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

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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>
 
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 ==
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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.
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~Horizon
  
 
==Video Solution (CREATIVE THINKING)==
 
==Video Solution (CREATIVE THINKING)==

Latest revision as of 01:33, 7 December 2024

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 $\boxed{\text{Thursday} \ \textbf{(C)}}$

Solution 3

As 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, and Roy buys a can on Monday morning, he eats $\frac{7}{12}$ cans each day. As Roy has 6 cans and his cat eats $\frac{7}{12}$ cans each day, and we know that $6$ cans of food is $\frac{72}{12}$, 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

https://youtu.be/P9g1U4TQlUo

~savannahsolver

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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