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

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.
_________________

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.

Hi,
The equation is \(|ps-pt| > p(s-t)\).
or \(p|s-t| > p(s-t)\)

Since all numbers are positive , p cannot be 0..

Thus two cases..
1) \(|s-t|\) will never be smaller than \(s-t\) as |s-t| 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
a) NO if s>t or s=t and
b) 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
_________________

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!
_________________

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

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?

|x-y| is the distance and distance is always positive, so |4-6|=|-2|=2, but x-y=4-6=-2.
2 is not equal to-2
_________________

Solution:

Question Stem Analysis:

We need to determine whether |ps - pt| > p(s - t). Notice that ps - pt = p(s - t). Since |x| > x if and only if x is negative, we need to determine whether p(s - t) is negative. Lastly, since p is positive, we really need to determine whether s - t is negative.

Statement One Alone:

Since we don’t know anything about t (except that it’s positive), statement one alone is not sufficient.

Statement Two Alone:

Since s < t, then s - t < 0, i.e., s - t is negative. Therefore, p(s - t) < 0 and |ps - pt| > p(s - t). Statement two alone is sufficient.

Answer: B
_________________

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.

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.
_________________

Hi BrentGMATPrepNow

I am Curios to understand why squaring on both sides of the original equation doesn't work to solve this question. Can you please help explain?

We can only square inequality when both sides are non-negative, which is not the case for |s - t| > (s - t). By the ways |s - t| > (s - t) is true ONLY when s - t < 0, so we cannot square.

When dealing with inequalities, it's important to remember that we can only square both sides of the inequality if both sides are non-negative. However, this condition is not met in cases such as |s - t| > (s - t) because the inequality |s - t| > (s - t) is only true when the value of s - t is negative. Therefore, attempting to square both sides of the inequality in this case would not be valid.
_________________

|ps-pt| => |p(s-t)|
|p(s-t)| > p(s-t)
if s< t then only the above condition will hold. if s=t then both side will become 0 hence inequality will not hold. if s > t then then both the side on inequality will become equal hence the condition will not hold.

Hence B is the answer