CAT 2019 | DILR Set 1
DIRECTIONS for the question: Go through the graph and the information given below and answer the question that follows.
A large store has only three departments, Clothing, Produce, and Electronics. The following
figure shows the percentages of revenue and cost from the three departments for the years
2016, 2017 and 2018. The dotted lines depict percentage levels. So for example, in 2016, 50%
of store’s revenue came from its Electronics department while 40% of its costs were incurred
in the Produce department.
In this setup, Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost × 100%.
It is known that
- The percentage profit for the store in 2016 was 100%.
- The store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.
- There was no profit from the Electronics department in 2017.
- In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department.
1. What was the percentage profit of the store in 2018? (type in box)
The information in the given triangles is summarized in following table:
.
As Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost
It is known that
1. The percentage profit for the store in 2016 was 100%.it means that half of revenue is cost and half the revenue is profit. Now let revenue in 2016 is 100 so cost in 2016 is 50.
2.store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.so revenue in 2107 is 200 and cost in 2108 is 150.
3.There was no profit from the Electronics department in 2017.from this we can find the cost in 2017 shown below:
No profit means revenue and cost are equal . as revenue in the Electronics department in 2017 is 30% of 200 which is equal to cost in the Electronics department in 2017 which further is40% of total cost.
40% of total cost in 2017=30% of 200 =60
So total cost in 2017=60/40%=150
4. In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department from this we can find the total revenue in 2018 as shown below
as the cost incurred in the Produce department in 2018 is 50% of 100 which is equal to revenue from the Clothing department in2018which further is 40% of total revenue.
40% of total revenue in 2018=50% of 100 =50
So total revenue in 2018= 50/40%=125
Now whole solution is summarized as below:
Total revenue in 2018 is 125 and total cost =100
Hence % profit==25%
2. What was the ratio of revenue generated from the Produce department in 2017 to that in 2018?
- 4 : 3
- 9 : 16
- 8 : 5
- 16 : 9
The information in the given triangles is summarized in following table:
As Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost
It is known that
1. The percentage profit for the store in 2016 was 100%.it means that half of revenue is cost and half the revenue is profit. Now let revenue in 2016 is 100 so cost in 2016 is 50.
2. store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.so revenue in 2107 is 200 and cost in 2108 is 150.
3. There was no profit from the Electronics department in 2017. From this we can find the cost in 2017 shown below:
No profit means revenue and cost are equal. as revenue in the Electronics department in 2017 is 30% of 200 which is equal to cost in the Electronics department in 2017 which further is40% of total cost.
40% of total cost in 2017=30% of 200 =60
So total cost in 2017=60/40%=125
4. In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department from this we can find the total revenue in 2018 as shown below
as the cost incurred in the Produce department in 2018 is 50% of 100 which is equal to revenue from the Clothing department in2018which further is 40% of total revenue.
40% of total revenue in 2018=50% of 100 =50
So total revenue in 2018= 50/40%=125
Now whole solution is summarized as below:
Required ratio=40% of 200:40% of 125=80:50=8:5
3. What percentage of the total profits for the store in 2016 was from the Electronics department? (type in box)
The information in the given triangles is summarized in following table:
As Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost
It is known that
1. The percentage profit for the store in 2016 was 100%.it means that half of revenue is cost and half the revenue is profit. Now let revenue in 2016 is 100 so cost in 2016 is 50.
2.store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.so revenue in 2107 is 200 and cost in 2108 is 150.
3.There was no profit from the Electronics department in 2017.from this we can find the cost in 2017 shown below:
No profit means revenue and cost are equal . as revenue in the Electronics department in 2017 is 30% of 200 which is equal to cost in the Electronics department in 2017 which further is40% of total cost.
40% of total cost in 2017=30% of 200 =60
So total cost in 2017= 60/40% =150
4. In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department
from this we can find the total revenue in 2018 as shown below
as the cost incurred in the Produce department in 2018 is 50% of 100 which is equal to revenue from the Clothing department in2018which further is 40% of total revenue.
40% of total revenue in 2018=50% of 100 =50
So total revenue in 2018= 50/40%= 125
Now whole solution is summarized as below:
Total profit in 2016=100-50=50
Profit in 2016 from Electronics dept=50% of 100-30% of 50=50-15=35
Hence required %= 35/50* 100 =70%
4. What was the approximate difference in profit percentages of the store in 2017 and 2018?
- 8.3
- 15.5
- 25.0
- 33.3
The information in the given triangles is summarized in following table:
As Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost
It is known that
1. The percentage profit for the store in 2016 was 100%.it means that half of revenue is cost and half the revenue is profit. Now let revenue in 2016 is 100 so cost in 2016 is 50.
2.store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.so revenue in 2107 is 200 and cost in 2108 is 150.
3.There was no profit from the Electronics department in 2017.from this we can find the cost in 2017 shown below:
No profit means revenue and cost are equal . as revenue in the Electronics department in 2017 is 30% of 200 which is equal to cost in the Electronics department in 2017 which further is40% of total cost.
40% of total cost in 2017=30% of 200 =60
So total cost in 2017= 60/40% =150
4. In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department
from this we can find the total revenue in 2018 as shown below
as the cost incurred in the Produce department in 2018 is 50% of 100 which is equal to revenue from the Clothing department in2018which further is 40% of total revenue.
40% of total revenue in 2018=50% of 100 =50
So total revenue in 2018= 50/40%=125
Now whole solution is summarized as below:
hence required difference=33.33%-25%=8.3
CAT 2019 | DILR Set 2
DIRECTIONS for the question: Go through the graph and the information given below and answer the question that follows.
Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.
There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
5.What is the total amount of money (in rupees) in the three pouches kept in the first column of the second row? (type in box)
Now there were two important points that had to be kept in mind while solving this block were that
(i) As it is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Hence the sum of nine pouches in any row or column should be a multiple of 9.
(ii)In any of nine slots of 3 × 3 grid minimum and maximum amount should be kept in mind while placing the amount in third pouch.
The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table below
It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.so amount of money kept in the third pouch should have been 1 and also the maximum and minimum amount of money kept in second column of the second row is(1,1) so amount of money kept in the third pouch here should also be 1.
Now further money in the first column in seven of nine pouches is 6+8+4=18 .also no pouch is empty and sum of all in pouches any column or row is a multiple of 9 . so in remaining two pouches in colomn 1 the sum should be 9 making total sum as 27 in first colomn. (we cannot make sum 36 or next multiple of 9 as it will violate max and min range given) .so third pouch in colomn 1 of row 1 is 4 and colomn 2 of row 5.
Further moving in same way and keeping all condition in mind we get the following solution
As shown the required sum is 13
6. How many pouches contain exactly one coin? (type in box)
Now there were two important points that had to be kept in mind while solving this block were that
(i) As it is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Hence the sum of nine pouches in any row or column should be a multiple of 9.
(ii)In any of nine slots of 3 × 3 grid minimum and maximum amount should be kept in mind while placing the amount in third pouch..
The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table below
It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.so amount of money kept in the third pouch should have been 1 and also the maximum and minimum amount of money kept in second column of the second row is(1,1) so amount of money kept in the third pouch here should also be 1.
Now further money in the first column in seven of nine pouches is 6+8+4=18 .also no pouch is empty and sum of all in pouches any column or row is a multiple of 9 . so in remaining two pouches in column1 the sum should be 9 making total sum as 27 in first column. (we cannot make sum 36 or next multiple of 9 as it will violate max and min range given) .so third pouch in column 1 of row 1 is 4 and column 1 of row 2 is 5
Further moving in same way and keeping all condition in mind we get the following solution
As shown 8 pouches contain exactly one coin
7. What is the number of slots for which the average amount (in rupees) of its three pouches is an integer? (type in box)
Now there were two important points that had to be kept in mind while solving this block were that
(i) As it is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Hence the sum of nine pouches in any row or column should be a multiple of 9.
(ii) In any of nine slots of 3 × 3 grid minimum and maximum amount should be kept in mind while placing the amount in third pouch.
The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table below
It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.so amount of money kept in the third pouch should have been 1 and also the maximum and minimum amount of money kept in second column of the second row is(1,1) so amount of money kept in the third pouch here should also be 1.
Now further money in the first column in seven of nine pouches is 6+8+4=18 .also no pouch is empty and sum of all in pouches any column or row is a multiple of 9 . so in remaining two pouches in column1 the sum should be 9 making total sum as 27 in first column. (we cannot make sum 36 or next multiple of 9 as it will violate max and min range given) .so third pouch in column1 of row 1 is 4 and column 2 of row 5.
Further moving in same way and keeping all condition in mind we get the following solution
average amount (in rupees) of its three pouches will be an integer in the slot in which sum of amount is multiple of 3 which is there in two slots i.e.column 3 of row 1where sum is 6 and column two of row two where sum is 3
8. The number of slots for which the total amount in its three pouches strictly exceeds Rs. 10 is (type in box)
Now there were two important points that had to be kept in mind while solving this block were that
(i)As it is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Hence the sum of nine pouches in any row or column should be a multiple of 9.
(ii) In any of nine slots of 3 × 3 grid minimum and maximum amount should be kept in mind while placing the amount in third pouch..
The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table below
It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.so amount of money kept in the third pouch should have been 1 and also the maximum and minimum amount of money kept in second column of the second row is(1,1) so amount of money kept in the third pouch here should also be 1. Now further money in the first column in seven of nine pouches is 6+8+4=18 .also no pouch is empty and sum of all in pouches any column or row is a multiple of 9 . so in remaining two pouches in colomn 1 the sum should be 9 making total sum as 27 in first colomn. (we cannot make sum 36 or next multiple of 9 as it will violate max and min range given) .so third pouch in colomn 1 of row 1 is 4 and colomn 2 of row 5.
Further moving in same way and keeping all condition in mind we get the following solution
As shown above ,the number of slots for which the total amount in its three pouches strictly exceeds Rs. 10 is 3
CAT 2019 | DILR Set 3
DIRECTIONS for the question: Go through the graph and the information given below and answer the question that follows.
Students in a college are discussing two proposals –
A: a proposal by the authorities to introduce dress code on campus, and
B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.
A student does not necessarily support either of the two proposals.
In an upcoming election for student union president, there are two candidates in fray: Sunita and Ragini. Every student prefers one of the two candidates.
A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.
- 250 students supported proposal A and 250 students supported proposal B.
- Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.
- Among those who preferred Ragini, 30% supported proposal A.
- 20% of those who supported proposal B preferred Sunita.
- 40% of those who did not support proposal B preferred Ragini.
- Every student who preferred Sunita and supported proposal B also supported proposal A.
- Among those who preferred Ragini, 20% did not support any of the proposals.
9. Among the students surveyed who supported proposal A, what percentage preferred Sunita for student union president? (type in box)
From second point we have d + e = 160 --(1) From third point we have a + b = 90 --(2) From fourth point we have e + f = 50 --(3)
From fifth point we have (150 + a +x)2/5 = a + x
⇒ 300+2a+2x=5a+5x ⇒ 100 = a + x ............(4)
From point six, we have, f = 0
From point seven, we have x = 20% of 300 = 60
Now as f = 0, (3) ⇒ e = 50 ∴ (1) ⇒ d = 110
⇒ y = 200 - (110 + 50) = 40 From(4),weget100=a+ 60⇒a=40
From (2) we get b = 50
∴ c = 300 - (40 + 50 + 60) = 150 So we have
10. What percentage of the students surveyed who did not support proposal A preferred Ragini as student union president? (type in box)
From second point we have d + e = 160 --(1) From third point we have a + b = 90 --(2) From fourth point we have e + f = 50 --(3)
From fifth point we have (150 + a +x)2/5 = a + x
⇒ 300+2a+2x=5a+5x ⇒ 100 = a + x ............(4)
From point six, we have, f = 0
From point seven, we have x = 20% of 300 = 60
Now as f = 0, (3) ⇒ e = 50 ∴ (1) ⇒ d = 110
⇒ y = 200 - (110 + 50) = 40 From(4),weget100=a+ 60 ⇒a=40
From (2) we get b = 50
∴ c = 300 - (40 + 50 + 60) = 150 So we have
Students who did not support A = 150 + 60+40=250
11. What percentage of the students surveyed who supported both proposals A and B preferred Sunita as student union president?
- 25
- 50
- 20
- 40
From second point we have d + e = 160 --(1) From third point we have a + b = 90 --(2) From fourth point we have e + f = 50 --(3)
From fifth point we have (150 + a +x)2/5 = a + x
⇒ 300+2a+2x=5a+5x ⇒ 100 = a + x ............(4)
From point six, we have, f = 0
From point seven, we have x = 20% of 300 = 60
Now as f = 0, (3) ⇒ e = 50 ∴ (1) ⇒ d = 110
⇒ y = 200 - (110 + 50) = 40 From(4),weget100=a+ 60 ⇒a=40
From (2) we get b = 50
∴ c = 300 - (40 + 50 + 60) = 150
So we have
Students who supported both proposals = 50 + 50 = 100 ∴ required %age
12. How many of the students surveyed supported proposal B, did not support proposal A and preferred Ragini as student union president?
- 150
- 210
- 200
- 40
From second point we have d + e = 160 --(1) From third point we have a + b = 90 --(2) From fourth point we have e + f = 50 --(3)
From fifth point we have (150 + a +x)2/5 = a + x
⇒ 300+2a+2x=5a+5x ⇒ 100 = a + x ............(4)
From point six, we have, f = 0
From point seven, we have x = 20% of 300 = 60
Now as f = 0, (3) ⇒ e = 50 ∴ (1) ⇒ d = 110
⇒ y = 200 - (110 + 50) = 40 From(4),weget100=a+ 60 ⇒a=40
From (2) we get b = 50
∴ c = 300 - (40 + 50 + 60) = 150
So we have
Students who supported both proposals = 50 + 50 = 100 150 students supported proposal B only supported Ragini
CAT 2019 | DILR Set 4
DIRECTIONS for the question: Read the information given below and answer the question that follows.
In the table below the check marks indicate all languages spoken by five people: Paula, Quentin, Robert, Sally and Terence. For example, Paula speaks only Chinese and English.
These five people form three teams, Team 1, Team 2 and Team 3. Each team has either 2 or 3 members. A team is said to speak a particular language if at least one of its members speak that language.
The following facts are known.
- Each team speaks exactly four languages and has the same number of members.
- English and Chinese are spoken by all three teams, Basque and French by exactly two teams and the other languages by exactly one team.
- None of the teams include both Quentin and Robert.
- Paula and Sally are together in exactly two teams.
- Robert is in Team 1 and Quentin is in Team 3.
13. Who among the following four is not a member of Team 2?
- Sally
- Terence
- Paula
- Quentin
14. Who among the following four people is a part of exactly two teams?
- Quentin
- Robert
- Paula
- Sally
15. Who among the five people is a member of all teams?
- Paula
- No one
- Sally
- Terence
16. Apart from Chinese and English, which languages are spoken by Team 1?
- Basque and Dutch
- Arabic and French
- Basque and French
- Arabic and Basque
CAT 2019 | DILR Set 5
DIRECTIONS for the question: Read the information given below and answer the question that follows.
Three doctors, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges Rs. 100/-. Dr. Kane sees each patient for 15 minutes and charges Rs. 200/-, while Dr. Wayne sees each patient for 25 minutes and charges Rs. 300/-.
The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.
The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.
On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.
17. What is the maximum number of patients that the clinic can cater to on any single day?
- 30
- 12
- 15
- 31
Maximum number of patients can be catered on single day when The queue is never empty and all doctors work to full capacity. The clinic is open from 9 a.m. to 11.30 a.m i.e. for 150 minutes every day. Maximum number of patients that can be seen by Dr. Ben are 150/10=15 Maximum number of patients that can be seen by Dr. Kane are 150/15=10 Maximum number of patients that can be seen by Dr. Dr. Wayne are 150/25=6 So the maximum number of patients that the clinic can cater to on any single day are=15+10+6=31
18. The queue is never empty on one particular Saturday. Which of the three doctors would earn the maximum amount in consultation charges on that day?
- Dr. Ben
- Dr. Wayne
- Both Dr. Wayne and Dr. Kane
- Dr. Kane
The queue is never empty on one particular Saturday it means all the doctor are working to their full capacity. (i) Maximum number of patients that can be seen by Dr. Ben are 150/10=15. As charges of of Dr. Ben are 100/- So maximum amount in consultation charges earned by Dr. Ben are 15×100 =1500/- (ii) Maximum number of patients that can be seen by Dr. Kane are 150/15=10 As charges of of Dr. Kane are 200/- So maximum amount in consultation charges earned by Dr. Kane are 10×200=2000/- (iii)Maximum number of patients that can be seen by Dr. Dr. Wayne are 150/25=6 As charges of of Dr. Wayne are 300/- So maximum amount in consultation charges earned by Dr. Ben are 6×300 =1800/- Hence among three doctors Dr. Kane would earn the maximum amount in consultation charges on Saturday.
19. Mr. Singh visited the clinic on Monday, Wednesday, and Friday of a particular week, arriving at 8:50 a.m. on each of the three days. His token number was 13 on all three days. On which day was he at the clinic for the maximum duration?
- Monday
- Wednesday
- Same duration on all three days
- Friday
Mr. Singh who is having token no 13 will be in clinic for the maximum duration on the on which he will be attended by Dr. Wayne
The movement of patients having token number number 1-13 on each given day is shown below
As shown above Mr. Singh will be in clinic for maximum duration on Monday
20. On a slow Thursday, only two patients are waiting at 9 a.m. After that two patients keep arriving at exact 15 minute intervals starting at 9:15 a.m. – i.e. at 9:15 a.m., 9:30 a.m., 9:45 a.m. etc. Then the total duration in minutes when all three doctors are simultaneously free is
- 0
- 15
- 10
- 30
As shown above token number 11,12 will have same movement as of token number 3 and 4 and the same sequence will follow between 10:11 and between 11:0-11;30.
Hence there is no time duration in which all the three doctors are simultaneously free.
CAT 2019 | DILR Set 6
DIRECTIONS for the question: Analyse the graph/s given below and answer the question that follows.
To compare the rainfall data, India Meteorological Department (IMD) calculated the Long Period Average (LPA) of rainfall during period June-August for each of the 16 states. The figure given below shows the actual rainfall (measured in mm) during June-August, 2019 and the percentage deviations from LPA of respective states in 2018. Each state along with its actual rainfall is presented in the figure.
21. If a ‘Heavy Monsoon State’ is defined as a state with actual rainfall from June-August, 2019 of 900 mm or more, then approximately what percentage of ‘Heavy Monsoon States’ have a negative deviation from respective LPAs in 2019?
- 57.14
- 14.29
- 75.00
- 42.86
There are seven states( Mizoram, Sikkim, Maharashtra ,Goa, Arunachal, Kerala and Meghalaya) which are under Heavy Monsoon State’ as per given criterion out of which three (Arunachal, Kerala and Meghalaya) have a negative deviation from respective LPA. Hence Required% = 3/7*100 = 42.86%
22. If a ‘Low Monsoon State’ is defined as a state with actual rainfall from June-August, 2019 of 750 mm or less, then what is the median ‘deviation from LPA’ (as defined in the Y-axis of the figure) of ‘Low Monsoon States’?
- -30%
- -20%
- 10%
- -10%
There are nine states(Gujarat , Karnataka, Rajasthan , MP, Assam, WN, Jharkhand, Delhi and Manipur) which are under ‘Low Monsoon State’ as per given criterion and their respective ‘deviation from LPA’ are 30,20,15,10,-10,-30,-35,-40 and -60 res. Hence required median is -10.
23. What is the average rainfall of all states that have actual rainfall of 600 mm or less in 2019 and have a negative deviation from LPA?
- 500 mm
- 460 mm
- 367 mm
- 450 mm
QNo:- 57 ,Correct Answer:- B Explanation:- states that have actual rainfall of 600 mm or less in 2019 and have a negative deviation from LPA are Assam, WB, Jharkhand, Delhi and Manipur and their respective rainfall are 600,600,400,300 and 400 Hence Required average = 600+600+400+300+400 / 5 = 2300 / 5 = 460 mm
24. The LPA of a state for a year is defined as the average rainfall in the preceding 10 years considering the period of June-August. For example, LPA in 2018 is the average rainfall during 2009-2018 and LPA in 2019 is the average rainfall during 2010-2019. It is also observed that the actual rainfall in Gujarat in 2019 is 20% more than the rainfall in 2009. The LPA of Gujarat in 2019 is closest to
- 490 mm
- 505 mm
- 525 mm
- 475 mm
The actual rainfall in Gujarat in 2019 is 20% more than the rainfall in 2009. So If the actual rainfall in 2009 = x mm Then the actual rainfall in 2019= 1.2x mm Actual rainfall in 2019= 600mm Then, actual rainfall in 2009 = 500mm As deviation is +25% so average 2009 - 2018 is 600/1.25 = 480 LPA 2019 = (480×10 - 500 + 600 )/10 = 490mm
CAT 2019 | DILR Set 7
DIRECTIONS for the question: Read the information given below and answer the question that follows.
The first year students in a business school are split into six sections. In 2019 the Business Statistics course was taught in these six sections by Annie, Beti, Chetan, Dave, Esha, and Fakir. All six sections had a common midterm (MT) and a common endterm (ET) worth 100 marks each. ET contained more questions than MT. Questions for MT and ET were prepared collectively by the six faculty members. Considering MT and ET together, each faculty member prepared the same number of questions.
Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5-mark questions preceded the 10-mark questions, and all the 15- mark questions followed the 10-mark questions.
The following additional facts are known.
- Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.
- Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT
- All questions prepared by a faculty member appeared consecutively in MT as well as ET.
- Chetan prepared the third question in both MT and ET; and Esha prepared the eighth question in both.
- Fakir prepared the first question of MT and the last one in ET. Dave prepared the last question of MT and the first one in ET.
25. The second question in ET was prepared by:
- Chetan
- Beti
- Dave
- Esha
If we broadly see the block two important tasks are to be done
(i)to find the number of questions in each categories of 5 marks ,10 marks and 15 marks
For both MT and ET
i. To allot each question number the faculty that has made that question for both ET and MT
As minimum the number of questions in each categories of 5 marks ,10 marks and 15 marks for both MT and ET are given. Also ET contained more questions than MT. Now considering all these facts total number of questions categories wise for both MT and ET are given below:
For MT there are two possible cases
Further it is given that Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT ,Also considering MT and ET together, each faculty member prepared the same number of questions
Total number of questions are 13+11=24 so each faculty made 4 questions. So keeping in mind all this fact following table gives us the number of question made by each faculty in MT and ET are
Now the information given is “All questions prepared by a faculty member appeared consecutively in MT as well as ET.”
This information will help us to narrow down the cases. Fakir prepared the first question of MT so he will also solve the second one
Chetan prepared the third question in both MT and ET. So considering MT he will solve the forth one. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.it means ist five questions in MT are of 5 marks this eliminates the second possible case for MT. now filling the faculty name consecutively we come to the conclusion as follows:
26. How many 50 mark questions were there in MT and ET combined?
- 13
- 10
- Cannot be determined
- 12
If we broadly see the block two important tasks are to be done
(i)to find the number of questions in each categories of 5 marks ,10 marks and 15 marks
For both MT and ET
i. To allot each question number the faculty that has made that question for both ET and MT
As minimum the number of questions in each categories of 5 marks ,10 marks and 15 marks for both MT and ET are given. Also ET contained more questions than MT. Now considering all these facts total number of questions categories wise for both MT and ET are given below:
For MT there are two possible cases
Further it is given that Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT ,Also considering MT and ET together, each faculty member prepared the same number of questions
Total number of questions are 13+11=24 so each faculty made 4 questions. So keeping in mind all this fact following table gives us the number of question made by each faculty in MT and ET are
Now the information given is “All questions prepared by a faculty member appeared consecutively in MT as well as ET.”
This information will help us to narrow down the cases. Fakir prepared the first question of MT so he will also solve the second one
Chetan prepared the third question in both MT and ET. So considering MT he will solve the forth one. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.it means ist five questions in MT are of 5 marks this eliminates the second possible case for MT. now filling the faculty name consecutively we come to the conclusion as follows:
27. Who prepared 15-mark questions for MT and ET?
- Only Beti, Dave, Esha and Fakir
- Only Esha and Fakir
- Only Dave, Esha and Fakir
- Only Dave and Fakir
If we broadly see the block two important tasks are to be done
(i)to find the number of questions in each categories of 5 marks ,10 marks and 15 marks
For both MT and ET
i. To allot each question number the faculty that has made that question for both ET and MT
As minimum the number of questions in each categories of 5 marks ,10 marks and 15 marks for both MT and ET are given. Also ET contained more questions than MT. Now considering all these facts total number of questions categories wise for both MT and ET are given below:
For MT there are two possible cases
Further it is given that Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT ,Also considering MT and ET together, each faculty member prepared the same number of questions
Total number of questions are 13+11=24 so each faculty made 4 questions. So keeping in mind all this fact following table gives us the number of question made by each faculty in MT and ET are
Now the information given is “All questions prepared by a faculty member appeared consecutively in MT as well as ET.”
This information will help us to narrow down the cases. Fakir prepared the first question of MT so he will also solve the second one
Chetan prepared the third question in both MT and ET. So considering MT he will solve the forth one. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.it means ist five questions in MT are of 5 marks this eliminates the second possible case for MT. now filling the faculty name consecutively we come to the conclusion as follws:
Only Dave, Esha and Fakir prepared 15-mark questions for MT and ET
28. Which of the following questions did Beti prepare in ET?
- Fourth question
- Seventh question
- Tenth question
- Ninth question
If we broadly see the block two important tasks are to be done
(i)to find the number of questions in each categories of 5 marks ,10 marks and 15 marks
For both MT and ET
i. To allot each question number the faculty that has made that question for both ET and MT
As minimum the number of questions in each categories of 5 marks ,10 marks and 15 marks for both MT and ET are given. Also ET contained more questions than MT. Now considering all these facts total number of questions categories wise for both MT and ET are given below:
For MT there are two possible cases
Further it is given that Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT ,Also considering MT and ET together, each faculty member prepared the same number of questions
Total number of questions are 13+11=24 so each faculty made 4 questions. So keeping in mind all this fact following table gives us the number of question made by each faculty in MT and ET are
Now the information given is “All questions prepared by a faculty member appeared consecutively in MT as well as ET.”
This information will help us to narrow down the cases. Fakir prepared the first question of MT so he will also solve the second one
Chetan prepared the third question in both MT and ET. So considering MT he will solve the forth one. Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.it means ist five questions in MT are of 5 marks this eliminates the second possible case for MT. now filling the faculty name consecutively we come to the conclusion as follws:
Among given options Tenth question was prepared by Beti in ET
CAT 2019 | DILR Set 8
DIRECTIONS for the question: Go through the graph and the information given below and answer the question that follows.
Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2,…, players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10, 1 and 2 in Round 7 and so on.
The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.
The following information is known about Rounds 1 through 6:
- Gordon did not score consecutively in any two rounds.
- Eric and Fatima both scored in a round.
The following information is known about Rounds 7 through 10:
- Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
- Joshin scored in Round 7, while Amita scored in Round 10.
- No player scored in all the four rounds.
29. What were the scores of Chen, David, and Eric respectively after Round 3?
- 3, 6, 3
- 3, 0, 3
- 3, 3, 3
- 3, 3, 0
If we broadly see the block two important tasks are to be done
(i) to find the break up of points of each player after round 6 and between round 7-10
(ii) To allot each match its Ist , second and third winner
Round 1-6
now after round 6 we need six 7’s, six 3’s and six 1’s as in each round there will one first ,one second and one seven position.to balance that we need to reject other possible breakups of Chen, David. Eric, Fatima, Gordon(shown yellow).now final break up of scores for each playrer after round 6 is shown in table given below
The next task is to now find the top three players of each round. As we know that
Amita will be playing in first round and sixth round . so her scores 7 and 1 could be only in these rounds. As Joshin has scored two 7.s and in first 6 round he is playing in nly 5th band 6th round . so both 7 scored by him are in these two rounds. So score 7 scored by Amita will be for round 1 . proceeding in this way we will reach the
Now we will do the same process for round 7-10
Round 7-10
now after round 7-10 we need four 7’s, four 3’s and four 1’s as in each round there will one first ,one second and one seven position.
Further it is given that
Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds. So apart from Chen , other one should be Ikea as she is having a option of three scores only so she will have to be settled down with7+7+1. So we will get the following table
As Only two players scored in three consecutive rounds in this stage and they are Chen and Ikea having scores 1,1,1 and 7,7 and 1. Further Ikea which is not in 10th round will have scores in 7th 8th and 9th round and chen would have scored in 8th 9th and 10th round. Proceeding in this way we get the following table for round 7-10.
As shown above the scores of Chen, David, and Eric respectively after Round 3 are
3, 3, 3
30. Which three players were in the last three positions after Round 4?
- Bala, Ikea, Joshin
- Bala, Hansa, Ikea
- Hansa, Ikea, Joshin
- Bala, Chen, Gordon
If we broadly see the block two important tasks are to be done
(i) to find the break up of points of each player after round 6 and between round 7-10
(ii) To allot each match its Ist , second and third winner
Round 1-6
now after round 6 we need six 7’s, six 3’s and six 1’s as in each round there will one first ,one second and one seven position.to balance that we need to reject other possible breakups of Chen, David. Eric, Fatima, Gordon(shown yellow).now final break up of scores for each playrer after round 6 is shown in table given below
The next task is to now find the top three players of each round. As we know that
Amita will be playing in first round and sixth round . so her scores 7 and 1 could be only in these rounds. As Joshin has scored two 7.s and in first 6 round he is playing in nly 5th band 6th round . so both 7 scored by him are in these two rounds. So score 7 scored by Amita will be for round 1 . proceeding in this way we will reach the
now after round 7-10 we need four 7’s, four 3’s and four 1’s as in each round there will one first ,one second and one seven position.
Further it is given that
Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds. So apart from Chen , other one should be Ikea as she is having a option of three scores only so she will have to be settled down with7+7+1. So we will get the following table
As Only two players scored in three consecutive rounds in this stage and they are Chen and Ikea having scores 1,1,1 and 7,7 and 1. Further Ikea which is not in 10th round will have scores in 7th 8th and 9th round and chen would have scored in 8th 9th and 10th round. Proceeding in this way we get the following table for round 7-10
As shown in table last three positions after Round 4 are of Hansa, Ikea, Joshin with scores 1,1 and 0
31. Which player scored points in maximum number of rounds?
- Ikea
- Joshin
- Chen
- Amita
If we broadly see the block two important tasks are to be done
(i) to find the break up of points of each player after round 6 and between round 7-10
(ii) To allot each match its Ist , second and third winner
Round 1-6
now after round 6 we need six 7’s, six 3’s and six 1’s as in each round there will one first ,one second and one seven position.to balance that we need to reject other possible breakups of Chen, David. Eric, Fatima, Gordon(shown yellow).now final break up of scores for each playrer after round 6 is shown in table given below
The next task is to now find the top three players of each round. As we know that
Amita will be playing in first round and sixth round . so her scores 7 and 1 could be only in these rounds. As Joshin has scored two 7.s and in first 6 round he is playing in nly 5th band 6th round . so both 7 scored by him are in these two rounds. So score 7 scored by Amita will be for round 1 . proceeding in this way we will reach the
now after round 7-10 we need four 7’s, four 3’s and four 1’s as in each round there will one first ,one second and one seven position.
Further it is given that
Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds. So apart from Chen , other one should be Ikea as she is having a option of three scores only so she will have to be settled down with7+7+1. So we will get the following table
As Only two players scored in three consecutive rounds in this stage and they are Chen and Ikea having scores 1,1,1 and 7,7 and 1. Further Ikea which is not in 10th round will have scores in 7th 8th and 9th round and chen would have scored in 8th 9th and 10th round. Proceeding in this way we get the following table for round 7-10
As shown Ikea scored points in five rounds which was maximum in number .
32. Which players scored points in the last round?
- Amita, Chen, Eric
- Amita, Chen, David
- Amita, Bala, Chen
- Amita, Eric, Joshin
If we broadly see the block two important tasks are to be done
(i) to find the break up of points of each player after round 6 and between round 7-10
(ii) To allot each match its Ist , second and third winner
Round 1-6
now after round 6 we need six 7’s, six 3’s and six 1’s as in each round there will one first ,one second and one seven position.to balance that we need to reject other possible breakups of Chen, David. Eric, Fatima, Gordon(shown yellow).now final break up of scores for each playrer after round 6 is shown in table given below
The next task is to now find the top three players of each round. As we know that
Amita will be playing in first round and sixth round . so her scores 7 and 1 could be only in these rounds. As Joshin has scored two 7.s and in first 6 round he is playing in nly 5th band 6th round . so both 7 scored by him are in these two rounds. So score 7 scored by Amita will be for round 1 . proceeding in this way we will reach the
now after round 7-10 we need four 7’s, four 3’s and four 1’s as in each round there will one first ,one second and one seven position.
Further it is given that
Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds. So apart from Chen , other one should be Ikea as she is having a option of three scores only so she will have to be settled down with7+7+1. So we will get the following table
As Only two players scored in three consecutive rounds in this stage and they are Chen and Ikea having scores 1,1,1 and 7,7 and 1. Further Ikea which is not in 10th round will have scores in 7th 8th and 9th round and chen would have scored in 8th 9th and 10th round. Proceeding in this way we get the following table for round 7-10
As shown Amita, Chen, Eric scored points in the last round.
