When the KJ Reservoir (KJR) was built, a single, well-known species of fish was introduced for its role in the...
GMAT Multi Source Reasoning : (MSR) Questions
When the KJ Reservoir (KJR) was built, a single, well-known species of fish was introduced for its role in the new ecosystem. KJR is a popular destination for a public fishing season that extends from June through August each year. Each May 1, the area's wildlife management team generates an estimate of the number of these fish in the reservoir. The team has determined that the fish's population should be no less than 1,500 fish at the start of the fishing season—less than 1,500, and the team acquires and releases enough fish to make up the difference.
The fish of this species reproduce once a year in mid-September. Due to environmental factors, of the species's total population at the end of September, 80% will survive until the following May. It is estimated that KJR cannot support a healthy, reproductive fish population greater than 2,000.
Based on the information provided, which one of the following was the team's estimate of KJR's fish population?
OWNING THE DATASET
Understanding Source A: Text Source - Wildlife Management Report for KJ Reservoir
| Information from Dataset | Analysis |
|---|---|
| "a single, well-known species of fish was introduced for its role in the new ecosystem" |
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| "public fishing season that extends from June through August each year" |
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| "Each May 1, the area's wildlife management team generates an estimate" |
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| "fish's population should be no less than 1,500 fish at the start of the fishing season" |
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| "less than 1,500, and the team acquires and releases enough fish to make up the difference" |
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| "reproduce once a year in mid-September" |
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| "80% will survive until the following May" |
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| "KJR cannot support a healthy, reproductive fish population greater than 2,000" |
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Summary: KJR manages a single fish species with annual assessments every May 1, maintaining the population between 1,500-2,000 fish through restocking when needed.
Understanding Source B: Text Source - Research Methodology Description
| Information from Dataset | Analysis |
|---|---|
| "capture a number (M) of individual animals...Each is marked, released" |
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| "A number (N) of individual animals are then captured in a second group" |
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| "some of which will have been captured in both groups, provided both...groups are sufficiently large" |
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| "number of fish common to both groups/N = M/P" |
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Summary: Source B describes the capture-recapture method for estimating animal populations, which is the technique used by KJR's wildlife team for their annual May 1 fish count.
Understanding Source C: Combined Source - Fish Reproduction Model and Data
Text Analysis:
| Information from Dataset | Analysis |
|---|---|
| "predict the annual increase in fish population in KJR due to reproduction" |
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| "fish population at the start of September (P)" |
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| "maximum supported healthy fish population (C)" |
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| "population increase = rP(1 - P/C)" |
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| "May 1 last year...marked the 60 fish comprising the capture group" |
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| "recapture group had 100 fish, including exactly 5 from the capture group" |
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Chart Analysis:
- Chart shows population increase vs. starting population
- Key patterns observed: Parabolic curve peaking at \(\mathrm{P} = 1,000\) with increase of 50 fish
- Inference: Maximum growth occurs at half the carrying capacity (1,000 is half of 2,000)
- Inference: Zero growth at P = 0 and P = 2,000 (confirms carrying capacity)
- Linkage to Source A: At minimum threshold (1,500), population increase is 37 fish - still healthy growth
- Linkage to Source A: Model validates that 1,500 minimum keeps population in sustainable growth zone
Summary: Source C provides both the mathematical model for predicting fish reproduction and real data showing how the capture-recapture method was used last year, with the chart confirming that KJR's management thresholds keep the fish population in a healthy growth range.
Overall Summary
The three sources reveal a complete fish management system at KJ Reservoir:
- Source A establishes the management framework (1,500-2,000 fish limits, May assessments, September breeding)
- Source B explains the capture-recapture method used for the May 1 population counts
- Source C shows how this method was applied (with 60 marked fish and 100 recaptured) and provides a reproduction model that validates the 1,500-fish minimum threshold
The logistic growth model demonstrates that maximum reproduction occurs at 1,000 fish (50% of capacity), confirming that the management strategy maintains the population well within sustainable limits.
Question Analysis
The question asks for the wildlife management team's calculated estimate of the total number of fish in KJR based on their capture-recapture study. This requires:
- Using the team's actual estimate from their data
- Applying the capture-recapture methodology
- Providing a numerical calculation result
Connecting to Our Analysis
Our analysis provides all necessary components for this calculation. Source C contains the specific capture-recapture data (60 marked fish, 100 recaptured fish, 5 found in common), while Source B provides the standard capture-recapture formula. These sources work together to enable a complete calculation, and the analysis confirms we can answer this question using the available data.
Extracting Relevant Findings
The capture-recapture method uses the fundamental relationship expressed in the formula: \(\frac{\mathrm{common}}{\mathrm{N}} = \frac{\mathrm{M}}{\mathrm{P}}\), where M represents marked fish (60), N represents recaptured fish (100), common represents fish found in both groups (5), and P represents the total population estimate we need to calculate.
Using our data: M = 60, N = 100, and 5 common fish. We can solve for P by rearranging the formula to: \(\mathrm{P} = \frac{\mathrm{M} \times \mathrm{N}}{\mathrm{common}} = \frac{60 \times 100}{5} = 1200\)
Individual Statement/Option Evaluations
Statement 1 Evaluation: 1000
- Testing the equation: \(\frac{5}{100} = \frac{60}{1000}\)
- Calculation shows: \(0.05 \neq 0.06\)
- The equation does not balance
- Result: INCORRECT
Statement 2 Evaluation: 1200
- Testing the equation: \(\frac{5}{100} = \frac{60}{1200}\)
- Calculation shows: \(0.05 = 0.05\)
- The equation balances perfectly
- Result: CORRECT
Statement 3 Evaluation: 1400
- Testing the equation: \(\frac{5}{100} = \frac{60}{1400}\)
- Calculation shows: \(0.05 \neq 0.043\)
- The equation does not balance
- Result: INCORRECT
Statement 4 Evaluation: 1500
- Testing the equation: \(\frac{5}{100} = \frac{60}{1500}\)
- Calculation shows: \(0.05 \neq 0.04\)
- The equation does not balance
- Result: INCORRECT
Statement 5 Evaluation: 2000
- Testing the equation: \(\frac{5}{100} = \frac{60}{2000}\)
- Calculation shows: \(0.05 \neq 0.03\)
- The equation does not balance
- Result: INCORRECT
Systematic Checking
Verification confirms our calculation and contextual understanding:
- Formula correctly applied: \(\frac{5}{100} = \frac{60}{\mathrm{P}}\) yields P = 1200
- This represents the May 1 population estimate from the previous year
- 1200 falls between the minimum threshold (1500) and maximum capacity (2000)
- This estimate would trigger restocking since it is below the 1500 minimum threshold
Final Answer
1200
1000
1200
1400
1500
2000