Difference between revisions of "2021 April MIMC 10 Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | 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)} | + | 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)} 5:33}</math>. |
Latest revision as of 12:28, 26 April 2021
Okestima is reading a 
page book. He reads a page every 
minutes, and he pauses 
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 
minutes. If he starts to read at 
, when does he finish the book?
Solution
To begin, we can start by find the amount of time Okestima takes to read 
pages. 
gives us 
minutes. After he comes back from break, he reads every page in two minutes. 
minutes. Add the additional minutes from break, he would finish the book at 
.
