Alejandra is designing a game of chance. For one part of the game, a player is to randomly choose 3...
GMAT Two Part Analysis : (TPA) Questions
Alejandra is designing a game of chance. For one part of the game, a player is to randomly choose \(\mathrm{3}\) marbles, without replacement, from a box containing \(\mathrm{B}\) blue marbles, \(\mathrm{R}\) red marbles, and no other marbles. Alejandra correctly determined the positive integers \(\mathrm{B}\) and \(\mathrm{R}\) so that the number of possible selections in which \(\mathrm{2}\) blue marbles and \(\mathrm{1}\) red marble are chosen is twice the number of possible selections in which \(\mathrm{2}\) red marbles and \(\mathrm{1}\) blue marble are chosen.
The positive integers \(\mathrm{B}\) and \(\mathrm{R}\) that Alejandra determined must be such that \(\mathrm{B}\) is the number that is \(\mathrm{1}\) _ the number that is \(\mathrm{2}\) _ \(\mathrm{R}\). Based on the information provided, select for \(\mathrm{1}\) and for \(\mathrm{2}\) the options that create the most accurate statement. Make only two selections, one in each column.
1
2
2 more than
1 more than
1 less than
2 less than
half
twice
Phase 1: Owning the Dataset
Visual Representation
We have a box with:
- B blue marbles
- R red marbles
- Total: \(\mathrm{B + R}\) marbles
Box visualization:
[BBBBB...B RRRR...R] B blues R reds
Given Information
- Player chooses 3 marbles without replacement
- Alejandra found B and R such that:
- Ways to choose (2 blue, 1 red) = 2 × Ways to choose (2 red, 1 blue)
Phase 2: Understanding the Question
We need to find how B relates to R. The question asks us to fill in:
"B is the number that is [1] the number that is [2] R"
Setting Up Combinations
- Ways to choose 2 blue, 1 red: \(\mathrm{C(B,2) \times C(R,1) = \frac{B(B-1)}{2} \times R}\)
- Ways to choose 2 red, 1 blue: \(\mathrm{C(R,2) \times C(B,1) = \frac{R(R-1)}{2} \times B}\)
Phase 3: Finding the Answer
Apply the Given Condition
\(\mathrm{C(B,2) \times C(R,1) = 2 \times C(R,2) \times C(B,1)}\)
Substituting:
\(\frac{\mathrm{B(B-1)}}{2} \times \mathrm{R} = 2 \times \frac{\mathrm{R(R-1)}}{2} \times \mathrm{B}\)
Simplifying:
\(\frac{\mathrm{B(B-1)R}}{2} = \mathrm{R(R-1)B}\)
Multiplying both sides by 2:
\(\mathrm{B(B-1)R = 2R(R-1)B}\)
Since B, R > 0, we can divide by BR:
\(\mathrm{B-1 = 2(R-1)}\)
Expanding:
\(\mathrm{B-1 = 2R-2}\)
\(\mathrm{B = 2R-1}\)
Interpreting the Result
\(\mathrm{B = 2R - 1}\) means:
- B is 1 less than 2R
- In other words: B is 1 less than the number that is twice R
Phase 4: Solution
Final Answer:
- Statement 1: "1 less than"
- Statement 2: "twice"
This makes the complete statement: "B is the number that is 1 less than the number that is twice R"
We can verify: If R = 3, then \(\mathrm{B = 2(3) - 1 = 5}\), and indeed 5 is 1 less than 6, which is twice 3.