2015 CEMC Gauss (Grade 8) Problems/Problem 14

Problem

Carolyn has a $$5$ bill, a $$10$ bill, a $$20$ bill, and a $$50$ bill in her wallet. She closes her eyes and removes one of the four bills from her wallet. What is the probability that the total value of the three bills left in her wallet is greater than $$70$ ?

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$\textbf{(A)}\ 0.5 \qquad\textbf{(B)}\ 0.25 \qquad\textbf{(C)}\ 0.75 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 0$

Solution 1

We can look at different cases for the dollar bill that was removed, and then see what the sum would be. We also know that there are $4$ total possibilities of dollar bills that she can remove.

First, if Carolyn removes a $10$ dollar bill, the total value remaining would then be $5 + 20 + 50 = 75$ . This is more than $70$ . We can also conclude that removing the $5$ dollar bill would also result in a total value greater than $70$ from this because the value of the $5$ dollar bill is less than the value of the $10$ dollar bill, so adding the remaining values will result in a value larger than $70$ .

Second, if Carolyn removes a $20$ dollar bill, the total value remaining would be $5 + 10 + 50 = 65$ . This is less than $70$ . We can also conclude that removing the $50$ dollar bill would also result in a total value less than $70$ using similar logic.

Of the possibilities above, we can see that there were $2$ where the total value was greater than $70$ . Thus, the probability is:

$\frac{2}{4} = \frac{2 \times 25}{4 \times 25} = \frac{50}{100} = \boxed {\textbf {(A) } 0.5}$

~anabel.disher

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2015 CEMC Gauss (Grade 8) Problems/Problem 14

Problem

Solution 1