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

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bagdbmba
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If |p−6|+p=6, which of the following must be true? [#permalink] New post   Updated on: 26 Jun 2013, 07:51
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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 and Can we write |p−6| as (p-6) and -(p-6)( I mean when modulus is withdrawn) ?
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E

Originally posted by bagdbmba on 26 Jun 2013, 07:32.
Last edited by bagdbmba on 26 Jun 2013, 07:51, 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, 07:43
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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, 08:04
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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Bunuel-Can we write |p−6| as (p-6) and -(p-6)( I mean when modulus is withdrawn) ?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 \(|p-6|=p-6\).
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post   13 Aug 2014, 11:16
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or simply |p−6|+p=6
|p−6|=6-p
means 6-p >=0 therefore 6>=p or p<= 6
hope it helps
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post   30 Jun 2013, 18:08
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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   09 Jul 2013, 23:13
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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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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post   10 Jul 2013, 11:43
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WholeLottaLove wrote:
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?


No. Because p<6 and p=6 means p<=6.
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post   26 Jun 2013, 07:57
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, 08:31
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, 08:34
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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, 08:43
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, 08:47
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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   01 Jul 2013, 14:39
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, 14:42
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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, 23:07
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   10 Jul 2013, 09:27
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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If |p−6|+p=6, which of the following must be true? [#permalink] New post   13 Jun 2015, 12:34
Bunuel wrote:
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.


Dear Bunuel,

if you substitute p=0 in the given equation then also you get LHS=RHS.

am i missing out anything here?
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post   13 Jun 2015, 12:45
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arshu27 wrote:
Bunuel wrote:
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.


Dear Bunuel,

if you substitute p=0 in the given equation then also you get LHS=RHS.

am i missing out anything here?


The question is "which of the following must be true", NOT "which of the following could be true".

Check other Must or Could be True Questions HERE.
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post   13 Jun 2015, 13:19
Hi Bunuel,

if we substitute p=0 then the answer must be 6. There can be no other answer right?
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Re: If |p−6|+p=6, which of the following must be true? [#permalink] New post   14 Jun 2015, 11:30
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arshu27 wrote:
Hi Bunuel,

if we substitute p=0 then the answer must be 6. There can be no other answer right?


Not sure I understand what you mean...

p = 0, as well as any other value of p less than or equal 6 satisfies |p − 6| + p = 6.
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Re: If |p−6|+p=6, which of the following must be true? [#permalink]
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