2006 AMC 8 Problems/Problem 15


Problems 14, 15 and 16 involve Mrs. Reed's English assignment.

A Novel Assignment

The students in Mrs. Reed's English class are reading the same 760-page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in 45 seconds and Chandra reads a page in 30 seconds.

Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?

$\textbf{(A)}\ 425\qquad\textbf{(B)}\ 444\qquad\textbf{(C)}\ 456\qquad\textbf{(D)}\ 484\qquad\textbf{(E)}\ 506$

Solution 1

Same as the previous problem, we only use the information we need. Note that it's not just Chandra reads half of it and Bob reads the rest since they have different reading rates. In this case, we set up an equation and solve.

Let $x$ be the number of pages that Chandra reads.

$30x = 45(760-x)$ Distribute the $45$

$30x = 45(760) - 45x$ Add $45x$ to both sides

$75x = 45(760)$ Divide both sides by $15$ to make it easier to solve

$5x = 3(760)$ Divide both sides by $5$

$x = 3(152) = \boxed{\textbf{(C)} 456}$

Solution 2 (a bit faster)

Chandra and Bob read at a rate of $30:45$ seconds per page, respectively. Simplifying that gets us Bob reads $2$ pages for every $3$ pages that Chandra reads. Therefore Chandra should read $\frac{3}{2+3}=\frac{3}{5}$ of the book. $\frac{3}{5}\cdot760$=$\boxed{\textbf{(C)} 456}$

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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