# Free DI and LR - 01 Practice Test - CAT

### Question 1

**Which team was ranked third in Group A after the first round?**

#### SOLUTION

Solution :D

Teams in Group A has following points, 17,12,11,7,0 in some order.

Initially its given that West Indies won 0 games. So if WI drew all the matches, it would end up with 8 points (4 draw). Other possibilities are,

DLPoints316224132040So we have only one matching value ie 0 points.

WDLPointsAustralia1127England21112South Africa31017West Indies0040New Zealand13011Hence answer option D

### Question 2

**Which team was ranked fourth in Group B after the first round?**

#### SOLUTION

Solution :A

Teams in Group B has the following points 20,9,7,6,4 in some order.

Initially its given that India lost 0 match. So possible combinations for India are,

WDPoints4020311722141311048 WDLPointsPakistan0316India40020Sri Lanka1127Zimbabwe0224Bangladesh1219

### Question 3

**How many matches in the first round resulted in a draw?**

#### SOLUTION

Solution :B

Total no. of teams in draw is 14. So no. of matches ending in draw is 142=7

Hence answer option B

### Question 4

**Which of the following teams did not lose a single match in the tournament?**

#### SOLUTION

Solution :C

Team with 20,17,12,9 points are India , South Africa, England and Bangladesh respectively.

After the second round, India has 12 points, SA has 2 points, England has 6 points and Bangladesh has 7.

Given that Team4 has 3 draws. Therefore it has 6 points after the second round. Hence this team is England.

Rest all teams have 1 draw in account which is against England. Its given that Team 1 has 0 win and Team 3 has 1 loss. Therefore only Team 2 has the possiblity of securing 12 points. Hence team 2 is India.

Team 1 has 0 win and 1 draw. Maximum possible points for this team is 2. Hence this team is SA.

WDLPointsTeamTeam10122South AfricaTeam221012IndiaTeam31117BangladeshTeam40306England

### Question 5

**Which of the following teams was the runner-up in the tournament?**

#### SOLUTION

Solution :B

Team with 20,17,12,9 points are India , South Africa, England and Bangladesh respectively.

After the second round, India has 12 points, SA has 2 points, England has 6 points and Bangladesh has 7.

Given that Team4 has 3 draws. Therefore it has 6 points after the second round. Hence this team is England.

Rest all teams have 1 draw in account which is against England. Its given that Team 1 has 0 win and Team 3 has 1 loss. Therefore only Team 2 has the possiblity of securing 12 points. Hence team 2 is India.

Team 1 has 0 win and 1 draw. Maximum possible points for this team is 2. Hence this team is SA.

WDLPointsTeamTeam10122South AfricaTeam221012IndiaTeam31117BangladeshTeam40306England

### Question 6

What is the ‘Overall Score’ of the Congressman/ Congressmen who stood first in the test?

#### SOLUTION

Solution :A

Option A is the correct answer.

### Question 7

Which of the following statements is false?

#### SOLUTION

Solution :D

Option D is the correct answer.

### Question 8

The number of Congressmen who definitely scored 0 marks in each of the three fields is:

#### SOLUTION

Solution :A

The number of Congressmen who scored 0 in the Honesty, Accessibility and Efficiency fields is 18, 25 and 31 respectively. As 18 is the least number, 18 Congressmen definitely scored 0 in all the three fields .

### Question 9

Based on the information given above, the total number of bugs found by MnD Ltd. is equal to

#### SOLUTION

Solution :

Let the number of bugs in Edusoft, Elevate and Financo solution alone be a, b and c respectively. Number of bugs in all the solution = x and number of bugs in Edusoft and Elevate solutions alone be d, Elevate and Financo solutions alone be g and Number of bugs in Edusoft and Financo solutions alone e.

Acc to passage,

C = 2x............eq(i)

a=3x -4...........eq(ii)

b=4+x....................eq(iii)

It is given that Total number of bugs in Edusoft solution is 24,

Hence,

3x -4+x+d+e = 24.......eq(iv)

4x +d+e=28

d:e:f=1:2:3 (in some order)

From last time of the passage

⇒ 3x - 4 =14

⇒ x =6This implies d + e = 4, d =1 or 3 or e =1 or 3

Total number of bugs = 14+12+10+6+2+(d+e)

= 48

### Question 10

**As a corporate policy, MnD Ltd. had decided that no bug would be found in more than two solutions and, so, those found in three solutions should be withdrawn from exactly one solution. The number of bugs withdrawn from the Elevate solution is two less than the number of bugs withdrawn from the Financo solution which, in turn, is two less than the number of bugs withdrawn from the Edusoft solution. Now, which of the following statements is/are definitely true? **

The number of bugs found in the Elevate solution is greater than 21.

The number of bugs found in the Financo solution is 3 more than that found in the Edusoft solution.

The number of bugs found in the Edusoft solution cannot be less than 18.

Both [1] and [2]

#### SOLUTION

Solution :C

Let the number of bugs in Edusoft, Elevate and Financo solution alone be a, b and c respectively. Number of bugs in all the solution = x and number of bugs in Edusoft and Elevate solutions alone be d, Elevate and Financo solutions alone be g and Number of bugs in Edusoft and Financo solutions alone e.

Acc to passage,

C = 2x............eq(i)

a=3x -4...........eq(ii)

b=4+x....................eq(iii)

It is given that Total number of bugs in Edusoft solution is 24,

Hence,

3x -4+x+d+e = 24.......eq(iv)

4x +d+e=28

d:e:f=1:2:3 (in some order)

From last time of the passage

⇒ 3x - 4 =14

⇒ x =6This implies d + e = 4, d =1 or 3 or e = 1 or 3

Total number of bugs = 14+12+10+6+2+(d+e)

= 48

### Question 11

**With reference to the above question, which of the following statements are definitely false?**

#### SOLUTION

Solution :D

Let the number of bugs in Edusoft, Elevate and Financo solution alone be a, b and c respectively. Number of bugs in all the solution = x and number of bugs in Edusoft and Elevate solutions alone be d, Elevate and Financo solutions alone be g and Number of bugs in Edusoft and Financo solutions alone e.

Acc to passage,

C = 2x............eq(i)

a=3x -4...........eq(ii)

b=4+x....................eq(iii)

It is given that Total number of bugs in Edusoft solution is 24,

Hence,

3x -4+x+d+e = 24.......eq(iv)

4x +d+e=28

d:e:f=1:2:3 (in some order)

From last time of the passage

⇒ 3x - 4 =14

⇒ x =6This implies d + e = 4, d =1 or 3 or e =1 or 3

Total number of bugs = 14+12+10+6+2+(d+e)

= 48

### Question 12

**After the withdrawal of bugs, as indicated in the previous question, some new bugs are discovered. Each of them belongs to only one solution in a manner such that the number of bugs found in each of the three solutions become identical.So, the minimum number of bugs which were discovered is equal to:**

#### SOLUTION

Solution :A

The number of bugs involved in the three solutions, after withdrawal are 20, 19 and 21. These numbers being consecutive can be made equal by adding a minimum of 3 new bugs.