1997 AJHSME Problems/Problem 13

Problem

Three bags of jelly beans contain 26, 28, and 30 beans. The ratios of yellow beans to all beans in each of these bags are $50\%$ , $25\%$ , and $20\%$ , respectively. All three bags of candy are dumped into one bowl. Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?

$\text{(A)}\ 31\% \qquad \text{(B)}\ 32\% \qquad \text{(C)}\ 33\% \qquad \text{(D)}\ 35\% \qquad \text{(E)}\ 95\%$

Solution

In bag $A$ , there are $26$ jellybeans, and $50\%$ are yellow. That means there are $26\times 50\% = 26\times 0.50 = 13$ yellow jelly beans in this bag.

In bag $B$ , there are $28$ jellybeans, and $25\%$ are yellow. That means there are $28\times 25\% = 28\times 0.25 = 7$ yellow jelly beans in this bag.

In bag $C$ , there are $30$ jellybeans, and $20\%$ are yellow. That means there are $30\times 20\% = 30\times 0.20 = 6$ yellow jelly beans in this bag.

In all three bags, there are $13 + 7 + 6 = 26$ yellow jelly beans in total , and $26 + 28 + 30 = 84$ jelly beans of all types in total.

Thus, $\frac{26}{84} = \frac{26}{84}\cdot 100\% = \frac{13 }{ 42} \cdot 100\% = 30.9\%$ of all jellybeans are yellow. Thus, the correct answer is $\boxed{A}$

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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 AJHSME/AMC 8 Problems and Solutions

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1997 AJHSME Problems/Problem 13

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