This is a must be true question. In other words, what facts are always true given the conditions? In this case, we're not focusing on what could be true, like the scenarios you gave - but what can never happen, no matter what we try to do?

Since \(p>e\) and the maximum amount of \(p+e\) is \(20\), then it's possible to have bought one eraser at a minimum. It also states that this person bought pencils and erasers, so we can assume \(e>0\).

\(e < 10\), since if it were \(10\), \(p\) could be no more than \(10\), since the max they bought at the store is \(20\). But this contradicts \(p>e\) (we did not buy the same amount or more erasers than pencils), so it follows that \(9\) is the maximum.

So we get \(0 < e < 10\).

\(p\) can be a maximum of \(19\) using the same reasoning for the minimum number of erasers bought being \(1\).

To find \(p's\) minimum we need to maximize \(e\) while staying in the constraints of \(p>e\). To do that let's bring \(p\) and \(e\) as close as possible.

Since \(p>e\), let \(p=e+1\), or \(e=p-1\).

From here:

\(p+e≥12\)

\(p+(p-1)≥12\)

\(2p-1≥12\)

\(2p≥13\)

\(p≥6.5\)

Since were dealing with whole pencils and not halves, this really means:

\(p≥7\)

So our minimum number of pencils bought is 7, and put that together with the maximum to get:

\(6<p<20\)

Using both:

\(6<p<20\)

\(0<e<10\)

We can see how B and C \(can\) happen. But looking for what definitely cannot happen, we land on A and D, since it must be true that they cannot occur.

Thus, A and D are the solutions.

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In essence, must means always true no matter what. Could means can happen though not all the time.

let pencils=P and erasers=E
(P>E)^(P+E>=12)v(P+E=<20), where ^ conjuntive and v implies disjunctive logical connectors

analyzing answer choices:
a)Roger bought no fewer than 7 pencils. (P>E)^(P+E>=12) True
b)Roger bought no more than 12 pencils. (P>E)^(P+E>=12)v(P+E=<20) False
c)Roger bought no fewer than 6 erasers. (P>E)^(P+E>=12)v(P+E=<20) False
d)Roger bought no more than 9 erasers. (P>E)^(P+E=<20) True

Answers are a, d