CAT 2017 Slot 1 — DILR questions with answers

Q35MCQTeam Selection & Scheduling

1. A10:17
2. B10:10
3. C10:15
4. D10:20

Q36MCQTeam Selection & Scheduling

1. A10:35
2. B10:22
3. C10:25
4. D10:17

Q37MCQTeam Selection & Scheduling

1. A10:10
2. B10:12
3. C10:15
4. D10:17

Q38MCQTeam Selection & Scheduling

1. A7
2. B10
3. C15
4. D23

Q39MCQTables & Caselets

1. A37%
2. B6%
3. C79%
4. D56%

Answer and solution

Answer: (A) 37%

S has $200 \times 50 = 10000$ surveyed kids. G: condition 1 says 60% of the kids in S were in G, which is 6000. O: the table shows 300 kids in S who were not in school and whose mothers dropped out. Condition 1 also says every kid in S whose mother completed primary education was in school. So S has exactly 300 kids in O. P: $10000 - 6000 - 300 = 3700$ kids, which is 37% of the kids from S. The completed table for all 30000 kids shows this in the S row: Option C (79%) is the share in P only among kids whose mothers completed primary education. S has $10000 - 5700 = 4300$ such kids, of whom $6000 - 5100 = 900$ are in G and none in O, so 3400 are in P, and $\frac{3400}{4300} \approx 0.79$. The question asks about all kids from S. Hence, option A (37%).

Q40MCQTables & Caselets

1. A300
2. B1200
3. C1050
4. D1500

Answer and solution

Answer: (A) 300

We need the kids in W whose mothers completed primary education and who were not in school (O). All regions: 8% of 30000 kids, that is 2400, were in O. The table shows 1800 of them had mothers who dropped out, so $2400 - 1800 = 600$ O kids had mothers who completed primary education. NE: condition 2 says half of NE's O kids had mothers who dropped out. The table gives 300 such kids, so NE has 600 O kids, and 300 of them had mothers who completed primary education. S: condition 1 says every kid in S whose mother completed primary education was in school, so S has none. W: $600 - 300 - 0 = 300$. The completed table for kids whose mothers completed primary education shows this in the O column: Option B (1200) is the number of W's O kids whose mothers dropped out, straight from the table, and option D (1500) counts all O kids in W. Neither is the group asked for. Hence, option A (300).

Q41MCQTables & Caselets

1. A6000
2. B5250
3. C6750
4. D6300

Answer and solution

Answer: (A) 6000

Out-of-school (O) kids: 8% of 30000 is 2400. NE has 600, since the table's 300 is half of them (condition 2). S has only the table's 300, because in S every kid whose mother completed primary education was in school. W has $2400 - 600 - 300 = 1500$. G before the follow-up: G in NE equals G in W, G in S is 6000 (60% of 10000), and G in all is 16500 (55% of 30000). So G in W is $\frac{16500 - 6000}{2} = 5250$. The completed table for all kids shows these counts: Half of the 2400, that is 1200, joined G. The rates 25%, 100% and 50% go to the three regions in some order. W cannot be the 100% region, because 1500 alone exceeds 1200. If NE were, the total would be $600 + 375 + 150 = 1125$ or $600 + 750 + 75 = 1425$. So S is the 100% region. Then NE at 25% and W at 50% give $150 + 750 + 300 = 1200$, while the other order gives $300 + 375 + 300 = 975$. So 750 of W's O kids joined G. G in W now: $5250 + 750 = 6000$. Option B (5250) forgets the new joiners. Hence, option A (6000).

Q42MCQTables & Caselets

1. A94.7%
2. B89.5%
3. C93.4%
4. DCannot be determined from the given information

Answer and solution

Answer: (A) 94.7%

First find which region sent all its out-of-school (O) kids to G. O kids: 8% of 30000 is 2400. NE has 600, since the table's 300 is half of them (condition 2). S has 300, because in S every kid whose mother completed primary education was in school. So W has 1500, as the completed table for all kids shows: Half of 2400, that is 1200, joined G. W cannot be the 100% region, since $1500 > 1200$. If NE were, the total would be $600 + 375 + 150 = 1125$ or $600 + 750 + 75 = 1425$. So S is the 100% region, with NE at 25% and W at 50%: $150 + 750 + 300 = 1200$. The other order gives 975. In S, 5700 kids had mothers who dropped out: 5100 in G, 300 in P and 300 in O (from the table). All 300 of those O kids joined G, so G now has $5100 + 300 = 5400$ of them. So the share is $\frac{5400}{5700} \approx 0.947$, that is 94.7%. Option B (89.5%) is $\frac{5100}{5700}$, the share before the follow-up. Option D fails because only one region-to-rate match works. Hence, option A (94.7%).

Q43MCQSet Theory

1. A3 or 10
2. B10
3. C5
4. D7 or 10

Answer and solution

Answer: (A) 3 or 10

Label the 200 as in the diagram: a, b, c are those at or above the 80th percentile in P only, M only and C only, with $a = b = c$; d, e, f are those in exactly P and M, P and C, M and C; g is all three; n is 0. As $d + e + f = 150$, we get $3a + g = 50$, so $a \le 16$. Those below 80 in P, C and M number $2a + f$, $2a + d$ and $2a + e$, in the ratio 4:2:1. With $2a + e = k$, these are $4k$, $2k$, $k$; adding gives $6a + 150 = 7k$. Only $a = 3$ or $a = 10$ makes $k$ whole. The separate test is for those at or above 80 in P who are not in the AET. Among the 200, the AET takes everyone at or above 80 in two or more sections, so only those at or above 80 in P alone are left: $a$. So the number is 3 or 10. Option B (10) keeps only one case, and the 7 in option D fails because $6 \times 7 + 150 = 192$ is not a multiple of 7. Hence, option A (3 or 10).

Q44TITASet Theory

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Answer and solution

Answer: 60

Label the 200 as in the diagram: a, b, c are those at or above the 80th percentile in P only, M only and C only, with $a = b = c$; d is P and M only, e is P and C only, f is M and C only; g is all three; n is 0. Then $d + e + f = 150$, so $3a + g = 50$ and $a \le 16$. The numbers below 80 in P, C and M are $2a + f$, $2a + d$ and $2a + e$, in the ratio 4:2:1. Put $2a + e = k$, $2a + d = 2k$ and $2a + f = 4k$. Adding gives $6a + 150 = 7k$, so $a = 3$ ($k = 24$) or $a = 10$ ($k = 30$). Then $g = 50 - 3a$ is 41 or 20. Since g is a multiple of 5, $g = 20$, so $a = 10$ and $k = 30$. This gives $d = 2k - 2a = 60 - 20 = 40$. The candidates at or above 80 in both P and M are those in P and M only plus those in all three: $d + g = 40 + 20 = 60$. The answer is 60.

Q45TITASet Theory

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Answer and solution

Answer: 170

Use the regions of the diagram for the 200: a, b, c are those at or above the 80th percentile only in P, only in M and only in C, with $a = b = c$ (condition 3); d, e, f are those in exactly two sections; g is all three; n is 0 (condition 1). Condition 2 gives $d + e + f = 150$, so $3a + g = 50$ and $a \le 16$. The numbers below 80 in P, C and M are $2a + f$, $2a + d$ and $2a + e$, in the ratio 4:2:1. Setting $2a + e = k$ and adding gives $6a + 150 = 7k$, so $a = 3$ or $a = 10$, and $g = 50 - 3a$ is 41 or 20. Since g is a multiple of 5, $g = 20$. The AET needs at or above the 90th percentile overall, which all 200 are, and at or above the 80th percentile in at least two sections. So the shortlist is everyone in exactly two sections plus everyone in all three: $d + e + f + g = 150 + 20 = 170$. The answer is 170.

Q46MCQSet Theory

1. A299
2. B310
3. C321
4. D330

Answer and solution

Answer: (A) 299

Label the 200 as in the diagram: a, b, c are those at or above the 80th percentile in P only, M only and C only, with $a = b = c$; d, e, f are those in exactly P and M, P and C, M and C; g is all three; n is 0. Then $d + e + f = 150$ and $3a + g = 50$. Those below 80 in P, C and M, $2a + f$, $2a + d$ and $2a + e$, are in the ratio 4:2:1. With $2a + e = k$, adding gives $6a + 150 = 7k$, so $a = 3$ or $10$. If $a = 3$: $k = 24$, $e = 18$, $d = 42$, $g = 41$, so $a + d + e + g = 104$ are at or above 80 in P. If $a = 10$: $k = 30$, $e = 10$, $d = 40$, $g = 20$, giving 80. Only 104 exceeds 100, so $a = 3$. The $d + e + g = 101$ AET candidates at or above 80 in P use their AET score. The other $400 - 101 = 299$ take the separate test. Option D (330) is $400 - 70$, from the ruled-out case. Hence, option A (299).

Q47TITABar & Line Charts

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Q48TITABar & Line Charts

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Q49MCQBar & Line Charts

1. A14
2. B15
3. C16
4. D17

Q50MCQBar & Line Charts

1. A0
2. B1
3. C2
4. D3

Q51MCQTeam Selection & Scheduling

1. A(2,1)
2. B(1,0)
3. C(0,0)
4. D(1,1)

Q52MCQTeam Selection & Scheduling

1. A(0,2)
2. B(0,3)
3. C(1,2)
4. D(1,3)

Answer and solution

Answer: (A) (0,2)

Teams have 4 members, except the challenging team, which has 5. In month 1 the SE counts fall by one from T1 to T5 and total 10, so they are 4, 3, 2, 1, 0, and the RE counts are 1, 1, 2, 3, 4. Month 1 to 2: T1 gives 2 SE to T2 and gets 1 RE back, leaving T1 with 2 SE and 2 RE. Rule b applies, since T1 has SE and T5 has RE, so T1 sends 1 SE to T5 for 1 RE. Now T1 has (1 SE, 3 RE) and T5 has (1 SE, 3 RE). Month 2 to 3: rule a moves SE from T2 to T3 and does not touch T1 or T5. Rule b applies again, since T1 has 1 SE and T5 has 3 RE. So T1 has (0 SE, 4 RE) and T5 has (2 SE, 2 RE). The table shows every team in month 3: So in month 3, T1 has 0 SE and T5 has 2 SE. T1 has lost its last SE, which rules out options C and D, and T5 gained exactly one SE at each change, so it has 2, not 3, which rules out option B. Hence, option A ((0,2)).

Q53MCQTeam Selection & Scheduling

1. A140
2. B150
3. C170
4. D200

Q54MCQTeam Selection & Scheduling

1. AAneek worked only in teams T1, T2, T3, and T4.
2. BAneek worked only in teams T1,T2, T4, and T5.
3. CAneek worked only in teams T2,T3, T4, and T5.
4. DAneek worked only in teams T1,T3, T4, and T5.

Q55MCQLogical Puzzles

1. A3
2. B5
3. C7
4. D8

Answer and solution

Answer: (C) 7

For each individual, count who can reach them: look left, right, up and down, and count a person only if their platform is lower and every platform in between is lower still (condition (iii)). The grid below gives each platform's height with, in brackets, the number of individuals who can reach it: Exactly seven individuals are reached by just one person (positions given as row, column): the 2 at (1, 3), by the 1 on its left; the 7 at (3, 1), by the 3 below it; the 4 at (3, 3), by the 3 above it; the 5 at (3, 5), by the 2 below it; the 3 at (4, 1), by the 1 below it; the 2 at (4, 5), by the 1 on its left; and the 3 at (5, 4), by the 1 above it. Option A (3) is what you get by ignoring condition (iii). For example, the 6 at (1, 1) is lower than the 7 at (3, 1) but cannot reach it, because the 9 between them is higher than 6. Counting such blocked people adds reachers and leaves only 3 individuals with a single reacher. Hence, option C (7).

Q56MCQLogical Puzzles

1. AThey can be reached by all the individuals in their own row and column.
2. BThey can be reached by at least 4 individuals.
3. CThey can be reached by at least one individual.
4. DThey cannot be reached by anyone.

Q57MCQLogical Puzzles

1. Athe last row.
2. Bthe fourth row.
3. Cthe fourth column.
4. Dthe middle column.

Q58MCQLogical Puzzles

1. AEach row has an individual who can be reached by 5 or more individuals.
2. BEach row has an individual who cannot be reached by anyone.
3. CEach row has at least two individuals who can be reached by an equal number of individuals.
4. DAll individuals at the height of 9 m can be reached by at least 5 individuals.

Q59MCQNetworks & Routes

1. A45
2. B90
3. C180
4. D135

Q60MCQNetworks & Routes

1. A54
2. B120
3. C96
4. D60

Q61TITANetworks & Routes

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Q62TITANetworks & Routes

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Q63TITANetworks & Routes

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Q64MCQNetworks & Routes

1. A1
2. B0.1
3. C0.2
4. D0.9

Q65TITANetworks & Routes

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Q66MCQNetworks & Routes

1. A26
2. B32
3. C29.9
4. D30