<ol>
<li>The town library donated some books to Mr. Steven's third-grade class. If each student takes 4 books, there will be 20 books left. If 3 students do not take a book and the rest of the students take 5 books each, there will be no books left. How many books were donated to the class?</li>
</ol>
<p>Answer was C. 160; I kept making 2 equations based on each situation:
books/(students-3)=5 and (books-20)/students=4</p>
<p>Unless I keep screwing up my math I'm getting some weird negative fractions when trying to plug these into each other and solving. Any ideas on how to solve this?</p>
<ol>
<li>For all positive integers w and y, where w>y, let the operation @ be defined by w@y = (2^w+y)/(2^w-y). For how many positive integers w is w@1 equal to 4?</li>
</ol>
<p>Answer was E. More than four; I kept using properties of logs to solve this thing, and kept getting either y=1 (which wouldn't work since w must be greater) or 0 on both sides (both sides of the equation w+1=w+1 cancel out).
Any ideas?</p>
<p>Use the following system of equations for # 19:</p>
<p>y=# of books
X=number of students</p>
<p>y-4x=20
y-5(x-3)=0</p>
<p>go through and solve as you normally would it works.</p>
<h1>20</h1>
<p>if you divide out the exponents you get 2^(2y) cuz the w’s always cancel out. This means any number can be placed into the w cuz it always winds up cancelling anyways.</p>
<p>So the equation is ALWAYS true. In other words, no matter what number you plug in for w, you will always get 4.</p>
<p>Some notes:</p>
<p>(1) Let me just isolate the important computation in case you’re getting bogged down by the parentheses: (w+1)-(w-1)=w+1-w+1=2.</p>
<p>[Don’t forget to distribute that minus sign!]</p>
<p>(2) Another way to do this (but more time consuming) is to just plug in 1, 2, 3, 4, and 5 for w and you will see that the answer is 4 in each case. For example, if w=4 we get: 4@1 = 2^5/2^3 = 32/8 = 4.</p>