1)Problem
Pablo buys popsicles for his friends. The store sells single popsicles for $$1$ each, $3$-popsicle boxes for $$2$ each, and $5$-popsicle boxes for $$3$. What is the greatest number of popsicles that Pablo can buy with $$8$?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{©}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$
Solution
$$3$ boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying $2$, we have $$2$ left. We cannot buy a third $$3$ box, so we opt for the $$2$ box instead (since it has a higher popsicles/dollar ratio than the $$1$ pack). We’re now out of money. We bought $5+5+3=13$ popsicles, so the answer is $\boxed{\textbf{(D) }13}$.
2)Mia is helping her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{©}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$
Solution
Every $30$ seconds $3-2=1$ toys are put in the box, so after $27\cdot30$ seconds there will be $27$ toys in the box. Mia’s mom will then put $3$ toys into to the box and we have our total amount of time to be $27\cdot30+30=840$ seconds, which equals $14$ minutes. $\boxed{(\textbf{B})\ 14}$
3)The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{©}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$
Solution
Let the two real numbers be $x,y$. We are given that $x+y=4xy,$ and dividing both sides by $xy$, $\frac{x}{xy}+\frac{y}{xy}=4.$
[\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{© } 4}.]
Note: we can easily verify that this is the correct answer; for example, 1/2 and 1/2 work, and the sum of their reciprocals is 4.