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Founder  Joined: 18 Apr 2015
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p + |k| > |p| + k [#permalink]
Expert's post 00:00

Question Stats: 66% (00:57) correct 33% (00:58) wrong based on 140 sessions
$$p + |k| > |p| + k$$

 Quantity A Quantity B p k

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E
[Reveal] Spoiler: OA

_________________ Manager  Joined: 06 Jun 2018
Posts: 94
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Re: p + |k| > |p| + k [#permalink]
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Carcass wrote:
$$p + |k| > |p| + k$$

 Quantity A Quantity B p k

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

Given ,

$$p + |k| > |p| + k$$

Both sides include the same variable but p + |k| is greater. Why ?

let assume some values.

p = 5 / -5

k = - 100

In this case and only in this case inequality is true.

Thus p is greater than k.

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Re: p + |k| > |p| + k [#permalink]
Can someone explain some more? GRE Instructor Joined: 10 Apr 2015
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Re: p + |k| > |p| + k [#permalink]
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Expert's post
Carcass wrote:
$$p + |k| > |p| + k$$

 Quantity A Quantity B p k

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

Warning: This is a long solution, but there are some key properties/strategies that some students may find useful

Useful properties:
#1: If x is POSITIVE, then |x| = x
#2: If x is NEGATIVE, then |x| = -x

For example, if x = 3, then |x| = |3| = 3 = x
Conversely, if x = -3, then |x| = |-3| = 3 = -(-3) = -x

Okay, first notice that p and k cannot be equal.
IF it were the case that p = k, then we can replace p with k to get: $$k + |k| > |k| + k$$
This makes no sense. So, we can be sure that p and k are not equal

Now let's examine 4 possible cases:

case i: p is POSITIVE and k is POSITIVE
Applying property #1, we get: $$p + k > p + k$$
This makes no sense. $$p + k = p + k$$
So, case i is impossible.

case ii: p is POSITIVE and k is NEGATIVE
Applying properties #1 and 2, we get: $$p + (-k) > p + k$$
Subtract p from both sides of the inequality to get: $$-k > k$$
Add k to both sides to get: $$0 > 2k$$
Since k is NEGATIVE in this case, the inequality $$0 > 2k$$ is true.
So, case ii is possible

case iii: p is NEGATIVE and k is POSITIVE
Applying properties #1 and 2, we get: $$p + k > (-p) + k$$
Subtract k from both sides of the inequality to get: $$p > -p$$
Add p to both sides to get: $$2p > 0$$
Since p is NEGATIVE in this case, the inequality $$2p > 0$$ is NOT true.
So, case iii is impossible.

case iv: p is NEGATIVE and k is NEGATIVE
Applying property #2, we get: $$p + (-k) > (-p) + k$$
Add p to both sides to get: $$2p - k > k$$
Add k to both sides to get: $$2p > 2k$$
Divide both sides by 2 to get: $$p > k$$
This tells us that, if p is NEGATIVE and k is NEGATIVE, then $$p > k$$
case iv is possible

At this point, we can see that there are only two possible cases, and for each case, we can be certain that $$p > k$$

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com  Intern Joined: 18 Apr 2020
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Re: p + |k| > |p| + k [#permalink]
1
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another way could be to square both sides of the inequality

you would be left with P|K| > |P|K.....this can only be true when P>K.....

Therefore A..

Cheers. Re: p + |k| > |p| + k   [#permalink] 25 May 2020, 09:44
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