Difference between revisions of "2014 AMC 10A Problems/Problem 2"

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Because the number of days is greater than 10 and less than 11, the cat will finish eating in on the 11th day, which is equal to <math>10</math> days after Monday, or <math>\boxed{\textbf{(C)}\ \text{Thursday}}</math>
 
Because the number of days is greater than 10 and less than 11, the cat will finish eating in on the 11th day, which is equal to <math>10</math> days after Monday, or <math>\boxed{\textbf{(C)}\ \text{Thursday}}</math>
  
(Solution/revised by bestwillcui1)
 
  
 
==See Also==
 
==See Also==

Revision as of 06:57, 19 February 2014

Problem

Roy's cat eats $\frac{1}{3}$ of a can of cat food every morning and $\frac{1}{4}$ of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing $6$ cans of cat food. On what day of the week did the cat finish eating all the cat food in the box?

$\textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}$ (Error compiling LaTeX. Unknown error_msg)

Solution

Each day, the cat eats $\dfrac13+\dfrac14=\dfrac7{12}$ of a can of cat food. Therefore, the cat food will last for $\dfrac{6}{\dfrac7{12}}=\dfrac{72}7$ days, which is greater than $10$ days but less than $11$ days.


Because the number of days is greater than 10 and less than 11, the cat will finish eating in on the 11th day, which is equal to $10$ days after Monday, or $\boxed{\textbf{(C)}\ \text{Thursday}}$


See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 AMC 10 Problems and Solutions

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