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CAT 2018 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 · 20
Verbal Ability · 4

DILR

Data Interpretation & Logical Reasoning

22 questions · 12 MCQ · 10 TITA

Logical Reasoning · 12
Data Interpretation · 10

QA

Quantitative Ability

22 questions · 14 MCQ · 8 TITA

Arithmetic · 8
Algebra · 7
Geometry · 3
Modern Math · 2
Number System · 2

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

The complexity of modern problems often precludes any one person from fully understanding them. Factors contributing to rising obesity levels, for example, include transportation systems and infrastructure, media, convenience foods, changing social norms, human biology and psychological factors. . . . The multidimensional or layered character of complex problems also undermines the principle of meritocracy: the idea that the ‘best person' should be hired. There is no best person. When putting together an oncological research team, a biotech company such as Gilead or Genentech would not construct a multiple-choice test and hire the top scorers, or hire people whose resumes score highest according to some performance criteria. Instead, they would seek diversity. They would build a team of people who bring diverse knowledge bases, tools and analytic skills.

Believers in a meritocracy might grant that teams ought to be diverse but then argue that meritocratic principles should apply within each category. Thus the team should consist of the 'best' mathematicians, the ‘best' oncologists, and the 'best' biostatisticians from within the pool. That position suffers from a similar flaw. Even with a knowledge domain, no test or criteria applied to individuals will produce the best team. Each of these domains possesses such depth and breadth, that no test can exist. Consider the field of neuroscience. Upwards of 50,000 papers were published last year covering various techniques, domains of enquiry and levels of analysis, ranging from molecules and synapses up through networks of neurons. Given that complexity, any attempt to rank a collection of neuroscientists from best to worst, as if they were competitors in the 50-metre butterfly, must fail. What could be true is that given a specific task and the composition of a particular team, one scientist would be more likely to contribute than another. Optimal hiring depends on context. Optimal teams will be diverse.

Evidence for this claim can be seen in the way that papers and patents that combine diverse ideas tend to rank as high-impact. It can also be found in the structure of the so-called random decision forest, a state-of-the-art machine-learning algorithm. Random forests consist of ensembles of decision trees. If classifying pictures, each tree makes a vote: is that a picture of a fox or a dog? A weighted majority rules. Random forests can serve many ends. They can identify bank fraud and diseases, recommend ceiling fans and predict online dating behaviour. When building a forest, you do not select the best trees as they tend to make similar classifications. You want diversity. Programmers achieve that diversity by training each tree on different data, a technique known as bagging. They also boost the forest ‘cognitively' by training trees on the hardest cases – those that the current forest gets wrong. This ensures even more diversity and accurate forests.

Yet the fallacy of meritocracy persists. Corporations, non-profits, governments, universities and even preschools test, score and hire the ‘best’. This all but guarantees not creating the best team. Ranking people by common criteria produces homogeneity. . . . That's not likely to lead to breakthroughs.

Question 1 · MCQ

Which of the following conditions, if true, would invalidate the passage's main argument?

If top-scorers possessed multidisciplinary knowledge that enabled them to look at a problem from several perspectives.
If assessment tests were made more extensive and rigorous.
If it were proven that teams characterised by diversity end up being conflicted about problems and take a long time to arrive at a solution.
If a new machine-learning algorithm were developed that proved to be more effective than the random decision forest.

A.
1
✓ Correct
B.
2
C.
3
D.
4

Solution

The main argument is that meritocracy fails because selecting the 'best' by common criteria creates homogeneity, while diverse teams are essential for complex problems. Option 1 would invalidate this: if top-scorers already possess multidisciplinary knowledge, then selecting them could directly yield diverse perspectives within individuals, undermining the need for team diversity.

Question 2 · MCQ

The author critiques meritocracy for all the following reasons EXCEPT that:

an ideal team comprises of best individuals from diverse fields of knowledge.
modern problems are multifaceted and require varied skill-sets to be solved.
criteria designed to assess merit are insufficient to test expertise in any field of knowledge.
diversity and context-specificity are important for making major advances in any field.

A.
1
✓ Correct
B.
2
C.
3
D.
4

Solution

The author explicitly rebuts the idea that teams should consist of the 'best' individuals from each field (para 2). Options 2, 3, and 4 are all reasons he uses to critique meritocracy, but 1 is a view he opposes.

Question 3 · MCQ

Which of the following conditions would weaken the efficacy of a random decision forest?

If a large number of decision trees in the ensemble were trained on data derived from easy cases.
If the types of decision trees in each ensemble of the forest were doubled.
If a large number of decision trees in the ensemble were trained on data derived from easy and hard cases.
If the types of ensembles of decision trees in the forest were doubled.

A.
1
✓ Correct
B.
2
C.
3
D.
4

Solution

The passage notes that boosting forest accuracy by training trees on the hardest cases ensures diversity and high performance. Training many trees only on easy cases would neglect hard cases, reduce diversity in handling difficult data, and weaken overall efficacy.

Question 4 · MCQ

On the basis of the passage, which of the following teams is likely to be most effective in solving the problem of rising obesity levels?

A team comprised of nutritionists, psychologists, urban planners and media personnel, who have each scored a distinction in their respective subject tests.
A team comprised of nutritionists, psychologists, urban planners and media personnel, who have each performed well in their respective subject tests.
A specialised team of nutritionists from various countries, who are also trained in the machine-learning algorithm of random decision forest.
A specialised team of top nutritionists from various countries, who also possess some knowledge of psychology.

A.
1
B.
2
✓ Correct
C.
3
D.
4

Solution

The passage argues that selecting the absolute 'best' (e.g., distinction holders) leads to homogeneity and fails to leverage diversity. A team with diverse fields who have all performed well—but not necessarily been top-ranked—aligns better with the principle that optimal teams are diverse and context-dependent.

Question 5 · MCQ

Which of the following best describes the purpose of the example of neuroscience?

Unlike other fields of knowledge, neuroscience is an exceptionally complex field, making a meaningful assessment of neuroscientists impossible.
In narrow fields of knowledge, a meaningful assessment of expertise has always been possible.
Neuroscience is an advanced field of science because of its connections with other branches of science like oncology and biostatistics.
In the modern age, every field of knowledge is so vast that a meaningful assessment of merit is impossible.

A.
1
B.
2
C.
3
D.
4
✓ Correct

Solution

The neuroscience example (50,000 papers, multiple levels of analysis) demonstrates the immense breadth of a single field, making ranking impossible. It supports the general claim that in the modern age, every field is too vast for meaningful merit assessment.

DILR · Data Interpretation & Logical Reasoning

Set

The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.

A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly with another. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates.

The following facts are known about the outlet on a particular day:

The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
The amounts of L used by the outlet to buy A and B are in the ratio 5:3.
The amounts of L the outlet received from the sales of A and B are in the ratio 5:9.
The outlet received 88000 units of L by selling A during the day.
The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.

Question 1 · TITA (type in the answer)

How many units of currency A did the outlet buy on that day?

Answer: 1200

Solution

Let the base exchange rate of C w.r.t. L be r L per C. Then base rates: A = 100r, B = 120r, C = r. Buy rates (5% below base): A: 0.95 × 100r = 95r, B: 114r, C: 0.95r. Sell rates (10% above base): A: 110r, B: 132r, C: 1.10r.

From (4), SA = 88000 L → units of A sold = 88000/(110r) = 800/r. From (3), SA:SB = 5:9 → SB = 158400 L → units of B sold = 158400/(132r) = 1200/r.

Let BA : BB = 5 : 3. Let BA = 5x, BB = 3x. Then units of A bought = 5x/(95r) = x/(19r); units of B bought = 3x/(114r) = x/(38r).

From (5) and (6), B's start 4800 and end 4800, so bought = sold: x/(38r) = 1200/r → x = 45600.

For A: start 2500 + x/(19r) – 800/r = 3300. Substituting x = 45600 gives 45600/(19r) = 800 + 800/r → r = 2.

Hence bought A = 45600/(19×2) = 45600/38 = 1200.

Question 2 · MCQ

How many units of currency C did the outlet sell on that day?

A.
19000
✓ Correct
B.
3000
C.
6000
D.
22000

Solution

From Q1, r = 2, so buy rate C = 0.95 × 2 = 1.9 L/C, sell rate C = 1.10 × 2 = 2.2 L/C. Let Y be the (equal) amount of L used to buy C and received from selling C (by fact 1). Then units bought = Y/1.9, units sold = Y/2.2.

Net change in C = +3000 (start 48000, end 51000). So Y/1.9 – Y/2.2 = 3000.

\(\frac{Y}{1.9} - \frac{Y}{2.2} = 3000 \Rightarrow Y\left(\frac{10}{19} - \frac{10}{22}\right) = 3000 \Rightarrow 10Y \cdot \frac{3}{418} = 3000 \Rightarrow Y = 41800\).

Units of C sold = 41800 / 2.2 = 19000. Option A corresponds to 19000.

Question 3 · TITA (type in the answer)

What was the base exchange rate of currency B with respect to currency L on that day?

Answer: 240

Solution

Let the base exchange rates per unit be \(R_A = 100k\), \(R_B = 120k\), \(R_C = k\).
Buying rates: \(0.95R_A = 95k\), \(0.95R_B = 114k\), \(0.95R_C = 0.95k\).
Selling rates: \(1.10R_A = 110k\), \(1.10R_B = 132k\), \(1.10R_C = 1.10k\).

Given \(S_A = 88000\):
\(110k \cdot v_A = 88000 \Rightarrow v_A = \frac{800}{k}\).

From net changes:
\(\Delta A = +800 \Rightarrow u_A - v_A = 800\);
\(\Delta B = 0 \Rightarrow u_B = v_B\);
\(\Delta C = +3000 \Rightarrow u_C - v_C = 3000\).

Buying and selling amount ratios:
\(\frac{95k\,u_A}{114k\,u_B} = \frac{5}{3} \Rightarrow u_A = 2u_B\);
\(\frac{110k\,v_A}{132k\,v_B} = \frac{5}{9} \Rightarrow v_B = \frac{3}{2}v_A\).

Using \(u_B = v_B = \frac{3}{2} \cdot \frac{800}{k} = \frac{1200}{k}\).
Then \(u_A = 2 \cdot \frac{1200}{k} = \frac{2400}{k}\).
Now \(u_A - v_A = \frac{2400}{k} - \frac{800}{k} = \frac{1600}{k} = 800 \Rightarrow k = 2\).

Base rate of B wrt L = \(120k = 240\).

Answer: 240.

Question 4 · MCQ

What was the buying exchange rate of currency C with respect to currency L on that day?

A.
0.95
B.
1.10
C.
1.90
✓ Correct
D.
2.20

Solution

From Q3, \(k = 2\).
Base rate of C wrt L = \(R_C = k = 2\).
Buying exchange rate is 5% below base:
\(0.95 \times 2 = 1.90\).

Hence option 3 (1.90).

QA · Quantitative Ability

Question 1 · TITA (type in the answer)

Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit's dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y.

Answer: 4000

Solution

Let Gopal borrow \(X\) and add \(Y\).
Interest paid to Ankit: \(0.08X\). Interest from Ishan: \(0.10(X+Y)\).
Net interest retained = \(0.10(X+Y) - 0.08X\).
Given this equals Ankit's interest: \(0.10(X+Y) - 0.08X = 0.08X\).
\(0.02X + 0.10Y = 0.08X\) → \(0.10Y = 0.06X\) → \(X = \frac{5}{3}Y\).
If loan is \(X+2Y\), net interest = \(0.10(X+2Y) - 0.08X\).
Increase = \(0.10Y = 150\) → \(Y = 1500\), \(X = 2500\).
Hence \(X+Y = 4000\).

Question 2 · TITA (type in the answer)

On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The difference between their speeds, in km per hour, is

Answer: 50

Solution

Question 3 · TITA (type in the answer)

The arithmetic mean of \(x, y\) and \(z\) is 80, and that of \(x, y, z, u\) and \(v\) is 75, where \(u=(x+y)/2\) and \(v=(y+z)/2\). If \(x \ge z\), then the minimum possible value of \(x\) is

Answer: 105

Solution

Given: \(\frac{x+y+z}{3}=80 \Rightarrow x+y+z=240\).
Also \(\frac{x+y+z+u+v}{5}=75\) with \(u=\frac{x+y}{2}, v=\frac{y+z}{2}\).
Substituting: \(x+y+z+\frac{x+y}{2}+\frac{y+z}{2}=375\).
Multiply by 2: \(3x+4y+3z=750\).
Using \(y=240-x-z\), we get \(x+z=210\) and \(y=30\).
Since \(x \ge z\) and \(x+z=210\), minimum \(x\) occurs when \(x=z=105\).
Thus minimum possible value of \(x\) is 105.

Question 4 · TITA (type in the answer)

The smallest integer \(n\) such that \(n^3 - 11n^2 + 32n - 28 > 0\) is

Answer: 8

Solution

Factorise: \(n^3-11n^2+32n-28\).
By inspection, \(n=2\) and \(n=7\) are roots. Polynomial becomes \((n-2)^2(n-7)\).
We need \((n-2)^2(n-7) > 0\).
\((n-2)^2 \ge 0\), and is zero only at \(n=2\).
Thus for positivity, require \(n-7 > 0 \Rightarrow n > 7\).
The smallest integer \(n > 7\) is 8.

Question 5 · TITA (type in the answer)

Let \(f(x)=\text{max}\{5x, 52-2x^2\}\), where \(x\) is any positive real number. Then the minimum possible value of \(f(x)\) is

Answer: 20

Solution

To minimise \(\max\{5x,\,52-2x^2\}\) for \(x>0\), find where \(5x = 52-2x^2\).

\(2x^2+5x-52=0\) ⇒ \(x = \frac{-5+\sqrt{25+416}}{4}=4\) (positive root).

At \(x=4\), both are \(20\); for \(x<4\) the quadratic is larger, for \(x>4\) the linear term is larger, both >20.

\(\therefore\) minimum value = 20.

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