# Free DI and LR - 30 Practice Test - CAT

### Question 1

Which team does not qualify for the finals?

#### SOLUTION

Solution :A

All teams score two points at the end of the first round, hence, the RRD needs to be calculated for each team

After the 6 matches,

Net RRD for A = -0.32

Net RRD for B = +1.04

Net RRD for C= -0.72. hence, C will not proceed to the finals. Answer is option (a)

### Question 2

Which team displayed the best performance in the first round (Best performance is defined as the team standing first by the criteria points + RRD)?

#### SOLUTION

Solution :B

Option B is the correct answer.

### Question 3

The team which failed to make it to the finals, would have made it to the finals if it had won in _____ overs in the last match it played

#### SOLUTION

Solution :B

C does not make it to the finals (Refer answer for 1

^{st}question of this caselet). Its last match is against A. this is the 6^{th}match. C’s Net RRD lags that of A by 0.40. hence, in the final match; run rate of C should exceed that of A by ≥ 0.41.In that match, A has a run rate of 6.12. C should have a run rate of 6.52or more. Let the number of overs required to achieve this run rate be x.

307/x=6.52, giving x= 47.08 overs. This is approximately 47 overs. Hence, if C had won at 47 overs or earlier, it could have qualified for the finals. Answer is option (b)

### Question 4

Which team made the best comeback to reach the finals?

#### SOLUTION

Solution :C

Option c

it can be observed that B made the best comeback after losing its first two matches and then winning its next two matches to make it to the finals. Answer is option (c)

### Question 5

What is the highest value of Net Runs per wicket (NRW) for a team? NRW is calculated as thedifference between runs per wicket scored by the team and runs per wicket conceded by the team?

#### SOLUTION

Solution :D

answer is option (d) and the team which has this high RRD is team A. Let us see how

Total Runs scored by team A = 263+290+185+306=1044

Wickets lost= 5+3+10+4=22

Run per wicket =1044/22 = 47.45

Total runs conceded = 258+260+188+307 = 1013

Wickets taken = 10+10+2+8= 30

Runs per wicket conceded = 1013/30= 33.77

Net runs per wicket = 47.45- 33.77 = 13.6

Other two teams will have a lower value than this.

C= 30.8 & B=12.3

### Question 6

A team wither wins by scoring more runs ( if it wins by batting first) or by wickets ( if it wins after batting second). The margin by which a team wins is calculated by converting all victories to Final Victory Margin (FVM). If a team batting second wins, the FVM is calculated as = (10- wickets lost by winning team)*5 + deliveries to spare when the match was won. If the team batting first wins, the FVM will be the difference between the runs scored by it and its opponents. What is the highest FVM in any match?

#### SOLUTION

Solution :D

### Question 7

As is the normal scenario, there is a legal betting going on. The “number 1 choice” to win the final is the team which had first position in the first round standings. The betting agency accepts bets as 2 to 1 for “number 1 choice” to win the final and 4 to 1 for “number 2 choice” to win the final. A standard fee of Rs.1000 is paid upfront by every person who wants to bet on either of the two teams winning. A total of 10000 people bet. At the end of the match, “number 2 choice” was the winner. The agency had an income of 50 lakhs. How many people betted correctly?

#### SOLUTION

Solution :E

Total income of the betting agency = 1000*10000 = 100,00,000

The number 2 choice has won. Hence, if “x” people have voted for the “number 2 choice”, these people will together win 4000x (since betting on “number 2 choice” at at odds 4:1).

Net profit of the agency = 5000000=10000000-4000x

x= 1250. option (e)

### Question 8

**Find the least possible number of W’s in all the baskets put together.**

#### SOLUTION

Solution :D

To find the minimum number of Ws, give the surplus over 20 in each basket to W to get the minimum. Thus the minimum is 9+8+8+5+1=31.

### Question 9

**Which of the following statement is definitely not true?**

#### SOLUTION

Solution :A

If the number of M's in basket 4 is 1, then a total of 20 cannot be obtained. Thus, this is definitely false.

### Question 10

**If the number of Gs in basket 4 is 6, then the number of O’s in basket 4 is**

#### SOLUTION

Solution :A

Number of Os will be 2

### Question 11

**In how many baskets did Ming find the numbers of M’s correctly**

#### SOLUTION

Solution :A

In basket 1 and basket 3

### Question 12

**The person who won the max number of bets is**

#### SOLUTION

Solution :B

From the question, none of them win / loss equal number of bets. Thus, the number of wins are 4,3,2,1,0 in some order . It is also known that The number of bets won by D is equal to the number of bets lost by A. thus the only possibilities for this are

A D Wins (case 1)13Wins (case 2)04But it is also given that the number of bets lost by C> number of bets lost by A. Thus, A ≠ 0

A wins 1 bet and D wins 3. Thus C wins 0. Also,

It is also known that

PersonNo. of WinsNo. of LossesA13B40C04D31E22

Thus, the outcome of each bet can also be arrived at. A table can be drawn up as followsIn the table above every + means a gain and every - means a loss.

For example: +17 in the last row means, E won 17 from A and similarly in the first row -17 means A lost 17 to E.

Taking into account the following conditions:

(vii) E won Rs.17 against A and the stake involved in the bet between C and D is Rs.4. The gain of A is equal to the loss of D.

And we already know that A has only one gain (against C) and D only one loss (against B). Lets assume that value to be x.

Now, we know that

(iii)The stake of any bet is an integer amount (in Rs) and is neither less than Rs.4 nor more than Rs.21

(iv)Neither the gain of any person who won atleast one bet nor the loss of any person who lost at least one bet isless than Rs.20 or more than Rs.40.

Hence we can write

Eqn1: 4 ≤ x ≤ 21

Eqn2: 20 ≤ x ≤40

From the above 2 eqns we get: 20 ≤ x ≤ 21

So x can have a value of either 20 or 21.

Step 2Now take stake (A vs D) = y

In the first row, the sum of all the stakes (neglecting them as gains or losses) should be equal to the total stake i.e. 55. So I get stake(A vs B) = (38-y). Similarly we can get, stake(D vs E) = 51-(x+y)

Again assume stake (B vs C) = z and henceforth fill in the stakes for (C vs E).

Now we have all the cells filled in the table

ABCDETotalStakeA−38−(x+y)+x−y−1755B+38−(x+y)+z+x+y−z−137C−x−z−4−36−(x+z)40D+y−x+4+51−(x+y)55E+17−y−z−1+36−(x+z)−51−(x+y)53

Step 3Adding all stakes of E: 17 + y - z - 1 + 36 - (x+z) + 51 - (x+y) = 53

⇒ 103 - 2 (x + z) = 53

⇒ 2 (x + z) = 50

⇒ x + z = 25

Now z can have two values 4 or 5. But, (ii)For any person the stakes of any 2 bets were different. And since C has 2 stakes - both “z” and “4”, therefore z is not equal to 4.

z = 5, and hence x = 20.

This answers the next question. The stake (in Rs) in the bet between B and C is (b) 5

Now total loss of A = 38 - (x + y) + y + 17

=55 - x

=55 - 20 = 35

### Question 13

**The person who won the same number of bets as he lost is**

#### SOLUTION

Solution :D

From the question, none of them win / loss equal number of bets. Thus, the number of wins are 4,3,2,1,0 in some order . It is also known that The number of bets won by D is equal to the number of bets lost by A. thus the only possibilities for this are

A D Wins (case 1)13Wins (case 2)04But it is also given that the number of bets lost by C> number of bets lost by A. Thus, A ≠ 0

A wins 1 bet and D wins 3. Thus C wins 0. Also,

It is also known that

PersonNo. of WinsNo. of LossesA13B40C04D31E22

Thus, the outcome of each bet can also be arrived at. A table can be drawn up as followsIn the table above every + means a gain and every - means a loss.

For example: +17 in the last row means, E won 17 from A and similarly in the first row -17 means A lost 17 to E.

Taking into account the following conditions:

(vii) E won Rs.17 against A and the stake involved in the bet between C and D is Rs.4. The gain of A is equal to the loss of D.

And we already know that A has only one gain (against C) and D only one loss (against B). Lets assume that value to be x.

Now, we know that

(iii)The stake of any bet is an integer amount (in Rs) and is neither less than Rs.4 nor more than Rs.21

(iv)Neither the gain of any person who won atleast one bet nor the loss of any person who lost at least one bet isless than Rs.20 or more than Rs.40.

Hence we can write

Eqn1: 4 ≤ x ≤ 21

Eqn2: 20 ≤ x ≤40

From the above 2 eqns we get: 20 ≤ x ≤ 21

So x can have a value of either 20 or 21.

Step 2Now take stake (A vs D) = y

In the first row, the sum of all the stakes (neglecting them as gains or losses) should be equal to the total stake i.e. 55. So I get stake(A vs B) = (38-y). Similarly we can get, stake(D vs E) = 51-(x+y)

Again assume stake (B vs C) = z and henceforth fill in the stakes for (C vs E).

Now we have all the cells filled in the table

ABCDETotalStakeA−38−(x+y)+x−y−1755B+38−(x+y)+z+x+y−z−137C−x−z−4−36−(x+z)40D+y−x+4+51−(x+y)55E+17−y−z−1+36−(x+z)−51−(x+y)53

Step 3Adding all stakes of E: 17 + y - z - 1 + 36 - (x+z) + 51 - (x+y) = 53

⇒ 103 - 2 (x + z) = 53

⇒ 2 (x + z) = 50

⇒ x + z = 25

Now z can have two values 4 or 5. But, (ii)For any person the stakes of any 2 bets were different. And since C has 2 stakes - both “z” and “4”, therefore z is not equal to 4.

z = 5, and hence x = 20.

This answers the next question. The stake (in Rs) in the bet between B and C is (b) 5

Now total loss of A = 38 - (x + y) + y + 17

=55 - x

=55 - 20 = 35

### Question 14

**The stake (in Rs) in the bet between B and C is**

#### SOLUTION

Solution :B

From the question, none of them win / loss equal number of bets. Thus, the number of wins are 4,3,2,1,0 in some order . It is also known that The number of bets won by D is equal to the number of bets lost by A. thus the only possibilities for this are

A D Wins (case 1)13Wins (case 2)04But it is also given that the number of bets lost by C> number of bets lost by A. Thus, A ≠ 0

A wins 1 bet and D wins 3. Thus C wins 0. Also,

It is also known that

PersonNo. of WinsNo. of LossesA13B40C04D31E22

Thus, the outcome of each bet can also be arrived at. A table can be drawn up as followsIn the table above every + means a gain and every - means a loss.

For example: +17 in the last row means, E won 17 from A and similarly in the first row -17 means A lost 17 to E.

Taking into account the following conditions:

(vii) E won Rs.17 against A and the stake involved in the bet between C and D is Rs.4. The gain of A is equal to the loss of D.

And we already know that A has only one gain (against C) and D only one loss (against B). Lets assume that value to be x.

Now, we know that

(iii)The stake of any bet is an integer amount (in Rs) and is neither less than Rs.4 nor more than Rs.21

(iv)Neither the gain of any person who won atleast one bet nor the loss of any person who lost at least one bet isless than Rs.20 or more than Rs.40.

Hence we can write

Eqn1: 4 ≤ x ≤ 21

Eqn2: 20 ≤ x ≤40

From the above 2 eqns we get: 20 ≤ x ≤ 21

So x can have a value of either 20 or 21.

Step 2Now take stake (A vs D) = y

In the first row, the sum of all the stakes (neglecting them as gains or losses) should be equal to the total stake i.e. 55. So I get stake(A vs B) = (38-y). Similarly we can get, stake(D vs E) = 51-(x+y)

Again assume stake (B vs C) = z and henceforth fill in the stakes for (C vs E).

Now we have all the cells filled in the table

ABCDETotalStakeA−38−(x+y)+x−y−1755B+38−(x+y)+z+x+y−z−137C−x−z−4−36−(x+z)40D+y−x+4+51−(x+y)55E+17−y−z−1+36−(x+z)−51−(x+y)53

Step 3Adding all stakes of E: 17 + y - z - 1 + 36 - (x+z) + 51 - (x+y) = 53

⇒ 103 - 2 (x + z) = 53

⇒ 2 (x + z) = 50

⇒ x + z = 25

Now z can have two values 4 or 5. But, (ii)For any person the stakes of any 2 bets were different. And since C has 2 stakes - both “z” and “4”, therefore z is not equal to 4.

z = 5, and hence x = 20.

This answers the next question. The stake (in Rs) in the bet between B and C is (b) 5

Now total loss of A = 38 - (x + y) + y + 17

=55 - x

=55 - 20 = 35

### Question 15

**What is the total loss (in Rs) of A?**

#### SOLUTION

Solution :C

But it is also given that the number of bets lost by C> number of bets lost by A. Thus, A ≠ 0

A wins 1 bet and D wins 3. Thus C wins 0. Also,

It is also known that

Thus, the outcome of each bet can also be arrived at. A table can be drawn up as followsIn the table above every + means a gain and every - means a loss.

Taking into account the following conditions:

Now, we know that

Hence we can write

Eqn1: 4 ≤ x ≤ 21

Eqn2: 20 ≤ x ≤40

From the above 2 eqns we get: 20 ≤ x ≤ 21

So x can have a value of either 20 or 21.

Step 2Now take stake (A vs D) = y

Again assume stake (B vs C) = z and henceforth fill in the stakes for (C vs E).

Now we have all the cells filled in the table

ABCDETotalStakeA−38−(x+y)+x−y−1755B+38−(x+y)+z+x+y−z−137C−x−z−4−36−(x+z)40D+y−x+4+51−(x+y)55E+17−y−z−1+36−(x+z)−51−(x+y)53

Step 3Adding all stakes of E: 17 + y - z - 1 + 36 - (x+z) + 51 - (x+y) = 53

⇒ 103 - 2 (x + z) = 53

⇒ 2 (x + z) = 50

⇒ x + z = 25

z = 5, and hence x = 20.

This answers the next question. The stake (in Rs) in the bet between B and C is (b) 5

Now total loss of A = 38 - (x + y) + y + 17

=55 - x

=55 - 20 = 35

This answers the question - What is the total loss (in Rs) of A?