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If p,s, and t are positive integer, is |ps-pt| > p(s-t)

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

need help to solve this one.

Do not forget to mention the source of the question.

it is simple after you realise that as p is positive, you can rewrite the given inequality as |ps-pt| > p(s-t) = p|s-t| > p(s-t) , you can also remove 'p' from both the sides , leaving you with

is |s-t| > s-t which is very clearly obtainable if you know about the relative relation between t and s (side note, not required for this question: for |s-t| > s-t to be true, this means that s-t<0 or s<t)

Statement 1, gives no information about relation between t and s , hence not sufficient.

Statement 2, gives the exact relation we need to answer the question and as such is sufficient.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If p,s, and t are positive integer, is |ps-pt| > p(s-t)?

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

When you modify the condition and the question, |ps-pt| > p(s-t)?--> ps-pt<0?, therefore p(s-t)<0?. Since p>0, you need to find out s-t<0?. In 2), s-t<0, which is yes and sufficient. Therefore, the answer is B.

-> Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
_________________

If p,s, and t are positive integer, is |ps-pt| > p(s-t)

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

need help to solve this one.

Hi, the eq is |ps-pt| > p(s-t).. or pls-tl > p(s-t) since all numbers are positive , p cannot be 0.. two instances.. 1) ls-tl will never be smaller than s-t as ls-tl will always be positive while s-t will depend on relative values of s and t.. 2) if s and t are equal both will be 0..

so our answer for |ps-pt| > p(s-t) will be NO if s>t or s=t.. and it will be YES if s<t... so we are just looking for the relation between s and t

lets see the statements.. (1) p < s... nothing between s and t.. insuff

(2) s < t answer is YES.. suff ans B
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Re: If p,s, and t are positive integers, is |ps-pt| > p(s-t) [#permalink]

Show Tags

06 Oct 2016, 10:19

Engr2012 wrote:

ajayvyavahare wrote:

If p,s, and t are positive integer, is |ps-pt| > p(s-t)

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

need help to solve this one.

Do not forget to mention the source of the question.

it is simple after you realise that as p is positive, you can rewrite the given inequality as |ps-pt| > p(s-t) = p|s-t| > p(s-t) , you can also remove 'p' from both the sides , leaving you with

is |s-t| > s-t which is very clearly obtainable if you know about the relative relation between t and s (side note, not required for this question: for |s-t| > s-t to be true, this means that s-t<0 or s<t)

Statement 1, gives no information about relation between t and s , hence not sufficient.

Statement 2, gives the exact relation we need to answer the question and as such is sufficient.

B is the correct answer.

Hope this helps.

I did exactly what you did in this question but picked E. I was thinking |x| is x or -x, so |s - t| can be equal to s-t or -(s-t). Plugging in some numbers for s and t: 3 - 5= -2 and -3 + 5 = 2, so |s-t| could be equal to 2 or -2 which would be insufficient. Could you help me to understand why this is wrong? Tks

If p,s, and t are positive integers, is |ps-pt| > p(s-t) [#permalink]

Show Tags

04 Jan 2017, 13:12

AJ1012 wrote:

If p,s, and t are positive integer, is |ps-pt| > p(s-t)

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

need help to solve this one.

Excellent question, this one looks a lot tricker than it actually is. Took me for a ride until I saw the "only positive integers". So let's look at |ps - pt| > P(s-t) or re-write it as | p(s-t)| > p (s-t).

Now let's consider a possibility where P(s-t) is greater than | p(s-t)| , think about this one. I can't come up with one. So there's no way | p(s-t | < p (s-t). We know that |p(s-t)| is always positive. So if p (s-t) is negative then, lets call P (s-t) x, so |x| > x is equal when both positive and |x| < x when x is negative correct?

Since all of them are positive, to make P (s-t) negative, T has to be bigger than S. That's the only option. No other way to make it negative since they're all positive. T > s is necessary. 1) insufficient. 2) sufficient !

Here's a easy way: re-write : |p (s-t)| > p(s-t) let call inside X. |x| > x only if X is negative. so if all are positive, (s-t) has to be negative. for that T has to be bigger than S. Bam!
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Re: If p,s, and t are positive integers, is |ps-pt| > p(s-t) [#permalink]

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06 Jan 2017, 09:30

in the problem we can take p as common as: |p(s-t)| > p(s-t) so now we need to look at the values of s and t and which tells only statement 2 so its suff. to get a exact yes. hence answer is B

Re: If p,s, and t are positive integers, is |ps-pt| > p(s-t) [#permalink]

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14 Aug 2017, 00:51

1

This post received KUDOS

hi all, I still am unable to derive the relationship between the |s-t| > s-t. Can anyone please share an easier way to crack this. Thanks.
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Re: If p,s, and t are positive integers, is |ps-pt| > p(s-t) [#permalink]

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22 Sep 2017, 17:38

AnubhavK wrote:

hi all, I still am unable to derive the relationship between the |s-t| > s-t. Can anyone please share an easier way to crack this. Thanks.

Modulus is always positive or 0.

Hence |s-t| will always be positive or 0. In this problem it can be zero since s can be equal to t (there is nothing that says that these numbers are unique)

On the other hand, s-t will be positive if s>t and negative otherwise.

Statement 2 clearly tells us that s<t. This means that the RHS will be negative. Since LHS, i.e Modulus of any value is always positive, we can answer this question using statement B alone.
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