CAT 2018 Slot 1 — DILR questions with answers

Q35MCQLogical Puzzles

1. AChitra and Daisy
2. BAdriana and Bandita
3. CBandita and Chitra
4. DAdriana and Daisy

Answer and solution

Answer: (C) Bandita and Chitra

Z has two students, both female. Clue 2 speaks of both the male students from Y and the female student from Y, so Y has two males and one female, and X has the other two males and one female. Barun is in Y and Chetan in X, so Amit and Deb take one place each in X and Y. Daisy is in the same institute as Amit, a male, so she is not in Z. She minors in Operations, so she cannot be Y's female, who majors in Operations. Hence Daisy and Amit are in X, Deb is in Y, and Adriana, who is with Deb, is Y's female. The two students left, Bandita and Chitra, form institute Z, as the table built from the clues shows: Every other option names Adriana or Daisy. Option D (Adriana and Daisy) fails because Adriana is in Y with Deb and Daisy is in X with Amit. Hence, option C (Bandita and Chitra).

Q36MCQLogical Puzzles

1. AOperations
2. BFinance
3. CMarketing
4. DCannot be determined uniquely from the given information

Answer and solution

Answer: (B) Finance

Z has two students, both female. Clue 2 speaks of both the male students from Y and the female student from Y, so Y has two males and one female, and X has the other two males and one female. Barun is in Y and Chetan in X, so Amit and Deb take one place each in X and Y. Daisy is with Amit, so she is not in Z. She minors in Operations, so she cannot be Y's female, who majors in Operations. So Daisy and Amit are in X, and Deb is the second male in Y. Both males from Y minor in Finance, so Deb minors in Finance. His major is then Marketing: he cannot major in his minor, and Barun is the only male who majors in Operations. The table built from the clues shows this: Option D (cannot be determined) fails because the clues fix Deb's institute, and with it his minor. Hence, option B (Finance).

Q37MCQLogical Puzzles

1. AMarketing
2. BOperations
3. CCannot be determined uniquely from the given information
4. DFinance

Answer and solution

Answer: (D) Finance

Z has two students, both female. Clue 2 speaks of both the male students from Y and the female student from Y, so Y has two males and one female, and X has the other two males and one female. Barun is in Y and Chetan in X, so Amit and Deb take one place each in X and Y. Daisy is with Amit, so she is not in Z. She minors in Operations, so she cannot be Y's female, who majors in Operations. So Amit and Daisy are in X, and Deb is in Y. Two males major in Finance. Barun majors in Operations, and Deb, as a male from Y, minors in Finance and so cannot also major in it. The two male Finance majors must therefore be Chetan and Amit, as the table built from the clues shows: Option B (Operations) fails because only one male majors in Operations, and that is Barun. Hence, option D (Finance).

Q38MCQLogical Puzzles

1. AFinance
2. BCannot be determined uniquely from the given information
3. COperations
4. DMarketing

Answer and solution

Answer: (C) Operations

Three of the four female students minor in Marketing. Daisy minors in Operations, so the other three, Adriana, Bandita and Chitra, all minor in Marketing, as the table built from the clues shows: A student's major and minor are different subjects, so Bandita cannot major in Marketing. Exactly one female student majors in Finance. If that student is Chitra, Bandita cannot major in Finance either. The only subject left for Bandita is Operations. Option B (cannot be determined) is the tempting choice, but the Marketing minor and the single female Finance major leave Bandita just one possible major. Hence, option C (Operations).

Q39TITALogical Puzzles

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

Q40TITALogical Puzzles

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

Q41MCQLogical Puzzles

1. A12
2. B10
3. C13
4. D16

Q42MCQLogical Puzzles

1. A900
2. B800
3. C750
4. D1400

Q43TITALogical Puzzles

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

Answer and solution

Answer: 4

Any $2 \times 2$ block of cells contains four cells that all touch one another, horizontally, vertically or diagonally. A $3 \times 3$ matrix contains such blocks, and the four cells of a block need four different numerals. So at least 4 numerals are needed. Four numerals are enough. Fill the three rows as 1 2 1, then 3 4 3, then 1 2 1: Within each row, neighbouring cells differ. The first and third rows use only 1 and 2, while the middle row uses only 3 and 4, so no cell touches a cell with the same numeral in the row above or below. The answer is 4.

Q44TITALogical Puzzles

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

Answer and solution

Answer: 4

Any $2 \times 2$ block of cells contains four cells that all touch one another, so those four cells need four different numerals. Hence a $5 \times 5$ matrix needs at least 4 numerals. Four are enough. Fill rows 1, 3 and 5 as 1 2 1 2 1, and rows 2 and 4 as 3 4 3 4 3: Within a row, neighbouring cells alternate, so they differ. Neighbouring rows use different pairs of numerals, 1 and 2 against 3 and 4, so no cell touches a cell with the same numeral above, below or diagonally. Rows 2 and 4 do not touch each other, so they can reuse 3 and 4. The answer is 4.

Q45MCQLogical Puzzles

1. A4
2. B16
3. C9
4. D25

Answer and solution

Answer: (A) 4

Without any mistake, a $5 \times 5$ matrix needs 4 numerals, and 4 are enough, for example with rows alternating 1 2 1 2 1 and 3 4 3 4 3: Allowing one mistake can only keep this number the same or lower it. It cannot lower it to 3. Every $2 \times 2$ block has four cells that all touch one another, so with only 3 numerals each block must contain a pair of touching cells with the same numeral. A $5 \times 5$ matrix has 16 such blocks, and one pair of touching cells lies in at most two of them. One allowed mistake can therefore cover at most 2 blocks, leaving at least 14 that break the rule. So the minimum is still 4. Options B, C and D (16, 9 and 25) are more numerals than the 4 that already suffice. Hence, option A (4).

Q46MCQLogical Puzzles

1. A25
2. B4
3. C16
4. D9

Answer and solution

Answer: (D) 9

Take the centre cell of the $5 \times 5$ matrix. Its 8 neighbours must all differ from one another under the new rule, and each must differ from the centre cell under the original rule. So the central $3 \times 3$ block needs 9 different numerals, as below, and at least 9 are required. Nine are enough. Fill row 1 as 1 2 3 1 2, row 2 as 4 5 6 4 5 and row 3 as 7 8 9 7 8, then repeat rows 1 and 2 as rows 4 and 5. Two cells with the same numeral are then 3 apart in their row numbers or in their column numbers. Two cells that share a neighbour are at most 2 apart in both, and touching cells at most 1 apart, so no rule is broken. So the minimum is 9. Option B (4) is the answer for the original rule only, and 16 or 25 use more numerals than needed. Hence, option D (9).

Q47MCQLogical Puzzles

1. AThe contamination level at P20 was recorded as medium.
2. BThe contamination level at P13 was recorded as low.
3. CThe contamination level at P12 was recorded as high.
4. DThe contamination level at P10 was recorded as high.

Answer and solution

Answer: (D) The contamination level at P10 was recorded as high.

Only P7 and P8 are equal neighbours, and P6 is the only low up to P10, so P1–P5, with three highs, read H-M-H-M-H. P16–P20 have no high, so they alternate low and medium, with at least 2 lows; with P6, $L \ge 3$, so $H = 2L \ge 6$. If P7 = P8 = M, then P9 = H and P10 = M: 4 highs, plus at most 3 in P11–P15. As $H$ is even, $H = 6$ and $L = 3$, so P11–P15 need 2 highs, 3 mediums and no low: M-H-M-H-M, which repeats P10 = M. So P7 = P8 = H, and P1–P10 read H-M-H-M-H-L-H-H-M-H. So P10 is always high. All six arrangements that fit the clues are shown below: The others need not hold. P11–P20 = M-H-M-H-M-L-M-L-M-L fits every clue (8 highs, 4 lows), yet P20 is low and P13 medium, so A and B fail. P11–P20 = L-M-H-M-H-M-L-M-L-M also fits, yet P12 is medium, so C fails. Hence, option D (The contamination level at P10 was recorded as high.).

Q48MCQLogical Puzzles

1. AAt least 8
2. BMore than 4
3. CExactly 8
4. DAt most 9

Answer and solution

Answer: (C) Exactly 8

Only P7 and P8 are equal neighbours, and P6 is the only low up to P10, so P1–P5, with three highs, read H-M-H-M-H. P16–P20 have no high, so they alternate low and medium, with at least 2 lows; with P6, $L \ge 3$, so $H = 2L \ge 6$. If P7 = P8 = M, then P9 = H and P10 = M: 4 highs, plus at most 3 in P11–P15. As $H$ is even, $H = 6$ and $L = 3$, so P11–P15 need 2 highs, 3 mediums and no low: M-H-M-H-M, which repeats P10 = M. So P7 = P8 = H, and P1–P10 read H-M-H-M-H-L-H-H-M-H. Up to P10 there are 6 highs; P11 cannot be high (P10 is), so P12–P15 add at most 2. So $H \le 8$ and $L = H/2 \le 4$. Up to P10 there are 3 mediums, and P11–P20 hold at most 5, as no two neighbours there match, so $M \le 8$. As $H + L + M = 20$, all three bounds are reached: exactly 8 mediums, as in each arrangement below. Options A, B and D are also true, but only C pins the number down. Hence, option C (Exactly 8).

Q49MCQLogical Puzzles

1. AThe contamination level at P12 was recorded as high.
2. BThe contamination level at P15 was recorded as medium.
3. CThe contamination level at P18 was recorded as low.
4. DThe contamination level at P14 was recorded as medium.

Answer and solution

Answer: (D) The contamination level at P14 was recorded as medium.

Only P7 and P8 are equal neighbours, and P6 is the only low up to P10, so P1–P5, with three highs, read H-M-H-M-H. P16–P20 have no high, so they alternate low and medium, with at least 2 lows; with P6, $L \ge 3$, so $H = 2L \ge 6$. If P7 = P8 = M, then P9 = H and P10 = M: 4 highs, plus at most 3 in P11–P15. As $H$ is even, $H = 6$ and $L = 3$, so P11–P15 need 2 highs, 3 mediums and no low: M-H-M-H-M, which repeats P10 = M. So P7 = P8 = H, and P1–P10 read H-M-H-M-H-L-H-H-M-H. Let P11 be low. Counting P6, P11 and at least 2 lows in P16–P20, $L \ge 4$, so $H \ge 8$. Only P12–P15 can add highs, at most 2, so $H = 8$ and $L = 4$. Then P16–P20 hold just 2 lows (M-L-M-L-M), and P12–P15, with no low, alternate high and medium; P15 cannot be medium beside P16 = M, so they read M-H-M-H. This is the first arrangement below: So P14 is medium, while P12 is medium (A fails), P15 is high (B fails) and P18 is medium (C fails). Hence, option D (The contamination level at P14 was recorded as medium.).

Q50MCQLogical Puzzles

1. AContamination levels at P13 and P17 were recorded as the same.
2. BContamination levels at P11 and P16 were recorded as the same.
3. CContamination level at P14 was recorded to be higher than that at P15.
4. DContamination levels at P10 and P14 were recorded as the same.

Answer and solution

Answer: (B) Contamination levels at P11 and P16 were recorded as the same.

Only P7 and P8 are equal neighbours, and P6 is the only low up to P10, so P1–P5, with three highs, read H-M-H-M-H. P16–P20 have no high, so they alternate low and medium, with at least 2 lows; with P6, $L \ge 3$, so $H = 2L \ge 6$. If P7 = P8 = M, then P9 = H and P10 = M: 4 highs, plus at most 3 in P11–P15. As $H$ is even, $H = 6$ and $L = 3$, so P11–P15 need 2 highs, 3 mediums and no low: M-H-M-H-M, which repeats P10 = M. So P7 = P8 = H, and P1–P10 read H-M-H-M-H-L-H-H-M-H. Let P15 be medium. Then P16 is not medium, so P16–P20 read L-M-L-M-L; with P6, $L \ge 4$, so $H \ge 8$. As P11 cannot be high (P10 is), P12–P15 add at most 2 highs, so $H = 8$, $L = 4$, and P11–P15, with no low, read M-H-M-H-M: So P11 is medium and P16 low, and B must be false. The rest hold: P13 and P17 are both medium (A), P14 is high, above P15 (C), and P10 and P14 are both high (D). Hence, option B (Contamination levels at P11 and P16 were recorded as the same.).

Q51MCQPie & Scatter Charts

1. A38.46
2. B22.22
3. C28.57
4. D50.00

Answer and solution

Answer: (C) 28.57

December 2016 is not shown, but October, November and December 2016 form an arithmetic progression. October is 100 and the fourth quarter of 2016 totals 360, so with common difference $d$: $100 + (100 + d) + (100 + 2d) = 360$, giving $d = 20$. So December 2016 $= 140$. December 2017 is not shown either. The fourth quarter of 2017 totals 500, with October 150 and November 170, so December 2017 $= 500 - 150 - 170 = 180$. The chart with the missing figures filled in: Percentage increase $= \frac{180 - 140}{140} \times 100 = \frac{40}{140} \times 100 \approx 28.57\%$. Option B (22.22) is the trap: it divides the increase by the 2017 figure, $\frac{40}{180}$, instead of by the 2016 base. Hence, option C (28.57).

Q52MCQPie & Scatter Charts

1. AQ2
2. BQ1
3. CQ4
4. DQ3

Answer and solution

Answer: (B) Q1

Compare each quarter of 2017 with the same quarter of 2016. The chart below has the missing figures filled in. Q1: 2016 $= 80 + 60 + 100 = 240$; 2017 $= 120 + 100 + 160 = 380$. Growth $= \frac{140}{240} \approx 58.3\%$. Q2: 2016 $= 150$ (given); 2017 $= 60 + 75 + 65 = 200$. Growth $= \frac{50}{150} \approx 33.3\%$. Q3: 2016 $= 75 + 120 + 55 = 250$; 2017 $= 220$ (given). Sales fell. Q4: 2016 $= 360$; 2017 $= 500$ (both given). Growth $= \frac{140}{360} \approx 38.9\%$. Option C (Q4) is the closest rival: it rose by 140 sets, the same as Q1, but from a larger base of 360 against 240, so its percentage growth is smaller. Hence, option B (Q1).

Q53MCQPie & Scatter Charts

1. AQ2 of 2017
2. BQ4 of 2017
3. CQ2 of 2016
4. DQ1 of 2017

Answer and solution

Answer: (A) Q2 of 2017

Only quarterly totals are needed. The chart below has the missing figures filled in. 2016: Q1 $= 80 + 60 + 100 = 240$, Q2 $= 150$, Q3 $= 75 + 120 + 55 = 250$, Q4 $= 360$. 2017: Q1 $= 120 + 100 + 160 = 380$, Q2 $= 60 + 75 + 65 = 200$, Q3 $= 220$, Q4 $= 500$. Sales fall from the previous quarter only twice: Q2 2016: $\frac{240 - 150}{240} \times 100 = 37.5\%$ decrease. Q2 2017: $\frac{380 - 200}{380} \times 100 \approx 47.4\%$ decrease. Every other quarter shows an increase, including Q1 2017 ($360 \to 380$) and Q4 2017 ($220 \to 500$), so options D and B cannot be the answer. Option C (Q2 of 2016) is the closest rival, but its fall is smaller. Hence, option A (Q2 of 2017).

Q54MCQPie & Scatter Charts

1. AMarch of 2017
2. BOctober of 2017
3. CMarch of 2016
4. DOctober of 2016

Answer and solution

Answer: (B) October of 2017

Read the monthly figures from the chart and compare each option with the month before it. The chart below has the missing figures filled in. March 2017: February 2017 $= 100$, March $= 160$, a rise of $\frac{60}{100} = 60\%$. October 2017: September 2017 $= 70$, October $= 150$, a rise of $\frac{80}{70} \approx 114.3\%$. March 2016: February 2016 $= 60$, March $= 100$, a rise of $\frac{40}{60} \approx 66.7\%$. October 2016: September 2016 $= 55$, October $= 100$, a rise of $\frac{45}{55} \approx 81.8\%$. Option D (October of 2016) is the closest rival, but only October 2017 more than doubles the previous month's sales. Hence, option B (October of 2017).

Q55MCQLogical Puzzles

1. AIn the teaching committee the number of educationalists is equal to the number of politicians
2. BIn the administration committee the number of bureaucrats is equal to the number of educationalists
3. CThe size of the research committee is less than the size of the teaching committee
4. DThe size of the research committee is less than the size of the administration committee

Answer and solution

Answer: (C) The size of the research committee is less than the size of the teaching committee

Let the administration committee have $4x$ bureaucrats, so research and teaching have $3x$ each (75% of $4x$). Let research have $y$ educationalists; as $y$ is the average of the other two and teaching has fewer, teaching has $y - d$ and administration $y + d$, with $d \ge 1$. Let there be $5z$ politicians: $3z$ in administration, $z$ in teaching and $z$ in research. In total, $10x + 3y + 5z = 24$. If $x \ge 2$, then $3y + 5z \le 4$, impossible since every type is present. So $x = 1$ and $3y + 5z = 14$, solved only by $y = 3$, $z = 1$. Teaching has $3 - d$ educationalists, so $d = 1$ or $2$. Sizes: research $3 + 3 + 1 = 7$, teaching $7 - d$ (6 or 5), administration $10 + d$ (11 or 12). In the table, entries like 6/5 cover $d = 1$ and $d = 2$. Research (7) is always bigger than teaching, so C must be false. A can hold: with $d = 2$, teaching has 1 educationalist and 1 politician. B can hold: with $d = 1$, administration has 4 bureaucrats and 4 educationalists. D always holds. Hence, option C (The size of the research committee is less than the size of the teaching committee).

Q56TITALogical Puzzles

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

Answer and solution

Answer: 4

Let the administration committee have $4x$ bureaucrats, so research and teaching have $3x$ each (75% of $4x$). Let research have $y$ educationalists; as $y$ is the average of the other two and teaching has fewer, teaching has $y - d$ and administration $y + d$, with $d \ge 1$. Let there be $5z$ politicians: $3z$ in administration, $z$ in teaching and $z$ in research. In total, $10x + 3y + 5z = 24$. If $x \ge 2$, then $3y + 5z \le 4$, impossible since every type is present. So $x = 1$ and $3y + 5z = 14$, solved only by $y = 3$, $z = 1$. So the administration committee has $4x = 4$ bureaucrats, and research and teaching have 3 each. The educationalists split as 3 (research), $3 - d$ (teaching) and $3 + d$ (administration) with $d = 1$ or $2$. Both values give three different committee sizes (7, 6, 11 or 7, 5, 12), but neither changes the bureaucrat count. In the table, entries like 6/5 cover $d = 1$ and $d = 2$. The answer is 4.

Q57TITALogical Puzzles

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

Answer and solution

Answer: 3

Let the administration committee have $4x$ bureaucrats, so research and teaching have $3x$ each (75% of $4x$). Let research have $y$ educationalists; as $y$ is the average of the other two and teaching has fewer, teaching has $y - d$ and administration $y + d$, with $d \ge 1$. Let there be $5z$ politicians: $3z$ in administration, $z$ in teaching and $z$ in research. In total, $10x + 3y + 5z = 24$. If $x \ge 2$, then $3y + 5z \le 4$, impossible since every type is present. So $x = 1$ and $3y + 5z = 14$, solved only by $y = 3$, $z = 1$. So the research committee has $y = 3$ educationalists. Teaching has $3 - d$ and administration $3 + d$, with $d = 1$ or $2$ since teaching needs at least one. The value of $d$ changes those two committees but not research. In the table, entries like 2/1 cover $d = 1$ and $d = 2$. The answer is 3.

Q58MCQLogical Puzzles

1. AThe size of the teaching committee
2. BThe size of the research committee
3. CThe total number of bureaucrats in the three committees
4. DThe total number of educationalists in the three committees

Answer and solution

Answer: (A) The size of the teaching committee

Let the administration committee have $4x$ bureaucrats, so research and teaching have $3x$ each (75% of $4x$). Let research have $y$ educationalists; as $y$ is the average of the other two and teaching has fewer, teaching has $y - d$ and administration $y + d$, with $d \ge 1$. Let there be $5z$ politicians: $3z$ in administration, $z$ in teaching and $z$ in research. In total, $10x + 3y + 5z = 24$. If $x \ge 2$, then $3y + 5z \le 4$, impossible since every type is present. So $x = 1$ and $3y + 5z = 14$, solved only by $y = 3$, $z = 1$. Teaching has $3 - d$ educationalists, so $d = 1$ or $2$, giving sizes: research $3 + 3 + 1 = 7$, teaching $7 - d$, administration $10 + d$. In the table, entries like 6/5 cover $d = 1$ and $d = 2$. Research is always 7, the bureaucrats total $4 + 3 + 3 = 10$, and the educationalists total $3 + (3 - d) + (3 + d) = 9$. Only the teaching committee, 6 or 5, changes with $d$. D is the tempting one: the educationalists in teaching and administration change, but their total stays 9. Hence, option A (The size of the teaching committee).

Q59MCQSet Theory

1. AMust be at least 100
2. BCannot be more than 800
3. CMust be between 450 and 725
4. DMust be between 400 and 800

Answer and solution

Answer: (C) Must be between 450 and 725

No satellite serving B, C or S also serves O. Let $10x$ satellites serve only B, so only C and only S have $3x$ each. Let $p$ serve only B and C, $q$ only B and S, $r$ only C and S, and 100 all three; O then has $r$ satellites. C and S are equal in size, so $p = q$. B is twice C, so $10x + 2p + 100 = 2(3x + p + r + 100)$, giving $r = 2x - 50$. The total, $16x + 2p + 2r + 100 = 1600$, then gives $p = 800 - 10x$. The diagram shows every region in terms of $x$. As $r \ge 0$ and $p \ge 0$, $25 \le x \le 80$. Also $n(C) = n(S) = 3x + p + r + 100 = 850 - 5x$. As $x$ runs from 25 to 80, $850 - 5x$ runs from 725 down to 450, and both ends are possible. So the number serving C must be between 450 and 725. Options A, B and D are also true but looser; D's range of 400 to 800 includes values that cannot occur. Hence, option C (Must be between 450 and 725).

Q60MCQSet Theory

1. A250
2. B100
3. C500
4. D200

Answer and solution

Answer: (A) 250

No satellite serving B, C or S also serves O. Let $10x$ satellites serve only B, so only C and only S have $3x$ each. Let $p$ serve only B and C, $q$ only B and S, $r$ only C and S, and 100 all three; O then has $r$ satellites. C and S are equal in size, so $p = q$. B is twice C, so $10x + 2p + 100 = 2(3x + p + r + 100)$, giving $r = 2x - 50$. The total, $16x + 2p + 2r + 100 = 1600$, then gives $p = 800 - 10x$. The diagram shows every region in terms of $x$. As $r \ge 0$ and $p \ge 0$, $25 \le x \le 80$. Satellites serving only B number $10x$, which is least when $x$ is least. $x = 25$ is allowed (then $r = 0$ and $p = 550$), giving $10 \times 25 = 250$. Smaller answers such as 200 or 100 need $x < 25$, which makes $r = 2x - 50$ negative. Hence, option A (250).

Q61MCQSet Theory

1. AAt most 475
2. BExactly 475
3. CNo conclusion is possible based on the given information
4. DAt least 475

Answer and solution

Answer: (A) At most 475

No satellite serving B, C or S also serves O. Let $10x$ satellites serve only B, so only C and only S have $3x$ each. Let $p$ serve only B and C, $q$ only B and S, $r$ only C and S, and 100 all three; O then has $r$ satellites. C and S are equal in size, so $p = q$. B is twice C, so $10x + 2p + 100 = 2(3x + p + r + 100)$, giving $r = 2x - 50$. The total, $16x + 2p + 2r + 100 = 1600$, then gives $p = 800 - 10x$. The diagram shows every region in terms of $x$. As $r \ge 0$ and $p \ge 0$, $25 \le x \le 80$. Also $n(C) = n(S) = 3x + p + r + 100 = 850 - 5x$. Now O has at least 100 satellites: $2x - 50 \ge 100$, so $x \ge 75$, and $75 \le x \le 80$. Then $n(S) = 850 - 5x$ goes from 475 (at $x = 75$) down to 450 (at $x = 80$). So the number serving S is at most 475. 'Exactly 475' holds only at $x = 75$, and 'at least 475' fails for any larger $x$. Hence, option A (At most 475).

Q62MCQSet Theory

1. AThe number of satellites serving B exclusively is exactly 250
2. BThe number of satellites serving B is more than 1000
3. CThe number of satellites serving C cannot be uniquely determined
4. DAll 1600 satellites serve B or C or S

Answer and solution

Answer: (C) The number of satellites serving C cannot be uniquely determined

No satellite serving B, C or S also serves O. Let $10x$ satellites serve only B, so only C and only S have $3x$ each. Let $p$ serve only B and C, $q$ only B and S, $r$ only C and S, and 100 all three; O then has $r$ satellites. C and S are equal in size, so $p = q$. B is twice C, so $10x + 2p + 100 = 2(3x + p + r + 100)$, giving $r = 2x - 50$. The total, $16x + 2p + 2r + 100 = 1600$, then gives $p = 800 - 10x$. As $r \ge 0$ and $p \ge 0$, $25 \le x \le 80$. Satellites in at least two of B, C and S number $2p + r + 100 = 1650 - 18x$. Setting this to 1200 gives $x = 25$: only-B $= 250$, only-C $=$ only-S $= 75$, $p = q = 550$, $r = 0$, and O is empty. So A and D hold, and so does B, as $n(B) = 250 + 1100 + 100 = 1450$. But $n(C) = 75 + 550 + 0 + 100 = 725$ is fixed, so C must be false. Hence, option C (The number of satellites serving C cannot be uniquely determined).

Q63MCQMissing Value Tables

1. ABoth 2 and 3
2. BOnly 1
3. COnly 2
4. DBoth 1 and 2

Answer and solution

Answer: (D) Both 1 and 2

Composite $= 2 \times \text{DI} + \text{WE} + \text{GA}$, and anyone with under 14 marks (70%) in two sections is disqualified. Jatin has WE 16 and GA 14, so his 100% section is DI and his composite is $40 + 16 + 14 = 70$. Indu's is then 60. Indu's WE is 8. If her 20 were in DI, her GA would be $60 - 40 - 8 = 12$, two sections below 14, but she was recruited. So her GA is 20 and her DI is $\frac{60 - 8 - 20}{2} = 16$. Danish has the same GA as Indu, 20, so his composite is $2 \times 8 + 15 + 20 = 51$. The completed table below shows these values; the WE marks of Ajay and Geeta in it are not needed here. Statement 1: 70 is more than 51, so it is true. Statement 2: Indu's DI of 16 is less than Chetna's 19, so it is true. Statement 3: Jatin's GA of 14 is less than Indu's 20, so it is false. Option A (Both 2 and 3) fails because of statement 3. Hence, option D (Both 1 and 2).

Q64MCQMissing Value Tables

1. ABala scored same as Jatin in DI
2. BHarini’s composite score was less than that of Falak
3. CBala’s composite score was less than that of Ester
4. DChetna scored more than Bala in DI

Answer and solution

Answer: (A) Bala scored same as Jatin in DI

Composite $= 2 \times \text{DI} + \text{WE} + \text{GA}$, and no two candidates share a composite. Jatin has WE 16 and GA 14, so his 20 is in DI and his composite is 70; Indu's is 60. Indu's WE is 8, and a DI of 20 would leave her GA at 12, two sections below 14, although she was recruited. So her GA is 20 (and her DI 16). The completed table below lists every known mark and composite. A: if Bala had Jatin's DI of 20, his composite would be $40 + 9 + 11 = 60$, the same as Indu's. That is not allowed, so A must be false. The others can be true. Harini shares Indu's GA of 20, so she scores $30 + \text{WE}$; with a WE of 12 she scores 42, below Falak's 47 (B). Bala scores $2 \times \text{DI} + 20$; with a DI of 10 he scores 40, below Ester's 58 (C), and 10 is below Chetna's 19 (D). All composites stay different. Hence, option A (Bala scored same as Jatin in DI).

Q65TITAMissing Value Tables

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

Answer and solution

Answer: 13

Composite $= 2 \times \text{DI} + \text{WE} + \text{GA}$. Jatin has WE 16 and GA 14, so his 20 is in DI and he scores 70; Indu scores 60. Indu's WE is 8, and a DI of 20 would leave her GA at 12, two sections below 14, although she was recruited. So her GA is 20 and her DI is $\frac{60 - 8 - 20}{2} = 16$. The completed table below shows every known DI mark. Chetna scores $2 \times 19 + 4 + 12 = 54$. Bala has WE 9 and GA 11, so with a DI of $x$ he scores $2x + 20$. This must be below 54, so $x < 17$, i.e. $x \le 16$. Apart from Ajay and Danish, everyone's DI marks differ. 16 is Indu's, 15 is Falak's and 14 is Geeta's, so Bala's highest possible DI is 13. That gives him a composite of 46, which does not clash with any other candidate. The answer is 13.

Q66TITAMissing Value Tables

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

Answer and solution

Answer: 14

Composite $= 2 \times \text{DI} + \text{WE} + \text{GA}$; under 14 marks (70%) in two sections disqualifies. Jatin's 20 must be in DI (WE 16, GA 14), so he scores 70 and Indu 60. Indu's WE is 8, and a DI of 20 would leave her GA at 12 and disqualify her, so her GA is 20. Danish therefore scores 51, and Harini $30 + \text{WE}$. Geeta is recruited but scores $34 + \text{WE} \le 53$, as her WE is below Ajay's, so Jatin, Indu and Ester (58), who all qualify, are the other recruits. Danish qualifies with 51 and is not recruited, so Geeta must beat 51: her WE is 18 or 19. Ajay scores $32 + \text{WE}$ with WE above Ester's 18. If Geeta had 18 (52), Ajay would score 51 or 52 and tie someone, so Geeta has 19 and Ajay 20. Now the WE marks 20, 19, 18, 16, 15, 9, 8, 7 and 4 are taken. Harini's best free mark is 17, but that gives her 47, equal to Falak's. The next free mark is 14, giving 44, which clashes with no one. The answer is 14.