Practice Questions for Routes & Network | Logical Reasoning (LR) and Data Interpretation (DI) - CAT PDF Download

Q1: Points E, F, G, and H lie respectively on the sides AB, BC, CD, and DA of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of the length of EB to that of CG is
(a) 2: 5
(b) 4: 9
(c) 1: 3
(d) 3: 8
Ans: (c)

Sol: 

Tip 1: Draw a rough diagram

Tip 2: Use Pythagoras' theorem to find the side of the internal square.
If the side of the ABCD is ‘a, then take the measurements as shown in the figure.
GF² = GC² + GF²
GF² = (a-x)² + x²
GF² = a² + x² – 2ax + x²
GF² = a² + 2x² – 2ax
Tip 3: Take the ratio areas of squares
Ar(ABCD)/Ar(EFGH) = 100/62.5 = 8/5
So, a²/ (a² + 2x2 – 2ax) = 8/5
⇒ 5a²= 8(a² + 2x2 – 2ax)
⇒ 16x² +3a² – 16ax= 0
On solving, we get x = a/4 or x = 3a/4
EB < CG, so x must be a/4 and CG = 3a/4
So, EB/CG = (a/4)/(3a/4) = 1: 3

Q2: With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y + 1) or to (x + 1, y), is
Ans: 3920

Sol:
Case 1:
Dimension of the rectangle with ends of diagonal being (1,1) and (4,6) = 3*5
[3 ways in horizontal direction and 5 ways in vertical direction]
Thus, the total number of paths from (1,1) to (4,6) = (5+3)! /5!3! =56
Case 2:
Dimension of the rectangle with ends of diagonal being (8,10) and (4,6) = 4*4
[4 ways in horizontal direction and 4 ways in vertical direction]
Thus, the number of paths from (4,6) to (8,10) = (4+4)! /4!4! =70
So, the total number of paths from (1,1) to (8,10) via (4,6) = 56*70 =3920

Q3: Two trains A and B were moving in opposite directions, their speeds being in the ratio 5: 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of the length of train A to that of train B is
(a) 3 : 2
(b) 5 : 3
(c) 2 : 3
(d) 2 : 1
Ans: (a)

Sol: Let the speed of train A is 5x.
The speed of train B is 3x.
And Lengths of train A and train B are L₁ and L₂ respectively.
Situation 1:
The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.
It implies that the L₂ distance will be covered in 46 seconds with the speed of (5x+ 3x= 8x)
L₂ = 46(8x)
Situation 2:
Trains took another 69 seconds for the rear ends of the trains to cross each other.
It implies that the L1 distance is covered in 69 sec.
L₁ = 69(8x)
L₁/L₂ = 69(8x)/[46(8x)] = 3/2
L₁: L₂ = 3 : 2

Q4: A new airline company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with. The flight duration between any pair of cities will be less than one hour. To start operations, the company has to decide on a daily schedule.
The underlying principle that they are working on is the following:
Any person staying in any of these 10 cities should be able to make a trip to any other city in the morning and should be able to return by the evening of the same day
If the underlying principle is to be satisfied in such a way that the journey between any two cities can be performed using only direct (non-stop) flights, then the minimum number of direct flights to be scheduled is:
(a) 45
(b) 90
(c) 180
(d) 135
Ans: (c)

Sol: The question is simply asking to connect 2 dots out of 10 dots or two stations out of 10 stations this could be done by selecting 2 points out of 10 points that could be done by ¹⁰C₂ which is 45 but 45 will not be the answer because consider the person is travelling from city A to city B and at the same time an another person must be travelling from city B to city A 
So, we need to multiply 45 by 2 so final answer must be 45*2=90 but this is for morning only so we need to multiply this 90 by 2 for evening also so the final answer must be 90*2=180

Q5: Suppose three of the ten cities are to be developed as hubs. A hub is a city which is connected with every other city by direct flights each way, both in the morning as well as in the evening. The only direct flights which will be scheduled are originating and/or terminating in one of the hubs. Then the minimum number of direct flights that need to be scheduled so that the underlying principle of the airline to serve all the ten cities is met without visiting more than one hub during one trip is:
(a) 54
(b) 120
(c) 96
(d) 60
Ans: (c)

Sol: Consider hubs first there will be 4 flights between two hubs (2 at morning and 2 at evening).
So, there will be 3 hub cities and each route will have 4 flights so we will have a total of 12 cities between 3 hubs
Now consider a direct flight from 3 hub cities to 7 non-hub cities is = ³C₁ * ⁷C₁*4 = 84 
So, the total number of flights will be = 84 + 12 = 96