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If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  09 Sep 2020, 07:37
Genoa2000 wrote:
Hanzel101 wrote:
Hi all!

Can somebody please help me in understanding why this approach is incorrect?

|ps - pt| > p(s - t)

^^ I decided to solve this further in the Q stem to get a clearer picture. I used |x| = root x^2

and converted L.H.S into root form, squared on both sides but all the variables cancel out.

What am I doing wrong?

VeritasKarishma Bunuel chetan2u egmat


You are dealing with an inequality! You cannot cancel stuff out as if you were solving an equation


Although it is true that one must refrain from cancelling variables, especially when dealing with inequalities, in this very question, we can go on and cancel 'p' on b.s. because p can neither be 0 nor be -ve as the question itself says that p is a positive integer.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  10 Oct 2020, 14:44
BrentGmatPrepNow wrote " if x = 5 and y = 2, then we get: |5 - 2| = 5 - 2. In this case |x - y| = x - y
CONVERSELY, if x = 4 and y = 6, then we get: |4 - 6| = 4 - 6. In this case |x - y| ≠ x - y"

My question is if x = 5 and y = 2, then we get: |5 - 2| = 5 - 2. In this case |x - y| = x - y
Then why not if x = 4 and y = 6, then we get: |4 - 6| = 4 - 6. In this case |x - y| = x - y ? since |x - y| can be thought as the DISTANCE between x and y on the number line?
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  09 Nov 2020, 09:54
If p, s, and t are positive integer, is \(|ps - pt| > p(s - t) \)?

If p, s, and t are positive, we can re-write the question as

\(p|s-t| > p(s-t)\) ?
\(|s-t| > s-t?\)

This is true if s < t. Lets look at the statements:

(1) p < s

Tells us nothing about s relative to t. INSUFFICIENT.

(2) s < t

Gives us directly what we're looking for. SUFFICIENT.

Answer is B.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  31 Dec 2020, 14:44
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AJ1012 wrote:
If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?

(1) p < s
(2) s < t


|ps - pt| > p(s - t)

=|p(s - t)| > p(s - t); If we know s and t then we can figure out whether |p(s - t)| > p(s - t)? Because LHS is always positive

(1) t is not given. Insufficient.

(2) \(s-t<0\); Sufficient.

The answer is B
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If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  10 Aug 2021, 02:50
Please see below for the full solution. The problem simplifies after opening up the modulus sign and a little bit of reasoning.



Apologies, had linked to incorrect video initially.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  18 Aug 2021, 07:23
p, s, and t are positive integer
|ps - pt| > p(s - t)
|s - t| > (s - t)
Now we just need relation between s and t.

(1) p < s . Insufficient
No relation between s and t

(2) s < t . Sufficient
Relation t is greater than s.

Hence, OA is (B).
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If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  26 Aug 2021, 06:53
chetan2u sir

If it were not given that p,s, and t are +ve intergers, but only that they are some number not equal to zero. Is the below approach valid?

Q: If p, s, and t are integers, is |ps - pt| > p(s - t)?

We can re arrange the ineq. as:

Case: 1
ps - pt > ps - pt
0 > 0
It is not possible, hence we can discard this case.

Case: 2
ps - pt > -(ps - pt)
ps - pt > pt - ps
2ps - 2pt > 0
ps - pt > 0
p(s-t)>0

In this case we will check whether both, p and (s-t), are either +ve or -ve.
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If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  26 Aug 2021, 20:07
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rocky620 wrote:
chetan2u sir

If it were not given that p,s, and t are +ve intergers, but only that they are some number not equal to zero. Is the below approach valid?

Q: If p, s, and t are integers, is |ps - pt| > p(s - t)?

We can re arrange the ineq. as:

Case: 1
ps - pt > ps - pt
0 > 0
It is not possible, hence we can discard this case.

Case: 2
ps - pt > -(ps - pt)
ps - pt > pt - ps
2ps - 2pt > 0
ps - pt > 0
p(s-t)>0

In this case we will check whether both, p and (s-t), are either +ve or -ve.


You are correct in the approach but there is an error in the red portion.

You have to take negative of the mod.
So -(ps-pt)>ps-pt
p(t-s)>0

You can also look at it in the following manner
1) If ps>pt
Both sides will be positive and equal…….Answer will be NO
2) If pt >ps
|ps-pt|>0, but ps-pt <0…. Answer will be YES
3) If ps=pt
0=0…. Answer will be NO

So if we can find the relation between ps and Pt, we will get our answer.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  27 Aug 2021, 05:40
AJ1012 wrote:
If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?


Since p is positive, the given inequality becomes |ps-pt| > p(s-t) = p|s-t| > p(s-t) , p can be cancelled

is |s-t| > s-t this is the reduced equation

1) p < s
Clearly insufficient since there are no relationship that we can determine through S and t

(2) s < t[
This definitely provides the equation between s and t

Clearly sufficient therefore IMO B
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink] New post  29 Sep 2021, 00:43
Given P,S,T >0

|P(S-T)| >P(S-T)

Let assume P(S-T) =X

Then Question is asking |x| > x ? it is only possible when X<0

PS-PT<0
PS<PT (Given P>0)

S>T?

Statement 2 IS Sufficient on its own ?
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ? [#permalink]
29 Sep 2021, 00:43
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