Is p > q ? (1) p + r > q + s and r > s (2) p + s < q + r and s > r

Target question: Is p > q ?

Statement 1: p + r > q + s and r > s
Whenever we're given two inequalities in which the inequality symbols are facing the same direction, we might be able to gain some valuable insights by ADDING the inequalities (see the video below for more on this)

So we can take the following inequalities...
p + r > q + s
r > s
.... and ADD for them to get: p + 2r > q + 2s
This doesn't help us answer the target question.
We can demonstrate this by considering the following values of p, q, r and s that satisfy statement 1:
Case a: p = 1, q = 0, r = 5 and s = 1. In this case, the answer to the target question is YES, p is greater than q
Case b: p = 0, q = 1, r = 5 and s = 1. In this case, the answer to the target question is NO, p is not greater than q
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p + s < q + r and s > r
Since the two inequality symbols aren't facing the same direction, we can't add the inequalities...YET!
One way to get the inequality symbols facing the same direction is to take the second inequality, s > r, and multiply both sides by -1 to get: -s < -r [Note: Since I multiplied both sides by a NEGATIVE value, I REVERSED the direction of the inequality symbol]

At this point, we can take the following inequalities...
p + s < q + r
-s < -r
...and ADD them to get: p < q, which means the definitive answer to the target question is NO, p is not greater than q
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

RELATED VIDEO

Can someone clarify this question in a better way? Also, why are we not using the number substitution method in solving both 1 and 2?

Number substitution (aka testing values) will only yield definitive results when a statement is not sufficient.
More here: https://www.gmatprepnow.com/articles/da ... lug-values

Target question: Is p > q?

Statement 1: p + r > q + s and r > s
Important: We must resist the temptation to subtract the second inequality from the first inequality (to incorrectly conclude that p > q).
If we have two inequalities in which the inequality symbols are facing the same direction, we can ADD those inequalities, but we can't subtract them.
For more on this, watch the video below.

If we take...
p + r > q + s
r > s

... and ADD them, we get: p + 2r > q + 2s
There are several values of p, q, r and s that satisfy the inequality p + 2r > q + 2s. Here are two:
Case a: p = 0, q = 1, r = 10 and s = 0. In this case, the answer to the target question is NO, p is not greater than q
Case b: p = 1, q = 0, r = 10 and s = 0. In this case, the answer to the target question is YES, p is greater than q
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p + s < q + r and s > r
In this case, we have 2 inequalities, BUT the inequality symbols are not facing the same direction, which means we can't add them...YET!
Take the inequality s > r, and multiply both sides by -1 to get: -s < -r [notice that, since we multiplied both sides of the inequality by a NEGATIVE value, we had to reverse the direction of the inequality symbol]
We now have the following two inequalities:
p + s < q + r
-s < -r

Now that the inequality symbols are facing the same direction, we can ADD the inequalities to get: p > q. Perfect!
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

RELATED VIDEO

Rephrased question:

Is: p - q > 0?

Statement 1
p - q > s - r

And we are also given: r > s
Which is the same as:
s - r < 0

So the value of (s - r) = some -negative value

And statement 1 tells us that: (p - q) > that -negative value

So it is possible for:

p - q = -negative
OR
p - q = +positive

S1 NOT Sufficient

Statement 2: tells us that

p - q < r - s …… (inequality I)

And we are also given: s > r
Or
r - s < 0…… (inequality II)

Using the Transitive Property of Inequalities for the one stated above in I and II we can therefore say:

p - q < r - s < 0

Definite NO: the value of (p-q) must be less than < 0 and can NOT be greater than > 0

p - q < 0 —- s2 sufficient alone

*B*

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Both options have different relations between r and s, one says that r>s and other says r<s. Experts, is that possible? Isn't the question incorrect?