1990 AJHSME Problems/Problem 14
Contents
[hide]Problem
A bag contains only blue balls and green balls. There are blue balls. If the probability of drawing a blue ball at random from this bag is , then the number of green balls in the bag is
Solution 1
The total number of balls in the bag must be , so there are green balls
Solution 2
If b= number of blue balls in the bag and g = number of green balls in the bag then b/(b+g) = 1/4. Substituting b=6 and solving for g we get g=18, or B
- goldenn
See Also
1990 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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