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.
18 Sep 2021, 07:14
66% (01:28) correct
