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

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

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New post 18 Dec 2015, 13:11
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If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?

(1) p < s
(2) s < t
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 19 Dec 2015, 06:39
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.


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 integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 18 Dec 2015, 13:18
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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.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 19 Dec 2015, 05:43
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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.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?  [#permalink]

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

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New post 04 Jan 2017, 13:12
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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

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 integer, is |ps - pt| > p(s - t) ?  [#permalink]

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

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New post 14 Aug 2017, 00:51
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 integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 22 Sep 2017, 05:21
If p,s, and t are positive integer, is |ps-pt| > p(s-t)
=> Find is |p(s-t)| > p(s-t)

As we know |x| > x only when x is -ve . When x is positive |x| =x

So here we have to check if product p*(s-t) is -ve or not. => either p is -ve or (s-t) is -ve. If both -ve, than product will become +ve => |x|=x

given in question stem that all variables are +ve. So we have to find if (s-t) is -ve or not. => we have to find is s<t or not.

(1) p < s
This just tell value of p is less than s . Not sufficient.

(2) s < t
This directly gives us equation that we are looking for.
=> s<t => (s-t) is -ve => p(s-t) is -ve => |p(s-t)| > p(s-t)
Sufficient

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

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

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New post 15 Mar 2018, 14:41
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.



so we have :

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

Now for L.H.S to be greater than R.H.S, 'p' and '(s-t)' should be of opposite signs (as L.H.S is in mod and will eventually become +ve because of mod)
Thus, for 'p' and '(s-t)' to be of opposite signs , 'p' needs to be +ve and (s-t) needs to be -ve.(p can't be negative as the question says p,s and t are +ve integers)
Now:
from 1) p<s----->we get nothing...thus insufficient.

from 2) s<t ===> (s-t)<0 . (as s,t are postive integers)
thus sufficient .
Answer B.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 04 Apr 2018, 08:23
ENGRTOMBA2018 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.


Though this is a old post i think there is an error here |ps-pt| can't be written as p|s-t| (as i have highlighted above) but instead |ps-pt| = l P(s-t) l
members plz correct me if i am wrong here
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 04 Apr 2018, 09:12
cruiseav wrote:
ENGRTOMBA2018 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.


Though this is a old post i think there is an error here |ps-pt| can't be written as p|s-t| (as i have highlighted above) but instead |ps-pt| = l P(s-t) l
members plz correct me if i am wrong here


Hi cruiseav

Note, here p,s & t are positive. so p can be moved out of modulus function and we can divide both sides of the inequality by p as this will not impact the inequality.
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 22 Oct 2018, 11:58
AJ1012 wrote:
If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?

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


Given that all are positive integers.

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

What will provide us an answer is the relationship between s and t

From statement 1)

Knowing anything about p will not help us.

From statement 2)

Knowing that t > s

will give us a clear answer p|(-)| > p (-)

Since the p is positive it can be cancelled without changing the inequality direction

+ve > -ve

Answer choice B
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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?  [#permalink]

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New post 21 Jan 2019, 11:28
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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


Target question: Is |ps - pt| > p(s - t) ?
In other words, Is |ps - pt| > ps - pt?
This is a good candidate for rephrasing the target question.

KEY CONCEPT: |x - y| can be thought as the DISTANCE between x and y on the number line.
For example, |3 - 10| = the DISTANCE between 3 and 10 on the number line.
And |6 - 1| = the DISTANCE between 6 and 1 on the number line.

IMPORTANT: We can also find the distance 6 and 1 on the number line by simply subtracting 6 - 1 to get 5, so why do we need absolute values? Can't we just conclude that |x - y| = x - y?
Great questions, me!
For SOME values of x and y, it's true that |x - y| = x - y, and for other values it is NOT the case that |x - y| = x - y
For example, if x = 5 and y = 2, then we get: |5 - 2| = 5 - 2. In this case |x - y| = x - y
Likewise, if x = 11 and y = 3, then we get: |11 - 3| = 11 - 3. In this case |x - y| = x - y
And, if x = 7 and y = 7, then we get: |7 - 7| = 7 - 7. 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
Likewise, if x = 5 and y = 20, then we get: |5 - 20| = 5 - 20. In this case |x - y| x - y
And, if x = 0 and y = 1, then we get: |0 - 1| = 0 - 1. In this case |x - y| x - y

Notice the |x - y| = x - y IS true when x > y, and |x - y| = x - y is NOT true when x < y

If x < y, then |x - y| = some POSITIVE value, and x - y = some NEGATIVE value.
This means that, if x < y, then |x - y| > x - y
The target question asks Is |ps - pt| > ps - pt?

According to our conclusion above, if ps > pt, then |ps - pt| = ps - pt and . . .
if ps < pt, then |ps - pt| > ps - pt

This means we can REPHRASE the target question....
REPHRASED target question: Is ps < pt?

We can make things even easier, if we notice that, since p is POSITIVE, we can safely take the inequality ps < pt and divide both sides by p to get: s < t?.

RE-REPHRASED target question: Is s < t?

At this point, it will be very easy to analyze the answer choices....

Statement 1: p < s
Since there's no information about t, there's no way to answer the RE-REPHRASED target question with certainty.
Statement 1 is SUFFICIENT

Statement 2: s < t
Perfect!
The answer to the RE-REPHRASED target question is YES, s IS less than t
Since we can answer the RE-REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: B

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

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New post 17 Aug 2019, 20:40
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?

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Re: If p, s, and t are positive integer, is |ps - pt| > p(s - t) ?   [#permalink] 17 Aug 2019, 20:40
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