2010 AMC 10A Problems/Problem 4: Difference between revisions
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51.5 | == Problem 4 == | ||
A book that is to be recorded onto compact discs takes <math>412</math> minutes to read aloud. Each disc can hold up to <math>56</math> 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? | |||
<math> | |||
\mathrm{(A)}\ 50.2 | |||
\qquad | |||
\mathrm{(B)}\ 51.5 | |||
\qquad | |||
\mathrm{(C)}\ 52.4 | |||
\qquad | |||
\mathrm{(D)}\ 53.8 | |||
\qquad | |||
\mathrm{(E)}\ 55.2 | |||
</math> | |||
==Solution== | |||
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>. | |||
Revision as of 15:43, 20 December 2010
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 
.
