CAT 2020 Slot 3 — DILR questions with answers

Q27MCQLogical Puzzles

1. APatient 14
2. BPatient 8
3. CPatient 6
4. DPatient 2

Q28MCQLogical Puzzles

1. AVial B
2. BVial E
3. CVial C
4. DVial H

Q29MCQLogical Puzzles

1. AVials A and E positive, vials C and D negative
2. BVial B positive, vials C, F and H negative
3. CVials A and G positive, vials D and E negative
4. DVials B and D positive, vials F and H negative

Q30MCQLogical Puzzles

1. A{5,6,7,8}
2. B{4,5,6,7}
3. C{4,5,6,7,8}
4. D{4,5}

Q31MCQTables & Caselets

1. A15th
2. B16th
3. C13th
4. D14th

Answer and solution

Answer: (A) 15th

Orders booked on a day are the rise in cumulative bookings. An order booked on day $d$ that is not delivered is declared lost at the end of day $d+2$, so it appears in the rise in cumulative lost on day $d+2$. 14th: booked $249-219=30$, lost $106-94=12$. 15th: booked $277-249=28$, lost $118-106=12$. 16th: booked $302-277=25$, lost $120-118=2$. 13th: lost $94-92=2$. The table has no 12th row, so find the 13th's bookings from deliveries. The 27 orders delivered on the 14th are the 13th's next-day orders plus the 12th's 6 two-day orders, so 21 of the 13th's orders came the next day. With 8 two-day orders and 2 lost, the 13th had $21+8+2=31$ bookings. Fractions lost: 13th $\frac{2}{31}\approx 0.06$; 14th $\frac{12}{30}=0.40$; 15th $\frac{12}{28}\approx 0.43$; 16th $\frac{2}{25}=0.08$. The 14th is the close runner-up. Both days lost 12 orders, but the 14th had more bookings (30 against 28), so its fraction is smaller. Hence, option A (15th).

Q32MCQTables & Caselets

1. A12th
2. B15th
3. C13th
4. D14th

Answer and solution

Answer: (C) 13th

The 14th and 15th come straight from the cumulative column: $249-219=30$ and $277-249=28$. There is no row before the 13th, so count the 12th's and 13th's bookings as next-day deliveries plus two-day deliveries plus lost orders. An order booked on day $d$ that is not delivered is lost at the end of day $d+2$. So the 12th lost $92-91=1$ order (the rise on the 14th), and the 13th lost $94-92=2$ (the rise on the 15th). Deliveries on a day are the previous day's next-day orders plus the two-day orders from two days earlier. The two-day orders of the 11th, 12th and 13th were 4, 6 and 8. On the 13th, 11 orders were delivered, so the 12th had $11-4=7$ next-day orders. On the 14th, 27 were delivered, so the 13th had $27-6=21$. Bookings: 12th $=7+6+1=14$; 13th $=21+8+2=31$. So the counts are 14 (12th), 31 (13th), 30 (14th) and 28 (15th). The 14th is closest, one short of the 13th. Hence, option C (13th).

Q33MCQTables & Caselets

1. A15th
2. B16th
3. C13th
4. D14th

Answer and solution

Answer: (D) 14th

Deliveries on a day are the previous day's next-day orders plus the two-day orders from two days earlier. Each day's bookings split into next-day, two-day and lost orders. An order booked on day $d$ that is lost appears in the rise in cumulative lost on day $d+2$. From the table, the 14th, 15th and 16th had 30, 28 and 25 bookings, and lost 12, 12 and 2 orders ($106-94$, $118-106$, $120-118$). Two-day orders for the 12th and 13th are given as 6 and 8. 13th: next-day $=27-6=21$ (delivered on the 14th), two-day $=8$. Ratio $\frac{21}{8}\approx 2.6$. 14th: next-day $=23-8=15$, two-day $=30-15-12=3$. Ratio $\frac{15}{3}=5$. 15th: next-day $=11-3=8$, two-day $=28-8-12=8$. Ratio $1$. 16th: next-day $=21-8=13$, two-day $=25-13-2=10$. Ratio $1.3$. The 13th has the next-highest ratio, but 2.6 is well below the 14th's 5. Hence, option D (14th).

Q34MCQTables & Caselets

1. A15th
2. B13th
3. C14th
4. D16th

Answer and solution

Answer: (C) 14th

First find, for each booking day, $x$ (orders delivered the next day) and $y$ (orders delivered the day after). Deliveries on a day are the previous day's next-day orders plus the two-day orders from two days earlier. Bookings split into $x$, $y$ and lost orders, and orders booked on day $d$ that are lost appear in the rise in cumulative lost on day $d+2$. 13th: $x=27-6=21$ (delivered on the 14th, less the 12th's 6 two-day orders), $y=8$ (given). 14th: $x=23-8=15$. It had 30 bookings and lost 12, so $y=3$. 15th: $x=11-3=8$. It had 28 bookings and lost 12, so $y=8$. 16th: $x=21-8=13$. It had 25 bookings and lost 2, so $y=10$. Average time $=\frac{x+2y}{x+y}=1+\frac{y}{x+y}$: 13th $\frac{37}{29}\approx 1.28$; 14th $\frac{21}{18}\approx 1.17$; 15th $\frac{24}{16}=1.5$; 16th $\frac{33}{23}\approx 1.43$. The 13th is the runner-up at 1.28. The 14th is lower because only 3 of its 18 delivered orders took two days. Hence, option C (14th).

Q35MCQMissing Value Tables

1. A49
2. B84
3. C98
4. D126

Answer and solution

Answer: (C) 98

Row Y holds the teak trees. By clue 3, if Y2 has $x$ trees then Y3 has $2x$ and Y4 has $4x$. With Y1 $=21$, teak $=21+7x$. By clue 9, mango $=2(21+7x)=42+14x$. The total is 205, so pine $=205-3(21+7x)=142-21x$. Row Z already has 9 and 28, and each of its other two plots needs at least 3 trees, since every plot holds a non-zero multiple of 3 or 4. So $142-21x\ge 43$, which gives $x\le 4$. $x$ itself must be a multiple of 3 or 4, so $x=3$ or $x=4$. With $x=3$, Y4 would hold 12, the same as X1, but no two plots have the same number of trees. So $x=4$. Teak $=49$, mango $=42+56=98$ and pine $=58$ (check: $98+49+58=205$). Option A (49) is the number of teak trees, not mango trees. Hence, option C (98).

Q36MCQMissing Value Tables

1. A50, 69, 30, 56
2. B54, 57, 34, 60
3. C44, 87, 24, 50
4. D60, 39, 40, 66

Answer and solution

Answer: (A) 50, 69, 30, 56

Plot counts are even, Dipti has at least 2 (clue 6), Chitra at least 2 (X1, Z2), and Abha and Bina each have more than Dipti. So Abha, Bina, Chitra and Dipti have 4, 4, 2 and 2 of the 12 plots, and Chitra has only X1 and Z2. Dipti's two plots adjoin in one row without touching X1 or Z2, so they are X3 and X4. Corner Z4 is not Abha's (clue 5), so it is Bina's. Bina also needs a plot in row X (X2) and one in column 3 (Y3). Abha has Y1, Y2, Y4 and Z3. By clue 3, Y2, Y3 and Y4 hold $x$, $2x$ and $4x$, so teak $=21+7x$, mango $=42+14x$ (clue 9) and pine $=205-3(21+7x)=142-21x$. Row Z holds $9+28$ plus at least 3 in each other plot, so $x\le 4$. Since $x$ is a multiple of 3 or 4, and $x=3$ would repeat 12 (Y4 and X1), $x=4$. Abha $=21+4+16+9=50$. By clue 1, Chitra $=50-20=30$ and Dipti $=50+6=56$. Bina $=205-50-30-56=69$. Every option satisfies clue 1 and adds to 205, so the deciding figure is Abha's 50, which only option A has. Hence, option A (50, 69, 30, 56).

Q37MCQMissing Value Tables

1. A18
2. B30
3. C21
4. D15

Answer and solution

Answer: (A) 18

Plot counts are even, Dipti has at least 2 (clue 6), Chitra at least 2 (X1, Z2), and Abha and Bina each have more than Dipti. So Abha, Bina, Chitra and Dipti have 4, 4, 2 and 2 of the 12 plots, and Chitra has only X1 and Z2. Dipti's two plots adjoin in one row without touching X1 or Z2, so they are X3 and X4. Corner Z4 is not Abha's (clue 5), so it is Bina's. Bina also needs a plot in row X (X2) and one in column 3 (Y3). Abha has Y1, Y2, Y4 and Z3. By clue 3, Y2, Y3 and Y4 hold $x$, $2x$ and $4x$, so teak $=21+7x$, mango $=42+14x$ (clue 9) and pine $=205-3(21+7x)=142-21x$. Row Z holds $9+28$ plus at least 3 in each other plot, so $x\le 4$. Since $x$ is a multiple of 3 or 4, and $x=3$ would repeat 12 (Y4 and X1), $x=4$. Abha $=21+4+16+9=50$, so by clue 1 Chitra has $50-20=30$ trees. Chitra's plots are X1 (12 mango) and Z2, her only pine plot, which holds $30-12=18$. Option B (30) is Chitra's total, not her pine trees. Hence, option A (18).

Q38MCQMissing Value Tables

1. ADipti, 6 trees
2. BBina, 3 trees
3. CBina, 4 trees
4. DAbha, 4 trees

Answer and solution

Answer: (B) Bina, 3 trees

Plot counts are even, Dipti has at least 2 (clue 6), Chitra at least 2 (X1, Z2), and Abha and Bina each have more than Dipti. So Abha, Bina, Chitra and Dipti have 4, 4, 2 and 2 of the 12 plots, and Chitra has only X1 and Z2. Dipti's two plots adjoin in one row without touching X1 or Z2, so they are X3 and X4. Corner Z4 is not Abha's (clue 5), so it is Bina's. Bina also needs a plot in row X (X2) and one in column 3 (Y3). Abha has Y1, Y2, Y4 and Z3. By clue 3, Y2, Y3 and Y4 hold $x$, $2x$ and $4x$, so teak $=21+7x$, mango $=42+14x$ (clue 9) and pine $=205-3(21+7x)=142-21x$. Row Z holds $9+28$ plus at least 3 in each other plot, so $x\le 4$. Since $x$ is a multiple of 3 or 4, and $x=3$ would repeat 12 (Y4 and X1), $x=4$. So pine $=58$. Abha $=21+4+16+9=50$, so Chitra $=30$ and Z2 $=30-12=18$. Then Z1 $=58-9-28-18=3$. Every plot holds a non-zero multiple of 3 or 4, so 3 is the least possible, and Z1 is Bina's. Options C and D point to a 4-tree plot; the only one is Y2, which is Abha's. Hence, option B (Bina, 3 trees).

Q39MCQMissing Value Tables

1. AChitra got 12 mango trees
2. BBina got 32 pine trees.
3. CAbha got 41 teak trees.
4. DDipti got 56 mango trees

Answer and solution

Answer: (B) Bina got 32 pine trees.

Plot counts are even, Dipti has at least 2 (clue 6), Chitra at least 2 (X1, Z2), and Abha and Bina each have more than Dipti. So Abha, Bina, Chitra and Dipti have 4, 4, 2 and 2 of the 12 plots, and Chitra has only X1 and Z2. Dipti's two plots adjoin in one row without touching X1 or Z2, so they are X3 and X4. Corner Z4 is not Abha's (clue 5), so it is Bina's. Bina also needs a plot in row X (X2) and one in column 3 (Y3). Abha has Y1, Y2, Y4 and Z3. By clue 3, Y2, Y3 and Y4 hold $x$, $2x$ and $4x$, so teak $=21+7x$, mango $=42+14x$ (clue 9) and pine $=205-3(21+7x)=142-21x$. Row Z holds $9+28$ plus at least 3 in each other plot, so $x\le 4$. Since $x$ is a multiple of 3 or 4, and $x=3$ would repeat 12 (Y4 and X1), $x=4$. So pine $=58$. Abha $=21+4+16+9=50$, so Chitra $=30$, Dipti $=56$, Z2 $=30-12=18$ and Z1 $=58-37-18=3$. Bina's pine plots are Z1 and Z4: $3+28=31$, not 32, so B is false. The others are true: Chitra's mango plot X1 has 12; Abha's teak is $21+4+16=41$; Dipti's plots are both mango, 56 in all. Hence, option B (Bina got 32 pine trees.).

Q40MCQMissing Value Tables

1. A4
2. B3
3. CCannot be determined
4. D2

Answer and solution

Answer: (A) 4

Plot counts are even, and Abha and Bina each have more than Dipti, so Abha, Bina, Chitra and Dipti have 4, 4, 2 and 2 plots; Chitra has only X1 and Z2. Dipti's adjoining pair, clear of Chitra's plots, is X3 and X4. Corner Z4 is Bina's (clue 5), and Bina also needs X2 (row X) and Y3 (column 3). Abha has Y1, Y2, Y4 and Z3. By clue 3, Y2, Y3, Y4 hold $x$, $2x$, $4x$, so mango $=42+14x$ and pine $=142-21x$. Row Z needs at least $9+28+3+3=43$ pine, so $x\le 4$, and $x=3$ would repeat 12, so $x=4$: mango 98, pine 58. Abha $=21+4+16+9=50$, so Chitra $=30$ and Dipti $=56$. Then Z2 $=30-12=18$, Z1 $=58-37-18=3$ and X2 $=98-12-56=30$. X3 and X4 add to 56, and one of them is the 32-tree plot of clue 2, so they are 32 and 24, in an unknown order. Column 1 $=12+21+3=36$ and column 2 $=30+4+18=52$. Column 3 is X3 $+8+9$ and column 4 is X4 $+16+28$: 49 and 68, or 41 and 76. Column 4 is largest either way, so option C fails. Hence, option A (4).

Q41MCQGames & Tournaments

1. AHi, Lo, Lo, Hi
2. BHi, Lo, Lo, Lo
3. CHi, Hi, Lo, Lo
4. DLo, Lo, Lo, Hi

Answer and solution

Answer: (A) Hi, Lo, Lo, Hi

A round's points total $-4$ if all bid Hi, $+4$ if all bid Lo, and $0$ otherwise. Arun's 6 points in three rounds, at most 3 a round, mean he gained every round. Dipak's $(D_1, D_2, D_3)$ total 2 with $D_1 > D_3 > D_2$: $(3, -3, 2)$, $(3, -2, 1)$ or $(2, -1, 1)$. A 3 in round 1 makes Dipak the lone Hi, costing Arun 1, so Dipak scored $2, -1, 1$. Round 1: Dipak's $+2$ means two Hi bids, his and Arun's; Bankim and Charu bid Lo. Round 2: with Dipak on $-1$, Arun gains only as the lone Hi ($+3$). Round 3: all four need $+1$ (Bankim: $-2-(-2)-(-1)=1$), so all bid Lo. Rounds 4–6 give Arun $+1$, Bankim $+1$, Charu and Dipak $-3$ each. By clue 4, exactly one of them is Arun's other lone-Hi round ($+3$, others $-1$). The other two give Bankim $+2$ and the rest $-2$, total $-4$: one all-Hi round and one with only Bankim on Hi. The table calls rounds 4–6 Rx, Ry, Rz, order unknown. In round 1 Arun and Dipak bid Hi, and Bankim and Charu bid Lo. Option B fails: with Arun alone on Hi, Dipak would lose 1, not gain 2. Hence, option A (Hi, Lo, Lo, Hi).

Q42TITAGames & Tournaments

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

Answer and solution

Answer: 4

A round's points total $-4$ if all bid Hi, $+4$ if all bid Lo, and $0$ otherwise. Arun's 6 points in three rounds, at most 3 a round, mean he gained every round. Dipak's $(D_1, D_2, D_3)$ total 2 with $D_1 > D_3 > D_2$: $(3, -3, 2)$, $(3, -2, 1)$ or $(2, -1, 1)$. A 3 in round 1 makes Dipak the lone Hi, costing Arun 1, so Dipak scored $2, -1, 1$. Round 1: Dipak's $+2$ means two Hi bids, his and Arun's; Bankim and Charu bid Lo. Round 2: with Dipak on $-1$, Arun gains only as the lone Hi ($+3$). Round 3: all four need $+1$ (Bankim: $-2-(-2)-(-1)=1$), so all bid Lo. Rounds 4–6 give Arun $+1$, Bankim $+1$, Charu and Dipak $-3$ each. By clue 4, exactly one of them is Arun's other lone-Hi round ($+3$, others $-1$). The other two give Bankim $+2$ and the rest $-2$, total $-4$: one all-Hi round and one with only Bankim on Hi. The table calls rounds 4–6 Rx, Ry, Rz, order unknown. Arun bid Hi in rounds 1 and 2, in his later lone-Hi round and in the all-Hi round. He bid Lo in round 3 and in Bankim's lone-Hi round. The answer is 4.

Q43TITAGames & Tournaments

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

Answer and solution

Answer: 4

A round's points total $-4$ if all bid Hi, $+4$ if all bid Lo, and $0$ otherwise. Arun's 6 points in three rounds, at most 3 a round, mean he gained every round. Dipak's $(D_1, D_2, D_3)$ total 2 with $D_1 > D_3 > D_2$: $(3, -3, 2)$, $(3, -2, 1)$ or $(2, -1, 1)$. A 3 in round 1 makes Dipak the lone Hi, costing Arun 1, so Dipak scored $2, -1, 1$. Round 1: Dipak's $+2$ means two Hi bids, his and Arun's; Bankim and Charu bid Lo. Round 2: with Dipak on $-1$, Arun gains only as the lone Hi ($+3$). Round 3: all four need $+1$ (Bankim: $-2-(-2)-(-1)=1$), so all bid Lo. Rounds 4–6 give Arun $+1$, Bankim $+1$, Charu and Dipak $-3$ each. By clue 4, exactly one of them is Arun's other lone-Hi round ($+3$, others $-1$). The other two give Bankim $+2$ and the rest $-2$, total $-4$: one all-Hi round and one with only Bankim on Hi. The table calls rounds 4–6 Rx, Ry, Rz, order unknown. Bankim bid Lo in rounds 1, 2 and 3 and in Arun's later lone-Hi round. He bid Hi in the all-Hi round and in his own lone-Hi round. The answer is 4.

Q44TITAGames & Tournaments

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

Answer and solution

Answer: 2

A round's points total $-4$ if all bid Hi, $+4$ if all bid Lo, and $0$ otherwise. Arun's 6 points in three rounds, at most 3 a round, mean he gained every round. Dipak's $(D_1, D_2, D_3)$ total 2 with $D_1 > D_3 > D_2$: $(3, -3, 2)$, $(3, -2, 1)$ or $(2, -1, 1)$. A 3 in round 1 makes Dipak the lone Hi, costing Arun 1, so Dipak scored $2, -1, 1$. Round 1: Dipak's $+2$ means two Hi bids, his and Arun's; Bankim and Charu bid Lo. Round 2: with Dipak on $-1$, Arun gains only as the lone Hi ($+3$). Round 3: all four need $+1$ (Bankim: $-2-(-2)-(-1)=1$), so all bid Lo. Rounds 4–6 give Arun $+1$, Bankim $+1$, Charu and Dipak $-3$ each. By clue 4, exactly one of them is Arun's other lone-Hi round ($+3$, others $-1$). The other two give Bankim $+2$ and the rest $-2$, total $-4$: one all-Hi round and one with only Bankim on Hi. The table calls rounds 4–6 Rx, Ry, Rz, order unknown. All four bid alike only in round 3 (all Lo) and in the all-Hi round. Every other round mixes Hi and Lo. The answer is 2.

Q45TITAGames & Tournaments

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

Answer and solution

Answer: 1

A round's points total $-4$ if all bid Hi, $+4$ if all bid Lo, and $0$ otherwise. Arun's 6 points in three rounds, at most 3 a round, mean he gained every round. Dipak's $(D_1, D_2, D_3)$ total 2 with $D_1 > D_3 > D_2$: $(3, -3, 2)$, $(3, -2, 1)$ or $(2, -1, 1)$. A 3 in round 1 makes Dipak the lone Hi, costing Arun 1, so Dipak scored $2, -1, 1$. Round 1: Dipak's $+2$ means two Hi bids, his and Arun's; Bankim and Charu bid Lo. Round 2: with Dipak on $-1$, Arun gains only as the lone Hi ($+3$). Round 3: all four need $+1$ (Bankim: $-2-(-2)-(-1)=1$), so all bid Lo. Rounds 4–6 give Arun $+1$, Bankim $+1$, Charu and Dipak $-3$ each. By clue 4, exactly one of them is Arun's other lone-Hi round ($+3$, others $-1$). The other two give Bankim $+2$ and the rest $-2$, total $-4$: one all-Hi round and one with only Bankim on Hi. The table calls rounds 4–6 Rx, Ry, Rz, order unknown. Dipak scored $2, -1, 1$ in rounds 1–3 and $-1$ in each of rounds 4–6, so he gained exactly 1 point only in round 3. The answer is 1.

Q46MCQGames & Tournaments

1. ASecond
2. BThird
3. CFourth
4. DFirst

Answer and solution

Answer: (A) Second

A round's points total $-4$ if all bid Hi, $+4$ if all bid Lo, and $0$ otherwise. Arun's 6 points in three rounds, at most 3 a round, mean he gained every round. Dipak's $(D_1, D_2, D_3)$ total 2 with $D_1 > D_3 > D_2$: $(3, -3, 2)$, $(3, -2, 1)$ or $(2, -1, 1)$. A 3 in round 1 makes Dipak the lone Hi, costing Arun 1, so Dipak scored $2, -1, 1$. Round 1: Dipak's $+2$ means two Hi bids, his and Arun's; Bankim and Charu bid Lo. Round 2: with Dipak on $-1$, Arun gains only as the lone Hi ($+3$). Round 3: all four need $+1$ (Bankim: $-2-(-2)-(-1)=1$), so all bid Lo. Rounds 4–6 give Arun $+1$, Bankim $+1$, Charu and Dipak $-3$ each. By clue 4, exactly one of them is Arun's other lone-Hi round ($+3$, others $-1$). The other two give Bankim $+2$ and the rest $-2$, total $-4$: one all-Hi round and one with only Bankim on Hi. The table calls rounds 4–6 Rx, Ry, Rz, order unknown. Arun was surely the lone Hi in round 2. His other lone-Hi round may be any of rounds 4–6, so option C is not definite. In round 1 Dipak also bid Hi; round 3 was all Lo. Hence, option A (Second).

Q47MCQSet Theory

1. A64
2. B50
3. C200
4. D80

Answer and solution

Answer: (C) 200

Let each of the four exactly-three regions hold $x$ schools (fact 4), and let only F1 and F2 hold $b$. From the data: only F2 is 30, only F2 and F3 is 26 (fact 6), only F2 and F4 is 45 (fact 7), and all four is 40. F2 lies in three of the four triples, so $30 + b + 26 + 45 + 3x + 40 = 313$, giving $b + 3x = 172$. F1 and F2 together lie in two triples, so $b + 2x + 40 = 162$, giving $b + 2x = 122$. Subtracting, $x = 50$ and $b = 22$. The fully solved diagram: So the schools with exactly three facilities number $4x = 4 \times 50 = 200$. Option B (50) is the count for just one of the four triples, not all four. Hence, option C (200).

Q48MCQSet Theory

1. A185
2. B95
3. C45
4. D85

Answer and solution

Answer: (A) 185

Let each of the four exactly-three regions hold $x$ schools (fact 4), and let only F1 and F2 hold $b$. From the data: only F2 is 30, only F2 and F3 is 26 (fact 6), only F2 and F4 is 45 (fact 7), and all four is 40. F2 lies in three of the four triples, so $30 + b + 26 + 45 + 3x + 40 = 313$, giving $b + 3x = 172$. F1 and F2 together lie in two triples, so $b + 2x + 40 = 162$, giving $b + 2x = 122$. Subtracting, $x = 50$ and $b = 22$. The fully solved diagram: Schools with both F2 and F4 are: only F2 and F4 (45), the two triples that contain both, F1F2F4 and F2F3F4 ($50 + 50$), and all four (40). That is $45 + 100 + 40 = 185$. Option D (85) leaves out the two triples. Hence, option A (185).

Q49TITASet Theory

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

Answer and solution

Answer: 42

Let each of the four exactly-three regions hold $x$ schools (fact 4), and let only F1 and F2 hold $b$. From the data: only F2 is 30, only F2 and F3 is 26 (fact 6), only F2 and F4 is 45 (fact 7), and all four is 40. F2 lies in three of the four triples, so $30 + b + 26 + 45 + 3x + 40 = 313$, giving $b + 3x = 172$. F1 and F2 together lie in two triples, so $b + 2x + 40 = 162$, giving $b + 2x = 122$. Subtracting, $x = 50$ and $b = 22$. Let only F1 and F3 be $c$ and only F1 and F4 be $d$. Only F1 is 25, only F4 is 20 and only F3 and F4 is 24 (fact 7). F1 and F4 have equal totals (fact 9): $25 + 22 + c + d + 3x + 40 = 20 + 45 + 24 + d + 3x + 40$. Cancelling $d$ and $3x$ leaves $87 + c = 129$, so $c = 42$. The answer is 42.

Q50TITASet Theory

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

Answer and solution

Answer: 20

Let each exactly-three region hold $x$ schools (fact 4) and only F1 and F2 hold $b$. Only F2 is 30, only F2 and F3 is 26 (fact 6), only F2 and F4 is 45 (fact 7), and all four is 40. F2 lies in three triples: $30 + b + 26 + 45 + 3x + 40 = 313$, so $b + 3x = 172$. F1 and F2 share two triples: $b + 2x + 40 = 162$, so $b + 2x = 122$. Hence $x = 50$ and $b = 22$. Let only F1 and F3 be $c$ and only F1 and F4 be $d$. F1 and F4 have equal totals (fact 9); with only F1 25, only F4 20 and only F3 and F4 24: $25 + 22 + c + d + 3x + 40 = 20 + 45 + 24 + d + 3x + 40$, so $87 + c = 129$ and $c = 42$. All 600 schools: none 80; one facility $25 + 30 + 26 + 20 = 101$; two facilities $22 + 42 + d + 26 + 45 + 24 = 159 + d$; three 200; four 40. So $580 + d = 600$ and $d = 20$. The answer is 20.