Two amateur wrestling squads—Squad A and Squad B—competed against each other. Each squad had the same (nonzero) number of its members who were contestants. In the competition, each contestant in Squads A and B wrestled against exactly one contestant in the other squad, and no non-contestant member of either squad wrestled. Let x be the proportion of Squad A's members who were contestants in the competition and let y be the proportion of Squad B's members who were contestants in the competition. In terms of x and y only, what is the proportion of the total number of members in the two squads who were contestants in the competition?

Two amateur wrestling squads—Squad A and Squad B—competed against each other. Each squad had the same (nonzero) number of its...

GMAT Multi Source Reasoning : (MSR) Questions

Source: Mock

Multi Source Reasoning

Conditions

MEDIUM

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Notes

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Problem Statement

Concise Solution

Table

Two amateur wrestling squads—Squad A and Squad B—competed against each other. Each squad had the same (nonzero) number of its members who were contestants. In the competition, each contestant in Squads A and B wrestled against exactly one contestant in the other squad, and no non-contestant member of either squad wrestled. Let \(\mathrm{x}\) be the proportion of Squad A's members who were contestants in the competition and let \(\mathrm{y}\) be the proportion of Squad B's members who were contestants in the competition. In terms of \(\mathrm{x}\) and \(\mathrm{y}\) only, what is the proportion of the total number of members in the two squads who were contestants in the competition?

Ques. 1/3

Which one of the following procedures, when applied to the last mathematical expression in the Concise Solution tab, gives a mathematical expression that answers the question asked in the Problem Statement tab?

Multiply both the numerator and denominator by a.

Multiply both the numerator and denominator by y.

Divide both the numerator and denominator by a.

Subtract a and then divide the result \(\mathrm{a + (x/y)a}\).

Divide by \(\mathrm{a + (x/y)a}\) and then subtract a from the result.

Solution

OWNING THE DATASET

Understanding Source A: Text - Problem Statement

Information from Dataset	Analysis
""Two amateur wrestling squads—Squad A and Squad B—competed against each other. Each squad had the same (nonzero) number of its members who were contestants.""	Two squads with equal numbers of contestants competing Not all squad members necessarily compete Inference: The equal contestant requirement is a key constraint that will shape the mathematical relationship
""In the competition, each contestant in Squads A and B wrestled against exactly one contestant in the other squad""	One-to-one matching between contestants Inference: This one-to-one matching forces equal contestant numbers from each squad
""no non-contestant member of either squad wrestled""	Clear distinction between contestants and non-contestants Inference: Only designated contestants participate in matches
""Let (mathrm{x}) be the proportion of Squad A's members who were contestants...let (mathrm{y}) be the proportion of Squad B's members who were contestants""	(mathrm{x}) and (mathrm{y}) are fractions between 0 and 1 Inference: These represent participation rates within each squad
""what is the proportion of the total number of members in the two squads who were contestants in the competition?""	Seeking a combined participation rate across both squads Inference: Answer should be expressed in terms of (mathrm{x}) and (mathrm{y}) only

Summary: Source A sets up a wrestling competition where equal numbers from two squads compete, and asks for a formula to find the overall participation rate in terms of individual squad participation rates (mathrm{x}) and (mathrm{y}).

Understanding Source B: Text - Concise Solution

Information from Dataset	Analysis
""Let (mathrm{a}) and (mathrm{b}) be the number of members of Squad A and Squad B, respectively, and let (mathrm{a'}) and (mathrm{b'}) be the number of contestants""	Introduces notation: (mathrm{a, b}) for total members; (mathrm{a', b'}) for contestants Inference: Distinguishes between membership and participation
""Clearly, (mathrm{a' = xa}) and (mathrm{b' = yb})""	Contestant counts equal total members times participation rates Inference: Direct application of proportion definitions Linkage to Source A: This mathematically expresses the (mathrm{x}) and (mathrm{y}) proportions defined in the problem
""Also, (mathrm{xa = yb})""	Critical equation showing equal contestant numbers Inference: This allows relating squad sizes: (mathrm{b = frac{x}{y}a}) Linkage to Source A: This equation captures the ""same number of contestants"" constraint from the problem statement
""the ratio of the total number of contestants...is (frac{mathrm{xa+yb}}{mathrm{a+b}})""	Formula for overall participation rate Inference: Numerator is total contestants, denominator is total members Linkage to Source A: This directly answers the question posed in the problem
""which is equal to (frac{mathrm{2xa}}{mathrm{a+(frac{x}{y})a}})""	Simplified form using the constraint (mathrm{xa = yb}) Inference: Answer expressed solely in terms of (mathrm{x, y}), and (mathrm{a}) (which cancels out)

Summary: Source B derives the mathematical solution showing that the overall participation rate equals (frac{mathrm{xa+yb}}{mathrm{a+b}}), which can be simplified using the equal-contestants constraint from Source A.

Understanding Source C: Table

What the table shows: A 5×5 grid displaying calculated values of (frac{mathrm{xa+yb}}{mathrm{a+b}}) for different combinations of (mathrm{x}) and (mathrm{y}) values
Key patterns observed:
- Diagonal entries (where (mathrm{x=y})) show values: (frac{1}{4}), (frac{1}{3}), (frac{1}{2}), (frac{2}{3}), (frac{3}{4})
- ""n/p"" entries appear at opposite corners: ((frac{1}{4}, frac{3}{4}))) and ((frac{3}{4}, frac{1}{4})))
- All values range between (frac{1}{4}) and (frac{3}{4})
Linkage insights:
- To Source A: The table provides practical answers to the problem's question for specific (mathrm{x}) and (mathrm{y}) values
- To Source B: The table contains pre-calculated results using the formula (frac{mathrm{xa+yb}}{mathrm{a+b}}), and the ""n/p"" entries occur where the constraint (mathrm{xa=yb}) cannot be satisfied with reasonable squad sizes
Key findings:
- When participation rates are equal ((mathrm{x=y})), the overall rate equals that same rate
- Certain (mathrm{x,y}) combinations marked ""n/p"" cannot satisfy the equal-contestants constraint with integer squad sizes

Summary: Source C provides a lookup table for the formula derived in Source B, showing that certain participation rate combinations are impossible due to the equal-contestants constraint from Source A.

Overall Summary

The wrestling competition problem requires equal numbers of contestants from each squad, creating the mathematical constraint (mathrm{xa = yb})
This constraint allows us to derive a formula for overall participation rate: (frac{mathrm{xa+yb}}{mathrm{a+b}})
The lookup table reveals:
- When squads have equal participation rates, the overall rate equals that same rate
- Certain combinations (like (frac{1}{4}) and (frac{3}{4})) are impossible because they can't satisfy the equal-contestants requirement with integer squad sizes
- The overall participation rate is always bounded by the individual squad rates

Question Analysis

Goal: Find which mathematical operation transforms the expression (frac{2xa}{a+frac{x}{y}a}) into an expression involving only x and y variables, eliminating 'a'
Key constraints:
- Must eliminate variable 'a'
- Result must be expressed only in terms of x and y
- Result should represent the proportion of contestants
Answer type needed: Identification of the correct algebraic manipulation procedure

Connecting to Our Passage Analysis

Analyzing each proposed procedure to determine whether it simplifies the expression to one involving only x and y, without 'a'
Can be answered from analysis alone through algebraic manipulation of the given expression

Statement Evaluations

Statement 1 Analysis

""Multiply both the numerator and denominator by a.""

Calculation: (frac{2xa cdot a}{(a + frac{x}{y}a) cdot a} = frac{2xa^2}{a^2 + frac{x}{y}a^2})
Result: Still contains 'a' and fails to eliminate the variable
Conclusion: INCORRECT

Statement 2 Analysis

""Multiply both the numerator and denominator by y.""

Calculation: (frac{2xa cdot y}{(a + frac{x}{y}a) cdot y} = frac{2xay}{ay + xa} = frac{2xy}{x + y})
Result: Successfully removes 'a' from expression
Conclusion: CORRECT

Statement 3 Analysis

""Divide both the numerator and denominator by a.""

Calculation: (frac{2xa / a}{(a + frac{x}{y}a) / a} = frac{2x}{1 + frac{x}{y}} = frac{2xy}{x + y})
Result: Successfully removes 'a' from expression
Conclusion: CORRECT

Statement 4 Analysis

""Subtract a and then divide the result a + (x/y)a.""

This statement is unclear and does not represent a valid algebraic operation on the given expression
Conclusion: INCORRECT

Statement 5 Analysis

""Divide by a + (x/y)a and then subtract a from the result.""

This statement describes a sequence of operations that would not eliminate 'a' from the expression
Conclusion: INCORRECT

Verification

Options 2 and 3 both simplify the expression to (frac{2xy}{x + y}), eliminating 'a'
Option 1 keeps 'a' in the expression and therefore fails the requirement
Option 3 (division by 'a') is the most straightforward and standard method to remove the variable

Answer

Option 3

Answer Choices Explained

Multiply both the numerator and denominator by a.

Multiply both the numerator and denominator by y.

Divide both the numerator and denominator by a.

Subtract a and then divide the result \(\mathrm{a + (x/y)a}\).

Divide by \(\mathrm{a + (x/y)a}\) and then subtract a from the result.

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