CMI Data Science Chapter-wise PYQs: 150 Previous Year Questions with Solutions (2019 to 2026)
150 CMI Data Science previous year questions from 8 papers (2019 to 2026), sorted chapter-wise across 79 chapters with answers and step-by-step solutions.
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CMI Data Science Chapter-wise PYQs: 150 Previous Year Questions with Solutions (2019 to 2026)
150 CMI Data Science previous year questions from 8 papers (2019 to 2026), sorted chapter-wise across 79 chapters with answers and step-by-step solutions.
CMI Data Science Past Papers Archive by Year and Slot
CMI Data Science Previous Year Questions with Solutions
School Level Mathematics: Solved PYQs
Q1.(CAT 2023) Find the limit
\[
\lim_{x\to\infty}\left(x-x\cos\frac{1}{\sqrt{x}}\right)
\]
Answer: none
Solution: Insight: This is an $\infty - \infty$ indeterminate form at infinity; factoring $x$ and substituting $t = 1/\sqrt{x}$ converts it to the standard limit $(1-\cos t)/t^2 = 1/2$. Exam route: 1. Factor out $x$: $x(1 - \cos(1/\sqrt{x}))$. 2. Substitute $t = 1/\sqrt{x}$, so $t \to 0^+$ and $x = 1/t^2$. 3. The expression becomes $(1 - \cos t)/t^2$. 4. Apply the standard limit $\lim_{t\to 0} (1-\cos t)/t^2 = 1/2$. Learning route: This is a limits-at-infinity question with an $\infty - \infty$ indeterminate form, recognisable because both $x$ and $x\cos(1/\sqrt{x})$ grow without bound as $x \to \infty$. The trigger for the substitution method is the $1/\sqrt{x}$ inside the cosine, which shrinks to zero as $x$ grows. Step 1: Never split $\infty - \infty$ into two separate limits. Instead, factor out $x$: $$x - x\cos\frac{1}{\sqrt{x}} = x\left(1 - \cos\frac{1}{\sqrt{x}}\right)$$ This converts the form from $\infty - \infty$ to $\infty \cdot 0$, which is still indeterminate but easier to handle. Step 2: Substitute $t = 1/\sqrt{x}$. As $x \to \infty$, $t \to 0^+$, and $x = 1/t^2$: $$x\left(1 - \cos\frac{1}{\sqrt{x}}\right) = \frac{1}{t^2}(1 - \cos t) = \frac{1 - \cos t}{t^2}$$ Step 3: Apply the standard limit. Using $1 - \cos t = 2\sin^2(t/2)$ and $\lim_{u\to 0}\frac{\sin u}{u} = 1$: $$\lim_{t\to 0}\frac{1-\cos t}{t^2} = \lim_{t\to 0}\frac{2\sin^2(t/2)}{t^2} = \lim_{t\to 0}\frac{2\cdot(t/2)^2}{t^2} = \frac{1}{2}$$ Wrong path: A tempting mistake is to split the limit: $\lim_{x\to\infty} x - \lim_{x\to\infty} x\cos(1/\sqrt{x})$. This yields $\infty - \infty$, which is invalid because you can only split limits if both individual limits exist and are finite.
Q2.(CAT 2020) Let \(f(x)\) be a real-valued function all of whose derivatives exist. Recall that a point \(x_0\) in the domain is called an <b>inflection point</b> of \(f(x)\) if the second derivative \(f''(x)\) changes sign at \(x_0\). Given the function
\[
f(x)=\frac{x^5}{20}-\frac{x^4}{2}+3x+1,
\]
which of the following statements are true?
\(x_0=0\) is not an inflection point.
\(x_0=6\) is the only inflection point.
\(x_0=0\) and \(x_0=6\), both are inflection points.
The function does not have an inflection point.
Answer: ["A","B"]
Solution: Key idea: This is a "statement_truth" question requiring the systematic identification and verification of inflection points by checking the sign change of the second derivative. Step 1: Find the first derivative. $f'(x) = \frac{d}{dx}\left(\frac{x^5}{20} - \frac{x^4}{2} + 3x + 1\right) = \frac{x^4}{4} - 2x^3 + 3$. Step 2: Find the second derivative. $f''(x) = \frac{d}{dx}\left(\frac{x^4}{4} - 2x^3 + 3\right) = x^3 - 6x^2$. Step 3: Find candidate points where $f''(x) = 0$. $x^3 - 6x^2 = x^2(x - 6) = 0 \implies x = 0$ or $x = 6$. Step 4: Verify sign change at $x = 0$. For $x < 0$, $x^2 > 0$ and $x - 6 < 0$, so $f''(x) < 0$. For $0 < x < 6$, $x^2 > 0$ and $x - 6 < 0$, so $f''(x) < 0$. Since the sign does not change, $x = 0$ is NOT an inflection point. Step 5: Verify sign change at $x = 6$. For $0 < x < 6$, $f''(x) < 0$. For $x > 6$, $x^2 > 0$ and $x - 6 > 0$, so $f''(x) > 0$. Since the sign changes from negative to positive, $x = 6$ IS an inflection point. Step 6: Evaluate options. Option A is true ($x_0=0$ is not an inflection point). Option B is true ($x_0=6$ is the only one). Options C and D are false. Answer: Options A and B.
Q3.(CAT 2021) For a non-zero real number \(a\), the inverse of \(J=\begin{pmatrix}a&1&0\\0&a&1\\0&0&a\end{pmatrix}\) is
Solution: Key idea: This is a matrix inverse problem for a special upper triangular matrix, recognizable because it has a constant diagonal $a$ and constant superdiagonal $1$. Step 1: Decompose the matrix as $J = aI + N$, where $N = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$. Step 2: Observe that $N$ is nilpotent. Specifically, $N^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ and $N^3 = 0$. Step 3: Use the finite geometric series expansion for the inverse: $(aI + N)^{-1} = a^{-1}(I + a^{-1}N)^{-1} = a^{-1}(I - a^{-1}N + a^{-2}N^2)$. Step 4: Substitute $I$, $N$, and $N^2$ into the expansion: $J^{-1} = \begin{pmatrix} a^{-1} & 0 & 0 \\ 0 & a^{-1} & 0 \\ 0 & 0 & a^{-1} \end{pmatrix} - \begin{pmatrix} 0 & a^{-2} & 0 \\ 0 & 0 & a^{-2} \\ 0 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & a^{-3} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} a^{-1} & -a^{-2} & a^{-3} \\ 0 & a^{-1} & -a^{-2} \\ 0 & 0 & a^{-1} \end{pmatrix}$. Answer: B
Q4.(CAT 2024) Starting with the number \(n=1\), we generate a sequence of numbers. In the second step we replace 1 by either \(2=2\times 1\) or \(3=2\times 1+1\). In general, replace \(n\) by either \(2n\) or \(2n+1\) to get a new value for \(n\). Which of the following numbers can be obtained as values of \(n\) in this fashion?
7
10
15
2026
Answer: ["A","C"]
Solution: Key idea: The operations $n \to 2n$ and $n \to 2n+1$ correspond to appending a binary digit ($0$ or $1$) to the right of the binary representation of $n$.
Since we start with $n=1$ (which is $(1)_2$), any number generated by this process will have a binary representation that starts with $1$ and consists only of the digits appended during the steps. Crucially, since every positive integer has a unique binary representation starting with $1$, **every positive integer** can be generated by this process.
Let's verify this for each option by converting to binary and checking if it can be reached from $1$ by appending bits.
Option D: 2026 Any positive integer $N$ can be written in binary. The process of generating $N$ from $1$ corresponds exactly to reading the binary digits of $N$ from left to right (excluding the leading 1 which is our start state) and applying $2n$ for '0' and $2n+1$ for '1'. Since 2026 is a positive integer, it has a binary representation. $2026 = 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 = (11111101010)_2$. It starts with 1. We can reach it by following the bits after the first one. So, 2026 is obtainable.
Wait, let me re-read the question carefully. "Which of the following numbers can be obtained...?" Usually, in such MSQ questions, if all are correct, all should be selected. Let's double check if there's a constraint I missed. "Starting with n=1... replace n by either 2n or 2n+1". This generates the set of all positive integers. Proof: By strong induction. Base case: 1 is in the set. Assume all integers $< k$ are in the set. If $k$ is even, $k=2m$, then $m < k$, so $m$ is in the set, and we can get $k$ from $m$ by $2m$. If $k$ is odd, $k=2m+1$, then $m < k$, so $m$ is in the set, and we can get $k$ from $m$ by $2m+1$. Thus all positive integers are reachable.
Therefore, 7, 10, 15, and 2026 are all obtainable.
Answer: A, B, C, D
Q5.(CAT 2024) Which of the following statements is/are true for real numbers \(x,y\)?
If \(x^2=y^2\) then \(x=y\).
If \(x^3=y^3\) then \(x=y\).
If \(x<y\) then \(x^2<y^2\).
If \(x<y\) then \(x^3<y^3\).
Answer: ["B","D"]
Solution: Key idea: This is a universal statement evaluation question involving real number properties. We must test each implication for all real numbers $x, y$, looking for counterexamples to disprove false statements. Step 1: Analyze Option A: If $x^2 = y^2$ then $x = y$. Counterexample: Let $x = 1$ and $y = -1$. Then $1^2 = (-1)^2 = 1$, but $1 \neq -1$. Thus, $x = \pm y$. Statement A is False. Step 2: Analyze Option B: If $x^3 = y^3$ then $x = y$. The function $f(t) = t^3$ is strictly increasing for all real $t$. Therefore, it is one-to-one (injective). If $x^3 = y^3$, taking the cube root of both sides yields $x = y$. Statement B is True. Step 3: Analyze Option C: If $x < y$ then $x^2 < y^2$. Counterexample: Let $x = -2$ and $y = 1$. Then $-2 < 1$, but $(-2)^2 = 4$ and $1^2 = 1$. Here $4 > 1$, so $x^2 > y^2$. Statement C is False. Step 4: Analyze Option D: If $x < y$ then $x^3 < y^3$. Since $f(t) = t^3$ is strictly increasing on $\mathbb{R}$, $x < y$ implies $f(x) < f(y)$, i.e., $x^3 < y^3$. Statement D is True. Answer: Options B and D are true.
Discrete Mathematics: Solved PYQs
Q1.(CAT 2020) How many squares are there on a \(7\times 7\) chessboard?
49
204
203
140
Answer: ["D"]
Solution: Key idea: this is a grid enumeration question asking for the total number of squares of all sizes in an $n \times n$ grid. The trigger is "how many squares", which implies counting $1 \times 1$, $2 \times 2$, up to $n \times n$ squares, not just the unit cells.
Step 1: A $k \times k$ square on a $7 \times 7$ board is uniquely determined by the position of its top-left corner. Step 2: The top-left corner can be placed in $(7 - k + 1) = (8 - k)$ horizontal positions and $(8 - k)$ vertical positions. Thus, there are $(8 - k)^2$ squares of size $k \times k$. Step 3: Sum over all possible sizes $k$ from 1 to 7: Total squares $= \sum_{k=1}^{7} (8-k)^2 = 7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2$. Step 4: Calculate the sum: $49 + 36 + 25 + 16 + 9 + 4 + 1 = 140$. (This matches the standard formula $\frac{n(n+1)(2n+1)}{6}$ for $n=7$, which gives $\frac{7 \times 8 \times 15}{6} = 140$). Step 5: Match with options. Option A (49) counts only the $1 \times 1$ cells. Option B (204) is the sum for an $8 \times 8$ board. Option D (140) is the correct total for a $7 \times 7$ board.
Answer: ["D"]
Q2.(CAT 2022) A relation \(R\) on the set \(A=\{a,b,c,d\}\) is defined by reading the columns of the following table from top to bottom. If a column in the table reads \((x,y,1)\) it means \(x\) is related to \(y\) in \(R\). If a column in the table reads \((x,y,0)\) it means \(x\) is not related to \(y\).<br/><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 560 95" width="560" height="95"><rect x="30" y="10" width="500" height="70" fill="white" stroke="black"/><line x1="30" y1="35" x2="530" y2="35" stroke="black"/><line x1="30" y1="58" x2="530" y2="58" stroke="black"/><line x1="155" y1="10" x2="155" y2="80" stroke="black"/><line x1="280" y1="10" x2="280" y2="80" stroke="black"/><line x1="405" y1="10" x2="405" y2="80" stroke="black"/><g font-size="15" text-anchor="middle" font-family="serif"><text x="45" y="28">a</text><text x="75" y="28">b</text><text x="105" y="28">c</text><text x="135" y="28">d</text><text x="170" y="28">a</text><text x="200" y="28">b</text><text x="230" y="28">c</text><text x="260" y="28">d</text><text x="295" y="28">a</text><text x="325" y="28">b</text><text x="355" y="28">c</text><text x="385" y="28">d</text><text x="420" y="28">a</text><text x="450" y="28">b</text><text x="480" y="28">c</text><text x="510" y="28">d</text><text x="45" y="51">a</text><text x="75" y="51">a</text><text x="105" y="51">a</text><text x="135" y="51">a</text><text x="170" y="51">b</text><text x="200" y="51">b</text><text x="230" y="51">b</text><text x="260" y="51">b</text><text x="295" y="51">c</text><text x="325" y="51">c</text><text x="355" y="51">c</text><text x="385" y="51">c</text><text x="420" y="51">d</text><text x="450" y="51">d</text><text x="480" y="51">d</text><text x="510" y="51">d</text><text x="45" y="74">0</text><text x="75" y="74">0</text><text x="105" y="74">1</text><text x="135" y="74">0</text><text x="170" y="74">1</text><text x="200" y="74">0</text><text x="230" y="74">0</text><text x="260" y="74">0</text><text x="295" y="74">1</text><text x="325" y="74">1</text><text x="355" y="74">0</text><text x="385" y="74">1</text><text x="420" y="74">0</text><text x="450" y="74">0</text><text x="480" y="74">1</text><text x="510" y="74">0</text></g></svg><br/>For instance, from the fifth column we have, \((a,b)\in R\), and from the second we have \((b,a)\notin R\).<br/>Another relation \(S\) on the set \(A\) is defined as: for any \(x,y\in A\), the pair \((x,y)\) is in \(S\) if and only if there exists \(z\in A\) such that both \((x,z)\in R\) and \((z,y)\in R\) hold.<br/>Which of the following pairs are in \(S\)?
\((a,a)\)
\((b,b)\)
\((c,c)\)
\((d,d)\)
Answer: ["A","C"]
Solution: Key idea: This is a **Relation Composition via Matrix/Table** problem. Recognise it because relation $S$ is defined as "exists $z$ such that $(x,z) \in R$ and $(z,y) \in R$", which is precisely the definition of $R^2$ or $R \circ R$. Step 1: Decode the table into relation $R$. The table columns represent pairs $(row\_label, col\_label, value)$. Reading the bottom row (values) against the top two rows (labels): Col 1: $(a,a,0)$ Col 2: $(b,a,0)$ Col 3: $(c,a,1) \implies (c,a) \in R$ Col 4: $(d,a,0)$ Col 5: $(a,b,1) \implies (a,b) \in R$ Col 6: $(b,b,0)$ Col 7: $(c,b,0)$ Col 8: $(d,b,0)$ Col 9: $(a,c,1) \implies (a,c) \in R$ Col 10: $(b,c,1) \implies (b,c) \in R$ Col 11: $(c,c,0)$ Col 12: $(d,c,1) \implies (d,c) \in R$ Col 13: $(a,d,0)$ Col 14: $(b,d,0)$ Col 15: $(c,d,0)$ Col 16: $(d,d,0)$ So $R = \{(c,a), (a,b), (a,c), (b,c), (d,c)\}$.
Step 2: Compute $S = R \circ R$. We need pairs $(x,y)$ where a path $x \to z \to y$ exists in $R$. Let's trace paths of length 2 from each starting node: - From $a$: $a \to b \to c$ (since $(a,b) \in R, (b,c) \in R$) $\implies (a,c) \in S$ $a \to c \to a$ (since $(a,c) \in R, (c,a) \in R$) $\implies (a,a) \in S$ - From $b$: $b \to c \to a$ (since $(b,c) \in R, (c,a) \in R$) $\implies (b,a) \in S$ - From $c$: $c \to a \to b$ (since $(c,a) \in R, (a,b) \in R$) $\implies (c,b) \in S$ $c \to a \to c$ (since $(c,a) \in R, (a,c) \in R$) $\implies (c,c) \in S$ - From $d$: $d \to c \to a$ (since $(d,c) \in R, (c,a) \in R$) $\implies (d,a) \in S$
So $S = \{(a,c), (a,a), (b,a), (c,b), (c,c), (d,a)\}$.
Q3.(CAT 2020) It is mid-semester exam week at CMI and first-year students from both M.Sc. Data Science (DS) and M.Sc. Computer Science (CS) have their exams scheduled for Monday from 10 a.m. to 1 p.m. in Lecture Hall 1. The first row in Lecture Hall 1 has six seats. In how many different ways can three M.Sc. DS students - Anish, Binish and Finish - and three M.Sc. CS students - Ramesh, Suresh, and Ragesh - be seated in this row, in such a way that two students from the same course do not sit next to each other?
36
48
72
96
Answer: ["C"]
Solution: Key idea: This is an alternating arrangement problem, recognizable by the condition "two students from the same course do not sit next to each other". Step 1: We have 3 DS students and 3 CS students. To ensure no two students from the same course sit together, they must strictly alternate. Step 2: There are exactly two valid alternating patterns for 6 seats: Pattern 1: DS - CS - DS - CS - DS - CS Pattern 2: CS - DS - CS - DS - CS - DS Step 3: For Pattern 1, the 3 DS students can be arranged in their 3 seats in $3! = 6$ ways. The 3 CS students can be arranged in their 3 seats in $3! = 6$ ways. Total for Pattern 1 = $6 \times 6 = 36$. Step 4: Similarly, for Pattern 2, the arrangements = $3! \times 3! = 36$. Step 5: Total valid arrangements = $36 + 36 = 72$. Answer: 72
Q4.(CAT 2023) A perfect shuffle of a deck of cards divides the deck into two equal parts and then interleaves the cards from each half, starting with the first card of the first half.<br/>For instance, if we shuffle a deck of cards containing 10 cards arranged \([1,2,3,4,5,6,7,8,9,10]\), we first create two equal decks with cards \([1,2,3,4,5]\) and \([6,7,8,9,10]\) and then interleave them to get a new deck \([1,6,2,7,3,8,4,9,5,10]\).<br/>We start with the deck \([8,1,4,5,3,6,2,7]\) and keep shuffling. Which card(s) will never appear next to 5?
1
2
7
8
Answer: ["D"]
Solution: Key idea: This is a permutation cycle decomposition question, recognizable because it asks about the long-term behavior of a deterministic rearrangement (shuffle). Step 1: Understand the shuffle mapping. For an 8-card deck, the perfect shuffle maps positions as follows: $p \to 2p-1$ if $p \le 4$, and $p \to 2(p-4)$ if $p > 4$. Step 2: Find the cycles of positions. - Position 1 maps to 1. (Cycle: {1}) - Position 8 maps to 8. (Cycle: {8}) - Position 2 $\to$ 3 $\to$ 5 $\to$ 2. (Cycle: {2, 3, 5}) - Position 4 $\to$ 7 $\to$ 6 $\to$ 4. (Cycle: {4, 6, 7}) Step 3: Track the cards in these cycles. Initial deck: `[8, 1, 4, 5, 3, 6, 2, 7]` at positions 1 to 8. - Cycle {1} always contains card 8. - Cycle {8} always contains card 7. - Cycle {2, 3, 5} contains cards {1, 4, 3}. - Cycle {4, 6, 7} contains cards {5, 6, 2}. Step 4: Determine adjacencies for card 5. Card 5 is always in Cycle {4, 6, 7}. - If 5 is at pos 4, neighbors are pos 3 and 5 (both in Cycle {2, 3, 5}, cards {1, 3, 4}). - If 5 is at pos 6, neighbors are pos 5 (Cycle {2, 3, 5}) and pos 7 (Cycle {4, 6, 7}). - If 5 is at pos 7, neighbors are pos 6 (Cycle {4, 6, 7}) and pos 8 (Cycle {8}, card 7). Step 5: Check the options. Card 8 is fixed at position 1. For 8 to be next to 5, 5 would need to be at position 2. However, 5 only visits positions 4, 6, and 7. Thus, 8 can never be next to 5. Answer: 8
Q5.(CAT 2022) A ternary tree starts with a single root node at the top of the tree. Each node in the tree can have up to three nodes as its children. No node in the tree is the child of two different nodes. A node which has no children is called a leaf node.<br/>The children of a node are drawn below it, connected by edges. The level of a node \(v\) in the ternary tree is the number of edges in the (unique) path from the root node to \(v\). Thus, for instance, the root node is at level 0, and each child of the root node is at level 1.<br/>A complete ternary tree is a ternary tree in which (i) each non-leaf node has exactly three children, and (ii) all leaf nodes are at the same level. This latter level is called the height of the complete ternary tree. The complete ternary trees of heights 0, 1, and 2, respectively are shown in the figure below, where we use the symbol \(\otimes\) to denote a node.<br/><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 560 180" width="560" height="180"><g fill="none" stroke="black" stroke-width="1.5"><text x="70" y="65" font-size="20" text-anchor="middle">⊗</text><text x="205" y="35" font-size="20" text-anchor="middle">⊗</text><line x1="205" y1="45" x2="165" y2="95"/><line x1="205" y1="45" x2="205" y2="95"/><line x1="205" y1="45" x2="245" y2="95"/><text x="165" y="115" font-size="20" text-anchor="middle">⊗</text><text x="205" y="115" font-size="20" text-anchor="middle">⊗</text><text x="245" y="115" font-size="20" text-anchor="middle">⊗</text><text x="405" y="25" font-size="20" text-anchor="middle">⊗</text><line x1="405" y1="35" x2="325" y2="85"/><line x1="405" y1="35" x2="405" y2="85"/><line x1="405" y1="35" x2="485" y2="85"/><text x="325" y="105" font-size="20" text-anchor="middle">⊗</text><text x="405" y="105" font-size="20" text-anchor="middle">⊗</text><text x="485" y="105" font-size="20" text-anchor="middle">⊗</text><line x1="325" y1="110" x2="295" y2="145"/><line x1="325" y1="110" x2="325" y2="145"/><line x1="325" y1="110" x2="355" y2="145"/><line x1="405" y1="110" x2="375" y2="145"/><line x1="405" y1="110" x2="405" y2="145"/><line x1="405" y1="110" x2="435" y2="145"/><line x1="485" y1="110" x2="455" y2="145"/><line x1="485" y1="110" x2="485" y2="145"/><line x1="485" y1="110" x2="515" y2="145"/></g><g font-size="20" text-anchor="middle"><text x="295" y="165">⊗</text><text x="325" y="165">⊗</text><text x="355" y="165">⊗</text><text x="375" y="165">⊗</text><text x="405" y="165">⊗</text><text x="435" y="165">⊗</text><text x="455" y="165">⊗</text><text x="485" y="165">⊗</text><text x="515" y="165">⊗</text></g></svg><br/>What is the total number of nodes in a complete ternary tree of height 9?
\(2^{11}-1\)
\(\frac{2^{10}+2}{3}\)
\(\frac{3^{10}-1}{2}\)
\(\frac{3^{10}+1}{2}\)
Answer: ["C"]
Solution: Key idea: The total number of nodes in a complete $k$-ary tree of height $h$ is the sum of a geometric progression. Step 1: Identify the number of nodes at each level. In a complete ternary tree, level $k$ has exactly $3^k$ nodes. Step 2: The tree has height 9, meaning the levels range from $k=0$ (the root) to $k=9$. Step 3: The total number of nodes is the sum of nodes at all levels: $\sum_{k=0}^{9} 3^k$. Step 4: Apply the geometric series sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$, where $a=1$, $r=3$, and the number of terms is $10$ (from 0 to 9). Step 5: Calculate the sum: $\frac{1 \cdot (3^{10} - 1)}{3 - 1} = \frac{3^{10}-1}{2}$. Answer: Option C.
Probability Theory: Solved PYQs
Q1.(CAT 2021) Fifteen telephones are received at a service center. Of these, 5 are mobile, 6 are cordless, and 4 are wired. These 15 phones are randomly numbered from 1 to 15 to establish the order in which they are serviced. Which of the following statement(s) is/are correct?
The probability that among the first 3 serviced, the first and third are mobile and the second is not, is \(\frac{5\times 10\times 4}{15\times 14\times 13}\).
The probability that the first four serviced are all the wired phones, is \(\frac{1}{\binom{15}{4}}\).
The probability that after servicing ten of these phones, only one of the three types remain to be serviced, is \(\frac{\binom{6}{5}}{\binom{15}{5}}\).
The probability that two phones of each type are among the first six serviced, is \(\frac{\binom{5}{2}+\binom{6}{2}+\binom{4}{2}}{\binom{15}{6}}\).
Answer: ["A","B"]
Solution: Key idea: this is a *sampling without replacement* question with a *random ordering* of 15 items split into three types (5M, 6C, 4W). The probability of any ordered type-pattern is computed by multiplying the changing fractions, or equivalently by counting favourable permutations against total permutations.
**Option A.** "First and third are mobile, second is not." - P(1st is M) = 5/15. - Given that, 14 phones remain, 10 of which are not M. P(2nd is not M) = 10/14. - Given that, 13 phones remain, 4 of which are M. P(3rd is M) = 4/13. - Product = (5·10·4)/(15·14·13). Option A is **correct**.
**Option B.** "First four serviced are all wired." - P = (4/15)·(3/14)·(2/13)·(1/12) = 4!/(15·14·13·12) = 24/32760 = 1/1365. - Also 1/ C(15,4) = 1/1365 (choosing which 4 positions the 4 wired phones occupy among the first 4 is forced). Option B is **correct**.
**Option C.** "After servicing 10 phones, only one type remains." - The 5 unserviced phones must all be of one type. Only M has 5 phones, so the 5 unserviced must be exactly the 5 mobiles. - P = C(5,5)/C(15,5) = 1/3003. Option C gives C(6,5)/C(15,5) = 6/3003, which is wrong (6 cordless cannot fit into 5 slots). Option C is **wrong**.
**Option D.** "Two of each type among the first six." - Favourable count = C(5,2)·C(6,2)·C(4,2) (choose 2 M, 2 C, 2 W independently), divided by C(15,6). - Option D writes a sum in the numerator instead of a product. Option D is **wrong**.
Answer: A, B.
Q2.(CAT 2021) Common Description:
<b>Description for following two questions:</b> In the 2019-2020 season of the English Premier League (EPL), 380 matches were played in a home and away format. The figure below describes the number of goals scored by the home team and the away team against the number of matches played. For example, the home team scored one goal in 125 matches.<br/><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 760 300" width="760" height="300"><g fill="none" stroke="black" stroke-width="1"><line x1="60" y1="240" x2="310" y2="240"/><line x1="60" y1="240" x2="60" y2="35"/><line x1="450" y1="240" x2="700" y2="240"/><line x1="450" y1="240" x2="450" y2="35"/></g><g fill="white" stroke="black"><rect x="85" y="139" width="28" height="101"/><rect x="125" y="91" width="28" height="149"/><rect x="165" y="122" width="28" height="118"/><rect x="205" y="184" width="28" height="56"/><rect x="245" y="221" width="28" height="19"/><rect x="285" y="230" width="20" height="10"/><rect x="475" y="93" width="28" height="147"/><rect x="515" y="87" width="28" height="153"/><rect x="555" y="142" width="28" height="98"/><rect x="595" y="201" width="28" height="39"/><rect x="635" y="230" width="20" height="10"/><rect x="665" y="235" width="16" height="5"/><rect x="690" y="238" width="12" height="2"/></g><g font-size="11" font-family="serif" text-anchor="middle"><text x="99" y="135">84</text><text x="139" y="87">125</text><text x="179" y="118">99</text><text x="219" y="180">47</text><text x="259" y="217">16</text><text x="295" y="226">8</text><text x="489" y="89">123</text><text x="529" y="83">128</text><text x="569" y="138">82</text><text x="609" y="197">33</text><text x="645" y="226">8</text><text x="673" y="231">4</text><text x="696" y="234">1</text><text x="185" y="285">(a)</text><text x="575" y="285">(b)</text><text x="185" y="270">Number of goals scored by the home team</text><text x="575" y="270">Number of goals scored by the away team</text></g></svg>
Which of the following statement(s) is/are correct?
In more than 50% matches, the home team scored at most one goal.
In more than 10% matches, the away team scored more than two goals.
In more than 15% matches, the home team scored three or more goals.
In more than 90% matches, the away team scored less than three goals.
Answer: ["A","B","C"]
Solution: Insight: This is an empirical probability question, recognizable because it provides frequency data from a bar chart and asks you to verify statements about proportions and cumulative percentages. Exam route: Extract frequencies from the chart for the specific conditions in each option, compute the cumulative sums, and divide by the total matches $N = 380$ to check the claimed percentages. Learning route: Step 1: Extract the data. The problem states $N = 380$ matches. Home team goals: 0:84, 1:125, 2:99, 3:47, 4:16, 5:8. (Sum = 379, we use $N=380$ as the denominator per the problem statement). Away team goals: 0:123, 1:128, 2:82, 3:33, 4:8, 5:4, 6:1. (Sum = 379).
Step 2: Verify Statement A. "Home team scored at most one goal" means 0 or 1 goal. Count = $84 + 125 = 209$. Percentage = $\frac{209}{380} \times 100 \approx 55.0\%$. Since $55.0\% > 50\%$, Statement A is TRUE.
Step 3: Verify Statement B. "Away team scored more than two goals" means 3, 4, 5, or 6 goals. Count = $33 + 8 + 4 + 1 = 46$. Percentage = $\frac{46}{380} \times 100 \approx 12.1\%$. Since $12.1\% > 10\%$, Statement B is TRUE.
Step 4: Verify Statement C. "Home team scored three or more goals" means 3, 4, or 5 goals. Count = $47 + 16 + 8 = 71$. Percentage = $\frac{71}{380} \times 100 \approx 18.7\%$. Since $18.7\% > 15\%$, Statement C is TRUE.
Step 5: Verify Statement D. "Away team scored less than three goals" means 0, 1, or 2 goals. Count = $123 + 128 + 82 = 333$. Percentage = $\frac{333}{380} \times 100 \approx 87.6\%$. Since $87.6\%$ is NOT more than $90\%$, Statement D is FALSE.
Final Answer: A, B, and C are correct.
Q3.(CAT 2021) Common Description:
Description for following two questions: A Non-Banking Finance Corporation (NBFC) declares fixed annual rates of simple interest on their auto and housing loans each year. The rates of interest offered by the company differ from year to year depending on the variation in macro economic indicators like inflation, RBI’s repo rate etc. The annual rates of interest offered by the company for the Auto and Housing sectors over the years are shown in the figure.
<br/>
<svg xmlns="http://www.w3.org/2000/svg" width="520" height="300" viewBox="0 0 520 300">
<rect x="0" y="0" width="520" height="300" fill="white"/>
<line x1="80" y1="230" x2="430" y2="230" stroke="black"/>
<line x1="80" y1="50" x2="80" y2="230" stroke="black"/>
<text x="70" y="232" text-anchor="end" font-size="10">0</text>
<text x="70" y="172" text-anchor="end" font-size="10">5</text>
<text x="70" y="112" text-anchor="end" font-size="10">10</text>
<text x="70" y="52" text-anchor="end" font-size="10">15</text>
<rect x="115" y="98" width="13" height="132" fill="white" stroke="black"/>
<line x1="116" y1="227" x2="127" y2="216" stroke="black"/>
<line x1="116" y1="214" x2="127" y2="203" stroke="black"/>
<line x1="116" y1="201" x2="127" y2="190" stroke="black"/>
<line x1="116" y1="188" x2="127" y2="177" stroke="black"/>
<line x1="116" y1="175" x2="127" y2="164" stroke="black"/>
<line x1="116" y1="162" x2="127" y2="151" stroke="black"/>
<line x1="116" y1="149" x2="127" y2="138" stroke="black"/>
<line x1="116" y1="136" x2="127" y2="125" stroke="black"/>
<line x1="116" y1="123" x2="127" y2="112" stroke="black"/>
<line x1="116" y1="110" x2="127" y2="99" stroke="black"/>
<rect x="130" y="110" width="13" height="120" fill="white" stroke="black"/>
<text x="118" y="93" font-size="10">11</text>
<text x="130" y="105" font-size="10">10</text>
<text x="129" y="250" text-anchor="middle" font-size="10">2010</text>
<rect x="160" y="74" width="13" height="156" fill="white" s