Difference between revisions of "2021 April MIMC 10 Problems/Problem 2"

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==Solution==
 
==Solution==
To be Released on April 26th.
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To begin, we can start by find the amount of time Okestima takes to read <math>90</math> pages. <math>90\cdot \frac{2}{3}</math> gives us <math>60</math> minutes. After he comes back from break, he reads every page in two minutes. <math>60\cdot2=120</math> minutes. Add the additional <math>3</math> minutes from break, he would finish the book at <math>\fbox{\textbf{(C)} </math>5\colon33<math>}</math>.

Revision as of 13:28, 26 April 2021

Okestima is reading a $150$ page book. He reads a page every $\frac{2}{3}$ minutes, and he pauses $3$ minutes when he reaches the end of page 90 to take a break. He does not read at all during the break. After, he comes back with food and this slows down his reading speed. He reads one page in $2$ minutes. If he starts to read at $2:30$, when does he finish the book?

$\textbf{(A)} ~4:33 \qquad\textbf{(B)} ~5:30 \qquad\textbf{(C)} ~5:33 \qquad\textbf{(D)} ~6:30 \qquad\textbf{(E)} ~7:33$

Solution

To begin, we can start by find the amount of time Okestima takes to read $90$ pages. $90\cdot \frac{2}{3}$ gives us $60$ minutes. After he comes back from break, he reads every page in two minutes. $60\cdot2=120$ minutes. Add the additional $3$ minutes from break, he would finish the book at $\fbox{\textbf{(C)}$ (Error compiling LaTeX. Unknown error_msg)5\colon33$}$ (Error compiling LaTeX. Unknown error_msg).