CMI Data Science 2023 Question Paper with Solutions: 21 Questions, Answer Key & Section-wise Analysis

CMI Data Science 2023 previous year paper: 21 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

21 questions · 48 marks · 1.6800000000000002 minutes. School Level Mathematics: 12 · Discrete Mathematics: 5 · Programming: 2 · Probability Theory: 2

Free sample questions from CMI Data Science 2023 Question Paper

Question 1 · School Level Mathematics · 2023 MSQ

Consider the following single variable, real-valued functions:

$f (x) = \frac{x ^{4} + 1}{x ^{3} - 1}, g (x) = \frac{x ^{3} - x}{x ^{3} + x} .$

Which of the following is/are true?

A.
The function $f$ has a removable discontinuity at 1 and $g$ has it at 0.
B.
Only $g$ is continuous everywhere.
C.
As $x$ tends to $\infty$ $f (x)$ also tends to $\infty$ .
D.
The range of $g$ is bounded.

Correct Answer:

["C","D"]

Step-by-Step Solution

Key idea: This is a multi-statement question testing continuity, discontinuities, and limits at infinity for rational functions. We must analyze the domain, factor the expressions, and check the behavior at critical points and infinity.

Step 1: Analyze Option A. $f (x) = \frac{x ^{4} + 1}{x ^{3} - 1}$ . At $x = 1$ , the denominator is $1^{3} - 1 = 0$ , but the numerator is $1^{4} + 1 = 2 \neq = 0$ . This is a non-removable (infinite) discontinuity, not removable. False.

Step 2: Analyze Option B. $g (x) = \frac{x ^{3} - x}{x ^{3} + x} = \frac{x ( x ^{2} - 1 )}{x ( x ^{2} + 1 )}$ . For $x \neq = 0$ , $g (x) = \frac{x ^{2} - 1}{x ^{2} + 1}$ . At $x = 0$ , the original function is undefined, but the limit exists. Thus, $g$ has a removable discontinuity at $x = 0$ and is not continuous everywhere. False.

Step 3: Analyze Option C. As $x \to \infty$ , $f (x) = \frac{x ^{4} + 1}{x ^{3} - 1} \approx \frac{x ^{4}}{x ^{3}} = x$ . Since $x \to \infty$ , $f (x) \to \infty$ . True.

Step 4: Analyze Option D. For $x \neq = 0$ , $g (x) = \frac{x ^{2} - 1}{x ^{2} + 1} = 1 - \frac{2}{x ^{2} + 1}$ . Since $x^{2} > 0$ for $x \neq = 0$ , we have $x^{2} + 1 > 1$ , so $0 < \frac{2}{x ^{2} + 1} < 2$ . Therefore, $- 1 < 1 - \frac{2}{x ^{2} + 1} < 1$ . The range of $g$ is $(- 1, 1)$ , which is bounded. True.

Answer: C, D

Question 2 · School Level Mathematics · 2023 MSQ

Let $n$ be a positive integer such that

$\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{n}$

is an integer. Which of the following is/are true?

A.
$n$ has to be even.
B.
$3$ does not divide $n$
C.
$7$ divides $n$
D.
$5$ does not divide $n$

Correct Answer:

["A","C","D"]

Step-by-Step Solution

Key idea: This is a Diophantine condition problem where a sum of fractions must equal an integer. We use bounding arguments to find the unique value of $n$ .

Step 1: Simplify the constant part of the sum.

Let $S = \frac{1}{2} + \frac{1}{3} + \frac{1}{7}$ .

The common denominator is $42$ .

$S = \frac{21}{42} + \frac{14}{42} + \frac{6}{42} = \frac{41}{42}$

Step 2: Set up the equation.

We are given that $\frac{41}{42} + \frac{1}{n} = K$ , where $K$ is an integer.

$\frac{1}{n} = K - \frac{41}{42}$

Step 3: Bound the possible values of $K$ .

Since $n$ is a positive integer, $\frac{1}{n} > 0$ , so $K > \frac{41}{42}$ .

Also, since $n \geq 1$ , $\frac{1}{n} \leq 1$ .

Thus, $K = \frac{41}{42} + \frac{1}{n} \leq \frac{41}{42} + 1 = \frac{83}{42} \approx 1.97$ .

The only integer $K$ satisfying $\frac{41}{42} < K \leq 1.97$ is $K = 1$ .

Step 4: Solve for $n$ .

$\frac{41}{42} + \frac{1}{n} = 1 ⟹ \frac{1}{n} = 1 - \frac{41}{42} = \frac{1}{42} ⟹ n = 42$

Step 5: Evaluate the options for $n = 42$ .

A: " $n$ has to be even." -> $42$ is even. (True)

B: " $3$ does not divide $n$ ." -> $42 = 3 \times 14$ , so 3 divides $n$ . The statement is False.

C: " $7$ divides $n$ ." -> $42 = 7 \times 6$ , so 7 divides $n$ . (True)

D: " $5$ does not divide $n$ ." -> $42$ is not divisible by 5. (True)

Answer: ["A", "C", "D"]

Question 3 · School Level Mathematics · 2023 SUB

Let $f$ be the following function defined on integers:

$f (n) = \frac{1}{3 ^{n} + 3}$

Compute $3 [f (- 5) + f (- 4) + f (- 3) + \dots + f (4) + f (5) + f (6)]$

Correct Answer:

none

Step-by-Step Solution

Insight: This is a symmetric-summation puzzle — the sum runs over a consecutive integer range and the function has the form $\frac{c}{c ^{n} + c}$ . The endpoints $- 5$ and $6$ add to $1$ , which tells you to pair index $n$ with $1 - n$ . Each pair sums to exactly $1$ .

Exam route:

Step 1: Define $g (n) = 3 f (n) = \frac{3}{3 ^{n} + 3}$ . We need $S = \sum_{n = - 5}^{6} g (n)$ .

Step 2: Endpoints $a = - 5$ , $b = 6$ give $a + b = 1$ . Pair index $n$ with $1 - n$ .

Step 3: Compute $g (1 - n) = \frac{3}{3 ^{1 - n} + 3}$ .

Step 4: Multiply numerator and denominator by $3^{n}$ : $g (1 - n) = \frac{3 \cdot 3 ^{n}}{3 + 3 \cdot 3 ^{n}}$ .

Step 5: Factor $3$ from the denominator: $3 + 3 \cdot 3^{n} = 3 (3 + 3^{n})$ .

Step 6: Cancel $3$ : $g (1 - n) = \frac{3 ^{n}}{3 ^{n} + 3}$ .

Step 7: Add the pair: $g (n) + g (1 - n) = \frac{3 + 3 ^{n}}{3 ^{n} + 3} = 1$ .

Step 8: Count terms: $6 - (- 5) + 1 = 12$ terms, forming $6$ pairs.

Step 9: Total $= 6 \times 1 = 6$ .

Verification: pair $(- 5, 6), (- 4, 5), (- 3, 4), (- 2, 3), (- 1, 2), (0, 1)$ — six disjoint pairs covering all 12 indices. Each pair sums to $1$ . ✓

Learning route:

This is a symmetric-summation puzzle. The sum is over a consecutive integer range and the function has the form $\frac{c}{c ^{n} + c}$ . The endpoints of the summation range add up to a specific value, which tells you how to pair the indices.

Step 1: Identify the symmetry. The endpoints are $- 5$ and $6$ . Their sum is $1$ . This means we pair index $n$ with index $1 - n$ .

Step 2: Evaluate the pair sum. Let $g (n) = \frac{3}{3 ^{n} + 3}$ . We compute $g (1 - n)$ by multiplying top and bottom by $3^{n}$ , then factor $3$ out of the denominator. The result is $g (1 - n) = \frac{3 ^{n}}{3 ^{n} + 3}$ .

Step 3: Add the pair: $g (n) + g (1 - n) = \frac{3 + 3 ^{n}}{3 ^{n} + 3} = 1$ .

Step 4: Count the pairs. Twelve terms form six pairs, so the total is $6$ .

Trap: A common trap is trying to evaluate each term individually, or to pair $n$ with $- n$ (which does not produce a constant sum here — the correct partner is $1 - n$ ).

Question 4 · Discrete Mathematics · 2023 SUB

There are 5 friends

A, B, C, D, E

. Because of a heated argument not all of them are on speaking terms anymore. The array below describes who is talking to whom. A value of 1 indicates that the pair is on speaking terms, while a 0 indicates that they are not.

A gossip route is an ordered sequence of all the five friends such that every person hears the gossip from exactly one friend (with whom they are on speaking terms) and shares it with exactly one other friend (with whom they are on speaking terms). The friend who starts the gossip doesn’t hear it from anyone else and the fifth friend doesn’t pass it on to anybody.
Answer the following questions:
(a) Suppose

A

starts spreading gossip along a gossip route and

E

hears it the last. Who is the third person to hear it?
(b) List all the gossip routes from

B

E

.
(c) There is another dispute between the friends. As a result,

A

is now not speaking with

B

and

D

. List all the gossip routes starting from

A

Correct Answer:

none

Step-by-Step Solution

Insight: A "gossip route" visiting every friend exactly once is a Hamiltonian path in the undirected graph defined by the adjacency matrix.

Exam route: Translate the matrix to a graph, identify the missing edges, and use endpoint neighbors to systematically build valid paths by casework.

Learning route:

Step 1: Graph Translation. The matrix is symmetric with dashes on the diagonal (no self-loops). The only off-diagonal 0s are at $(B, D)$ , $(D, B)$ , $(C, E)$ , and $(E, C)$ . Thus, the graph is a complete graph $K_{5}$ minus the edges ${B, D}$ and ${C, E}$ .

Step 2: Part (a) – Paths from A to E. A path has the form $A \to x \to y \to z \to E$ . The 4th person ( $z$ ) must be a neighbor of $E$ . Neighbors of $E$ are $A, B, D$ . Since $A$ is the start, $z$ must be $B$ or $D$ .

If $z = B$ : Path is $A \to x \to y \to B \to E$ . Remaining vertices for $x, y$ are $C, D$ . Sequence $(C, D)$ fails because edge ${D, B}$ is missing. Sequence $(D, C)$ gives $A \to D \to C \to B \to E$ . All edges exist. Valid.
If $z = D$ : Path is $A \to x \to y \to D \to E$ . Remaining vertices for $x, y$ are $B, C$ . Sequence $(C, B)$ fails because edge ${B, D}$ is missing. Sequence $(B, C)$ gives $A \to B \to C \to D \to E$ . All edges exist. Valid.

In both valid paths, the third person ( $y$ ) is $C$ .

Step 3: Part (b) – Paths from B to E. Path form: $B \to x \to y \to z \to E$ . $z$ must be a neighbor of $E$ (excluding $B$ ), so $z \in {A, D}$ .

If $z = A$ : Remaining ${C, D}$ . $B \to C \to D \to A \to E$ is valid. ( $B \to D$ is missing).
If $z = D$ : Remaining ${A, C}$ . $B \to A \to C \to D \to E$ and $B \to C \to A \to D \to E$ are both valid.

Total routes: $B \to C \to D \to A \to E$ , $B \to A \to C \to D \to E$ , $B \to C \to A \to D \to E$ .

Step 4: Part (c) – A not speaking with B and D. A's neighbors are now only ${C, E}$ .

If path starts $A \to C$ : Next can be $B$ or $D$ .
$A \to C \to B \to E \to D$ (valid, since $B$ connects to $E$ , $E$ connects to $D$ ).
$A \to C \to D \to E \to B$ (valid, since $D$ connects to $E$ , $E$ connects to $B$ ).
If path starts $A \to E$ : Next can be $B$ or $D$ .
$A \to E \to B \to C \to D$ (valid).
$A \to E \to D \to C \to B$ (valid).

Total routes: 4.

Question 5 · Discrete Mathematics · 2023 MSQ

Solitaire Tic-Tac-Toe is a new game on the market. Instead of adding X’s and O’s to an empty

3 \times 3

grid, you start with a

3 \times 3

grid in which every position already has an X or an O. In each move, you select a row, column or diagonal and reverse all the entries along the chosen line: all X’s become O’s and all O’s become X’s.
For instance, here is a sequence of possible moves.

Given an arrangement

S

, we want to explore all arrangements of the grid that we can generate starting with

S

. What is the smallest number

m

such that each such arrangement can be reached using at most

m

moves, starting with

S

A.
8
B.
16
C.
64
D.
256

Correct Answer:

["A"]

Step-by-Step Solution

Key idea: This is a grid transformation invariant question, recognizable by the operation of flipping rows, columns, or diagonals.

Step 1: Identify the available moves. There are 3 rows, 3 columns, and 2 diagonals, making a total of 8 distinct moves.

Step 2: Note the properties of the moves. Each move is its own inverse (flipping a line twice returns it to the original state, since X becomes O and O becomes X). Also, the moves commute (the order of flips does not matter).

Step 3: Determine the maximum number of moves needed. Since applying any move twice is redundant, any reachable arrangement can be achieved by applying a subset of the 8 available moves, each at most once.

Step 4: Conclude the upper bound. The size of any such subset is at most 8. Therefore, every reachable arrangement can be reached in at most 8 moves. (While a tighter bound of 5 exists due to linear dependencies among the moves, 8 is the smallest valid upper bound among the given options and follows directly from the 8 distinct moves).

Answer: 8

Question 6 · Discrete Mathematics · 2023 MSQ

A perfect shuffle of a deck of cards divides the deck into two equal parts and then interleaves the cards from each half, starting with the first card of the first half.
For instance, if we shuffle a deck of cards containing 10 cards arranged

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

, we first create two equal decks with cards

[1, 2, 3, 4, 5]

and

[6, 7, 8, 9, 10]

and then interleave them to get a new deck

[1, 6, 2, 7, 3, 8, 4, 9, 5, 10]

.
We start with the deck

[8, 1, 4, 5, 3, 6, 2, 7]

and keep shuffling. Which card(s) will never appear next to 5?

A.
1
B.
2
C.
7
D.
8

Correct Answer:

["D"]

Step-by-Step Solution

Key idea: This is a permutation cycle decomposition question, recognizable because it asks about the long-term behavior of a deterministic rearrangement (shuffle).

Step 1: Understand the shuffle mapping. For an 8-card deck, the perfect shuffle maps positions as follows: $p \to 2 p - 1$ if $p \leq 4$ , and $p \to 2 (p - 4)$ if $p > 4$ .

Step 2: Find the cycles of positions.

Position 1 maps to 1. (Cycle: {1})
Position 8 maps to 8. (Cycle: {8})
Position 2 $\to$ 3 $\to$ 5 $\to$ 2. (Cycle: {2, 3, 5})
Position 4 $\to$ 7 $\to$ 6 $\to$ 4. (Cycle: {4, 6, 7})

Step 3: Track the cards in these cycles.

Initial deck: [8, 1, 4, 5, 3, 6, 2, 7] at positions 1 to 8.

Cycle {1} always contains card 8.
Cycle {8} always contains card 7.
Cycle {2, 3, 5} contains cards {1, 4, 3}.
Cycle {4, 6, 7} contains cards {5, 6, 2}.

Step 4: Determine adjacencies for card 5. Card 5 is always in Cycle {4, 6, 7}.

If 5 is at pos 4, neighbors are pos 3 and 5 (both in Cycle {2, 3, 5}, cards {1, 3, 4}).
If 5 is at pos 6, neighbors are pos 5 (Cycle {2, 3, 5}) and pos 7 (Cycle {4, 6, 7}).
If 5 is at pos 7, neighbors are pos 6 (Cycle {4, 6, 7}) and pos 8 (Cycle {8}, card 7).

Step 5: Check the options. Card 8 is fixed at position 1. For 8 to be next to 5, 5 would need to be at position 2. However, 5 only visits positions 4, 6, and 7. Thus, 8 can never be next to 5.

Answer: 8

Question 7 · Programming · 2023 MSQ

Consider the following pseudocodes of two functions, where $u % 2$ denotes the remainder when $u$ is divided by 2. The function $ab s (v)$ returns the absolute value of an integer $v$ .

function foo1(u):

if u % 2 == 0 AND abs(u) > 0:

u = u + foo2(u-1)

return u

function foo2(v):

if v % 2 == 1 AND abs(v) > 1:

v = v + foo1(v-1)

return v

Which of the following is/are true?

A.
$f oo 1$ , when called on a positive integer $u$ , returns $u (u + 1) /2$
B.
$f oo 2$ , when called on an odd positive integer $v$ , returns $v (v + 1) /2$
C.
$f oo 2 (- 13)$ goes into infinite recursion
D.
$f oo 1 (14)$ returns 105

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This is a mutual recursion evaluation question. We need to trace the functions for small values to find a mathematical pattern, and carefully evaluate the conditions for negative numbers using standard C-style remainder rules.

Step 1: Trace foo1(u) for small positive even u.

foo1(2): 2 % 2 == 0 and abs(2) > 0. u = 2 + foo2(1).

foo2(1): 1 % 2 == 1 but abs(1) > 1 is False. Returns 1.

foo1(2) = 2 + 1 = 3. Note: $2 (3) /2 = 3$ .

Step 2: Trace foo2(v) for small positive odd v.

foo2(3): 3 % 2 == 1 and abs(3) > 1. v = 3 + foo1(2) = 3 + 3 = 6. Note: $3 (4) /2 = 6$ .

Step 3: Generalize the pattern.

By induction, foo1(u) for positive even u sums all integers from 1 to u, yielding $u (u + 1) /2$ .

foo2(v) for positive odd v sums all integers from 1 to v, yielding $v (v + 1) /2$ .

Step 4: Evaluate Option A.

For an odd positive integer like u=3, foo1(3) checks 3 % 2 == 0, which is false, and returns 3. But $3 (4) /2 = 6 \neq = 3$ . So Option A is false.

Step 5: Evaluate Option B.

For any odd positive integer v, it follows the pattern $v (v + 1) /2$ . Option B is true.

Step 6: Evaluate Option D.

foo1(14) where 14 is a positive even integer. It returns $14 (15) /2 = 105$ . Option D is true.

Step 7: Evaluate Option C.

For foo2(-13), we check -13 % 2. In standard C-style programming remainder, -13 % 2 = -1. The condition -1 == 1 is false. The function returns -13 immediately without recursion. Option C is false.

Answer: B, D

Question 8 · Programming · 2023 MSQ

Consider the pseudocode below, where $n % 6$ denotes the remainder when $n$ is divided by 6. The notation $n //2$ stands for integer division, i.e., $15//2 = 7, 36//2 = 18, 49//2 = 24$ .

function sixer(n):

count = 0

while n > 0:

if n % 6 == 0:

n = n - 1

else:

n = n//2

count = count + 1

return count

Which of the following is true when sixer(n) is invoked with sufficiently large $n$ , say $n \geq 1 0^{6}$ ?

A.
The value of count is approximately $n /6$
B.
The value of count is approximately $n /2$
C.
The value of count is approximately $lo g_{2} (n)$
D.
The value of count is approximately $(lo g_{2} (n))^{2}$

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: This is a while-loop iteration count question, recognisable because it asks for the approximate number of steps to reduce $n$ to $0$ using conditional halving and subtracting.

Step 1: Analyze the loop behavior. If $n$ is a multiple of $6$ , the loop executes $n = n - 1$ , making it no longer a multiple of $6$ .

Step 2: In the very next iteration, since $n$ is not a multiple of $6$ , the else branch executes: $n = n //2$ .

Step 3: Thus, every time $n$ is a multiple of $6$ , it takes exactly $2$ steps to approximately halve $n$ . If $n$ is not a multiple of $6$ , it takes $1$ step to halve $n$ .

Step 4: In all cases, $n$ is halved in at most $2$ steps. Therefore, the number of iterations is bounded between $lo g_{2} (n)$ and $2 lo g_{2} (n)$ .

Step 5: This means the count grows logarithmically with $n$ . Among the given options, "approximately $lo g_{2} (n)$ " is the only one that correctly describes this logarithmic growth rate.

Answer: C

Question 9 · Probability Theory · 2023 MSQ

Let $A, B, C$ be events such that $P (A) = P (B) = P (C) = 0.5, P (A \cap B) = 0.3, P (A \cap C) = 0$ . Which of the following is/are true?

A.
$P (A \cup B) = 0.75$
B.
$P (A \cup C) = 1$
C.
$P (B \cap C) = 0.2$
D.
$P (B \cup C) = 0.9$

Correct Answer:

["B"]

Step-by-Step Solution

Key idea: This is an event algebra question requiring the Addition Rule and bounding using the fact that the probability of any event cannot exceed 1.

Step 1: Evaluate Option A using the Addition Rule. $P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.5 + 0.5 - 0.3 = 0.7$ . The option claims 0.75, so A is false.

Step 2: Evaluate Option B. Since $P (A \cap C) = 0$ , the events are mutually exclusive. $P (A \cup C) = P (A) + P (C) - 0 = 0.5 + 0.5 = 1$ . This is always true, so B is correct.

Step 3: Evaluate Options C and D using the universal bound $P (A \cup B \cup C) \leq 1$ .

We know $P (A \cup B \cup C) = P (A \cup B) + P (C) - P ((A \cup B) \cap C)$ .

Since $A \cap C = \emptyset$ , $(A \cup B) \cap C = B \cap C$ .

Thus, $P (A \cup B \cup C) = 0.7 + 0.5 - P (B \cap C) = 1.2 - P (B \cap C)$ .

Because $P (A \cup B \cup C) \leq 1$ , we must have $1.2 - P (B \cap C) \leq 1 ⟹ P (B \cap C) \geq 0.2$ .

This means $P (B \cap C)$ could be 0.2, 0.3, etc., so it is not necessarily exactly 0.2 (Option C is false).

Consequently, $P (B \cup C) = P (B) + P (C) - P (B \cap C) = 1 - P (B \cap C) \leq 0.8$ . Option D claims 0.9, which is impossible (Option D is false).

Answer: Only B is necessarily true.

Question 10 · Probability Theory · 2023 MSQ

Consider the set $S = {0, 1, 2, \dots, n - 1}$ such that $n \geq 2$ . Let $x$ be an element sampled uniformly at random from $S$ . Let $y$ denote $(x + 2^{n})$ mod $n$ . Which of the following statements is/are true?

A.
$Pr [y = ⌊ n - 1 ⌋] = \frac{⌊ n - 1 ⌋}{n}$
B.
$Pr [x = a, y = b] = Pr [x = a] \cdot Pr [y = b]$ for all $a, b \in S$
C.
$y$ is always an element of $S$
D.
$Pr [y = ⌊ lo g_{2} (n - 1)⌋] = 1/ n$

Correct Answer:

["C","D"]

Step-by-Step Solution

Key idea: Modular arithmetic preserves uniform distribution, but creates deterministic dependence.

Step 1: Recognize that x is uniformly distributed over S = {0, 1, ..., n-1}.

Step 2: The transformation y = (x + 2^n) mod n is a bijection on S because adding a constant and taking modulo n simply permutes the elements of S.

Step 3: Therefore, y is also uniformly distributed over S. This means P[y = k] = 1/n for any k in S.

Step 4: Evaluate Option 1: It claims P[y = k] = k/n, which is false (it is 1/n).

Step 5: Evaluate Option 2: It claims x and y are independent. But y is a deterministic function of x, so they are perfectly dependent. P[x=a, y=b] is 1/n if b = (a+2^n) mod n, and 0 otherwise. This does not equal (1/n)*(1/n) = 1/n^2. False.

Step 6: Evaluate Option 3: By definition of modulo n, the result is always in {0, 1, ..., n-1}, which is exactly S. True.

Step 7: Evaluate Option 4: For n >= 2, n-1 >= 1, so log_2(n-1) >= 0. Also, for n >= 2, n-1 < 2^{n-1}, so log_2(n-1) < n-1. Thus, floor(log_2(n-1)) is an integer between 0 and n-2, which is in S. Since y is uniform on S, the probability is 1/n. True.

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