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

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KJ Reservoir

Methodology

Fish Reproduction

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.

Ques. 1/3

Based on the information provided, which one of the following was the team's estimate of KJR's fish population?

1000

1200

1400

1500

2000

Solution

OWNING THE DATASET

Understanding Source A: Text Source - KJ Reservoir Management Information

Information from Dataset	Analysis
""When the KJ Reservoir (KJR) was built, a single, well-known species of fish was introduced for its role in the new ecosystem""	The reservoir was artificially created and stocked with just one type of fish Key finding: KJR is an artificial reservoir that contains only one fish species deliberately chosen for ecological purposes
""public fishing season that extends from June through August each year""	People can fish at the reservoir for 3 months during summer Key finding: Fish are removed from the population during this period
""Each May 1, the area's wildlife management team generates an estimate of the number of these fish""	Annual fish count happens on May 1st Key finding: Annual population assessment is timed just before fishing season starts
""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""	Minimum required: 1,500 fish before fishing starts If below 1,500, managers add more fish Key finding: 1,500 is the minimum threshold with active management through fish stocking to ensure adequate population for fishing season
""fish of this species reproduce once a year in mid-September""	Fish breed only once annually in September Key finding: Single annual breeding cycle that occurs after fishing season ends
""80% will survive until the following May""	20% of fish die between September and May Key finding: 20% mortality rate over ~7.5 months showing natural attrition between breeding and next assessment
""KJR cannot support a healthy, reproductive fish population greater than 2,000""	Maximum sustainable population is 2,000 fish Key finding: 2,000 is the carrying capacity - environmental limit to population size

Summary: KJR is an actively managed artificial reservoir with one fish species, where managers maintain populations between 1,500-2,000 fish through annual May assessments and stocking when needed.

Understanding Source B: Text Source - Population Estimation Methodology

Information from Dataset	Analysis
""capture a number (M) of individual animals of the species (the capture group) without doing harm. Each is marked, released, and allowed to disperse""	Researchers catch and mark M animals, then release them Non-harmful sampling method Key finding: M represents first sample size, marking allows identification in future captures
""A number (N) of individual animals are then captured in a second group (the recapture group)""	Second group of N animals caught later Key finding: Two-stage sampling process with time gap allowing population mixing
""some of which will have been captured in both groups, provided both the capture group and recapture group are sufficiently large""	Some animals will be caught both times Key finding: Method relies on recapturing some marked individuals, sample sizes must be adequate for reliability
""number of fish common to both groups/N = M/P""	Formula uses proportion of marked fish in second sample P = total population estimate Key finding: P = (M × N) ÷ (number common to both groups) Connection to Source A: This method is what the wildlife management team uses for their May 1 annual assessments

Summary: Source B describes the mark-recapture method for estimating populations, which is the same technique used by KJR's management team for their annual May 1 fish counts.

Understanding Source C: Mixed Source - Fish Reproduction Model and Data

Text Analysis

Information from Dataset	Analysis
""predict the annual increase in fish population in KJR due to reproduction""	Model calculates how many new fish are born each year Key finding: Model specifically for reproductive growth, doesn't account for mortality or fishing losses Connection to Source A: This model predicts the September reproduction mentioned in Source A
""P = fish population at the start of September C = maximum supported healthy fish population r = constant chosen to fit the model to the data""	P = population before breeding C = carrying capacity r = model constant Connection to Source A: C appears to be the 2,000 fish carrying capacity from Source A
""On May 1 last year, KJR's wildlife management team marked the 60 fish comprising the capture group. The recapture group had 100 fish, including exactly 5 from the capture group""	Real example: M = 60, N = 100, 5 marked fish recaptured Key calculation: Using the formula from Source B: P = (60 × 100) ÷ 5 = 1,200 fish Connection to Source B: Direct application of the mark-recapture method Connection to Source A: Since 1,200 < 1,500 minimum, fish were added before fishing season
""population increase = rP(1 - P/C)""	Mathematical formula for reproduction Key finding: Population increase = rP(1 - P/C) - Logistic growth model where growth decreases as P approaches C

Graph Analysis

Chart/Table Analysis:
- Shows relationship between fish population (x-axis) and population increase (y-axis)
- Key patterns observed: Inverted parabola peaking at P = 1,000 with maximum increase of 50 fish
- Zero growth occurs at P = 0 and P = 2,000
- Specific point marked: (400, 32) showing 32 fish increase when population is 400
- Key finding: Growth is highest at mid-range populations and decreases toward carrying capacity
- Connection to Source A: The zero growth at 2,000 confirms the carrying capacity limit; at the 1,500 minimum threshold, reproduction adds 37 fish
- Connection to Source B: The 1,200 fish calculated from last year's mark-recapture would produce 48 new fish according to this model

Summary: Source C provides the reproduction model and a real example showing last year's population was 1,200 fish (below minimum), demonstrating how all three sources work together in the complete management system.

Overall Summary

The KJR fish management system uses mark-recapture counting every May 1 to estimate population
Maintains a minimum of 1,500 fish through stocking
Manages reproduction using a model that peaks at 1,000 fish
With a carrying capacity of 2,000 fish and 80% annual survival rate
The reservoir operates in the upper half of the growth curve (1,500-2,000 range) where reproduction naturally slows, requiring active management to maintain adequate fishing stocks

Question Analysis

Understanding the Question

What the question is asking: What number of fish did the wildlife management team calculate when they did their population estimate using the mark-recapture method?
Key constraints:
- Must be the team's estimate (not actual population)
- Based on information provided in the sources
- Looking for a specific numerical result
Answer type needed: Numerical calculation

Connecting to Our Passage Analysis

The analysis directly contains the calculation result, which states 'Last year's May 1 population was about 1,200 fish (quick calculation: 60 × 100 ÷ 5)'
Can answer from analysis alone: YES - The analysis explicitly provides the calculation and result

Solution Analysis

Finding the Answer

Evaluating each numerical option against the mark-recapture calculation from the analysis
Our calculation: The team's estimate was calculated using the mark-recapture formula with M = 60, N = 100, and 5 fish common to both groups

Option Evaluations

Statement 1 Analysis

""1000""

What this suggests: Is the team's estimate 1000 fish?
Why this is wrong: The calculation yields 60 × 100 ÷ 5 = 1200, not 1000
Result: INCORRECT

Statement 2 Analysis

""1200""

What this suggests: Is the team's estimate 1200 fish?
Why this is correct: The calculation yields 60 × 100 ÷ 5 = 1200, which matches exactly
Result: CORRECT

Statement 3 Analysis

""1400""

What this suggests: Is the team's estimate 1400 fish?
Why this is wrong: The calculation yields 1200, not any of these higher values
Result: INCORRECT

Statement 4 Analysis

""1500""

What this suggests: Is the team's estimate 1500 fish?
Why this is wrong: The calculation yields 1200, not any of these higher values
Result: INCORRECT

Statement 5 Analysis

""2000""

What this suggests: Is the team's estimate 2000 fish?
Why this is wrong: The calculation yields 1200, not any of these higher values
Result: INCORRECT

Verification Check

Source C provides the specific numbers: M = 60 marked fish, N = 100 recaptured fish, 5 fish common to both groups
Source B provides the formula: common/N = M/P
Solving: 5/100 = 60/P ⟹ P = 60 × 100 ÷ 5 = 1200
The analysis confirms this calculation and identifies it as 'last year's May 1 population'
This was the team's official estimate using their standard methodology

Answer

1200

Answer Choices Explained

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