Mock Test 1 for CMI Data Science: 40 Questions with Solutions & Analysis

Attempt the Mock Test 1 for CMI Data Science: 40 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

40 questions · 85 marks · 149.59 minutes. School Level Mathematics: 14 · Probability Theory: 10 · Discrete Mathematics: 10 · Programming: 6

Free sample questions from Mock Test 1

Question 1 · School Level Mathematics · 2024 MSQ

Suppose that $A$ is an $n \times n$ matrix with $n \geq 6$ . For an $n \times 1$ vector $b$ consider the equations

$A x = b, for an n \times 1 vector x (1)$

$A^{T} x = b for an n \times 1 vector x (2)$

where $A^{T}$ is the transpose of the matrix $A$ . Which of the following statements are correct?

A.
If equation (1) admits a solution for all $b$ , then $A^{- 1}$ exists.
B.
If equation (1) admits a solution for all $b$ , then equation (2) also admits a solution for all $b$ .
C.
If equation (1) admits a solution for some $b$ , then $A^{- 1}$ exists.
D.
If equation (1) admits a solution for some $b$ , then equation (2) also admits a solution for that $b$ .

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: This question tests the relationship between the solvability of $A x = b$ and the invertibility of $A$ , and how this relates to $A^{T}$ .

Step 1: Analyze Option A. "If equation (1) admits a solution for all $b$ , then $A^{- 1}$ exists."

If $A x = b$ has a solution for every $b \in R^{n}$ , then the column space of $A$ is all of $R^{n}$ . This means $A$ has full rank ( $n$ ). For a square matrix, full rank implies invertibility. So, $A^{- 1}$ exists. Option A is Correct.

Step 2: Analyze Option B. "If equation (1) admits a solution for all $b$ , then equation (2) also admits a solution for all $b$ ."

From Step 1, if (1) is solvable for all $b$ , then $A$ is invertible.

If $A$ is invertible, then $A^{T}$ is also invertible (since $det (A^{T}) = det (A) \neq = 0$ ).

If $A^{T}$ is invertible, then $A^{T} x = b$ has a unique solution $x = (A^{T})^{- 1} b$ for every $b$ .

So, Option B is Correct.

Step 3: Analyze Option C. "If equation (1) admits a solution for some $b$ , then $A^{- 1}$ exists."

Solvability for some $b$ does not imply full rank. $A$ could be singular, and $b$ just happens to be in the column space.

Counterexample: $A = (1000)$ , $b = (10)$ . $A x = b$ has solution $x = (10)$ . But $A$ is not invertible. Option C is Incorrect.

Step 4: Analyze Option D. "If equation (1) admits a solution for some $b$ , then equation (2) also admits a solution for that $b$ ."

Using the same counterexample: $A = (1000)$ , $b = (10)$ .

$A x = b$ has a solution.

Check $A^{T} x = b$ . Since $A$ is symmetric, $A^{T} = A$ . So $A^{T} x = b$ is the same system, which has a solution.

Wait, let's pick a non-symmetric singular matrix.

Let $A = (0010)$ . $A^{T} = (0100)$ .

Let $b = (10)$ .

$A x = b ⟹ (0010) (x_{1} x_{2}) = (10) ⟹ x_{2} = 1, 0 = 0$ . Solution exists ( $x = (01)$ ).

Now check $A^{T} x = b ⟹ (0100) (x_{1} x_{2}) = (10) ⟹ 0 = 1$ . No solution.

So, solvability of $A x = b$ does not imply solvability of $A^{T} x = b$ for the same $b$ if $A$ is singular. Option D is Incorrect.

Answer: Options A and B are correct.

Question 2 · School Level Mathematics SUB

Let $f (x) = x^{4} - 8 x^{3} + 18 x^{2}$ . Let $L$ be the tangent line to the curve $y = f (x)$ at its inflection point with the smaller $x$ -coordinate. Find the area of the finite region bounded by $y = f (x)$ and $L$ .

Correct Answer:

none

Step-by-Step Solution

Key idea: This is a geometry and inflection point synthesis question. The tangent at an inflection point intersects the curve with multiplicity at least 3, which greatly simplifies the area calculation.

Step 1: Find the inflection points.

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

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

Inflection points at $x = 1$ and $x = 3$ . The smaller is $x = 1$ .

Step 2: Find the tangent line $L$ at $x = 1$ .

$f (1) = 1 - 8 + 18 = 11$ .

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

Tangent line: $y = 16 (x - 1) + 11 = 16 x - 5$ .

Step 3: Find intersection points of $f (x)$ and $L$ .

$f (x) - L (x) = x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 5$ .

Since $x = 1$ is both a tangency point and an inflection point, $f (x) - L (x)$ has a root of multiplicity at least 3 at $x = 1$ . Factor out $(x - 1)^{3}$ :

$x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 5 = (x - 1)^{3} (x - 5)$ .

Verification: $(x - 1)^{3} (x - 5) = (x^{3} - 3 x^{2} + 3 x - 1) (x - 5) = x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 5$ . Correct.

Intersection points: $x = 1$ and $x = 5$ .

Step 4: Determine which function is on top.

For $x \in (1, 5)$ : $(x - 1)^{3} > 0$ and $(x - 5) < 0$ , so $f (x) - L (x) < 0$ . The line $L$ is above the curve.

Step 5: Compute the area.

$A = \int_{1}^{5} [L (x) - f (x)] d x = \int_{1}^{5} - (x - 1)^{3} (x - 5) d x$ .

Substitute $u = x - 1$ , $d u = d x$ , limits $0$ to $4$ , and $x - 5 = u - 4$ :

$A = \int_{0}^{4} - u^{3} (u - 4) d u = \int_{0}^{4} (4 u^{3} - u^{4}) d u = [u^{4} - \frac{u ^{5}}{5}]_{0}^{4} = 256 - \frac{1024}{5} = \frac{256}{5}$ .

Answer: The area is $\frac{256}{5}$ .

Question 3 · School Level Mathematics SUB

When $x^{100}$ is divided by $x^{2} - 3 x + 2$ , the remainder is of the form $a x + b$ . The value of $a + b$ is:

Correct Answer:

Step-by-Step Solution

Key idea: This is a polynomial remainder question using the Remainder Theorem. Since the divisor is quadratic, the remainder is linear: $R (x) = a x + b$ . We can find $a$ and $b$ by evaluating at the roots of the divisor.

Step 1: Factor the divisor.

$x^{2} - 3 x + 2 = (x - 1) (x - 2)$ . Roots are $x = 1$ and $x = 2$ .

Step 2: Write the division algorithm.

$x^{100} = (x^{2} - 3 x + 2) \cdot Q (x) + a x + b$

$x^{100} = (x - 1) (x - 2) \cdot Q (x) + a x + b$

Step 3: Substitute $x = 1$ (to eliminate $Q (x)$ ).

$1^{100} = 0 \cdot Q (1) + a (1) + b$

$1 = a + b$ ... (equation 1)

Step 4: Substitute $x = 2$ .

$2^{100} = 0 \cdot Q (2) + a (2) + b$

$2^{100} = 2 a + b$ ... (equation 2)

Step 5: Solve the system.

Subtract equation 1 from equation 2:

$2^{100} - 1 = (2 a + b) - (a + b) = a$

So $a = 2^{100} - 1$ .

From equation 1: $b = 1 - a = 1 - (2^{100} - 1) = 2 - 2^{100}$ .

Step 6: Compute $a + b$ .

$a + b = (2^{100} - 1) + (2 - 2^{100}) = 1$ .

Answer: 1

Common trap: Trying to perform polynomial long division of $x^{100}$ by a quadratic, which is computationally infeasible. The Remainder Theorem shortcut avoids this entirely.

Question 4 · Probability Theory · 2020 SUB

Two friends Amar and Prem wish to meet at a theme party between 5 p.m. and 6 p.m. (They are said to meet if they are in the room at the same time or if one of them leaves as the other enters.) Once they enter, they stay for exactly 20 minutes.
(a) Fill in the blanks:
i. The latest time by which any one of them can enter is ________;
ii. If Amar and Prem are to meet, then their entry times can be separated by at most an interval of ______ minutes.
(b) What is the probability that Amar and Prem will meet? (Hint: Plot the arrival times on the x- and y-axes.)

Correct Answer:

0.56

Step-by-Step Solution

Key idea: This is a classic geometric probability meeting problem, recognisable from the setup: two people arrive uniformly at random in a time interval, stay for a fixed duration, and meet if their intervals overlap.

Step 1: Interpret the meeting condition.

The party runs from 5 p.m. to 6 p.m., a 60-minute window. Each stays for exactly 20 minutes. They meet if their 20-minute intervals overlap. This happens if and only if the difference in their arrival times is at most 20 minutes. Let X and Y be their arrival times in minutes after 5 p.m. They meet if |X - Y| <= 20.

Step 2: Model the sample space.

Since arrivals are uniform and independent over the 60-minute window, the sample space is a 60x60 square in the XY-plane. The total area is 60 * 60 = 3600.

Step 3: Compute the probability geometrically.

The region where they do not meet is where |X - Y| > 20. This consists of two right triangles in the corners of the square. Each triangle has legs of length 60 - 20 = 40.

Area of one triangle = (1/2) 40 40 = 800.

Total non-meeting area = 2 * 800 = 1600.

Meeting area = Total area - Non-meeting area = 3600 - 1600 = 2000.

Probability = Meeting area / Total area = 2000 / 3600 = 5/9.

Step 4: Round to two decimal places.

5/9 = 0.555... which rounds to 0.56.

Question 5 · Probability Theory SUB

Let $X$ be a discrete random variable with $E [X] = 3$ and $Var (X) = 2$ . Define $Y = X^{2} - 3 X + 5$ . Find $E [Y]$ .

Correct Answer:

Step-by-Step Solution

Key idea: This is a non-linear transformation question. You need $E [g (X)]$ where $g (X) = X^{2} - 3 X + 5$ . Since $g$ is non-linear, you cannot simply plug in $E [X]$ ; you must use LOTUS and the relationship between $E [X^{2}]$ , $E [X]$ , and $Var (X)$ .

Step 1: Use linearity of expectation.

$E [Y] = E [X^{2} - 3 X + 5] = E [X^{2}] - 3 E [X] + 5$

Step 2: Find $E [X^{2}]$ from the given information.

We know $Var (X) = E [X^{2}] - (E [X])^{2}$ .

So $E [X^{2}] = Var (X) + (E [X])^{2} = 2 + 3^{2} = 2 + 9 = 11$ .

Step 3: Substitute back.

$E [Y] = 11 - 3 (3) + 5 = 11 - 9 + 5 = 7$

Answer: $E [Y] = 7$

Question 6 · Probability Theory SUB

5 balls are placed independently and uniformly at random into 5 bins. Let $X$ be the number of bins that contain exactly one ball. Find $E [X]$ .

Correct Answer:

2.048

Step-by-Step Solution

Key idea: This is a standard occupancy model question, recognizable because it asks for the expected number of bins satisfying a specific capacity condition (exactly one ball).

Why it applies: We can define an indicator variable for each bin being in the target state, and use linearity of expectation to sum their probabilities.

Step 1: Define indicators. Let $I_{j}$ be 1 if bin $j$ contains exactly 1 ball, for $j = 1, \dots, 5$ .

Step 2: Calculate the probability for one bin. For bin $j$ to have exactly 1 ball, we must choose which of the 5 balls lands in it (5 choices), that ball must land in bin $j$ (prob 1/5), and the other 4 balls must NOT land in bin $j$ (prob $(4/5)^{4}$ ).

$P (I_{j} = 1) = 5 \times (1/5) \times (4/5)^{4} = (4/5)^{4} = 256/625$ .

Step 3: Apply linearity. The total number of bins with exactly 1 ball is $X = \sum_{j = 1}^{5} I_{j}$ .

$E [X] = \sum_{j = 1}^{5} E [I_{j}] = 5 \times (256/625) = 256/125 = 2.048$ .

Trap: Forgetting the combinatorial factor of 5 for which ball is the single one, or using $(1/5)^{1} (4/5)^{4}$ without multiplying by 5, which gives the probability for a specific ball being the only one, not the bin having exactly one ball.

Answer: 2.048

Question 7 · Discrete Mathematics · 2025 SUB

Three hostel friends Amar, Prem and Raj are suspected of breaking a window. They made the following statements when questioned by the warden:
• Amar: I did not break it. Prem is lying.
• Prem: Amar is telling the truth. Raj broke the window.
• Raj: I did not break it. Either Amar is telling the truth or Prem is telling the truth.
You know that exactly one of them lied and the other two told the truth. Then, who broke the window? Justify.

Correct Answer:

none

Step-by-Step Solution

Insight: The global lock "exactly one liar" guarantees at least two truth-tellers, so any disjunction of two speakers' truth-values is automatically $T$ . This collapses Raj's OR clause before any casework begins.

Exam route: Translate all three statements via $X ⟺ P$ . Use $(A \lor P) = T$ (from the global constraint) to simplify Raj's claim to $R ⟺ \neg r$ . Test the three single-liar cases; only "Prem is the liar" survives, forcing $a = F$ , $r = F$ , $p = T$ . Prem broke the window.

Learning route:

This is an exactly-one-liar puzzle with compound statements, recognisable by the phrase "exactly one of them lied" combined with AND/OR connectives inside the spoken claims.

Step 1: Assign boolean variables. Let $A, P, R \in {T, F}$ represent truth-teller status ( $T$ = truth-teller, $F$ = liar). Let $a, p, r \in {T, F}$ represent "broke the window" ( $T$ = broke it). Exactly one of ${a, p, r}$ is $T$ .

Step 2: Translate each statement via the core rule $X ⟺ P$ .

Amar: "I did not break it AND Prem is lying" $⟹ A ⟺ (\neg a \land \neg P)$ .
Prem: "Amar is telling the truth AND Raj broke the window" $⟹ P ⟺ (A \land r)$ .
Raj: "I did not break it AND (Amar is telling the truth OR Prem is telling the truth)" $⟹ R ⟺ (\neg r \land (A \lor P))$ .

Step 3: Apply the global constraint. Exactly one of ${A, P, R}$ is $F$ , so at least two are $T$ . In every legal assignment, $(A \lor P) = T$ (the only way $A \lor P = F$ is if both $A = F$ and $P = F$ , which would be two liars).

Step 4: Simplify Raj's statement. $R ⟺ (\neg r \land T) = \neg r$ . So $R ⟺ \neg r$ .

Step 5: Case-split on the single liar.

Case 1: $A = F, P = T, R = T$ . Prem gives $T ⟺ (F \land r) = T ⟺ F$ . Contradiction. Dead.
Case 2: $A = T, P = T, R = F$ . Amar gives $T ⟺ (\neg a \land \neg T) = T ⟺ F$ . Contradiction. Dead.
Case 3: $A = T, P = F, R = T$ . Amar gives $T ⟺ (\neg a \land T) ⟹ \neg a = T$ , so $a = F$ . Raj gives $T ⟺ \neg r ⟹ r = F$ . Since exactly one broke the window and $a = F, r = F$ , we get $p = T$ . Verify Prem (liar): $F ⟺ (T \land F) = F ⟺ F = T$ . Consistent.

Wrong path: A student might not simplify Raj's OR clause and instead test Case 2 ( $A = T, P = T, R = F$ ) and conclude "Raj is the liar, so Raj broke the window." The break occurs at not checking Amar's equation in this case: $T ⟺ (\neg a \land F) = T ⟺ F$ , which is a contradiction. The case is dead regardless of who broke the window.

Generalization: In every case, substitute into ALL translated equations, not just the ones that seem relevant. One contradiction kills the entire case.

Verification: With $A = T, P = F, R = T$ and $a = F, p = T, r = F$ : Amar says "I did not break it" (true) AND "Prem is lying" (true) — both true, $A = T$ checks out. Prem says "Amar is telling the truth" (true) AND "Raj broke the window" (false) — conjunction false, $P = F$ checks out. Raj says "I did not break it" (true) AND "Either Amar or Prem is telling the truth" (true) — both true, $R = T$ checks out.

Answer: Prem broke the window.

Question 8 · Discrete Mathematics SUB

In a survey of 70 people about two social media platforms $P$ and $Q$ , it is found that 44 people use $P$ , 38 use $Q$ , and the number of people who use only $P$ is exactly 4 more than the number who use neither platform. How many people use exactly one platform?

Correct Answer:

Step-by-Step Solution

Key idea: This is a 2-set problem where a region is linked to the "neither" category, recognisable because the condition relates "P only" to "neither" rather than to another region within the union.

Step 1: Let both $= b$ . Then:

$P$ only $= 44 - b$
$Q$ only $= 38 - b$
Union $= 44 + 38 - b = 82 - b$
Neither $= 70 - (82 - b) = b - 12$

Step 2: Apply the condition: $P$ only $=$ Neither $+ 4$ .

$44 - b = (b - 12) + 4 = b - 8$

$44 + 8 = 2 b ⟹ b = 26$

Step 3: Compute regions.

$P$ only $= 44 - 26 = 18$
$Q$ only $= 38 - 26 = 12$
Neither $= 26 - 12 = 14$
Union $= 56$

Step 4: Exactly one $= P$ only $+ Q$ only $= 18 + 12 = 30$ .

Step 5: Verify. $18 + 12 + 26 + 14 = 70$ . $✓$ $P$ only $= 18 = 14 + 4$ . $✓$

Answer: 30

Question 9 · Discrete Mathematics MSQ

Let $A = {1, 2, 3, 4}$ and $B = {1, 2, 3, 4, 5, 6}$ . How many injective functions $f : A \to B$ satisfy $f (1) < f (2)$ and $f (3) < f (4)$ ?

A.
60
B.
75
C.
90
D.
120

Correct Answer:

Step-by-Step Solution

Key idea: This is a "counting injections with ordering constraints" problem. The constraints $f (1) < f (2)$ and $f (3) < f (4)$ remove the freedom to permute within pairs. We count by first choosing the image set, then counting valid arrangements.

Step 1: Choose the image set.

Since $f$ is injective, we choose 4 distinct elements from $B$ . Let the chosen set be ${b_{1}, b_{2}, b_{3}, b_{4}}$ with $b_{1} < b_{2} < b_{3} < b_{4}$ .

Number of ways: $(4 6) = 15$ .

Step 2: Count valid assignments for each image set.

We must assign ${b_{1}, b_{2}, b_{3}, b_{4}}$ to $f (1), f (2), f (3), f (4)$ such that $f (1) < f (2)$ and $f (3) < f (4)$ .

Choose which 2 of the 4 values go to positions ${1, 2}$ : $(2 4) = 6$ ways.
The remaining 2 values go to positions ${3, 4}$ .
Within each pair, the ordering constraint forces the assignment: the smaller value goes to the first position, the larger to the second.

So there are exactly 6 valid assignments per image set.

Step 3: Multiply.

Total = $15 \times 6 = 90$ .

Answer: 90

Question 10 · Programming · 2020 SUB

For any string str, length(str) returns the length of the string, append(str1,str2) concatenates str1 with another string str2, and trim(str) removes any spaces that exist at the end of the string str. The function reverse(str, i, j) reverses the part of the string from position i to position j. Assume that position 0 refers to the first character in the string. What does the following pseudo-code do?
def manipulate(string str)
{
    reverse(str, 0, length(str)-1);
    append(str, ‘ ’);
    n = length(str);
    j = 0;

    for (i = 0; i < n; i=i+1) {
        if (str[i] is ‘ ’) {
            reverse(str, j, i-1);
            j = i + 1;
        }
    }

    trim(str);
    return str;
}

Correct Answer:

0.00

Step-by-Step Solution

Insight: This is the classic three-step word reversal algorithm; a global reverse followed by localized word reverses restores word spelling but keeps word order reversed.

Exam route:

Step 1: reverse(str, 0, length(str)-1) reverses the entire string. For example, "hello world" becomes "dlrow olleh".

Step 2: The code appends a space to ensure the last word is processed. It then scans the string for spaces. Whenever it finds a space at index i, it reverses the segment from j to i-1, which corresponds to a single word.

Step 3: Reversing the whole string reverses the order of the words and also reverses the characters within each word. The loop then reverses each word individually, restoring the characters within each word to their original order, while keeping the words in the reversed sequence.

Step 4: trim(str) removes the extra space added at the beginning.

Step 5: The algorithm reverses the sequence of words in the string, leaving the characters inside each word unchanged.

Defect: The original question asks for a descriptive answer, but the platform requires a NAT numeric answer. The placeholder "0.00" is used.

Learning route:

The algorithm leverages the property that reversing a sequence twice restores it. By reversing the entire string, we achieve the desired word order reversal but corrupt the internal spelling of each word. The subsequent loop identifies word boundaries (spaces) and reverses each word segment individually, fixing the spelling without altering the newly established word order.

Question 11 · 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. What do the following function calls return? Briefly explain your answer.
(a) SecondBest([2,2,3], 3)
(b) SecondBest([3,2,2], 3)

Correct Answer:

none

Step-by-Step Solution

Insight: This is an algorithm tracing question, recognizable because it asks for the return value of specific function calls. The method applies here because we must simulate the code execution step-by-step to see if the flawed logic coincidentally produces the correct output.

Exam route: For (a) $A = [2, 2, 3]$ : first=2, second=2. $i = 2 \to A [2] = 3$ . Condition (3 >= 2) and (3 >= 2) is True. second becomes 2, first becomes 3. Returns 2. For (b) $A = [3, 2, 2]$ : first=3, second=2. $i = 2 \to A [2] = 2$ . Condition (2 >= 3) is False. Else: 2 >= 2 is True, second becomes 2. Returns 2.

Learning route:

Step 1: This is an algorithm tracing question. We apply faithful execution, simulating the code exactly as written.

Step 2: Trace (a) SecondBest([2, 2, 3], 3).

Initialization: first = 2, second = 2.
Loop $i = 2$ : $A [2] = 3$ .
Check (3 >= 2) and (3 >= 2) $\to$ True.
Update: second = first (so second = 2), first = 3.
End loop. first = 3, second = 2.
Final check: 3 != 2 is True. Returns 2.

Step 3: Trace (b) SecondBest([3, 2, 2], 3).

Initialization: first = 3, second = 2.
Loop $i = 2$ : $A [2] = 2$ .
Check (2 >= 3) and (2 >= 2) $\to$ False.
Else branch: 2 >= 2 $\to$ True. second = 2.
End loop. first = 3, second = 2.
Final check: 3 != 2 is True. Returns 2.

Wrong path: A student might assume the function returns None for (a) because of duplicate initialization, or 3 for (b) by confusing first and second. Both are incorrect because the specific values trigger different branches that coincidentally yield the correct second largest value (2) in both cases.

Generalization: Flawed code can still produce correct outputs for specific inputs. Tracing must be mechanical, not assumed.

Question 12 · Programming MSQ

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

```

function f(n):

if n == 0: return 1

if n == 1: return 2

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

```

Assertion (A): $f (10)$ is an even number.

Reason (R): $f (n)$ is even for all even $n \geq 0$ .

Which of the following is correct?

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: This is an assertion-reason question about a Fibonacci-like sequence. The trap is misreading the base cases, which changes the entire sequence.

Step 1: Trace $f (n)$ to find $f (10)$ :

$f (0) = 1$
$f (1) = 2$
$f (2) = f (1) + f (0) = 2 + 1 = 3$
$f (3) = f (2) + f (1) = 3 + 2 = 5$
$f (4) = f (3) + f (2) = 5 + 3 = 8$
$f (5) = f (4) + f (3) = 8 + 5 = 13$
$f (6) = f (5) + f (4) = 13 + 8 = 21$
$f (7) = f (6) + f (5) = 21 + 13 = 34$
$f (8) = f (7) + f (6) = 34 + 21 = 55$
$f (9) = f (8) + f (7) = 55 + 34 = 89$
$f (10) = f (9) + f (8) = 89 + 55 = 144$

Step 2: Evaluate Assertion A:

$f (10) = 144$ , which is even.
Therefore, A is true.

Step 3: Evaluate Reason R:

R claims $f (n)$ is even for all even $n \geq 0$ .
Check $n = 0$ : $f (0) = 1$ , which is odd.
Therefore, R is false.

Step 4: Conclusion:

A is true, but R is false.

Answer: C

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Mock Test 1 for CMI Data Science: 40 Questions with Solutions & Analysis

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Step-by-Step Solution

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