CAT 2017 Question Paper Slot 2

From the table we can say that number of students who opted for E2 after reshuffle = 5 + 34 + 6 + 3 + 5 + 7 + 16 = 76. 

It is given us that the number of students in E2 increased by 30 after the change process. Hence, we can say that the number of students who were enrolled in E2 before reshuffle = 76 - 30 = 46. 

It is given that before the change process there were 10 more students in E2 than in E3. Therefore, the number of students who were enrolled in E3 before reshuffle = 46 - 10 = 36. 

Number of students who moved from E1 to all other electives are known. Therefore, the number of students who were enrolled in E1 before reshuffle =  9 + 5 + 10 + 1 + 4 + 2 = 31. 

It is given that before the change process there were 6 more students in E1 than in E4. Therefore, the number of students who were enrolled in E4 before reshuffle = 31 - 6 = 25. 

Also, it is given that E4 had 2 more students than E6 before reshuffle. Therefore, the number of students who were enrolled in E6 before reshuffle =  25 - 2 = 23. 

All the students from E7 moved to one of electives among E1 to E6.  Therefore, the number of students who were enrolled in E7 before reshuffle = 4 + 16 + 30 + 5 + 5 + 41 = 101. 

Except E5 we know the number of students who were enrolled in all electives. We also know that there were total 300 students who opted for exactly 1 elective. 

Hence, the the number of students who were enrolled in E7 before reshuffle = 300 - (46+36+31+25+23+101) = 38. 

For each elective, the number of students who were enrolled before reshuffle will be same as sum of the number of students who moved from that elective to another elective including no movement cases. 

For elective E2,

Number of students who moved to E1 + 34 + 8 + Number of students who moved to E4 + 2 + 2 = 46 

i.e. Number of students who moved from to E1 =  Number of students who moved from to E4 = 0

For elective E4,

Number of students who moved to E1 + 3 + 2 + 14 + Number of students who moved to E5 + 4 = 25 

i.e. Number of students who moved from to E1 =  Number of students who moved from to E5 = 1     {As the remaining blanks can be filled by either 0 or 1}

For elective E6,

Number of students who moved to E1 + 7 + 3 + Number of students who moved to E4 + 2 + 9 = 23

i.e. Number of students who moved from to E1 =  Number of students who moved from to E4 = 1     {As the remaining blanks can be filled by either 0 or 1}

It is given that after the reshuffle, the number of students in E4 was 3 more than that in E1. As of now the number of students enrolled in E4 after reshuffle = 1 + 0 + E3 to E4 + 14 +  E5 to E4 + 1 + 5 = 21 + {E3 to E4} + {E5 to E4}

Also, the number of students enrolled in E1 after reshuffle = 9 + 0 + 2 + 1 + E5 to E1 + 1 + 4  = 17 + E5 to E1. 

Hence, it is possible only when  E5 to E1 = 1 and E3 to E4 = E5 to E4 = 0. 

Remaining blank places can be filled easily as we know the total sum of each row. 

Therefore, the number of students who moved from E3 to E5 = the number of students who moved from E5 to E3 = the number of students who moved from E5 to E6 = 1.

We are asked the largest change in its enrollment as a percentage of its original enrollment for all 6 electives but as we can see there are only 4 electives. Hence, we will check only for E1, E2, E3 and E6. 

The percentage change in the number of students for E1 = $$\dfrac{18-31}{31}\times 100$$ $$\approx$$ 42 %

The percentage change in the number of students for E2 = $$\dfrac{76-46}{46}\times 100$$ $$\approx$$ 65 % 

The percentage change in the number of students for E3 = $$\dfrac{79-36}{36}\times 100$$ $$\approx$$ 119 % 

The percentage change in the number of students for E6 = $$\dfrac{61-23}{23}\times 100$$ $$\approx$$ 165 % 

We can see that the percent change in the number of student for E6 is the largest. Therefore, option D is the correct answer. 

From the table we can say that number of students who opted for E2 after reshuffle = 5 + 34 + 6 + 3 + 5 + 7 + 16 = 76. 

It is given us that the number of students in E2 increased by 30 after the change process. Hence, we can say that the number of students who were enrolled in E2 before reshuffle = 76 - 30 = 46. 

It is given that before the change process there were 10 more students in E2 than in E3. Therefore, the number of students who were enrolled in E3 before reshuffle = 46 - 10 = 36. 

Number of students who moved from E1 to all other electives are known. Therefore, the number of students who were enrolled in E1 before reshuffle =  9 + 5 + 10 + 1 + 4 + 2 = 31. 

It is given that before the change process there were 6 more students in E1 than in E4. Therefore, the number of students who were enrolled in E4 before reshuffle = 31 - 6 = 25. 

Also, it is given that E4 had 2 more students than E6 before reshuffle. Therefore, the number of students who were enrolled in E6 before reshuffle =  25 - 2 = 23. 

All the students from E7 moved to one of electives among E1 to E6.  Therefore, the number of students who were enrolled in E7 before reshuffle = 4 + 16 + 30 + 5 + 5 + 41 = 101. 

Except E5 we know the number of students who were enrolled in all electives. We also know that there were total 300 students who opted for exactly 1 elective. 

Hence, the the number of students who were enrolled in E7 before reshuffle = 300 - (46+36+31+25+23+101) = 38. 

For each elective, the number of students who were enrolled before reshuffle will be same as sum of the number of students who moved from that elective to another elective including no movement cases. 

For elective E2,

Number of students who moved to E1 + 34 + 8 + Number of students who moved to E4 + 2 + 2 = 46 

i.e. Number of students who moved from to E1 =  Number of students who moved from to E4 = 0

For elective E4,

Number of students who moved to E1 + 3 + 2 + 14 + Number of students who moved to E5 + 4 = 25 

i.e. Number of students who moved from to E1 =  Number of students who moved from to E5 = 1     {As the remaining blanks can be filled by either 0 or 1}

For elective E6,

Number of students who moved to E1 + 7 + 3 + Number of students who moved to E4 + 2 + 9 = 23

i.e. Number of students who moved from to E1 =  Number of students who moved from to E4 = 1     {As the remaining blanks can be filled by either 0 or 1}

It is given that after the reshuffle, the number of students in E4 was 3 more than that in E1. As of now the number of students enrolled in E4 after reshuffle = 1 + 0 + E3 to E4 + 14 +  E5 to E4 + 1 + 5 = 21 + {E3 to E4} + {E5 to E4}

Also, the number of students enrolled in E1 after reshuffle = 9 + 0 + 2 + 1 + E5 to E1 + 1 + 4  = 17 + E5 to E1. 

Hence, it is possible only when  E5 to E1 = 1 and E3 to E4 = E5 to E4 = 0. 

Remaining blank places can be filled easily as we know the total sum of each row. 

Therefore, the number of students who moved from E3 to E5 = the number of students who moved from E5 to E3 = the number of students who moved from E5 to E6 = 1.

We can see from the table that number of students enrolled in E1 dropped to 18. Hence, all the students who moved from E1 to any other elective will have to re-enroll in E1. 

We can see that the number of students who enrolled for E1 prior to reshuffle = 31. Out of these 31 students, 9 students didn't move to any other elective whereas remaining 22 students moved to other electives. Hence, all these 22 students have to re-enroll in E1. 

Therefore, the total number of students in E1 post re-enrollment = 18 + 22 = 40 which is shown in the table. 

Therefore, the sequence of electives in decreasing order of their final enrollments = E2, E3, E6, E5, E1, E4. 

Hence, option A is the correct answer. 

Neeta received least amount in bank deposits implies she received highest amount in property and the vice-versa for Geeta. The assets are 3 flats worth 90 lakh,a house worth 50 lakh, and a deposit worth 70 lakh. The total value of assets is 210 lakhs. They are divided equally, so each will receive assets worth 70 lakh. 

No one daughter can get 3 flats as the total value of asset will be 90 lakhs which is greater than actual share. 

All three daughters can't get 1-1 flat each as well. In that case, the daughter who owns 1 flat and the house will have assets worth 30+50 = 80 lakhs which is more than the actual share. Hence, we can conclude the one of the three daughter gets 2 flats and 10 lakhs bank deposit. 

Out of the remaining two daughters, one will get the house and bank deposit worth 20 lakhs and the other one must have 1 flat and 40 lakhs in bank deposit. On the basis of bank distribution we can easily determine that property and bank deposits for each Neera, Seeta and Geeta. 
                         

From the table, we can see that Seeta must have received Rs. 20 lakh in bank deposits. Hence, option C is the correct answer. 

Neeta received least amount in bank deposits implies she received highest amount in property and the vice-versa for Geeta. The assets are 3 flats worth 90 lakh,a house worth 50 lakh, and a deposit worth 70 lakh. The total value of assets is 210 lakhs. They are divided equally, so each will receive assets worth 70 lakh. 

No one daughter can get 3 flats as the total value of asset will be 90 lakhs which is greater than actual share. 

All three daughters can't get 1-1 flat each as well. In that case, the daughter who owns 1 flat and the house will have assets worth 30+50 = 80 lakhs which is more than the actual share. Hence, we can conclude the one of the three daughter gets 2 flats and 10 lakhs bank deposit. 

Out of the remaining two daughters, one will get the house and bank deposit worth 20 lakhs and the other one must have 1 flat and 40 lakhs in bank deposit. On the basis of bank distribution we can easily determine that property and bank deposits for each Neera, Seeta and Geeta.                           

From the table, we can see that Neeta received 2 flats. 

Let the total number of gold coins with the old woman be '9n'.
Total value of the assets with the old woman = 50 + 3*30 + 70 + 9n = 210+9n.
We know that the assets have been distributed in the ratio 1:2:3.
Therefore, Neeta must have received 35+1.5n (by value), Seeta must have received 70+3n and Geeta must have received 105+4.5n. 
Further, it has been given that the gold coins distributed were in the ratio 2:3:4. 
Therefore, the number of gold coins with Neeta must be '2n', Seeta must be '3n' and Geeta must be '4n'. 
Seeta has '3n' gold coins. Therefore, the total value of the assets with her must be 70. Seeta could not have inherited all the flats. Therefore, Seeta must have received the house ( worth 50 lakh) and 20 lakh from bank deposits. 

We know that Geeta did not receive Rs. 30 lakh from the bank deposits. Therefore, Neeta must have received Rs. 30 lakh. 
The remaining 5 lakh must be contributed by the gold coins (Since there is no other asset worth 5 lakh).
=> 5 + 1.5n = 2n
=> 0.5n = 5
=> n = 10
The old-woman must have had 10*9 = 90 gold coins. Therefore, option B is the right answer. 

Let the total number of gold coins with the old woman be '9n'.
Total value of the assets with the old woman = 50 + 3*30 + 70 + 9n = 210+9n.
We know that the assets have been distributed in the ratio 1:2:3.
Therefore, Neeta must have received 35+1.5n (by value), Seeta must have received 70+3n and Geeta must have received 105+4.5n. 
Further, it has been given that the gold coins distributed were in the ratio 2:3:4. 
Therefore, the number of gold coins with Neeta must be '2n', Seeta must be '3n' and Geeta must be '4n'. 
Seeta has 3n gold coins. Therefore, the total value of the assets with her must be 70. Seeta could not have inherited all the flats. Therefore, Seeta must have received the house ( worth 50 lakh) and 20 lakh from bank deposits. 

We know that Geeta did not receive Rs. 30 lakh from the bank deposits. Therefore, Neeta must have received Rs. 30 lakh. 
The remaining 5 lakh must be contributed by the gold coins (Since there is no other asset worth 5 lakh).
=> 5 + 1.5n = 2n
=> 0.5n = 5
=> n = 10
The oldwoman must have had 10*9 = 90 gold coins.
Total assets = 210 + 90*1 = 300 lakh
Neeta has received 50 lakh in total, Seeta has received 100 lakh and Geeta has received 150 lakh.
Geeta must have received 90 lakh from 3 flats. Out of the remaining 60 lakh, 4*10 = 40 lakh has been contributed by the gold coins. Geeta must have received 150 - 90 - 40 = 20 lakh from bank deposits. Therefore, 20 is the right answer. 

Odd numbered dorms need either moderate or extensive repair.

Even numbered dorms need either light or extensive repair.

It has been given that dorms 4 to 9 all require different repairing costs. The dorms 3 and 9 should require moderate repair (going by the table). Dorm 7 costs the highest. Therefore, dorm 7 should require 6 crores to repair. Dorm 8 requires the least cost to repair. Therefore, dorm 8 should cost 1 crore to repair. We can eliminate these dorm numbers from other 2 lists. 

Dorms 4 to 9 cost different costs to repair. => Both dorms 5 and 9 cannot require the same cost of repair. Dorms 1 and 3 should require 3 crores to repair. 

Dorm 6 should require light repair (2 crores) since dorm 8 requires 1 crore to repair.
=> Dorm 4 requires 5 crore to repair. 

We can see that all options except option D are definitely true. Option D cannot be ascertained to be true. Dorm 10 can cost Rs. 1 crore or Rs. 6 crores to repair. Therefore, option D is the right answer.

Odd numbered dorms need either moderate or extensive repair.

Even numbered dorms need either light or extensive repair.

It has been given that dorms 4 to 9 all require different repairing costs. The dorms 3 and 9 should require moderate repair (going by the table). Dorm 7 costs the highest. Therefore, dorm 7 should require 6 crores to repair. Dorm 8 requires the least cost to repair. Therefore, dorm 8 should cost 1 crore to repair. We can eliminate these dorm numbers from other 2 lists. 

Dorms 4 to 9 cost different costs to repair. => Both dorms 5 and 9 cannot require the same cost of repair. Dorms 1 and 3 should require 3 crores to repair. 

Dorm 6 should require light repair (2 crores) since dorm 8 requires 1 crore to repair.
=> Dorm 4 requires 5 crore to repair. 

Cost = 3 + 3 + 6 + 3 + 4 = Rs.19 crores. Therefore, 19 is the correct answer.

Odd numbered dorms need either moderate or extensive repair.

Even numbered dorms need either light or extensive repair.

It has been given that dorms 4 to 9 all require different repairing costs. The dorms 3 and 9 should require moderate repair (going by the table). Dorm 7 costs the highest. Therefore, dorm 7 should require 6 crores to repair. Dorm 8 requires the least cost to repair. Therefore, dorm 8 should cost 1 crore to repair. We can eliminate these dorm numbers from other 2 lists. 

Dorms 4 to 9 cost different costs to repair. => Both dorms 5 and 9 cannot require the same cost of repair. Dorms 1 and 3 should require 3 crores to repair. 

Dorm 6 should require light repair (2 crores) since dorm 8 requires 1 crore to repair.
=> Dorm 4 requires 5 crore to repair. 

There are 3 dorms from 6 to 10 which are women's dorms.


It has been given that the cost of repairing the woman dorms add up to 20. Therefore, the distribution of the costs should be 6+6+5+3.
Dorm 4 is the dorm whose number is below 5 but is a woman's dorm. Therefore, dorm 9 should cost Rs.3 crores to repair. Dorm 8 cannot be a woman's dorm. Therefore, dorm 10 should be a woman's dorm and should cost Rs. 6 crore to repair.

Dorm 9 will cost Rs.3 crore to repair and hence, 3 is the correct answer.

Odd numbered dorms need either moderate or extensive repair.

Even numbered dorms need either light or extensive repair.

It has been given that dorms 4 to 9 all require different repairing costs. The dorms 3 and 9 should require moderate repair (going by the table). Dorm 7 costs the highest. Therefore, dorm 7 should require 6 crores to repair. Dorm 8 requires the least cost to repair. Therefore, dorm 8 should cost 1 crore to repair. We can eliminate these dorm numbers from other 2 lists. 

Dorms 4 to 9 cost different costs to repair. => Both dorms 5 and 9 cannot require the same cost of repair. Dorms 1 and 3 should require 3 crores to repair. 

Dorm 6 should require light repair (2 crores) since dorm 8 requires 1 crore to repair.
=> Dorm 4 requires 5 crore to repair. 

It has been given that the cost of repairing the woman dorms add up to 20. Therefore, the distribution of the costs should be 6+6+5+3.
Dorm 4 is the dorm whose number is below 5 but is a woman's dorm. Therefore, dorm 9 should cost Rs.3 crores to repair. Dorm 8 cannot be a woman's dorm. Therefore, dorm 10 should be a woman's dorm and should cost Rs. 6 crore to repair.

Hence, Option D is the right answer.