giatch,giatch wrote:|x+k| < |y+k|
1) k > 0
2) |x| < |y|
I think the answer is C.
What do you guys think?
Tell me what changes if k < 0 ?
Inequality
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Source: Beat The GMAT — Data Sufficiency |
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logitech
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Tell me what changes if k < 0 ?
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
This is how I approached the question. I'm not sure if i approached it correctly. I know that when you have an inequality with an absolute value, you must have two scenarios (+ve or -ve). Im not sure what you do with absolute values on both sides.
|x+k|<|y+k|
(1)k> 0
(2) |x|<|y|
a) (y+k) is +ve
x+k < y + k
x < y + k -k
x < y
b) (y+k) is -ve
x+k > -y - k
x > - y - 2k
(1) k>0
I dont think that tells much since dont know whether (y+k) is positive or negative (b/c the value of y in unknown).. so -> insufficient
from (2)
|x|<|y|
y is +ve
x< y
y is -ve
x> -y
but this mentions nothing about the value of k, so ->insufficient
Then, taking both together, i got x<y with k>0...since x is less than y and k is greater than 0...y+k will always be larger than x..
What am I doing wrong?
|x+k|<|y+k|
(1)k> 0
(2) |x|<|y|
a) (y+k) is +ve
x+k < y + k
x < y + k -k
x < y
b) (y+k) is -ve
x+k > -y - k
x > - y - 2k
(1) k>0
I dont think that tells much since dont know whether (y+k) is positive or negative (b/c the value of y in unknown).. so -> insufficient
from (2)
|x|<|y|
y is +ve
x< y
y is -ve
x> -y
but this mentions nothing about the value of k, so ->insufficient
Then, taking both together, i got x<y with k>0...since x is less than y and k is greater than 0...y+k will always be larger than x..
What am I doing wrong?
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logitech
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Okay I like the way you think that you should analyze the question for + and - 's but remember
this is actually X and Y question and whatever that k is, it is being added to both sides
And since they are in absolute values you dont really need the signs of X and Y but more like which one has bigger ABSOLUTE values.
So why dont you work on statement 2 with two scenarios:
1) k> 0
2) k< 0
this is actually X and Y question and whatever that k is, it is being added to both sides
And since they are in absolute values you dont really need the signs of X and Y but more like which one has bigger ABSOLUTE values.
So why dont you work on statement 2 with two scenarios:
1) k> 0
2) k< 0
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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vittalgmat
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IMO it is B.
The Q can be simplified by removing K.
|x +k| < |y+k| ?
-> k +|x| < k + |y|?
-> |x| < |y| ?
Now the statement is independent of k,
Stmt1 is useless. So insufficient.
Stmt 2 tells us exactly what we want to see
|x| < |y|. hence sufficient.
So B.
Conceptually, |x+k| and |y+k| can be picturized as shifting the origin.
|x+k| and |y+k| are both k units away from the origin 0, and the distance between them is |x+k -(y+k)| = |x -y|. When u remove |k| units from both, essentially , u r moving x and y from origin= k to origin =0. The absolute relative distance between x and y is still unchanged.
HT Helps.
The Q can be simplified by removing K.
|x +k| < |y+k| ?
-> k +|x| < k + |y|?
-> |x| < |y| ?
Now the statement is independent of k,
Stmt1 is useless. So insufficient.
Stmt 2 tells us exactly what we want to see
|x| < |y|. hence sufficient.
So B.
Conceptually, |x+k| and |y+k| can be picturized as shifting the origin.
|x+k| and |y+k| are both k units away from the origin 0, and the distance between them is |x+k -(y+k)| = |x -y|. When u remove |k| units from both, essentially , u r moving x and y from origin= k to origin =0. The absolute relative distance between x and y is still unchanged.
HT Helps.
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logitech
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Vital you kind of ruined the whole thingvittalgmat wrote:IMO it is B.
The Q can be simplified by removing K.
|x +k| < |y+k| ?
-> k +|x| < k + |y|?
-> |x| < |y| ?
Now the statement is independent of k,
Stmt1 is useless. So insufficient.
Stmt 2 tells us exactly what we want to see
|x| < |y|. hence sufficient.
So B.
Conceptually, |x+k| and |y+k| can be picturized as shifting the origin.
|x+k| and |y+k| are both k units away from the origin 0, and the distance between them is |x+k -(y+k)| = |x -y|. When u remove |k| units from both, essentially , u r moving x and y from origin= k to origin =0. The absolute relative distance between x and y is still unchanged.
HT Helps.
I want giatch to see that he does not need the Statement 1. But you gave an excellent explanation.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
haha..actually he did ruin it....i understand now...
i didnt think to approach the question like that...what confused me was the absolute value on both sides...when i was going through the OG, there werent any questions that covered a inequality like that...i came across this browsing through the net, so i was curious to how to tackle it.
i didnt think to approach the question like that...what confused me was the absolute value on both sides...when i was going through the OG, there werent any questions that covered a inequality like that...i came across this browsing through the net, so i was curious to how to tackle it.
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vittalgmat
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Sorry Logitech!!! 
I should not have interrupted your teaching.. !!!.
I liked your thought provoking questions... kinda solidifies shaky concepts.
Happy new year to you.
I should not have interrupted your teaching.. !!!.
I liked your thought provoking questions... kinda solidifies shaky concepts.
Happy new year to you.
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vittalgmat
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If u r hungry for some more 'absolute' ammunition, check this out.
https://en.wikipedia.org/wiki/Absolute_value
https://en.wikipedia.org/wiki/Absolute_value
Hi,
How can we interpret "|x + k| < |y + k|" as "(k + |x|) < (k + |y|)"?
Say, x = -3, y = -1, and k = 5. |x + k| => |-3 + 5| = 2, and |y + k| => |-1 + 5| = 4.
2 < 4, thus |x + k| < |y + k|.
Let's see whether the interpretation works,
5 + |-3| < 5 + |-1| => 8 < 6. No.
Please help.
Regards.
How can we interpret "|x + k| < |y + k|" as "(k + |x|) < (k + |y|)"?
Say, x = -3, y = -1, and k = 5. |x + k| => |-3 + 5| = 2, and |y + k| => |-1 + 5| = 4.
2 < 4, thus |x + k| < |y + k|.
Let's see whether the interpretation works,
5 + |-3| < 5 + |-1| => 8 < 6. No.
Please help.
Regards.
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logitech
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You are absolutely right.ifairo wrote:Hi,
How can we interpret "|x + k| < |y + k|" as "(k + |x|) < (k + |y|)"?
Say, x = -3, y = -1, and k = 5. |x + k| => |-3 + 5| = 2, and |y + k| => |-1 + 5| = 4.
2 < 4, thus |x + k| < |y + k|.
Let's see whether the interpretation works,
5 + |-3| < 5 + |-1| => 8 < 6. No.
Please help.
Regards.
|x+k| < |y+k|
2) |x| < |y| can not prove that |x+k| < |y+k|
On the other hand 1) k > 0 is meaningless too since we dont know what X and Y are.
Even combining the both statements:
|+4 | < | -5 |
and k =1
will give us NO
and
|+4 | < | +5 |
and k = 1
will give is YES
so here comes the answer: E
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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vittalgmat
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OOOOOOPsy daisy!! .
I messed it up!!! Dint I???
Afairo, U r absolutely right.
I stand corrected.
|a + b| <= |a| + |b|
My earlier theory works only if all of them are +ve, which implies that absolute is not needed at all in that case.
The answer is E.
I messed it up!!! Dint I???
Afairo, U r absolutely right.
I stand corrected.
|a + b| <= |a| + |b|
My earlier theory works only if all of them are +ve, which implies that absolute is not needed at all in that case.
The answer is E.
