2014 AMC 10A Problems/Problem 2

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?

$\textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}$

Solution 1

Each day, the cat eats $\dfrac13+\dfrac14=\dfrac7{12}$ of a can of cat food. Therefore, the cat food will last for $\dfrac{6}{\frac7{12}}=\dfrac{72}7$ 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 $\boxed{\textbf{(C)}\ \text{Thursday}}$

Solution 2

As above, the cat eats $\dfrac{7}{12}$ of a can of cat food. Therefore we can write a table:

$\begin{array}{c|c|c|c|c|c|c} \text{Mon} & \text{Tue} & \text{Wed} & \text{Thu} & \text{Fri} & \text{Sat} & \text{Sun} \\ \hline 7/12 & 14/12 & 21/12 & 28/12 & 35/12 & 42/12 & 49/12 \\ 56/12 & 63/12 & 70/12 & 77/12 & 84/12 & 91/12 & 98/12 \end{array}$

Since $6$ cans of cat food is just $\dfrac{72}{12}$ cans of cat food, then we see that the cat finished the sixth can of cat food on $\boxed{\text{Thursday} \ \textbf{(C)}}$

Video Solution (CREATIVE THINKING)

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Video Solution

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2014 AMC 10A (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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