CMI Data Science 2021 Question Paper with Solutions: 23 Questions, Answer Key & Section-wise Analysis

CMI Data Science 2021 previous year paper: 23 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

23 questions · 52 marks · 1.82 minutes. School Level Mathematics: 11 · Discrete Mathematics: 6 · Probability Theory: 5 · Programming: 1

Free sample questions from CMI Data Science 2021 Question Paper

Question 1 · School Level Mathematics · 2021 MSQ

The roots of the polynomial $p (x) = x^{4} - 2 x^{3} - 2 x^{2} + 8 x - 8$ are:

A.
$1, - 1, 2, 2 + 3 i$
B.
$1 + i, 1 - i, 2, - 2$
C.
$1, - 1 + i, 2, 2 + 3 i$
D.
$1 + i, - 1 + i, 2, - 2$

Correct Answer:

["B"]

Step-by-Step Solution

Key idea: This is a polynomial factorisation question, recognisable because it asks for all roots of a degree-4 polynomial with integer coefficients.

Step 1: Use the Rational Root Theorem. Possible rational roots are divisors of $- 8$ : $\pm 1, \pm 2, \pm 4, \pm 8$ .

Step 2: Test $x = 2$ : $p (2) = 16 - 16 - 8 + 16 - 8 = 0$ . So $(x - 2)$ is a factor.

Step 3: Divide $p (x)$ by $(x - 2)$ using synthetic division.

Coefficients: $1, - 2, - 2, 8, - 8$ .

Bring down 1. Multiply by 2: 2. Add to $- 2$ : 0. Multiply by 2: 0. Add to $- 2$ : $- 2$ . Multiply by 2: $- 4$ . Add to 8: 4. Multiply by 2: 8. Add to $- 8$ : 0.

Quotient: $x^{3} - 2 x + 4$ .

Step 4: Find roots of $q (x) = x^{3} - 2 x + 4$ . Test $x = - 2$ : $(- 2)^{3} - 2 (- 2) + 4 = - 8 + 4 + 4 = 0$ . So $(x + 2)$ is a factor.

Step 5: Divide $q (x)$ by $(x + 2)$ . Coefficients: $1, 0, - 2, 4$ .

Bring down 1. Multiply by $- 2$ : $- 2$ . Add to 0: $- 2$ . Multiply by $- 2$ : 4. Add to $- 2$ : 2. Multiply by $- 2$ : $- 4$ . Add to 4: 0.

Quotient: $x^{2} - 2 x + 2$ .

Step 6: Solve $x^{2} - 2 x + 2 = 0$ . Discriminant: $4 - 8 = - 4$ . Roots: $x = \frac{2 \pm 2 i}{2} = 1 \pm i$ .

Step 7: All four roots are $2, - 2, 1 + i, 1 - i$ .

Answer: Option B

Question 2 · School Level Mathematics · 2021 SUB

Let $S$ be the sphere given by the equation $x^{2} + y^{2} + z^{2} - 6 x + 4 y - 4 z = 272$ . What are the centre and radius of $S$ ? What is the minimum distance between $S$ and the plane $2 x + 2 y + z = 214$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a 3D coordinate geometry question requiring completing the square to find a sphere's center and radius, followed by finding the distance from a point to a plane. It is recognisable by the general second-degree equation in $x, y, z$ and the request for minimum distance to a plane.

Step 1: Convert the sphere equation to standard form.

The given equation is $x^{2} + y^{2} + z^{2} - 6 x + 4 y - 4 z = 272$ .

Group the terms by variable:

$(x^{2} - 6 x) + (y^{2} + 4 y) + (z^{2} - 4 z) = 272$

Complete the square for each group by adding $(b /2)^{2}$ :

$(x^{2} - 6 x + 9) + (y^{2} + 4 y + 4) + (z^{2} - 4 z + 4) = 272 + 9 + 4 + 4$

$(x - 3)^{2} + (y + 2)^{2} + (z - 2)^{2} = 289$

Comparing with $(x - h)^{2} + (y - k)^{2} + (z - l)^{2} = r^{2}$ :

Center $C = (3, - 2, 2)$

Radius $r = 289 = 17$

Step 2: Calculate the perpendicular distance from the center to the plane.

The plane equation is $2 x + 2 y + z - 214 = 0$ .

The distance $d$ from point $(x_{0}, y_{0}, z_{0})$ to plane $A x + B y + C z + D = 0$ is:

$d = \frac{∣ A x _{0} + B y _{0} + C z _{0} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$

Substitute $C (3, - 2, 2)$ :

Numerator: $∣2 (3) + 2 (- 2) + 1 (2) - 214∣ = ∣6 - 4 + 2 - 214∣ = ∣ - 210∣ = 210$

Denominator: $2^{2} + 2^{2} + 1^{2} = 4 + 4 + 1 = 9 = 3$

$d = \frac{210}{3} = 70$

Step 3: Determine the minimum distance between the sphere and the plane.

Compare $d$ and $r$ :

Since $d = 70$ and $r = 17$ , we have $d > r$ . This means the plane does not intersect the sphere; the sphere is entirely on one side of the plane.

The minimum distance between the sphere surface and the plane is:

$Min Distance = d - r = 70 - 17 = 53$

Answer: 53

Question 3 · School Level Mathematics · 2021 MSQ

Which of the following statement(s) is/are true?

A.
For $k = 69597$ and $x_{1}, x_{2}, \dots, x_{2 k} \in (0, \infty)$ , $\frac{x _{1} + x _{2} + \dots + x _{2 k}}{2 k} \leq (x_{1} x_{2} \dots x_{2 k})^{\frac{1}{2 k}}$ .
B.
For any three real numbers $x, y, z$ , $∣ x - z ∣ \leq ∣ x - y ∣ + ∣ y - z ∣$ .
C.
For $∣ r ∣ < 1$ , $\sum_{n = 0}^{\infty} 37 \cdot r^{n} = \frac{37}{1 - r}$ .
D.
$(n^{3} - n)$ is divisible by 3 for all positive integers $n$ .

Correct Answer:

["B","C","D"]

Step-by-Step Solution

Key idea: This is a classical-facts verification question, recognisable because each option is a disguised version of a standard result: AM-GM, triangle inequality, geometric series, or divisibility of consecutive integers.

Why this method applies: The fastest method is to identify the theorem behind each statement, then check the direction of the inequality and the hypotheses. A reversed direction or a missing hypothesis makes the statement false. Large constants are often distractors.

Step 1: Check option A.

The statement says that for positive real numbers,

$\frac{x _{1} + x _{2} + \dots + x _{2 k}}{2 k} \leq (x_{1} x_{2} \dots x_{2 k})^{1/ (2 k)} .$

This claims

$AM \leq GM .$

But the standard AM-GM inequality says

$AM \geq GM$

for positive real numbers, with equality only when all numbers are equal. Therefore the direction is reversed. Option A is FALSE. The particular value $k = 69597$ is irrelevant.

Step 2: Check option B.

The statement is

$∣ x - z ∣ \leq ∣ x - y ∣ + ∣ y - z ∣.$

Put

$a = x - y, b = y - z .$

Then

$a + b = x - z .$

The triangle inequality gives

$∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣.$

Substituting back,

$∣ x - z ∣ \leq ∣ x - y ∣ + ∣ y - z ∣.$

Hence option B is TRUE.

Step 3: Check option C.

For $∣ r ∣ < 1$ , the infinite geometric series formula gives

$n = 0 \sum \infty a r^{n} = \frac{a}{1 - r} .$

Here $a = 37$ , so

$n = 0 \sum \infty 37 r^{n} = \frac{37}{1 - r} .$

The required condition $∣ r ∣ < 1$ is already stated. Hence option C is TRUE.

Step 4: Check option D.

We need to show that

$n^{3} - n$

is divisible by 3 for all positive integers $n$ . Factorise:

$n^{3} - n = n (n^{2} - 1) = n (n - 1) (n + 1) .$

The three factors

$n - 1, n, n + 1$

are three consecutive integers. Among any three consecutive integers, exactly one is divisible by 3. Therefore their product is divisible by 3.

Hence option D is TRUE.

Common trap: Option A uses a huge number $k = 69597$ to distract the reader. The truth of AM-GM does not depend on such a constant; the direction of the inequality is the real issue.

Answer: Options B, C and D.

Question 4 · Discrete Mathematics · 2021 SUB

Food delivery agents Aman, Boni, Chan, Dong and Eman were assessed on five parameters P1 to P5. They received an integer rating between 1 and 5 for each parameter. None of them received the same rating in four or more parameters. Everyone received a score of 1 in P2 or P5. All of them received the same rating in at least two parameters. The partial information related to the ratings are given in the table below

Name Aman Boni Chan Dong Eman P1 42 P2 P3 5 P4 31 P5

Considering the above information, answer the following questions:
(a) What is the maximum average rating Boni could have achieved across all the five parameters?
(b) What is the minimum average rating Chan could have received across all the five parameters?

Correct Answer:

4.0

Step-by-Step Solution

Key idea: This is a Constraint Table / Logic Grid puzzle with numerical ratings. The triggers are the multiple agents, multiple parameters, integer ratings 1–5, and several overlapping constraints. We must use deductive elimination on the grid to find the maximum average for Boni and minimum average for Chan.

Given constraints:

(C1) Ratings are integers from 1 to 5.

(C2) No agent received the same rating in 4 or more parameters. (So each agent has at most 3 identical ratings.)

(C3) Everyone received a score of 1 in P2 or P5 (or both).

(C4) All agents received the same rating in at least two parameters.

(C5) Fixed entries from the table: Aman P4=3, Boni P1=4, Chan P3=5, Dong P1=2, Eman P4=1.

Part (a): Maximum average rating for Boni.

Step 1: Boni has P1 = 4 fixed. To maximise the average, maximise the sum of P2+P3+P4+P5.

Step 2: By (C3), Boni must have a 1 in P2 or P5. To maximise the total, place the 1 in exactly one of these (say P2 = 1) and make the other as large as possible.

Step 3: Try P2 = 1, P3 = 5, P4 = 5, P5 = 5. Sum = 4+1+5+5+5 = 20. Average = 4.0.

Step 4: Check (C2): ratings are {4,1,5,5,5}. The rating 5 appears 3 times (P3,P4,P5). That is at most 3, so (C2) is satisfied (not 4 or more).

Step 5: Check (C4): Boni needs the same rating in at least two parameters. Rating 5 appears in P3, P4, P5 — that is 3 parameters. Satisfied.

Step 6: Check (C3): P2 = 1. Satisfied.

Step 7: Can we do better? The maximum possible sum with one forced 1 and P1=4 is 4+1+5+5+5 = 20. We cannot exceed 5 per parameter. So 20 is the maximum sum.

Step 8: Maximum average for Boni = 20/5 = 4.0.

Part (b): Minimum average rating for Chan.

Step 1: Chan has P3 = 5 fixed. To minimise the average, minimise P1+P2+P4+P5.

Step 2: By (C3), Chan must have a 1 in P2 or P5. Set P2 = 1 (minimum possible).

Step 3: Try to set remaining parameters to 1 as well: P1 = 1, P4 = 1, P5 = 1. Ratings: {1,1,5,1,1}. Sum = 9. Average = 1.8.

Step 4: Check (C2): Rating 1 appears 4 times (P1,P2,P4,P5). This violates (C2) — no agent can have the same rating in 4 or more parameters.

Step 5: Reduce the count of 1s to at most 3. We already need P2 = 1 (from C3). Set two more to 1 and one to 2: e.g., P1 = 1, P2 = 1, P4 = 1, P5 = 2. Ratings: {1,1,5,1,2}. Count of 1s = 3. OK for (C2).

Step 6: Check (C4): Chan needs the same rating in at least two parameters. Rating 1 appears in P1, P2, P4 — that is 3 parameters. Satisfied.

Step 7: Sum = 1+1+5+1+2 = 10. Average = 2.0.

Step 8: Can we get sum = 9 while satisfying (C2)? We need four 1s and one 5, but four 1s violates (C2). What about P1=1, P2=1, P4=2, P5=1? Same issue — three 1s plus the 5 and a 2 gives sum 10. Any arrangement with at most three 1s and P3=5 gives minimum sum = 1+1+5+1+2 = 10 (three 1s, one 2, one 5).

Step 9: Minimum average for Chan = 10/5 = 2.0.

Answer: (a) 4.0 (b) 2.0

Question 5 · Discrete Mathematics · 2021 MSQ

A binary tree starts with a single root node at the top of the tree. Each node can have either a left child or a right child, or both, or neither. The children of a node are drawn below it, connected by edges. Here are the five possible binary trees with three nodes.

Note that the directions left and right of the children matter. In the second tree, the root has a left child that has a left child, while, in the fourth tree, the root has a left child that has a right child, and so on.
How many different binary trees can be constructed with four nodes?

A.
13
B.
15
C.
14
D.
30

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: The number of distinct binary trees with $n$ nodes is given by the $n$ -th Catalan number.

Step 1: Recognize that the problem asks for the number of structurally unique binary trees with $n = 4$ nodes, where left and right children are distinguished.

Step 2: Recall the formula for the $n$ -th Catalan number: $C_{n} = \frac{1}{n + 1} (n 2 n)$ .

Step 3: Substitute $n = 4$ into the formula: $C_{4} = \frac{1}{4 + 1} (4 2 \times 4) = \frac{1}{5} (4 8)$ .

Step 4: Calculate the binomial coefficient: $(4 8) = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$ .

Step 5: Divide by 5 to get the final count: $C_{4} = \frac{70}{5} = 14$ .

Answer: 14.

Question 6 · Discrete Mathematics · 2021 MSQ

In order to select a debating team to represent a school, 7 students from class XII and 13 students from class XI were shortlisted and were undergoing trials. The coach had to select a team of 5 students, out of which at least two students should be from each class. One out of the 5 students was to be named as team leader, who was to be from class XII. Two teams with the same members but different leaders are considered to be two different teams. The number of different teams the coach can select is

A.
$2 \times ((2 7) \times (3 13)) + ((3 7) \times (2 13))$
B.
$(2 7) \times (3 13) + (3 7) \times (2 13)$
C.
$2 \times (2 7) \times (3 13) + 3 \times (3 7) \times (2 13)$
D.
$3 \times (2 7) \times (3 13) + 2 \times (3 7) \times (2 13)$

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: This is a committee selection with constraints and a designated leader. The trigger is "at least two from each class" (requiring cases) combined with "leader from class XII" (requiring multiplication by the number of eligible leaders in each case).

Step 1: Determine valid team compositions.

Team size is 5 with at least 2 from each class. The only possibilities are:

Case A: 2 from XII + 3 from XI
Case B: 3 from XII + 2 from XI

Step 2: Count Case A (2 from XII, 3 from XI).

Choose members: $(2 7) \times (3 13)$ .

Choose leader: the leader must be from class XII. There are exactly 2 class XII students in this team, so there are 2 choices for leader.

Case A total: $2 \times (2 7) \times (3 13)$ .

Step 3: Count Case B (3 from XII, 2 from XI).

Choose members: $(3 7) \times (2 13)$ .

Choose leader: the leader must be from class XII. There are 3 class XII students, so 3 choices.

Case B total: $3 \times (3 7) \times (2 13)$ .

Step 4: Add the cases.

Total $= 2 (2 7) (3 13) + 3 (3 7) (2 13)$ .

Step 5: Identify the matching option.

This matches option C exactly.

Common trap: Option B forgets to multiply by the number of leader choices. Option D swaps the multipliers (3 and 2), which would correspond to the leader being from class XI instead.

Answer: C

Question 7 · Probability Theory · 2021 MSQ

Fifteen telephones are received at a service center. Of these, 5 are mobile, 6 are cordless, and 4 are wired. These 15 phones are randomly numbered from 1 to 15 to establish the order in which they are serviced. Which of the following statement(s) is/are correct?

A.
The probability that among the first 3 serviced, the first and third are mobile and the second is not, is $\frac{5 \times 10 \times 4}{15 \times 14 \times 13}$ .
B.
The probability that the first four serviced are all the wired phones, is $\frac{1}{( 4 15 )}$ .
C.
The probability that after servicing ten of these phones, only one of the three types remain to be serviced, is $\frac{( 5 6 )}{( 5 15 )}$ .
D.
The probability that two phones of each type are among the first six serviced, is $\frac{( 2 5 ) + ( 2 6 ) + ( 2 4 )}{( 6 15 )}$ .

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: this is a sampling without replacement question with a random ordering of 15 items split into three types (5M, 6C, 4W). The probability of any ordered type-pattern is computed by multiplying the changing fractions, or equivalently by counting favourable permutations against total permutations.

Option A. "First and third are mobile, second is not."

P(1st is M) = 5/15.
Given that, 14 phones remain, 10 of which are not M. P(2nd is not M) = 10/14.
Given that, 13 phones remain, 4 of which are M. P(3rd is M) = 4/13.
Product = (5·10·4)/(15·14·13). Option A is correct.

Option B. "First four serviced are all wired."

P = (4/15)·(3/14)·(2/13)·(1/12) = 4!/(15·14·13·12) = 24/32760 = 1/1365.
Also 1/ C(15,4) = 1/1365 (choosing which 4 positions the 4 wired phones occupy among the first 4 is forced). Option B is correct.

Option C. "After servicing 10 phones, only one type remains."

The 5 unserviced phones must all be of one type. Only M has 5 phones, so the 5 unserviced must be exactly the 5 mobiles.
P = C(5,5)/C(15,5) = 1/3003. Option C gives C(6,5)/C(15,5) = 6/3003, which is wrong (6 cordless cannot fit into 5 slots). Option C is wrong.

Option D. "Two of each type among the first six."

Favourable count = C(5,2)·C(6,2)·C(4,2) (choose 2 M, 2 C, 2 W independently), divided by C(15,6).
Option D writes a sum in the numerator instead of a product. Option D is wrong.

Answer: A, B.

Question 8 · Probability Theory · 2021 MSQ

The proportion of visitors to an e-commerce website in a week who would purchase a product follows the beta distribution with the probability density function

$f (x) = \frac{Γ ( α + β )}{Γ ( α ) Γ ( β )} x^{α - 1} (1 - x)^{β - 1} for 0 < x < 1,$

where $Γ$ is the gamma function, $α > 0, β > 0$ and $(α + β) > 2$ . The mean and mode of the beta distribution are $E (X) = \frac{α}{α + β}$ and $M o d e (X) = \frac{α - 1}{α + β - 2}$ . From historical data the mean and mode of the proportion of buyers are estimated as $\frac{1}{4}$ and $\frac{1}{5}$ respectively. Which of the following statement(s) is/are correct?

A.
$α = 3$ and $β = 9$ .
B.
The probability that in a given week the proportion of visitors who buy the product is between 10% and 20% is given by $\frac{Γ ( α + β )}{Γ ( α ) Γ ( β )} [\int_{0}^{0.2} x^{α - 1} (1 - x)^{β - 1} d x - \int_{0}^{0.1} x^{α - 1} (1 - x)^{β - 1} d x] .$
C.
The mean proportion of buyers can be calculated as follows $\int_{0}^{1} \frac{Γ ( α + β )}{Γ ( α ) Γ ( β )} x^{α - 1} (1 - x)^{β - 1} d x .$
D.
The probability that in a given week the proportion of visitors who buy the product is less than 20% is given by $1 - \int_{0.2}^{1} \frac{Γ ( α + β )}{Γ ( α ) Γ ( β )} x^{α} (1 - x)^{β} d x .$

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: This is a parameter estimation and PDF recognition question for the Beta distribution. We must solve a system of equations for the parameters and carefully inspect the integral forms for probability and expectation.

Step 1: Use the given mean and mode to find $α$ and $β$ .

We are given $E (X) = \frac{α}{α + β} = \frac{1}{4}$ . This implies $4 α = α + β$ , so $β = 3 α$ .

We are also given $M o d e (X) = \frac{α - 1}{α + β - 2} = \frac{1}{5}$ .

Substitute $β = 3 α$ into the mode equation:

$\frac{α - 1}{α + 3 α - 2} = \frac{1}{5} ⟹ \frac{α - 1}{4 α - 2} = \frac{1}{5}$ .

Cross-multiplying gives $5 (α - 1) = 4 α - 2 ⟹ 5 α - 5 = 4 α - 2 ⟹ α = 3$ .

Then $β = 3 (3) = 9$ . Since $α + β = 12 > 2$ , this is valid. Option A is correct.

Step 2: Evaluate the probability statement in Option B.

The probability that the proportion is between 10% and 20% is $P (0.1 < X < 0.2) = \int_{0.1}^{0.2} f (x) d x$ .

By the properties of definite integrals, $\int_{0.1}^{0.2} f (x) d x = \int_{0}^{0.2} f (x) d x - \int_{0}^{0.1} f (x) d x$ .

Substituting the Beta PDF $f (x)$ , this exactly matches the expression in Option B. Option B is correct.

Step 3: Evaluate the mean calculation in Option C.

The expected value is defined as $E (X) = \int_{0}^{1} x \cdot f (x) d x$ .

Option C presents $\int_{0}^{1} f (x) d x$ , which is missing the $x$ multiplier. This integral simply equals 1 (the total probability), not the mean. Option C is incorrect.

Step 4: Evaluate the probability statement in Option D.

The probability $P (X < 0.2) = 1 - P (X \geq 0.2) = 1 - \int_{0.2}^{1} f (x) d x$ .

Option D uses the integrand $x^{α} (1 - x)^{β}$ , which has incorrect exponents (it should be $x^{α - 1} (1 - x)^{β - 1}$ ). Thus, it does not represent the valid PDF. Option D is incorrect.

Answer: Options A and B are correct.

Question 9 · Probability Theory · 2021 SUB

Common Description: Description for following two questions: A Non-Banking Finance Corporation (NBFC) declares fixed annual rates of simple interest on their auto and housing loans each year. The rates of interest offered by the company differ from year to year depending on the variation in macro economic indicators like inflation, RBI’s repo rate etc. The annual rates of interest offered by the company for the Auto and Housing sectors over the years are shown in the figure.
In 2013, a customer took a housing loan and a car loan. The total loan amount was ₹60 lakhs. The interest paid by the customer after one year was ₹7.36 lakhs. What was the housing loan amount?

Correct Answer:

12.5

Step-by-Step Solution

Key idea: This is a comparative trend analysis question, recognizable because it requires extracting data points from a dual-bar chart and calculating a statistical measure (average) over specific years.

Step 1: Identify the Auto loan rates (left bar in each pair) for the years 2010 to 2013 from the chart description.

2010: 11%
2011: 13%
2012: 12%
2013: 14%

Step 2: Sum the Auto loan rates: $11 + 13 + 12 + 14 = 50$ .

Step 3: Divide by the number of years (4) to find the average: $50/4 = 12.5$ .

Answer: 12.5

Question 10 · Programming · 2021 MSQ

Consider the following code, in which A is an array indexed from 0.

function foo(A,n) {

m = A[0];

x = 0;

for i = 0 to n-1 {

x = x + A[i];

if (m < x) {

m = x;

}

if (x < 0) {

x = 0;

}

return(m);

}

If $A = [- 12, - 3, 5, 10, 8, - 16, - 23, 12, - 5, 7]$ , what will foo(A,10) return?

A.
23
B.
17
C.
-17
D.
35

Correct Answer:

["A"]

Step-by-Step Solution

Insight: This is Kadane's algorithm, which finds the maximum contiguous subarray sum by discarding negative running sums.

Exam route:

Step 1: Initialize m = A[0] = -12, x = 0.

Step 2: Trace the loop:

i=0, A[0]=-12: x = 0 + (-12) = -12. m < x (-12 < -12) is False. x < 0 is True -> x = 0.
i=1, A[1]=-3: x = 0 + (-3) = -3. m < x (-12 < -3) is True -> m = -3. x < 0 is True -> x = 0.
i=2, A[2]=5: x = 0 + 5 = 5. m < x (-3 < 5) is True -> m = 5. x < 0 is False.
i=3, A[3]=10: x = 5 + 10 = 15. m < x (5 < 15) is True -> m = 15. x < 0 is False.
i=4, A[4]=8: x = 15 + 8 = 23. m < x (15 < 23) is True -> m = 23. x < 0 is False.
i=5, A[5]=-16: x = 23 + (-16) = 7. m < x (23 < 7) is False. x < 0 is False.
i=6, A[6]=-23: x = 7 + (-23) = -16. m < x (23 < -16) is False. x < 0 is True -> x = 0.
i=7, A[7]=12: x = 0 + 12 = 12. m < x (23 < 12) is False. x < 0 is False.
i=8, A[8]=-5: x = 12 + (-5) = 7. m < x (23 < 7) is False. x < 0 is False.
i=9, A[9]=7: x = 7 + 7 = 14. m < x (23 < 14) is False. x < 0 is False.

Step 3: Return m = 23.

Learning route: Kadane's algorithm relies on the insight that a negative running sum will only decrease the total of any future subarray. By resetting x to 0 whenever it drops below zero, we effectively start a new subarray at the next index. The variable m safely preserves the highest sum encountered, even if the array starts with negative numbers (hence m is initialized to A[0]).

Answer: 23

Section-wise Paper Structure

Free Solved Questions with Step-by-Step Solutions

Unlock All 23 Questions in Real Examination Mode

More CMI Data Science Previous Year Papers

Login to view the complete paper and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

CMI Data Science 2021 Question Paper with Solutions: 23 Questions, Answer Key & Section-wise Analysis

Paper breakdown

Free sample questions from CMI Data Science 2021 Question Paper

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Other CMI Data Science papers