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Re: Roger bought some pencils and erasers at the stationery store [#permalink]
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clary wrote:
I'm not understanding why option C) is wrong as Roger could buy 6 erasers and since pencils>erasers then 7 pencils can be considered which adds to 13 (which is between 12 and 20 range)
Also for option B) lets say he bought 12 pencils(max) and since pencils are greater lets say 8 erasers were bought which adds to 20 which is between 12 and 20 range.


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.

_______

In essence, must means always true no matter what. Could means can happen though not all the time.
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Re: Roger bought some pencils and erasers at the stationery store [#permalink]
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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

clary wrote:
Roger bought some pencils and erasers at the stationery store. If he bought more pencils than erasers, and the total number of the pencils and erasers he bought is between 12 and 20 (inclusive), which of the following statements must be true?

Indicate all such statements.

a)Roger bought no fewer than 7 pencils.
b)Roger bought no more than 12 pencils.
c)Roger bought no fewer than 6 erasers.
d)Roger bought no more than 9 erasers.
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