Can someone please explain..
What is the distance between x and y on the number line?
1) |x| - |y| = 5
(2) |x| + |y| = 11
thank you in advance
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clingymonkee
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gmatmachoman
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IMO Aclingymonkee wrote:Can someone please explain..
What is the distance between x and y on the number line?
1) |x| - |y| = 5
(2) |x| + |y| = 11
thank you in advance
st 1 helps to find the distance between the 2 parameters.
Say X =3, y= -2, takin Mod on both , we get 5.
One more to add on say X=12 & y = 7, we are just concerened about the distance between them SO A is suffice!
St 2 doesnt help much becox (x,y) can take any value say (6,5) or (0,11). the distance between them are not the same for all the sets of constants.
Pick A
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harshavardhanc
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we've to find |X-Y|.clingymonkee wrote:Can someone please explain..
What is the distance between x and y on the number line?
1) |x| - |y| = 5
(2) |x| + |y| = 11
thank you in advance
Statement 1: doesn't help. (X,Y) can be (7,2) (-7,2), (6,1),(6,-1) and infinite other possibilities.
Statement2 : doesn't help. same reasoning as above.
combo.
We can get |x|,|y|, and |x|^2 - |y|^2. Doesn't help either.
IMO E.
Regards,
Harsha
Harsha
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gmatmachoman
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harshavardhanc wrote:we've to find |X-Y|.clingymonkee wrote:Can someone please explain..
What is the distance between x and y on the number line?
1) |x| - |y| = 5
(2) |x| + |y| = 11
thank you in advance
Statement 1: doesn't help. (X,Y) can be (7,2) (-7,2), (6,1),(6,-1) and infinite other possibilities.
Statement2 : doesn't help. same reasoning as above.
combo.
We can get |x|,|y|, and |x|^2 - |y|^2. Doesn't help either.
harsha Bhai,
y shuld we find |x-Y| ??
Plz explain
IMO E.
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akhpad
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Statement 1 and 2 together
Case 1: x>0, y<0
x+y = 5
x-y = 11
magnitude of Distance = 11
Case 2: x<0, y<0
-x+y = 5
-x-y = 11
magnitude of Distance = 5
Case 3: x<0, y>0
-x-y = 5
-x+y = 11
magnitude of Distance = 11
Case 4: x>0, y>0
x-y = 5
x+y = 11
magnitude of Distance = 5
Distance is either 5 or 11
Not Sufficient
Answer: E
Case 1: x>0, y<0
x+y = 5
x-y = 11
magnitude of Distance = 11
Case 2: x<0, y<0
-x+y = 5
-x-y = 11
magnitude of Distance = 5
Case 3: x<0, y>0
-x-y = 5
-x+y = 11
magnitude of Distance = 11
Case 4: x>0, y>0
x-y = 5
x+y = 11
magnitude of Distance = 5
Distance is either 5 or 11
Not Sufficient
Answer: E
Last edited by akhpad on Wed Apr 21, 2010 8:59 am, edited 1 time in total.
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sk818020
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If you can figure out if x or y is positive or negative, then you could solve this. I got:
(1) |x| - |y| = 5
The only thing that this tells you is that x's distance from zero is 5 units greater than y's, or;
lxl = 5 + lyl
This doesn't provide any clues as to the positivity or negativity of x or y, thus, insufficient.
(2) |x| + |y| = 11
Again, this only tells your that when you add the x's and y's distance from zero that you will be 11 units away from zero.
This does not provide any clues about whether x or y is positive.
If you combine the statements you get:
From (1), lxl = 5 + lyl, and from (2), lxl + lyl = 11, thus,
5 + lyl + lyl = 11, and thus,
2lyl = 6
lyl = 3
y= +or-3, and thus,
lxl - 3 = 5
lxl = 8
x = +or-8
There are four possible solutions. Therefore, the answer must be E.
(1) |x| - |y| = 5
The only thing that this tells you is that x's distance from zero is 5 units greater than y's, or;
lxl = 5 + lyl
This doesn't provide any clues as to the positivity or negativity of x or y, thus, insufficient.
(2) |x| + |y| = 11
Again, this only tells your that when you add the x's and y's distance from zero that you will be 11 units away from zero.
This does not provide any clues about whether x or y is positive.
If you combine the statements you get:
From (1), lxl = 5 + lyl, and from (2), lxl + lyl = 11, thus,
5 + lyl + lyl = 11, and thus,
2lyl = 6
lyl = 3
y= +or-3, and thus,
lxl - 3 = 5
lxl = 8
x = +or-8
There are four possible solutions. Therefore, the answer must be E.
