DXC Technology Logical Reasoning Questions and Answers 2023

We can represent the matches In the tournament as follows:

Solution:

From the above diagram, we can see that if the player seeded 43 reached finals, then the player seeded 46 cannot reach the final, as they are in the same half of the draw and exactly one person from each half of the draw reaches the finals.

As the player seeded 12 is present in the top half of the draw, the lowest seeded player in the bottom half i.e.. the player seeded 63 can play with him in the final.

Any player, to win the tournament, needs to win six matches. If five of them are upsets, then the player seeded 32 can win the tournament.

Amit is not the tallest and Chris is not the shortest. Further, Amit is taller than Chris and Basheer is taller than Dharam. Since Chris is shorter than Amit but not the shortest,

Amit cannot be number 5.

Similarly Chris cannot be number 2.

Basheer cannot be the shortest, since he is taller than Dharam. Dharam cannot be 1.

The possibilities are

Amit (A) —> 2, 3

Basheer (B) —> 1, 2, 3, 4

Chris (C) —> 3, 4

Dharam (D) —> 2, 3, 4, 5

Esso (E) —> 1, 2, 3, 4, 5

Now A + E = C + D.

But C can take only 3 or 4, and neither A nor D can take 1.

Therefore, C+D= (3+2) or(2+3) is not possible, A+ E: (1 +4)

or (4 + 1), and A is neither 1 nor 4.

Therefore, C + D = 3 + 4 is possible.

Hence, A+E= 2+5-------- (1)

C + D = 4 + 3 is possible.

Hence, A + E = 2 + 5 -------- (2)

Therefore, Basheer is 1 in both the cases.

Basheer is the tallest.

Statement (1) is not true as the arrangement will be as follows:

Statement (2) is true, since Chris + Amit = 3 + 2 = 5

= Esso’s number.

Chris + Basheer =3+1=4=(2)^2 and Dharam + Esso =4+5=9=(3)^2.

Let us analyse the scores of Match-1 first:

Runs scored by Bimal, Dinu and Elan = 88 + 72 + 60 = 220

Hence, runs scored by Arpit and Chatur = 270 – 220 = 50

Also 10% of 270 = 27

So, both Arpit and Chatur can score a maximum of 27 runs but the sum of their scores should be 50.

Arpit’s score’s range in Match-1 is 23–27 and subsequently Chatur’s score’s range in Match-1 is 27–23.

Similarly Chatur and Dinu scored 30 runs each in Match-2.

In Match-3 even though 10% of 240 = 24, as Dinu scored 20 runs, both Arpit and Bimal can score a maximum of 19 runs, but the sum of their scores should be 32.

In Match-4, Chatur and Elan combined scored 200 – 53 – 52 – 56 = 39 runs. As 10% of 200 = 20, one of Chatur or Elan scores 20 runs and the other scores 19 runs.

The table can be re-written as :

 Maximum possible runs scored by Arpit in Match-1 = 27

Maximum possible runs scored by Arpit in Match-3 = 19

Maximum possible percentage contribution:

(27+100+19+53)/(270+300+240+200)x100% = 199/1010x100%
= 19.7%

Let us analyse the scores of Match-1 first:

Runs scored by Bimal, Dinu and Elan = 88 + 72 + 60 = 220

Hence, runs scored by Arpit and Chatur = 270 – 220 = 50

Also 10% of 270 = 27

So, both Arpit and Chatur can score a maximum of 27 runs but the sum of their scores should be 50.

Arpit’s score’s range in Match-1 is 23–27 and subsequently Chatur’s score’s range in Match-1 is 27–23.

Similarly Chatur and Dinu scored 30 runs each in Match-2.

In Match-3 even though 10% of 240 = 24, as Dinu scored 20 runs, both Arpit and Bimal can score a maximum of 19 runs, but the sum of their scores should be 32.

In Match-4, Chatur and Elan combined scored 200 – 53 – 52 – 56 = 39 runs. As 10% of 200 = 20, one of Chatur or Elan scores 20 runs and the other scores 19 runs.

The table can be re-written as :

Maximum possible total runs scored by Chatur in the four matches = 27 + 30 + 110 + 20 = 187.

In such a case minimum possible total runs scored by Arpit in the four matches = 23 + 100 + 13 + 53 = 189.

Difference = 189 – 187 = 2 (minimum possible) Subsequently total runs scored by Bimal in the four matches = 88 + 65 + 19 + 52 = 224.

Also, total runs scored by Elan in the four matches = 60 + 30 + 78 + 19 = 187

Absolute difference = 224 – 187 = 37

Let us analyse the scores of Match-1 first:

Runs scored by Bimal, Dinu and Elan = 88 + 72 + 60 = 220

Hence, runs scored by Arpit and Chatur = 270 – 220 = 50

Also 10% of 270 = 27

So, both Arpit and Chatur can score a maximum of 27 runs but the sum of their scores should be 50.

Arpit’s score’s range in Match-1 is 23–27 and subsequently Chatur’s score’s range in Match-1 is 27–23.

Similarly Chatur and Dinu scored 30 runs each in Match-2.

In Match-3 even though 10% of 240 = 24, as Dinu scored 20 runs, both Arpit and Bimal can score a maximum of 19 runs, but the sum of their scores should be 32.

In Match-4, Chatur and Elan combined scored 200 – 53 – 52 – 56 = 39 runs. As 10% of 200 = 20, one of Chatur or Elan scores 20 runs and the other scores 19 runs.

The table can be re-written as :

ndividual ranges for total score:

Arpit-> 189-199

Bimal-> 218-224

Chatur-> 182-187

Dinu-> 223

Elan-> 187-188

Least total will be of Chatur (Rank 5)

2nd least will be Elan (Rank 4)

Rank 3 must be of Arpit.

It is not possible to determine the exact ranks of Bimal and Dinu.