giatch wrote:|x+k| < |y+k|

1) k > 0
2) |x| < |y|

I think the answer is C.

What do you guys think?

giatch,

Tell me what changes if k < 0 ?

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?

wait..i didnt take into account whether x+k is positive or negative?

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

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.

hrm..
okay..using statement 2..

a)x<y
if k>0, then y+k> x+k

if k<0...we know that y is greater than x, but not by how much, but since a constant negative value of k is being added to both sides, then y+k is still > x+k

b)x>-y ...
actually, im not sure what to do here..

vittalgmat 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.

Vital you kind of ruined the whole thing I want giatch to see that he does not need the Statement 1.

But you gave an excellent explanation.

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.

Sorry Logitech!!!
I should not have interrupted your teaching.. !!!.
I liked your thought provoking questions... kinda solidifies shaky concepts.

Happy new year to you.

thank you both!

happy new yr!

If u r hungry for some more 'absolute' ammunition, check this out.

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.

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.

You are absolutely right.

|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

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.