Q1.(CAT 2025) The number of patients per shift \((X)\) consulting Dr. Gita in her past \(100\) shifts is shown in the figure. If the amount she earns is ₹ \(1000(X - 0.2)\), what is the average amount (in ₹) she has earned per shift in the past \(100\) shifts?<br/><br/>
Note: The figure shown is representative.<br/><br/>
<svg width="420" height="300" viewBox="0 0 420 300" xmlns="http://www.w3.org/2000/svg">
<line x1="70" y1="240" x2="360" y2="240" stroke="black"/>
<line x1="70" y1="240" x2="70" y2="40" stroke="black"/>
<line x1="70" y1="200" x2="360" y2="200" stroke="#999"/>
<line x1="70" y1="160" x2="360" y2="160" stroke="#999"/>
<line x1="70" y1="120" x2="360" y2="120" stroke="#999"/>
<line x1="70" y1="80" x2="360" y2="80" stroke="#999"/>
<line x1="70" y1="40" x2="360" y2="40" stroke="#999"/>
<rect x="100" y="160" width="35" height="80" fill="white" stroke="black"/>
<rect x="170" y="80" width="35" height="160" fill="white" stroke="black"/>
<rect x="240" y="120" width="35" height="120" fill="white" stroke="black"/>
<rect x="310" y="200" width="35" height="40" fill="white" stroke="black"/>
<text x="117" y="155" text-anchor="middle" font-size="12">20</text>
<text x="187" y="75" text-anchor="middle" font-size="12">40</text>
<text x="257" y="115" text-anchor="middle" font-size="12">30</text>
<text x="327" y="195" text-anchor="middle" font-size="12">10</text>
<text x="62" y="244" text-anchor="end" font-size="12">0</text>
<text x="62" y="204" text-anchor="end" font-size="12">10</text>
<text x="62" y="164" text-anchor="end" font-size="12">20</text>
<text x="62" y="124" text-anchor="end" font-size="12">30</text>
<text x="62" y="84" text-anchor="end" font-size="12">40</text>
<text x="62" y="44" text-anchor="end" font-size="12">50</text>
<text x="117" y="260" text-anchor="middle" font-size="12">5</text>
<text x="187" y="260" text-anchor="middle" font-size="12">6</text>
<text x="257" y="260" text-anchor="middle" font-size="12">7</text>
<text x="327" y="260" text-anchor="middle" font-size="12">8</text>
<text x="215" y="285" text-anchor="middle" font-size="13">Number of patients per shift (X)</text>
<text x="20" y="145" text-anchor="middle" font-size="13" transform="rotate(-90 20 145)">Number of shifts</text>
</svg>
6,100
6,300
6,000
6,500
Answer: A
Solution: Key idea: This is a weighted average and linear transformation question, recognisable because it gives a frequency distribution (bar chart) and asks for the average of a linear function of the variable. Step 1: Calculate the total number of patients across all shifts. From the chart: (5 * 20) + (6 * 40) + (7 * 30) + (8 * 10) = 100 + 240 + 210 + 80 = 630. Step 2: Calculate the average number of patients per shift, \bar{X}. \bar{X} = 630 / 100 = 6.3. Step 3: Apply the linear transformation for earnings. The average earnings per shift is 1000(\bar{X} - 0.2). Step 4: Substitute \bar{X}: 1000(6.3 - 0.2) = 1000(6.1) = 6100. Answer: A
Q2.(CAT 2026) The number of bijections \(f(\cdot)\) from the set \(S = \{1, 2, 3, 4\}\) to itself such that \(f(f(n)) = n\), for all \(n \in S\), is __________ . (Answer in integer)
Answer: 10.00
Solution: Insight: The condition $f(f(n)) = n$ defines an involution, meaning the permutation consists entirely of 1-cycles (fixed points) and 2-cycles (swaps). Exam route: Use the involution recurrence $I(n) = I(n-1) + (n-1)I(n-2)$. With $I(1)=1$ and $I(2)=2$, we get $I(3) = 2 + 2(1) = 4$, and $I(4) = 4 + 3(2) = 10$. Learning route: We classify the bijections by their cycle structure for $n=4$: 1. Zero swaps (4 fixed points): There is exactly $\binom{4}{0} = 1$ way. 2. One swap (2 fixed points): Choose 2 elements to swap out of 4. This is $\binom{4}{2} = 6$ ways. 3. Two swaps (0 fixed points): Choose 2 elements for the first swap ($\binom{4}{2} = 6$), and the remaining 2 form the second swap ($\binom{2}{2} = 1$). Since the two swaps are indistinguishable, we divide by $2!$. This gives $\frac{6 \times 1}{2} = 3$ ways. Total involutions = $1 + 6 + 3 = 10$.
Common Trap: Forgetting to divide by $2!$ in the two-swap case leads to $6 \times 1 = 6$ ways, incorrectly totaling $1 + 6 + 6 = 13$. Verification: The recurrence $I(4) = I(3) + 3I(2) = 4 + 3(2) = 10$ perfectly matches the manual enumeration.
Solution: Key idea: This is a systematic counting question on a matrix, recognisable because it provides a grid of numerical values and asks for the count of cells satisfying a specific inequality condition. Step 1: Identify the condition. We need to count cells where the pixel intensity U \leq 4. Step 2: Examine each row systematically. - Row 1: 0, 1, 0, 2 (All 4 values are \leq 4) -> Count = 4 - Row 2: 4, 7, 3, 3 (4, 3, 3 are \leq 4) -> Count = 3 - Row 3: 5, 5, 4, 4 (4, 4 are \leq 4) -> Count = 2 - Row 4: 6, 7, 3, 2 (3, 2 are \leq 4) -> Count = 2 Step 3: Sum the counts. Total = 4 + 3 + 2 + 2 = 11. Answer: C
Q4.(CAT 2024) The sum of the following infinite series is<br/>\[2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \cdots\]
\(\frac{11}{3}\)
\(\frac{7}{2}\)
\(\frac{13}{4}\)
\(\frac{9}{2}\)
Answer: B
Solution: Insight: The series is a mix of a constant, a geometric series with ratio 1/2, and another with ratio 1/3. Exam route: Group the terms by their denominators' patterns. Sum the two infinite geometric series separately using $S = a/(1-r)$ and add to the initial constant. Learning route: The given series is $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots$ Observe the denominators after the first term: 2, 4, 8, 16... are powers of 2. 3, 9, 27... are powers of 3. We can split the series into three parts: $S = 2 + \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right) + \left( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \right)$ The first bracket is a geometric series with $a = 1/2$ and $r = 1/2$. Its sum is $\frac{1/2}{1 - 1/2} = 1$. The second bracket is a geometric series with $a = 1/3$ and $r = 1/3$. Its sum is $\frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = 1/2$. Total sum $S = 2 + 1 + 1/2 = 3.5 = 7/2$. Correct option is B.
Q5.(CAT 2026) The product of the digits of a three-digit number is 70. The sum of the digits of this three-digit number is _____
12
14
16
18
Answer: B
Solution: Insight: The product of three single digits is 70, so factorise 70 into three single-digit factors first; the sum follows immediately. Exam route: $70 = 2 \times 5 \times 7$. These are the only three single-digit factors (any other factorisation like $1 \times 7 \times 10$ uses a non-digit). Sum $= 2 + 5 + 7 = 14$. Learning route: Step 1: Prime factorise $70 = 2 \times 5 \times 7$. Step 2: We need three single digits $a, b, c \in \{1, 2, \dots, 9\}$ such that $a \times b \times c = 70$. Step 3: Since $70 = 2 \times 5 \times 7$, and all three are single digits, the only valid triple is $\{2, 5, 7\}$. Any attempt to introduce a 1 (e.g. $1 \times 5 \times 14$) forces a factor $\geq 10$, which is not a digit. Step 4: Sum $= 2 + 5 + 7 = 14$. Trap: Using $1 \times 7 \times 10$ gives sum 18, but 10 is not a single digit. Verification: $2 \times 5 \times 7 = 70$ ✓, and $2 + 5 + 7 = 14$ ✓.
Programming, Data Structures and Algorithms: Solved PYQs
Q1.(CAT 2026) Consider the given Python program.<br/>
<br/>
def fun(L, i=0):<br/>
if i >= len(L)-1:<br/>
return 0<br/>
if L[i] > L[i+1]:<br/>
L[i+1], L[i] = L[i], L[i+1]<br/>
return 1+fun(L, i+1)<br/>
else:<br/>
return fun(L, i+1)<br/>
<br/>
data = [5, 3, 4, 1, 2]<br/>
count = 0<br/>
for _ in range(len(data)):<br/>
count += fun(data)<br/>
print(count)<br/>
<br/>
The output of the program is __________ . (Answer in integer)
Answer: 8.00
Solution: Insight: The inner function `fun` performs one left-to-right pass of adjacent swaps, returning the number of swaps. The outer loop calls it `n` times, which is exactly the Bubble Sort algorithm. The total number of swaps in Bubble Sort equals the number of inversions in the initial array. Exam route: Count the inversions in `[5, 3, 4, 1, 2]`. 5 is greater than 3, 4, 1, 2 (4 inversions). 3 is greater than 1, 2 (2 inversions). 4 is greater than 1, 2 (2 inversions). Total = 4 + 2 + 2 = 8. Learning route: Pass 1: `[5, 3, 4, 1, 2]` -> 5 bubbles to the end. Swaps: (5,3), (5,4), (5,1), (5,2). Count = 4. Array: `[3, 4, 1, 2, 5]`. Pass 2: `[3, 4, 1, 2, 5]` -> 4 bubbles to index 3. Swaps: (4,1), (4,2). Count = 2. Array: `[3, 1, 2, 4, 5]`. Pass 3: `[3, 1, 2, 4, 5]` -> 3 bubbles to index 2. Swaps: (3,1), (3,2). Count = 2. Array: `[1, 2, 3, 4, 5]`. Pass 4 & 5: Already sorted, 0 swaps. Total count = 4 + 2 + 2 + 0 + 0 = 8.
Q2.(CAT 2024) Consider the following Python function:<br/>def fun(D, s1, s2):<br/> if s1 < s2:<br/> D[s1], D[s2] = D[s2], D[s1]<br/> fun(D, s1+1, s2-1)<br/>What does this Python function fun() do? Select the ONE appropriate option <br/>below.
It finds the smallest element in D from index s1 to s2, both inclusive.
It performs a merge sort in-place on this list D between indices s1 and s2, both inclusive.
It reverses the list D between indices s1 and s2, both inclusive.
It swaps the elements in D at indices s1 and s2, and leaves the remaining elements unchanged.
Answer: C
Solution: Insight: The function swaps the elements at `s1` and `s2`, then recursively calls itself with `s1+1` and `s2-1`. This is the standard two-pointer recursive reversal pattern. Exam route: Recognize the two pointers moving inwards (`s1+1`, `s2-1`) and swapping elements. This reverses the subarray from `s1` to `s2`. Learning route: Base case: `s1 < s2` is false (i.e., `s1 >= s2`), it stops. Recursive step: swaps `D[s1]` and `D[s2]`, then moves pointers inward. This exactly reverses the segment `D[s1...s2]` in place.
Q3.(CAT 2024) Match the items in Column 1 with the items in Column 2 in the following table:<br/><br/><table><tr><td>Column 1</td><td>Column 2</td></tr><tr><td>(p) First In First Out</td><td>(i) Stacks</td></tr><tr><td>(q) Lookup Operation</td><td>(ii) Queues</td></tr><tr><td>(r) Last In First Out</td><td>(iii) Hash Tables</td></tr></table>
(p) − (ii), (q) − (iii), (r) − (i)
(p) − (ii), (q) − (i), (r) − (iii)
(p) − (i), (q) − (ii), (r) − (iii)
(p) − (i), (q) − (iii), (r) − (ii)
Answer: A
Solution: Key idea: This is an ADT property matching question, recognizable because it asks to pair fundamental access patterns (FIFO, LIFO, Lookup) with their corresponding data structures. Step 1: Analyze "First In First Out" (p). This is the defining property of a Queue, where the first element added is the first to be removed. So, (p) matches with (ii). Step 2: Analyze "Last In First Out" (r). This is the defining property of a Stack, where the most recently added element is the first to be removed. So, (r) matches with (i). Step 3: Analyze "Lookup Operation" (q). Hash Tables are specifically designed to provide fast, average-case $O(1)$ key-based lookup operations. So, (q) matches with (iii). Step 4: Combine the matches: (p) - (ii), (q) - (iii), (r) - (i). Answer: Option A
Q4.(CAT 2024) Consider sorting the following array of integers in ascending order using an in-place <br/>Quicksort algorithm that uses the last element as the pivot.<br/>\[60 \quad 70 \quad 80 \quad 90 \quad 100\]<br/>The minimum number of swaps performed during this Quicksort is ______.
Answer: 15
Solution: Key idea: This is a Quicksort trace problem, recognizable by the request to count the exact number of swaps for a specific array and pivot choice. We must simulate the standard Lomuto partition scheme.
Step 1: Identify the partition scheme. The problem specifies an "in-place Quicksort algorithm that uses the last element as the pivot". This corresponds to the standard Lomuto partition scheme.
Step 2: Trace the first partition on the array $[60, 70, 80, 90, 100]$. - Pivot $x = 100$. - Initialize $i = p - 1 = -1$. - Loop $j$ from $0$ to $3$: - $j=0$: $A[0]=60 \le 100$. Increment $i$ to $0$. Swap $A[0]$ with $A[0]$. (1 swap) - $j=1$: $A[1]=70 \le 100$. Increment $i$ to $1$. Swap $A[1]$ with $A[1]$. (1 swap) - $j=2$: $A[2]=80 \le 100$. Increment $i$ to $2$. Swap $A[2]$ with $A[2]$. (1 swap) - $j=3$: $A[3]=90 \le 100$. Increment $i$ to $3$. Swap $A[3]$ with $A[3]$. (1 swap) - End of loop. Swap $A[i+1]$ with $A[r]$, which is Swap $A[4]$ with $A[4]$. (1 swap) Total swaps in this step = $5$.
Step 3: Analyze the recursive calls. The pivot $100$ is now in its correct final position. The left subarray is $[60, 70, 80, 90]$ (size 4), and the right subarray is empty.
Step 4: Repeat the process for the remaining subarrays. - For size 4 ($[60, 70, 80, 90]$), pivot is $90$. By the same logic, it requires $4$ swaps. - For size 3 ($[60, 70, 80]$), pivot is $80$. Requires $3$ swaps. - For size 2 ($[60, 70]$), pivot is $70$. Requires $2$ swaps. - For size 1 ($[60]$), pivot is $60$. Requires $1$ swap.
Step 5: Calculate the total number of swaps. Total swaps = $5 + 4 + 3 + 2 + 1 = 15$.
Solution: Insight: A topological sort is valid if and only if for every directed edge $u \to v$, vertex $u$ appears before vertex $v$ in the linear ordering. Exam route: Extract all directed edges from the graph diagram. Check each given option to see if it violates any of these precedence constraints. Eliminate options with violations. Learning route: Step 1: Identify the vertices and directed edges from the SVG diagram. The edges are: $P \to Q$, $R \to Q$, $Q \to S$, $Q \to V$, $S \to U$, and $V \to T$. Step 2: List the precedence constraints derived from these edges: - $P$ must appear before $Q$. - $R$ must appear before $Q$. - $Q$ must appear before $S$ and $V$. - $S$ must appear before $U$. - $V$ must appear before $T$. Step 3: Evaluate each option against these constraints. - Option A ("P Q R S T U V"): $Q$ appears before $R$. This violates the constraint $R \to Q$. Invalid. - Option B ("P R Q V S U T"): $P, R$ are before $Q$; $Q$ is before $V$ and $S$; $S$ is before $U$; $V$ is before $T$. All constraints are satisfied. Valid. - Option C ("P Q R S V U T"): $Q$ appears before $R$. This violates the constraint $R \to Q$. Invalid. - Option D ("P R Q S V T U"): $P, R$ are before $Q$; $Q$ is before $S$ and $V$; $S$ is before $U$; $V$ is before $T$. All constraints are satisfied. Valid. Step 4: Conclude that options B and D are the valid topological orderings.
Analytical Aptitude: Solved PYQs
Q1.(CAT 2026) Rishi and Swathi are students of Class 5. Pavan and Tanvi are students of Class 4. Rishi and Pavan are boys. Swathi and Tanvi are girls. The four students played a total of three games of chess. The games were played one after another. A player who lost a game did not participate in any more games. It was observed that:<br/>
(i) the first game was the only game where two students of the same class played against each other,<br/>
(ii) the students of Class 5 won more games than the students of Class 4, and<br/>
(iii) the boys won two games and the girls won one game.<br/>
The student who did not lose any game is __________.
Pavan
Rishi
Swathi
Tanvi
Answer: D
Solution: Key idea: This is a constraint-based sequencing and elimination question, recognizable by the sequential games, elimination rules, and conditions on wins/classes. Step 1: Understand the game format. There are 4 students and 3 games. A loser is eliminated. For all 4 students to play, the format must be: Game 1 (A vs B), Game 2 (Winner of G1 vs C), Game 3 (Winner of G2 vs D). Step 2: Use condition (i). Game 1 is the ONLY game where two students of the same class played. The same-class pairs are (Rishi, Swathi) in Class 5 and (Pavan, Tanvi) in Class 4. Thus, Game 1 must be either (R, S) or (P, T). Step 3: Test if Game 1 is (P, T). If Class 4 plays each other in G1, Class 4 wins G1. Condition (ii) states Class 5 won more games than Class 4. Since there are 3 games total, Class 5 must win 2 and Class 4 must win exactly 1. This means Class 5 must win G2 and G3. If Class 5 wins G2, the winner is R or S. Then Game 3 would be (R or S) vs the remaining (S or R), which is a same-class game. This violates condition (i). Thus, Game 1 cannot be (P, T). Step 4: Test if Game 1 is (R, S). Game 1 is R vs S. Class 5 wins G1. To satisfy the quotas (Class 5 wins 2, Class 4 wins 1; Boys win 2, Girls win 1), Rishi (Boy) must win G1. If Swathi won, we would need 2 Boy wins from the remaining games, forcing Class 4 to win 2 games, violating (ii). So Rishi wins G1. Step 5: Determine Game 2 and 3. Rishi plays Game 2 against Pavan or Tanvi. If Rishi loses G2, the Class 4 student wins, and Class 4 would end up winning 2 games total (violating ii). Thus, Rishi must win G2. Step 6: Now Rishi has won G1 and G2 (2 Boy wins, 2 Class 5 wins). The quotas for Boys and Class 5 are fully met. Therefore, Game 3 must be won by a Girl from Class 4 to satisfy the remaining quotas (1 Girl win, 1 Class 4 win). Step 7: Game 3 is Rishi vs Tanvi. Tanvi must win. Since Tanvi won her only game and didn't play before, she never lost any game. Answer: D
Q2.(CAT 2025) Let \(p\) and \(q\) be any two propositions. Consider the following propositional statements.<br/>
\(S_1 : p \rightarrow q, S_2 : \neg p \land q, S_3 : \neg p \lor q, S_4 : \neg p \lor \neg q,\)<br/>
where \(\land\) denotes conjunction (AND operation), \(\lor\) denotes disjunction (OR operation), and \(\neg\) denotes negation (NOT operation). Which one of the following options is correct?<br/>
(Note: \(\equiv\) denotes logical equivalence)
\(S_1 \equiv S_3\)
\(S_2 \equiv S_3\)
\(S_2 \equiv S_4\)
\(S_1 \equiv S_4\)
Answer: A
Solution: Insight: $p \to q$ is logically equivalent to $\neg p \lor q$ by material implication. Exam route: Recall the material implication law $p \to q \equiv \neg p \lor q$. Compare $S_1 = p \to q$ with $S_3 = \neg p \lor q$. They are identical in meaning. Learning route: Step 1: Write out each statement clearly. $S_1 : p \to q$ $S_2 : \neg p \land q$ $S_3 : \neg p \lor q$ $S_4 : \neg p \lor \neg q$ Step 2: Apply the material implication equivalence to $S_1$. $p \to q \equiv \neg p \lor q$ Step 3: Compare the result with $S_3$. $S_1 \equiv \neg p \lor q \equiv S_3$ Step 4: Verify with a truth table to be absolutely sure. $p$ | $q$ | $S_1$ | $S_3$ T | T | T | T T | F | F | F F | T | T | T F | F | T | T The columns for $S_1$ and $S_3$ match exactly. Answer: $S_1 \equiv S_3$, which is option A. Common trap: Students sometimes confuse $\neg p \lor q$ with $\neg p \land q$ or $\neg p \lor \neg q$. Remember that implication becomes a disjunction ($\lor$), not a conjunction ($\land$).
Q3.(CAT 2026) Assume that a Creative \((C)\) person will Succeed \((S)\) if the person is also Disciplined \((D)\), but will not succeed otherwise. Now, consider the following statements:<br/>
<br/>
(i) \(C \land S \Leftrightarrow D\)<br/>
(ii) \(C \Rightarrow (S \Longleftrightarrow D)\)<br/>
(iii) \(C \Leftrightarrow ((D \Rightarrow S) \lor \neg S)\)<br/>
<br/>
Which of the following options is correct?
Both (i) and (ii) are TRUE
Only (ii) is TRUE
Both (ii) and (iii) are TRUE
Only (iii) is TRUE
Answer: B
Solution: Insight: "A creative person will succeed if disciplined, but not otherwise" means: if creative, then (succeed iff disciplined). Exam route: Translate the English sentence step by step. "Creative person will succeed if disciplined" gives $C \to (D \to S)$. "But not otherwise" adds $C \to (\neg D \to \neg S)$. Together: $C \to (S \iff D)$. Check which option matches. Learning route: Step 1: Parse the sentence structure. "A Creative (C) person will Succeed (S) if the person is also Disciplined (D), but will not succeed otherwise." The phrase "but not otherwise" is the key. It means the condition is both necessary and sufficient. Step 2: Translate the "if" part. "will succeed if disciplined" $\implies D \to S$ (within the context of being creative). Step 3: Translate the "but not otherwise" part. "will not succeed otherwise" $\implies \neg D \to \neg S$ (within the context of being creative). Step 4: Combine 2 and 3. $(D \to S) \land (\neg D \to \neg S) \equiv S \iff D$. Step 5: Apply the context. The whole rule applies to creative people, so: $C \to (S \iff D)$. Step 6: Evaluate the given statements. (i) $C \land S \iff D$: This means $(C \land S) \iff D$, which is not equivalent. For example, if $C = F, D = F$, the original gives $T$ (vacuously), but (i) gives $F \iff F = T$. Wait, let me check $C = F, S = T, D = F$: original gives $T$, (i) gives $(F \land T) \iff F = F \iff F = T$. Let me check $C = F, S = F, D = T$: original gives $T$, (i) gives $(F \land F) \iff T = F \iff T = F$. Not equivalent. So (i) is FALSE. (ii) $C \to (S \iff D)$: This is exactly what we derived. TRUE. (iii) $C \iff ((D \to S) \lor \neg S)$: Simplify the right side. $(D \to S) \lor \neg S = (\neg D \lor S) \lor \neg S = \neg D \lor (S \lor \neg S) = \neg D \lor T = T$. So (iii) becomes $C \iff T$, which means $C$ must be true. This is not equivalent to $C \to (S \iff D)$. FALSE. Answer: Only (ii) is TRUE, which is option B.
Q4.(CAT 2025) Weight of a person can be expressed as a function of their age. The function usually varies from person to person. Suppose this function is identical for two brothers, and it monotonically increases till the age of \(50\) years and then it monotonically decreases. Let \(a_1\) and \(a_2\) (in years) denote the ages of the brothers and \(a_1 < a_2\).<br/>
Which one of the following statements is correct about their age on the day when they attain the same weight?
\(a_1 < a_2 < 50\)
\(a_1 < 50 < a_2\)
\(50 < a_1 < a_2\)
Either \(a_1 = 50\) or \(a_2 = 50\)
Answer: B
Solution: Key idea: This is a monotonicity-based logical deduction question, recognizable by the function's trend (increasing then decreasing) and the equality of outputs for two different inputs. Step 1: Understand the function's behavior. The weight function $W(a)$ monotonically increases for $a \le 50$ and monotonically decreases for $a > 50$. This creates a single peak at $a = 50$. Step 2: Analyze the condition $W(a_1) = W(a_2)$ with $a_1 < a_2$. Step 3: Evaluate the positions of $a_1$ and $a_2$ relative to the peak. If both $a_1 < a_2 \le 50$, they lie on the monotonically increasing segment. For a strictly monotonic function, $W(a_1) < W(a_2)$, so they cannot be equal. If both $50 \le a_1 < a_2$, they lie on the monotonically decreasing segment. Similarly, $W(a_1) > W(a_2)$, so they cannot be equal. Step 4: The only way for $W(a_1) = W(a_2)$ with $a_1 \neq a_2$ is if one age is on the increasing slope and the other is on the decreasing slope. Step 5: Since $a_1 < a_2$, it must be that $a_1 < 50$ and $a_2 > 50$. Step 6: This matches the condition $a_1 < 50 < a_2$. Answer: B
Q5.(CAT 2024) Let \(x\) and \(y\) be two propositions. Which of the following statements is a tautology<br/>/are tautologies?
\((\neg x \land y ) \implies (y \implies x)\)
\((x \land \neg y ) \implies (\neg x \implies y)\)
\((\neg x \land y ) \implies (\neg x \implies y)\)
\((x \land \neg y ) \implies (y \implies x)\)
Answer: ["B","C","D"]
Solution: Insight: Convert each implication to disjunction form and simplify. A tautology simplifies to $T$. Exam route: For each option, replace $A \to B$ with $\neg A \lor B$, then simplify using De Morgan's and absorption. If the result is $T$, it's a tautology. Learning route: Step 1: Recall $A \to B \equiv \neg A \lor B$. Step 2: Evaluate option A: $(\neg x \land y) \to (y \to x)$. $= \neg(\neg x \land y) \lor (\neg y \lor x)$ $= (x \lor \neg y) \lor (\neg y \lor x)$ $= x \lor \neg y$ This is NOT always true (false when $x = F, y = T$). So A is not a tautology. Step 3: Evaluate option B: $(x \land \neg y) \to (\neg x \to y)$. $= \neg(x \land \neg y) \lor (x \lor y)$ $= (\neg x \lor y) \lor (x \lor y)$ $= \neg x \lor y \lor x \lor y$ $= (\neg x \lor x) \lor (y \lor y)$ $= T \lor y$ $= T$ This IS a tautology. Step 4: Evaluate option C: $(\neg x \land y) \to (\neg x \to y)$. $= \neg(\neg x \land y) \lor (x \lor y)$ $= (x \lor \neg y) \lor (x \lor y)$ $= x \lor \neg y \lor x \lor y$ $= x \lor (\neg y \lor y)$ $= x \lor T$ $= T$ This IS a tautology. Step 5: Evaluate option D: $(x \land \neg y) \to (y \to x)$. $= \neg(x \land \neg y) \lor (\neg y \lor x)$ $= (\neg x \lor y) \lor (\neg y \lor x)$ $= (\neg x \lor x) \lor (y \lor \neg y)$ $= T \lor T$ $= T$ This IS a tautology. Answer: B, C, D are tautologies.
Linear Algebra: Solved PYQs
Q1.(CAT 2025) Which of the following statements is/are correct?
\(\mathbb{R}^n\) has a unique set of orthonormal basis vectors
\(\mathbb{R}^n\) does not have a unique set of orthonormal basis vectors
Linearly independent vectors in \(\mathbb{R}^n\) are orthonormal
Orthonormal vectors \(\mathbb{R}^n\) are linearly independent
Answer: ["B","D"]
Solution: Insight: Orthonormal bases are not unique (any rotation works), but orthonormality strictly guarantees linear independence. Exam route: Evaluate each option. A is false because we can rotate the standard basis. B is true. C is false because independent vectors need not be orthogonal. D is true because orthonormal vectors are always independent. Learning route: 1. An orthonormal basis for $\mathbb{R}^n$ requires vectors to be mutually orthogonal and unit length. The standard basis is one, but any rotation of it (e.g., in $\mathbb{R}^2$, using $(\cos \theta, \sin \theta)$ and $(-\sin \theta, \cos \theta)$) is another. Thus, it is not unique (A is false, B is true). 2. For C, consider vectors $(1, 0)$ and $(1, 1)$ in $\mathbb{R}^2$. They are linearly independent, but their dot product is $1 \neq 0$, so they are not orthogonal. Thus, independent vectors are not necessarily orthonormal. 3. For D, let $v_1, \dots, v_k$ be orthonormal. Suppose $\sum c_i v_i = 0$. Taking the dot product with $v_j$ gives $c_j (v_j \cdot v_j) = 0 \implies c_j = 0$. Thus, they are linearly independent.
Q2.(CAT 2025) The sum of the elements in each row of \(A \in \mathbb{R}^{n \times n}\) is \(1\). If \(B = A^3 - 2A^2 + A\), which one of the following statements is correct (for \(x \in \mathbb{R}^n\))?
The equation \(Bx = 0\) has no solution
The equation \(Bx = 0\) has exactly two solutions
The equation \(Bx = 0\) has infinitely many solutions
The equation \(Bx = 0\) has a unique solution
Answer: C
Solution: Key idea: The row sum condition directly gives an eigenvalue and its eigenvector, which can be evaluated in the matrix polynomial. Step 1: The statement "sum of the elements in each row of $A$ is $1$" means that if we multiply $A$ by the all-ones column vector $\mathbf{1} = [1, 1, \dots, 1]^T$, the result is $\mathbf{1}$. Thus, $A\mathbf{1} = 1 \cdot \mathbf{1}$. Step 2: This implies $\lambda = 1$ is an eigenvalue of $A$, and $\mathbf{1}$ is a corresponding non-zero eigenvector. Step 3: We are given $B = A^3 - 2A^2 + A$. We can factor this polynomial as $B = A(A^2 - 2A + I) = A(A - I)^2$. Step 4: Evaluate $B\mathbf{1}$. Since $(A - I)\mathbf{1} = A\mathbf{1} - \mathbf{1} = \mathbf{1} - \mathbf{1} = \mathbf{0}$, we have $(A - I)^2\mathbf{1} = (A - I)\mathbf{0} = \mathbf{0}$. Step 5: Therefore, $B\mathbf{1} = A(A - I)^2\mathbf{1} = A\mathbf{0} = \mathbf{0}$. Step 6: Since $\mathbf{1}$ is a non-zero vector and $B\mathbf{1} = \mathbf{0}$, the homogeneous system $Bx = 0$ has a non-trivial solution. Any homogeneous system with a non-trivial solution has infinitely many solutions. Answer: The equation $Bx = 0$ has infinitely many solutions.
Q3.(CAT 2025) Let \(A = I_n + xx^T\), where \(I_n\) is the \(n \times n\) identity matrix and \(x \in \mathbb{R}^n\), \(x^T x = 1\). Which of the following options is/are correct?