Free DI and LR - 30 Practice Test - CAT 

 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)

Option B is the correct answer.

 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)

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)

 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

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)

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. 

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

Number of Os will be 2

In basket 1 and basket 3

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

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

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

It is also known that

$\begin{array}{ccc}Person& No.ofWins& No.ofLosses\\ A& 1& 3\\ B& 4& 0\\ C& 0& 4\\ D& 3& 1\\ E& 2& 2\end{array}$
Thus, the outcome of each bet can also be arrived at. A table can be drawn up as follows

In 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 $\le$ x $\le$ 21

Eqn2: 20 $\le$ x $\le$40

From the above 2 eqns we get: 20 $\le$ x $\le$ 21

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

Step 2

Now 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

Step 3

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

$\Rightarrow$ 103 - 2 (x + z) = 53

$\Rightarrow$ 2 (x + z) = 50

$\Rightarrow$ 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

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

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

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

It is also known that

$\begin{array}{ccc}Person& No.ofWins& No.ofLosses\\ A& 1& 3\\ B& 4& 0\\ C& 0& 4\\ D& 3& 1\\ E& 2& 2\end{array}$
Thus, the outcome of each bet can also be arrived at. A table can be drawn up as follows

In 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 $\le$ x $\le$ 21

Eqn2: 20 $\le$ x $\le$40

From the above 2 eqns we get: 20 $\le$ x $\le$ 21

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

Step 2

Now 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

Step 3

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

$\Rightarrow$ 103 - 2 (x + z) = 53

$\Rightarrow$ 2 (x + z) = 50

$\Rightarrow$ 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

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

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

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

It is also known that

$\begin{array}{ccc}Person& No.ofWins& No.ofLosses\\ A& 1& 3\\ B& 4& 0\\ C& 0& 4\\ D& 3& 1\\ E& 2& 2\end{array}$
Thus, the outcome of each bet can also be arrived at. A table can be drawn up as follows

In 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 $\le$ x $\le$ 21

Eqn2: 20 $\le$ x $\le$40

From the above 2 eqns we get: 20 $\le$ x $\le$ 21

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

Step 2

Now 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

Step 3

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

$\Rightarrow$ 103 - 2 (x + z) = 53

$\Rightarrow$ 2 (x + z) = 50

$\Rightarrow$ 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

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

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

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

It is also known that

$\begin{array}{ccc}Person& No.ofWins& No.ofLosses\\ A& 1& 3\\ B& 4& 0\\ C& 0& 4\\ D& 3& 1\\ E& 2& 2\end{array}$
Thus, the outcome of each bet can also be arrived at. A table can be drawn up as follows

In 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 $\le$ x $\le$ 21

Eqn2: 20 $\le$ x $\le$40

From the above 2 eqns we get: 20 $\le$ x $\le$ 21

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

Step 2

Now 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

Step 3

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

$\Rightarrow$ 103 - 2 (x + z) = 53

$\Rightarrow$ 2 (x + z) = 50

$\Rightarrow$ 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

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