Difference between revisions of "1990 AJHSME Problems/Problem 14"

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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==
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==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>
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==Solution 2==
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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 20:58, 15 September 2019

Problem

A bag contains only blue balls and green balls. There are $6$ blue balls. If the probability of drawing a blue ball at random from this bag is $\frac{1}{4}$, then the number of green balls in the bag is

$\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36$

Solution 1

The total number of balls in the bag must be $4\times 6=24$, so there are $24-6=18$ green balls $\rightarrow \boxed{\text{B}}$

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 (ProblemsAnswer KeyResources)
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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