Difference between revisions of "2014 AMC 8 Problems/Problem 3"
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− | Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days | + | ==Problem== |
+ | Isabella had a week to read a book for a school assignment. She read an average of <math>36</math> pages per day for the first three days and an average of <math>44</math> pages per day for the next three days. She then finished the book by reading <math>10</math> pages on the last day. How many pages were in the book? | ||
+ | |||
<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== | ||
+ | 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. However, she read <math>10</math> more pages on the last day, for a total of <math>240+10=\boxed{\textbf{(B)}~250}</math> pages. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2014|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 10:08, 2 December 2015
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. However, she read more pages on the last day, for a total of pages.
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.