Beat the McKinsey Game In Just 3 Hours
89% Success – Or It’s FREE*

Case 1

A research team is planning to plant new banana trees on an island as part of a repopulation plan. Below is a table showing the number of banana trees for the different areas of the Island.

Question:
For which year and which region there was the highest percentage increase in banana trees?

Answer:
Y13-14 in Kaya Forest

—

Explanation

The increase in each area and year is reported below:

Sunrise Hill:

• Y11-Y12: (63-60)/60 = 3/60
• Y12-Y13: (68-63)/63 = 5/63
• Y13-Y14: (71-68)/68 = 3/68

Kaya Forest:

• Y11-Y12: (49-45)/45 = 4/45
• Y12-Y13: (53-49)/49 = 4/49
• Y13-Y14: (59-53)/53 = 6/53

Blue Lagoon:

• Y11-Y12: (55-50)/50 = 5/50
• Y12-Y13: (59-55)/55 = 4/55
• Y13-Y14: (65-59)/59 = 6/59

North Beach:

• Y11-Y12: (49-45)/45 = 4/45
• Y12-Y13: (54-49)/49 = 5/49
• Y13-Y14: (60-54)/54 = 6/54

Note that we don’t need to do all the calculations, as if a fraction has

1. The same numerator and greater denominator than another or
2. Lower numerator and the same denominator than another or
3. Lower numerator and greater denominator than another

it cannot be the solution. Therefore, we can eliminate the following (the reference used to determine whether the fraction is smaller could be different from what is reported)

Sunrise Hill:

• Y11-Y12: (63-60)/60 = 3/60 (smaller than Kaya Forest #1)
• Y12-Y13: (68-63)/63 = 5/63 (smaller than North Beach #2)
• Y13-Y14: (71-68)/68 = 3/68 (smaller than Sunrise Hill #1)

Kaya Forest:

• Y11-Y12: (49-45)/45 = 4/45
• Y12-Y13: (53-49)/49 = 4/49 (smaller than Kaya Forest #1)
• Y13-Y14: (59-53)/53 = 6/53

Blue Lagoon:

• Y11-Y12: (55-50)/50 = 5/50 (smaller than North Beach #2)
• Y12-Y13: (59-55)/55 = 4/55 (smaller than Blue Lagoon #1)
• Y13-Y14: (65-59)/59 = 6/59 (smaller than Kaya Forest #3)

North Beach:

• Y11-Y12: (49-45)/45 = 4/45
• Y12-Y13: (54-49)/49 = 5/49
• Y13-Y14: (60-54)/54 = 6/54 (smaller than Kaya Forest #3)

If we now calculate the remaining, we get

• 4/45= 8.9%
• 5/49 = 10.2%
• 6/53 = 11.32%

Thus, Kaya Forest Y13-14 is the one with highest growth

Case 2

Researchers have analyzed the main source of food of Red Monkeys in Kaya Forest for different areas.

Question:
Which is the fruit with the highest median and highest mean consumption?

Answer:
Mean: Drupes
Median: Figs

—

Explanation

The mean is average in a series. Therefore, the mean for the different fruits is the following:

• Bananas: (4+3+2+5+5)/5 =3.8
• Mangos: (2+5+2+3+4)/5 = 3.2
• Figs: (1+1+5+6+7)/5 = 4
• Drupes: (7+5+4+2+3)/5 = 4.2

The median is middle value in a series. Therefore, the median for the different fruits is the following

• Bananas: 2,3,4,4,5 -> 4
• Mangos: 2,2,3,4,5 -> 3
• Figs: 1,1,5,6,7 -> 5
• Drupes: 2,3,4,5,7 -> 4

Case 3

On the Maruvi Plains, Grass Gazelles and Striped Antelopes both graze on Wild Grass. In Year 5, a drought sharply reduced Wild Grass, forcing both species into direct competition. Rainfall recovered in Year 8, leading to regrowth of Wild Grass.

Question:
What could be the best graph type to visualize how drought conditions affected competition between the two species and their recovery afterward?

Answer:
Line graph of Grass Gazelle and Striped Antelope populations, and bar graph of annual rainfall levels.

—

Explanation

• Bar graph of total grazing events by both species, and line graph of grassland size –> while line graph of grassland size can show the variation in Wild Grass, total grazing events do not directly measure population impact
• Stacked area graph of the combined population of both species, and bar graph of rainfall –> stacked area graph can show how Gazelle and Antelope populations change relative to each other, but makes it difficult to determine the absolute number of each population at any point in time
• Line graph of Wild Grass coverage and bar graph of Grass Gazelle population –> Striped Antelopes are not represented in this graph at all
• Line graph of Grass Gazelle and Striped Antelope populations, and bar graph of annual rainfall levels –> Line graphs are perfect to clearly visualize population changes over time, while the bar graph allows us to easily understand the drought intensity/ recovery at any point of time

Case 4

The research team has identified the following data in South Beach:

Question:
Which information is true based on the data provided? You can select multiple options.

Answer:
A lower number of crocodiles is correlated with an above-average number of frogs the following year

—

Explanation

• A lower number of crocodiles is correlated with an above-average number of frogs the following year –> The data show that sentence #1 is correct, as when crocodiles decrease (Y12), the following year frogs grow more than usual (+10 instead of +5 in Y13).
• Each crocodile eats exactly 5 frogs –> There are no data sustaining sentence #2, thus we cannot say that sentence #2 is true based on the data provided.
• There is no correlation between Phytons’ and Crocodiles’ evolution –> There is an inverse correlation between Phytons and Crocodiles, thus we cannot say that sentence #3 is true based on the data provided.
• Phytons eat both crocodiles and frogs –> There are no data sustaining sentence #4, thus we cannot say that sentence #4 is true based on the data provided.

Case 5

In Evermist Forest, rangers track the average body mass of three species of deer: Red Deer, Roe Deer, and Fallow Deer.

• The average weight of a Red Deer is 220 kg, and there are 40 of them.
• The average weight of a Roe Deer is half that of a Red Deer, and there are 50% more Roe Deer than Red Deer.
• The average weight of a Fallow Deer is 25% lighter than a Roe Deer, and there are 20 more Fallow Deer than Red Deer.

Question:
What is the average weight of a deer in the Evermist Forest?

Answer:
127.19kg

—

Explanation

Weight of each species:

Red Deer = 220kg

Roe Deer = half of Red Deer = 220 * 0.5 = 110kg

Fallow Deer = 25% lighter than Roe Deer = 110 * (1-0.25) = 110 * 0.75 = 82.5kg

Population counts:

Red Deer = 40

Roe Deer = 50% more than Red Deer = 40 * 1.5 = 60

Fallow Deer = 20 more than Red Deer = 40 + 20 = 60

Total weights:

Red Deer = 40 * 220 = 8,800kg

Roe Deer = 60 * 110 = 6,600kg

Fallow Deer = 60 * 82.5kg = 4,950kg

Total weight = 8,800 + 6,600 + 4,950 = 20,350kg

Total deer = 40 + 60 + 60 = 160

Average weight = 20,350 / 160 = 127.19kg

Case 6

In February, Snow Owls were relocated to the Fjordwood Range as part of conservation efforts. The Owls came from Norvant, Gravik, and Eastreach. The table below shows how much each region in Norvant contributed in terms of Snow Owls.

The relocated Owls from Gravik were equal to the total from Norvant, while those from Eastreach were double the number from Gravik.

Question:
Post-relocation, what is the probability that an Owl, chosen at random, is from Hrymin, Salvek, or Lorka?

Answer:
13.39%

—

Explanation

Total owls from Norvant = 28 + 31 + 19 + 25 + 37 = 140

Total owls from Gravik = 140 (same as Norvant)

Total owls from Eastreach = double of Gravik = 2 * 140 = 280

Total owls from Norvant, Gravik, and Eastreach = 140 + 140 + 280 = 560

Total owls from Hrymin, Salvek, and Lorka = 31 + 19 + 25 = 75

Probability of owls from Hrymin, Salvek, or Lorka = 75 / 560 = 0.1339 = 13.39%

Case 7

Conservationists recorded the number of Loggerhead turtle hatchlings emerging across 4 beaches in a year.

Question:
Which statement(s) are supported by the data? You can select multiple options.

Answer:

• Shell Beach has the most suitable environment for turtles laying eggs.
• Loggerhead turtle hatchlings are less likely to emerge in cold weather.

—

Explanation

• Shell Beach has the most suitable environment for turtles laying eggs –> Supported; Shell Beach consistently record the highest hatchling numbers across all seasons.
• Loggerhead turtle hatchlings are less likely to emerge in cold weather –> Supported; Winter has the lowest hatchlings emerging across all beaches
• Coral Beach is the primary nesting ground of Loggerheads –> Not supported; Shell Beach is actually the most popular out of the 4 beaches

Case 8

Scientists have been tracking the daily calorie intake of polar bears in the Frostpeak region over the past 4 years. The following shows their collected data:

Question:
Which formula best describes the relationship shown in the data?

Answer:
C = (5,000 * s * z) / i

—

Explanation

By adding the numbers in the equation, we get the following:

Y1 = (5,000 * 1.0 * 1.0) / 1.0 = 5,000

Y2 = (5,000 * 1.4 * 1.2) / 2.8 = 3,000

Y3 = (5,000 * 1.2 * 1.1) / 2.0 = 3,300

Y4 = (5,000 * 0.8 * 0.8) / 1.6 = 2,000

Thus, the 4^th formula is correct.

Case 9

Researchers in the Sareth Highlands have been studying a group of 60 Mountain Hares. These Hares have been tagged and tested beforehand, with 12 of them carrying a specific parasite. The researchers decided to randomly choose 2 hares for testing (without replacement).

Question:
Assuming no new Hares were infected with the parasite, what is the probability that at least one of the two hares tested carries the parasite?

Answer:
36.27%

—

Explanation

The question asks for the probability that at least one of the two hares carries the parasite; this means it could be either:

1. Hare 1 has the parasite but Hare 2 is parasite-free
2. Hare 1 is parasite-free but Hare 2 has the parasite
3. Hares 1 and 2 both have the parasite

It is not necessary to calculate the probability of all 3 and sum them up; instead, we can simply find the probability that both Hares are parasite-free and deduct this probability from 100%.

Therefore:

Heathy Hares = 60 – 12 = 48

Probability that the first pick is parasite-free = 48/60

If the first Hare is parasite-free, we are now left with 47 healthy Hares and 59 Hares in total; therefore, probability of the second Hare being parasite-free is now 47/59

Probability of both being parasite-free = 48/60 * 47/59 = 0.6373 = 63.73%

Probability of AT LEAST ONE hare being infected = 1 – 0.6373 = 0.3627 = 36.27%

Case 10

Environmental scientists are studying river trout populations across the Aelwin River system.

If a sampling site is located downstream of an industrial zone, there is a 30% chance that mercury levels will exceed the safe limit. When this happens, about 8% of the trout in that section are contaminated, though contamination can never affect more than 400 trout in a section.

Sites that are upstream of industrial zones face no contamination risk.

You are given a monitoring report with the following additional variables:

• Sampling site coordinates
• Section fish population
• Date of sampling
• Water temperature at site
• Industrial zone coordinates
• Industrial activity rating

Question:
Which variables are needed to calculate the total number of contaminated trout across sites?

Answer:

• 30% chance of mercury levels exceeding safe limit when downstream
• 8% of trout contaminated when mercury levels exceed safe limit
• Contamination cannot affect more than 400 trout
• Sampling site coordinates
• Section fish population
• Industrial zone coordinates

—

Explanation

The total number of contaminated trout = sum of contaminated trout in each section

If upstream of an industrial zone, contaminated trout in section = 0

If downstream of an industrial zone, contaminated trout in section = population in section * % chance that mercury levels exceed safe limit * % in section contaminated if mercury levels exceed limit

With the above formula in mind, we can determine what variables we need:

1. Sampling site and industrial zone coordinates – determines whether a site is downstream or upstream
2. Section fish population – as highlighted in the formula above
3. 30% chance mercury levels exceed limit – as highlighted in the formula above
4. 8% contaminated if mercury levels exceed limit – as highlighted in the formula above
5. 400 trout limit – acts as a cap on the effected trout be site

The remaining factors were not stated to impact contaminated trout populations.

Case 11

In the cold North Frigia Sea, Krill populations support both Penguins and Sea Lions. Krills are very small shrimp-like animals that are very numerous (up to 30,000 animals per cubic meters in some locations). In Year 3, warming seas started reducing krill abundance. By Year 8, a marine reserve was established to protect krill.

Question:
What is the best way to visualize the relationship between krill abundance and predator populations over time?

Answer:
Dual-axis line graph with krill on one axis and combined predator population on the other.

—

Explanation

• Line graph of krill, Penguin, and Sea Lion populations all on the same axis –> Krill greatly outnumbers Penguins and Sea Lions so using the same axis would make the graph less readable
• Scatter plot of predator population vs. krill abundance –> Scatterplots remove the time dimension altogether and therefore cannot show trends over time
• Dual-axis line graph with krill on one axis and predator population on the other –> Allows direct comparison when there are different scales (krill vs. predator populations)
• Stacked bar graph of Penguin vs. Sea Lion populations, and line graph of krill –> Stacked bar graph can work but is not optimal as it makes individual species trends harder to read

Case 12

The following table shows the results of a study on Savannah Elephants in Africa:

Question:
Which formula best describes the relationship shown in the data?

Answer:
P = P_n-1 + 50d – 50(p-1) + 50(v-1)

—

Explanation

Note: P_n-1 denotes the previous year’s population

By adding the numbers in the equation, we get the following:

Y2 = 500 + 50*1.0 – 50*(1-1) + 50*(1.2-1) = 500 + 50 – 0 + 10 = 560

Y3 = 560 + 50*1.2 – 50*(1.1-1) + 50*(1.5-1) = 560 + 60 – 5 + 25 = 640

Y4 = 640 + 50*0.8 – 50(1.5-1) + 50*(1.5-1) = 640 + 40 – 25 + 25 = 680

Thus, the 4^th formula is correct.