2010 AMC 10A Problems/Problem 4

Problem 4

A book that is to be recorded onto compact discs takes $412$ minutes to read aloud. Each disc can hold up to $56$ minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?

$\mathrm{(A)}\ 50.2 \qquad \mathrm{(B)}\ 51.5 \qquad \mathrm{(C)}\ 52.4 \qquad \mathrm{(D)}\ 53.8 \qquad \mathrm{(E)}\ 55.2$

Solution

Assuming that there are fractions of compact discs, it would take $412/56 ~= 7.357$ CDs to have equal reading time. However, since the number of discs must be a whole number, there are at least 8 CDs, in which case there would be $412/8 = 51.5$ minutes of reading on each of the 8 discs. The answer is $\boxed{B}$ .

Solution 2

We look at the options, and see which one is divisible by 412. We try A, and see that you don't get a whole number. Next, we try B. We see that the B is divisible by 412 and find that $\boxed{B}$ is our answer.

-ILoveMath31415926535

Video Solution

https://youtu.be/C1VCk_9A2KE?t=196

~IceMatrix

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2010 AMC 10A Problems/Problem 4

Contents

Problem 4

Solution

Solution 2

Video Solution

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