Mock Test 6 for CMI Data Science: 40 Questions with Solutions & Analysis

Attempt the Mock Test 6 for CMI Data Science: 40 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

40 questions · 81 marks · 155.645 minutes. School Level Mathematics: 17 · Probability Theory: 10 · Discrete Mathematics: 10 · Programming: 3

Free sample questions from Mock Test 6

Question 1 · School Level Mathematics SUB

Let $σ$ be a permutation of ${1, 2, 3, 4, 5}$ . Suppose $σ$ has exactly 2 fixed points.

How many such permutations are even?

Correct Answer:

Step-by-Step Solution

Key idea: We must count the permutations with exactly 2 fixed points and determine the parity of their cycle structures.

Step 1: Choose the fixed points.

There are $(2 5) = 10$ ways to choose which 2 elements are fixed points.

Step 2: Analyze the remaining elements.

The remaining 3 elements must form a permutation with NO fixed points (a derangement).

Let's find the derangements of 3 elements.

The only way to permute 3 elements with no fixed points is a single 3-cycle.

For example, if the elements are ${a, b, c}$ , the derangements are $(ab c)$ and $(a c b)$ .

There are exactly $D_{3} = 2$ such derangements.

Step 3: Determine the parity.

A permutation with 2 fixed points and one 3-cycle has the cycle structure $(x) (y) (ab c)$ .

The sign is the product of the signs of the cycles.

Sign of 1-cycle is $+ 1$ .

Sign of 3-cycle is $(- 1)^{3 - 1} = (- 1)^{2} = + 1$ .

Therefore, EVERY such permutation is even.

Step 4: Calculate the total count.

Since all valid permutations are even, we just multiply the number of ways to choose the fixed points by the number of derangements.

Total = $10 \times 2 = 20$ .

Answer: 20

Question 2 · School Level Mathematics SUB

Let $f (x) = \int_{0}^{x} (t^{2} - 1) e^{- t^{2}} d t$ .

Find the number of inflection points of the curve $y = f (x)$ on $R$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a "find_number" question involving an integral function. We must use the Fundamental Theorem of Calculus to find $f^{''} (x)$ and then analyze its sign changes.

Step 1: Find the first derivative using the Fundamental Theorem of Calculus.

$f^{'} (x) = (x^{2} - 1) e^{- x^{2}}$ .

Step 2: Find the second derivative using the product rule and chain rule.

$f^{''} (x) = \frac{d}{d x} [(x^{2} - 1)] \cdot e^{- x^{2}} + (x^{2} - 1) \cdot \frac{d}{d x} [e^{- x^{2}}]$

$f^{''} (x) = 2 x e^{- x^{2}} + (x^{2} - 1) (- 2 x) e^{- x^{2}}$

$f^{''} (x) = e^{- x^{2}} [2 x - 2 x (x^{2} - 1)] = e^{- x^{2}} [2 x - 2 x^{3} + 2 x] = e^{- x^{2}} (4 x - 2 x^{3})$ .

Step 3: Factor $f^{''} (x)$ to find candidate points.

$f^{''} (x) = 2 x e^{- x^{2}} (2 - x^{2})$ .

The candidate points are where $f^{''} (x) = 0$ , which are $x = 0$ , $x = 2$ , and $x = - 2$ .

Step 4: Test for sign changes of $f^{''} (x)$ around these points. Note that $e^{- x^{2}} > 0$ for all $x$ , so the sign of $f^{''} (x)$ depends entirely on $x (2 - x^{2})$ .

For $x < - 2$ : $x < 0$ and $(2 - x^{2}) < 0$ , so $f^{''} (x) > 0$ .
For $- 2 < x < 0$ : $x < 0$ and $(2 - x^{2}) > 0$ , so $f^{''} (x) < 0$ . (Sign changes at $x = - 2$ )
For $0 < x < 2$ : $x > 0$ and $(2 - x^{2}) > 0$ , so $f^{''} (x) > 0$ . (Sign changes at $x = 0$ )
For $x > 2$ : $x > 0$ and $(2 - x^{2}) < 0$ , so $f^{''} (x) < 0$ . (Sign changes at $x = 2$ )

Step 5: Since $f^{''} (x)$ changes sign at all three candidate points, there are exactly 3 inflection points.

Answer: 3

Question 3 · School Level Mathematics MSQ

Let $f (x)$ be a polynomial of degree 4 with leading coefficient 1. Suppose the inflection points of $f (x)$ are at $x = 1$ and $x = 3$ . If $f (x)$ has exactly three distinct local extrema, which of the following statements MUST be true?

A.
$f^{'} (3) < 0$
B.
$f^{'} (1) > 0$
C.
$f^{'} (1) - f^{'} (3) = 16$
D.
$f^{'} (0) > 0$

Correct Answer:

["A","B","C"]

Step-by-Step Solution

Key idea: This is a reverse engineering and parameter analysis question. We reconstruct the derivatives from the given inflection points and use the condition for three local extrema to constrain the unknown constant.

Step 1: Determine $f^{''} (x)$ . Since $f (x)$ is degree 4 with leading coefficient 1, $f^{''} (x)$ is quadratic with leading coefficient $4 \times 3 = 12$ . Given inflection points at $x = 1$ and $x = 3$ :

$f^{''} (x) = 12 (x - 1) (x - 3) = 12 x^{2} - 48 x + 36$ .

Step 2: Integrate to find $f^{'} (x)$ .

$f^{'} (x) = 4 x^{3} - 24 x^{2} + 36 x + C$ for some constant $C$ .

Step 3: Evaluate $f^{'} (x)$ at the critical points of $f^{'} (x)$ , which are the roots of $f^{''} (x)$ : $x = 1$ and $x = 3$ .

$f^{'} (1) = 4 - 24 + 36 + C = 16 + C$ .

$f^{'} (3) = 108 - 216 + 108 + C = C$ .

Step 4: Apply the three-extrema condition. For $f (x)$ to have three distinct local extrema, $f^{'} (x)$ (a cubic with positive leading coefficient) must have three distinct real roots. This requires its local maximum to be positive and its local minimum to be negative.

Since $f^{''} (x)$ changes from positive to negative at $x = 1$ , $f^{'} (x)$ has a local maximum at $x = 1$ . So $f^{'} (1) > 0$ , giving $C > - 16$ .

Since $f^{''} (x)$ changes from negative to positive at $x = 3$ , $f^{'} (x)$ has a local minimum at $x = 3$ . So $f^{'} (3) < 0$ , giving $C < 0$ .

Step 5: Evaluate each option using $- 16 < C < 0$ .

Option A: $f^{'} (3) = C < 0$ . True.

Option B: $f^{'} (1) = 16 + C > 0$ . True.

Option C: $f^{'} (1) - f^{'} (3) = (16 + C) - C = 16$ . True, independent of $C$ .

Option D: $f^{'} (0) = C < 0$ . So $f^{'} (0) > 0$ is false.

Answer: Options A, B, and C.

Question 4 · Probability Theory · 2023 MSQ

The mean and variance of a data set of size $n \geq 100$ are 1.1 and 4.3 respectively. It was then discovered that one data point was wrongly recorded as -0.9 and should be ignored. Then we can conclude that the mean $μ$ and variance $σ^{2}$ of the $n - 1$ data points satisfy

A.
$σ^{2} < 4.3$
B.
$σ^{2} > 4.3$
C.
$μ > 1.1$
D.
$μ < 1.1$

Correct Answer:

["B","C"]

Step-by-Step Solution

Key idea: When a data point is removed, the new mean and variance can be evaluated using the original sum and sum of squares, or by analyzing the removed point's distance from the mean.

Step 1: The original dataset has size $n$ , mean $\overset{x}{ˉ} = 1.1$ , and variance $σ^{2} = 4.3$ .

Step 2: The original sum of the data is $\sum x_{i} = 1.1 n$ . The removed point is $x_{k} = - 0.9$ .

Step 3: The new sum is $1.1 n - (- 0.9) = 1.1 n + 0.9$ . The new size is $n - 1$ .

Step 4: The new mean is $μ = \frac{1.1 n + 0.9}{n - 1}$ . To compare with $1.1$ , we evaluate $μ - 1.1 = \frac{1.1 n + 0.9 - 1.1 ( n - 1 )}{n - 1} = \frac{2.0}{n - 1}$ .

Step 5: Since $n \geq 100$ , $\frac{2.0}{n - 1} > 0$ , which means $μ > 1.1$ . Thus, the third option is correct and the fourth is incorrect.

Step 6: For the variance, the change depends on the squared distance of the removed point from the mean. The exact formula for the new variance is $σ^{'2} = \frac{n}{n - 1} σ^{2} - \frac{n}{( n - 1 ) ^{2}} (\overset{x}{ˉ} - x_{k})^{2}$ .

Step 7: We check the sign of $σ^{'2} - σ^{2} = \frac{( n - 1 ) σ ^{2} - n ( x ˉ - x _{k} ) ^{2}}{( n - 1 ) ^{2}}$ .

Step 8: Substitute $σ^{2} = 4.3$ and $(\overset{x}{ˉ} - x_{k})^{2} = (1.1 - (- 0.9))^{2} = 2. 0^{2} = 4.0$ .

Step 9: The numerator becomes $(n - 1) (4.3) - n (4.0) = 4.3 n - 4.3 - 4.0 n = 0.3 n - 4.3$ .

Step 10: Since $n \geq 100$ , $0.3 n - 4.3 \geq 30 - 4.3 = 25.7 > 0$ . Therefore, $σ^{'2} > σ^{2} = 4.3$ . Thus, the second option is correct and the first is incorrect.

Answer: Options B and C.

Question 5 · Probability Theory · 2021 MSQ

Common Description: Description for following two questions: In the 2019-2020 season of the English Premier League (EPL), 380 matches were played in a home and away format. The figure below describes the number of goals scored by the home team and the away team against the number of matches played. For example, the home team scored one goal in 125 matches.
Which of the following statement(s) is/are correct?

A.
The average number of goals scored by the home team over 380 matches is less than 1.5, and for away teams are less than 1.
B.
In the whole season of the EPL, a total of 1034 goals were scored.
C.
The probability that away team failed to score any goal is 22.11%.
D.
The probability that home team scored more than 3 goals is 6.6%.

Correct Answer:

["B"]

Step-by-Step Solution

Insight: This is an empirical expected value and probability question, requiring computation of means, totals, and probabilities from frequency data, while also recognizing real-world dataset facts or chart discrepancies.

Exam route: Compute the sample mean for home and away teams, check total goals, and calculate specific probabilities to eliminate false options.

Learning route:

Step 1: Analyze the data. The problem explicitly states $N = 380$ matches were played.

Step 2: Verify Statement A. "Average number of goals... home team < 1.5, away team < 1".

Home team mean: $\overset{ˉ}{X}_{home} = \frac{0 ( 84 ) + 1 ( 125 ) + 2 ( 99 ) + 3 ( 47 ) + 4 ( 16 ) + 5 ( 8 )}{380} = \frac{568}{380} \approx 1.49$ . (This is $< 1.5$ ).

Away team mean: $\overset{ˉ}{X}_{away} = \frac{0 ( 123 ) + 1 ( 128 ) + 2 ( 82 ) + 3 ( 33 ) + 4 ( 8 ) + 5 ( 4 ) + 6 ( 1 )}{380} = \frac{449}{380} \approx 1.18$ .

Since $1.18$ is NOT less than $1$ , Statement A is FALSE. (Note: Even using the real-world away total of 466, the mean is $\approx 1.22$ , still not $< 1$ ).

Step 3: Verify Statement B. "Total of 1034 goals were scored".

This is a factual statistic for the 2019-2020 English Premier League season ( $568$ home goals $+ 466$ away goals $= 1034$ total goals). Despite the chart's minor summation discrepancy (summing to $1017$ ), this statement about the "whole season" is factually TRUE and the only defensible choice.

Step 4: Verify Statement C. "Probability that away team failed to score any goal is 22.11%".

$P (Away = 0) = \frac{123}{380} \times 100 \approx 32.4%$ .

The value $22.11%$ actually corresponds to the HOME team ( $\frac{84}{380} \approx 22.1%$ ). Statement C is FALSE.

Step 5: Verify Statement D. "Probability that home team scored more than 3 goals is 6.6%".

$P (Home > 3) = P (Home \in {4, 5}) = \frac{16 + 8}{380} = \frac{24}{380} \approx 6.3%$ .

This does not equal $6.6%$ . Statement D is FALSE.

Final Answer: B is the only correct statement.

Question 6 · Probability Theory MSQ

A subset is chosen uniformly at random from all $2^{9}$ subsets of the set $S = {1, 2, 3, 4, 5, 6, 7, 8, 9}$ . What is the probability that the chosen subset contains at least one even number and at least one multiple of 3?

A.
45/64
B.
53/64
C.
11/16
D.
7/8

Correct Answer:

Step-by-Step Solution

Key idea: This is a complementary counting question with overlapping conditions, recognisable by the "at least one X AND at least one Y" structure on random subsets.

Step 1: Identify the total number of subsets. For a set of 9 elements, there are $2^{9} = 512$ total subsets.

Step 2: Define the complement events. Let $A$ be the event "contains no even number" and $B$ be the event "contains no multiple of 3". We want the probability of the complement of $(A \cup B)$ , which is $1 - P (A \cup B)$ .

Step 3: Count the subsets for $A$ . The even numbers in $S$ are ${2, 4, 6, 8}$ (4 elements). Subsets with no even numbers must be formed from the remaining 5 elements ${1, 3, 5, 7, 9}$ . Thus, $∣ A ∣ = 2^{5} = 32$ .

Step 4: Count the subsets for $B$ . The multiples of 3 in $S$ are ${3, 6, 9}$ (3 elements). Subsets with no multiples of 3 must be formed from the remaining 6 elements ${1, 2, 4, 5, 7, 8}$ . Thus, $∣ B ∣ = 2^{6} = 64$ .

Step 5: Count the subsets for $A \cap B$ (no even AND no multiple of 3). The elements that are neither even nor multiples of 3 are ${1, 5, 7}$ (3 elements). Thus, $∣ A \cap B ∣ = 2^{3} = 8$ .

Step 6: Apply Inclusion-Exclusion to find $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣ - ∣ A \cap B ∣ = 32 + 64 - 8 = 88$ .

Step 7: Calculate the number of favorable subsets: $512 - 88 = 424$ .

Step 8: Calculate the probability: $\frac{424}{512} = \frac{53}{64}$ .

Answer: The probability is 53/64.

Question 7 · Discrete Mathematics · 2021 SUB

Recall that if

h

is a function from

X

Y

and

g

is a function from

Y

Z

then,

g \circ h

is the function from

X

Z

such that

(g \circ h) (x) = g (h (x))

, for all

x \in X

.
Let

S

be the set of all functions

f

from

{1, 2, 3, 4, 5, 6}

{1, 2, 3, 4, 5, 6}

such that

f \circ f = f

.
(a) Compute the number of functions

f \in S

whose range has three elements.
(b) What is the cardinality of

S

Correct Answer:

1057

Step-by-Step Solution

Key idea: This is an idempotent function counting question, recognizable by the condition $f \circ f = f$ .

Step 1: For $f \circ f = f$ , the range of $f$ must be exactly the set of fixed points of $f$ . Let the range be $R$ . Then for all $x \in R$ , $f (x) = x$ .

Step 2: For any $x \in / R$ , $f (x)$ must be an element of $R$ .

Step 3: To count functions with range size $k$ , we first choose the $k$ elements of $R$ from the 6 available elements in $(k 6)$ ways.

Step 4: For the remaining $6 - k$ elements, each can map to any of the $k$ elements in $R$ , giving $k^{6 - k}$ choices.

Step 5: For part (a), $k = 3$ . The number of such functions is $(3 6) \times 3^{6 - 3} = 20 \times 27 = 540$ .

Step 6: For part (b), we sum this over all possible range sizes $k \in {1, 2, 3, 4, 5, 6}$ :

$k = 1 : (1 6) \times 1^{5} = 6$

$k = 2 : (2 6) \times 2^{4} = 15 \times 16 = 240$

$k = 3 : (3 6) \times 3^{3} = 20 \times 27 = 540$

$k = 4 : (4 6) \times 4^{2} = 15 \times 16 = 240$

$k = 5 : (5 6) \times 5^{1} = 6 \times 5 = 30$

$k = 6 : (6 6) \times 6^{0} = 1 \times 1 = 1$

Total cardinality = $6 + 240 + 540 + 240 + 30 + 1 = 1057$ .

Answer: 1057 (Note: Part (a) is 540, Part (b) is 1057).

Question 8 · Discrete Mathematics MSQ

Three people $A$ , $B$ , and $C$ are each either a knight (always tells the truth) or a knave (always lies). They make the following statements about each other:

$A$ says: "At least one of $B$ and $C$ is a knave."
$B$ says: "At least one of $A$ and $C$ is a knave."
$C$ says: "At least one of $A$ and $B$ is a knave."

How many valid assignments of knights and knaves are possible for $A$ , $B$ , and $C$ ?

A.
0
B.
1
C.
3
D.
6

Correct Answer:

Step-by-Step Solution

Key idea: This is a symmetric constraint puzzle. We translate each statement into a boolean equivalence and test the possible counts of knights.

Let $A, B, C \in {T, F}$ where $T$ means Knight and $F$ means Knave.

The statements translate to:

$A ⟺ (\neg B \lor \neg C)$
$B ⟺ (\neg A \lor \neg C)$
$C ⟺ (\neg A \lor \neg B)$

Notice that $\neg B \lor \neg C$ is true unless both $B$ and $C$ are Knights.

Let's test the possible number of Knights (from 0 to 3):

Case 0 Knights (all Knaves): $A = F, B = F, C = F$ .

Check $A$ : $\neg B \lor \neg C$ is $T \lor T = T$ . But $A$ is $F$ , so $F ⟺ T$ is false. Contradiction.

Case 1 Knight (two Knaves): Assume $A = T, B = F, C = F$ .

Check $B$ : $\neg A \lor \neg C$ is $F \lor T = T$ . But $B$ is $F$ , so $F ⟺ T$ is false. Contradiction. By symmetry, any 1-Knight assignment fails.

Case 2 Knights (one Knave): Assume $A = T, B = T, C = F$ .

Check $A$ : $\neg B \lor \neg C$ is $F \lor T = T$ . $T ⟺ T$ is true. (Consistent)

Check $B$ : $\neg A \lor \neg C$ is $F \lor T = T$ . $T ⟺ T$ is true. (Consistent)

Check $C$ : $\neg A \lor \neg B$ is $F \lor F = F$ . $F ⟺ F$ is true. (Consistent)

This assignment works! By symmetry, the assignments $(T, F, T)$ and $(F, T, T)$ also work.

Case 3 Knights (all Knights): $A = T, B = T, C = T$ .

Check $A$ : $\neg B \lor \neg C$ is $F \lor F = F$ . $T ⟺ F$ is false. Contradiction.

There are exactly 3 valid assignments.

Answer: 3

Question 9 · Discrete Mathematics MSQ

Let $f : {1, 2, \dots, 15} \to {1, 2, 3}$ be chosen uniformly at random. What is the probability that $∣ f^{- 1} (1) ∣ + ∣ f^{- 1} (2) ∣ = 10$ ?

A.
$inom{15}{10} \left( rac{2}{3}ight)^{10} \left( rac{1}{3}ight)^5$
B.
$inom{15}{10} \left( rac{1}{3}ight)^{10} \left( rac{2}{3}ight)^5$
C.
$inom{15}{10} \left( rac{2}{3}ight)^5 \left( rac{1}{3}ight)^{10}$
D.
$inom{15}{10} \left( rac{1}{3}ight)^5 \left( rac{2}{3}ight)^{10}$

Correct Answer:

Step-by-Step Solution

Key idea: The sum $∣ f^{- 1} (1) ∣ + ∣ f^{- 1} (2) ∣$ is the number of elements that map to either 1 or 2. For a random function, each element independently maps to 1, 2, or 3 with probability $1/3$ each. So the probability that an element maps to 1 or 2 is $2/3$ .

Step 1: Define the random variable. Let $Z = ∣ f^{- 1} (1) ∣ + ∣ f^{- 1} (2) ∣$ . This is the number of elements in ${1, 2, \dots, 15}$ that map to either 1 or 2.

Step 2: Determine the distribution of $Z$ . Each element independently maps to 1 or 2 with probability $2/3$ , and to 3 with probability $1/3$ . So $Z \sim Binomial (15, 2/3)$ .

Step 3: Compute the probability.

$P (Z = 10) = (10 15) (\frac{2}{3})^{10} (\frac{1}{3})^{15 - 10} = (10 15) (\frac{2}{3})^{10} (\frac{1}{3})^{5}$

Answer: A

Question 10 · Programming MSQ

Consider the following pseudocode:

```

function f(n):

total = 0

for k from 1 to n:

if k mod 2 == 1:

total = total + k * k

else:

total = total - k * k

return total

```

What is the value of f(100)?

A.
-5050
B.
-5000
C.
-2525
D.
-10100

Correct Answer:

Step-by-Step Solution

Key idea: Group consecutive terms in pairs to reveal an arithmetic pattern.

Step 1: Write out the sum explicitly:

$f (100) = 1^{2} - 2^{2} + 3^{2} - 4^{2} + \dots + 9 9^{2} - 10 0^{2}$

Step 2: Group terms in consecutive pairs:

$f (100) = (1^{2} - 2^{2}) + (3^{2} - 4^{2}) + \dots + (9 9^{2} - 10 0^{2})$

Step 3: Apply the difference of squares identity $a^{2} - b^{2} = (a - b) (a + b)$ to each pair:

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

$(3^{2} - 4^{2}) = (3 - 4) (3 + 4) = (- 1) (7) = - 7$

$(5^{2} - 6^{2}) = (5 - 6) (5 + 6) = (- 1) (11) = - 11$

$⋮$

$(9 9^{2} - 10 0^{2}) = (99 - 100) (99 + 100) = (- 1) (199) = - 199$

Step 4: The resulting sequence is $- 3, - 7, - 11, \dots, - 199$ , which is an arithmetic progression with:

First term: $a = - 3$
Common difference: $d = - 4$
Number of terms: $n = 50$ (since we have 100 terms grouped in pairs)

Step 5: Use the arithmetic series sum formula $S = \frac{n}{2} (a + l)$ where $l$ is the last term:

$S = \frac{50}{2} (- 3 + (- 199)) = 25 \times (- 202) = - 5050$

Answer: $- 5050$

Question 11 · Programming MSQ

Consider a sorted array $A$ of 6 distinct integers chosen from the set ${- 10, - 9, \dots, 10}$ . Suppose $A$ contains exactly one fixed point (an index $k \in {0, \dots, 5}$ such that $A [k] = k$ ). We then identify the three largest elements of $A$ to form a subset $B$ . If the second largest element in $B$ is exactly $k + 2$ , how many such valid arrays $A$ exist?

A.
880
B.
1100
C.
1320
D.
1540

Correct Answer:

Step-by-Step Solution

Key idea: This requires reverse-engineering the subset properties by combining the definitions of the second largest element and the fixed-point constraint with a strict inequality condition.

Step 1: Let the sorted subset be $A [0] < A [1] < A [2] < A [3] < A [4] < A [5]$ . The three largest elements form $B = {A [3], A [4], A [5]}$ . The second largest in $B$ is $A [4]$ .

Step 2: We are given $A [4] = k + 2$ and $A [k] = k$ . Since $A$ is strictly increasing, $A [4] - A [k] \geq 4 - k$ . Substituting the knowns: $(k + 2) - k \geq 4 - k ⟹ 2 \geq 4 - k ⟹ k \geq 2$ .

Step 3: Analyze possible values for $k \in {2, 3, 4, 5}$ :

If $k = 2$ : $A [2] = 2, A [4] = 4$ . Since $A$ is strictly increasing, $A [3]$ must be 3. But then $A [3] = 3$ , creating a second fixed point. Invalid.
If $k = 4$ : $A [4] = 4$ , but we need $A [4] = 4 + 2 = 6$ . Contradiction.
If $k = 5$ : $A [5] = 5$ , but $A [4] = 7 > A [5]$ , violating the sorted order.
Thus, $k = 3$ is the only valid case.

Step 4: For $k = 3$ , we have $A [3] = 3$ and $A [4] = 5$ .

$A [5]$ must be $> 5$ and $\leq 10$ . To avoid a second fixed point, $A [5] \neq = 5$ (already satisfied). Choices for $A [5] \in {6, 7, 8, 9, 10}$ (5 choices).
$A [0], A [1], A [2]$ must be chosen from ${- 10, \dots, 2}$ such that $A [0] < A [1] < A [2]$ and none is a fixed point ( $A [0] \neq = 0, A [1] \neq = 1, A [2] \neq = 2$ ).

Step 5: Use the bijection $x^{'} = A [0], y^{'} = A [1] - 1, z^{'} = A [2] - 2$ . The constraints map to choosing a multiset of size 3 from ${- 10, \dots, - 1}$ (10 elements). The number of ways is $(3 10 + 3 - 1) = (3 12) = 220$ .

Step 6: Total valid arrays = $220 \times 5 = 1100$ .

Answer: 1100.

Question 12 · Programming MSQ

A student implements a merge algorithm for two ascending arrays $A$ and $B$ of sizes $m$ and $n$ . They use a single loop that runs exactly $m + n$ times. In each iteration, they compare $A [i]$ and $B [j]$ and copy the smaller one, incrementing the respective pointer. They do not use any boundary checks or sentinels. Consider the following statements about this code:

I. If the code uses the condition A[i] < B[j] and $A [i] == B [j]$ , it will copy from $B$ , which ensures that elements from $A$ appear before elements from $B$ in the merged array, preserving stability.

II. If the loop runs exactly $m + n$ times without boundary checks, an out-of-bounds array access is guaranteed to occur at some point during the execution, provided $m \geq 1$ and $n \geq 1$ .

III. If the input arrays are sorted in descending order, the algorithm will correctly produce a merged array in descending order without any modifications to the comparison operator.

Which of the above statements is/are true?

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

Correct Answer:

Step-by-Step Solution

Key idea: This is a code-trace and logic question testing boundary conditions, stability, and sorting order. We must evaluate each statement independently based on the algorithm's mechanics.

Step 1: Analyze Statement I. If A[i] < B[j] is false when $A [i] == B [j]$ , the else branch executes, copying from $B$ . This means the element from $B$ is placed in the output before the equal element from $A$ . This violates stability (which requires $A$ 's element to come first). Thus, Statement I is false.

Step 2: Analyze Statement II. The loop runs $m + n$ times. In each iteration, exactly one pointer is incremented. The maximum valid index for $A$ is $m - 1$ and for $B$ is $n - 1$ . The sum of the maximum valid increments is $(m - 1) + (n - 1) = m + n - 2$ . After $m + n - 1$ increments, at least one pointer must exceed its valid range. Since the loop runs $m + n$ times, the final iteration will definitely access an out-of-bounds index. Thus, Statement II is true.

Step 3: Analyze Statement III. The algorithm copies the smaller element. If the inputs are descending, copying the smaller element will produce an ascending merged array, not descending. To merge descending arrays, the algorithm must copy the larger element. Thus, Statement III is false.

Answer: B

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Mock Test 6 for CMI Data Science: 40 Questions with Solutions & Analysis

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