Algebra & Number Theory, Functions & Calculus, Combinatorics & Induction… Test for CMI Data Science: 40 Questions with Solutions & Analysis

Attempt the Algebra & Number Theory, Functions & Calculus, Combinatorics & Induction… test for CMI Data Science: 40 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

40 questions · 88 marks · 131.11 minutes. School Level Mathematics: 20 · Probability Theory: 9 · Discrete Mathematics: 7 · Programming: 4

Free sample questions from Algebra & Number Theory, Functions & Calculus, Combinatorics & Induction…

Question 1 · School Level Mathematics · 2026 SUB

Common Description:

Questions (3) and (4) are based on the following information

Let $f : R \to R$ be a continuous function that satisfies

$f (x + y) = f (x) \cdot f (y), \forall x, y \in R,$

and that $f (0) = 1$ . The following two questions are based on this information.

Assuming that $f (x) = (f (1))^{x}$ holds for all real $x$ , determine all values of $f (1)$ for which $f$ is strictly increasing.

Correct Answer:

none

Step-by-Step Solution

Insight: With $f (x) = (f (1))^{x}$ granted, the problem reduces to asking when $b^{x}$ (with $b = f (1) > 0$ ) is strictly increasing, which happens exactly when $b > 1$ .

Exam route: Set $b = f (1)$ . Positivity forces $b > 0$ . The exponential $b^{x}$ is strictly increasing iff $b > 1$ .

Learning route:

This is an exponential-type functional equation question, recognisable by $f (x + y) = f (x) f (y)$ with $f (0) = 1$ and the explicit assumption $f (x) = (f (1))^{x}$ .

Step 1: Let $b = f (1)$ . The problem grants $f (x) = b^{x}$ for all real $x$ .

Step 2: Prove $b > 0$ . From $f (x) = f (x /2 + x /2) = f (x /2)^{2} \geq 0$ . If $f (a) = 0$ for some $a$ , then $1 = f (0) = f (a + (- a)) = f (a) f (- a) = 0$ , contradiction. Hence $f (x) > 0$ for all $x$ , so $b = f (1) > 0$ .

Step 3: Classify monotonicity of $b^{x}$ for $b > 0$ :

$b > 1$ : $x_{1} < x_{2} ⟹ b^{x_{1}} < b^{x_{2}}$ , so $f$ is strictly increasing.
$b = 1$ : $f (x) = 1$ for all $x$ , constant, not strictly increasing.
$0 < b < 1$ : $x_{1} < x_{2} ⟹ b^{x_{1}} > b^{x_{2}}$ , so $f$ is strictly decreasing.

Step 4: Conclude $f$ is strictly increasing if and only if $f (1) > 1$ .

Wrong path: Writing $f (1) \geq 1$ . The boundary $f (1) = 1$ gives the constant function $f (x) = 1$ , which is not strictly increasing.

Verification: $f (1) = 2 ⟹ f (x) = 2^{x}$ . $f (1) = 2 < f (2) = 4$ : strictly increasing. $f (1) = 1 ⟹ f (x) = 1$ : constant, not strictly increasing. Confirmed.

Question 2 · School Level Mathematics · 2022 MSQ

A stick is fixed on the ground at an angle of 60 degrees as shown in Figure 1. The part of the stick which is above the ground—which is the part that is shown in Figure 1—has a length of 10 metres. At a certain time

T_{1}

during the day, the sun is directly overhead at this location and its rays strike the ground at an angle of 90 degrees. This is shown in Figure 2. At a certain later time

T_{2}

the sun’s rays strike the ground at an angle of 60 degrees as shown in Figure 3. While Figures 2 and 3 show only the sun’s rays for the sake of clarity, the stick is present at this location at times

T_{1}

and

T_{2}

.

What are the lengths of the shadows of the stick at times

T_{1}

and

T_{2}

, respectively?

A.
$53$ metres and $103$ metres
B.
$5$ metres and $103$ metres
C.
$5$ metres and $10$ metres
D.
$53$ metres and $10$ metres

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: This is a trigonometric projection question, recognisable because it involves finding shadow lengths cast by a tilted object under varying sun angles.

Step 1: Resolve the stick's position into coordinates. Let the base of the stick be at the origin $(0, 0)$ . The stick has length $L = 10$ m and is tilted at $6 0^{\circ}$ to the ground.

Step 2: The top of the stick has horizontal coordinate $x_{t o p} = 10 cos 6 0^{\circ} = 10 \times \frac{1}{2} = 5$ m, and vertical coordinate $y_{t o p} = 10 sin 6 0^{\circ} = 10 \times \frac{3}{2} = 53$ m.

Step 3: At time $T_{1}$ , the sun is directly overhead ( $9 0^{\circ}$ ). The rays are vertical, so the shadow is simply the horizontal projection of the stick. Shadow length = $x_{t o p} = 5$ m.

Step 4: At time $T_{2}$ , the sun's rays strike the ground at $6 0^{\circ}$ . The ray passing through the top of the stick forms a right triangle with the ground. The additional horizontal distance from the top's vertical projection to the shadow tip is $\frac{y _{t o p}}{t a n 6 0 ^{\circ}}$ .

Step 5: Calculate this additional distance: $\frac{5 3}{3} = 5$ m.

Step 6: Since the stick leans in the direction the shadow is cast, the total shadow length is the sum of the horizontal projection and the additional distance: $x_{t o p} + 5 = 5 + 5 = 10$ m.

Answer: 5 metres and 10 metres (Option C).

Question 3 · School Level Mathematics SUB

Let $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ . The tangent line to the graph of $y = f (x)$ at $x = 1$ intersects the graph at exactly one other point. What is the x-coordinate of this other point?

Correct Answer:

Step-by-Step Solution

Key idea: The intersection of a curve and its tangent line creates a polynomial equation where the point of tangency is a repeated root. We can use Vieta's formulas (sum of roots) to find the other intersection point without fully expanding the line equation.

Step 1: Let the tangent line be $y = m x + c$ . The x-coordinates of the intersections are the roots of $f (x) - (m x + c) = 0$ .

Step 2: Write the difference polynomial: $P (x) = x^{3} - 6 x^{2} + 9 x + 1 - m x - c = x^{3} - 6 x^{2} + (9 - m) x + (1 - c)$ .

Step 3: Since the line is tangent to the curve at $x = 1$ , $x = 1$ is a root of $P (x)$ with multiplicity at least 2. Let the roots be $x_{1}, x_{2}, x_{3}$ . We know $x_{1} = 1$ and $x_{2} = 1$ .

Step 4: By Vieta's formulas, the sum of the roots of a cubic $A x^{3} + B x^{2} + C x + D = 0$ is $- B / A$ .

Here, the sum of the roots is $x_{1} + x_{2} + x_{3} = - (- 6) /1 = 6$ .

Step 5: Substitute the known roots: $1 + 1 + x_{3} = 6 ⟹ 2 + x_{3} = 6 ⟹ x_{3} = 4$ .

Step 6: The other intersection point has an x-coordinate of 4.

Answer: 4

Question 4 · Probability Theory · 2022 SUB

Common Description: Description for the next four questions:
Dasholytics Inc runs an online analytics dashboard. The company employs three machines—Server 1, Server 2, and Server 3—to serve the data for the dashboard. At any point in time one of these three machines has the job of serving the data, and the other two are kept in standby mode. Dasholytics uses a server scheduler (software) that decides which of the three machines should serve the data at any given point of time. The scheduler switches between data servers without disrupting the data feed to the dashboard.
The machine which is serving the data sometimes fails to do so; in this case an alert is sent out to the Dasholytics team and they investigate and fix the problem. The time for which a machine fails to serve data is accounted as service outage caused by that machine. For technical reasons the server scheduler is deactivated when there is such an outage: the outage gets over only when the issue with the server is fixed and it is put back online.
The figures below describe the server usage and outage statistics as compiled over the last one year (365 days). Please use this information to answer the questions that follow.

Figure 1: Figure (a) shows the total service outage caused by each server, as a percentage of the total time that it served data. Figure (b) shows the total time that each server served data, as a percentage of the total duration over which this information was collected. An alert that a service outage has been caused by one of the three servers was sent out to the Dasholytics team. What is the probability that this outage was caused by Server 1?

Correct Answer:

0.14

Step-by-Step Solution

Insight: Given that an outage occurred, we need the posterior probability that Server 1 caused it, which is a direct application of Bayes' theorem.

Exam route: P(S1|O) = P(O|S1)P(S1) / P(O) = 0.004 / 0.028 = 1/7 ≈ 0.14.

Learning route:

We need P(Server 1 | Outage).
By Bayes' theorem, P(S1|O) = [P(O|S1) * P(S1)] / P(O).
The numerator is the joint probability calculated in P2: 0.004.
The denominator is the total outage probability calculated in P4/P1: 0.028.
P(S1|O) = 0.004 / 0.028 = 4/28 = 1/7 ≈ 0.1428...
Rounded to 2 decimal places, this is "0.14".

Wrong path: Answering 0.40 (the prior probability of Server 1) commits the base rate fallacy, ignoring the new evidence (the outage).

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.

What is the expected number of new maxima in a list?

Correct Answer:

none

Step-by-Step Solution

Insight: This is an expected value of a sum of indicators question. The total number of records $R_{n}$ is the sum of indicator variables $I_{i}$ , and linearity of expectation gives the answer directly without needing independence.

Exam route:

Define $I_{i} = 1 {position i is a new maximum}$ .
Write $R_{n} = \sum_{i = 1}^{n} I_{i}$ .
Apply linearity of expectation: $E [R_{n}] = \sum_{i = 1}^{n} P (I_{i} = 1)$ .
Substitute $P (I_{i} = 1) = 1/ i$ to get $\sum_{i = 1}^{n} 1/ i = H_{n}$ .

Learning route:

Step 1. For each position $i \in {1, 2, \dots, n}$ , define the indicator random variable $I_{i} = 1$ if $a_{i} > a_{j}$ for all $j < i$ , and $0$ otherwise.

Step 2. The total number of new maxima is $R_{n} = \sum_{i = 1}^{n} I_{i}$ .

Step 3. By linearity of expectation, $E [R_{n}] = \sum_{i = 1}^{n} E [I_{i}] = \sum_{i = 1}^{n} P (I_{i} = 1)$ .

Step 4. From the single-position result, $P (I_{i} = 1) = 1/ i$ .

Step 5. Therefore, $E [R_{n}] = \sum_{i = 1}^{n} \frac{1}{i} = H_{n}$ , the $n$ -th harmonic number.

Common trap: Attempting to compute the full distribution of $R_{n}$ using Stirling numbers, or trying to evaluate covariances because the indicators seem dependent. Linearity of expectation bypasses all dependence concerns.

Verification: For $n = 3$ , $H_{3} = 1 + 1/2 + 1/3 = 11/6$ . Enumerating all 6 permutations of ${1, 2, 3}$ : record counts are 3, 2, 2, 2, 1, 1, averaging to $11/6$ . Matches.

Question 6 · Probability Theory · 2022 SUB

Consider the following code, in which

A

is an integer array of length

n

indexed from 0, and

x

is an integer.
function foo(A,x,n) {
    found = False;

    while(found != True) {
        i = randInt(0,n);

        if (A[i] == x) {
            found = True;
        }
    }

    return(i);
}
Here, randInt(0,n) returns an integer picked uniformly at random from the range

{0, 1, \dots, (n - 1)}

.
All integers present in array

A

are distinct, and integer

x

is present in array

A

. Suppose we make the call foo(

A, x, n

). What is the expected number of times that the call randInt(0, n) is made from within this call to foo(

A, x, n

Correct Answer:

10.00

Step-by-Step Solution

Insight: The algorithm performs independent uniform random sampling with replacement until the target is found, which perfectly matches the Geometric distribution.

Exam route: Identify the success probability $p = 1/10$ per trial. The expected number of trials for a geometric distribution is simply the reciprocal, $1/ p = 10$ .

Learning route:

Step 1: The array $A$ has length 10 and contains distinct integers. The target $x$ is present exactly once.

Step 2: The function randInt(0,10) returns an integer uniformly at random from the set ${0, 1, \dots, 9}$ .

Step 3: The probability of selecting the correct index in any single call is $p = 1/10$ .

Step 4: The while loop repeats until the correct index is found. Because randInt is called fresh each time, each call is an independent trial.

Step 5: The number of trials until the first success follows a Geometric distribution with success probability $p$ .

Step 6: The expected value of a Geometric distribution (counting the successful trial) is $1/ p$ .

Step 7: Therefore, the expected number of calls to randInt is $1/ (1/10) = 10$ .

Wrong path: A student might incorrectly assume the algorithm samples without replacement (like a standard deterministic linear search), leading them to believe the expected number of calls is $(10 + 1) /2 = 5.5$ . This is a conceptual error, as randInt is called independently each time, meaning sampling is strictly with replacement.

Generalization: When an algorithm repeatedly samples uniformly at random with replacement until a specific condition is met, the expected number of trials is simply the reciprocal of the probability of success in a single trial.

Verification: If the array had length 2, the probability of guessing correctly is $1/2$ . The expected number of guesses is $1/ (1/2) = 2$ . This matches our formula.

Answer: 10.00

Question 7 · Discrete Mathematics · 2026 SUB

How many binary strings of length 8 contain exactly three 1s such that no two 1s are adjacent.

Correct Answer:

none

Step-by-Step Solution

Insight: The phrase "no two 1s are adjacent" is the signature trigger for the gap method — place all 0s first, then choose gaps to insert the 1s.

Exam route: $n = 8$ , $k = 3$ . Zeros $= 8 - 3 = 5$ . Gaps $= 5 + 1 = 6$ . Answer $= (3 6) = 20$ .

Learning route:

This is a non-adjacent selection problem expressed in binary-string language. The constraint "no two 1s are adjacent" means there must be at least one 0 between every pair of 1s. The gap method handles exactly this type of spacing restriction.

Step 1 — Identify the parameters. The string has length $n = 8$ and must contain exactly $k = 3$ ones. The remaining $n - k = 8 - 3 = 5$ positions are zeros.

Step 2 — Lay down the five 0s in a row and identify every available gap. A gap exists before the first 0, between each consecutive pair of 0s, and after the last 0:

$\land 0 \land 0 \land 0 \land 0 \land 0 \land$

The number of gaps is $(n - k) + 1 = 5 + 1 = 6$ .

Step 3 — Choose $k = 3$ of these 6 gaps and place exactly one 1 into each chosen gap. Since every pair of gaps is separated by at least one 0, no two 1s can ever end up adjacent. This one-to-one correspondence between gap-choices and valid strings guarantees the count is exact.

Step 4 — Compute the binomial coefficient:

$(3 6) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ .

Tempting wrong path: A student might ignore the adjacency constraint entirely and compute $(3 8) = 56$ . This counts all ways to place three 1s in eight positions, including those where 1s are adjacent. The adjacency constraint eliminates 36 of these arrangements, leaving only 20 valid strings.

Verification: We can verify by listing a few valid strings: 10101000, 01010100, 10010100. All have exactly three 1s with no two adjacent. The formula $(k n - k + 1) = (3 6) = 20$ is correct.

Question 8 · Discrete Mathematics · 2023 SUB

A domino is a sheet of paper with one number from

{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

printed on each half. For example, a

4 - 4

domino has 4 printed on both halves. A

0 - 4

domino has 0 printed on one half and 4 on the other half. The order is not important, so a

0 - 4

domino is the same as a

4 - 0

domino.
A complete set of dominos is a set of dominos containing exactly one copy of each different domino.
(a) How many dominos are there in a complete set of dominos?
(b) In how many ways can you select 2 dominos from a complete set so that at least one of the dominos has a 0 on it and at least one of the dominos has a 9 on it.

Correct Answer:

none

Step-by-Step Solution

Insight: Part (a) is the standard unordered-pairs-with-repetition model; part (b) is a collective-constraint selection cleanly solved by partitioning the dominos into a $2 \times 2$ feature table or by complementary counting.

Exam route:

(a) $n = 10$ symbols, unordered with repetition allowed: $(2 11) = 55$ .

(b) Partition into 4 cells: $A$ (0 only, size 9), $B$ (9 only, size 9), $C$ (both, size 1), $D$ (neither, size 36).

Valid pairs: $C$ paired with any other domino ( $1 \times 54 = 54$ ) plus one from $A$ and one from $B$ ( $9 \times 9 = 81$ ). Total = $54 + 81 = 135$ .

Learning route:

This is a two-phase domino question, recognisable because the problem explicitly defines dominos as unordered pairs with doubles allowed, then asks a selection question with a collective "at least one ... and at least one ..." constraint across two chosen dominos.

Phase 1 — Counting the complete set (part a).

The trigger words are "order is not important" and the explicit mention of doubles like $4 - 4$ . This means we need the unordered-pairs-with-repetition model from the Pair Model Checklist:

$Count = (2 n + 1)$

Here $n = 10$ symbols ${0, 1, \dots, 9}$ , so:

$(2 11) = \frac{11 \times 10}{2} = 55.$

Verification by splitting cases: $10$ doubles ( $0 - 0$ through $9 - 9$ ) plus $(2 10) = 45$ non-doubles gives $10 + 45 = 55$ . Confirmed.

Phase 2 — Selecting 2 dominos with a collective constraint (part b).

The trigger is "at least one domino has a 0 and at least one has a 9" applied to a pair of dominos. This is a collective constraint across the pair, not on each domino individually.

Partition the 55 dominos by whether they contain 0 and whether they contain 9:

$A$ : contains 0 but not 9 → ${0 - 0, 0 - 1, \dots, 0 - 8}$ → size $9$ .
$B$ : contains 9 but not 0 → ${9 - 9, 9 - 1, \dots, 9 - 8}$ → size $9$ .
$C$ : contains both 0 and 9 → ${0 - 9}$ → size $1$ .
$D$ : contains neither → dominos from ${1, 2, \dots, 8}$ → $(2 9) = 36$ .

Check: $9 + 9 + 1 + 36 = 55$ . ✓

A valid pair of 2 dominos must collectively have at least one 0 and at least one 9. The only ways this happens:

One domino is from $C$ (it alone supplies both 0 and 9), paired with any of the remaining $54$ dominos. This gives $1 \times 54 = 54$ pairs.
One domino is from $A$ (supplies the 0) and the other is from $B$ (supplies the 9). This gives $9 \times 9 = 81$ pairs.

No other combination works: $A + A$ lacks 9, $B + B$ lacks 0, $A + D$ lacks 9, $B + D$ lacks 0, $D + D$ lacks both.

Total valid pairs = $54 + 81 = 135$ .

Verification by complementary counting:

Total pairs of 2 from 55: $(2 55) = 1485$ .

Pairs with no 0: choose 2 from dominos without 0 (formed from ${1, \dots, 9}$ , size 45) → $(2 45) = 990$ .

Pairs with no 9: choose 2 from dominos without 9 (formed from ${0, \dots, 8}$ , size 45) → $(2 45) = 990$ .

Pairs with neither 0 nor 9 (double-counted above): choose 2 from $D$ (size 36) → $(2 36) = 630$ .

By inclusion–exclusion, invalid = $990 + 990 - 630 = 1350$ .

Valid = $1485 - 1350 = 135$ . ✓

Final answers: (a) $55$ , (b) $135$ .

Question 9 · Discrete Mathematics MSQ

From a group of 10 people including persons $A$ , $B$ , and $C$ , a committee of 5 is to be formed such that person $A$ must serve on the committee, and persons $B$ and $C$ cannot both serve on the committee. From the committee, a chairperson is to be chosen. The number of ways to form the committee and choose the chairperson is

A.
480
B.
525
C.
560
D.
600

Correct Answer:

Step-by-Step Solution

Key idea: this is a multi-constraint committee question with a mandatory inclusion ( $A$ must serve) and a conditional exclusion ( $B$ and $C$ cannot both serve). Since $A$ must be on the committee, fix $A$ first, then choose the remaining 4 members from the other 9 people, subject to the $B$ - $C$ constraint. Finally, multiply by the number of chairperson choices.

Step 1: $A$ is fixed on the committee. We need to choose 4 more members from the remaining 9 people (including $B$ and $C$ ), such that $B$ and $C$ are not both on the committee.

Step 2: Use the complement method. Count total ways to choose 4 from 9, then subtract the invalid ways (both $B$ and $C$ are chosen).

Total ways to choose 4 from 9: $(4 9) = \frac{9 !}{4 ! 5 !} = 126$ .

Invalid ways (both $B$ and $C$ are chosen): $B$ and $C$ are fixed, so we need to choose 2 more from the remaining 7 people: $(2 7) = \frac{7 !}{2 ! 5 !} = 21$ .

Step 3: Subtract to get valid committees.

Valid committees = $126 - 21 = 105$ .

Step 4: Choose a chairperson from the 5 committee members.

Chairperson choices = $5$ .

Step 5: Multiply.

Total ways = $105 \times 5 = 525$ .

Alternative method (casework):

Case 1: $B$ is on the committee, $C$ is not. $A$ and $B$ are fixed. Choose 3 more from the 7 people (excluding $C$ ): $(3 7) = 35$ .
Case 2: $C$ is on the committee, $B$ is not. $A$ and $C$ are fixed. Choose 3 more from the 7 people (excluding $B$ ): $(3 7) = 35$ .
Case 3: Neither $B$ nor $C$ is on the committee. $A$ is fixed. Choose 4 more from the 7 people (excluding $B$ and $C$ ): $(4 7) = 35$ .

Total committees = $35 + 35 + 35 = 105$ .

Total ways = $105 \times 5 = 525$ .

Common trap: students often forget to fix $A$ first and use $(5 10)$ instead of $(4 9)$ , or they forget to multiply by 5 for the chairperson at the end.

Answer: 525

Question 10 · Programming · 2024 SUB

Common Description:

Questions 4 and 5 are based on the following description.

The following question appeared in a quiz:

“Write the code for a function SecondBest( $A, n$ ) that takes an array $A$ and a positive integer $n$ as arguments. The elements of $A$ are all integers, and $n$ is the number of elements in $A$ . The call SecondBest( $A, n$ ) should return the second largest element in $A$ . If $A$ has no second largest element, then the function should return the special value None.”

A student submitted the code below as the answer to this question. In the code the array $A$ is indexed from 0.

function SecondBest(A, n) {

if n == 1 {

return(None);

}

first = A[0];

second = A[1];

for i from 2 to (n-1) {

if (A[i] >= first) and (A[i] >= second) {

second = first;

first = A[i];

} else {

if A[i] >= second {

second = A[i];

}

if first != second {

return(second);

} else {

return(None);

}

This answer turned out to be wrong; this function gives the correct answer for some valid inputs, and wrong answers for other valid inputs. Answer the next two questions about this function.

Give one example of an input array $A$ with exactly 3 elements for which the call SecondBest( $A$ , 3) returns a wrong answer. What is this wrong answer? What is the correct answer?

Correct Answer:

none

Step-by-Step Solution

Insight: This is an algorithm tracing and counterexample construction question, recognizable because it asks for a specific input that causes a flawed function to return a wrong answer.

Exam route: Test $A = [10, 10, 5]$ . Initialization sets first = 10, second = 10. Loop $i = 2$ : $A [2] = 5$ . Condition (5 >= 10) and (5 >= 10) is False. Else branch: 5 >= 10 is False. Loop ends. Final check 10 != 10 is False, returning None. The correct second largest is 5.

Learning route:

Step 1: Identify the pattern. The function initializes first and second to $A [0]$ and $A [1]$ . If these are identical maximums, the final check first != second will fail.

Step 2: Construct $A = [10, 10, 5]$ .

Step 3: Initialize first = 10, second = 10.

Step 4: Loop $i = 2$ : $A [2] = 5$ . Check (5 >= 10) and (5 >= 10) $\to$ False.

Step 5: Else branch: check 5 >= 10 $\to$ False. second remains 10.

Step 6: End loop. first = 10, second = 10.

Step 7: Final check first != second $\to$ 10 != 10 is False. Returns None.

Step 8: The distinct elements are 10 and 5. The correct second largest is 5.

Wrong path: A student might assume the algorithm correctly updates second to 5, producing the answer 5. This breaks at Step 5, where the condition 5 >= 10 is strictly False, so second is never updated.

Generalization: When tracing flawed code, execute it exactly as written without assuming it implements the intended logic. Initialization with the first two elements creates a duplicate hazard if they are equal.

Question 11 · Programming · 2025 SUB

Common Description: Questions 10 and 11 are based on the following description.
The following question appeared in a quiz:
“Write the pseudocode for a function Closest(

A, n, x

) that takes an array

A

, a positive integer

n

, and an integer

x

as arguments. The elements of

A

are all integers less than

2^{64}

, and

n

is the number of elements in

A

. The call Closest(

A, n, x

) should return an integer

y

such that: (i)

y \neq = x

, (ii)

y

is present in

A

, and (iii) there is no

z \neq = x

A

where

∣ z - x ∣ < ∣ y - x ∣

holds. If

A

has no such element

y

, then the function should return the special value None.”
A student submitted the code below as the answer to this question. In the code the array

A

is indexed from 0, and MAXINT =

2^{64} - 1

. The call abs(

z

) returns the absolute value

∣ z ∣

of integer

z

.
function Closest(A, n, x) { minVal = MAXINT; for i from 0 to (n-1) { absDiff = abs(x - A[i]); if (absDiff < minVal) { minVal = absDiff; y = A[i]; } } if (minVal != MAXINT) { return(y); } else { return(None); } }
This answer turned out to be wrong; this function gives the correct answer for some valid inputs, and wrong answers for other valid inputs. Answer the next two questions about this function. What do the following function calls return?
(a) Closest([-10,2,10], 3, 8)
(b) Closest([0,-5,4], 3, 7)

Correct Answer:

none

Step-by-Step Solution

Insight: This is an algorithm tracing question for a closest value search, recognizable because it asks for the return value of specific function calls. The method applies here because we must track minVal and y by computing absolute differences step-by-step.

Exam route: For (a) $A = [- 10, 2, 10]$ , $x = 8$ : $i = 0 \to$ diff 18, $y = - 10$ . $i = 1 \to$ diff 6, $y = 2$ . $i = 2 \to$ diff 2, $y = 10$ . Returns 10. For (b) $A = [0, - 5, 4]$ , $x = 7$ : $i = 0 \to$ diff 7, $y = 0$ . $i = 1 \to$ diff 12 (no update). $i = 2 \to$ diff 3, $y = 4$ . Returns 4.

Learning route:

Step 1: This is an algorithm tracing question. We apply the linear scan method, computing absolute differences and updating trackers strictly on <.

Step 2: Trace (a) Closest([-10, 2, 10], 3, 8).

Init: minVal = MAXINT.
$i = 0$ : $A [0] = - 10$ . absDiff = $∣8 - (- 10) ∣ = 18$ . $18 < MAXINT \to$ minVal = 18, $y = - 10$ .
$i = 1$ : $A [1] = 2$ . absDiff = $∣8 - 2∣ = 6$ . $6 < 18 \to$ minVal = 6, $y = 2$ .
$i = 2$ : $A [2] = 10$ . absDiff = $∣8 - 10∣ = 2$ . $2 < 6 \to$ minVal = 2, $y = 10$ .
End loop. Returns $y = 10$ .

Step 3: Trace (b) Closest([0, -5, 4], 3, 7).

Init: minVal = MAXINT.
$i = 0$ : $A [0] = 0$ . absDiff = $∣7 - 0∣ = 7$ . $7 < MAXINT \to$ minVal = 7, $y = 0$ .
$i = 1$ : $A [1] = - 5$ . absDiff = $∣7 - (- 5) ∣ = 12$ . $12 < 7$ is False.
$i = 2$ : $A [2] = 4$ . absDiff = $∣7 - 4∣ = 3$ . $3 < 7 \to$ minVal = 3, $y = 4$ .
End loop. Returns $y = 4$ .

Wrong path: A student might miscalculate the absolute difference, e.g., computing $8 - (- 10)$ as $- 2$ instead of $18$ , or $7 - (- 5)$ as $2$ instead of $12$ . This leads to incorrect minVal updates.

Generalization: Absolute difference $∣ x - A [i] ∣$ is always non-negative. Careful arithmetic is required when negative numbers are involved in the array.

Question 12 · Programming MSQ

Match the flawed post-loop check of a SecondBest algorithm in Column P with its specific behavior when executed on the provided test array in Column Q. Which of the following matchings is completely correct?

Column P (Flawed Post-Loop Check)

P1: if n == 1: return NULL

P2: if A[0] == A[n-1]: return NULL

P3: if second_largest == -\infty: return 0

P4: if second_largest == -\infty: return n

Column Q (Behavior on Specific Test Array)

Q1: Fails to return NULL for $A = [4, 4]$ , because the length condition is not met.

Q2: Incorrectly returns NULL for $A = [5, 3, 5]$ , due to a structural boundary check mismatch.

Q3: Incorrectly returns $0$ for $A = [2, 2]$ , due to a value-domain mismatch.

Q4: Incorrectly returns $3$ for $A = [7, 7, 7]$ , due to a unit mismatch (returning length instead of NULL).

A.
P1-Q1, P2-Q2, P3-Q3, P4-Q4
B.
P1-Q2, P2-Q1, P3-Q4, P4-Q3
C.
P1-Q3, P2-Q4, P3-Q1, P4-Q2
D.
P1-Q4, P2-Q3, P3-Q2, P4-Q1

Correct Answer:

Step-by-Step Solution

Key idea: Each flawed post-loop check uses a condition that either mismatches the semantic requirement of "no second largest" or fails to trigger on specific edge cases.

Step 1: Analyze P1. The condition n == 1 only catches arrays of length 1. On $A = [4, 4]$ , $n = 2$ , so the condition is False. It fails to return NULL and incorrectly proceeds to return the sentinel $- \infty$ . This matches Q1.

Step 2: Analyze P2. On $A = [5, 3, 5]$ , $A [0] = 5$ and $A [2] = 5$ . The condition A[0] == A[n-1] is True, so it returns NULL. However, the array has a valid second largest element (3). This is a structural check that incorrectly triggers. This matches Q2.

Step 3: Analyze P3. On $A = [2, 2]$ , the algorithm correctly leaves second_largest as $- \infty$ . The condition -\infty == -\infty is True, so it returns $0$ . Returning $0$ is a value-domain mismatch, as $0$ is not the designated NULL indicator. This matches Q3.

Step 4: Analyze P4. On $A = [7, 7, 7]$ , second_largest is $- \infty$ . The condition is True, so it returns $n$ , which is $3$ . This is a unit mismatch, returning the array length instead of the required NULL indicator. This matches Q4.

Answer: The correct matching is P1-Q1, P2-Q2, P3-Q3, P4-Q4, which corresponds to Option A.

Section-wise Paper Structure

Free Solved Questions with Step-by-Step Solutions

Unlock All 40 Questions in Real Examination Mode

More CMI Data Science Test Series

Login to view the complete test 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

Algebra & Number Theory, Functions & Calculus, Combinatorics & Induction… Test for CMI Data Science: 40 Questions with Solutions & Analysis

Paper breakdown

Free sample questions from Algebra & Number Theory, Functions & Calculus, Combinatorics & Induction…

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

Step-by-Step Solution

Step-by-Step Solution

More CMI Data Science tests