Mock Test 8 for CMI Data Science: 40 Questions with Solutions & Analysis

Attempt the Mock Test 8 for CMI Data Science: 40 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

40 questions · 84 marks · 143.36 minutes. School Level Mathematics: 17 · Probability Theory: 10 · Discrete Mathematics: 10 · Programming: 3

Free sample questions from Mock Test 8

Question 1 · School Level Mathematics · 2022 MSQ

Which of the following plots correspond to a bijective function?

\frac{1}{2}

01101101(i)(ii)(iii)(iv)

A.
(i) and (ii)
B.
(ii) and (iii)
C.
(i) and (iv)
D.
(iii) and (iv)

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: A function is bijective if and only if it is both one-one (injective) and onto (surjective). Graphically, this means it must pass both the Vertical Line Test (for being a function) and the Horizontal Line Test (for being one-one and onto the given codomain).

Step 1: Analyze Plot (i). The graph is a straight line segment with strictly decreasing y-values as x increases. It passes the Vertical Line Test (it is a function) and the Horizontal Line Test (it is one-one). Since it spans the full y-range from 0 to 1, it is onto. Thus, it is bijective.

Step 2: Analyze Plot (ii). The curve bends back on itself (x decreases then increases). A vertical line can intersect it at more than one point. It fails the Vertical Line Test, so it is not a function.

Step 3: Analyze Plot (iii). The graph is a piecewise linear curve that goes up and down. A horizontal line (e.g., y = 0.5) would intersect it at multiple points. It fails the Horizontal Line Test, so it is not one-one, hence not bijective.

Step 4: Analyze Plot (iv). The graph is a smooth curve where x strictly increases and y strictly decreases (in standard math coordinates). It passes both the Vertical Line Test and the Horizontal Line Test, and spans the full codomain. Thus, it is bijective.

Answer: C

Question 2 · School Level Mathematics SUB

Let $x$ and $y$ be positive real numbers satisfying $x + y = 9$ . What is the minimum possible value of

$\frac{1}{x} + \frac{4}{y} ?$

Correct Answer:

Step-by-Step Solution

Key idea: This is a constrained optimization by single-variable reduction question, recognisable because we have an expression in two variables tied together by a linear constraint. We eliminate one variable and minimise a single-variable function.

Step 1: Use the constraint to reduce to one variable.

From $x + y = 9$ : $y = 9 - x$ , where $0 < x < 9$ .

Define $g (x) = \frac{1}{x} + \frac{4}{9 - x}$ for $x \in (0, 9)$ .

Step 2: Find critical points.

$g^{'} (x) = - \frac{1}{x ^{2}} + \frac{4}{( 9 - x ) ^{2}}$ .

Set $g^{'} (x) = 0$ :

$\frac{4}{( 9 - x ) ^{2}} = \frac{1}{x ^{2}}$

$4 x^{2} = (9 - x)^{2}$

$2 x = 9 - x$ (taking positive square roots since $x > 0$ and $9 - x > 0$ )

$3 x = 9$

$x = 3$ .

Step 3: Verify it is a minimum.

$g^{''} (x) = \frac{2}{x ^{3}} + \frac{8}{( 9 - x ) ^{3}}$ .

For $x \in (0, 9)$ : both terms are positive, so $g^{''} (x) > 0$ .

Therefore $x = 3$ gives a local minimum. Since $g (x) \to \infty$ as $x \to 0^{+}$ or $x \to 9^{-}$ , this is the global minimum.

Step 4: Compute the minimum value.

$g (3) = \frac{1}{3} + \frac{4}{9 - 3} = \frac{1}{3} + \frac{4}{6} = \frac{1}{3} + \frac{2}{3} = 1$ .

Common trap: Students who take the square root of $4 x^{2} = (9 - x)^{2}$ and write $2 x = \pm (9 - x)$ might consider the negative branch $2 x = - (9 - x)$ , giving $x = - 9$ , which is outside the domain. Only the positive branch is valid.

Answer: 1

Question 3 · School Level Mathematics SUB

The sum of all integer solutions to the inequality

$3 x + 4 > x$

Correct Answer:

Step-by-Step Solution

Key idea: This is a radical inequality that requires careful casework based on the sign of the right side and domain restrictions.

Step 1: Determine the domain:

For $3 x + 4$ to be defined: $3 x + 4 \geq 0$ , so $x \geq - \frac{4}{3} \approx - 1.33$ .

Step 2: Case 1 — When $x < 0$ :

If $x < 0$ , then the right side is negative: $x < 0$ .

Since $3 x + 4 \geq 0$ always, we have $3 x + 4 \geq 0 > x$ .

So the inequality holds for all $x$ in the domain with $x < 0$ .

Domain restriction: $x \geq - \frac{4}{3}$ .

So $x \in [- \frac{4}{3}, 0)$ .

Integer solutions in this range: $x = - 1$ .

Step 3: Case 2 — When $x \geq 0$ :

Both sides are non-negative, so we can square both sides:

$3 x + 4 > x^{2}$

$x^{2} - 3 x - 4 < 0$

Step 4: Factor the quadratic:

$x^{2} - 3 x - 4 = (x - 4) (x + 1) < 0$

This holds when $- 1 < x < 4$ .

Step 5: Combine with $x \geq 0$ :

$x \in [0, 4)$ .

Integer solutions: $x = 0, 1, 2, 3$ .

Step 6: Verify each integer solution:

$x = - 1$ : $3 (- 1) + 4 = 1 = 1 > - 1$ . ✓
$x = 0$ : $4 = 2 > 0$ . ✓
$x = 1$ : $7 \approx 2.65 > 1$ . ✓
$x = 2$ : $10 \approx 3.16 > 2$ . ✓
$x = 3$ : $13 \approx 3.61 > 3$ . ✓
$x = 4$ : $16 = 4 > 4$ . ✗ (not strictly greater)

Step 7: Sum all integer solutions:

$(- 1) + 0 + 1 + 2 + 3 = 5$

Answer: The sum is 5.

Question 4 · Probability Theory · 2023 SUB

Two friends

A

and

B

are playing a coin flipping game with a fair coin, based on the following rules:
• When

A

flips the coin
– If it is heads then

A

wins and the game ends.
– If it is tails then

B

gets to flip the coin.
• When

B

flips the coin
– If it is heads then

B

wins and the game ends.
– If it is tails then

A

gets to flip the coin.
• The flipping continues till someone gets heads.
Suppose

A

is the first player to flip the coin. What is the probability that

A

will win the game.

Correct Answer:

none

Step-by-Step Solution

Insight: This is a classic infinite alternating-turn game. The key is recognizing that after one full round of failures (A fails, B fails), the game resets to the exact same state.

Exam route:

Let $p = P (A wins)$ .

A wins immediately on the first flip with probability $1/2$ .

If A flips tails (probability $1/2$ ), B gets a turn. At this point, the game state is identical to the start, but with B as the first player. So $P (B wins from here) = p$ , meaning $P (A wins from here) = 1 - p$ .

Setting up the recursion:

$p = \frac{1}{2} + \frac{1}{2} (1 - p)$

$p = \frac{1}{2} + \frac{1}{2} - \frac{p}{2}$

$p + \frac{p}{2} = 1 ⟹ \frac{3 p}{2} = 1 ⟹ p = \frac{2}{3}$ .

Alternative route (geometric series):

A wins on turn 1 (prob $1/2$ ), or turn 3 (prob $(1/2)^{3}$ ), or turn 5 (prob $(1/2)^{5}$ ), etc.

$p = \frac{1}{2} + \frac{1}{2 ^{3}} + \frac{1}{2 ^{5}} + \dots = \frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{2}{3}$ .

Learning route:

Both recursive and series formulations yield the same result. The recursive approach is faster under exam pressure. The key insight is that the game is memoryless: after any double-failure, the state is identical to the start.

Question 5 · Probability Theory · 2025 SUB

There are 18 chocolates in a bag, of which 7 are green, 6 are blue, and 5 are red. We pick chocolates one at a time from the bag without replacement.
(a) What is the probability that the first and the third chocolate are green?
(b) What is the probability that after picking twelve chocolates, only chocolates of one colour remain in the bag?

Correct Answer:

none

Step-by-Step Solution

Insight: For specific positions with unspecified middle draws, symmetry allows ignoring the middle; for "remaining" items, count the leftover subset instead of the drawn sequence.

Exam route: (a) $P (1st G) = 7/18$ . Given 1st G, 17 remain and 6 are G, so $P (3rd G ∣ 1st G) = 6/17$ . Product = $7/51$ . (b) Total ways to leave 6 is $(6 18) = 18564$ . Favourable monochromatic leftovers: $(6 7) + (6 6) = 8$ . Probability = $8/18564 = 2/4641$ .

Learning route: This is a without-replacement question with unspecified middle draws and a "remaining items" part. The symmetry method applies because the 2nd draw is unconstrained, so the 3rd draw is just a random position among the remaining items after the 1st. For part (b), counting the leftover subset is simpler than tracking 12 ordered draws.

Step 1: For part (a), the first chocolate must be green: $P (1st G) = 7/18$ .

Step 2: After one green is removed, 17 remain and 6 are green. The third draw is one random position among these 17, so $P (3rd G ∣ 1st G) = 6/17$ .

Step 3: The second draw is unrestricted, so it does not need to be split into cases. Multiply: $(7/18) \times (6/17) = 42/306 = 7/51$ .

Step 4: For part (b), after 12 draws, 6 remain. It is easier to count the remaining 6-subset than the ordered sequence of 12 draws.

Step 5: Total ways to choose the 6 remaining chocolates is $(6 18) = 18564$ .

Step 6: The remaining 6 must be all one colour. The only possibilities are 6 blue (from the 6 available) or 6 green (from the 7 available). Red cannot have 6 remaining since there are only 5.

Step 7: Favourable outcomes = $(6 6) + (6 7) = 1 + 7 = 8$ .

Step 8: Probability = $8/18564 = 2/4641$ .

Question 6 · Probability Theory · 2022 SUB

Common Description: Description for the next two questions
The probability density function of a normal distribution with mean

μ

and variance

σ^{2}

is of the form

f (x) = \frac{1}{σ 2 π} e^{- 0.5 (\frac{x - μ}{σ})^{2}}, - \infty < x < \infty.

Let

X

be a random variable with mean

E (X) = μ

and variance

V a r (X) = σ^{2}

. Let

X_{1}, X_{2}, \dots, X_{n}

n

independently sampled values of

X

, and let

\overline{X} = \frac{1}{n} \sum_{i} X_{i}

be the sample mean. The central limit theorem states that if the sample size

n

is large enough, then

\overline{X}

approximately follows the normal distribution with mean

μ

and variance

σ^{2} / n

. That is,

\overline{X} \approx N (μ, σ^{2} / n)

. This in turn implies that

Y = \frac{n ( X - μ )}{σ} \approx N (0, 1) .

Let

Z

be a random variable that follows the normal distribution with mean 0 and variance 1. For any real number

a

let

Φ (a) = P (Z < a)

be the probability that

Z

takes values smaller than

a

. Then

Φ (a) = P (Z < a) = \int_{- \infty}^{a} \frac{e ^{- u^{2} /2}}{2 π} d u .

For solving the next two problems you may assume the following approximations:

Φ (- 2) = 0.02, Φ (- 1) = 0.16, Φ (0) = 0.5, Φ (1) = 0.84, Φ (2) = 0.98

.
The weekly number of sales

S

at a certain car dealership is known to follow a probability distribution with mean

E (X) = 10

and variance

V a r (X) = 144

. A performance audit picks a random sample of 36 weekly sales figures

S_{1}, S_{2}, \dots, S_{36}

from the last two years. They find that the sample mean is 10 and the sample variance is 144. Use this information to answer the next two questions. You may assume that

n = 36

is a large enough sample size. Check if the following statement(s) are correct. Briefly explain your reasons.
(a) The probability that the average number of sales in a week will be more than 8 but less than 14 is

\int_{8}^{14} \frac{1}{12 2 π} e^{- 0.5 (\frac{u - 10}{12})^{2}} d u

(b) The probability that a salesperson would be able to sell, on an average, ten or more products in a week is 50%.

Correct Answer:

Step-by-Step Solution

Key idea: This is a conceptual question testing whether you can distinguish the distribution of a single observation from the distribution of the sample mean, recognisable because it asks you to verify probability statements about "the average."

Step 1: Recall the setup.

Population mean $μ = 10$ , variance $σ^{2} = 144$ , so $σ = 12$ .

Sample size $n = 36$ . By CLT, $\overset{ˉ}{S} \approx N (10, 144/36) = N (10, 4)$ .

The standard deviation of $\overset{ˉ}{S}$ is $σ_{\overset{ˉ}{S}} = 4 = 2$ .

Step 2: Evaluate statement (a).

Statement (a) gives the integral:

$\int_{8}^{14} \frac{1}{12 2 π} e^{- 0.5 (\frac{u - 10}{12})^{2}} d u .$

This is the density of a normal distribution with mean $10$ and standard deviation $12$ .

But $12$ is the standard deviation of a single observation $S$ , not of the sample mean $\overset{ˉ}{S}$ .

The correct density for $\overset{ˉ}{S}$ uses standard deviation $2$ , not $12$ .

Therefore statement (a) is incorrect.

Step 3: Evaluate statement (b).

Statement (b) says: the probability of selling, on average, ten or more products is 50%.

This means $P (\overset{ˉ}{S} \geq 10)$ .

Since $\overset{ˉ}{S} \sim N (10, 4)$ , the distribution is symmetric about its mean $10$ .

$P (\overset{ˉ}{S} \geq 10) = 0.5 = 50%.$

Therefore statement (b) is correct.

Only statement (b) is correct, so exactly 1 statement is correct.

Common trap: accepting statement (a) because the mean and bounds look plausible. Always check whether the standard deviation in the density matches the standard error $σ / n$ , not the population $σ$ .

Answer: 1 (only statement (b) is correct).

Question 7 · Discrete Mathematics · 2020 SUB

Let R be the set of all binary relations on the set ${1, 2, 3}$ . Suppose a relation is chosen from R at random. What is the probability that the chosen relation is symmetric?

Correct Answer:

0.125

Step-by-Step Solution

Key idea: This is a Counting Symmetric Relations problem. Recognise it by "probability that a random relation is symmetric" on a finite set.

Step 1: Total number of binary relations on set of size $n$ .

A relation on ${1, 2, 3}$ is a subset of $A \times A$ .

$∣ A \times A ∣ = 3^{2} = 9$ .

Total relations = $2^{9} = 512$ .

Step 2: Count symmetric relations.

A relation $R$ is symmetric iff $(a, b) \in R ⟹ (b, a) \in R$ .

This constrains off-diagonal pairs to be chosen TOGETHER.

Partition the 9 positions in $A \times A$ :

Diagonal: $(1, 1), (2, 2), (3, 3)$ — 3 positions. Each can be independently included or not. $2^{3}$ choices.
Off-diagonal pairs: ${(1, 2), (2, 1)}, {(1, 3), (3, 1)}, {(2, 3), (3, 2)}$ — 3 pairs. Each pair must be both in or both out. $2^{3}$ choices.

Total symmetric relations = $2^{3} \times 2^{3} = 2^{6} = 64$ .

Step 3: Compute probability.

$P = \frac{64}{512} = \frac{2 ^{6}}{2 ^{9}} = 2^{- 3} = \frac{1}{8} = 0.125$ .

General formula: For set of size $n$ , probability = $2^{- n (n - 1) /2}$ .

Here $n = 3$ : $2^{- 3 (2) /2} = 2^{- 3} = 1/8$ .

Answer: 0.125

Question 8 · Discrete Mathematics SUB

Consider the following transformation rules on binary strings:

Replace any occurrence of "11" with "0"
Replace any occurrence of "00" with "1"
Replace any "0" with "11"
Replace any "1" with "00"

Two strings are in the same equivalence class if one can be transformed into the other using these rules.

How many distinct equivalence classes contain at least one string of length exactly 4?

Enter your answer as a single integer.

Correct Answer:

Step-by-Step Solution

Key idea: This is a counting problem based on the modular invariant. We need to find how many distinct values of the invariant $I (mod 3)$ can be achieved by strings of length 4.

Step 1: Recall the invariant

From the previous analysis, the invariant is $I = n_{1} - n_{0} (mod 3)$ . This invariant completely determines the equivalence class.

Step 2: Express $I$ in terms of $n_{1}$ for length 4

For a string of length 4, $n_{1} + n_{0} = 4 ⟹ n_{0} = 4 - n_{1}$ .

Substitute into $I$ :

$I = n_{1} - (4 - n_{1}) = 2 n_{1} - 4$ .

Step 3: Check all possible values of $n_{1}$

Since the length is 4, $n_{1}$ can be $0, 1, 2, 3, 4$ .

$n_{1} = 0 ⟹ I = - 4 \equiv 2 (mod 3)$
$n_{1} = 1 ⟹ I = - 2 \equiv 1 (mod 3)$
$n_{1} = 2 ⟹ I = 0 \equiv 0 (mod 3)$
$n_{1} = 3 ⟹ I = 2 \equiv 2 (mod 3)$
$n_{1} = 4 ⟹ I = 4 \equiv 1 (mod 3)$

Step 4: Count distinct classes

The possible values of $I (mod 3)$ are ${0, 1, 2}$ . All three possible residue classes modulo 3 are represented.

Therefore, there are exactly 3 distinct equivalence classes that contain a string of length 4.

Answer: 3

Common trap: Students might try to list all 16 strings of length 4 and manually group them, which is time-consuming and error-prone. Using the invariant makes it a simple algebraic check.

Question 9 · Discrete Mathematics MSQ

You encounter two inhabitants, $A$ and $B$ , on an island of Knights (always tell the truth) and Knaves (always lie).

$A$ says to $B$ : "If I were to ask you whether you are a Knight, you would say 'yes'."
$B$ says to $A$ : "If I were to ask you whether you are a Knave, you would say 'yes'."

Which of the following is correct?

A.
$A$ is a Knight and $B$ is a Knight.
B.
$A$ is a Knave and $B$ is a Knave.
C.
$A$ is a Knight and $B$ is a Knave.
D.
$A$ is a Knave and $B$ is a Knight.

Correct Answer:

Step-by-Step Solution

Key idea: This is a meta-statement puzzle involving counterfactuals ("If I asked you..."). The key is to determine what a person would actually say when asked a direct question, and then translate that into a logical equivalence.

Step 1: Analyze what a person $X$ would say when asked a question $Q$ .

If $X$ is a Knight, $X$ tells the truth, so $X$ says 'yes' iff $Q$ is true.
If $X$ is a Knave, $X$ lies, so $X$ says 'yes' iff $Q$ is false.

In both cases, $X$ says 'yes' iff $X ⟺ Q$ .

Step 2: Translate $A$ 's statement.

$A$ says to $B$ : "If I asked you whether you are a Knight, you would say 'yes'."

Here, the listener is $B$ , and the question $Q$ is "Are you a Knight?" (which is equivalent to $B = 1$ ).

The statement "You would say 'yes'" is logically equivalent to $B ⟺ B$ .

Since $B ⟺ B$ is a tautology (always true), $A$ is making a TRUE statement.

Therefore, $A$ must be a Knight.

Step 3: Translate $B$ 's statement.

$B$ says to $A$ : "If I asked you whether you are a Knave, you would say 'yes'."

Here, the listener is $A$ , and the question $Q$ is "Are you a Knave?" (which is equivalent to $\neg A$ ).

The statement "You would say 'yes'" is logically equivalent to $A ⟺ \neg A$ .

Since $A ⟺ \neg A$ is a contradiction (always false), $B$ is making a FALSE statement.

Therefore, $B$ must be a Knave.

Step 4: Verify the assignment.

$A$ is Knight, $B$ is Knave.

$A$ 's claim: If asked "Are you a Knight?", $B$ (Knave) would lie and say 'yes'. $A$ says $B$ would say 'yes', which is true. Consistent.
$B$ 's claim: If asked "Are you a Knave?", $A$ (Knight) would tell the truth and say 'no'. $B$ says $A$ would say 'yes', which is false. Consistent.

Answer: $A$ is a Knight and $B$ is a Knave.

Question 10 · Programming · 2021 MSQ

Consider the following code, in which A is an array indexed from 0. function foo(A,n) { x = 0; for i = 0 to n-1 { x = A[i]^x; } return(x); } Here,

a

b

represents the bitwise Exclusive OR function over variables

a

and

b

.
Given integers

a

and

b

, we write them in binary, padded by zeros to the left to make them of equal length. We then apply Exclusive OR to these binary representations bitwise. The operation

a

b

denotes the integer value obtained by performing bitwise Exclusive OR on the binary values of

a

and

b

. For example,

3

^4 = 011^

100 = 111 = 7

, and

9

^5 = 1001^

0101 = 1100 = 12

. The truth table for the Exclusive OR function is provided below.

If

A = [2, 3, 7, 2, 3]

, what will foo(A,5) return?

A.
0
B.
2
C.
3
D.
7

Correct Answer:

["D"]

Step-by-Step Solution

Insight: The code computes the cumulative XOR sum of all array elements, where pairs of identical numbers cancel out to 0.

Exam route: Group identical elements using commutativity: $(2 \oplus 2) \oplus (3 \oplus 3) \oplus 7 = 0 \oplus 0 \oplus 7 = 7$ .

Learning route:

Step 1: Recall the properties of the XOR operation. It is commutative ( $a \oplus b = b \oplus a$ ) and associative ( $(a \oplus b) \oplus c = a \oplus (b \oplus c)$ ).

Step 2: Crucially, any number XORed with itself is 0 ( $x \oplus x = 0$ ), and any number XORed with 0 is itself ( $x \oplus 0 = x$ ).

Step 3: The array is $A = [2, 3, 7, 2, 3]$ . The function computes $2 \oplus 3 \oplus 7 \oplus 2 \oplus 3$ .

Step 4: Rearrange the terms using commutativity: $(2 \oplus 2) \oplus (3 \oplus 3) \oplus 7$ .

Step 5: Simplify using the cancellation property: $0 \oplus 0 \oplus 7 = 7$ .

Answer: The function returns 7, which corresponds to option D.

Question 11 · Programming MSQ

Let $S (n)$ denote the sum of the digits of a positive integer $n$ . Which of the following equations has NO solution for a 2-digit integer $n$ ?

A.
$n - S (n) = 18$
B.
$n - S (n) = 27$
C.
$n - S (n) = 35$
D.
$n - S (n) = 72$

Correct Answer:

Step-by-Step Solution

Key idea: Model the 2-digit number algebraically to find a simplified expression for $n - S (n)$ , revealing a strict divisibility constraint.

Step 1: Let the 2-digit number be $n = 10 a + b$ , where $a \in {1, \dots, 9}$ and $b \in {0, \dots, 9}$ .

Step 2: The sum of the digits is $S (n) = a + b$ .

Step 3: Substitute these into the expression $n - S (n)$ :

$n - S (n) = (10 a + b) - (a + b)$

Step 4: Simplify the expression:

$n - S (n) = 10 a + b - a - b = 9 a$ .

Step 5: Analyze the result. The difference between any 2-digit number and its digit sum is always exactly $9 a$ , which means it must be a multiple of 9.

Step 6: Check the options for multiples of 9:

18 is $9 \times 2$ (valid, e.g., $a = 2$ , any $b$ , such as $n = 20$ ).
27 is $9 \times 3$ (valid, e.g., $a = 3$ , any $b$ , such as $n = 30$ ).
35 is NOT a multiple of 9 ( $35/9 = 3.88 \dots$ ). Thus, no integer $a$ can satisfy $9 a = 35$ .
72 is $9 \times 8$ (valid, e.g., $a = 8$ , any $b$ , such as $n = 80$ ).

Answer: $n - S (n) = 35$

Question 12 · Programming MSQ

Let $n$ be a 2-digit positive integer. Define the following functions:

$R_{10} (n)$ : the integer formed by reversing the base-10 digits of $n$ .
$V_{2} (n)$ : the integer formed by interpreting the base-2 representation of $n$ as a base-10 number.
$S_{10} (n)$ : the sum of the base-10 digits of $n$ .
$S_{2} (n)$ : the sum of the base-2 digits of $n$ .
$D_{2} (n)$ : the number of base-2 digits of $n$ .

Which of the following equations has NO solution for any 2-digit integer $n$ ?

A.
$D_{2} (n) = S_{10} (n)$
B.
$S_{10} (n) = 2 \cdot D_{2} (n)$
C.
$R_{10} (n) = V_{2} (n) (mod 10)$
D.
$S_{2} (n) = S_{10} (n)$

Correct Answer:

Step-by-Step Solution

Key idea: Compare the magnitude bounds and parity constraints of the base-10 reversed integer and the base-2 visual integer to identify a strict impossibility.

Step 1: Analyze the domain. $n$ is a 2-digit integer, so $10 \leq n \leq 99$ .

Step 2: Evaluate Option A: $D_{2} (n) = S_{10} (n)$ .

For $n = 13$ , $S_{10} (13) = 1 + 3 = 4$ . The binary representation of 13 is $110 1_{2}$ , which has $D_{2} (13) = 4$ digits. This equation has a solution.

Step 3: Evaluate Option B: $S_{10} (n) = 2 \cdot D_{2} (n)$ .

For $n = 77$ , $S_{10} (77) = 7 + 7 = 14$ . The binary representation of 77 is $100110 1_{2}$ ( $64 + 8 + 4 + 1$ ), which has $D_{2} (77) = 7$ digits. Since $2 \cdot 7 = 14$ , this equation has a solution.

Step 4: Evaluate Option D: $S_{2} (n) = S_{10} (n)$ .

For $n = 20$ , $S_{10} (20) = 2 + 0 = 2$ . The binary representation of 20 is $1010 0_{2}$ ( $16 + 4$ ). The sum of its bits is $S_{2} (20) = 1 + 0 + 1 + 0 + 0 = 2$ . This equation has a solution.

Step 5: Evaluate Option C: $R_{10} (n) = V_{2} (n) (mod 10)$ .

The value $V_{2} (n) (mod 10)$ is simply the last digit of the base-10 interpretation of the binary string, which is exactly the least significant bit of $n$ (i.e., $n (mod 2)$ ). Thus, $V_{2} (n) (mod 10) \in {0, 1}$ .

For the equation to hold, we must have $R_{10} (n) \in {0, 1}$ .

$R_{10} (n) = 0$ is impossible for any 2-digit number.
$R_{10} (n) = 1$ requires $n$ to end in 0 (i.e., $n \in {10, 20, \dots, 90}$ ).

However, all such numbers are even, meaning their least significant bit is 0. Thus, for these $n$ , $V_{2} (n) (mod 10) = 0$ .

This creates a contradiction: we need $V_{2} (n) (mod 10) = 1$ to match $R_{10} (n) = 1$ , but it is always 0.

Therefore, this equation has no solution.

Answer: $R_{10} (n) = V_{2} (n) (mod 10)$

Section-wise Paper Structure

Free Solved Questions with Step-by-Step Solutions

Unlock All 40 Questions in Real Examination Mode

More CMI Data Science Mock Tests

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

Mock Test 8 for CMI Data Science: 40 Questions with Solutions & Analysis

Paper breakdown

Free sample questions from Mock Test 8

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 mock tests