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

Question

If P + |k| > |P| + k then which of the following is true of Inequalities ? A. P=K B. P<=k C. Pk E. Can't determined

akumar5 · Accepted Answer

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

Krage17 · Answer

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.

manpreetsingh86 · Answer

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 P2+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

mejia401 · Answer

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: + is never greater than + . If P>0 then K must be less than zero, or else the statement would be invalid: + equals + . Remember absolute value of some expression is always non-negative.

akhilbajaj · Answer

Here's how i approached the problem. Inequality_Matrix.PNG 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.

kanigmat011 · Answer

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.

sagar2911 · Answer

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

arhumsid · Answer

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

nivivacious · Answer

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!

ManifestDreamMBA · Answer

P + |k| > |P| + k
P-k>|P|-|k|

We know |P|-|k|>=0
P-k>0
P>k

ShipraSeth · Answer

Why can not be k shifted to LHS and p at RHS?

Krunaal · Answer

It can be, then you will have,  |k| - k > |P| - P  => so LHS > RHS

Both k and P cannot be positive, because it will give 0, and LHS = RHS.

If P is positive or 0, then k needs to be negative for LHS > RHS => so P > k

If both are negative, then k < P for LHS > RHS.

Thus, in all the cases P > k

avellanjc · Answer

While this approach leads you to the correct multiple choice answer, it forces you to make additional inferences that makes you discard valid answer choices.

The final inequality P|k| > k|P| only works when P > 0 and k < 0

To prove , the ineq becomes : P (-k) > k (p)
 -pk > pk
 -   > -
 + > - `(this is true. Positive greater than neg)

While P positive and q negative, will always satisfy the ineq and answer choice D ( P> k), there are many other P> k that sastify the original inequality but are not positive and neg (as your conclusion)

Eg: p=-2 and K= -5 ( P> k)
Test in original ineq 
 P + |k| > |P| + k  
 (-2) + |-5| > |-2| + (-5) 
 -2 + 5 > 2 - 5 
  3 > -3 (This is TRUE)

This is a valid solution, but disregarded in squaring both sides method.

You can only square both sides when one is certian that both sides are nonnegative.Here both sides can be negative indeed.

Regards,

Juan C. Avellan

SwethaReddyL · Answer

Bunuel , is this a right approach for this question? we don't know the sign of k, how can we bring this to the other side without affecting the inequality sign?  KarishmaB

KarishmaB · Answer

|A| - |B| is not necessarily positive. A can easily be say 2 and B can be 3 in which case it becomes negative. 
So we can't start with the logic "We know |P|-|k|>=0". Though some other thought process could have been responsible for this statement of the commentator which may not have come across in the explanation.

Solving the question,

P + |k| > |P| + k
P - |P| > k - |k|

We know that A - |A| is either 0 (in case A is 0 or positive) or A - |A| is negative if A is negative. The smaller the value of A, the smaller the value of A - |A|

Since P - |P| > k - |k|, therefore, P > k. 
k is definitely negative and P is either positive or 0 or a negative value greater than k.

Hence (D)

egmat · Answer

Hi SwethaReddyL,

Great instinct to pause here, because the worry you have is exactly the right one to have  most  of the time. The good news is that this particular move is completely safe.

The key distinction:  an inequality's direction only flips when you  multiply or divide both sides by a negative number . It never flips when you  add or subtract  something from both sides - no matter what sign that something has.

Moving k to the other side is just subtracting k from both sides:

- Start: P + |k| > |P| + k
- Subtract k from both sides: P + |k| - k > |P|
- Subtract |k| from both sides: P - k > |P| - |k|

Every step here is an addition or subtraction of the same quantity from both sides. That's like sliding both ends of a see-saw by the same amount - the balance doesn't change, so the > stays >. You never had to know whether k was positive or negative, because that only matters for multiplying/dividing.

So the prior approach is valid. From P - k > |P| - |k|, and since |P| >= |k| is  not  guaranteed, the clean finish is to note |P| - |k| can be shown to be <= P - k, landing you on P > k.

Lock the rule in with two quick ones:

- Is  5  >  3  still true after you subtract  10  from both sides? ( -5  >  -7  - yes.)
- Is  5  >  3  still true after you multiply both sides by  -2 ? ( -10  >  -6  - no, it flips.)

Same idea every time:  add/subtract is sign-safe; multiply/divide by a negative is what flips it.

Answer: D

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

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