2014 AMC 10A Problem 2

P1-10~2 minPrint

Linear Equations & Word Problems Clocks, Calendars & Time Fractions & Decimals Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

One day uses 7/12 of a can. Ten days use 70/12 of the 72/12 in the box, so the last bit is eaten on the 11th morning.

Solution

It is easiest to count food in twelfths of a can, because $12$ works for both $\frac13$ and $\frac14$ .

Morning meal: $\frac13 = \frac{4}{12}$ of a can. Evening meal: $\frac14 = \frac{3}{12}$ of a can. One whole day uses $\frac{4}{12} + \frac{3}{12} = \frac{7}{12}$ of a can.
The box holds $6$ cans, which is $6 \cdot 12 = 72$ twelfths.
Each day uses $7$ twelfths. After $10$ days the cat has eaten $70$ twelfths, and $72 - 70 = 2$ twelfths are left.
Day $11$ starts with a morning meal of $4$ twelfths. Only $2$ are left, so the food runs out during the morning of day $11$ .
Now name the day. Day $1$ is Monday. Seven days later, day $8$ , is Monday again. Then day $9$ is Tuesday, day $10$ is Wednesday, and day $11$ is Thursday.

The answer is $\boxed{\textbf{(C)}\ \text{Thursday}}$ .

Why this works

Two meals a day become one daily amount once you add them. Dividing the supply by the daily amount tells you how many full days are covered, and the leftover tells you whether the food runs out on the next morning or the next evening. To turn a day number into a weekday, remember that the weekday repeats every $7$ days.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Each full day the cat eats
$\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$
of a can. Measure everything in twelfths: the box holds $6 = \frac{72}{12}$ cans.

After $10$ full days the cat has eaten $\frac{70}{12}$ cans, leaving $\frac{2}{12}$ . That is less than the morning portion of $\frac{4}{12}$ , so the food runs out during the morning of day $11$ .

Day $1$ is Monday, so day $8$ is Monday again and day $11$ is Thursday.

The answer is $\boxed{\textbf{(C)}\ \text{Thursday}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 11: Decimals and Fractions
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Prealgebra, Chapter 6: Decimals (§6.3, §6.4)