Difference between revisions of "2010 AMC 10A Problems/Problem 4"
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Assuming that there were fractions of compact discs, it would take <math>412/56 ~= 7.357</math> CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have <math>412/8 = 51.5</math> minutes on each of the 8 discs. The answer is <math>\boxed{B}</math>. | Assuming that there were fractions of compact discs, it would take <math>412/56 ~= 7.357</math> CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have <math>412/8 = 51.5</math> minutes on each of the 8 discs. The answer is <math>\boxed{B}</math>. | ||
| + | ==Video Solution== | ||
| + | https://youtu.be/C1VCk_9A2KE?t=196 | ||
| + | |||
| + | ~IceMatrix | ||
== See Also == | == See Also == | ||
Revision as of 05:18, 26 May 2020
Contents
Problem 4
A book that is to be recorded onto compact discs takes 
minutes to read aloud. Each disc can hold up to 
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?
Solution
Assuming that there were fractions of compact discs, it would take 
CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have 
minutes on each of the 8 discs. The answer is 
.
Video Solution
https://youtu.be/C1VCk_9A2KE?t=196
~IceMatrix
See Also
| 2010 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
