Mock Test 7 for CMI Data Science: 40 Questions with Solutions & Analysis

Attempt the Mock Test 7 for CMI Data Science: 40 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

40 questions · 84 marks · 147.49 minutes. School Level Mathematics: 17 · Probability Theory: 10 · Discrete Mathematics: 10 · Programming: 3

Free sample questions from Mock Test 7

Question 1 · School Level Mathematics · 2025 SUB

Consider the following polynomial of positive degree $n$

$P_{n} (x) = 1 + 2 x + 3 x^{2} + \dots + (n + 1) x^{n} .$

Show that there is no real number $r$ such that $P_{n} (r) = 0$ when $n$ is even.

Correct Answer:

none

Step-by-Step Solution

Insight: Multiply by $(1 - x)$ to telescope the arithmetico-geometric series into a geometric sum plus one tail term, then exploit the parity of $n$ to force every piece positive when $x < 0$ .

Exam route:

$x \geq 0$ : all terms non-negative, constant is 1 $⟹ P_{n} (x) \geq 1 > 0$ .
$x < 0$ : multiply by $(1 - x) > 0$ .
$(1 - x) P_{n} (x) = 1 + x + \dots + x^{n} - (n + 1) x^{n + 1}$ .
$n$ even $⟹ n + 1$ odd $⟹ x^{n + 1} < 0 ⟹ - (n + 1) x^{n + 1} > 0$ .
Geometric sum $\frac{1 - x ^{n + 1}}{1 - x} > 0$ . Both parts positive $⟹ P_{n} (x) > 0$ .

Learning route:

This is a sign-analysis / bounding question, recognisable because it asks to "show there is no real number $r$ " making the polynomial zero. The trigger is the phrase "no real root" combined with a polynomial whose coefficients are all positive — you cannot just read off the sign for negative $x$ because the terms alternate.

Step 1 — Non-negative $x$ .

When $x \geq 0$ , every term $(k + 1) x^{k}$ is non-negative and the constant term is $1$ , so

$P_{n} (x) = 1 + 2 x + 3 x^{2} + \dots + (n + 1) x^{n} \geq 1 > 0.$

No root exists in $[0, \infty)$ .

Step 2 — Negative $x$ : the telescoping trick.

For $x < 0$ the terms alternate in sign, so direct inspection fails. Multiply by $(1 - x)$ , which is strictly positive when $x < 0$ and therefore preserves the sign of $P_{n} (x)$ :

$(1 - x) P_{n} (x) = k = 0 \sum n (k + 1) x^{k} - k = 0 \sum n (k + 1) x^{k + 1}$

$= 1 + k = 1 \sum n ((k + 1) - k) x^{k} - (n + 1) x^{n + 1} = 1 + x + x^{2} + \dots + x^{n} - (n + 1) x^{n + 1} .$

Step 3 — Exploit the parity of $n$ .

Since $n$ is even, $n + 1$ is odd. For $x < 0$ , $x^{n + 1} < 0$ , so the tail term $- (n + 1) x^{n + 1}$ is strictly positive.

The remaining part is a geometric series:

$1 + x + \dots + x^{n} = \frac{1 - x ^{n + 1}}{1 - x} .$

Since $x^{n + 1} < 0$ , the numerator $1 - x^{n + 1} > 1$ . The denominator $1 - x > 1$ . Thus, the fraction is strictly positive.

Step 4 — Conclusion.

We have $(1 - x) P_{n} (x) = positive + positive > 0$ . Since $1 - x > 0$ , we can divide by it without flipping any signs to get $P_{n} (x) > 0$ for all $x < 0$ . Combined with Step 1, $P_{n} (x) > 0$ for all real $x$ , so it has no real roots.

Question 2 · School Level Mathematics SUB

Let $f (x) = (x^{2} - 4) e^{x /2}$ . How many local extrema does $f (x)$ have?

Correct Answer:

Step-by-Step Solution

Key idea: This is a local extrema counting question for a product of a polynomial and an exponential. Factor the derivative completely and count sign changes.

Step 1: Find the first derivative using the product rule.

$f^{'} (x) = 2 x \cdot e^{x /2} + (x^{2} - 4) \cdot \frac{1}{2} e^{x /2}$ .

Factor out $e^{x /2}$ :

$f^{'} (x) = e^{x /2} (2 x + \frac{x ^{2} - 4}{2}) = e^{x /2} (\frac{4 x + x ^{2} - 4}{2}) = \frac{e ^{x /2}}{2} (x^{2} + 4 x - 4)$ .

Step 2: Find the critical points.

Since $e^{x /2} > 0$ , set $x^{2} + 4 x - 4 = 0$ .

Using the quadratic formula: $x = \frac{- 4 \pm 16 + 16}{2} = \frac{- 4 \pm 32}{2} = - 2 \pm 22$ .

These are two distinct real roots, both simple (multiplicity 1).

Step 3: Count sign changes.

Since both roots are simple, $f^{'} (x)$ changes sign at each root.

Therefore, $f (x)$ has exactly 2 local extrema.

Answer: 2

Question 3 · School Level Mathematics MSQ

Let $f (x)$ be a polynomial of odd degree $n \geq 3$ and $g (x)$ be a polynomial of even degree $m \geq 2$ , both with real coefficients. Which of the following statements is ALWAYS true?

A.
The graphs of $y = f (x)$ and $y = g (x)$ must intersect in at least one point.
B.
If the leading coefficients of $f (x)$ and $g (x)$ are equal, the degree of $f (x) - g (x)$ is strictly less than $max (n, m)$ .
C.
The equation $f (x) = g (x)$ can have exactly $max (n, m) + 1$ distinct real solutions.
D.
If $n > m$ , the graphs of $y = f (x)$ and $y = g (x)$ must intersect in at least one point.

Correct Answer:

Step-by-Step Solution

Key idea: The number of intersections is governed by the roots of the difference polynomial $h (x) = f (x) - g (x)$ . The degree and parity of $h (x)$ dictate whether it must have at least one real root.

Step 1: Let $h (x) = f (x) - g (x)$ . Since $n$ is odd and $m$ is even, $n \neq = m$ . Therefore, the highest degree term in $h (x)$ comes entirely from whichever polynomial has the larger degree. The leading terms cannot cancel each other out because they have different powers of $x$ .

Step 2: This means the degree of $h (x)$ is exactly $max (n, m)$ . This immediately makes statement B FALSE, because the degree never drops regardless of the leading coefficients.

Step 3: A polynomial of degree $k$ can have at most $k$ distinct real roots. Thus, $h (x)$ can have at most $max (n, m)$ roots. Statement C claims it can have $max (n, m) + 1$ roots, which is impossible. So C is FALSE.

Step 4: If $n < m$ , then $de g (h) = m$ , which is even. An even degree polynomial can be strictly positive (e.g., $x^{2} + 1$ ), meaning it might have 0 real roots. Thus, the graphs do not HAVE to intersect. Statement A is FALSE.

Step 5: If $n > m$ , then $de g (h) = n$ , which is odd. A fundamental property of polynomials with real coefficients is that every odd-degree polynomial has at least one real root (its ends go to opposite infinities, so it must cross the x-axis by the Intermediate Value Theorem). Thus, $h (x)$ has at least one real root, meaning the graphs MUST intersect in at least one point. Statement D is TRUE.

Answer: D

Question 4 · Probability Theory · 2019 SUB

Suppose $X$ is a continuous distribution with probability density function

$f (x) = k (x - x^{2}), 0 \leq x \leq 1,$

where $k$ is the normalizing constant. Find the value of $k$ and the expected value of the distribution.

Correct Answer:

6.00

Step-by-Step Solution

Key idea: This is a normalization and expectation question for a continuous random variable with a quadratic density function. It is recognisable because the PDF is given up to a constant k, and both the normalizing constant and the expected value are requested.

Step 1: Find the normalizing constant k.

For any valid PDF f(x), the total area under the curve over its support must be 1:

\int_0^1 f(x) dx = 1

Substituting f(x) = k(x - x^2),

\int_0^1 k(x - x^2) dx = k \int_0^1 (x - x^2) dx = 1

Compute the integral:

\int_0^1 (x - x^2) dx = [x^2/2 - x^3/3]_0^1 = 1/2 - 1/3 = 1/6

So, k * (1/6) = 1 => k = 6.

Step 2: Compute the expected value E[X].

By definition, E[X] = \int_0^1 x f(x) dx = \int_0^1 x * 6(x - x^2) dx = 6 \int_0^1 (x^2 - x^3) dx

= 6 [x^3/3 - x^4/4]_0^1 = 6(1/3 - 1/4) = 6(1/12) = 0.5.

Since the NAT format requires a single numeric answer and k is the primary normalizing constant requested first, the answer is 6.00.

Question 5 · Probability Theory · 2026 SUB

Common Description:

Question (6) and (7) are based on the following information

A list $(a_{1}, \dots, a_{n})$ is an arrangement of natural numbers $1, 2, \dots, n$ in a random order, with all orderings equally likely. We say that a position $i$ is a new maximum if $a_{i} > a_{j}$ for all $j < i$ . For example, in the list

$(3, 1, 4, 2, 5)$

positions 1,3,5 are new maxima. Note that position 1 is always a new maximum. Now answer the two questions below based on this information.

For a given $i$ , determine the probability that position $i$ is a new maximum.

Correct Answer:

none

Step-by-Step Solution

Insight: This is a record probability question, recognisable because it asks for the probability that a specific position sets a new running maximum in a random permutation. The key is to isolate the first $i$ elements and use symmetry.

Exam route:

Focus only on the first $i$ elements $a_{1}, \dots, a_{i}$ .
The event "position $i$ is a new maximum" means $a_{i} > a_{j}$ for all $j < i$ , which is exactly $a_{i} = max (a_{1}, \dots, a_{i})$ .
By symmetry, the largest of these $i$ distinct values is equally likely to be at any of the $i$ positions.
Therefore, the probability it lands at position $i$ is $1/ i$ .

Learning route:

Step 1. The full permutation $(a_{1}, \dots, a_{n})$ is drawn uniformly from all $n!$ orderings of ${1, 2, \dots, n}$ .

Step 2. The condition ${a_{i} > a_{j} for all j < i}$ depends only on the relative order of the first $i$ entries.

Step 3. The set of values ${a_{1}, \dots, a_{i}}$ forms some $i$ -element subset of ${1, \dots, n}$ . Conditional on this subset, all $i!$ relative orderings are equally likely.

Step 4. Among the $i!$ orderings, exactly $(i - 1)!$ place the largest value at position $i$ (the remaining $i - 1$ values can be arranged freely in the first $i - 1$ positions).

Step 5. The probability is $\frac{( i - 1 )!}{i !} = \frac{1}{i}$ .

Common trap: Answering $1/ n$ . This mistake comes from confusing the running maximum with the global maximum. The question asks for the probability of beating the previous $i - 1$ elements, not all $n$ elements.

Verification: For $i = 1$ , probability is $1/1 = 1$ (position 1 is always a record). For $i = 2$ , probability is $1/2$ (position 2 is a record iff $a_{2} > a_{1}$ , which happens in half of all permutations). Both match intuition.

Question 6 · Probability Theory · 2023 SUB

Burger Paradise plans to expand to other cities and is considering opening a new branch in Coimbatore. They have identified three potential locations,

A, B

and

C

, and are evaluating the profitability of each location. Based on their research, they estimate that the fixed cost for opening a new location will be ₹50,00,000, ₹70,00,000 and ₹65,00,000 at

A, B

and

C

, respectively. They also estimate the variable cost per customer to be ₹85 and the average revenue per customer to be ₹150 at all three locations. They expect to serve 100,000, 150,000 and 130,000 customers in the first year at locations

A, B

and

C

, respectively
Which location in Coimbatore is expected to be the most profitable in the first year of operation, and what is the expected profit at this location?

Correct Answer:

none

Step-by-Step Solution

Insight: Three locations share identical per-customer economics but differ in both fixed costs and expected volumes — the "lowest fixed cost wins" shortcut fails; you must compute full profit at each location's own expected volume.

Exam route:

$C M = 150 - 85 = 65$ .

Profit $= (Q \times 65) - F C$ for each.

A: $(1, 00, 000 \times 65) - 50, 00, 000 = 15, 00, 000$ .

B: $(1, 50, 000 \times 65) - 70, 00, 000 = 27, 50, 000$ .

C: $(1, 30, 000 \times 65) - 65, 00, 000 = 19, 50, 000$ .

Location B is the most profitable with an expected profit of ₹27,50,000.

Learning route:

This is a multi-alternative Cost-Volume-Profit comparison question, recognisable because three locations share identical per-customer economics (same selling price, same variable cost) but differ in both fixed costs and expected customer volumes. The key trigger is that expected volumes are different ( $1, 00, 000$ vs $1, 50, 000$ vs $1, 30, 000$ ), so we cannot use the "lowest fixed cost wins" shortcut — that shortcut only holds when expected volumes are identical across all alternatives.

Step 1: Compute the shared Contribution Margin per customer.

Since selling price $P = 150$ and variable cost $V C = 85$ are the same at all three locations:

$C M = P - V C = 150 - 85 = ₹65 per customer$

Step 2: Apply the profit equation $Profit = (Q \times C M) - F C$ at each location's own expected volume.

Location A ( $Q = 1, 00, 000$ , $F C = 50, 00, 000$ ):

$Profit_{A} = (1, 00, 000 \times 65) - 50, 00, 000 = 65, 00, 000 - 50, 00, 000 = ₹15, 00, 000$

Location B ( $Q = 1, 50, 000$ , $F C = 70, 00, 000$ ):

$Profit_{B} = (1, 50, 000 \times 65) - 70, 00, 000 = 97, 50, 000 - 70, 00, 000 = ₹27, 50, 000$

Location C ( $Q = 1, 30, 000$ , $F C = 65, 00, 000$ ):

$Profit_{C} = (1, 30, 000 \times 65) - 65, 00, 000 = 84, 50, 000 - 65, 00, 000 = ₹19, 50, 000$

Step 3: Compare and conclude.

$₹27, 50, 000 > ₹19, 50, 000 > ₹15, 00, 000$ , so Location B yields the highest profit.

Common trap: A student who applies the "lowest fixed cost wins" shortcut picks Location A (FC = ₹50,00,000) and reports ₹15,00,000. This is wrong because that shortcut requires identical expected volumes across all alternatives. Here, B's much higher volume (1,50,000 vs 1,00,000) more than compensates for its higher fixed cost.

Verification: For Location B, total revenue $= 1, 50, 000 \times 150 = ₹2, 25, 00, 000$ . Total cost $= 70, 00, 000 + (1, 50, 000 \times 85) = 70, 00, 000 + 1, 27, 50, 000 = ₹1, 97, 50, 000$ . Profit $= 2, 25, 00, 000 - 1, 97, 50, 000 = ₹27, 50, 000$ . ✓

Answer: Location B is expected to be the most profitable, with an expected profit of ₹27,50,000.

Question 7 · Discrete Mathematics · 2021 MSQ

There are two longest subsequences, not necessarily contiguous, common to the strings “ARTIFICIAL” and “INTELLIGENCE”. They are “IIC” and “TIC” which are of length three.
Consider two strings S1 = “CORONAVIRUS” and S2 = “SARSCOVID”. Let

x

be the length of a longest common subsequence between S1 and S2 and let

y

be the number of such longest common subsequences of length

x

between S1 and S2. What is

x + 5 y

A.
13
B.
15
C.
14
D.
16

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: Use dynamic programming to find the length of the Longest Common Subsequence (LCS), then backtrack to count the number of distinct LCS strings of that maximum length.

Step 1: Identify the strings S1 = "CORONAVIRUS" and S2 = "SARSCOVID".

Step 2: Eliminate characters that cannot be part of a long common subsequence. 'N', 'U', 'D' appear in only one string. 'S' appears at the very end of S1 but early in S2, meaning no characters can follow 'S' in both strings simultaneously. Thus, 'S' cannot be part of any LCS of length > 1.

Step 3: The viable common characters are C, O, R, A, V, I. List their valid index pairs (i, j) where i is the index in S1 and j is the index in S2:

C: (0, 4)
O: (1, 5), (3, 5)
R: (2, 2), (8, 2)
A: (5, 1)
V: (6, 6)
I: (7, 7)

Step 4: Find the longest sequence of strictly increasing pairs (both i and j must increase).

Path 1: R(2, 2) $\to$ O(3, 5) $\to$ V(6, 6) $\to$ I(7, 7) gives the string "ROVI".

Path 2: C(0, 4) $\to$ O(1, 5) $\to$ V(6, 6) $\to$ I(7, 7) gives the string "COVI".

Path 3: C(0, 4) $\to$ O(3, 5) $\to$ V(6, 6) $\to$ I(7, 7) also gives "COVI" (same string, different 'O' in S1).

Step 5: The maximum length is $x = 4$ . The distinct strings of this length are "ROVI" and "COVI", so $y = 2$ .

Step 6: Calculate $x + 5 y = 4 + 5 (2) = 14$ .

Answer: 14.

Question 8 · Discrete Mathematics SUB

In a class of 56 students, each student plays at least one of two games: Chess ( $C$ ) or Carrom ( $R$ ). The number of students playing only Chess is twice the number playing only Carrom. If the number playing Chess exceeds the number playing Carrom by 8, how many students play only Carrom?

Correct Answer:

Step-by-Step Solution

Key idea: This is a 2-set problem with two constraints linking the exclusive regions, recognisable because "only Chess" and "only Carrom" are related by a ratio, and the set-size difference gives a second equation.

Step 1: Define variables. Let Carrom only $= y$ . Then Chess only $= 2 y$ (given). Let both $= b$ .

Step 2: Express set sizes.

$∣ C ∣ = 2 y + b, ∣ R ∣ = y + b$

$∣ C ∣ - ∣ R ∣ = (2 y + b) - (y + b) = y$

Given $∣ C ∣ - ∣ R ∣ = 8$ , so $y = 8$ .

Step 3: Use the total. Since everyone plays at least one:

$2 y + y + b = 56 ⟹ 3 y + b = 56$

$3 (8) + b = 56 ⟹ b = 32$

Step 4: Verify. Chess only $= 16$ , Carrom only $= 8$ , Both $= 32$ . Sum $= 56$ . $✓$

$∣ C ∣ = 48$ , $∣ R ∣ = 40$ . Difference $= 8$ . $✓$

Answer: 8

Question 9 · Discrete Mathematics SUB

Let $A = {1, 2, 3, 4}$ and $B = {1, 2, 3}$ . How many surjective functions $f : A \to B$ satisfy the condition $f (1) = 1$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a constrained surjection counting problem. We fix one mapping and count the valid completions that ensure surjectivity.

Step 1: Analyze the constraint.

We are given $f (1) = 1$ .

The remaining domain elements are ${2, 3, 4}$ .

The codomain is ${1, 2, 3}$ .

For $f$ to be surjective, the image of $f$ must be ${1, 2, 3}$ .

Since $1$ is already in the image (via $f (1) = 1$ ), the remaining elements ${2, 3, 4}$ must map to ${1, 2, 3}$ such that the values $2$ and $3$ are covered. The value $1$ may or may not be hit again by ${2, 3, 4}$ .

Step 2: Define the sub-problem.

Let $g$ be the restriction of $f$ to ${2, 3, 4}$ .

$g : {2, 3, 4} \to {1, 2, 3}$ .

Condition: ${1, 2, 3} \subseteq Image (f) = {1} \cup Image (g)$ .

This implies ${2, 3} \subseteq Image (g)$ .

Note: It is NOT required that $1 \in Image (g)$ , because $1$ is already provided by $f (1)$ .

So, we need to count functions $g : {2, 3, 4} \to {1, 2, 3}$ such that $2 \in Image (g)$ AND $3 \in Image (g)$ .

Step 3: Use Inclusion-Exclusion.

Total functions from ${2, 3, 4}$ to ${1, 2, 3}$ is $3^{3} = 27$ .

Let $P_{2}$ be the property that $2 \in / Image (g)$ .

Let $P_{3}$ be the property that $3 \in / Image (g)$ .

We want to exclude functions satisfying $P_{2}$ or $P_{3}$ .

Number of functions where $2 \in / Image (g)$ :

The image is a subset of ${1, 3}$ . There are $2^{3} = 8$ such functions.

Number of functions where $3 \in / Image (g)$ :

The image is a subset of ${1, 2}$ . There are $2^{3} = 8$ such functions.

Number of functions where BOTH $2 \in / Image (g)$ AND $3 \in / Image (g)$ :

The image is a subset of ${1}$ . There is $1^{3} = 1$ such function (the constant function $g (x) = 1$ ).

By Inclusion-Exclusion, the number of invalid functions (missing 2 or missing 3) is:

$∣ P_{2} \cup P_{3} ∣ = ∣ P_{2} ∣ + ∣ P_{3} ∣ - ∣ P_{2} \cap P_{3} ∣ = 8 + 8 - 1 = 15$ .

Step 4: Calculate valid functions.

Valid functions = Total - Invalid

Valid = $27 - 15 = 12$ .

Wait, let me re-evaluate.

Is it possible that I missed something?

Let's list them manually to be sure for small numbers.

Domain ${2, 3, 4}$ , Codomain ${1, 2, 3}$ . Must hit 2 and 3.

Case 1: Image is exactly ${2, 3}$ .

Surjections from 3-set to 2-set ${2, 3}$ .

$2^{3} - 2 = 6$ .

Case 2: Image is exactly ${1, 2, 3}$ .

Surjections from 3-set to 3-set.

$3! = 6$ .

Case 3: Image contains ${2, 3}$ and possibly others?

The codomain is only ${1, 2, 3}$ . So the image can be ${2, 3}$ or ${1, 2, 3}$ .

Are there other subsets containing 2 and 3? No.

So total = (Surjections onto ${2, 3}$ ) + (Surjections onto ${1, 2, 3}$ )?

NO. This logic is slightly flawed because "Surjections onto ${2, 3}$ " means the image is EXACTLY ${2, 3}$ . "Surjections onto ${1, 2, 3}$ " means image is EXACTLY ${1, 2, 3}$ .

These are disjoint cases.

Let's check the Inclusion-Exclusion result again.

IE calculated functions where Image(g) CONTAINS 2 and 3.

This allows Image(g) to be ${2, 3}$ OR ${1, 2, 3}$ .

My manual sum: $6 + 6 = 12$ .

My IE calculation: $12$ .

So the answer is 12?

Let me re-read the question carefully.

$f : A \to B$ surjective. $A = {1, 2, 3, 4}, B = {1, 2, 3}$ . $f (1) = 1$ .

Let's try another path.

Total surjections from $A$ to $B$ is $3! S (4, 3) = 6 \times 6 = 36$ .

By symmetry, does each element in $B$ appear as $f (1)$ equally often?

No, not necessarily.

Let's count surjections where $f (1) = 1$ .

Let $N_{k}$ be the number of surjections where $f (1) = k$ .

Sum $N_{1} + N_{2} + N_{3} = 36$ .

Due to symmetry of the codomain elements in the definition of surjection (before fixing $f (1)$ ), $N_{1} = N_{2} = N_{3}$ .

Therefore, $N_{1} = 36/3 = 12$ .

Why did I get 14 in the initial thought block? I made an arithmetic error in the scratchpad or misremembered a standard result.

Let's double check the symmetry argument.

The set of all surjections is invariant under permutation of the codomain labels.

If we permute the codomain labels such that $1 \to 2$ , the number of surjections mapping $1 \to 1$ becomes the number of surjections mapping $1 \to 2$ in the new labeling. Since the structure is identical, the counts must be equal.

Total surjections = 36.

There are 3 possible values for $f (1)$ .

Each value is equally likely among the set of all surjections.

So count is $36/3 = 12$ .

Let's re-verify the manual count.

Restriction $g : {2, 3, 4} \to {1, 2, 3}$ .

Condition: $2 \in I m (g)$ and $3 \in I m (g)$ .

Subcase A: $1 \in / I m (g)$ . Then $I m (g) = {2, 3}$ .

Number of surjections from 3-set to 2-set is $2^{3} - 2 = 6$ .

Subcase B: $1 \in I m (g)$ . Then $I m (g) = {1, 2, 3}$ .

Number of surjections from 3-set to 3-set is $3! = 6$ .

Total = $6 + 6 = 12$ .

The answer is definitely 12.

Answer: 12

Question 10 · Programming MSQ

Consider the following pseudocode:

```

function h(n):

total = 0

for i from 1 to n:

total = 0

for j from 1 to i:

total = total + j

return total

```

Which of the following cannot be the output of h(n) for any positive integer n?

A.
1
B.
3
C.
6
D.
20

Correct Answer:

Step-by-Step Solution

Key idea: This is an accumulator reset question, recognisable because the variable total is set to 0 inside the outer loop. This means total is reset in every iteration, so only the last iteration matters.

Step 1: Analyze the inner loop.

In each iteration of the outer loop (for a given i), total is reset to 0, then:

total = 1 + 2 + ... + i = i(i+1)/2

This is the i-th triangular number.

Step 2: Determine the final output.

Since total is reset in every iteration, the final value is from the last iteration (i = n):

h(n) = n(n+1)/2

Step 3: Check which options are triangular numbers.

1 = 1(2)/2 ✓ (n = 1)
3 = 2(3)/2 ✓ (n = 2)
6 = 3(4)/2 ✓ (n = 3)
20: Is 20 a triangular number? Solve n(n+1)/2 = 20 → n² + n - 40 = 0.

Discriminant = 1 + 160 = 161. √161 ≈ 12.69, not an integer.

So 20 is not a triangular number.

Answer: D

Common trap: Not realizing total is reset, and thinking h(n) accumulates across iterations. If you think total accumulates, you might compute h(4) = 1 + 3 + 6 + 10 = 20, which is wrong. The correct h(4) = 10.

Question 11 · Programming MSQ

Consider the following three recursive functions defined for non-negative integers $n$ :

```python

def f(n):

if n == 0: return 1

return f(n-1) + 2*n - 1

def g(n):

if n == 0: return 10

if n % 4 == 0: return g(n-1) + 20

return g(n-1) - 5

def h(n):

if n == 0: return 0

return h(n-1) + 3*n

```

Which of the following correctly ranks the values of $f (12)$ , $g (12)$ , and $h (12)$ in increasing order?

A.
$g (12) < f (12) < h (12)$
B.
$f (12) < g (12) < h (12)$
C.
$h (12) < f (12) < g (12)$
D.
$g (12) < h (12) < f (12)$

Correct Answer:

Step-by-Step Solution

Key idea: This is a chapter-level estimation question that requires synthesizing closed-form mathematical equivalents for each sequence. The trap is ignoring the conditional branch in $g (n)$ and assuming uniform linear growth.

Step 1: Formulate $f (n)$ . The function adds $2 n - 1$ at each step, which is the sequence of odd numbers.

$f (n) = 1 + \sum_{i = 1}^{n} (2 i - 1) = 1 + n^{2}$ .

For $n = 12$ : $f (12) = 1 + 144 = 145$ .

Step 2: Formulate $g (n)$ . The function adds 20 when $n$ is a multiple of 4, and subtracts 5 otherwise.

Let $k = ⌊ n /4 ⌋$ be the number of multiples of 4 up to $n$ .

$g (n) = 10 + 20 k - 5 (n - k) = 10 + 25 k - 5 n$ .

For $n = 12$ : $k = 3$ . $g (12) = 10 + 25 (3) - 5 (12) = 10 + 75 - 60 = 25$ .

Step 3: Formulate $h (n)$ . The function adds $3 n$ at each step, which is 3 times the sum of the first $n$ integers.

$h (n) = 3 \times \frac{n ( n + 1 )}{2}$ .

For $n = 12$ : $h (12) = 3 \times \frac{12 \times 13}{2} = 3 \times 78 = 234$ .

Step 4: Compare the values: $25 < 145 < 234$ .

Therefore, $g (12) < f (12) < h (12)$ .

Answer: A

Question 12 · Programming MSQ

Consider the correct algorithm for finding the second largest distinct element, which maintains two variables largest and second_largest, both initialized to $- \infty$ . During a left-to-right scan of an array, the pair (largest, second_largest) transitions through a sequence of states. Which of the following state transitions is strictly impossible during the execution of this correct algorithm on any valid array?

A.
$(- \infty, - \infty) o (5, - \infty) o (8, 5)$
B.
$(5, - \infty) o (5, 3) o (8, 5)$
C.
$(5, 3) o (8, 5) o (8, 6)$
D.
$(8, 5) o (8, 7) o (9, 7)$

Correct Answer:

Step-by-Step Solution

Key idea: The correct algorithm enforces a strict invariant: when a new maximum is found, the old maximum is explicitly pushed down to become the new second_largest.

Step 1: Analyze the update rules.

If $A [i] > largest$ : second_largest becomes the old largest, and largest becomes $A [i]$ .
If $second_largest < A [i] < largest$ : only second_largest is updated to $A [i]$ .

Step 2: Evaluate Option A. $(- \infty, - \infty) read 5 (5, - \infty) read 8 (8, 5)$ . Valid.

Step 3: Evaluate Option B. $(5, - \infty) read 3 (5, 3) read 8 (8, 5)$ . Valid.

Step 4: Evaluate Option C. $(5, 3) read 8 (8, 5) read 6 (8, 6)$ . Valid (since $5 < 6 < 8$ ).

Step 5: Evaluate Option D. $(8, 5) read 7 (8, 7)$ . This is valid. However, the next transition is $(8, 7) read 9 (9, 7)$ .

Step 6: Check the rule for reading 9. Since $9 > 8$ , the algorithm must execute: second_largest = largest (which is 8), then largest = 9. The resulting state must be $(9, 8)$ , not $(9, 7)$ .

Answer: Option D describes an impossible state transition because it discards the old maximum instead of pushing it down.

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