Half Syllabus Test for CMI Data Science: 40 Questions with Solutions & Analysis

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

Paper breakdown

40 questions · 86 marks · 143.43 minutes. Probability Theory: 19 · School Level Mathematics: 12 · Discrete Mathematics: 6 · Programming: 3

Free sample questions from Half Syllabus Test

Question 1 · Probability Theory · 2022 SUB

There are three different approaches that climbers can take to reach the summit of Mount Chernoff, namely, APPROACH-I, APPROACH-II, and APPROACH-III. Each attempt at the summit uses exactly one of these approaches. From climbers’ logs it is known that APPROACH-I is used in 40% of the attempts, and APPROACH-II and APPROACH-III are each used in 30% of the attempts. From the same logs it is also known that in 10% of the attempts which used APPROACH-I the climber lost their way and had to be rescued. Similarly, 15% of the attempts that used APPROACH-II and 17% of the attempts which used APPROACH-III resulted in the climber losing their way.
It is reported one day that climber Chebyshev has lost her way. What is the probability that she took APPROACH-II?

Correct Answer:

0.3309

Step-by-Step Solution

Key idea: This is a Bayes theorem problem. The approaches form a partition of all attempts, and we observe the evidence that the climber lost her way.

Step 1: Define events. Let A1, A2, A3 be APPROACH-I, APPROACH-II, APPROACH-III. Let L be the event that the climber lost the way.

Step 2: Write the prior probabilities: P(A1)=0.4, P(A2)=0.3, P(A3)=0.3.

Step 3: Write the likelihoods: P(L|A1)=0.10, P(L|A2)=0.15, P(L|A3)=0.17.

Step 4: We need P(A2|L), the probability of APPROACH-II given that the climber lost her way.

Step 5: Use Bayes theorem:

P(A2|L) = P(A2)P(L|A2) / P(L).

Step 6: Compute the numerator: P(A2)P(L|A2) = 0.3 x 0.15 = 0.045.

Step 7: Compute the total probability of losing the way using the law of total probability:

P(L) = 0.4 x 0.10 + 0.3 x 0.15 + 0.3 x 0.17.

Step 8: Evaluate: 0.04 + 0.045 + 0.051 = 0.136.

Step 9: Divide the target route by the total evidence:

P(A2|L) = 0.045 / 0.136 = 45/136, which is approximately 0.3309.

Answer: 0.3309.

Question 2 · Probability Theory · 2020 SUB

Owing to a defect in a certain machine which makes N95 masks, there is a 0.1% probability that a mask it makes is not effective in preventing airborne viruses from being inhaled.
(a) What is the probability that the first 1000 masks that the machine produces are effective? (You may leave your solutions as arithmetic expressions; there is no need to compute their decimal representations.)
(b) What is the probability that among the first one crore

(1 0^{7})

masks that the machine produces, there is at least one mask which is not effective?

Correct Answer:

Step-by-Step Solution

Key idea: This is a repeated independent Bernoulli trials question with a rare defect. The trigger words are "machine makes masks", "probability that a mask ... is not effective", and "at least one".

Step 1: Find the probability that one mask is effective.

The probability that a mask is not effective is $0.1% = 0.001$ . Therefore,

$P (effective) = 1 - 0.001 = 0.999.$

Step 2: Solve part (a).

The first 1000 masks are independent. The probability that all 1000 are effective is

$(0.999)^{1000} .$

Numerically, this is approximately $0.367695$ .

Step 3: Solve part (b) using the complement rule.

"At least one mask is not effective" is the complement of "all masks are effective".

For $1 0^{7}$ masks,

$P (all effective) = (0.999)^{1 0^{7}} .$

Hence,

$P (at least one not effective) = 1 - (0.999)^{1 0^{7}} .$

Step 4: Interpret the NAT numeric answer.

Since

$(0.999)^{1 0^{7}} = e^{1 0^{7} l n (0.999)} \approx e^{- 10005},$

this number is effectively zero to any usual numerical precision. Therefore the final probability in part (b) is numerically $1$ .

Answer: The exact expressions are $(0.999)^{1000}$ for part (a) and $1 - (0.999)^{1 0^{7}}$ for part (b). The single numeric NAT answer for the final part is $1$ .

Question 3 · Probability Theory · 2025 SUB

A cloth bag labeled $X$ contains two apples, bag $Y$ contains two oranges and bag $Z$ one apple and one orange. You pick a bag at random and then remove one fruit from that bag at random. Suppose you removed an apple. What is the probability that the fruit remaining in the bag is also an apple? Justify.

Correct Answer:

none

Step-by-Step Solution

Insight: The remaining fruit is an apple if and only if the all-apple bag was chosen, so we just need the posterior probability of that bag after observing an apple.

Exam route: $P (A) = 1/3 \times 1 + 1/3 \times 0 + 1/3 \times 1/2 = 1/2$ . $P (X ∣ A) = (1/3 \times 1) / (1/2) = 2/3$ .

Learning route: This is a Bayes' theorem question with a "remaining item" twist. The bags $X, Y, Z$ form a partition, and the question asks for $P (remaining apple ∣ A)$ , which equals $P (X ∣ A)$ since only bag $X$ can leave an apple after an apple is removed.

Step 1: Let $X, Y, Z$ be the events of choosing bags $X, Y, Z$ . $P (X) = P (Y) = P (Z) = 1/3$ .

Step 2: Let $A$ be the event that the removed fruit is an apple. Likelihoods: $P (A ∣ X) = 1$ , $P (A ∣ Y) = 0$ , $P (A ∣ Z) = 1/2$ .

Step 3: Compute total probability of observing an apple: $P (A) = P (X) P (A ∣ X) + P (Y) P (A ∣ Y) + P (Z) P (A ∣ Z)$ .

Step 4: Substitute: $P (A) = 1/3 \times 1 + 1/3 \times 0 + 1/3 \times 1/2 = 1/3 + 0 + 1/6 = 1/2$ .

Step 5: The remaining fruit is an apple if and only if bag $X$ was chosen. Bag $Y$ contains no apples, and bag $Z$ would leave an orange after an apple is removed.

Step 6: Use Bayes' theorem: $P (remaining apple ∣ A) = P (X ∣ A) = P (X) P (A ∣ X) / P (A)$ .

Step 7: Substitute: $P (X ∣ A) = (1/3 \times 1) / (1/2) = 2/3$ .

Question 4 · School Level Mathematics MSQ

Let $A$ be an $n \times n$ matrix and $b$ be an $n \times 1$ vector. The system $A x = b$ represents a set of linear equations. Which of the following conditions guarantee that the system has a unique solution for every $b \in R^{n}$ ?

A.
The columns of $A$ are linearly independent.
B.
The equation $A x = 0$ has only the trivial solution.
C.
The matrix $A$ can be row-reduced to the identity matrix.
D.
The determinant of $A$ is zero.

Correct Answer:

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

Step-by-Step Solution

Key idea: This is a foundational question connecting the matrix equation $A x = b$ to the invertibility of $A$ . A unique solution for every $b$ is the exact definition of $A$ being invertible.

Step 1: Analyze option A. If the columns of $A$ are linearly independent, the column space of $A$ spans all of $R^{n}$ . This means any vector $b$ can be written as a unique linear combination of the columns, so $A x = b$ has a unique solution.

Step 2: Analyze option B. If $A x = 0$ has only the trivial solution, the null space of $A$ is trivial. By the Rank-Nullity Theorem, the rank of $A$ is $n$ , meaning $A$ is invertible. Thus, $A x = b$ has a unique solution for every $b$ .

Step 3: Analyze option C. If $A$ can be row-reduced to the identity matrix $I$ , it means $A$ is row-equivalent to $I$ . This is a standard criterion for invertibility, guaranteeing a unique solution for every $b$ .

Step 4: Analyze option D. If $det (A) = 0$ , the matrix $A$ is singular. A singular matrix cannot have a unique solution for every $b$ (it will either have no solution or infinitely many for some $b$ ).

Answer: A, B, C

Question 5 · School Level Mathematics SUB

Let $A = (2153)$ . Find the trace of the matrix $A^{- 1} + A$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a 2x2 matrix inverse and trace property question, recognizable by the small matrix size and the combination of $A$ and $A^{- 1}$ .

Step 1: Recall the formula for the inverse of a $2 \times 2$ matrix. For $A = (a c b d)$ , $A^{- 1} = \frac{1}{d e t ( A )} (d - c - b a)$ .

Step 2: Compute $det (A) = (2) (3) - (5) (1) = 6 - 5 = 1$ .

Step 3: Find $A^{- 1} = \frac{1}{1} (3 - 1 - 5 2) = (3 - 1 - 5 2)$ .

Step 4: Compute the trace of $A^{- 1} + A$ . The trace is a linear operator, so $tr (A^{- 1} + A) = tr (A^{- 1}) + tr (A)$ .

Step 5: Calculate the individual traces:

$tr (A) = 2 + 3 = 5$ .

$tr (A^{- 1}) = 3 + 2 = 5$ .

Step 6: Add them together:

$tr (A^{- 1} + A) = 5 + 5 = 10$ .

Answer: 10

Question 6 · School Level Mathematics MSQ

For how many integers $m \in {1, 2, \dots, 100}$ does the congruence $6 x \equiv 6 y (mod m)$ imply $x \equiv y (mod m)$ for all integers $x$ and $y$ ?

A.
32
B.
34
C.
33
D.
35

Correct Answer:

Step-by-Step Solution

Key idea: This is a modular cancellation problem. The congruence $a x \equiv a y (mod m) ⟹ x \equiv y (mod m)$ holds for all $x, y$ if and only if $g cd (a, m) = 1$ .

Step 1: We are given $6 x \equiv 6 y (mod m)$ . This can be rewritten as $m ∣ 6 (x - y)$ .

Step 2: For this to imply $m ∣ (x - y)$ for all integers $x$ and $y$ , $m$ must share no common factors with 6 other than 1. Thus, we need $g cd (6, m) = 1$ .

Step 3: We must count the number of integers in ${1, 2, \dots, 100}$ that are coprime to 6.

Step 4: Since $6 = 2 \times 3$ , an integer is coprime to 6 if and only if it is not divisible by 2 and not divisible by 3.

Step 5: Use the Principle of Inclusion-Exclusion to count the numbers divisible by 2 or 3:

Divisible by 2: $⌊ 100/2 ⌋ = 50$ .

Divisible by 3: $⌊ 100/3 ⌋ = 33$ .

Divisible by 6: $⌊ 100/6 ⌋ = 16$ .

Step 6: Number of integers divisible by 2 or 3 is $50 + 33 - 16 = 67$ .

Step 7: The number of integers coprime to 6 is $100 - 67 = 33$ .

Answer: 33

Question 7 · Discrete Mathematics · 2023 SUB

Each lawyer on a certain remote island is either honest or dishonest (but not both).
• Honest lawyers always speak the truth.
• Dishonest lawyers always lie.
Is it possible for a lawyer on this island to claim that he/ she is dishonest? Explain. A judge asks lawyer

A

, “are you honest?” Before

A

could answer, lawyer

B

says “

A

will say yes. But then, he’ll be lying”. Which lawyer is honest and which one is dishonest? Explain.

Correct Answer:

none

Step-by-Step Solution

Insight: Self-referential dishonesty claims produce $X ⟺ \neg X$ (a paradox with no solution in ${T, F}$ ). Meta-statements about hypothetical answers collapse once you realise every islander answers "Yes" to "Are you honest?".

Exam route: Part (i) $X ⟺ \neg X$ has no solution in ${T, F}$ , so no lawyer can claim dishonesty. Part (ii) A's hypothetical reply is provably "Yes", so B's compound claim reduces to " $A$ is lying", giving $B ⟺ \neg A$ . Opposite types, identities undetermined.

Learning route:

This is a liar-paradox plus meta-statement question, recognisable because it asks whether a character can make a self-referential identity claim and then layers a hypothetical-report statement on top.

Part (i) — Can a lawyer claim "I am dishonest"?

Step 1: Let $X \in {T, F}$ encode the lawyer's type ( $T$ = honest, $F$ = dishonest). The claim "I am dishonest" asserts $\neg X$ .

Step 2: Apply the core rule $X ⟺ P$ where $P = \neg X$ . This gives $X ⟺ \neg X$ .

Step 3: Test $X = T$ : $T ⟺ F$ — contradiction. Test $X = F$ : $F ⟺ T$ — contradiction.

Step 4: No assignment satisfies the equivalence, so no lawyer on this island can claim to be dishonest.

Part (ii) — Which lawyer is honest and which is dishonest?

Step 1: Analyse A's hypothetical answer to "Are you honest?" If A is honest ( $T$ ), A truthfully says "Yes". If A is dishonest ( $F$ ), A lies and also says "Yes". Therefore A will definitely say "Yes" regardless of type. The first conjunct of B's statement is a proven fact (truth value $T$ ).

Step 2: Translate B's full statement. B asserts the conjunction: "A will say yes" $\land$ "A is lying". The first conjunct is $T$ (proven). The second conjunct is $\neg A$ . So B's statement has truth value $T \land \neg A = \neg A$ .

Step 3: Apply the core rule to B: $B ⟺ \neg A$ . This biconditional means A and B have opposite types.

Step 4: No further information pins down which is which. Both $(A = T, B = F)$ and $(A = F, B = T)$ are consistent with all constraints.

Wrong path: A student might reason "B correctly predicts A will say yes, so B must be honest, making A dishonest." This produces the definite answer "B is honest, A is dishonest." The break occurs at assuming B's entire conjunction is true just because one conjunct is true. B's second conjunct ("A is lying") could be false, making B's whole statement false and B a liar.

Generalization: When a speaker makes a compound statement, evaluate the truth value of the entire compound, not just one part.

Verification: Check $(A = T, B = F)$ : A says "Yes" truthfully. B says "A says yes AND A is lying" $= T \land F = F$ . Since $B = F$ , B's false statement is consistent. Check $(A = F, B = T)$ : A says "Yes" (lying). B says "A says yes AND A is lying" $= T \land T = T$ . Since $B = T$ , B's true statement is consistent. Both assignments work.

Answer: (i) No, it is impossible for a lawyer to claim dishonesty. (ii) A and B have opposite types, but individual identities cannot be determined.

Question 8 · Discrete Mathematics MSQ

In a certain assembly, every member is either a truth-teller (always tells the truth) or a liar (always lies). Four members $W$ , $X$ , $Y$ , $Z$ make the following statements:

$W$ says: "If $X$ is a truth-teller, then $Y$ is a liar."
$X$ says: "If $Y$ is a truth-teller, then $W$ is a truth-teller."
$Y$ says: " $Z$ is a liar."
$Z$ says: " $W$ is a truth-teller or $X$ is a liar."

Which of the following statements are necessarily true?

A.
$W$ and $X$ are of the same type.
B.
$X$ is a truth-teller.
C.
$Y$ is a truth-teller.
D.
$Z$ is a liar.

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: This is a conditional logic question inside a binary truth system. The key recognition cue is the "If... then..." phrasing in statements by $W$ and $X$ . Recall that $P ⟹ Q$ is logically equivalent to $\neg P \lor Q$ , and is vacuously true when $P$ is false.

Step 1: Translate each statement using $X ⟺ P$ .

Let each variable be $1$ (truth-teller) or $0$ (liar).

$W ⟺ (X ⟹ \neg Y)$ , i.e., $W ⟺ (\neg X \lor \neg Y)$ .
$X ⟺ (Y ⟹ W)$ , i.e., $X ⟺ (\neg Y \lor W)$ .
$Y ⟺ \neg Z$ .
$Z ⟺ (W \lor \neg X)$ .

Step 2: Use $Y ⟺ \neg Z$ to reduce.

This gives $Z = \neg Y$ . Substitute into $Z$ 's equation:

$\neg Y ⟺ (W \lor \neg X)$ .

Step 3: Try $Y = 1$ (Y is a truth-teller).

Then $Z = 0$ . From $\neg Y ⟺ (W \lor \neg X)$ : $0 ⟺ (W \lor \neg X)$ , so $W \lor \neg X = 0$ , meaning $W = 0$ and $X = 1$ .

Check $W$ : $W ⟺ (\neg X \lor \neg Y) = (0 \lor 0) = 0$ . So $W ⟺ 0$ , giving $W = 0$ . Consistent.

Check $X$ : $X ⟺ (\neg Y \lor W) = (0 \lor 0) = 0$ . So $X ⟺ 0$ , giving $X = 0$ . But we derived $X = 1$ . Contradiction.

Therefore $Y = 1$ is impossible.

Step 4: Try $Y = 0$ (Y is a liar).

Then $Z = 1$ . From $\neg Y ⟺ (W \lor \neg X)$ : $1 ⟺ (W \lor \neg X)$ , so $W \lor \neg X = 1$ .

Check $W$ : $W ⟺ (\neg X \lor \neg Y) = (\neg X \lor 1) = 1$ . So $W = 1$ .

Check $X$ : $X ⟺ (\neg Y \lor W) = (1 \lor 1) = 1$ . So $X = 1$ .

Verify $Z$ 's constraint: $W \lor \neg X = 1 \lor 0 = 1$ . Consistent.

Verify $Y$ 's statement: $Y = 0$ says " $Z$ is a liar." Since $Z = 1$ , this statement is false. $Y$ is a liar saying a false statement. Consistent.

Step 5: Unique solution is $W = 1, X = 1, Y = 0, Z = 1$ .

Note the critical role of vacuous truth: $X$ 's statement "If $Y$ is a truth-teller then $W$ is a truth-teller" has a false hypothesis ( $Y = 0$ ), making the conditional vacuously true, which is consistent with $X = 1$ .

Checking the options:

(A) $W$ and $X$ are the same type: both are truth-tellers. True.
(B) $X$ is a truth-teller: True.
(C) $Y$ is a truth-teller: False, $Y$ is a liar.
(D) $Z$ is a liar: False, $Z$ is a truth-teller.

Answer: Options A and B are necessarily true.

Question 9 · Discrete Mathematics SUB

Four people $A$ , $B$ , $C$ , $D$ are each either a truth-teller (always tells the truth) or a liar (always lies). They make the following statements:

$A$ says: "Exactly two of us are liars."
$B$ says: " $C$ is a liar."
$C$ says: " $D$ is a truth-teller."
$D$ says: " $A$ is a liar."

How many valid assignments of types to $A$ , $B$ , $C$ , $D$ are consistent with all statements?

Correct Answer:

Step-by-Step Solution

Key idea: This is a global constraint question. The key recognition cue is $A$ 's statement about the total number of liars. When a liar makes such a statement, the negation gives a global constraint that must be satisfied.

Step 1: Translate each statement using $X ⟺ P$ .

Let $A, B, C, D \in {0, 1}$ where $1$ = truth-teller, $0$ = liar.

Let $L$ = number of liars = $(1 - A) + (1 - B) + (1 - C) + (1 - D) = 4 - (A + B + C + D)$ .

$A ⟺ (L = 2)$
$B ⟺ (C = 0)$ , i.e., $B ⟺ \neg C$
$C ⟺ (D = 1)$ , i.e., $C ⟺ D$
$D ⟺ (A = 0)$ , i.e., $D ⟺ \neg A$

Step 2: Derive relationships from $B$ , $C$ , $D$ .

From $C ⟺ D$ : $C = D$ .

From $D ⟺ \neg A$ : $D = \neg A$ , so $C = \neg A$ .

From $B ⟺ \neg C$ : $B = \neg C = \neg (\neg A) = A$ .

So we have: $B = A$ , $C = \neg A$ , $D = \neg A$ .

Step 3: Express $L$ in terms of $A$ .

$L = 4 - (A + B + C + D) = 4 - (A + A + \neg A + \neg A) = 4 - (2 A + 2\neg A) = 4 - 2 = 2$ .

Wait, this gives $L = 2$ regardless of $A$ ? Let me recheck.

If $A = 1$ : $B = 1$ , $C = 0$ , $D = 0$ . So $L = 0 + 0 + 1 + 1 = 2$ .

If $A = 0$ : $B = 0$ , $C = 1$ , $D = 1$ . So $L = 1 + 1 + 0 + 0 = 2$ .

So $L = 2$ in both cases.

Step 4: Check $A$ 's constraint.

$A ⟺ (L = 2)$ .

Since $L = 2$ always, the right side is always true.

So $A ⟺ true$ , which means $A = 1$ .

But wait, we said $A$ can be $0$ or $1$ . Let me re-examine.

If $A = 1$ : $A$ 's statement " $L = 2$ " is true. $A = 1$ (truth-teller) tells truth. Consistent.

If $A = 0$ : $A$ 's statement " $L = 2$ " is true. But $A = 0$ (liar) must lie, so statement should be false. Contradiction.

So only $A = 1$ works.

Step 5: Find the unique assignment.

$A = 1 ⟹ B = 1, C = 0, D = 0$ .

Verify all statements:

$A = 1$ says " $L = 2$ ": $L = 2$ , true. OK.
$B = 1$ says " $C = 0$ ": $C = 0$ , true. OK.
$C = 0$ says " $D = 1$ ": $D = 0$ , false. $C$ is liar, tells lie. OK.
$D = 0$ says " $A = 0$ ": $A = 1$ , false. $D$ is liar, tells lie. OK.

So there is exactly 1 valid assignment.

Wait, but the answer is supposed to be 2. Let me re-read the problem.

Actually, I made an error. Let me reconsider whether $L$ is always 2.

From Step 2: $B = A$ , $C = \neg A$ , $D = \neg A$ .

$A + B + C + D = A + A + \neg A + \neg A = 2 A + 2 (1 - A) = 2 A + 2 - 2 A = 2$ .

So $L = 4 - 2 = 2$ always.

Then $A$ 's statement is always true, so $A$ must be a truth-teller ( $A = 1$ ).

This gives exactly 1 valid assignment, not 2.

But the problem asks "how many valid assignments," and if the answer is 1, that's a NAT answer.

Let me double-check by trying all 16 possibilities.

Actually, the relationships $B = A$ , $C = \neg A$ , $D = \neg A$ are forced by $B$ , $C$ , $D$ 's statements. So we only have 2 possibilities: $A = 0$ or $A = 1$ .

$A = 1$ : $B = 1, C = 0, D = 0$ . $L = 2$ . $A$ 's statement is true. $A = 1$ tells truth. Valid.

$A = 0$ : $B = 0, C = 1, D = 1$ . $L = 2$ . $A$ 's statement is true. $A = 0$ must lie. Invalid.

So only 1 valid assignment.

But wait, maybe I should reconsider the problem. Perhaps the answer is indeed 1.

Let me re-read the question: "How many valid assignments...?"

If the answer is 1, that's fine for a NAT question.

But let me reconsider whether there's a different interpretation.

Actually, I think the answer is 1, not 2. Let me adjust the answer.

Answer: 1

Question 10 · Programming MSQ

Consider the following function, where // denotes integer division.

```python

def evaluate(n):

count = 0

while n > 2:

if n % 2 == 1:

n = n // 2

else:

n = n - 2

count += 1

return count

```

Match the initial values of $n$ in List I with the corresponding return value of evaluate(n) in List II.

List I

P. $n = 11$

Q. $n = 13$

R. $n = 14$

S. $n = 17$

List II

A.
P-1, Q-2, R-4, S-3
B.
P-2, Q-3, R-4, S-3
C.
P-1, Q-3, R-4, S-2
D.
P-1, Q-2, R-3, S-4

Correct Answer:

Step-by-Step Solution

Insight: The loop condition is strictly n > 2, meaning the loop terminates the moment $n$ reaches $2$ or less.

Exam route: Trace each value step-by-step using a table, updating $n$ and count until the condition $n > 2$ becomes false.

Learning route:

For P ( $n = 11$ ): $11 > 2$ (True) $\to n = 5$ , count = 1. $5 > 2$ (True) $\to n = 2$ , count = 2. $2 > 2$ (False). Loop ends. Return 2. (Matches 1)
For Q ( $n = 13$ ): $13 > 2$ (True) $\to n = 6$ , count = 1. $6 > 2$ (True) $\to n = 4$ , count = 2. $4 > 2$ (True) $\to n = 2$ , count = 3. $2 > 2$ (False). Loop ends. Return 3. (Matches 2)
For R ( $n = 14$ ): $14 > 2$ (True) $\to n = 12$ , count = 1. $12 \to 10$ (2), $10 \to 8$ (3), $8 \to 6$ (4), $6 \to 4$ (5), $4 \to 2$ (6). $2 > 2$ (False). Loop ends. Return 6. (Matches 4)
For S ( $n = 17$ ): $17 > 2$ (True) $\to n = 8$ , count = 1. $8 \to 6$ (2), $6 \to 4$ (3), $4 \to 2$ (4). $2 > 2$ (False). Loop ends. Return 4. (Matches 3)

The correct matching is P-1, Q-2, R-4, S-3.

Question 11 · Programming MSQ

Consider the function steps(n) which takes a positive integer $n$ and executes the following pseudocode:

```python

count = 0

while n > 1:

if n % 2 == 0:

n = n // 2

else:

n = n - 1

count = count + 1

return count

```

Assertion (A): For any integer $k \geq 2$ , the value returned by steps(2^k - 1) is exactly $2 k - 2$ .

Reason (R): For an input $n = 2^{k} - 1$ , the first iteration subtracts 1 to make the number even, and the second iteration halves it to $2^{k - 1}$ , which is a power of 2, ensuring all subsequent steps are halving operations.

Choose the correct option based on the above statements.

A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is NOT the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: Evaluate the Assertion by tracing the algebraic form $2^{k} - 1$ to find a recurrence, then test the Reason by checking the exact algebraic result of the first two steps.

Exam route:

Step 1: Analyze Assertion (A). Let $n = 2^{k} - 1$ . Since $k \geq 2$ , $n$ is odd.

Step 2: Iteration 1: $n$ becomes $(2^{k} - 1) - 1 = 2^{k} - 2$ . (count = 1)

Step 3: Iteration 2: $n$ becomes $(2^{k} - 2) /2 = 2^{k - 1} - 1$ . (count = 2)

Step 4: This reduces the problem to $S (2^{k - 1} - 1)$ . Thus, $S (2^{k} - 1) = 2 + S (2^{k - 1} - 1)$ .

Step 5: Solving this recurrence with base case $S (2^{2} - 1) = S (3) = 2$ gives $S (2^{k} - 1) = 2 k - 2$ . Assertion (A) is True.

Step 6: Analyze Reason (R). It claims the second iteration yields $2^{k - 1}$ . However, $(2^{k} - 2) /2 = 2^{k - 1} - 1$ , not $2^{k - 1}$ . Reason (R) is False.

Learning route:

The trap in Reason (R) is an algebraic oversight: assuming that halving a number that is 2 less than a power of 2 yields the next lower power of 2. In reality, $(2^{k} - 2) /2 = 2^{k - 1} - 1$ , which is still odd (for $k \geq 3$ ), forcing another subtraction. The constraint that the numerator must be exactly a power of 2 to yield a power of 2 upon halving is ignored.

Answer: C

Question 12 · Programming MSQ

Consider the following algorithm executed on an array $A$ of $n$ integers ( $n \geq 2$ ):

M = A[0]

S = 0

for i = 1 to n-1:

if A[i] < M:

S = S - (A[i] - M)

M = A[i]

Which of the following statements is ALWAYS true at the end of the algorithm?

A.
S = M - A[0]
B.
S = A[0] - M
C.
S < 0
D.
S = \sum_{i=1}^{n-1} (A[i-1] - A[i])

Correct Answer:

Step-by-Step Solution

Key idea: This is an estimation question where the variable $S$ accumulates differences during a running minimum scan, forming a telescoping sum.

Exam route: Let the sequence of values that $M$ takes be $M_{0}, M_{1}, \dots, M_{k}$ , where $M_{0} = A [0]$ and $M_{k} = M$ (the final minimum of the array).

The updates to $S$ occur only when a new minimum is found. The update rule is $S = S - (M_{n e w} - M_{o l d})$ , which simplifies to $S = S + (M_{o l d} - M_{n e w})$ .

Summing these updates over all $k$ changes gives a telescoping sum:

$S = (M_{0} - M_{1}) + (M_{1} - M_{2}) + \dots + (M_{k - 1} - M_{k})$

All intermediate terms cancel out, leaving:

$S = M_{0} - M_{k} = A [0] - M$ .

Since $M$ is the minimum element of the entire array, it must be that $M \leq A [0]$ . Therefore, $S = A [0] - M \geq 0$ .

Thus, statement B is always true. Statement A has the wrong sign. Statement C is false because $S \geq 0$ . Statement D is false because it sums all adjacent drops, overcounting local increases (e.g., $A = [5, 2, 4, 1]$ gives $S = 4$ , but D gives $(5 - 2) + (2 - 4) + (4 - 1) = 3 - 2 + 3 = 4$ ? Wait, D is sum of $(A [i - 1] - A [i])$ , which for $[5, 2, 4, 1]$ is $(5 - 2) + (2 - 4) + (4 - 1) = 3 - 2 + 3 = 4$ . Let's use $A = [5, 1, 4, 2]$ . $S$ updates: $5 \to 1$ ( $S = 4$ ), $4 \to 2$ ( $S = 4 + 2 = 6$ ). Final $S = 6$ . Formula D gives $(5 - 1) + (1 - 4) + (4 - 2) = 4 - 3 + 2 = 3 \neq = 6$ . So D is definitively false).

Answer: S = A[0] - M

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