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If |p−6|+p=6, which of the following must be true?

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If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 06:32
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If |p−6|+p=6, which of the following must be true?

A. p=0
B. p=-6
C. p=6
D. p>-6
E. p<=6


Bunuel-Can you please explain! and Can we write |p−6| as (p-6) and -(p-6)( I mean when modulus is withdrawn) ?
[Reveal] Spoiler: OA

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Last edited by bagdbmba on 26 Jun 2013, 06:51, edited 1 time in total.
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 06:43
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debayan222 wrote:
If |p−6|+p=6, which of the following must be true?

A. p=0
B. p=-6
C. p=6
D. p>-6
E. p<=6


Bunuel-Can you please explain! and Can we write |p−6|= (p-6) and -(p-6)?


|p-6|+p=6 --> |p-6|=6-p --> |p-6|=-(p-6).

|x|=-x means, that x\leq{0}. So, p-6\leq{0} --> p\leq{6}.

Answer: E.
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 06:57
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Bunuel-Can we write |p−6| as (p-6) and -(p-6)( I mean when modulus is withdrawn) ?
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 07:04
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 07:31
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Bunuel wrote:
debayan222 wrote:
Bunuel-Can we write |p−6| as (p-6) and -(p-6)( I mean when modulus is withdrawn) ?


When p\leq{6}, then |p-6|=-(p-6)=6-p.
When p\geq{6}, then |p-6|=p-6.


So, only 6 satisfies both |p−6|= (p-6) and |p−6|=-(p-6)

Then why the answer is not ONLY 6 for the above qs?
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 07:34
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debayan222 wrote:
Bunuel wrote:
debayan222 wrote:
Bunuel-Can we write |p−6| as (p-6) and -(p-6)( I mean when modulus is withdrawn) ?


When p\leq{6}, then |p-6|=-(p-6)=6-p.
When p\geq{6}, then |p-6|=p-6.


So, only 6 satisfies both |p−6|= (p-6) and -(p-6)

Then why the answer is not 6 for the above qs?


These are not simultaneous equations.

If...
If...

Since given that |p-6|=-(p-6)=6-p, then p\leq{6}.
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 07:43
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Ah! got it... :) so the question itself already limits the scope of modulus.

Is it below 700 qs?

BTW,don't get the "If....If...." part in your above reply ! :o
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 26 Jun 2013, 07:47
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debayan222 wrote:
Ah! got it... :) so the question itself already limits the scope of modulus.

Is it below 700 qs?

BTW,don't get the highlighted part in your above reply ! :o


I think it's a sub-600 question, because it's about basic property (definition) of absolute value:

When x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)}.

When x\geq{0} then |x|=x, or more generally when some \ expression\geq{0} then |some \ expression|={some \ expression}.

P.S. What highlighted part?
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 30 Jun 2013, 17:08
If |p−6|+p=6, which of the following must be true?

There are two ways to solve, one of which is to just pick numbers.

The other way:
|p-6|+p=6
|p-6| = 6-p
|p-6| = -(p-6)
|x| = -x (i.e. the value inside the absolute value bars is negative)
So, (p-6) is negative when p≤6

(E)


A. p=0
B. p=-6
C. p=6
D. p>-6
E. p<=6
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 01 Jul 2013, 13:39
without plugging in numbers, how would you know that this |p-6| = -(p-6) would be p<=6 and not just p<6?
the way i'm solving it, this |x| = -x condition would be met when (p-6)<0 and therefore -(p-6)<0 yields p<6. but why also p=6?

thanks!
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 01 Jul 2013, 13:42
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nancerella wrote:
without plugging in numbers, how would you know that this |p-6| = -(p-6) would be p<=6 and not just p<6?
the way i'm solving it, this |x| = -x condition would be met when (p-6)<0 and therefore -(p-6)<0 yields p<6. but why also p=6?

thanks!


Because when x\leq{0} then |x|=-x, or more generally when some \ expression\leq{0} then |some \ expression|={-(some \ expression)}.

We have |p-6|=-(p-6), thus p-6\leq{0} --> p\leq{6}.
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 09 Jul 2013, 22:07
Why can't we (or how would we know not to) solve in the following manor:

If |p−6|+p=6, which of the following must be true?

|p−6|+p=6
p≥6
p-6+p=6
p=6

p<6
-(p-6)+p=6
-p+6+p=6
6=6

Thanks!
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 09 Jul 2013, 22:13
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WholeLottaLove wrote:
Why can't we (or how would we know not to) solve in the following manor:

If |p−6|+p=6, which of the following must be true?

|p−6|+p=6
p≥6
p-6+p=6
p=6

p<6
-(p-6)+p=6
-p+6+p=6
6=6

Thanks!


Actually we can. You got the correct result.

When p<6, you got that the equation holds for any value of p.
When p≥6, you got that the equation holds for p=6.

So, the equation holds for p<6 and p=6 --> p<=6.

Hope it's clear.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 10 Jul 2013, 08:27
Bunuel wrote:

So, the equation holds for p<6 and p=6 --> p<=6.

Hope it's clear.


because for p<6 and p≥6 the intersection is p ≤6, correct?
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post 10 Jul 2013, 10:43
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Re: If |p−6|+p=6, which of the following must be true?   [#permalink] 10 Jul 2013, 10:43
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