- Course Students
- Posts: 79
- Joined: Thu Jul 31, 2003 10:44 am
My pathway is similar to Jeff's, though I tend to try to set up explicit cases immediately as I rephrase the question (keeping it as a question) BEFORE looking at the statements. This works better for me, so that I don't get confused by the statements.
Is |x| = y - z?
This is rephrasable to a "two-case" question:
a) If x >= 0, is x = y - z?
b) If x =< 0, is -x = y - z?
(This comes from simply substituting, for |x|, a two part condition: (a) x, if x >=0 (b) -x, if x <= 0. You can do this for ANY absolute value expression: as you "drop the absolute-value fences," set up two cases.)
Now
(1) x + y = z
Rephrase to match one of my questions -- turns out to be #2. But I don't know which case I'm in, so insufficient.
(2) x < 0
Obviously insufficient.
Together, I know the case from statement 2 and the answer to the question from statement 1.
Is |x| = y - z?
This is rephrasable to a "two-case" question:
a) If x >= 0, is x = y - z?
b) If x =< 0, is -x = y - z?
(This comes from simply substituting, for |x|, a two part condition: (a) x, if x >=0 (b) -x, if x <= 0. You can do this for ANY absolute value expression: as you "drop the absolute-value fences," set up two cases.)
Now
(1) x + y = z
Rephrase to match one of my questions -- turns out to be #2. But I don't know which case I'm in, so insufficient.
(2) x < 0
Obviously insufficient.
Together, I know the case from statement 2 and the answer to the question from statement 1.
- Guest
thanks chris! i now understand :-) i have a minor question though as i am little bit confused about your "greater than or equal to" and "less than or equal to" zero. do both senarios require equal to zero??? please clarify.
christiancryan Wrote:My pathway is similar to Jeff's, though I tend to try to set up explicit cases immediately as I rephrase the question (keeping it as a question) BEFORE looking at the statements. This works better for me, so that I don't get confused by the statements.
Is |x| = y - z?
This is rephrasable to a "two-case" question:
a) If x >= 0, is x = y - z?
b) If x =< 0, is -x = y - z?
(This comes from simply substituting, for |x|, a two part condition: (a) x, if x >=0 (b) -x, if x <= 0. You can do this for ANY absolute value expression: as you "drop the absolute-value fences," set up two cases.)
Now
(1) x + y = z
Rephrase to match one of my questions -- turns out to be #2. But I don't know which case I'm in, so insufficient.
(2) x < 0
Obviously insufficient.
Together, I know the case from statement 2 and the answer to the question from statement 1.
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
- Pavan
another way to look at it
Hi,
|x| = x when x >= 0
-x when x < 0
lets assume x less than 0, then -x = y-z --> x+y = z (This is the first statement given to us, and second statement is our assumption that x < 0)
So choice is C is correct
if we try with x >0, then x = y-z ---> x+z = y-- This doesn't support either of the statement
|x| = x when x >= 0
-x when x < 0
lets assume x less than 0, then -x = y-z --> x+y = z (This is the first statement given to us, and second statement is our assumption that x < 0)
So choice is C is correct
if we try with x >0, then x = y-z ---> x+z = y-- This doesn't support either of the statement
9 posts Page 1 of 1