Daily Practice Questions Questions for CAT with Answers PDF

The various routes that can be followed are:
R1 - R3 - R5 - R6;
R1 - R3 - R5 - R2 - R6;
R1 - R3 - R4 - R2 - R6;
R1 - R3 - R4 - R2 - R5 - R6;
R1 - R4 - R3 - R5 - R2 - R6;
R1 - R4 - R3 - R5 - R6;
R1 - R4 - R2 - R5 - R6;
R1 - R4 - R2 - R6;
R1 - R2 - R4 - R3 - R5 - R6;
R1 - R2 - R5 - R6;
R1 - R2 - R6.
Thus, there are 11 ways in all.
Hence, answer option 2 is correct.

To travel from room R4 to room R5, a person must take the route - R4 - R2 - R1 - R3 - R5.
Thus, the person must pass through 3 rooms.
Hence, answer option 3 is correct.

In week-2, number of ways that a person can travel from room R1 to room R6 = 3 (R1R3R5R6, R1R4R2R6, R1R2R6)
In week-3, the possible routes are R1R2R6, R1R2R5R6, R1R2R4R3R5R6, R1R4R2R6, R1R4R2R5R6, R1R4R3R5R6, R1R4R3R5R2R6, R1R3R5R2R6, R1R3R5R6, R1R3R4R2R6, R1R3R4R2R5R6, i.e. 11 ways.
Required difference = 11 - 3 = 8

Option (1): In week-2, a person will take the route R2R6R5 and in week-3, a person will take the route R2R5.
Option (2): In week-2, a person will take the route R4R2R6 and in week-3, a person will take the same route.
Option (3): In week-2, a person will take the route R4R2R6R5/R4R1R3R5 and in week-3, a person will take the route R4R2R5/R4R3R5.
Option (4): In week-2, a person will take the route R3R1R4 and in week-3, a person will take the route R3R4. Hence, the route remains unchanged only in option 2.

Let a, b, c, and d be the number of boats travelling on the routes H1H2, H1H4, H3H2, and H5H4, respectively.
From (i), a + c = 2(b + d)
From (ii), c = 1/4(b + d)
From (iii), a + c = 2(1000 - a - b)
From (iv), a + c = 640 
⇒ b + d = 320
⇒ c = 80
⇒ a = 560
From (iii), 1000 - a - b = 320
⇒ a + b = 680
⇒ b = 120 and d = 200
The following diagram shows the number of boats that travelled along each waterway:

From the above diagram, it is clear that 320 boats travelled through H4.
Hence, answer option 2 is correct.

Let a, b, c, and d be the number of boats travelling on the routes H1H2, H1H4, H3H2, and H5H4, respectively.
From (i), a + c = 2(b + d)
From (ii), c = 1/4(b + d)
From (iii), a + c = 2(1000 - a - b)
From (iv), a + c = 640 
⇒ b + d = 320
⇒ c = 80
⇒ a = 560
From (iii), 1000 - a - b = 320
⇒ a + b = 680
⇒b = 120 and d = 200
The following diagram shows the number of boats that travelled along each waterway:

The number of boats which travelled along H1H2H5H6 is at least 440 - 80 = 360 (because 80 boats joined in at H2). 
The number of boats which travelled along H1H4H6 will be 120 (since the 120 boats which travelled from H1 to H4, must have gone to H6 from H4).
Similarly, the number of boats which travelled along H1H3H2H5H4H6 lies between 0 and 80.
The number of boats which travelled along H1H3H6 will be 240.
Thus, the maximum number of boats travelled through H1H2H5H6.
Hence, answer option 1 is correct.

Let a, b, c, and d be the number of boats travelling on the routes H1H2, H1H4, H3H2, and H5H4, respectively.
From (i), a + c = 2(b + d)
From (ii), c = 1/4(b + d)
From (iii), a + c = 2(1000 - a - b)
From (iv), a + c = 640 
⇒ b + d = 320
⇒ c = 80
⇒ a = 560
From (iii), 1000 - a - b = 320
⇒ a + b = 680
⇒ b = 120 and d = 200
The following diagram shows the number of boats that travelled along each waterway:

Out of the waterways given in the options, the highest number of boats travelled on the waterways connecting harbour H5 and harbour H6. 
Hence, answer option 1 is correct.

Let a, b, c, and d be the number of boats travelling on the routes H1H2, H1H4, H3H2, and H5H4, respectively.
From (i), a + c = 2(b + d)
From (ii), c = 1/4(b + d)
From (iii), a + c = 2(1000 - a - b)
From (iv), a + c = 640 
⇒ b + d = 320
⇒ c = 80
⇒ a = 560
From (iii), 1000 - a - b = 320
⇒ a + b = 680
⇒ b = 120 and d = 200
The following diagram shows the number of boats that travelled along each waterway:

320 boats travelled on the waterways connecting harbour H4 and harbour H6.
Hence, answer option 3 is correct.

The possible routes from CP1 to CP6 are:
CP1CP2CP5CP6, CP1CP2CP5CP4CP6, CP1CP2CP5CP4CP3CP6, CP1CP4CP6, CP1CP4CP5CP6, CP1CP4CP3CP6, CP1CP3CP6, CP1CP3CP4CP6, CP1CP3CP4CP5CP6
From (iii), Car-B is the only car which passes through both, check-point CP2 and check-point CP4, and from (ii), it also passes through check-point CP3. Hence, Car-B pass along CP1CP2CP5CP4CP3CP6.
From (i), Car-A and Car-D pass through CP1CP2CP5CP4CP6, CP1CP4CP6, or CP1CP3CP4CP6. No other car passes through these three routes. However, since we have already seen that only Car-B passes through both CP2 and CP4, Car-A and Car-D pass along CP1CP4CP6 and CP1CP3CP4CP6, not necessarily in the same order, while no other car passes along CP1CP2CP5CP4CP6. From (iv), since Car-F passes through CP5 and CP6 but not CP4, the only possibility for it is CP1CP2CP5CP6.
Since, one of Car-A and Car-D pass along CP1CP3CP4CP6, the other route with the same set of check-points, i.e. CP1CP4CP3CP6, cannot be taken by anyone else.
The remaining routes are CP1CP3CP6, CP1CP3CP4CP5CP6 and CP1CP4CP5CP6. But from (ii), Car-E does not travel through CP3, which means that the only possibility left for it is CP1CP4CP5CP6. Further from (iii), Car-C does not travel through CP5, which leaves CP1CP3CP6 as the only possibility for it. This leaves CP1CP3CP4CP5CP6 as the only possibility for Car-G.
The following table presents the possible routes that each car could pass along:

Car-F drives along CP1CP2CP5CP6.
Hence, answer option 1 is correct.

The total number of passengers that the metros have when leaving MS-1 is 210 (100 + 110). When they reach MS-5, the total number of passengers they have is 150 (70 + 80). Therefore, they drop-off 60 passengers at MS-2, MS-3 and MS-4. Since they drop-off at least 10 passengers at each location, they must drop-off 10 passengers at MS-2 and 10 at MS-4.
Therefore, the answer is option 1.

The total number of passengers that the metros have when leaving MS-1 is 210 (100 + 110). When they reach MS-5, the total number of passengers they have is 150 (70 + 80). Therefore, they drop-off 60 passengers at MS-2, MS-3 and MS-4. Since they drop-off at least 10 passengers at each location, they must drop-off 10 passengers at MS-2 and 10 at MS-4.
Since 10 passengers are dropped-off at MS-2, the two metros must carry 200 passengers to MS-3. They drop-off 40 passengers at MS-3, and carry 160 passengers in total from MS-3 to MS-4 and MS-3 to MS-5. Since 80 passengers are being carried from MS-3 to MS-5, 80 passengers must be carried from MS-3 to MS-4.
Hence, answer option 2 is correct.

The metros have a total of 130 passengers with them between MS5 and MS6. If the maximum number of passengers are to be dropped-off at MS-8, only ten passengers each must be dropped-off at MS-7 and MS-9. Therefore, the maximum number of passengers that can be dropped-off at MS-8 can be:
130 - 70 - 10 - 10 = 40 passengers
Hence, answer option 4 is correct.

The possible routes from CP1 to CP6 are:
CP1CP2CP5CP6, CP1CP2CP5CP4CP6, CP1CP2CP5CP4CP3CP6, CP1CP4CP6, CP1CP4CP5CP6, CP1CP4CP3CP6, CP1CP3CP6, CP1CP3CP4CP6, CP1CP3CP4CP5CP6
From (iii), Car-B is the only car which passes through both, check-point CP2 and check-point CP4, and from (ii), it also passes through check-point CP3. Hence, Car-B pass along CP1CP2CP5CP4CP3CP6.
From (i), Car-A and Car-D pass through CP1CP2CP5CP4CP6, CP1CP4CP6, or CP1CP3CP4CP6. No other car passes through these three routes. However, since we have already seen that only Car-B passes through both CP2 and CP4, Car-A and Car-D pass along CP1CP4CP6 and CP1CP3CP4CP6, not necessarily in the same order, while no other car passes along CP1CP2CP5CP4CP6. From (iv), since Car-F passes through CP5 and CP6 but not CP4, the only possibility for it is CP1CP2CP5CP6.
Since, one of Car-A and Car-D pass along CP1CP3CP4CP6, the other route with the same set of check-points, i.e. CP1CP4CP3CP6, cannot be taken by anyone else.
The remaining routes are CP1CP3CP6, CP1CP3CP4CP5CP6 and CP1CP4CP5CP6. But from (ii), Car-E does not travel through CP3, which means that the only possibility left for it is CP1CP4CP5CP6. Further from (iii), Car-C does not travel through CP5, which leaves CP1CP3CP6 as the only possibility for it. This leaves CP1CP3CP4CP5CP6 as the only possibility for Car-G.
The following table presents the possible routes that each car could pass along:

Car-E passes along CP1CP4CP5CP6.
Hence, answer option 2 is correct.

The possible routes from CP1 to CP6 are:
CP1CP2CP5CP6, CP1CP2CP5CP4CP6, CP1CP2CP5CP4CP3CP6, CP1CP4CP6, CP1CP4CP5CP6, CP1CP4CP3CP6, CP1CP3CP6, CP1CP3CP4CP6, CP1CP3CP4CP5CP6
From (iii), Car-B is the only car which passes through both, check-point CP2 and check-point CP4, and from (ii), it also passes through check-point CP3. Hence, Car-B pass along CP1CP2CP5CP4CP3CP6.
From (i), Car-A and Car-D pass through CP1CP2CP5CP4CP6, CP1CP4CP6, or CP1CP3CP4CP6. No other car passes through these three routes. However, since we have already seen that only Car-B passes through both CP2 and CP4, Car-A and Car-D pass along CP1CP4CP6 and CP1CP3CP4CP6, not necessarily in the same order, while no other car passes along CP1CP2CP5CP4CP6. From (iv), since Car-F passes through CP5 and CP6 but not CP4, the only possibility for it is CP1CP2CP5CP6.
Since, one of Car-A and Car-D pass along CP1CP3CP4CP6, the other route with the same set of check-points, i.e. CP1CP4CP3CP6, cannot be taken by anyone else.
The remaining routes are CP1CP3CP6, CP1CP3CP4CP5CP6 and CP1CP4CP5CP6. But from (ii), Car-E does not travel through CP3, which means that the only possibility left for it is CP1CP4CP5CP6. Further from (iii), Car-C does not travel through CP5, which leaves CP1CP3CP6 as the only possibility for it. This leaves CP1CP3CP4CP5CP6 as the only possibility for Car-G.
The following table presents the possible routes that each car could pass along:

Thus, car-G passes through a total of 5 check-points.
Hence, answer option 4 is correct.

The possible routes from CP1 to CP6 are:
CP1CP2CP5CP6, CP1CP2CP5CP4CP6, CP1CP2CP5CP4CP3CP6, CP1CP4CP6, CP1CP4CP5CP6, CP1CP4CP3CP6, CP1CP3CP6, CP1CP3CP4CP6, CP1CP3CP4CP5CP6
From (iii), Car-B is the only car which passes through both, check-point CP2 and check-point CP4, and from (ii), it also passes through check-point CP3. Hence, Car-B pass along CP1CP2CP5CP4CP3CP6.
From (i), Car-A and Car-D pass through CP1CP2CP5CP4CP6, CP1CP4CP6, or CP1CP3CP4CP6. No other car passes through these three routes. However, since we have already seen that only Car-B passes through both CP2 and CP4, Car-A and Car-D pass along CP1CP4CP6 and CP1CP3CP4CP6, not necessarily in the same order, while no other car passes along CP1CP2CP5CP4CP6. From (iv), since Car-F passes through CP5 and CP6 but not CP4, the only possibility for it is CP1CP2CP5CP6.
Since, one of Car-A and Car-D pass along CP1CP3CP4CP6, the other route with the same set of check-points, i.e. CP1CP4CP3CP6, cannot be taken by anyone else.
The remaining routes are CP1CP3CP6, CP1CP3CP4CP5CP6 and CP1CP4CP5CP6. But from (ii), Car-E does not travel through CP3, which means that the only possibility left for it is CP1CP4CP5CP6. Further from (iii), Car-C does not travel through CP5, which leaves CP1CP3CP6 as the only possibility for it. This leaves CP1CP3CP4CP5CP6 as the only possibility for Car-G.
The following table presents the possible routes that each car could pass along:

The correct combinations are CP5-4, CP4-5, CP3-4, and CP2-2. Only choice 4 is correct.

The possible routes from CP1 to CP6 are:
CP1CP2CP5CP6, CP1CP2CP5CP4CP6, CP1CP2CP5CP4CP3CP6, CP1CP4CP6, CP1CP4CP5CP6, CP1CP4CP3CP6, CP1CP3CP6, CP1CP3CP4CP6, CP1CP3CP4CP5CP6
From (iii), Car-B is the only car which passes through both, check-point CP2 and check-point CP4, and from (ii), it also passes through check-point CP3. Hence, Car-B pass along CP1CP2CP5CP4CP3CP6.
From (i), Car-A and Car-D pass through CP1CP2CP5CP4CP6, CP1CP4CP6, or CP1CP3CP4CP6. No other car passes through these three routes. However, since we have already seen that only Car-B passes through both CP2 and CP4, Car-A and Car-D pass along CP1CP4CP6 and CP1CP3CP4CP6, not necessarily in the same order, while no other car passes along CP1CP2CP5CP4CP6. From (iv), since Car-F passes through CP5 and CP6 but not CP4, the only possibility for it is CP1CP2CP5CP6.
Since, one of Car-A and Car-D pass along CP1CP3CP4CP6, the other route with the same set of check-points, i.e. CP1CP4CP3CP6, cannot be taken by anyone else.
The remaining routes are CP1CP3CP6, CP1CP3CP4CP5CP6 and CP1CP4CP5CP6. But from (ii), Car-E does not travel through CP3, which means that the only possibility left for it is CP1CP4CP5CP6. Further from (iii), Car-C does not travel through CP5, which leaves CP1CP3CP6 as the only possibility for it. This leaves CP1CP3CP4CP5CP6 as the only possibility for Car-G.
The following table presents the possible routes that each car could pass along:

Car-C passes along CP1CP3CP6.
Hence, answer option 3 is correct.

At MS-7, MS-8 and MS-9, the maximum number of passengers that can be dropped-off will be less than 70. Hence, the maximum number of passengers are dropped-off at MS-6, which is 70.

In order to protect all the club doors from being destroyed, each door must have enough bouncers to fight against all the competitor club bouncers that can attack that club door.
Number of bouncers at B-1 = 38 + 69 + 42 + 51 = 200
Number of bouncers at B-2 = 69 + 42 + 51 + 33 = 195
Number of bouncers at B-3 = 51 + 33 + 48 = 132
Number of bouncers at B-4 = 33 + 48 + 28 = 109
Number of bouncers at B-5 = 69 + 28 + 38 = 135
Total number of bouncers across all club doors = 200 + 195 + 132 + 109 + 135 = 771

If three club doors are to be saved from destruction, minimum number of bouncers will be required to protect B-3, B-4, and B-5.
Thus, total number of bouncers required = 132 + 109 + 135 = 376
Hence, answer option 4 is correct.

The total number of enemy bouncers that will attack the club door is 309.
Since the number of bouncers attacking each club door is already known, the number of defending bouncers will also be 309.Hence, a minimum of 618 bouncers will be injured if no club door is to be destroyed.