Mock Test 9 for CMI Data Science: 40 Questions with Solutions & Analysis

Attempt the Mock Test 9 for CMI Data Science: 40 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

40 questions · 82 marks · 159.81 minutes. School Level Mathematics: 17 · Probability Theory: 10 · Discrete Mathematics: 10 · Programming: 3

Free sample questions from Mock Test 9

Question 1 · School Level Mathematics SUB

Let $σ = (1223314554)$ and $τ = (1123324554)$ .

Define the operation $⋆$ on $S_{5}$ by $α ⋆ β = β \circ α$ (note the reverse order).

What is the value of $sgn (σ ⋆ τ)$ ?

Correct Answer:

-1

Step-by-Step Solution

Key idea: This tests the definition of composition order and the multiplicative property of the sign function.

Step 1: Understand the operation.

$σ ⋆ τ = τ \circ σ$ . This means we apply $σ$ first, then $τ$ .

Step 2: Use the property of the sign function.

$sgn (τ \circ σ) = sgn (τ) \cdot sgn (σ)$ .

We do not need to compute the resulting permutation explicitly if we can find the signs of $σ$ and $τ$ individually.

Step 3: Find $sgn (σ)$ .

$σ = (2, 3, 1, 5, 4)$ in one-line notation.

Inversions in $σ$ :

Pairs from ${2, 3, 1}$ : $(2, 1), (3, 1)$ . (2 inversions)
Pairs involving 5, 4: $(5, 4)$ . (1 inversion)
Check cross terms: $2 < 5, 2 < 4$ ; $3 < 5, 3 < 4$ ; $1 < 5, 1 < 4$ . No inversions.

Total inversions for $σ = 3$ .

$sgn (σ) = (- 1)^{3} = - 1$ .

Alternatively, look at cycles: $σ = (123) (45)$ .

Sign = $sgn (123) \cdot sgn (45) = (- 1)^{2} \cdot (- 1)^{1} = 1 \cdot - 1 = - 1$ .

Step 4: Find $sgn (τ)$ .

$τ = (1, 3, 2, 5, 4)$ in one-line notation.

Inversions in $τ$ :

$(3, 2)$ is an inversion.
$(5, 4)$ is an inversion.

Total inversions for $τ = 2$ .

$sgn (τ) = (- 1)^{2} = 1$ .

Alternatively, cycles: $τ = (1) (23) (45)$ .

Sign = $1 \cdot (- 1) \cdot (- 1) = 1$ .

Step 5: Calculate $sgn (σ ⋆ τ)$ .

$sgn (τ \circ σ) = sgn (τ) \cdot sgn (σ) = 1 \cdot (- 1) = - 1$ .

Answer: -1

Question 2 · School Level Mathematics SUB

Let $f (x) = x^{4} + k x^{3} + 6 x^{2}$ , where $k$ is a real constant.

The set of all values of $k$ for which the curve $y = f (x)$ has exactly two distinct inflection points is of the form $(- \infty, a) \cup (b, \infty)$ .

Find the value of $a + b$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a Parameter Analysis question for inflection points. It requires determining when the second derivative has two distinct real roots and changes sign at both.

Step 1: Compute derivatives.

$f^{'} (x) = 4 x^{3} + 3 k x^{2} + 12 x$

$f^{''} (x) = 12 x^{2} + 6 k x + 12 = 6 (2 x^{2} + k x + 2)$

Step 2: Analyze roots of $f^{''} (x) = 0$ .

For inflection points, $f^{''} (x)$ must change sign. Since $f^{''} (x)$ is a quadratic polynomial, it changes sign at a root if and only if the root is simple (multiplicity 1). Thus, we need two distinct real roots for $2 x^{2} + k x + 2 = 0$ .

Step 3: Apply discriminant condition.

Discriminant $Δ = k^{2} - 4 (2) (2) = k^{2} - 16$ .

For two distinct real roots, we require $Δ > 0$ :

$k^{2} - 16 > 0 ⟹ k^{2} > 16 ⟹ k \in (- \infty, - 4) \cup (4, \infty)$

Step 4: Verify sign change.

Since the leading coefficient is positive and roots are distinct, the quadratic necessarily changes sign at both roots. No further checking is needed for polynomials with simple roots.

Step 5: Identify $a$ and $b$ .

The set is $(- \infty, - 4) \cup (4, \infty)$ . Comparing with $(- \infty, a) \cup (b, \infty)$ , we have $a = - 4$ and $b = 4$ .

Step 6: Calculate sum.

$a + b = - 4 + 4 = 0$

Answer: 0

Question 3 · School Level Mathematics MSQ

Let $f (x)$ be a polynomial of degree 5 with real coefficients. Suppose $f (x)$ has exactly three distinct real roots, and $f^{''} (x)$ has exactly three distinct real roots. Which of the following statements MUST be true?

A.
$f^{'} (x)$ has exactly three distinct real roots.
B.
All roots of $f^{''} (x)$ lie strictly between the smallest and largest roots of $f (x)$ .
C.
$f^{'} (x)$ has exactly four distinct real roots.
D.
$f (x)$ has exactly five distinct real roots.

Correct Answer:

Step-by-Step Solution

Key idea: This is a polynomial roots and Rolle's theorem synthesis question. We must analyze the interlacing of roots across successive derivatives.

Step 1: Let the three distinct real roots of $f (x)$ be $r_{1} < r_{2} < r_{3}$ .

Step 2: Apply Rolle's Theorem to $f$ . Between $r_{1}$ and $r_{2}$ , there is at least one root of $f^{'}$ . Between $r_{2}$ and $r_{3}$ , there is at least one root of $f^{'}$ . Call these $s_{1}$ and $s_{2}$ with $r_{1} < s_{1} < s_{2} < r_{3}$ .

Step 3: Analyze $f^{'} (x)$ . Since $f$ is degree 5, $f^{'}$ is degree 4. It has at least 2 real roots. It can have 2 or 4 distinct real roots.

Step 4: Analyze $f^{''} (x)$ . We are told $f^{''}$ has exactly 3 distinct real roots $t_{1} < t_{2} < t_{3}$ . Since $f^{''}$ is a cubic, these are all its roots.

Step 5: Apply Rolle's Theorem to $f^{'}$ . Between any two distinct real roots of $f^{'}$ , there is a root of $f^{''}$ .

Case A: $f^{'}$ has exactly 2 distinct real roots $s_{1}, s_{2}$ . Then $f^{''}$ has at least one root in $(s_{1}, s_{2})$ . Since $f^{''}$ is cubic with 3 real roots and $f^{'}$ has only 2 real roots (meaning $f^{'}$ does not cross the axis outside $[s_{1}, s_{2}]$ ), all three roots of $f^{''}$ must lie in $(s_{1}, s_{2})$ . Thus $r_{1} < s_{1} < t_{1} < t_{2} < t_{3} < s_{2} < r_{3}$ .

Case B: $f^{'}$ has 4 distinct real roots. By Rolle's Theorem, $f^{''}$ has a root between each adjacent pair, giving exactly 3 roots of $f^{''}$ between the smallest and largest roots of $f^{'}$ . Since the roots of $f^{'}$ lie in $(r_{1}, r_{3})$ , so do the roots of $f^{''}$ .

Step 6: In both cases, $r_{1} < t_{1} < t_{3} < r_{3}$ . Option B is always true.

Options A and C are not necessarily true since $f^{'}$ can have 2 or 4 real roots. Option D contradicts the given information.

Answer: Option B.

Question 4 · Probability Theory SUB

Let $X$ and $Y$ be independent random variables, each uniformly distributed on ${1, 2, 3, 4, 5, 6}$ . Define $Z = max (X, Y) - min (X, Y)$ . Find $E [Z]$ . Express your answer as a decimal rounded to three decimal places.

Correct Answer:

1.944

Step-by-Step Solution

Key idea: This is a function of two independent RVs question. The expression $Z = max (X, Y) - min (X, Y)$ simplifies to $∣ X - Y ∣$ , the absolute difference.

Step 1: Find the PMF of $Z$ . Since $X, Y \in {1, ..., 6}$ , we have $Z \in {0, 1, 2, 3, 4, 5}$ .

For $Z = 0$ : $X = Y$ , so $P (Z = 0) = 6/36 = 1/6$ .

For $Z = k$ where $k \geq 1$ : We need $∣ X - Y ∣ = k$ . This means either $X - Y = k$ or $Y - X = k$ .

$X - Y = k$ : $X$ can be $k + 1, k + 2, ..., 6$ , giving $6 - k$ pairs.
$Y - X = k$ : Similarly, $6 - k$ pairs.
Total: $2 (6 - k)$ pairs.

So $P (Z = k) = 2 (6 - k) /36 = (6 - k) /18$ for $k = 1, 2, 3, 4, 5$ .

Step 2: Compute $E [Z]$ using LOTUS.

$E [Z] = k = 0 \sum 5 k \cdot P (Z = k) = 0 \cdot \frac{1}{6} + k = 1 \sum 5 k \cdot \frac{6 - k}{18}$

$= \frac{1}{18} k = 1 \sum 5 k (6 - k) = \frac{1}{18} k = 1 \sum 5 (6 k - k^{2})$

$= \frac{1}{18} [6 \cdot \frac{5 \cdot 6}{2} - \frac{5 \cdot 6 \cdot 11}{6}] = \frac{1}{18} [90 - 55] = \frac{35}{18}$

Step 3: Convert to decimal.

$35 \div 18 \approx 1.9444...$

Answer: $1.944$

Question 5 · Probability Theory SUB

Let $A$ , $B$ , and $C$ be events in a sample space such that they are pairwise independent, $P (A) = P (B) = P (C) = p$ , and $P (A \cap B \cap C) = 0$ . If $P (A \cup B \cup C) = \frac{3}{4}$ , find the value of $p$ .

Correct Answer:

0.5

Step-by-Step Solution

Key idea: this is an inclusion-exclusion question combined with the definition of pairwise independence. The conditions allow us to express the probability of the union entirely in terms of $p$ .

Step 1: recall the inclusion-exclusion formula for three events:

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

Step 2: use the given conditions to simplify the formula.

Since $P (A) = P (B) = P (C) = p$ , the sum of the singles is $3 p$ .

Since the events are pairwise independent, $P (A \cap B) = P (A) P (B) = p^{2}$ . Similarly, $P (B \cap C) = p^{2}$ and $P (A \cap C) = p^{2}$ . The sum of the pairs is $3 p^{2}$ .

We are given $P (A \cap B \cap C) = 0$ .

Step 3: substitute these into the formula:

$P (A \cup B \cup C) = 3 p - 3 p^{2} + 0 = 3 p - 3 p^{2}$ .

Step 4: set this equal to the given probability of the union and solve for $p$ :

$3 p - 3 p^{2} = \frac{3}{4}$ .

Divide by 3: $p - p^{2} = \frac{1}{4}$ .

Multiply by 4: $4 p - 4 p^{2} = 1 ⟹ 4 p^{2} - 4 p + 1 = 0$ .

Factor the quadratic: $(2 p - 1)^{2} = 0$ .

Therefore, $2 p - 1 = 0 ⟹ p = \frac{1}{2} = 0.5$ .

Answer: 0.5

Question 6 · Probability Theory SUB

Let $X$ be a continuous random variable with probability density function

$f (x) = {c x^{3}, 0, if 0 < x < a otherwise$

where $c$ is the normalizing constant. If the median of $X$ is exactly $1$ , find the value of $a^{4}$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a reverse-engineering question. Instead of finding a probability from a known PDF, you are given a property of the distribution (the median) and must find the unknown parameter of the support.

Step 1: Express the normalizing constant $c$ in terms of $a$ .

$\int_{0}^{a} c x^{3} d x = 1 ⟹ c [\frac{x ^{4}}{4}]_{0}^{a} = 1 ⟹ c \frac{a ^{4}}{4} = 1 ⟹ c = \frac{4}{a ^{4}}$

Step 2: Use the definition of the median.

The median $m$ satisfies $P (X \leq m) = 0.5$ . Here, $m = 1$ .

$\int_{0}^{1} c x^{3} d x = 0.5$

Step 3: Substitute $c$ and solve for $a$ .

$\frac{4}{a ^{4}} [\frac{x ^{4}}{4}]_{0}^{1} = 0.5$

$\frac{4}{a ^{4}} \cdot \frac{1}{4} = 0.5$

$\frac{1}{a ^{4}} = \frac{1}{2}$

Step 4: Isolate $a^{4}$ .

$a^{4} = 2$

Answer: 2

Question 7 · Discrete Mathematics · 2024 SUB

A supplier of art material has four reams of handmade paper, three boxes of acrylic colors and two printing blocks. The two artists in the shop want to buy one item each, but insist on having the same kind of art material. How many items does the supplier have to take out to be sure that the artists’ demand is met?

Correct Answer:

none

Step-by-Step Solution

Insight: "To be sure" signals a worst-case Pigeonhole guarantee — count categories (holes), not stock quantities.

Exam route:

Holes = 3 distinct kinds of material (paper, colors, blocks).
Goal = 2 of the same kind ( $m = 2$ ).
Worst case = draw $m - 1 = 1$ from each category = 3 items.
Next draw forces a match: $3 + 1 = 4$ .

Learning route:

This is a worst-case guarantee question, recognisable because the phrases "to be sure" and "insist on having the same kind" demand absolute certainty, not probability.
The tool is the Generalized Pigeonhole Principle in guarantee form: to guarantee $m$ items in one hole, you need $k (m - 1) + 1$ items, where $k$ is the number of holes.
Step-by-step:

Identify holes: The distinct categories are handmade paper, acrylic colors, and printing blocks. So $k = 3$ . The numbers 4, 3, and 2 are inventory levels — they only ensure you do not exhaust a category during the worst-case draw. They are not the number of holes.
Set the target: Both artists want the same kind, so $m = 2$ .
Build the worst case: Maximum bad luck means drawing exactly $m - 1 = 1$ item from every category without completing a pair. That uses $3 \times 1 = 3$ items.
Add one: The very next item drawn must duplicate one of the three kinds already drawn. Total = $3 + 1 = 4$ .

Tempting wrong path: A student might add the stock quantities $4 + 3 + 2 = 9$ , assuming they need to draw all items to be safe. Another wrong path is treating the stock levels as the number of holes. The exact quantities are distractors; only the number of categories matters.

Verification: If we draw 4 items, by PHP, at least $⌈ 4/3 ⌉ = 2$ items must be of the same kind. If we draw 3, we could have 1 of each, failing the condition. Thus, 4 is the minimum.

Question 8 · Discrete Mathematics · 2024 SUB

There are 7 elevators in a large shopping mall, each stopping at ground floor and at most six other floors. If at least 3 elevators stop at each floor and if it is possible to go from any floor to any other floor without changing elevators, what is the maximum number of floors in the mall?

Correct Answer:

none

Step-by-Step Solution

Insight: Incidence double counting with a hub node — the ground floor is served by every elevator but does not consume the "six other floors" capacity.

Defect: The phrase "without changing elevators" is a known typo in this PYQ for "by changing elevators" (i.e., the network is connected); taken literally, it would imply a single elevator covers all floors, capping the answer at 7. We proceed with the standard incidence double counting interpretation.

Exam route:

Let $n$ = number of upper floors. Total floors = $n + 1$ .
Connector side: $S = 7$ elevators, each with capacity $C = 6$ upper floors. Max incidences = $7 \times 6 = 42$ .
Element side: $n$ upper floors, each in $\geq K = 3$ elevators. Min incidences = $3 n$ .
Inequality: $3 n \leq 42 ⟹ n \leq 14$ .
Total floors = $14 + 1 = 15$ .

Learning route:

This is an incidence double counting question, recognisable by connectors (elevators) with a capacity cap, nodes (floors) with a minimum degree, and a request for the maximum node count.
The method works because every (elevator, upper-floor) pair is an incidence, and we can bound the total from above (elevator side) and below (floor side).
Step-by-step:

Let $n$ be the number of upper floors. Total floors in the mall = $n + 1$ (including ground floor).
Hub identification: The ground floor is explicitly listed as a stop for every elevator, but the capacity constraint says "at most six other floors." So the ground floor is a hub — it does not consume the capacity of 6.
Connector side: 7 elevators, each covering at most 6 upper floors. Total incidences $\leq 7 \times 6 = 42$ .
Element side: Each of the $n$ upper floors must be serviced by at least 3 elevators. Total incidences $\geq 3 n$ .
Combine: $3 n \leq 42 ⟹ n \leq 14$ .
Total floors = upper floors + ground floor = $14 + 1 = 15$ .

Tempting wrong path: A student might include the ground floor in the capacity, thinking each elevator covers 7 floors total, leading to $7 \times 7 = 49$ incidences and $n \leq 16$ , total 17. Another mistake is forgetting to add the ground floor back at the end, answering 14 instead of 15.

Verification: If $n = 14$ , we need $14 \times 3 = 42$ incidences. 7 elevators can provide exactly $7 \times 6 = 42$ incidences. This is perfectly balanced and achievable. Total floors = 15.

Question 9 · Discrete Mathematics SUB

In a survey of 90 people about two brands $X$ and $Y$ , the following is known:

The number who like brand $X$ is 4 times the number who like both brands.
The number who like brand $Y$ is 3 times the number who like both brands.
Exactly 18 people like neither brand.

How many people like exactly one brand?

Correct Answer:

Step-by-Step Solution

Key idea: This is a 2-set survey problem with multiplicative conditions relative to the intersection, recognisable because both set sizes are expressed as multiples of $∣ X \cap Y ∣$ .

Step 1: Let $∣ X \cap Y ∣ = t$ . Then $∣ X ∣ = 4 t$ and $∣ Y ∣ = 3 t$ .

Step 2: Compute the exclusive regions:

$X$ only $= ∣ X ∣ - ∣ X \cap Y ∣ = 4 t - t = 3 t$
$Y$ only $= ∣ Y ∣ - ∣ X \cap Y ∣ = 3 t - t = 2 t$

Step 3: Write the union:

$∣ X \cup Y ∣ = 3 t + 2 t + t = 6 t$

Step 4: Use the neither information:

$∣ X \cup Y ∣ + neither = 90$

$6 t + 18 = 90 ⟹ 6 t = 72 ⟹ t = 12$

Step 5: Compute exactly one:

$Exactly one = X only + Y only = 3 t + 2 t = 5 t = 5 (12) = 60$

Step 6: Verify. $∣ X ∣ = 48$ , $∣ Y ∣ = 36$ , both $= 12$ . Union $= 48 + 36 - 12 = 72$ . Neither $= 90 - 72 = 18$ . $✓$

Answer: 60

Question 10 · Programming MSQ

Consider the standard correct algorithm for finding the closest value to a target $x$ in an array $A$ , initialized with min_diff = \infty and closest_val = null. The algorithm uses a strict inequality (current_diff < min_diff) to update the trackers. Let $A$ be an array defined by $A = [18, 14, 6, 14, 6]$ . If the target is $x = 10$ , what is the final value of closest_val?

A.
14
B.
6
C.
18
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: The algorithm updates closest_val only when a strictly smaller absolute difference is found, retaining the first encountered value in case of a tie.

Step 1: Calculate the absolute difference from $x = 10$ for each element in $A$ .

$i = 0$ : $A [0] = 18$ , diff = $∣18 - 10∣ = 8$ . min_diff becomes 8, closest_val = 18.
$i = 1$ : $A [1] = 14$ , diff = $∣14 - 10∣ = 4$ . Since $4 < 8$ is True, min_diff becomes 4, closest_val = 14.
$i = 2$ : $A [2] = 6$ , diff = $∣6 - 10∣ = 4$ . Since $4 < 4$ is False, no update occurs.
$i = 3$ : $A [3] = 14$ , diff = $∣14 - 10∣ = 4$ . Since $4 < 4$ is False, no update occurs.
$i = 4$ : $A [4] = 6$ , diff = $∣6 - 10∣ = 4$ . Since $4 < 4$ is False, no update occurs.

Step 2: The final closest_val remains 14, as the ties at $i = 2, 3, 4$ are ignored due to the strict inequality.

Answer: 14.

Question 11 · Programming MSQ

Consider the following three modified bubble sort algorithms applied to the array $A = [3, 1, 4, 2]$ .

Algorithm X: Standard ascending bubble sort. Returns the final value of A[n-1].

Algorithm Y: Standard ascending bubble sort. Returns the final value of A[1].

Algorithm Z: Ascending bubble sort, but the inner loop runs for j from 0 to n-3 for all passes. Returns the total number of swaps performed.

Match the algorithms in List I with their returned values in List II.

List I

Algorithm X
Algorithm Y
Algorithm Z

List II

P. 1

Q. 2

R. 4

A.
1-P, 2-Q, 3-R
B.
1-R, 2-Q, 3-P
C.
1-Q, 2-R, 3-P
D.
1-R, 2-P, 3-Q

Correct Answer:

Step-by-Step Solution

Key idea: This is a language-to-math question testing the ability to translate modified loop bounds into exact array states and swap counts, identifying the boundary trap.

Step 1: Analyze Algorithm X. Standard ascending bubble sort fully sorts the array. The sorted array is $[1, 2, 3, 4]$ . It returns A[n-1], which is A[3] = 4. Matches R.

Step 2: Analyze Algorithm Y. Standard ascending bubble sort fully sorts the array to $[1, 2, 3, 4]$ . It returns A[1], which is 2. Matches Q.

Step 3: Analyze Algorithm Z. The inner loop runs j from 0 to n-3. For $n = 4$ , n-3 = 1. So j takes values 0 and 1. This means the algorithm only ever compares indices (0,1) and (1,2). It never compares indices (2,3).

Step 4: Trace Algorithm Z on $A = [3, 1, 4, 2]$ .

Pass 1: $j = 0 ⟹ 3 > 1$ (swap) $\to [1, 3, 4, 2]$ . $j = 1 ⟹ 3 > 4$ (no swap). Swaps = 1.
Pass 2: $j = 0 ⟹ 1 > 3$ (no). $j = 1 ⟹ 3 > 4$ (no). Swaps = 0.
Pass 3: No swaps.

The elements at indices 2 and 3 (which are 4 and 2) are never compared, so they are never swapped. The total number of swaps is exactly 1. Matches P.

Step 5: Combine matches: 1-R, 2-Q, 3-P.

Answer: B

Question 12 · Programming MSQ

Consider the following statements regarding the space complexity and bitwise properties of XOR accumulators:

Statement I: A 32-bit accumulator requires 32 bytes of auxiliary space.

Statement II: If the array elements are 64-bit integers, a 32-bit accumulator processing only the lowest 32 bits will yield the exact lowest 32 bits of the true 64-bit XOR sum.

Statement III: An 8-bit accumulator requires 1 byte of auxiliary space, and a 16-bit accumulator requires 2 bytes.

Which of the statements given above are true?

A.
Statement I and Statement II only
B.
Statement II and Statement III only
C.
Statement I and Statement III only
D.
Statement I, Statement II and Statement III

Correct Answer:

Step-by-Step Solution

Key idea: XOR operates bitwise and independently on each bit position. Space optimization must respect the physical unit of data (bits vs bytes).

Step 1: Evaluate Statement I. A 32-bit accumulator requires 32 bits of space. Since 8 bits = 1 byte, 32 bits = 4 bytes. The statement claims 32 bytes, which is a unit mismatch. False.

Step 2: Evaluate Statement II. XOR is bitwise independent. The lowest 32 bits of the XOR sum depend only on the lowest 32 bits of the inputs. Thus, a 32-bit accumulator processing only the lowest 32 bits will yield the exact lowest 32 bits of the true 64-bit XOR sum. True.

Step 3: Evaluate Statement III. An 8-bit accumulator requires 8 bits = 1 byte. A 16-bit accumulator requires 16 bits = 2 bytes. Both conversions are correct. True.

Answer: B

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Mock Test 9 for CMI Data Science: 40 Questions with Solutions & Analysis

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