Is (q - p)

Question

Is  (q - p) < 0 ?

(1)  (p^3 - q) < 0

(2)  (p^2 - q) > 0

toy9575 · Answer

there is nothing given if p,r are integers or positive integers or if one of them positive. regardless we need to know if q is a bigger negative number than p 
statement 1. all cases for p and q doesn't satisfy this for example if p is negative and q is positive or if q >p or vice versa
statement 2. this complicates even further with squaring p.even if p is negative squaring makes it positive
both together - doesn't say individually about q>p or if p is a greater negative number 
so E

IanStewart · Answer

I have no idea why there are parentheses throughout the question. We want to know if q < p. Using both Statements, we know:

S1: p^3 < q
S2: q < p^2

so we know p^3 < q < p^2.

Now, q < p^2 is automatically going to be true any time q is negative (since p^2 is at least zero). So if we pick numbers, and make q negative, we don't need to worry about how to ensure Statement 2 is true. We just need to make Statement 1 true. And we could have:

q = -3 and p = -2. Then both Statements are true, and q < p

or we could have

q = -3 and p = -4. Then both Statements are true, and q > p

So we can get two answers to the question, and the answer is E.

prakhar992 · Answer

It took me way more than 2 minutes. How to solve this question fast and confident?

TestPrepUnlimited · Answer

I would like to showcase two methods here, I will start with the plugging in number method and the 2nd method will simply be comparing the inequalities and attempting to simplify them. The second method will be much faster but requires good understanding on how to simplify inequalities. Method 1: Plugging in numbers When looking at variables in inequalities we have to ask ourselves a couple of questions: (i) Are the variables positive/negative? (ii) Are the size of the variables bigger than 1 or less than 1? It helps to consider how we can make p < q or the other way around. We want to identify the extreme/weird cases, jump to those, and cut any effort in finding the easy cases. Statement 1: p^3 < q . It should be quick to find a case where p^3 < q and p < q are both true, so let's focus on how we can make p > q and p^3 < q at the same time. We can have 0.5^3 < 0.2 which results in p > q . Since we have both sides of the answers this is insufficient. Statement 2: p^2 > q . This time we can find many cases where p > q can be true. So we want to focus on how we can get p < q from p^2 > q , and we could make a negative p positive q case that satisfies p^2 > q so this would also be insufficient. Combined: Let us try to recycle what we did in the earlier statements. p = 0.5 and q = 0.2 can satisfy both statements, hence p > q is possible. We could have the negative p positive q case from statement 2, so p < q is possible. Then combined this is still insufficient. Method 2: Simplifying inequalities We want to prove p < q or p > q. Statement 1: p^3 < q cannot be simplified. Hence we cannot prove p < q from this statement, then this is insufficient. Statement 2: p^2 > q cannot be simplified. Hence we cannot prove p < q from this statement, then this is insufficient. Combined: We have p^3 < q and q < p^2 . Combining the inequalities we can conclude p^3 < p^2 and p < 1 . Yet we have no information on q, hence we still don't know the relation between p and q. This is all we can conclude so combined still insufficient. Ans: E

TestPrepUnlimited · Answer

Hi  prakhar992  ,

I showed two methods which each have their merits. If you are good with number properties you may be able to come up with some good numbers to plug in for method 1, my favorite is still looking at if we're able to prove "q < p" with the second method since there is no need to search for numbers.

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Is (q - p) < 0?

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