Difference between revisions of "2014 AMC 8 Problems/Problem 3"
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<math>\textbf{(A) }240\qquad\textbf{(B) }250\qquad\textbf{(C) }260\qquad\textbf{(D) }270\qquad \textbf{(E) }280</math> | <math>\textbf{(A) }240\qquad\textbf{(B) }250\qquad\textbf{(C) }260\qquad\textbf{(D) }270\qquad \textbf{(E) }280</math> | ||
==Solution== | ==Solution== | ||
| + | Isabella read <math>3\cdot 36+3\cdot 44</math> pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as <math>3\cdot (36+44)=3\cdot 80</math>, which gives that she read <math>240</math> pages. On her last day, she read <math>10</math> more pages for a total of <math>240+10=\boxed{\textbf{(B)}~250}</math> pages. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/ggZxESzZmBQ | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/2x4OVpV87IY | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=2|num-a=4}} | {{AMC8 box|year=2014|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:17, 2 July 2023
Problem
Isabella had a week to read a book for a school assignment. She read an average of 
pages per day for the first three days and an average of 
pages per day for the next three days. She then finished the book by reading 
pages on the last day. How many pages were in the book?
Solution
Isabella read 
pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as 
, which gives that she read 
pages. On her last day, she read more pages for a total of 
pages.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
