a. $frac12n - tp$b. $frac12n + tp$c. $frac12np - t$d. $frac12p - tn$e. $frac12p + tn$

The answer offered is A. Any tips on how I can deal with this question?

The full variety of pencils is $12 imes n$. If tright here are $s$ students, and each one receives $p$ pencils, a total of $p imes s$ pencils are given, and hence the teacher is left via $12 imes n - p imes s$ pencils, which (we are told) is $t$.

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Then $$t = 12 imes n - p imes s$$

so the complete number of students ($s$) is given by

$$ s = frac12 imes n - t p$$

Let me understand if there somepoint you do not obtain.

**Hint:**Use these three relationships to settle for the number of students:$$ ewcommand sf<1>carolannpeacock.comsf ext#1eginalign sftotal number of pencils &= sfvariety of pencils offered to students + sfnumber of pencils left over\<0.1in> sffull variety of pencils &= sfnumber of boxes of pencils imes sfnumber of pencils per box\<0.1in>endalign$$$$ sfnumber of pencils provided to students= sfnumber of students imes sfvariety of pencils each student got$$

Hints:Let $s$ be the number of studentsHow many type of pencils are tbelow in total?How many pencils were offered out? What is the distinction in between the answers to (2) and (3)?If you set the number of pencil left over $t$ equal to the answer to (4), and also resolve for $s$, what carry out you get?

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