CAT 2025 Slot 1 — DILR questions with answers

Q25MCQTables & Caselets

1. A35%
2. B70%
3. C84%
4. D77%

Answer and solution

Answer: (B) 70%

Let $p,q,r,s$ be the tickets from A to B, C, D, E; $t,u,v$ from B to C, D, E; $w,x$ from C to D, E; and $y$ from D to E. Points 2 and 5 give $t=40$, $v=30$ and $p=u=y=0$. Point 4 gives $q=s>30$. D–E carries $s+30+x$ seats, of which $s+30$ are from A or B. By point 3, $s+30=\frac{4}{7}(s+30+x)$, so $4x=3s+90$. All counts are multiples of 10: $s=40,60,70,80$ give $x=52.5,67.5,75,82.5$, and $s=90$ gives $x=90$ but $210>200$ seats on D–E (larger $s$ is worse). So $q=s=50$ and $x=60$. C–D carries 95% of 200, i.e. 190 seats: $r+s+v+w+x=190$, so $r+w=50$. Only B–C is fuller, so it carries more than 190 seats, a multiple of 10 not above 200, i.e. 200: $q+r+s+t+v=200$ gives $170+r=200$, so $r=30$ and $w=20$. D–E carries $s + v + x + y = 50 + 30 + 60 + 0 = 140$ seats, so its occupancy factor is $\frac{140}{200} \times 100 = 70\%$. The other options would need 70, 168 or 154 seats on D–E. Hence, option B (70%).

Q26TITATables & Caselets

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 50

Let $p,q,r,s$ be the tickets from A to B, C, D, E; $t,u,v$ from B to C, D, E; $w,x$ from C to D, E; and $y$ from D to E. Points 2 and 5 give $t=40$, $v=30$ and $p=u=y=0$. Point 4 gives $q=s>30$. D–E carries $s+30+x$ seats, of which $s+30$ are from A or B. By point 3, $s+30=\frac{4}{7}(s+30+x)$, so $4x=3s+90$. All counts are multiples of 10: $s=40,60,70,80$ give $x=52.5,67.5,75,82.5$, and $s=90$ gives $x=90$ but $210>200$ seats on D–E (larger $s$ is worse). So $q=s=50$ and $x=60$. C–D carries 95% of 200, i.e. 190 seats: $r+s+v+w+x=190$, so $r+w=50$. Only B–C is fuller, so it carries more than 190 seats, a multiple of 10 not above 200, i.e. 200: $q+r+s+t+v=200$ gives $170+r=200$, so $r=30$ and $w=20$. The tickets from A to E number $s = 50$. The answer is 50.

Q27TITATables & Caselets

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 80

Let $p,q,r,s$ be the tickets from A to B, C, D, E; $t,u,v$ from B to C, D, E; $w,x$ from C to D, E; and $y$ from D to E. Points 2 and 5 give $t=40$, $v=30$ and $p=u=y=0$. Point 4 gives $q=s>30$. D–E carries $s+30+x$ seats, of which $s+30$ are from A or B. By point 3, $s+30=\frac{4}{7}(s+30+x)$, so $4x=3s+90$. All counts are multiples of 10: $s=40,60,70,80$ give $x=52.5,67.5,75,82.5$, and $s=90$ gives $x=90$ but $210>200$ seats on D–E (larger $s$ is worse). So $q=s=50$ and $x=60$. C–D carries 95% of 200, i.e. 190 seats: $r+s+v+w+x=190$, so $r+w=50$. Only B–C is fuller, so it carries more than 190 seats, a multiple of 10 not above 200, i.e. 200: $q+r+s+t+v=200$ gives $170+r=200$, so $r=30$ and $w=20$. Tickets from C go to D or E: $w + x = 20 + 60 = 80$. The answer is 80.

Q28TITATables & Caselets

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Answer and solution

Answer: 40

Let $p,q,r,s$ be the tickets from A to B, C, D, E; $t,u,v$ from B to C, D, E; $w,x$ from C to D, E; and $y$ from D to E. Points 2 and 5 give $t=40$, $v=30$ and $p=u=y=0$. Point 4 gives $q=s>30$. D–E carries $s+30+x$ seats, of which $s+30$ are from A or B. By point 3, $s+30=\frac{4}{7}(s+30+x)$, so $4x=3s+90$. All counts are multiples of 10: $s=40,60,70,80$ give $x=52.5,67.5,75,82.5$, and $s=90$ gives $x=90$ but $210>200$ seats on D–E (larger $s$ is worse). So $q=s=50$ and $x=60$. C–D carries 95% of 200, i.e. 190 seats: $r+s+v+w+x=190$, so $r+w=50$. Only B–C is fuller, so it carries more than 190 seats, a multiple of 10 not above 200, i.e. 200: $q+r+s+t+v=200$ gives $170+r=200$, so $r=30$ and $w=20$. Tickets to C: $q + t = 50 + 40 = 90$. Tickets to D: $r + u + w = 30 + 0 + 20 = 50$. The difference is $90 - 50 = 40$. The answer is 40.

Q29TITATables & Caselets

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 60

Let $p,q,r,s$ be the tickets from A to B, C, D, E; $t,u,v$ from B to C, D, E; $w,x$ from C to D, E; and $y$ from D to E. Points 2 and 5 give $t=40$, $v=30$ and $p=u=y=0$. Point 4 gives $q=s>30$. D–E carries $s+30+x$ seats, of which $s+30$ are from A or B. By point 3, $s+30=\frac{4}{7}(s+30+x)$, so $4x=3s+90$. All counts are multiples of 10: $s=40,60,70,80$ give $x=52.5,67.5,75,82.5$, and $s=90$ gives $x=90$ but $210>200$ seats on D–E (larger $s$ is worse). So $q=s=50$ and $x=60$. C–D carries 95% of 200, i.e. 190 seats: $r+s+v+w+x=190$, so $r+w=50$. Only B–C is fuller, so it carries more than 190 seats, a multiple of 10 not above 200, i.e. 200: $q+r+s+t+v=200$ gives $170+r=200$, so $r=30$ and $w=20$. Tickets covering exactly one segment are A to B, B to C, C to D and D to E: $p + t + w + y = 0 + 40 + 20 + 0 = 60$. The answer is 60.

Q30TITALogical Puzzles

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Q31TITALogical Puzzles

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Q32TITALogical Puzzles

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Q33TITALogical Puzzles

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Q34MCQLogical Puzzles

1. AOnly II
2. BOnly I
3. CBoth I and II
4. DNeither I nor II

Q35TITACircular Arrangement

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Answer and solution

Answer: 4

Only Aslam moves in Turn 1, and Davies is on Chair 2 after it, so Davies starts on Chair 2. Davies has empty chairs on both sides, so Chairs 1 and 3 are empty. Only Aslam and Bashir move in Turns 1 and 2, and Chhavi is on Chair 7 after Turn 2, so she starts on Chair 7. Aslam sits next to Chhavi. Chair 1 is empty, so Aslam is on Chair 6. Bashir also has empty chairs on both sides. The chairs left are 3, 4 and 5. Chair 3 is empty, and Chair 5 is next to Aslam on Chair 6, so Bashir cannot sit there. So Bashir is on Chair 4, with Chairs 3 and 5 empty beside him. The answer is 4.

Q36MCQCircular Arrangement

1. ABashir
2. BChhavi
3. CDavies
4. DNo one

Answer and solution

Answer: (D) No one

Davies is on Chair 2 after Turn 1, when only Aslam has moved, so he starts there, with Chairs 1 and 3 empty. Chhavi is on 7 after Turn 2, so she starts there. Aslam sits beside her with Chair 1 empty, so he is on 6, and Bashir, needing empty neighbours, is on 4. After Turn 2 all four sit together. If Aslam went counterclockwise to 5, Bashir could reach only 3 or 6, with no block. So Aslam goes clockwise to 1, and Bashir to 3 (going to 5 leaves a gap), giving the block 7, 1, 2, 3. Davies is on Chair 4 after Turn 5, which is Aslam's turn, so he gets there in Turn 4. Had Chhavi gone clockwise to Chair 4 in Turn 3, Davies could reach only 5 or 7. So in Turn 3 Chhavi goes counterclockwise to 6, and Davies moves to 4 in Turn 4. At the end of Turn 3, Aslam is on 1, Davies on 2, Bashir on 3 and Chhavi on 6. Chair 4 is empty: Bashir left it in Turn 2, and Davies (option C) arrives only in Turn 4. Hence, option D (No one).

Q37MCQCircular Arrangement

1. AChairs numbered 1, 2, 6, and 7
2. BChairs numbered 2, 3, 4, and 5
3. CChairs numbered 1, 2, 3, and 4
4. DChairs numbered 4, 5, 6, and 7

Answer and solution

Answer: (D) Chairs numbered 4, 5, 6, and 7

Davies is on Chair 2 after Turn 1, when only Aslam has moved, so he starts there, with Chairs 1 and 3 empty. Chhavi likewise starts on 7, Aslam beside her on 6 (Chair 1 is empty), and Bashir, needing empty neighbours, on 4. After Turn 2 all four sit together. Aslam moving to 5 leaves Bashir only 3 or 6, with no block; so Aslam goes clockwise to 1 and Bashir to 3 (5 leaves a gap), giving 7, 1, 2, 3. Davies must reach Chair 4 in Turn 4, since Turn 5 is Aslam's. Had Chhavi taken Chair 4 in Turn 3, Davies could reach only 5 or 7, so Chhavi goes to 6 and Davies to 4. After Turn 6 all four sit together again. Aslam moving clockwise to 2 leaves Bashir only 1 or 5, with no block; so Aslam goes counterclockwise to 7 and Bashir clockwise past Davies to 5. Chairs 4 (Davies), 5 (Bashir), 6 (Chhavi) and 7 (Aslam) are occupied. Options A, B and C each include Chair 1, 2 or 3, which are empty. Hence, option D (Chairs numbered 4, 5, 6, and 7).

Q38MCQCircular Arrangement

1. ADavies only
2. BChhavi and Davies
3. CAslam and Chhavi
4. DChhavi only

Answer and solution

Answer: (A) Davies only

Davies is on Chair 2 after Turn 1, when only Aslam has moved, so he starts there, with Chairs 1 and 3 empty. Chhavi likewise starts on 7, Aslam beside her on 6, and Bashir, needing empty neighbours, on 4. After Turn 2 all four sit together. Aslam moving to 5 leaves no block, so Aslam goes clockwise to 1 and Bashir to 3, giving 7, 1, 2, 3. Davies must reach Chair 4 in Turn 4, since Turn 5 is Aslam's. Had Chhavi taken Chair 4 in Turn 3, Davies could reach only 5 or 7, so Chhavi goes to 6 and Davies to 4. After Turn 6 all four sit together again. Aslam moving to 2 leaves Bashir only 1 or 5, with no block; so Aslam goes to 7 and Bashir clockwise to 5. In Turn 7 Chhavi leaves Chair 6 for Chair 1 (clockwise, past Aslam) or Chair 3 (counterclockwise, past Bashir and Davies). Either way Chair 6 is empty, so Bashir on 5 has Davies on 4 and an empty chair on 6. Options B, C and D need Chhavi or Aslam beside him, but Chhavi has left and Aslam is on 7. Hence, option A (Davies only).

Q39TITALogical Puzzles

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 7

Each question gets 4 replies, with at least one Yes (1 tap), one No (2) and one Maybe (3). Those three give 6 taps, and the fourth reply adds 1, 2 or 3, so every question receives 7, 8 or 9 taps. Alia's question received 9 taps. Badal, Dilshan and Ehsaan received equal numbers, say $p$ each, and Clive received $q$. The total is 40, so $9 + 3p + q = 40$, i.e. $3p + q = 31$. With $p$ and $q$ between 7 and 9: $p = 7$ needs $q = 10$, and $p = 9$ needs $q = 4$, both impossible. So $p = 8$ and $q = 7$. The table also shows Alia's own replies: she tapped 6 times with Yes to Clive and Dilshan, and a third Yes is not allowed, so her replies to Badal and Ehsaan are both No. Dilshan's No to Badal is given. Clive's question received 7 taps: Yes, Yes, No, Maybe. The answer is 7.

Q40MCQLogical Puzzles

1. ABadal and Dilshan
2. BClive and Ehsaan
3. CDilshan and Clive
4. DAlia and Badal

Answer and solution

Answer: (D) Alia and Badal

Every person answers the four questions asked by the others, tapping once for Yes, twice for No and three times for Maybe. The 40 taps heard in all are the taps given by the five people. Alia tapped 6, Dilshan 11 and Ehsaan 9, which is 26. So Badal and Clive tapped $40 - 26 = 14$ between them. No one said Yes more than twice, so at least two of each person's four answers take 2 or more taps. Everyone therefore taps at least $1 + 1 + 2 + 2 = 6$ times. Badal and Clive share 14 with each at least 6, so the split is 6 and 8, or 7 and 7. Clive tapped more than Badal, so Badal tapped 6 and Clive 8. In the table's last column, $s$ and $w$ are Badal's and Clive's totals: $s = 6$ and $w = 8$. The totals are Alia 6, Badal 6, Clive 8, Dilshan 11 and Ehsaan 9, so the only equal pair is Alia and Badal. The nearest trap, option B, pairs Clive (8) and Ehsaan (9), who differ by one; options A and C involve Dilshan's 11, which no one else matches. Hence, option D (Alia and Badal).

Q41MCQLogical Puzzles

1. AYes
2. BMaybe
3. CCannot be determined
4. DNo

Answer and solution

Answer: (D) No

Each question gets at least one Yes (1 tap), No (2) and Maybe (3), plus one more reply, so it receives 7, 8 or 9 taps. With Badal, Dilshan and Ehsaan on $p$ each and Clive on $q$, $9+3p+q=40$ forces $p=8$, $q=7$. So Ehsaan's question got Yes, No, No, Maybe; Clive's got Yes, Yes, No, Maybe. Alia tapped 6 with Yes to Clive and Dilshan; a third Yes is not allowed, so she said No to Badal and Ehsaan. Dilshan tapped 11 with No to Badal, so he said Maybe to the other three. So Badal and Clive gave Ehsaan one Yes and one No. Suppose Badal said No. Badal tapped at most 6 (he and Clive tapped $40-26=14$, Clive more), and his other replies are at least 1, 1 and 2 (Dilshan's one Yes came from Alia). So Badal said Yes to Alia and Clive, and No to Dilshan. Clive's Yeses and Maybe are then taken, so Ehsaan told Clive No, and Alia Maybe (fact 5 bars No; Badal gave her the one Yes). His remaining 4 taps go to Badal (Yes or Maybe) and Dilshan (No or Maybe), so he told Dilshan Maybe, matching Dilshan's reply and breaking fact 5. So Clive said No, which also rules out 'Cannot be determined'. Hence, option D (No).

Q42TITALogical Puzzles

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 6

There are $5 \times 4 = 20$ replies and 40 taps. If $Y$, $N$, $M$ count the Yes, No and Maybe replies, $Y + N + M = 20$ and $Y + 2N + 3M = 40$. Subtracting gives $N + 2M = 20$, so $Y = M$. Each question has at least one of each, so $Y$ is at least 5, which leaves three cases. To choose between them, look at each question. Its three compulsory replies give 6 taps and the fourth adds 1, 2 or 3, so it receives 7, 8 or 9 taps. Alia's got 9, Badal, Dilshan and Ehsaan got $p$ each, and Clive got $q$. From $9 + 3p + q = 40$, $3p + q = 31$, and only $p = 8$, $q = 7$ keeps both between 7 and 9. So the fourth reply is a Maybe for Alia (9 taps), a No for Badal, Dilshan and Ehsaan (8 each) and a Yes for Clive (7). The bottom row of the table lists these. Clive's question has 2 Yes replies and each of the other four has 1, so the total is $2 + 4 = 6$. The answer is 6.

Q43MCQRadar & Special Graphs

1. A8.5 Billion USD
2. B7.0 Billion USD
3. C16.0 Billion USD
4. D1.75 Billion USD

Answer and solution

Answer: (B) 7.0 Billion USD

An import tariff is the tariff percentage times the imports, so imports equal the tariff divided by the percentage. The radar chart gives the percentages and the bar chart the tariffs (each row is the country charging): For example, the US charges France 20% and collects 6 billion USD, so US imports from France are $6 \div 0.2 = 30$ billion USD. Doing this for every pair gives the imports table (each row imports from the column country). A country's exports to another equal that country's imports from it. Japan's exports to India equal India's imports from Japan. India charges Japan 50% and collects 3.5 billion USD, so this is $3.5 \div 0.5 = 7$ billion USD. Option C (16.0) is the flow the other way: Japan imports $8 \div 0.5 = 16$ from India. Option D (1.75) multiplies $3.5 \times 0.5$ instead of dividing. Hence, option B (7.0 Billion USD).

Q44MCQRadar & Special Graphs

1. AExports by France to Japan
2. BImports by France from India
3. CImports by US from France
4. DExports by Japan to UK

Answer and solution

Answer: (C) Imports by US from France

An import tariff is the tariff percentage times the imports, so imports equal the tariff divided by the percentage. The radar chart gives the percentages and the bar chart the tariffs (each row is the country charging): For example, the US charges France 20% and collects 6 billion USD, so US imports from France are $6 \div 0.2 = 30$ billion USD. Doing this for every pair gives the imports table (each row imports from the column country). A country's exports to another equal that country's imports from it. A: France's exports to Japan = Japan's imports from France = $3 \div 0.3 = 10$. B: France's imports from India = $6.5 \div 0.4 = 16.25$. C: US imports from France = $6 \div 0.2 = 30$. D: Japan's exports to UK = UK's imports from Japan = $6 \div 0.4 = 15$. The largest is 30. Option B is the next largest, but 16.25 is far below it. Hence, option C (Imports by US from France).

Q45MCQRadar & Special Graphs

1. ADeficit of 10.0 Billion USD
2. BSurplus of 15.0 Billion USD
3. CSurplus of 10.0 Billion USD
4. DDeficit of 15.0 Billion USD

Answer and solution

Answer: (D) Deficit of 15.0 Billion USD

An import tariff is the tariff percentage times the imports, so imports equal the tariff divided by the percentage. The radar chart gives the percentages and the bar chart the tariffs (each row is the country charging): For example, the US charges France 20% and collects 6 billion USD, so US imports from France are $6 \div 0.2 = 30$ billion USD. Doing this for every pair gives the imports table (each row imports from the column country). A country's exports to another equal that country's imports from it. India's trade surplus with UK = India's exports to UK − India's imports from UK. India's exports to UK = UK's imports from India = $3 \div 0.3 = 10$ billion USD. India's imports from UK = $5 \div 0.2 = 25$ billion USD. Surplus $= 10 - 25 = -15$, a deficit of 15 billion USD. Option B has the right size but the wrong sign: India buys more from the UK than it sells there. Hence, option D (Deficit of 15.0 Billion USD).

Q46MCQRadar & Special Graphs

1. ANeither France nor UK
2. BBoth France and UK
3. COnly France
4. DOnly UK

Answer and solution

Answer: (C) Only France

An import tariff is the tariff percentage times the imports, so imports equal the tariff divided by the percentage. The radar chart gives the percentages and the bar chart the tariffs (each row is the country charging): For example, the US charges France 20% and collects 6 billion USD, so US imports from France are $6 \div 0.2 = 30$ billion USD. Doing this for every pair gives the imports table (each row imports from the column country). A country's exports to another equal that country's imports from it. France: exports to US = US imports from France = 30; imports from US = $5.5 \div 0.3 \approx 18.33$. Surplus $\approx 30 - 18.33 = 11.67$ billion USD. UK: exports to US = US imports from UK = $3 \div 0.3 = 10$; imports from US = $2.5 \div 0.2 = 12.5$. Balance $= 10 - 12.5 = -2.5$, a deficit. So only France has a surplus with the US. Option B fails because the UK runs a deficit. Hence, option C (Only France).