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Difficulty: Very hardBasic Probability and Counting Methods

A box contains only red, blue, and green marbles. The ratio of red marbles to blue marbles is 2:32:3, and the ratio of blue marbles to green marbles is 3:53:5. If one marble is drawn at random from the box, the probability of drawing a red marble is PP. If 66 green marbles are added to the box and no other marbles are removed, the probability of drawing a red marble becomes QQ. Given that P−Q=3115P - Q = \frac{3}{115}, how many blue marbles were originally in the box?

  1. A
    3
  2. B
    8
  3. 12Answer
  4. D
    15
  5. E
    20

Answer

12
The correct answer is 12. By expressing the original counts of red, blue, and green marbles as 2k2k, 3k3k, and 5k5k, the total number of marbles is 10k10k. The initial probability of drawing a red marble is P=2k10k=15P = \frac{2k}{10k} = \frac{1}{5}. When 6 green marbles are added, the new total becomes 10k+610k + 6, yielding a new probability Q=2k10k+6=k5k+3Q = \frac{2k}{10k + 6} = \frac{k}{5k + 3}. Setting their difference to 3115\frac{3}{115} results in 15−k5k+3=3115\frac{1}{5} - \frac{k}{5k + 3} = \frac{3}{115}. Simplifying this rational equation gives 25k+15=11525k + 15 = 115, which solves to k=4k = 4. Therefore, the original number of blue marbles is 3k=3(4)=123k = 3(4) = 12.

Step-by-Step Solution

1
Define the number of red, blue, and green marbles in terms of a variable kk.
Let red marbles = 2k2k, blue marbles = 3k3k, and green marbles = 5k5k. The initial total number of marbles is 2k+3k+5k=10k2k + 3k + 5k = 10k.
The given ratios are red:blue = 2:3 and blue:green = 3:5, which combine to a continuous ratio of red:blue:green = 2:3:5.
2
Express the initial probability PP of drawing a red marble.
P=2k10k=15P = \frac{2k}{10k} = \frac{1}{5}.
Probability is the number of favorable outcomes (red marbles) divided by the total number of outcomes.
3
Express the new probability QQ of drawing a red marble after adding 6 green marbles.
The number of green marbles becomes 5k+65k + 6, so the new total is 10k+610k + 6. Thus, Q=2k10k+6Q = \frac{2k}{10k + 6}.
Adding 6 green marbles increases the total marble count by 6 while the number of red marbles remains 2k2k.
4
Set up the equation using the given difference P−Q=3115P - Q = \frac{3}{115} and solve for kk.
15−2k10k+6=3115\frac{1}{5} - \frac{2k}{10k + 6} = \frac{3}{115}
15−k5k+3=3115\frac{1}{5} - \frac{k}{5k + 3} = \frac{3}{115}
(5k+3)−5k5(5k+3)=3115\frac{(5k + 3) - 5k}{5(5k + 3)} = \frac{3}{115}
325k+15=3115\frac{3}{25k + 15} = \frac{3}{115}
25k+15=115  ⟹  25k=100  ⟹  k=425k + 15 = 115 \implies 25k = 100 \implies k = 4.
Solving the rational equation gives the multiplier kk that determines the actual number of marbles.
5
Calculate the original number of blue marbles.
Blue marbles = 3k=3(4)=123k = 3(4) = 12.
The question asks for the original number of blue marbles, which is represented by 3k3k.

Key Concept

Basic Probability and Counting Methods
Estimated Time:3m 0s
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Basic Probability and Counting Methods

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