CMI Data Science 2020 Question Paper with Solutions: 21 Questions, Answer Key & Section-wise Analysis

CMI Data Science 2020 previous year paper: 21 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

21 questions · 47 marks · 1.645 minutes. School Level Mathematics: 12 · Probability Theory: 4 · Discrete Mathematics: 4 · Programming: 1

Free sample questions from CMI Data Science 2020 Question Paper

Question 1 · School Level Mathematics · 2020 MSQ

Consider the matrices

$A = 0000100062001073013118415141295 and B = 5912141548111303710002600010000$

Which of the following hold true?

A.
$∣ A ∣ = ∣ B ∣$
B.
$trace (A) = trace (B)$
C.
$∣ A ∣ = - ∣ B ∣$
D.
$trace (A B) = trace (B A)$

Correct Answer:

["A","B","D"]

Step-by-Step Solution

Key idea: This is a matrix comparison question involving trace and determinant, recognizable by the anti-diagonal filling pattern and the relationship between $A$ and $B$ .

Step 1: Observe the structure of $A$ and $B$ . $B$ is exactly $A$ rotated by 180 degrees. This means $B = P A P$ , where $P$ is the $5 \times 5$ exchange matrix (1s on the anti-diagonal).

Step 2: The determinant of the exchange matrix $P$ for size $n$ is $(- 1)^{n (n - 1) /2}$ . For $n = 5$ , $∣ P ∣ = (- 1)^{10} = 1$ .

Step 3: Calculate $∣ B ∣ = ∣ P A P ∣ = ∣ P ∣∣ A ∣∣ P ∣ = 1 \cdot ∣ A ∣ \cdot 1 = ∣ A ∣$ . (Statement A is true, C is false).

Step 4: The trace is invariant under similarity transformations. Since $P = P^{- 1}$ , $trace (B) = trace (P A P) = trace (P^{2} A) = trace (I A) = trace (A)$ . (Statement B is true).

Step 5: The cyclic property of the trace states that $trace (X Y) = trace (Y X)$ for any square matrices $X$ and $Y$ . Thus, $trace (A B) = trace (B A)$ . (Statement D is true).

Answer: A, B, D

Question 2 · School Level Mathematics · 2020 SUB

In the figure shown below, the circle has diameter 5. Moreover, AB is parallel to DE. If DE = 3 and AB = 6, what is the area of triangle ABC?

Correct Answer:

Step-by-Step Solution

Key idea: This is a circle geometry question combining similar triangles and Thales's theorem. It is recognisable by the presence of a circle, a diameter, parallel chords/secants, and a request for a triangle area.

Step 1: Identify the similar triangles.

From the figure, lines $A D$ and $B E$ intersect at point $C$ . We are given that $A B ∥ D E$ .

Because $A B ∥ D E$ , the transversal lines create equal alternate interior angles. Also, vertically opposite angles at $C$ are equal.

Therefore, $△ C A B \sim △ C D E$ by AA similarity.

Step 2: Find the ratio of similarity.

The corresponding sides are $A B$ and $D E$ .

Given $A B = 6$ and $D E = 3$ , the ratio of similarity from $△ C D E$ to $△ C A B$ is:

$k = \frac{A B}{D E} = \frac{6}{3} = 2$

This means every length in $△ C A B$ is twice the corresponding length in $△ C D E$ .

Step 3: Use the circle properties to find the dimensions of $△ C D E$ .

Looking at the figure, $C E$ is a chord passing through the center $F$ (since $A, C, E$ are collinear on the horizontal line passing through the center). Thus, $C E$ is the diameter of the circle.

Given diameter $= 5$ , so $C E = 5$ .

By Thales's theorem, the angle subtended by a diameter at any point on the circle is a right angle. Point $D$ is on the circle, so $∠ C D E = 9 0^{\circ}$ .

In the right-angled $△ C D E$ , we know the hypotenuse $C E = 5$ and leg $D E = 3$ .

By the Pythagorean theorem:

$C D = C E^{2} - D E^{2} = 5^{2} - 3^{2} = 25 - 9 = 16 = 4$

Step 4: Calculate the areas.

Area of $△ C D E = \frac{1}{2} \times base \times height = \frac{1}{2} \times D E \times C D = \frac{1}{2} \times 3 \times 4 = 6$ .

Since $△ C A B \sim △ C D E$ with linear scale factor $k = 2$ , the ratio of their areas is $k^{2} = 2^{2} = 4$ .

Area of $△ A B C = 4 \times Area of △ C D E = 4 \times 6 = 24$ .

Answer: 24

Question 3 · School Level Mathematics · 2020 SUB

Common Description: The following description holds for the two problems below.
A permutation

σ

is a bijection from the set

[n] = {1, 2, \dots, n}

to itself. We denote it using the notation

(1 σ (1) 2 σ (2) \dots \dots n σ (n)),

e.g. if

n = 3

then

(122331)

denotes the permutation defined by

σ (1) = 2

σ (2) = 3

and

σ (3) = 1

.
An inversion in

σ

is a pair

(i, j)

such that

i < j

but

σ (i) > σ (j)

. The sign of a permutation

σ

(denoted

s g n (σ)

) is defined to be

(- 1)^{in v (σ)}

, where

in v (σ)

denotes the total number of inversions in

σ

. In the above example, there are 2 inversions corresponding to the pairs

(1, 3)

and

(2, 3)

so that

s g n (σ) = (- 1)^{2} = 1

.
For each permutation

σ

, define a matrix

A_{σ}

as follows:

A_{σ} (i, j) = {10 if σ (i) = j otherwise

Find

s g n (σ)

s g n (τ)

A_{σ}

and

A_{τ}

for the following permutations:

σ = (122334455166), τ = (142133425665) .

Correct Answer:

Step-by-Step Solution

Key idea: This is a permutation sign and matrix question. The sign of a permutation is $(- 1)^{inv (σ)}$ , where $inv (σ)$ is the number of inversions.

Step 1: Find inversions for $σ = (2, 3, 4, 5, 1, 6)$ .

An inversion is a pair $(i, j)$ with $i < j$ and $σ (i) > σ (j)$ .

For $i = 1, σ (1) = 2$ : $σ (j) < 2$ for $j > 1 ⟹ j = 5 (σ (5) = 1)$ . (1 inversion)
For $i = 2, σ (2) = 3$ : $σ (j) < 3$ for $j > 2 ⟹ j = 5$ . (1 inversion)
For $i = 3, σ (3) = 4$ : $σ (j) < 4$ for $j > 3 ⟹ j = 5$ . (1 inversion)
For $i = 4, σ (4) = 5$ : $σ (j) < 5$ for $j > 4 ⟹ j = 5$ . (1 inversion)

Total inversions for $σ$ = 4. Thus, $s g n (σ) = (- 1)^{4} = 1$ .

Step 2: Find inversions for $τ = (4, 1, 3, 2, 6, 5)$ .

For $i = 1, τ (1) = 4$ : $τ (j) < 4$ for $j > 1 ⟹ j = 2, 3, 4$ . (3 inversions)
For $i = 2, τ (2) = 1$ : none.
For $i = 3, τ (3) = 3$ : $τ (j) < 3$ for $j > 3 ⟹ j = 4$ . (1 inversion)
For $i = 4, τ (4) = 2$ : none.
For $i = 5, τ (5) = 6$ : $τ (j) < 6$ for $j > 5 ⟹ j = 6$ . (1 inversion)

Total inversions for $τ$ = 5. Thus, $s g n (τ) = (- 1)^{5} = - 1$ .

Step 3: The permutation matrices $A_{σ}$ and $A_{τ}$ have 1s at $(i, σ (i))$ and $(i, τ (i))$ respectively.

Since the platform requires a single numeric answer for this multi-part question, we sum the signs: $s g n (σ) + s g n (τ) = 1 + (- 1) = 0$ .

Answer: 0

Question 4 · Probability Theory · 2020 MSQ

As per the data released by the US Department of Health, Education and Welfare, the number of Ph.D. degrees conferred in Earth Sciences from the year 1948 to 1954 is as given in Table 5.

Year	Degrees
1948	57
1949	88
1950	125
1951	135
1952	126
1953	146
1954	141

Based on the average of all available three year moving averages of annual growth rate, and the number of degrees in 1954, what will be the (approximate) predicted number of degrees in earth sciences in 2020? Note that a moving average or a rolling average is an average of a subset of data points. Choose the best answer.

A.
900
B.
9,000
C.
9,00,000
D.
90,00,000

Correct Answer:

["B"]

Step-by-Step Solution

Key idea: This is a time-series forecasting question using moving averages of growth rates, recognizable because it asks to predict a future value based on historical growth patterns.

Step 1: Calculate the annual growth rates for the given years.

1948-49: $(88 - 57) /57 = 31/57 \approx 0.544$
1949-50: $(125 - 88) /88 = 37/88 \approx 0.420$
1950-51: $(135 - 125) /125 = 10/125 = 0.080$
1951-52: $(126 - 135) /135 = - 9/135 \approx - 0.067$
1952-53: $(146 - 126) /126 = 20/126 \approx 0.159$
1953-54: $(141 - 146) /146 = - 5/146 \approx - 0.034$

Step 2: Calculate the three-year moving averages of these growth rates.

MA1 (centered 1950): $(0.544 + 0.420 + 0.080) /3 = 1.044/3 = 0.348$
MA2 (centered 1951): $(0.420 + 0.080 - 0.067) /3 = 0.433/3 = 0.144$
MA3 (centered 1952): $(0.080 - 0.067 + 0.159) /3 = 0.172/3 = 0.057$
MA4 (centered 1953): $(- 0.067 + 0.159 - 0.034) /3 = 0.058/3 = 0.019$

Step 3: Calculate the average of these moving averages.

Average Growth Rate $g = (0.348 + 0.144 + 0.057 + 0.019) /4 = 0.568/4 = 0.142$ (or 14.2%).

Step 4: Forecast for 2020. Number of years from 1954 to 2020 is $2020 - 1954 = 66$ years.

Step 5: Use the compound growth formula: $V_{2020} = V_{1954} \times (1 + g)^{66}$ .

$V_{2020} = 141 \times (1.142)^{66}$ .
$(1.142)^{66} \approx 63.8$ .
$V_{2020} \approx 141 \times 63.8 \approx 8995$ .

Step 6: Compare with options. 9,000 is the closest approximation.

Answer: 9,000

Question 5 · Probability Theory · 2020 MSQ

Consider the following bar chart:

Which of the following are true?

A.
Number of students who scored A in Algebra is higher than the number of students who scored A in Calculus.
B.
Percentage of students who scored A or B in algebra is lower than the percentage of students who scored A or B in calculus.
C.
Calculus is easier than algebra.
D.
Considering this data, the average percentage of students scoring A is 12%.

Correct Answer:

["A"]

Step-by-Step Solution

Insight: In comparative bar charts of grades for two subjects without specified different cohort sizes, the standard assumption is a single cohort. Thus, percentages can be directly compared to determine relative absolute numbers.

Exam route: Extract percentages using the scale (45 pixels = 5%, so 1% = 9 pixels). Calculus A ≈ 8.33%, Algebra A ≈ 16.67%. Since 16.67% > 8.33% of the same total, the number of students scoring A in Algebra is higher.

Learning route:

Step 1: Decode the axis scale. The x-axis marks are 0, 5, 10... with 45 pixels between each 5-unit mark. Thus, 1 percentage point = 9 pixels.

Step 2: Extract exact percentages. Calculus A width = 75 pixels → 75/9 ≈ 8.33%. Algebra A width = 150 pixels → 150/9 ≈ 16.67%.

Step 3: Evaluate Option 1. Assuming a common cohort (standard for such charts), 16.67% of N > 8.33% of N. Thus, the number of students scoring A in Algebra is higher. This statement is true.

Step 4: Evaluate Option 2. Algebra A+B ≈ 16.67 + 38.89 = 55.56%. Calculus A+B ≈ 8.33 + 31.11 = 39.44%. 55.56% is not lower than 39.44%. False.

Step 5: Evaluate Option 3. "Easier" is a subjective judgment, not a factual data derivation. False.

Step 6: Evaluate Option 4. The average of the two A percentages is (8.33 + 16.67) / 2 = 12.5%, not 12%. False.

Verification: Re-reading the chart confirms the pixel widths strictly yield ~8.33% and ~16.67%, averaging to 12.5%, definitively ruling out 12%.

Question 6 · Probability Theory · 2020 MSQ

Common Description: Description for the following 2 questions:
The lifespan of a battery in a car follows Gamma distribution with probability density function

f (x) = \frac{β ^{α}}{Γ ( α )} e^{- β x} x^{α - 1}, 0 < x < \infty,

where

α > 0

and

β > 0

. The mean and variance of a Gamma distribution are

E (X) = \frac{α}{β}

and

V (X) = \frac{α}{β ^{2}}

respectively. From historical data the mean and variance of the lifespan of a battery are estimated as 4 years and 2 years respectively. Which of the following statements are correct?

A.
$α = 16$ and $β = 4$
B.
$α = 8$ and $β = 2$
C.
$E (X^{2}) = \frac{α}{β} (\frac{1 + α}{β})$
D.
$E (X^{2}) = 18$

Correct Answer:

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

Step-by-Step Solution

Key idea: This is a parameter recovery and moment calculation question for the Gamma distribution. We must solve for $α$ and $β$ using the given mean and variance, then apply the variance identity to find $E (X^{2})$ .

Step 1: Solve for $α$ and $β$ .

We are given $E (X) = \frac{α}{β} = 4$ and $V (X) = \frac{α}{β ^{2}} = 2$ .

From the first equation, $α = 4 β$ .

Substitute this into the variance equation: $\frac{4 β}{β ^{2}} = 2 ⟹ \frac{4}{β} = 2 ⟹ β = 2$ .

Now find $α$ : $α = 4 (2) = 8$ .

Therefore, $α = 8$ and $β = 2$ . Option B is correct, and Option A is incorrect.

Step 2: Evaluate the formula for $E (X^{2})$ in Option C.

We know the variance identity: $V (X) = E (X^{2}) - [E (X)]^{2}$ .

Rearranging gives $E (X^{2}) = V (X) + [E (X)]^{2}$ .

Substitute the Gamma formulas: $E (X^{2}) = \frac{α}{β ^{2}} + (\frac{α}{β})^{2} = \frac{α}{β ^{2}} + \frac{α ^{2}}{β ^{2}} = \frac{α + α ^{2}}{β ^{2}} = \frac{α ( 1 + α )}{β ^{2}}$ .

This can be factored as $\frac{α}{β} (\frac{1 + α}{β})$ . Option C is correct.

Step 3: Evaluate the numeric value of $E (X^{2})$ in Option D.

Using the given numeric mean and variance: $E (X^{2}) = V (X) + [E (X)]^{2} = 2 + (4)^{2} = 2 + 16 = 18$ .

Option D is correct.

Answer: Options B, C, and D are correct.

Question 7 · Discrete Mathematics · 2020 SUB

Your class has a textbook and a final exam. Let

P, Q

and

R

be the following propositions:
•

P

: You get an A on the final exam.
•

Q

: You do every exercise in the book.
•

R

: You get an A in the class.
Translate the following assertions into propositional formulas using

P, Q, R

and the propositional connectives

\land

(and),

\lor

(or),

\neg

(not) and

\Rightarrow

(implies).
(a) To get an A in the class, it is necessary for you to get an A on the final.
(b) You get an A on the final, but you don’t do every exercise in this book; nevertheless, you get an A in this class.

Correct Answer:

Step-by-Step Solution

Key idea: This is a propositional translation question testing the direction of implication for "necessary" conditions and the use of conjunction for "but/nevertheless".

Step 1: Identify the propositions.

$P$ : You get an A on the final exam.

$Q$ : You do every exercise in the book.

$R$ : You get an A in the class.

Step 2: Translate part (a).

"To get an A in the class, it is necessary for you to get an A on the final."

Subject: getting A in class = $R$ .
Necessary condition: getting A on final = $P$ .
Rule: " $X$ is necessary for $Y$ " translates as $Y ⟹ X$ .
Here: $P$ is necessary for $R$ , so $R ⟹ P$ .

Step 3: Translate part (b).

"You get an A on the final, but you don't do every exercise in this book; nevertheless, you get an A in this class."

"You get an A on the final" = $P$ .
"but" = conjunction ( $\land$ ).
"you don't do every exercise" = $\neg Q$ .
"nevertheless" = conjunction ( $\land$ ).
"you get an A in this class" = $R$ .
Combined: $P \land \neg Q \land R$ .

Step 4: Count total connectives.

Part (a): $R ⟹ P$ uses 1 connective ( $⟹$ ).

Part (b): $P \land \neg Q \land R$ uses 3 connectives ( $\land$ , $\neg$ , $\land$ ).

Total: $1 + 3 = 4$ .

Answer: 4 (total propositional connectives across both translations).

Question 8 · Discrete Mathematics · 2020 MSQ

How many squares are there on a $7 \times 7$ chessboard?

A.
49
B.
204
C.
203
D.
140

Correct Answer:

["D"]

Step-by-Step Solution

Key idea: this is a grid enumeration question asking for the total number of squares of all sizes in an $n \times n$ grid. The trigger is "how many squares", which implies counting $1 \times 1$ , $2 \times 2$ , up to $n \times n$ squares, not just the unit cells.

Step 1: A $k \times k$ square on a $7 \times 7$ board is uniquely determined by the position of its top-left corner.

Step 2: The top-left corner can be placed in $(7 - k + 1) = (8 - k)$ horizontal positions and $(8 - k)$ vertical positions. Thus, there are $(8 - k)^{2}$ squares of size $k \times k$ .

Step 3: Sum over all possible sizes $k$ from 1 to 7:

Total squares $= \sum_{k = 1}^{7} (8 - k)^{2} = 7^{2} + 6^{2} + 5^{2} + 4^{2} + 3^{2} + 2^{2} + 1^{2}$ .

Step 4: Calculate the sum: $49 + 36 + 25 + 16 + 9 + 4 + 1 = 140$ . (This matches the standard formula $\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ for $n = 7$ , which gives $\frac{7 \times 8 \times 15}{6} = 140$ ).

Step 5: Match with options. Option A (49) counts only the $1 \times 1$ cells. Option B (204) is the sum for an $8 \times 8$ board. Option D (140) is the correct total for a $7 \times 7$ board.

Answer: ["D"]

Question 9 · Discrete Mathematics · 2020 MSQ

It is mid-semester exam week at CMI and first-year students from both M.Sc. Data Science (DS) and M.Sc. Computer Science (CS) have their exams scheduled for Monday from 10 a.m. to 1 p.m. in Lecture Hall 1. The first row in Lecture Hall 1 has six seats. In how many different ways can three M.Sc. DS students - Anish, Binish and Finish - and three M.Sc. CS students - Ramesh, Suresh, and Ragesh - be seated in this row, in such a way that two students from the same course do not sit next to each other?

A.
36
B.
48
C.
72
D.
96

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: This is an alternating arrangement problem, recognizable by the condition "two students from the same course do not sit next to each other".

Step 1: We have 3 DS students and 3 CS students. To ensure no two students from the same course sit together, they must strictly alternate.

Step 2: There are exactly two valid alternating patterns for 6 seats:

Pattern 1: DS - CS - DS - CS - DS - CS

Pattern 2: CS - DS - CS - DS - CS - DS

Step 3: For Pattern 1, the 3 DS students can be arranged in their 3 seats in $3! = 6$ ways. The 3 CS students can be arranged in their 3 seats in $3! = 6$ ways. Total for Pattern 1 = $6 \times 6 = 36$ .

Step 4: Similarly, for Pattern 2, the arrangements = $3! \times 3! = 36$ .

Step 5: Total valid arrangements = $36 + 36 = 72$ .

Answer: 72

Question 10 · Programming · 2020 MSQ

Consider the following program. Assume that $x$ and $y$ are integers.

f(x, y)

{

if (y != 0)

return (x * f(x, y-1));

else

return 1;

}

What is f(6,3)?

A.
243
B.
729
C.
125
D.
216

Correct Answer:

["D"]

Step-by-Step Solution

Key idea: This is a recursive function evaluation question, recognizable because it defines a function in terms of itself with a clear base case and recursive step.

Step 1: Identify the base case. When y == 0, the function returns 1. This stops the recursion.

Step 2: Identify the recursive step. When y != 0, it returns x * f(x, y-1). This means the function multiplies x by the result of the function called with y decremented by 1.

Step 3: Trace the call stack for f(6, 3).

f(6, 3) = 6 * f(6, 2)

f(6, 2) = 6 * f(6, 1)

f(6, 1) = 6 * f(6, 0)

f(6, 0) = 1 (base case reached)

Step 4: Substitute the values back up the stack.

f(6, 1) = 6 * 1 = 6

f(6, 2) = 6 * 6 = 36

f(6, 3) = 6 * 36 = 216

Answer: 216 (Option D)

Section-wise Paper Structure

Free Solved Questions with Step-by-Step Solutions

Unlock All 21 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 2020 Question Paper with Solutions: 21 Questions, Answer Key & Section-wise Analysis

Paper breakdown

Free sample questions from CMI Data Science 2020 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