CMI Data Science Mock Tests

**Step 1: Initialize variables.** When `calculateSpecialSum(1524)` is called, the function first checks if $n < 0$. Since $1524 \not< 0$, it proceeds to initialize the variables: **Step 2: Execute the while loop for each digit.** The `while` loop continues as long as $n > 0$. The digits are processed from right to left (least significant to most significant). * **Iteration 1:** * Current $n = 1524$ * $digit = 1524 \% 10 = 4$ * Since $4 \% 2 \text{==} 0$ (even), * * * **Iteration 2:** * Current $n = 152$ * $digit = 152 \% 10 = 2$ * Since $2 \% 2 \text{==} 0$ (even), * * * **Iteration 3:** * Current $n = 15$ * $digit = 15 \% 10 = 5$ * Since $5 \% 2 \ne 0$ (odd), * * * **Iteration 4:** * Current $n = 1$ * $digit = 1 \% 10 = 1$ * Since $1 \% 2 \ne 0$ (odd), * * **Step 3: Return the final sum.** The loop terminates because $n$ is now $0$. The function returns the final value of $\text{current\_sum}$. **Answer:** $\boxed{-1476}$

We analyze each statement based on the properties of matrix transpose and multiplication. **Step 1: For any square matrix $A$, the matrix $A + A^T$ is symmetric.** To check if a matrix $M$ is symmetric, we must verify if $M^T = M$. Let $M = A + A^T$. We compute its transpose: Using the property that the transpose of a sum is the sum of the transposes, $(X+Y)^T = X^T + Y^T$, we get: Using the property $(X^T)^T = X$, we have: Since $M^T = M$, the matrix $A + A^T$ is always symmetric. Thus, this statement is **true**. **Step 2: If $A$ is a skew-symmetric matrix, then $A^2$ is a symmetric matrix.** Given that $A$ is skew-symmetric, we have $A^T = -A$. We need to check if $(A^2)^T = A^2$. Let's compute the transpose of $A^2$: Using the property of the transpose of a product, $(XY)^T = Y^T X^T$, we get: Now, substitute the skew-symmetric property $A^T = -A$: Since $(A^2)^T = A^2$, the matrix $A^2$ is symmetric. Thus, this statement is **true**. **Step 3: If $A$ and $B$ are symmetric matrices, then their product $AB$ is also a symmetric matrix.** Given that $A$ and $B$ are symmetric, we have $A^T = A$ and $B^T = B$. We check if the product $AB$ is symmetric by computing its transpose: Using the symmetric property of $A$ and $B$, we substitute $B^T=B$ and $A^T=A$: For $AB$ to be symmetric, we would need $(AB)^T = AB$, which implies $BA = AB$. However, matrix multiplication is not commutative in general. We can provide a counterexample. Let $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 4 \\ 4 & 5 \end{pmatrix}$. Both are symmetric. Since $AB \neq (AB)^T$, the product $AB$ is not symmetric. Thus, this statement is **false**. **Step 4: For any two matrices $A$ and $B$, the identity $(A-B)(A+B) = A^2 - B^2$ holds.** Let's expand the left-hand side using the distributive property of matrix multiplication: For this to be equal to $A^2 - B^2$, we must have $AB - BA = O$, where $O$ is the zero matrix. This means $AB = BA$, i.e., the matrices $A$ and $B$ must commute. Since this is not true for all pairs of matrices, the identity does not hold in general. Thus, this statement is **false**. Answer: \boxed{Statements 1 and 2 are true, while Statements 3 and 4 are false.}

**Step 1: Identify the total number of individuals and the primary constraint.** We have 7 distinct individuals to be seated around a circular table. The primary constraint is that Carol and David (the married couple) must always sit together. The secondary constraint is that Alice and Bob (the siblings) must never sit next to each other. **Step 2: Treat Carol and David as a single unit.** Since Carol and David must always sit together, we can treat them as a single block. Let's denote this block as (CD). The remaining 5 individuals are Alice, Bob, and 3 other distinct individuals (let's call them X, Y, Z). So, we are effectively arranging 6 units around the circular table: (CD), A, B, X, Y, Z. The number of ways to arrange $n$ distinct units around a circular table is $(n-1)!$. Here, $n=6$ units. So, the number of circular arrangements for these 6 units is: Within the (CD) block, Carol and David can be arranged in $2!$ ways (CD or DC). Therefore, the total number of arrangements where Carol and David are together is: **Step 3: Calculate arrangements where Carol and David are together AND Alice and Bob are together.** Now, we need to consider the arrangements from Step 2 where Alice and Bob are also seated together. If Alice and Bob are together, we treat them as another single block, (AB). So, we have 5 units to arrange: (CD), (AB), X, Y, Z. The number of ways to arrange these 5 units around a circular table is: Within the (CD) block, Carol and David can be arranged in $2!$ ways. Within the (AB) block, Alice and Bob can be arranged in $2!$ ways. Therefore, the total number of arrangements where Carol and David are together AND Alice and Bob are together is: **Step 4: Subtract the unwanted arrangements to find the final count.** We want the number of arrangements where Carol and David are together, but Alice and Bob are *not* together. This can be found by subtracting the cases where both pairs are together from the cases where only Carol and David are together: Answer: \boxed{144}

The solution involves applying the principles of probability for unions and intersections of events, and checking for independence. **Step 1: Calculate pairwise intersection probabilities.** Given that the events are pairwise independent, we calculate the probabilities of their pairwise intersections: We are also given $P(A \cap B \cap C) = 0.1$. **Step 2: Evaluate the probability of the union $P(A \cup B \cup C)$.** Using the Principle of Inclusion-Exclusion for three events: Substituting the known values: Thus, the first option is **correct**. **Step 3: Evaluate the probability that exactly one event occurs.** The event that exactly one of $A, B, C$ occurs is given by the sum of probabilities of the mutually exclusive events $(A \cap B^c \cap C^c)$, $(A^c \cap B \cap C^c)$, and $(A^c \cap B^c \cap C)$. The probability can be calculated using the formula: Thus, the second option is **correct**. **Step 4: Evaluate the probability that at least two events occur.** The event that at least two of $A, B, C$ occur is the union of $(A \cap B)$, $(A \cap C)$, and $(B \cap C)$. The probability is: The option states the probability is $0.25$. Thus, the third option is **incorrect**. **Step 5: Check for independence of $A$ and $B \cup C$.** For $A$ and $B \cup C$ to be independent, we must have $P(A \cap (B \cup C)) = P(A)P(B \cup C)$. First, calculate $P(B \cup C)$: Then, Next, calculate $P(A \cap (B \cup C))$: Since $P(A \cap (B \cup C)) = 0.25 \neq 0.29 = P(A)P(B \cup C)$, the events are not independent. Thus, the fourth option is **incorrect**. Answer: \boxed{\text{Statements 1 and 2 are correct; Statements 3 and 4 are incorrect.}}

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