Difference between revisions of "2014 AMC 10A Problems/Problem 2"
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<math> \textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}</math> | <math> \textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}</math> | ||
| − | == Solution == | + | == Solution 1 == |
Each day, the cat eats <math>\dfrac13+\dfrac14=\dfrac7{12}</math> of a can of cat food. Therefore, the cat food will last for <math>\dfrac{6}{\frac7{12}}=\dfrac{72}7</math> days, which is greater than <math>10</math> days but less than <math>11</math> days. | Each day, the cat eats <math>\dfrac13+\dfrac14=\dfrac7{12}</math> of a can of cat food. Therefore, the cat food will last for <math>\dfrac{6}{\frac7{12}}=\dfrac{72}7</math> days, which is greater than <math>10</math> days but less than <math>11</math> 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 <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> | ||
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| + | |||
| + | == Solution 2 == | ||
| + | As above, the cat eats <math>\dfrac{7}{12}</math> of a can of cat food. Therefore we can write a table: | ||
| + | |||
| + | <cmath>\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}</cmath> | ||
| + | |||
| + | Since <math>6</math> cans of cat food is just <math>\dfrac{72}{12}</math> cans of cat food, then we see that the cat finished the sixth can of cat food on <math>\boxed{\text{Thursday} \ \textbf{(C)}}</math> | ||
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| + | == Solution 3 == | ||
| + | As Roy's cat eats <math>\frac{1}{3}</math> of a can of cat food every morning and <math>\frac{1}{4}</math> of a can of cat food every evening, and Roy buys a can on Monday morning, he eats <math>\frac{7}{12}</math> cans each day. As Roy has 6 cans and his cat eats <math>\frac{7}{12}</math> cans each day, and we know that <math>6</math> cans of food is <math>\frac{72}{12}</math>, we multiply by 10/1, so we get 70/12, which is 10 days from Monday, which is Thursday. We finish the rest of the food on Thursday, so therefore Thursday is the final Answer. | ||
| + | |||
| + | ~Horizon | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/hrin8SPWauU | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | |||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 02:33, 7 December 2024
Contents
Problem
Roy's cat eats 
of a can of cat food every morning and 
of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing 
cans of cat food. On what day of the week did the cat finish eating all the cat food in the box?
Solution 1
Each day, the cat eats 
of a can of cat food. Therefore, the cat food will last for 
days, which is greater than 
days but less than 
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 days after Monday, or 
Solution 2
As above, the cat eats 
of a can of cat food. Therefore we can write a table:
![\[\begin{array}{c|c|c|c|c|c|c} \text{Mon} & \text{Tue} & \text{Wed} & \text{Thu} & \text{Fri} & \text{Sat} & \text{Sun} \\ \hline 7/12 & 14/12 & 21/12 & 28/12 & 35/12 & 42/12 & 49/12 \\ 56/12 & 63/12 & 70/12 & 77/12 & 84/12 & 91/12 & 98/12 \end{array}\]](http://latex.artofproblemsolving.com/5/d/1/5d176f4e30b8ffc587c98b671be8908357aa8c9b.png)
Since cans of cat food is just 
cans of cat food, then we see that the cat finished the sixth can of cat food on 
Solution 3
As Roy's cat eats of a can of cat food every morning and of a can of cat food every evening, and Roy buys a can on Monday morning, he eats 
cans each day. As Roy has 6 cans and his cat eats cans each day, and we know that cans of food is 
, we multiply by 10/1, so we get 70/12, which is 10 days from Monday, which is Thursday. We finish the rest of the food on Thursday, so therefore Thursday is the final Answer.
~Horizon
Video Solution (CREATIVE THINKING)
https://youtu.be/hrin8SPWauU
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
