1990 AJHSME Problems/Problem 14: Difference between revisions
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<math>\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36</math> | <math>\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36</math> | ||
==Solution== | ==Solution 1== | ||
The total number of balls in the bag must be <math>4\times 6=24</math>, so there are <math>24-6=18</math> green balls <math>\rightarrow \boxed{\text{B}}</math> | The total number of balls in the bag must be <math>4\times 6=24</math>, so there are <math>24-6=18</math> green balls <math>\rightarrow \boxed{\text{B}}</math> | ||
==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 | |||
==See Also== | ==See Also== | ||
Revision as of 19:58, 15 September 2019
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
See Also
| 1990 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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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