Difference between revisions of "2014 AMC 8 Problems/Problem 3"

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==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. 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.
 
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.
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==Video Solution==
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https://youtu.be/2x4OVpV87IY
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 +
~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}}

Revision as of 08:12, 27 April 2022

Problem

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 and an average of $44$ pages per day for the next three days. She then finished the book by reading $10$ pages on the last day. How many pages were in the book?


$\textbf{(A) }240\qquad\textbf{(B) }250\qquad\textbf{(C) }260\qquad\textbf{(D) }270\qquad \textbf{(E) }280$

Solution

Isabella read $3\cdot 36+3\cdot 44$ pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as $3\cdot (36+44)=3\cdot 80$, which gives that she read $240$ pages. However, she read $10$ more pages on the last day, for a total of $240+10=\boxed{\textbf{(B)}~250}$ pages.

Video Solution

https://youtu.be/2x4OVpV87IY

~savannahsolver

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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