0
3508

# Most Important CAT DILR Questions PDF

Download and practice Top-8 CAT DILR set questions with solutions PDF. This PDF covers the most important data interpretation and logical reasoning questions and answers. ## Set-1

Instructions

The below graph depicts the maximum and minimum score obtained by 6 students Arhya, Aarthi, Ashu, Aanya, Ananya, Aditi in 10 subjects numbered $S_1$,$S_2$,…………$S_{10}$. For any two consecutive subjects, the score may either remain the same or will change by 2.

Question 1: If all the students scored 26 in a Subject $S_{n}$. Then the maximum possible value of n is

Question 2: The maximum average score possible for a student be X and the minimum average score possible for a student be Y. Then for how many students the value of X-Y is an integer?

Question 3: How many students cannot have an average score equal to 26?

Question 4: In subject $S_6$, the sum of the maximum and the minimum scores of Ananya is

## Set-2

Instructions

Five racers A,B,C,D and E participated in the “Havana athletics Competition”. In the competition they will run five races Race1, Race2, Race3, Race4 and Race 5. Each race have different prize money for the racers who secure first three positions.

Following are the prize money of all the races for 1st, 2nd and 3rd positions respectively.

Following information is known about the competition:
1.All the racers secured a different positions in each race.
2.A won a total of 2100$in the competition. 3.E won a total of 500$ in the competition.
4.C won less than 800$in the competition. 5.Total money won by B was the highest. Question 5: How much money does B win in the competition? a) 2110 b) 2200 c) 2400 d) 2150 Question 6: Who secured 3rd position in the race 3? a) B b) C c) D d) A Question 7: Who secured 4th position in the race 2? a) A b) B c) D d) Cannot be determined Question 8: Which of the following statement is correct? a) D secured 2nd position in Race 1. b) C secured 2nd position in Race 4. c) D won 50$ in Race 5.

d) E won 100$in Race 4. ## Set-3 Instructions The total number of students in a school is 500. Each student likes at least 2 subjects out of Physics, Chemistry, Biology and English. Further it is known that: 1. The ratio of the number of students who like exactly four subjects to the number of students who like exactly three subjects is 1:5. 2. The number of students who like only Physics and Biology is 65, while no student likes only Chemistry and English. 3. The number of students who like Physics and exactly one more subject is 135. 4. The number of people who like Biology but not English is 185. 5. The number of students who like Biology, Chemistry and English but not Physics is same as the number of students who like Physics, Chemistry and English but not Biology. 6. The sum of the number of students who like only Physics and Chemistry and the number of students who like only Biology and English is 60. 7. The number of students who like Biology, Physics and English but not Chemistry is 70. 8. The number of students who like at least three subjects is 240. The number of students who like both Biology and English is 195. Question 9: What is the total number of students who like English? a) 270 b) 300 c) 290 d) 255 Question 10: What is the ratio of the number of students who like Biology to the number of students who like Chemistry? a) 4:3 b) 5:4 c) 5:3 d) 3:2 Question 11: What is the sum the number of students who like only Chemistry and Biology and the number of the students who like Only English and Physics? a) 150 b) 140 c) 135 d) 130 Question 12: What is the difference of the number of students who like Biology and the number of students who like English? a) 105 b) 100 c) 95 d) 90 ## Set-4 Instructions The first graph shows the salaries paid to individual team members (the legend depicting the three departments has been omitted on purpose) while the second shows the salaries paid to the team of a department. Each member from a department is paid the same salary. It is also given that the number of employees in each department over the years is a distinct multiple of 10, but less than 130. Question 13: X in the first graph represents which department? a) Marketing b) Finance c) Technology d) Cannot be determined Question 14: How many people belonged to the Finance team in 2016? Question 15: What is the total number of employees in all three departments in the year 2017? a) 130 b) 150 c) 200 d) 180 Question 16: Which of the following statements is correct? a) Marketing department has the most number of employees in the year 2018. b) Total number of employees across all departments in 2017 is less than that in 2019. c) Technology department has the most number of employees in the year 2018. d) Marketing department has the most number of employees in the year 2017. ## Set-5 Instructions Eight students are sitting around a circular table positioned 1-8. They are two types of students: Type 1 whose statements are True and Type 2 whose statements are False. They made eight statements in the order of their positions alternating between truth and false statements. Sandeep is positioned at 5 and Radha is one of the students. When asked about the positions, the following statements are made in the given order. Position 1: ‘I am Anu, and I am of Type 1student.’ Position 2. ‘Rinesh is Type 1 student.’ Position 3. ‘Sruthi sits opposite to me.’ Position 4. ‘Anu, a type 2 student, sits opposite to me.’ Position 5. ‘Arya and Deepika sit next to me.’ Position 6. ‘Rinesh sits opposite to me.’ Position 7. ‘Radhika sits opposite to me.’ Position 8. ‘Deepika is one of the neighbours of Anu.’ Question 17: Who is seated at position 6? a) Sandeep b) Deepika c) Arya d) Cannot be determined Question 18: How many arrangements are possible? Question 19: Who among the following can be seated opposite to Arya? a) Rinesh b) Deepthi c) Deepika d) Anu Question 20: Which of the following statements is definitely true? a) Arya is seated at position 6 b) Arya is Type 2 student c) Sandeep is one of the neighbours of Anu. d) Rinesh and Anu are Type 1 students. ## Set-6 Instructions Rinesh and Rinsha have 150 marbles on the table. Each of them takes turns to pick marbles. They can pick at least two and at most six marbles in each turn. It is known that both of them play intelligently. If after Rinesh’s turn, there is only one marble left on the table, then Rinsha will pick that one marble. The same is true for Rinesh as well. Question 21: Rinesh picks the first marble and the person who picks the last marble is the winner. What is the number of marbles that Rinesh should pick in his first turn to ensure his win? Question 22: Rinsha picks the first marble and the person who picks the last marble is the loser. What is the minimum number of marbles that Rinsha should pick in her first turn to ensure her win? Question 23: If the number of marbles left on the table after a person’s turn is less than the minimum number of marbles that a person can pick, then the game ends. Rinesh picks the first marble and the person who picks the last marble is the winner. What is the minimum number of marbles that Rinesh should pick in his first turn to ensure his win? a) 4 b) 5 c) 6 d) Irrespective of how many coins he picks. He is a loser Question 24: If the number of marbles left on the table after a person’s turn is less than the minimum number of marbles that a person can pick, then that person has to pick all the coins. Rinsha picks the first marble and the person who picks the last marble is the loser. What is the maximum number of marbles that Rinsha should pick in her first turn to ensure her win? a) 6 b) 5 c) 4 d) 3 ## Set-7 Instructions A 8*8*8 cube is numbered in such a way that in the lowermost layer the first row is numbered 1,2…..8 from left to right and the next row is numbered 9,10….16 from right to left and so on. The next layer is numbered such that cube numbered 65 is just above the cube numbered 1.This way the cubes are numbered till 512. Now the faces of the 8*8*8 cube with the cubes numbered 9 and 16 are painted in blue and the rest are painted red. Question 25: The sum of the numbers on the face diagonal of the cube face having numbers 8 and 505 on it can be: a) 2042 b) 2052 c) 2049 d) 2050 Question 26: The number of cubes which are painted only in blue is: Question 27: The sum of the numbers on the cubes which are painted on three faces is: a) 2488 b) 2824 c) 2052 d) 2992 Question 28: The number of cubes which have at least two sides painted is: a) 70 b) 80 c) 64 d) 72 ## Set-8 Instructions Four friends A, B, C, D live in four different cities among Delhi, Mumbai, Kolkata and Hyderabad not in any particular order. They are in four different years – 1st, 2nd, 3rd and 4th and prefer four different colours among Yellow, Blue, Green, Red in no particular order. Additionally, we know that 1. A, B, C and D do not prefer colours Yellow, Green, Blue and Red respectively. 2. A lives in Delhi and prefers the colour Blue. 3. A, B, C and D do not live in Mumbai, Delhi, Hyderabad and Kolkata respectively. 4. The people living in Mumbai, Delhi, Kolkata and Hyderabad do not study in 2nd, 3rd, 1st and 4th year respectively. 5. The people living in Mumbai, Delhi, Kolkata and Hyderabad do not prefer Blue, Green, Red and Yellow colours respectively. 6. A, B, C and D do not study in the 2nd, 1st, 3rd and 4th year respectively. 7. C prefers the colour yellow. 8. The people who study in 1st, 2nd, 3rd and 4th year do not prefer the colours Green, Yellow, Blue, and Red respectively. Question 29: For how many friends we can uniquely determine their cities? a) 1 b) 2 c) 0 d) 4 Question 30: The person who prefers blue is in which year? a) 1st b) 2nd c) 3rd d) 4th Question 31: Who among the following is from Kolkata? a) B b) C c) D d) Cannot be determined Question 32: Which of the following pair is definitely correct? a) A-4th year b) D-Mumbai c) D-Green d) B- Mumbai ## Answers & Solutions: 1) Answer: 7 2) Answer: 3 The minimum and maximum scores obtained by the students are The minimum avg value is obtained when they score minimum value in as many subjects as possible and the vice versa for maximum average value. Minimum value for Arhya : Scores should be as follows :34,32,30,28,26,24,22,20,20,20 i.e 20 should be the score for the maximum number of times. Minimum avg. value =25.6 For the maximum avg. Scores are as follows:20,22,24,26,28,30,32,34,34 i.e 34 should be the score for the maximum number of times. Maximum avg value =28.4 3 is the correct answer. 3) Answer: 2 The minimum and maximum scores obtained by the students are The minimum avg value is obtained when they score minimum value in many subjects and the vice versa for maximum average value. Minimum value for Arhya : Scores are as follows :34,32,30,28,26,24,22,20,20,20 Minimum avg.=25.6 For the maximum avg. Scores are as follows:20,22,24,26,28,30,32,34,34 Maximum avg value=28.4 Following the same procedure for all the six students,We get Now we have check for the students who cannot maintain an average of 26 Sum of the marks in 10 subjects is 260 Arhya: Min is 20 and Max is 34 Since she has to score the minimum and maximum score, all the scores in between them also will be scored. Marks scored in 8 subjects will range from 20 to 34 20+22+24+26+28+30+32+34+ X$_1$+X$_2$=260 (20+34)/2 x 8 + X = 260 216+X = 260 X=260-184=44 44 can be obtained in the rest two subjects either as 20,24 or 22, 22 Let’s check for Ananya Min score is 16 , Max is 30 Sum of scores in 8 subjects is 8 x avg of 16 and 30 + X = 260 8*(16+30)/2 + X=260 X=76 and in two subjects she can score a max of 60. Hence Ananya cannot maintain an average of 26. Aanya: Min score is 24 and Max score is 34 Sum of scores in 6 subjects= 6*29+X=260 X=86 Even if she scores 24 in the rest of the subjects she will notcross 86. Hence Aanya cannot maintain an average of 26 Aarthi: Min score is 18 and Maximum score is 32. 8*25+X=260 X=60 She can score 30 in the rest two subjects. Hence she can maintain an average of 26. Aasu: Min score is 22 and Maximum score is 32. 6*27+X=260 X=98 She can score 98 in the rest four subjects. Hence she can maintain an average of 26. Aditi: Min score is 20 and Maximum score is 30. 6*25+X=260 X=110 She can score 110 in the rest four subjects. Hence she can maintain an average of 26. Hence 2 is the correct answer. 4) Answer: 46 The minimum and maximum scores obtained by the students are The minimum avg value id obtained when they score minimum value in many subjects and the vice versa for maximum average value. 5) Answer (B) 1. E won 500$ from securing 1st  , 2nd and 3rd in Race 5 (100$), Race 4 (200$) and Race 2 (200$)respectively. (only case possible) 2. As E has secured 3rd position in Race 2, A can win 2100$ from securing 1st , 2nd and 3rd in Race 2 (800$), Race 3 (1000$) and Race 1 (300$) respectively. (only case possible) 3. Now if C earned less than 800$, it must have secured 1st, 2nd and 3rd in Race 4 (300$) , Race 2 (400$)  and Race 5 (10$) respectively. 4. In order to win the highest total amount in the competition B must win more than 2100$, He must secure 1st, 2nd and 3rd in Race 3 (1500$), Race 1 (600$) and Race 4 (100$) respectively. 5. Now the remaining blocks can be filled for position 1st, 2nd and 3rd. We cannot identify the 4th and 5th positions of any racer with the information provided. According to the table B won a total of =1500$ + 600$+ 100$=2200$Answer B 6) Answer (C) 1. E won 500$ from securing 1st  , 2nd and 3rd in Race 5 (100$), Race 4 (200$) and Race 2 (200$)respectively. (only case possible) 2. As E has secured 3rd position in Race 2, A can win 2100$ from securing 1st , 2nd and 3rd in Race 2 (800$), Race 3 (1000$) and Race 1 (300$) respectively. (only case possible) 3. Now if C earned less than 800$, it must have secured 1st, 2nd and 3rd in Race 4 (300$) , Race 2 (400$)  and Race 5 (10$) respectively. 4. In order to win the highest total amount in the competition B must win more than 2100$, He must secure 1st, 2nd and 3rd in Race 3 (1500$), Race 1 (600$) and Race 4 (100$) respectively. 5. Now the remaining blocks can be filled for position 1st, 2nd and 3rd. We cannot identify the 4th and 5th positions of any racer with the information provided. According to the table D came 3rd in the race 3. Answer: Option C 7) Answer (D) 1. E won 500$ from securing 1st  , 2nd and 3rd in Race 5 (100$), Race 4 (200$) and Race 2 (200$)respectively. (only case possible) 2. As E has secured 3rd position in Race 2, A can win 2100$ from securing 1st , 2nd and 3rd in Race 2 (800$), Race 3 (1000$) and Race 1 (300$) respectively. (only case possible) 3. Now if C earned less than 800$, it must have secured 1st, 2nd and 3rd in Race 4 (300$) , Race 2 (400$)  and Race 5 (10$) respectively. 4. In order to win the highest total amount in the competition B must win more than 2100$, He must secure 1st, 2nd and 3rd in Race 3 (1500$), Race 1 (600$) and Race 4 (100$) respectively. 5. Now the remaining blocks can be filled for position 1st, 2nd and 3rd. We cannot identify the 4th and 5th positions of any racer with the information provided. Answer D 8) Answer (C) 1. E won 500$ from securing 1st  , 2nd and 3rd in Race 5 (100$), Race 4 (200$) and Race 2 (200$)respectively. (only case possible) 2. As E has secured 3rd position in Race 2, A can win 2100$ from securing 1st , 2nd and 3rd in Race 2 (800$), Race 3 (1000$) and Race 1 (300$) respectively. (only case possible) 3. Now if C earned less than 800$, it must have secured 1st, 2nd and 3rd in Race 4 (300$) , Race 2 (400$)  and Race 5 (10$) respectively. 4. In order to win the highest total amount in the competition B must win more than 2100$, He must secure 1st, 2nd and 3rd in Race 3 (1500$), Race 1 (600$) and Race 4 (100$) respectively. 5. Now the remaining blocks can be filled for position 1st, 2nd and 3rd. We cannot identify the 4th and 5th positions of any racer with the information provided. We can check each statement from the above table. Option C is correct. 9) Answer (C) The number of students who like exactly 2 subjects = II The number of students who like exactly 3 subjects = III The number of students who like exactly 4 subjects = IV Hence II+III+IV=500, since each student likes at least 2 subject Also, from 1, III=5IV & from 8, III+IV=240, Hence, III=200, IV=40 The Venn Diagram for the following case is as follows: The total number of students =500 Hence, a+65+c+120-c+70+40+d+70-a+d+60-a = 500 => 2d-a = 75……(1) 70+40+d+60-a=195 d-a=25……(2) Subtracting (2) from (1), we get d=50, a=25 Also, the number of students who like exactly three subjects = III = 200 Hence, c+2d+70 = 200 => c+2d=130 =>c=130-2d=130-2*50=30 Hence the complete Venn diagram is as follows: The total number number of students who like English is 70+40+50+35+45+50 = 290 10) Answer (A) The number of students who like exactly 2 subjects = II The number of students who like exactly 3 subjects = III The number of students who like exactly 4 subjects = IV Hence II+III+IV=500, since each student likes at least 2 subject Also, from 1, III=5IV & from 8, III+IV=240, Hence, III=200, IV=40 The Venn Diagram for the following case is as follows: The total number of students =500 Hence, a+65+c+120-c+70+40+d+70-a+d+60-a = 500 => 2d-a = 75……(1) 70+40+d+60-a=195 d-a=25……(2) Subtracting (2) from (1), we get d=50, a=25 Also, the number of students who like exactly three subjects = III = 200 Hence, c+2d+70 = 200 => c+2d=130 =>c=130-2d=130-2*50=30 Hence the complete Venn diagram is as follows: The total number number of students who like Biology = 65+30+90+70+40+50+35= 380 The total number number of students who like Chemistry = 25+30+40+50+90+50 = 285 Hence the required ratio = 380/285 = 4/3 11) Answer (C) The number of students who like exactly 2 subjects = II The number of students who like exactly 3 subjects = III The number of students who like exactly 4 subjects = IV Hence II+III+IV=500, since each student likes at least 2 subject Also, from 1, III=5IV & from 8, III+IV=240, Hence, III=200, IV=40 The Venn Diagram for the following case is as follows: The total number of students =500 Hence, a+65+c+120-c+70+40+d+70-a+d+60-a = 500 => 2d-a = 75……(1) 70+40+d+60-a=195 d-a=25……(2) Subtracting (2) from (1), we get d=50, a=25 Also, the number of students who like exactly three subjects = III = 200 Hence, c+2d+70 = 200 => c+2d=130 =>c=130-2d=130-2*50=30 Hence the complete Venn diagram is as follows: The number of students who like only Chemistry and Biology = 90 The number of the students who like Only English and Physics = 45 => Required sum = 90+45=135 12) Answer (D) The number of students who like exactly 2 subjects = II The number of students who like exactly 3 subjects = III The number of students who like exactly 4 subjects = IV Hence II+III+IV=500, since each student likes at least 2 subject Also, from 1, III=5IV & from 8, III+IV=240, Hence, III=200, IV=40 The Venn Diagram for the following case is as follows: The total number of students =500 Hence, a+65+c+120-c+70+40+d+70-a+d+60-a = 500 => 2d-a = 75……(1) 70+40+d+60-a=195 d-a=25……(2) Subtracting (2) from (1), we get d=50, a=25 Also, the number of students who like exactly three subjects = III = 200 Hence, c+2d+70 = 200 => c+2d=130 =>c=130-2d=130-2*50=30 Hence the complete Venn diagram is as follows: The total number number of students who like English is 70+40+50+35+45+50 = 290 The total number number of students who like Biology is 65+70+30+40+90+50+35 = 380 Required difference = 380-290 = 90 13) Answer (B) We can make the following table of the approximate salaries of a team or individual salaries: We can make the following table for individual score: the values in the table are approximate We can make the following table for individual score: (approximate values) Let us begin by looking at the team salaries during the year 2016. The salary earned by the finance department is over 200 but less than 300. The salary earned by the technology department is a little over 300 and the salary earned by the marketing department is exactly 1100. The individual salaries earned by x, y and z are close to 6, 4.5 and 10 respectively. If the individual salary earned by the marketing department was 6, then the number of people would have to be$\frac{1100}{6}$which is close to 183. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Similarly, If the individual salary earned by the marketing department was 4.5, then the number of people would have to be$\frac{1100}{4.5}$which is close to 244. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Hence, the individual salary earned by the marketing department has to be 10. So, z must be the marketing department. The number of people in each team over the years can be any of the given numbers: 10,20,30,40,50,60,70,80,90,100,110,120. For the marketing department the individual salaries in 2016, 2017, 2018, 2019 are approximately 10, 7.5, 6.5, 2.5 For the marketing department, the team salaries are 1100, a little less than 400, close to 650, a little more than 200. Now, 7.5*50=375, 7.5*60=450 and 7.5*40=300. So, there must be 50 people in the marketing team in 2017. 6.5*100=650, (110 has already been used) and 6.5*90=585. So, there must be 100 people in the marketing team in 2018. 2.5*90=225, 2.5*80=200 and 2.5*100=250. So, there must be 90 people in the marketing team in 2019. In year 2019 individual salary of Y is 2, the highest score of the team can be only 2 x 120 = 240 << 275. Thus y must be the technology department. Divide the values of Technology department in the two tables to find the approximate number of employees. In 2016, 320/4.5= 71.11$\approx$70 In 2017, 75/3.5= 21.4$\approx$20 In 2018, 720/6= 120 In 2019, 20/2= 10 Similarly, we can divide the values of Finance department in the two tables to find the approximate number of employees: According to the data X in the first graph represents Finance department. 14) Answer: 40 We can make the following table of the approximate salaries of a team or individual salaries: We can make the following table for individual score: the values in the table are approximate We can make the following table for individual score: (approximate values) Let us begin by looking at the team salaries during the year 2016. The salary earned by the finance department is over 200 but less than 300. The salary earned by the technology department is a little over 300 and the salary earned by the marketing department is exactly 1100. The individual salaries earned by x, y and z are close to 6, 4.5 and 10 respectively. If the individual salary earned by the marketing department was 6, then the number of people would have to be$\frac{1100}{6}$which is close to 183. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Similarly, If the individual salary earned by the marketing department was 4.5, then the number of people would have to be$\frac{1100}{4.5}$which is close to 244. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Hence, the individual salary earned by the marketing department has to be 10. So, z must be the marketing department. The number of people in each team over the years can be any of the given numbers: 10,20,30,40,50,60,70,80,90,100,110,120. For the marketing department the individual salaries in 2016, 2017, 2018, 2019 are approximately 10, 7.5, 6.5, 2.5 For the marketing department, the team salaries are 1100, a little less than 400, close to 650, a little more than 200. Now, 7.5*50=375, 7.5*60=450 and 7.5*40=300. So, there must be 50 people in the marketing team in 2017. 6.5*100=650, (110 has already been used) and 6.5*90=585. So, there must be 100 people in the marketing team in 2018. 2.5*90=225, 2.5*80=200 and 2.5*100=250. So, there must be 90 people in the marketing team in 2019. In year 2019 individual salary of Y is 2, the highest score of the team can be only 2 x 120 = 240 << 275. Thus y must be the technology department. Divide the values of Technology department in the two tables to find the approximate number of employees. In 2016, 320/4.5= 71.11$\approx$70 In 2017, 75/3.5= 21.4$\approx$20 In 2018, 720/6= 120 In 2019, 20/2= 10 Similarly, we can divide the values of Finance department in the two tables to find the approximate number of employees: 40 people belonged to the Finance team in 2016. 15) Answer (B) We can make the following table of the approximate salaries of a team or individual salaries: We can make the following table for individual salaries: the values in the table are approximate We can make the following table for team salaries: (approximate values) Let us begin by looking at the team salaries during the year 2016. The salary earned by the finance department is over 200 but less than 300. The salary earned by the technology department is a little over 300 and the salary earned by the marketing department is exactly 1100. The individual salaries earned by x, y and z are close to 6, 4.5 and 10 respectively. If the individual salary earned by the marketing department was 6, then the number of people would have to be$\frac{1100}{6}$which is close to 183. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Similarly, If the individual salary earned by the marketing department was 4.5, then the number of people would have to be$\frac{1100}{4.5}$which is close to 244. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Hence, the individual salary earned by the marketing department has to be 10. So, z must be the marketing department. The number of people in each team over the years can be any of the given numbers: 10,20,30,40,50,60,70,80,90,100,110,120. For the marketing department the individual salaries in 2016, 2017, 2018, 2019 are approximately 10, 7.5, 6.5, 2.5 For the marketing department, the team salaries are 1100, a little less than 400, close to 650, a little more than 200. Now, 7.5*50=375, 7.5*60=450 and 7.5*40=300. So, there must be 50 people in the marketing team in 2017. 6.5*100=650, (110 has already been used) and 6.5*90=585. So, there must be 100 people in the marketing team in 2018. 2.5*90=225, 2.5*80=200 and 2.5*100=250. So, there must be 90 people in the marketing team in 2019. In year 2019 individual salary of Y is 2, the highest score of the team can be only 2 x 120 = 240 << 275. Thus y must be the technology department. Divide the values of Technology department in the two tables to find the approximate number of employees. In 2016, 320/4.5= 71.11$\approx$70 In 2017, 75/3.5= 21.4$\approx$20 In 2018, 720/6= 120 In 2019, 20/2= 10 Similarly, we can divide the values of Finance department in the two tables to find the exact number of employees: From the table: We can see that in the year 2017 the number of employees in the Finance, Technology and Marketing are 80, 20 and 50 respectively. =80+20+50=150 Hence option B is correct. 16) Answer (C) We can make the following table of the approximate salaries of a team or individual salaries: We can make the following table for individual score: the values in the table are approximate We can make the following table for individual score: (approximate values) Let us begin by looking at the team salaries during the year 2016. The salary earned by the finance department is over 200 but less than 300. The salary earned by the technology department is a little over 300 and the salary earned by the marketing department is exactly 1100. The individual salaries earned by x, y and z are close to 6, 4.5 and 10 respectively. If the individual salary earned by the marketing department was 6, then the number of people would have to be$\frac{1100}{6}$which is close to 183. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Similarly, If the individual salary earned by the marketing department was 4.5, then the number of people would have to be$\frac{1100}{4.5}$which is close to 244. This is far more than the maximum number of people in any team. (Has to be a multiple of 10 and less than 130, so the maximum people in any of the teams is 120) Hence, the individual salary earned by the marketing department has to be 10. So, z must be the marketing department. The number of people in each team over the years can be any of the given numbers: 10,20,30,40,50,60,70,80,90,100,110,120. For the marketing department the individual salaries in 2016, 2017, 2018, 2019 are approximately 10, 7.5, 6.5, 2.5 For the marketing department, the team salaries are 1100, a little less than 400, close to 650, a little more than 200. Now, 7.5*50=375, 7.5*60=450 and 7.5*40=300. So, there must be 50 people in the marketing team in 2017. 6.5*100=650, (110 has already been used) and 6.5*90=585. So, there must be 100 people in the marketing team in 2018. 2.5*90=225, 2.5*80=200 and 2.5*100=250. So, there must be 90 people in the marketing team in 2019. In year 2019 individual salary of Y is 2, the highest score of the team can be only 2 x 120 = 240 << 275. Thus y must be the technology department. Divide the values of Technology department in the two tables to find the approximate number of employees. In 2016, 320/4.5= 71.11$\approx$70 In 2017, 75/3.5= 21.4$\approx\$20

In 2018, 720/6= 120

In 2019, 20/2= 10

Similarly, we can divide the values of Finance department in the two tables to find the approximate number of employees:

From the table we can find out that Technology department has the most, 120 employee in the year 2018

The question statement states that students make the eight statements in the order of the positions alternating between Truth and False responses.
Here we will have two cases either the first statement is true or it is False.

Case 1: Let us assume the first statement is False.
The sequence of statements made by students in position P1 to P8 will be as follows: F, T, F, T, F, T, F, T
Lets try to find out if this pattern of response will satisfy the circular arrangement.
The statement by the student at P1 is :
‘I am Anu, and I am a Type 1 student’,
Here the statement made by the person at P1 is false, so either Anu cannot be seated at P1 or she is a type 2 student or both.
The statement by the student at P2 is :
Rinesh is Type 1 student and is True.
The statement by the student at P3 is :
Sruthi sits opposite to me, which is false. So Sruthi is not seated at P7.
The statement by the student at P4 is :
‘Anu, a type 2 student, is seated opposite to me,’ is True. But this statement is impossible as type 1 and type 2 students alternate leading to all type 1 students sitting opposite each other and all type 2 students sitting opposite each other. Thus, a type 2 student can’t be seated opposite someone who is telling the truth.
Hence this case is negated.

Case 2: Let us consider the first statement is True.
The sequence of statements made by students in position P1 to P8 will be as follows: T, F, T, F, T, F, T, F
From Statement 1: Anu is at P1, and she is a Type 1 student.
From Statement 2: Rinesh is not Type 1 student. Since there are only two types of students, Rinesh should be a Type 2 student.
From Statement 3: Sruthi is seated at P7.
From Statement 4: Anu is not seated at P8.
From Statement 5: P5 is occupied by Sandeep and Deepika and Arya sit either side of him.
From statement 6: Rinesh is not seated at P2.
From statement 7: Radhika is seated at P3.
From statement 8: Deepika is one of the neighbours of Anu.

Arya or Deepika can be seated at position 6.
Hence D is the correct answer.

The question statement states that students make the eight statements in the order of the positions alternating between Truth and False responses.
Here we will have two cases either the first statement is true or it is False.

Case 1: Let us assume the first statement is False.
The sequence of statements made by students in position P1 to P8 will be as follows: F, T, F, T, F, T, F, T
Lets try to find out if this pattern of response will satisfy the circular arrangement.
The statement by the student at P1 is :
‘I am Anu, and I am a Type 1 student’,
Here the statement made by the person at P1 is false, so either Anu cannot be seated at P1 or she is a type 2 student or both.
The statement by the student at P2 is :
Rinesh is Type 1 student and is True.
The statement by the student at P3 is :
Sruthi sits opposite to me, which is false. So Sruthi is not seated at P7.
The statement by the student at P4 is :
‘Anu, a type 2 student, is seated opposite to me,’ is True. But this statement is impossible as type 1 and type 2 students alternate leading to all type 1 students sitting opposite each other and all type 2 students sitting opposite each other. Thus, a type 2 student can’t be seated opposite someone who is telling the truth.
Hence this case is negated.

From the above case, we can say that 2 arrangements are possible.

The question statement states that students make the eight statements in the order of the positions alternating between Truth and False responses.
Here we will have two cases either the first statement is true or it is False.

Case 1: Let us assume the first statement is False.
The sequence of statements made by students in position P1 to P8 will be as follows: F, T, F, T, F, T, F, T
Lets try to find out if this pattern of response will satisfy the circular arrangement.
The statement by the student at P1 is :
‘I am Anu, and I am a Type 1 student’,
Here the statement made by the person at P1 is false, so either Anu cannot be seated at P1 or she is a type 2 student or both.
The statement by the student at P2 is :
Rinesh is Type 1 student and is True.
The statement by the student at P3 is :
Sruthi sits opposite to me, which is false. So Sruthi is not seated at P7.
The statement by the student at P4 is :
‘Anu, a type 2 student, is seated opposite to me,’ is True. But this statement is impossible as type 1 and type 2 students alternate leading to all type 1 students sitting opposite each other and all type 2 students sitting opposite each other. Thus, a type 2 student can’t be seated opposite someone who is telling the truth.
Hence this case is negated.

Case 2: Let us consider the first statement is True.
The sequence of statements made by students in position P1 to P8 will be as follows: T, F, T, F, T, F, T, F
From Statement 1: Anu is at P1, and she is a Type 1 student.
From Statement 2: Rinesh is not Type 1 student. Since there are only two types of students, Rinesh should be a Type 2 student.
From Statement 3: Sruthi is seated at P7.
From Statement 4: Anu is not seated at P8.
From Statement 5: P5 is occupied by Sandeep and Deepika and Arya sit either side of him.
From statement 6: Rinesh is not seated at P2.
From statement 7: Radhika is seated at P3.
From statement 8: Deepika is one of the neighbours of Anu.

Either Radha or Rinesh can be seated opposite to Arya.

Hence A is the correct answer.

Case 2: Let us consider the first statement is True.
The sequence of statements made by students in position P1 to P8 will be as follows: T, F, T, F, T, F, T, F
From Statement 1: Anu is at P1, and she is a Type 1 student.
From Statement 2: Rinesh is not Type 1 student. Since there are only two types of students, Rinesh should be a Type 2 student.
From Statement 3: Sruthi is seated at P7.
From Statement 4: Anu is not seated at P8.
From Statement 5: P5 is occupied by Sandeep and Deepika and Arya sit either side of him.
From statement 6: Rinesh is not seated at P2.
From statement 7: Radhika is seated at P3.
From statement 8: Deepika is one of the neighbours of Anu.

Let’s analyze the options one by one.

Option A: Arya or Deepika can be seated at P6. Hence this statement is not definitely true.

Option B: Arya is definitely a Type 2 student. Hence B is the correct answer.

Option C:Sandeep is not one of the neighbours of Anu. Hence this statement is incorrect.

Option D: Rinesh is Type 2 student and Anu is Type 1 student.

Let’s determine which are losing and winning positions.

If there are 1 to 6 coins on the table, then it is a winning position, because a person can just pick them up and win.

If there are 7 or 8 coins on the table then it is a losing position. Even if you pick anything from 2-6, you will leave your opponent in a winning position.

If there are 9-14 coins on the table then it is a winning position. For 9 you can pick 2 leaving the opponent with 7 which is a losing position. For 10, you pick 3 and so on. For 14, you pick 6 leaving 8 on the table which is also a losing position. Hence, all of 9-14 are winning positions.

If there 15 or 16 coins on the table, then it is a losing position. Even if you pick anything from 2-6, you will leave your opponent in a winning position.

Hence, we can see a pattern emerging:

1-6 : win

7,8 : lose

9-14: win

15, 16: lose

17-22: win

23, 24:  lose

So all 8x-1 and 8x positions are losing positions. So if Rinesh has to win, he has to leave his opponent in a losing position. So he should leave 8k coins on the table such that 8k<150. The nearest multiple of 8 which is less than 150 is 144. Hence, if Rinesh leaves 144 coins on the table, Rinsha will lose. So he must pick 150-144= 6 coins

Since both of them play intelligently and the person who picks the last marble is the loser,

If it is Rinsha’s turn to pick the marble, she will ensure that after he picks the marbles, the number of marbles left on the table is either 1 or 2.

Since both of them play intelligently, Rinsha will always ensure to leave 8x+1 or 8x+2 marbles on the table after her turn.
Ultimately, we left with 150-(8x+1) or 150-(8x+2) i.e.5,4 marbles respectively on the table.
But there is a condition that if there are less than 2 marbles on the table, then the opponent has to pick the marble.

So Rinsha picks 4 marbles and will leave 2 marbles for Rinesh to pick.

Since we have to find the minimum number of marbles Rinsha should pick, so Rinsha will pick 4 marbles.

Since both of them play intelligently, and the person who picks the last marble is the winner, If it is Rinesh’s turn to pick the coin, he will
ensure that after he picks the marbles, the number of marbles left on the table is a multiple of 8.
In a single round, 8 marbles will be picked up both of them.
Ultimately, we left with 150-8x i.e.6 marbles on the table.
Since after a person’s turn if there are less than 2 marbles on the table, then the game ends. If Rinesh picks five marbles, then there will be one marble on the table, the game ends, Rinesh will be the winner.
If Rinesh picks six marbles, there he will try to make the number of marbles after his turn to be a multiple of 8, and he will be the winner.

Since both of them play intelligently, and the person who picks the last marble is the winner, If it is Rinesh’s turn to pick the coin, he will
ensure that after he picks the marbles, the number of marbles left on the table is a multiple of 8.
In a single round, 8 marbles will be picked up both of them.
Ultimately, we left with 150-8x i.e.6 marbles on the table.
Since after a person’s turn if there are less than 2 marbles on the table,then he has to pick all the marbles. If Rinsha picks five marbles, then there will be one marble on the table,Rinsha has to pick that marble as well. Rinesh will be the winner.
If Rinsha picks four marbles, then there will be 2 marbles which Rinesh has to pick.
If Rinsha picks three marbles, then even if Rinesh picks two marbles in his turn since there are less than 2 marbles on the table, Rinesh has to pick the third leftover marble aswell.

In the lowermost layer,

First row is numbered as follows 1,2,3…8 from left to right.

Next row is numbered as follows 9,10,11…16 from right to left

So the numbering in lower most layer is as follows The faces of the cube with cubes numbered 9 and 16 are painted blue and the rest in red.
So there will be three faces which are painted red and three in blue.

We can classify the cubes as either inside the cube, on the face but not on the edge, on the edge but not in the corners and 8 cubes in the corners.

In our case, all the 3 faces painted blue will have cubes of the 2nd type painted only in blue.

Number of cubes of the faces = 6*6 on each face

Since there are 3 faces = 6*6*3=108

Also, the two edges between the 3 blue faces will also have cubes painted only in blue.

On each edge, we have to subtract 2 cubes at the corner which have one of their faces painted in red

So number of cubes on the edges = 6*2=12

Number of cubes which are painted only in blue = 108+12=120

The cubes at the corners will have three faces painted.

Numbers on the cubes can be found out by  using the lowermost layer

The numbers on the corner cubes=1,8,57,64

The numbers on the corner cubes on the top most layer=1+7*64,8+7*64,57+7*64,64+7*64

Total sum=(1+8+57+64)*2+7*64*4

=2052

The cube is of dimensions 8*8*8. The cubes on the edges and corners will have at least 2 faces painted.
The number of cubes on the edges =6(Since the two cornered ones will have all the three faces painted)
There are 12 edges
Therefore the number of cubes which have two sides painted=12*6=72
Number of cubes which have three faces painted =8 (All the cubes at the corners)
Total number of cubes =72+8=80

With the initial data we can form the following table.

In the above table, data in Grey section cannot be the part of the boxes. For example -A lives in Delhi and prefers Blue while “hyderabad” in the 3rd column cannot be the city where C lives.

According to statement 1 B cannot have green color. Thus B prefers red while D prefers Green.

According to statement 5 One who lives in Kolkata do not prefers red color. Hence C lives in kolkata.

According to statement 4 and 8 One who lives in Kolkata is not in 1st year and One who prefers yellow is not in 2nd year. Thus C is in 4th year.

There will be two cases to resolve this .

Case 1

Case 2

Only for A and C we can uniquely determine the cities.

With the initial data we can form the following table.

In the above table, data in Grey section cannot be the part of the boxes. For example -A lives in Delhi and prefers Blue while “hyderabad” in the 3rd column cannot be the city where C lives.

According to statement 1 B cannot have green color. Thus B prefers red while D prefers Green.

According to statement 5 One who lives in Kolkata do not prefers red color. Hence C lives in kolkata.

According to statement 4 and 8 One who lives in Kolkata is not in 1st year and One who prefers yellow is not in 2nd year. Thus C is in 4th year.

There will be two cases to resolve this .

Case 1

Case 2

The person in blue is in 1st year in both the cases.

With the initial data we can form the following table.

In the above table, data in Grey section cannot be the part of the boxes. For example -A lives in Delhi and prefers Blue while “hyderabad” in the 3rd column cannot be the city where C lives.

According to statement 1 B cannot have green color. Thus B prefers red while D prefers Green.

According to statement 5 One who lives in Kolkata do not prefers red color. Hence C lives in kolkata.

According to statement 4 and 8 One who lives in Kolkata is not in 1st year and One who prefers yellow is not in 2nd year. Thus C is in 4th year.

There will be two cases to resolve this .

Case 1

Case 2

From the above table C is from Kolkata.

With the initial data we can form the following table.

In the above table, data in Grey section cannot be the part of the boxes. For example -A lives in Delhi and prefers Blue while “hyderabad” in the 3rd column cannot be the city where C lives.

According to statement 1 B cannot have green color. Thus B prefers red while D prefers Green.

According to statement 5 One who lives in Kolkata do not prefers red color. Hence C lives in kolkata.

According to statement 4 and 8 One who lives in Kolkata is not in 1st year and One who prefers yellow is not in 2nd year. Thus C is in 4th year.

There will be two cases to resolve this .

Case 1

Case 2

In the above question.

Mumbai and Hyderabad cannot be uniquely.

D is paired with Green in both the cases

Option C is correct.