Difference between revisions of "2010 AMC 10A Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | Assuming that there | + | Assuming that there are 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 must be a whole number, there are at least 8 CDs, in which case there would be <math>412/8 = 51.5</math> minutes of reading on each of the 8 discs. The answer is <math>\boxed{B}</math>. |
| + | ==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 <math>\boxed{B}</math> is our answer. | ||
| + | |||
| + | -ILoveMath31415926535 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/C1VCk_9A2KE?t=196 | ||
| + | |||
| + | ~IceMatrix | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2010|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2010|ab=A|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:38, 16 November 2022
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 are fractions of compact discs, it would take 
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 
minutes of reading on each of the 8 discs. The answer is 
.
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 is our answer.
-ILoveMath31415926535
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. 
