CAT 2017 Slot 1 Question Paper

In G1, we have three cities namely A, B, C. Person living in any of these three cities should be able to travel to other city once in the morning nad one in the night.  Therefore, a total of 4 flights are required between a pair of cities.
A ---> B (Morning flight)
A ---> B (Evening flight)
B ---> A (Morning flight)
B ---> A (Evening flight)

Number of flights between the cities of G1 = 3c2*4 = 12
Between cities in A and any city in G2 = 3*4 = 12
Between B and any city in G3 = 2*4 = 8
Between C and any city in G4 = 2*4 = 8
Total = 12*2 + 8*2 = 40

The cities those were a part of G2 will be shifted to either G3 or G4 but that will not have any impact on the number of the total flights from G1. The only reduction which will take place due to the number of flights shutting down from A to C.
Hence, the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose = 4 

Alternate method : 

Let us determine the number of flights under new conditions.
Flights between A and B = 4
Between B and any city in G3 = 4*4 = 16
Between C and B = 4
Between C and any city in G4 = 4*4 = 12
Hence, total flights = 36
Thus, reduction = 40 - 36 = 4.

Since there are two routes i.e A-M-B and A-N-B and four cars, then 2 cars must be allowed to take each route. In case if one car tried to break rule, then its travel time will increase. Now assume that on route A-M-B three cars are allowed and on route A-N-B one car is allowed, then one car running on A-M-B can break the rule and reduce its travel time. Hence, two cars must be allowed on each route.

Since, two cars are allowed on each route, The A-M part and N-B will take same time. The difference will be in travelling M-B part and A-N part, and that difference is 0.1 minute. In route M-B car will take 20+0.9 = 20.9 min and in route A-N car will take 20+1 = 21 min. So difference = 0.1 min

Case 1:  Let us assume 1 car takes AMB route, 3 cars take ANB route
Then travel time of AMB will be A-M + M-B = 6 + 20 = 26
Then travel time of ANB will be A-N + N-B = (20+2) + (6+3*2) = 34

Now, one car(A) travelling on ANB broke the rule and decided to move on AMB route. then

Case 2: Let us assume 2 cars take AMB route, 2 cars take ANB route
Then travel time of AMB will be A-M + M-B =(6+3) + (20+0.9) = 29.9
Then travel time of ANB will be A-N + N-B = (20+1) + (6+3) = 30

Since the car A reduced its time from 34 to 29.9, the case-1 route is not optimal, hence case 1 is invalid.
Now, one car(B) travelling on ANB broke the rule and decided to move on AMB route. then, case

Case 3:  Let us assume 3 cars take AMB route, 1 car take ANB route

Then travel time of AMB will be A-M + M-B = (6+3*2) + (20+0.9*2) = 33.8
Then travel time of ANB will be A-N + N-B = (20) + (6) = 29

Now, one car(C) travelling on AMB broke the rule and decided to move on AMNB route. then
Case 4:  Let us assume 2 cars take AMB route, 1 car take AMNB route and other take ANB route. 

Then travel time of AMB will be A-M + M-B = (6+3*2) + (20+0.9) = 32.9
Then travel time of AMNB will be A-M + M-N + N-B = (6+3*2) + (7) + (6+3) = 28
Then travel time of ANB will be A-N + N-B = (20) + (6+3) = 29

Since this car(C) reduced its time from 33.8 to 28, the route is not optimal, hence case 3 is invalid.
Since this car(B) reduced its time from 30 to 28, the route is not optimal, hence case 2 is invalid.
Now, one car(D) travelling on AMB broke the rule and decided to move on AMNB route. then

Case 5:   Let us assume 1 car takes AMB route, 2 cars take AMNB route and other takes ANB route.

Then, the portion A-M will be travelled by 3 cars, M-B by one car, M-N by 2 cars, A-N by 1 car and N-B by 3 cars. 

Then, travel time of AMB will be A-M + M-B = (6+3*2) + (20) = 32
Then, travel time of AMNB will be A-M + M-N + N-B = (6+3*2) + (7+1) + (6+3*2) = 32
Then, travel time of ANB will be A-N + N-B = (20) + (6+3*2) = 32

It is clear that the car D reduced its time from 32.9 min to 32 min if it broke the rule. Hence, case 4 is invalid.
In this arrangement of case 5, no car can improve their travel time by changing their path
Hence, the optimal allocation will be to order 2 cars on A-M-N-B route.

From the previous question we have found that

1 car take AMB route, 2 cars take AMNB route and other take ANB route. 

Then the portion A-M will be travelled by 3 cars, M-B by one car, M-N by 2 cars, A-N by 1 car and N-B by 3 cars. 

Then travel time of AMB will be A-M + M-B = (6+3*2) + (20) = 32

Then travel time of AMNB will be A-M + M-N + N-B = (6+3*2) + (7+1) + (6+3*2) = 32

Then travel time of ANB will be A-N + N-B = (20) + (6+3*2) = 32

The minimum travel time from A to B is 32 min.

Let Arun's current age be A. Hence, Barun's current age is 2.5A
Let Arun's age be half of Barun's age after X years.
Therefore, 2*(X+A) = 2.5A + X
Or, X = 0.5A
Hence, Barun's age increased by 0.5A/2.5A = 20%

Let the rate of work of a person be x units/day. Hence, the total work = 120x.

It is given that one first day, one person works, on the second day two people work and so on.

Hence, the work done on day 1, day 2,... will be x, 2x, 3x, ... respectively.

The sum should be equal to 120x.

$$120x = x* \frac{n(n+1)}{2}$$

$$n^2 + n - 240 = 0$$

n = 15 is the only positive solution.

Hence, it takes 15 days to complete the work.

It is given that the maximum weight limit is 630. The lightest person's weight is 53 Kg and the heaviest person's weight is 57 Kg.

In order to have maximum people in the lift, all the remaining people should be of the lightest weight possible, which is 53 Kg.

Let there be n people.

53 + n(53) + 57 = 630

n is approximately equal to 9.8. Hence, 9 people are possible.

Therefore, a total of 9 + 2 = 11 people can use the elevator.

We see that the man saves 20 minutes by changing his speed from 12 Km/hr to 15 Km/hr.

Let d be the distance

Hence,

$$\frac{d}{12} - \frac{d}{15} = \frac{1}{3}$$

$$\frac{d}{60} = \frac{1}{3}$$

d = 20 Km.