If P + |k| > |P| + k then which of the following is true of Inqual

given that P + |k| > |P| + k.................1

lets square both sides : p^2+k^2+2p|k| > p^2+k^2+2k|p|

so, p|k|>k|p| ;
Clearly, the above statement is valid only if p>k

Answer is D

I am neither of two you asked to help, but I am retaking the GMAT soon, so here is how I approached this problem:

Assuming that P and k and both sides have the absolute value, the only way we will have an inequality is when when one or more numbers have negative signs, and thus change their value when an absolute value is taken. We already know that P on the left side equals |P| on the right side because the left side of the inequality is larger (and therefore, there is no way P is negative). So, what truly shifts the equation here is the sign flip when we take an absolute value of k. In other words, k must be negative, otherwise |k| would be equal to k and P+|k| would be equal to |P|+k. Hence, P (which is positive) must be larger than k (which is negative).
I know it's somewhat conceptual and may not be easy to follow, but that's how I looked at it.

let's assume that p=k and k>0

then P + |k| > |P| + k becomes

k+k>k+k , 2k>2k, which is false. hence this inequality doesn't hold good.

lets try p=k, and k<0

-k+k>k-k or 0>0 which is false. hence option a is out.

similarly we can reject option b, because of the equality sign

let's try option C

P<k

let's pick numbers, say p=2 and k=8
then we have

2+8>2+8
which doesn't hold valid.

lets say p=-1 and k=1

then -1+1>1+1
0>2 , which is false.

say p=-2 and k=-1

then, -2+1> 2-1
-1>1, which is again false. hence option c is out.

let's try option D

P>k

p=3 and k=-1
3+1>3-1
4>2, bingo.

hence answer is option D

There are four potential scenarios.
1. P<0 and K<0
2. P<0 and K>0
3. P>0 and K<0
4. P>0 and K >0

Since |P| + K is less than zero, then P must be greater than zero, or else the statement would be invalid: [negative] + [positive] is never greater than [positive] + [positive]. If P>0 then K must be less than zero, or else the statement would be invalid: [positive] + [positive] equals [positive] + [positive].

Remember absolute value of some expression is always non-negative.

Here's how i approached the problem.

For case 3 and 4, the inequality given in the problem is true and in both the cases, P>K.
Answer D.

Please note: the question clearly states that, "If P + |k| > |P| + k is true, which of the following inequalities is true" and not the other way round. So you should not expect all the values that satisfy the right answer choice to satisfy the given statement as well.

Hope it helps.

HI,

p + |K| > |p|+ K

Can be written as
p-|p|>k -|K|

now think of this logically
This can be possible onlly if both p and k are negative
because if they will be postive
p-|p| will be zero , same the case with k-|k|

so p and k are negative
and now for them to hold this inequality
P>K
Try plugging in some numbers and you will get this.

This is how I solved:
P + |k| > |P| + k
=> P - |P| > k - |k|
=> Now, if P & K is positive, then |P| = P & |k| = k => P -P > k - k => 0 > 0 => Invalid scenario
This can only be possible only if both P & k are negative
=> P - (-P) > k - (-k)
=> 2P > 2k
=> P > k

Option D

This approach is interesting, and i think the most efficient way to approach this problem. But your assumption here is wrong my friend-

Here's how P can be negative:
p = -1, k = -2
p + |k| > |p| + k
-1 + |-2| > |-1| + (-2)
1 > -1

This is true for any negative value of p when k<p.

So we cannot conclude that p is non-negative. However, what we can certainly conclude from the given inequality is- k has to be negitive

|Any number| \(\geq\) That number

Now, the given inequality:
p + |k| > |p| + k
The |k| on LHS can be greater than or equal to k on RHS

Case 1:
|k| = k, which means that k is positive
in this case, to hold the ineq. true p > |p|. But this cant be true. Hence K is negative.

Case 2:
|k| > k, which means that k is negative
in this case, to hold the ineq. true, p on the LHS can be equal to (p is positive) or less than (p is negative) |p| on RHS.

So we can safely conclude from the given ineq. that
p can be +ve or -ve, but
k will always be -ve

I got confused over this a lot, and so this is how I approached it. Placing it here so it can help someone:

Given : p + |k| > |p| + k

1) LHS: Assume p + |k| > 0
This can be true in three ways:
a) |k| is bigger but k is negative. p is smaller but is positive (if you want to plug numbers: p=3, k=-4)
b) |k| is bigger but k is negative. p is smaller and is negative (if you want to plug numbers: p=-3, |k| =4)
c) p is bigger and positive. k is smaller but |k| is smaller and positive (if you want to plug numbers: p=4, |k| = 3)

2) RHS: Assume |p| + k < 0
This can be true only in two ways
a) p is positive but smaller than k which is negative and bigger (if you want to plug numbers: p=3, k=-4)
b) p is negative but |p| is positive and bigger but k is smaller and negative (if you want to plug numbers: p = -4, k=-3)

Now that we have understood scenarios, lets stay true to statement p + |k| > |p| + k

1a) 3+4>3-4 (agrees)
1b) -3+4>3-4 (does not agree)
1c) 4+3>4-3 (agrees)

2 a) 3+4>3-4 (agrees)
2 b) -4+3>4+3 (does not agree)

only option that must be true is p>k

Hope this helps!