CAT 2022 Slot 1 — DILR questions with answers

Q25MCQLogical Puzzles

1. A3
2. B2
3. C1
4. DCannot be determined

Answer and solution

Answer: (C) 1

Four different positive totals summing to 12: $(1,2,3,6)$ or $(1,2,4,5)$. Bimla scored in Match 1 and three other consecutive matches, so she has 4; Harita has more, so 5. Amla and Sarita have 1 and 2 in some order. Harita scored in exactly 3 matches, including 4 and 8. Even matches had one goal each, so she scored 1 in each and 3 in an odd match. Bimla's run avoids Matches 4 and 8, so it is 5, 6, 7. Let Matches 3 and 7 have $a$ goals each, and Matches 1 and 5 have $b$ and $c$; odd matches hold $12-4=8$ goals, so $2a+b+c=8$. Harita's 3 goals cannot be in Match 7 ($2a\ge8$ leaves nothing for $b,c$), Match 3 ($a=3$, $b=c=1$: Matches 3 and 7 tie as highest) or Match 5 (it would be highest). So they are in Match 1: $b\ge4$, so $a=1$, $b=4$, $c=2$. Amla and Sarita score one goal each in Matches 2, 3 and 5. So Match 7 had only Bimla's goal. Option B (2) is Match 5's total, and option D fails because the number of goals in every match, including Match 7, is fixed. Hence, option C (1).

Q26MCQLogical Puzzles

1. A5, 1, 0, 1
2. B3, 1, 2, 1
3. C3, 2, 1, 2
4. D4, 1, 2, 1

Answer and solution

Answer: (D) 4, 1, 2, 1

Four different positive totals summing to 12: $(1,2,3,6)$ or $(1,2,4,5)$. Bimla scored in Match 1 and three other consecutive matches, so she has 4; Harita has more, so 5. Amla and Sarita have 1 and 2 in some order. Harita scored in exactly 3 matches, including 4 and 8. Even matches had one goal each, so she scored 1 in each and 3 in an odd match. Bimla's run avoids Matches 4 and 8, so it is 5, 6, 7. Let Matches 3 and 7 have $a$ goals each, and Matches 1 and 5 have $b$ and $c$; odd matches hold $12-4=8$ goals, so $2a+b+c=8$. Harita's 3 goals cannot be in Match 7 ($2a\ge8$ leaves nothing for $b,c$), Match 3 ($a=3$, $b=c=1$: Matches 3 and 7 tie as highest) or Match 5 (it would be highest). So they are in Match 1: $b\ge4$, so $a=1$, $b=4$, $c=2$. Amla and Sarita score one goal each in Matches 2, 3 and 5. So Matches 1, 3, 5 and 7 had 4, 1, 2 and 1 goals. Option B (3, 1, 2, 1) counts only Harita's 3 goals in Match 1 and misses Bimla's. Hence, option D (4, 1, 2, 1).

Q27MCQLogical Puzzles

1. AStatement-1 only
2. BStatement-2 only
3. CBoth the statements
4. DNone of the statements

Answer and solution

Answer: (C) Both the statements

Four different positive totals summing to 12: $(1,2,3,6)$ or $(1,2,4,5)$. Bimla scored in Match 1 and three other consecutive matches, so she has 4; Harita has more, so 5. Amla and Sarita have 1 and 2 in some order. Harita scored in exactly 3 matches, including 4 and 8. Even matches had one goal each, so she scored 1 in each and 3 in an odd match. Bimla's run avoids Matches 4 and 8, so it is 5, 6, 7. Let Matches 3 and 7 have $a$ goals each, and Matches 1 and 5 have $b$ and $c$; odd matches hold $12-4=8$ goals, so $2a+b+c=8$. Harita's 3 goals cannot be in Match 7 ($2a\ge8$ leaves nothing for $b,c$), Match 3 ($a=3$, $b=c=1$: Matches 3 and 7 tie as highest) or Match 5 (it would be highest). So they are in Match 1: $b\ge4$, so $a=1$, $b=4$, $c=2$. Amla and Sarita score one goal each in Matches 2, 3 and 5. Statement 1: Matches 2, 3 and 5 each have exactly one goal from Amla or Sarita, so they never scored together. Statement 2: Harita scored only in Matches 1, 4 and 8, where Sarita did not. Both are true, so option A fails. Hence, option C (Both the statements).

Q28MCQLogical Puzzles

1. ANone of the statements
2. BStatement-1 only
3. CBoth the statements
4. DStatement-2 only

Answer and solution

Answer: (A) None of the statements

Four different positive totals summing to 12: $(1,2,3,6)$ or $(1,2,4,5)$. Bimla scored in Match 1 and three other consecutive matches, so she has 4; Harita has more, so 5. Amla and Sarita have 1 and 2 in some order. Harita scored in exactly 3 matches, including 4 and 8. Even matches had one goal each, so she scored 1 in each and 3 in an odd match. Bimla's run avoids Matches 4 and 8, so it is 5, 6, 7. Let Matches 3 and 7 have $a$ goals each, and Matches 1 and 5 have $b$ and $c$; odd matches hold $12-4=8$ goals, so $2a+b+c=8$. Harita's 3 goals cannot be in Match 7 ($2a\ge8$ leaves nothing for $b,c$), Match 3 ($a=3$, $b=c=1$: Matches 3 and 7 tie as highest) or Match 5 (it would be highest). So they are in Match 1: $b\ge4$, so $a=1$, $b=4$, $c=2$. Amla and Sarita score one goal each in Matches 2, 3 and 5. Statement 1: every match had at least one goal, so it is true. Statement 2: Bimla scored in 4 matches, Harita in 3, and Amla and Sarita in 1 and 2, all different, so it is true. Neither is false; option D fails because Statement 2 holds. Hence, option A (None of the statements).

Q29MCQLogical Puzzles

1. AStatement-2 only
2. BNone of the statements
3. CStatement-1 only
4. DBoth the statements

Answer and solution

Answer: (B) None of the statements

Four different positive totals summing to 12: $(1,2,3,6)$ or $(1,2,4,5)$. Bimla scored in Match 1 and three other consecutive matches, so she has 4; Harita has more, so 5. Amla and Sarita have 1 and 2 in some order. Harita scored in exactly 3 matches, including 4 and 8. Even matches had one goal each, so she scored 1 in each and 3 in an odd match. Bimla's run avoids Matches 4 and 8, so it is 5, 6, 7. Let Matches 3 and 7 have $a$ goals each, and Matches 1 and 5 have $b$ and $c$; odd matches hold $12-4=8$ goals, so $2a+b+c=8$. Harita's 3 goals cannot be in Match 7 ($2a\ge8$ leaves nothing for $b,c$), Match 3 ($a=3$, $b=c=1$: Matches 3 and 7 tie as highest) or Match 5 (it would be highest). So they are in Match 1: $b\ge4$, so $a=1$, $b=4$, $c=2$. Amla and Sarita score one goal each in Matches 2, 3 and 5. Harita scored in 3 matches, so Sarita scored in 2 (2 goals) and Amla in 1. Amla scored in one match only, so Statement 1 is false. Sarita's matches may be 2 and 3, or 2 and 5, or 3 and 5, so Statement 2 need not hold; option A fails. Hence, option B (None of the statements).

Q30TITASet Theory

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

Answer and solution

Answer: 50

Of the 6 singers, 4 are boys and 2 girls; of the 10 dancers, 6 are boys and 4 girls. With $x$ boys, $0.6x$ want the 2-day event, and $70\%$ of them, $0.42x$, are neither singers nor dancers. That leaves $0.18x$ singers or dancers. All 4 male singers want the 3-day event, so the 2-day event too. The 2 dancers wanting the 3-day event must be boys, as no girl wants it. So $0.18x-4$ male dancers want the 2-day event, which is 2 to 6: $6\le0.18x\le10$, so $33.3\le x\le55.6$. $0.18x=\frac{9x}{50}$ must be a whole number, so $x$ is a multiple of 50. The only such value in range is $x=50$. Then 40 boys want the 1-day event and 30 the 2-day event (21 neither, 4 singers, 5 dancers). The answer is 50.

Q31MCQSet Theory

1. AOnly I
2. BNeither I nor II
3. COnly II
4. DBoth I and II

Answer and solution

Answer: (C) Only II

Boys include 4 singers and 6 dancers, since 2 singers and 4 dancers are girls. With $x$ boys, $0.6x$ want the 2-day event and $0.42x$ of these are neither, leaving $0.18x$: the 4 male singers (they want the 3-day event) plus $0.18x-4$ male dancers. The 2 dancers wanting the 3-day event are boys, so this is 2 to 6, giving $6\le0.18x\le10$. As $0.18x=\frac{9x}{50}$ is whole, $x=50$: 40 boys want the 1-day event and 30 the 2-day event, 5 of them dancers. I: the 40 boys wanting the 1-day event are the 4 singers, the 5 dancers above, possibly the sixth dancer, and 30 or 31 boys who are neither. Nothing fixes which, so I cannot be determined. II: every girl wants the 1-day event, so all 4 female dancers do. II is determined. Option D fails because I is not fixed. Hence, option C (Only II).

Q32MCQSet Theory

1. A7/10
2. B7/13
3. C9/13
4. D2/3

Answer and solution

Answer: (B) 7/13

Boys include 4 singers and 6 dancers, since 2 singers and 4 dancers are girls. With $x$ boys, $0.6x$ want the 2-day event and $0.42x$ of these are neither, leaving $0.18x$: the 4 male singers (they want the 3-day event) plus $0.18x-4$ male dancers. The 2 dancers wanting the 3-day event are boys, so this is 2 to 6, giving $6\le0.18x\le10$. As $0.18x=\frac{9x}{50}$ is whole, $x=50$: 40 boys want the 1-day event and 30 the 2-day event, 5 of them dancers. Let $a$ girls want the 2-day event, with $0<a<15$ as some girls do and some do not. $0.6a$ of them are neither, so $a$ is a multiple of 5; the other $0.4a$ are $s$ singers and $d$ dancers. Statement 5 gives $4+s=(5+d)+1$, so $s=d+2$. Only 2 girls sing, so $s=2$, $d=0$, $0.4a=2$ and $a=5$. So $30+5=35$ of the $50+15=65$ students want the 2-day event: $\frac{35}{65}=\frac{7}{13}$. Option C would need 45 such students. Hence, option B (7/13).

Q33MCQSet Theory

1. A5 or 6
2. B6
3. C5
4. D4 or 6

Answer and solution

Answer: (A) 5 or 6

Boys include 4 singers and 6 dancers, since 2 singers and 4 dancers are girls. With $x$ boys, $0.6x$ want the 2-day event and $0.42x$ of these are neither, leaving $0.18x$: the 4 male singers (they want the 3-day event) plus $0.18x-4$ male dancers. The 2 dancers wanting the 3-day event are boys, so this is 2 to 6, giving $6\le0.18x\le10$. As $0.18x=\frac{9x}{50}$ is whole, $x=50$: 40 boys want the 1-day event and 30 the 2-day event, 5 of them dancers. Anyone wanting the 2-day event also wants the 1-day event, so these 5 male dancers want the 1-day event. The sixth male dancer may or may not: the other boys wanting the 1-day event can be 30 or 31 who are neither, and no condition decides it. So the number is 5 or 6. Options B and C each claim a single value the data does not fix. Hence, option A (5 or 6).

Q34MCQSet Theory

1. A2
2. B1
3. C0
4. DCannot be determined

Answer and solution

Answer: (C) 0

Boys include 4 singers and 6 dancers, since 2 singers and 4 dancers are girls. With $x$ boys, $0.6x$ want the 2-day event and $0.42x$ of these are neither, leaving $0.18x$: the 4 male singers (they want the 3-day event) plus $0.18x-4$ male dancers. The 2 dancers wanting the 3-day event are boys, so this is 2 to 6, giving $6\le0.18x\le10$. As $0.18x=\frac{9x}{50}$ is whole, $x=50$: 40 boys want the 1-day event and 30 the 2-day event, 5 of them dancers. Let $a$ girls want the 2-day event, with $0<a<15$ as some girls do and some do not. $0.6a$ of them are neither, so $a$ is a multiple of 5; the other $0.4a$ are $s$ singers and $d$ dancers. Statement 5 gives $4+s=(5+d)+1$, so $s=d+2$. Only 2 girls sing, so $s=2$, $d=0$, $0.4a=2$ and $a=5$. So no female dancer wants the 2-day event, and option B (1) is ruled out by $s=d+2\le2$. Hence, option C (0).

Q35TITALogical Puzzles

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

Answer and solution

Answer: 2

In thousands: $390=2\cdot3\cdot5\cdot13$, $210=2\cdot3\cdot5\cdot7$, $165=3\cdot5\cdot11$, $77=7\cdot11$ and $66=2\cdot3\cdot11$. All six values occur, so each interviewer uses a different one. Value 3 occurs 4 times; 2, 5 and 11 three times each; 7 twice; 13 once. Fathima gave 4 tokens, so her value is 3, and Qahira, whose amount alone lacks a 3, got Rs.77,000. Adhara gave one token, so her value is 13 and Pragnyaa got Rs.390,000. Rasheeda has the most tokens (four), so she got Rs.210,000. Esther gave her none, so Esther's value is not 2, 3, 5 or 7: it is 11. Dhanavi gave to Qahira ($7\times11$), so her value is 7, and Bithi and Chhaya have 2 and 5 in some order. Smera and Tantra got Rs.165,000 and Rs.66,000 in some order. Qahira's Rs.77,000 $=7\times11$ came from Dhanavi and Esther only. The answer is 2.

Q36MCQLogical Puzzles

1. APragnyaa
2. BRasheeda
3. CQahira
4. DTantra

Answer and solution

Answer: (A) Pragnyaa

In thousands: $390=2\cdot3\cdot5\cdot13$, $210=2\cdot3\cdot5\cdot7$, $165=3\cdot5\cdot11$, $77=7\cdot11$ and $66=2\cdot3\cdot11$. All six values occur, so each interviewer uses a different one. Value 3 occurs 4 times; 2, 5 and 11 three times each; 7 twice; 13 once. Fathima gave 4 tokens, so her value is 3, and Qahira, whose amount alone lacks a 3, got Rs.77,000. Adhara gave one token, so her value is 13 and Pragnyaa got Rs.390,000. Rasheeda has the most tokens (four), so she got Rs.210,000. Esther gave her none, so Esther's value is not 2, 3, 5 or 7: it is 11. Dhanavi gave to Qahira ($7\times11$), so her value is 7, and Bithi and Chhaya have 2 and 5 in some order. Smera and Tantra got Rs.165,000 and Rs.66,000 in some order. Dhanavi (7) gave only to Qahira and Rasheeda. Pragnyaa's amount has both 2 and 5, so she got Bithi's token whatever Bithi's value, and it has no 7. Rasheeda (option B) got Dhanavi's token; Qahira has neither 2 nor 5; Bithi's value is the one in Smera's amount, which Tantra's amount lacks. Hence, option A (Pragnyaa).

Q37TITALogical Puzzles

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

Answer and solution

Answer: 3

In thousands: $390=2\cdot3\cdot5\cdot13$, $210=2\cdot3\cdot5\cdot7$, $165=3\cdot5\cdot11$, $77=7\cdot11$ and $66=2\cdot3\cdot11$. All six values occur, so each interviewer uses a different one. Value 3 occurs 4 times; 2, 5 and 11 three times each; 7 twice; 13 once. Fathima gave 4 tokens, so her value is 3, and Qahira, whose amount alone lacks a 3, got Rs.77,000. Adhara gave one token, so her value is 13 and Pragnyaa got Rs.390,000. Rasheeda has the most tokens (four), so she got Rs.210,000. Esther gave her none, so Esther's value is not 2, 3, 5 or 7: it is 11. Dhanavi gave to Qahira ($7\times11$), so her value is 7, and Bithi and Chhaya have 2 and 5 in some order. Smera and Tantra got Rs.165,000 and Rs.66,000 in some order. Chhaya's value is 2 or 5. Value 2 occurs in 390, 210 and 66; value 5 in 390, 210 and 165. Either way she awarded 3 tokens. The answer is 3.

Q38TITALogical Puzzles

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

Answer and solution

Answer: 3

In thousands: $390=2\cdot3\cdot5\cdot13$, $210=2\cdot3\cdot5\cdot7$, $165=3\cdot5\cdot11$, $77=7\cdot11$ and $66=2\cdot3\cdot11$. All six values occur, so each interviewer uses a different one. Value 3 occurs 4 times; 2, 5 and 11 three times each; 7 twice; 13 once. Fathima gave 4 tokens, so her value is 3, and Qahira, whose amount alone lacks a 3, got Rs.77,000. Adhara gave one token, so her value is 13 and Pragnyaa got Rs.390,000. Rasheeda has the most tokens (four), so she got Rs.210,000. Esther gave her none, so Esther's value is not 2, 3, 5 or 7: it is 11. Dhanavi gave to Qahira ($7\times11$), so her value is 7, and Bithi and Chhaya have 2 and 5 in some order. Smera and Tantra got Rs.165,000 and Rs.66,000 in some order. Smera got Rs.165,000 ($3\times5\times11$) or Rs.66,000 ($2\times3\times11$). Each is a product of three values, so either way she received 3 tokens. The answer is 3.

Q39MCQLogical Puzzles

1. ANeither (a) nor (b)
2. BOnly (b)
3. COnly (a)
4. DBoth (a) and (b)

Answer and solution

Answer: (D) Both (a) and (b)

In thousands: $390=2\cdot3\cdot5\cdot13$, $210=2\cdot3\cdot5\cdot7$, $165=3\cdot5\cdot11$, $77=7\cdot11$ and $66=2\cdot3\cdot11$. All six values occur, so each interviewer uses a different one. Value 3 occurs 4 times; 2, 5 and 11 three times each; 7 twice; 13 once. Fathima gave 4 tokens, so her value is 3, and Qahira, whose amount alone lacks a 3, got Rs.77,000. Adhara gave one token, so her value is 13 and Pragnyaa got Rs.390,000. Rasheeda has the most tokens (four), so she got Rs.210,000. Esther gave her none, so Esther's value is not 2, 3, 5 or 7: it is 11. Dhanavi gave to Qahira ($7\times11$), so her value is 7, and Bithi and Chhaya have 2 and 5 in some order. Smera and Tantra got Rs.165,000 and Rs.66,000 in some order. Bithi gave Smera a token, so Bithi's value is 5 if Smera got Rs.165,000 and 2 if she got Rs.66,000, with Chhaya taking the other. Both arrangements meet every condition, so Tantra could have got Rs.66,000 or Rs.165,000. Options B and C each miss a valid case. Hence, option D (Both (a) and (b)).

Q40MCQNetworks & Routes

1. A9:11 am
2. B9:06 am
3. C9:01 am
4. D9:13 am

Q41MCQNetworks & Routes

1. A11:12 am
2. B11:22 am
3. C11:07 am
4. D11:28 am

Q42MCQNetworks & Routes

1. A11:39 pm
2. B11:49 am
3. C11:35 pm
4. D11:43 pm

Q43TITANetworks & Routes

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

Q44TITANetworks & Routes

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