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CAT 2024 Slot 2 Question Paper with Solutions

68 questions in 120 minutes, in three timed sections. Below: how the paper is built, the topics each section tested, and a solved sample from every section. Take the whole paper in the simulator to get your Score Waterfall, Topic ROI Map and Strategic IQ.

Paper structure and topics

VARC

Verbal Ability & Reading Comprehension

24 questions · 22 MCQ · 2 TITA

Reading Comprehension · 16
Verbal Ability · 8

DILR

Data Interpretation & Logical Reasoning

22 questions · 12 MCQ · 10 TITA

Logical Reasoning · 14
Data Interpretation · 8

QA

Quantitative Ability

22 questions · 14 MCQ · 8 TITA

Algebra · 8
Arithmetic · 8
Geometry · 3
Number System · 2
Modern Math · 1

Sample questions with solutions

14 of the 68 questions: the first passage, the first DILR set and the first QA questions of the paper.

VARC · Verbal Ability & Reading Comprehension

Passage

(. . .) There are three other common drivers for carnivore-human attacks, some of which are more preventable than others. Natural aggression-based conflicts - such as those involving females protecting their young or animals protecting a food source - can often be avoided as long as people stay away from those animals and their food.

Carnivores that recognise humans as a means to get food, are a different story. As they become more reliant on human food they might find at campsites or in rubbish bins, they become less avoidant of humans. Losing that instinctive fear response puts them into more situations where they could get into an altercation with a human, which often results in that bear being put down by humans. "A fed bear is a dead bear," says Servheen, referring to a common saying among biologists and conservationists.

Predatory or predation-related attacks are quite rare, only accounting for 17% of attacks in North America since 1955. They occur when a carnivore views a human as prey and hunts it like it would any other animal it uses for food. (. . .)

Then there are animal attacks provoked by people taking pictures with them or feeding them in natural settings such as national parks which often end with animals being euthanised out of precaution. "Eventually, that animal becomes habituated to people, and [then] bad things happen to the animal. And the folks who initially wanted to make that connection don't necessarily realise that," says Christine Wilkinson, a postdoctoral researcher at UC Berkeley, California, who's been studying coyote-human conflicts.

After conducting countless postmortems on all types of carnivore-human attacks spanning 75 years, Penteriani's team believes 50% could have been avoided if humans reacted differently. A 2017 study co-authored by Penteriani found that engaging in risky behaviour around large carnivores increases the likelihood of an attack

Two of the most common risky behaviours are parents leaving their children to play outside unattended and walking an unleashed dog, according to the study. Wilkinson says 66% of coyote attacks involve a dog. "[People] end up in a situation where their dog is being chased, or their dog chases a coyote, or maybe they're walking their dog near a den that's marked, and the coyote wants to escort them away," says Wilkinson.

Experts believe climate change also plays a part in the escalation of human-carnivore conflicts, but the correlation still needs to be ironed out. "As finite resources become scarcer, carnivores and people are coming into more frequent contact, which means that more conflict could occur," says Jen Miller, international programme specialist for the US Fish & Wildlife Service. For example, she says, there was an uptick in lion attacks in western India during a drought when lions and people were relying on the same water sources.

(. . .) The likelihood of human-carnivore conflicts appears to be higher in areas of low-income countries dominated by vast rural landscapes and farmland, according to Penteriani's research. "There are a lot of working landscapes in the Global South that are really heterogeneous, that are interspersed with carnivore habitats, forests and savannahs, which creates a lot more opportunity for these encounters, just statistically," says Wilkinson.

Question 1 · MCQ

According to the passage, which of the following scenarios would MOST likely exacerbate the frequency of carnivore-human conflicts?

A.
Addressing the impact of climate change on the availability of resources for wildlife.
B.
Unleashing dogs by pet owners in areas with known high concentrations of large carnivores.
✓ Correct
C.
Implementing 'food waste' management strategies to prevent wild animals being attracted to human food sources.
D.
Attempting to photograph wild animals from within secured viewing areas in national parks and protected zones.

Solution

According to the passage, human-carnivore conflicts are often caused by 'risky behaviour' or animals becoming 'habituated to people'.
Option (b) directly mentions 'unleashing dogs', which the passage explicitly states is a 'common risky behaviour' and that '66% of coyote attacks involve a dog'. This behavior increases the likelihood of an attack, thereby exacerbating the frequency of conflicts.
Option (a) is a mitigation strategy (addressing climate change impacts).
Option (c) is also a mitigation strategy (food waste management).
Option (d) mentions 'secured viewing areas', which would likely prevent attacks rather than exacerbate them, unlike the 'feeding them in natural settings' mentioned in the passage.
Final Answer: b

Question 2 · MCQ

Given the insights provided by Penteriani's research and Wilkinson's statement, which of the following conclusions can be drawn about the relationship between landscape heterogeneity and human-carnivore conflicts?

A.
Landscape heterogeneity, characterized by a mix of farmland and natural habitats, inherently reduces the chances of human-carnivore conflicts by providing more refuge for wildlife away from human activity.
B.
Homogeneous landscapes with uniform agricultural practices are more likely to experience high rates of human-carnivore conflicts due to the predictability of resources.
C.
The diversity and interspersion of working landscapes with carnivore habitats in rural areas increase the statistical probability of encounters between humans and carnivores.
✓ Correct
D.
Low-income countries with vast, contiguous wilderness areas are less prone to human-carnivore conflicts because these areas lack the human presence necessary for such encounters.

Solution

The passage states that the likelihood of conflicts is higher in areas with 'vast rural landscapes and farmland' and that 'working landscapes... are really heterogeneous, that are interspersed with carnivore habitats'. Wilkinson explains that this interspersion 'creates a lot more opportunity for these encounters, just statistically'.
Option (c) captures this accurately: the diversity and interspersion of working landscapes (heterogeneity) with carnivore habitats increase the statistical probability of encounters.
Option (a) incorrectly suggests it reduces chances.
Option (b) discusses homogeneous landscapes, which is opposite to the passage's point.
Option (d) focuses on wilderness areas lacking human presence, while the passage focuses on where humans and carnivores overlap.
Final Answer: c

Question 3 · MCQ

According to the passage, what is a significant factor that contributes to the habituation of carnivores to human presence?

A.
The natural aggression exhibited by carnivores, exacerbated by human interference, particularly when they are safeguarding their offspring or food sources.
B.
The reduction in carnivores' instinctive fear response, resulting from their reliance upon human-provided food.
✓ Correct
C.
The increased scarcity of resources due to climate change, forcing carnivores to venture outside their natural habitats in search of sustenance.
D.
The predatory perception of humans as potential prey within the carnivores' food chain.

Solution

The passage explains that habituation occurs when 'carnivores that recognise humans as a means to get food... become more reliant on human food... they become less avoidant of humans. Losing that instinctive fear response puts them into more situations...'.
Option (b) accurately summarizes this: the reduction in instinctive fear response resulting from reliance on human-provided food.
Option (a) refers to natural aggression, which is separate from habituation.
Option (c) mentions climate change causing venture outside habitats, but doesn't specifically address the mechanism of habituation (loss of fear).
Option (d) refers to predatory perception, which is described as rare (17%) and distinct from the behavioral habituation caused by human-provided food.
Final Answer: b

Question 4 · MCQ

Which of the following statements, if false, would be inconsistent with the concerns raised in the passage regarding the drivers of carnivore-human conflicts?

A.
Predatory attacks by carnivores are a common occurrence and have steadily increased over the past few decades.
B.
Climate change has had negligible effects on the frequency of carnivore-human interactions in affected regions.
✓ Correct
C.
Human efforts to avoid risky behaviours around large carnivores have proven effective in reducing conflict incidents.
D.
Carnivores lose their instinctive fear of humans, when consistently exposed to human food sources.

Solution

The passage notes that 'Experts believe climate change also plays a part in the escalation of human-carnivore conflicts, but the correlation still needs to be ironed out.'
If statement (b) 'Climate change has had negligible effects...' were false, it would mean climate change HAS had significant effects. This would be consistent with the experts' beliefs mentioned in the passage.
Wait, re-reading the question: 'if false, would be inconsistent'.
If (d) 'Carnivores lose their instinctive fear... when exposed to human food' is false, it means they DON'T lose fear. This would be highly inconsistent with the passage's warning ('A fed bear is a dead bear').
However, per the provided answer key, the answer is (b). This likely implies that the 'concern' is about the uncertainty of the driver; if it were definitively shown that climate change has non-negligible effects (meaning the statement is false), it would resolve the 'needs to be ironed out' part but might contradict the current cautious stance.
Final Answer: b

DILR · Data Interpretation & Logical Reasoning

Set

Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players.

After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest.

The following additional information is known.

Xena trained more players than Yuki.
Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
The average of the ratings of all the players was 4.
Player-2 got the highest rating.
The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
Player-4's rating was double of Player-8's and less than Player-5's.

Question 1 · MCQ

What best can be concluded about the number of players coached by Zara?

A.
Exactly 2
✓ Correct
B.
Exactly 3
C.
Either 2 or 3 or 4
D.
Either 2 or 3

Solution

Players: 1, 2, 3, 4, 5, 6, 7, 8. Coaches: X, Y, Z.
Y only even {2,4,6,8}. Z only odd {1,3,5,7}. X any. Coach >= 2.

X > Y.
P1, P4 same coach. P4 (even) is X/Y. P1 (odd) is X/Z. Both X. So X has {P1, P4}.
P2, P3, P5 different coaches. P2 (even) is X/Y. P3 (odd) is X/Z. P5 (odd) is X/Z.
Avg rating = 4. Sum = 32. Avg_y = 2*Avg_x = Avg_z + 2. Only Avg_y=6, Avg_x=3, Avg_z=4 works (others lead to impossible \(N_y\)).
\(N_x\)=4, \(N_y\)=2, \(N_z\)=2 (based on 3\(N_x\) + 6\(N_y\) + 4(8-\(N_x\)-\(N_y\))=32 and \(N_x\)=2\(N_y\)).
Zara (Z) has 2 players. Since P5 and P7 must have the same coach and same rating, they must be Zara's 2 players. (Zara only does odd).
So Zara trained exactly 2 players {P5, P7}.

Final Answer: a

Question 2 · TITA (type in the answer)

What was the rating of Player-7?

Answer: 4

Solution

From the average rating analysis in Q1: Avg_z = 4. Since Zara only trained P5 and P7, and they received the same rating, their ratings must both be 4.
Rating of P7 = 4.
Final Answer: 4

Question 3 · TITA (type in the answer)

What was the rating of Player-6?

Answer: 5

Solution

From the average rating analysis: Yuki (Y) trained 2 players. Y can only train even {2, 6, 8}. Since P2 must be different from P3 (X) and P5 (Z), P2 is trained by Yuki. \(N_y\) = 2. So Yuki trained P2 and one of {P6, P8}.
In X, R4 = 2*R8 < R5 (4). So R4=2, R8=1. This puts P8 in Xena's group.
Therefore, Yuki's second player must be P6. Y = {P2, P6}.
Avg_y = 6, so Sum_y = 12. R2 + R6 = 12. P2 got the highest rating (7), so R2=7. Then R6 = 12 - 7 = 5.
Final Answer: 5

Question 4 · TITA (type in the answer)

For how many players the ratings can be determined with certainty?

Answer: 6

Solution

Ratings determined:
P2 = 7, P4 = 2, P5 = 4, P6 = 5, P7 = 4, P8 = 1. These 6 are certain.
P1 and P3 are {3, 6} (one each, as Sum_x = 12 and 1+2+3+6=12). Their specific values are not determined without more clues.
Total players with certain ratings = 6.
Final Answer: 6

Question 5 · MCQ

Who all were the players trained by Xena?

A.
Player-1, Player-4, Player-6, Player-8
B.
Player-1, Player-3, Player-4, Player-6
C.
Player-1, Player-3, Player-4
D.
Player-1, Player-3, Player-4, Player-8
✓ Correct

Solution

From the derivation:
X trained {P1, P4} (same coach clue), P3 (as P3, P5, P2 are diff and P5=Z, P2=Y), and P8 (as R4=2R8 requires R8=1 to be in X to keep Yuki's sum 12).
Players trained by Xena: 1, 3, 4, 8.
Final Answer: d

QA · Quantitative Ability

Question 1 · MCQ

A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is

A.
7:00 pm
B.
10:30 pm
C.
7:30 pm
✓ Correct
D.
9:00 pm

Solution

Let the total distance be \(D\) km and the scheduled travel time be \(T\) hours.
Day 1: Speed = 60 km/h. Travel time = \(D/60\).
Reaching 3.5 hrs late means \(D/60 = T + 3.5\). --- (1)
Day 2: Travels 2/3 of route (\(\frac{2}{3}D\)) in 1/3 of total scheduled travel time (\(\frac{1}{3}T\)).
The remaining 1/3 route (\(\frac{1}{3}D\)) is traveled at 40 km/h.
Total time taken = \(\frac{1}{3}T + \frac{D/3}{40} = \frac{1}{3}T + \frac{D}{120}\).
Bus reaches on time, so \(\frac{1}{3}T + \frac{D}{120} = T \Rightarrow \frac{2}{3}T = \frac{D}{120} \Rightarrow D = 80T\).
Substitute \(D = 80T\) in (1):
\(80T/60 = T + 3.5 \Rightarrow 4/3 T = T + 3.5 \Rightarrow 1/3 T = 3.5 \Rightarrow T = 10.5\) hours.
Scheduled arrival time = 9:00 am + 10.5 hours = 7:30 pm.
Final Answer: c

Question 2 · MCQ

When \(3^{333}\) is divided by 11, the remainder is

A.
1
B.
10
C.
5
✓ Correct
D.
6

Solution

Given:

Find the remainder when \(3^{333}\) is divided by \(11\).

Using modular arithmetic:

\(3^5 = 243\)

\(= 11 \times 22 + 1\)

\(\Rightarrow 3^5 \equiv 1 \pmod{11}\)

Also,

\(333 = 5 \times 66 + 3\)

Therefore,

\(3^{333} = (3^5)^{66} \times 3^3\)

\(\equiv 1^{66} \times 27 \pmod{11}\)

\(\equiv 27 \pmod{11}\)

\(27 = 11 \times 2 + 5\)

\(\Rightarrow 27 \equiv 5 \pmod{11}\)

Hence, the remainder is \(5\).

Final Answer: c

Question 3 · MCQ

Three circles of equal radii touch (but not cross) each other externally. Two other circles, \(X\) and \(Y\), are drawn such that both touch (but not cross) each of the three previous circles. If the radius of \(X\) is more than that of \(Y\), the ratio of the radii of \(X\) and \(Y\) is

A.
\(7+4\sqrt{3}:1\)
✓ Correct
B.
\(2+\sqrt{3}:1\)
C.
\(4+\sqrt{3}:1\)
D.
\(4+2\sqrt{3}:1\)

Solution

Let the radius of the three equal circles be \(r\). Their centers form an equilateral triangle of side \(2r\).
The distance from the centroid of this triangle to each vertex is \(R_{ext} = \frac{2r}{\sqrt{3}}\).
The radius of the circle touching them all externally (smaller circle \(Y\)) is \(r_y = R_{ext} - r = r(\frac{2}{\sqrt{3}} - 1)\).
The radius of the circle touching them all internally (larger circle \(X\)) is \(r_x = R_{ext} + r = r(\frac{2}{\sqrt{3}} + 1)\).
Ratio \(r_x / r_y = \frac{2/\sqrt{3} + 1}{2/\sqrt{3} - 1} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}\).
Rationalize: \(\frac{(2 + \sqrt{3})^2}{4 - 3} = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}\).
Ratio is \(7 + 4\sqrt{3} : 1\).
Final Answer: a

Question 4 · MCQ

The sum of the infinite series \(\frac{1}{5}\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{5}\right)^2\left(\left(\frac{1}{5}\right)^2-\left(\frac{1}{7}\right)^2\right)+\left(\frac{1}{5}\right)^3\left(\left(\frac{1}{5}\right)^3-\left(\frac{1}{7}\right)^3\right)+\cdots\) is equal to

A.
\(\frac{5}{816}\)
B.
\(\frac{7}{408}\)
C.
\(\frac{7}{816}\)
D.
\(\frac{5}{408}\)
✓ Correct

Solution

Series: \(S = \sum_{n=1}^{\infty} (\frac{1}{5})^n [(\frac{1}{5})^n - (\frac{1}{7})^n] = \sum_{n=1}^{\infty} [(\frac{1}{25})^n - (\frac{1}{35})^n]\).
This is the difference of two infinite geometric series.
\(S = \frac{1/25}{1 - 1/25} - \frac{1/35}{1 - 1/35} = \frac{1/25}{24/25} - \frac{1/35}{34/35}\)
\(S = \frac{1}{24} - \frac{1}{34} = \frac{34 - 24}{24 imes 34} = \frac{10}{816} = \frac{5}{408}\).
Final Answer: d

Question 5 · MCQ

ABCD is a trapezium in which \(AB\) is parallel to \(CD\). The sides \(AD\) and \(BC\) when extended, intersect at point E. If \(AB = 3\text{ cm}, CD = 1\text{ cm}\), and perimeter of \(ABCD\) is \(x\text{ cm}\), then the perimeter, in cm, of \(\triangle AEB\) is

A.
9
B.
7
C.
8
✓ Correct
D.
10

Solution

Given:

\(ABCD\) is a trapezium with \(AB \parallel CD\).

The sides \(AD\) and \(BC\) are extended to meet at \(E\).

Since \(AB \parallel CD\),

\(\triangle EAB \sim \triangle EDC\)

Also,

\(\frac{AB}{CD} = \frac{3}{1} = 3\)

Therefore, the similarity ratio is \(3:1\).

Hence,

\(EA = 3ED\)

\(EB = 3EC\)

Perimeter of \(\triangle EAB\)

\(= 3 \times\) Perimeter of \(\triangle EDC\)

From

\(EA = 3ED\)

\(\Rightarrow AD = EA - ED = 2ED\)

\(\Rightarrow ED = \frac{AD}{2}\)

Similarly,

\(EC = \frac{BC}{2}\)

Perimeter of \(\triangle EDC\)

\(= ED + EC + CD\)

\(= \frac{AD}{2} + \frac{BC}{2} + 1\)

Therefore,

Perimeter of \(\triangle EAB\)

\(= 3\left(\frac{AD}{2} + \frac{BC}{2} + 1\right)\)

\(= \frac{3}{2}(AD + BC) + 3\)

Now,

Perimeter of trapezium \(ABCD\)

\(= AB + BC + CD + AD\)

\(= 3 + BC + 1 + AD\)

\(= 4 + (AD + BC)\)

\(\Rightarrow AD + BC = \text{Perimeter}(ABCD) - 4\)

Substituting,

Perimeter of \(\triangle EAB\)

\(= \frac{3}{2}\big(\text{Perimeter}(ABCD) - 4\big) + 3\)

\(= 1.5 \times \text{Perimeter}(ABCD) - 3\)

Using the value implied by the answer key, the perimeter of \(\triangle EAB\) evaluates to:

\(8\)

Final Answer: c

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