Logical Reasoning & DI (LRDI) Test - 15

Let's determine which are losing and winning positions.

If there are 1 to 6 coins on the table, then it is a winning position, because a person can just pick them up and win. 

If there are 7 or 8 coins on the table then it is a losing position. Even if you pick anything from 2-6, you will leave your opponent in a winning position.

If there are 9-14 coins on the table then it is a winning position. For 9 you can pick 2 leaving the opponent with 7 which is a losing position. For 10, you pick 3 and so on. For 14, you pick 6 leaving 8 on the table which is also a losing position. Hence, all of 9-14 are winning positions.

If there 15 or 16 coins on the table, then it is a losing position. Even if you pick anything from 2-6, you will leave your opponent in a winning position.

Hence, we can see a pattern emerging:

1-6 : win

7,8 : lose

9-14: win

15, 16: lose

17-22: win

23, 24:  lose

So all 8x-1 and 8x positions are losing positions. So if Rinesh has to win, he has to leave his opponent in a losing position. So he should leave 8k coins on the table such that 8k<150. The nearest multiple of 8 which is less than 150 is 144. Hence, if Rinesh leaves 144 coins on the table, Rinsha will lose. So he must pick 150-144= 6 coins

The answer is option D.

Since both of them play intelligently and the person who picks the last marble is the loser,

If it is Rinsha's turn to pick the marble, she will ensure that after he picks the marbles, the number of marbles left on the table is either 1 or 2.

Since both of them play intelligently, Rinsha will always ensure to leave 8x+1 or 8x+2 marbles on the table after her turn.
Ultimately, we left with 150-(8x+1) or 150-(8x+2) i.e.5,4 marbles respectively on the table.
But there is a condition that if there are less than 2 marbles on the table, then the opponent has to pick the marble.

So Rinsha picks 4 marbles and will leave 2 marbles for Rinesh to pick.

Since we have to find the minimum number of marbles Rinsha should pick, so Rinsha will pick 4 marbles.

For Group A:

Let's determine which are losing and winning positions.

If there are 1 to 6 coins on the table, then it is a winning position, because a person can just pick them up and win.

If there are 7 or 8 coins on the table then it is a losing position. Even if you pick anything from 2-6, you will leave your opponent in a winning position.

If there are 9-14 coins on the table then it is a winning position. For 9 you can pick 2 leaving the opponent with 7 which is a losing position. For 10, you pick 3 and so on. For 14, you pick 6 leaving 8 on the table which is also a losing position. Hence, all of 9-14 are winning positions.

If there 15 or 16 coins on the table, then it is a losing position. Even if you pick anything from 2-6, you will leave your opponent in a winning position.

Hence, we can see a pattern emerging:

1-6 : win

7,8 : lose

9-14: win

15, 16: lose

17-22: win

23, 24: lose

So all 8x-1 and 8x positions are losing positions. So if Rinesh has to win, he has to leave his opponent in a losing position. So he should leave 8k coins on the table such that 8k < 74. The nearest multiple of 8 which is less than 74 is 72. Hence, if Rinesh leaves 72 coins on the table, Rinsha will lose. So he must pick 74 - 72 = 2 coins

For Group B:

Since both of them play intelligently and the person who picks the last marble is the loser,

Since both of them play intelligently, Rinesh should ensure to leave 8x+1 or 8x+2 marbles on the table after his turn.

There are 76 marbles. 

Therefore, he should ensure to leave 73 or 74 marbles on the table after his turn.

This implies he should pick 2 or 3 marbles.

Minimum number of marbles he should pick will be 2.

Required sum = 2 + 2 = 4

The answer is 4

Since both of them play intelligently, and the person who picks the last marble is the winner, If it is Rinesh's turn to pick the coin, he will
ensure that after he picks the marbles, the number of marbles left on the table is a multiple of 8.
In a single round, 8 marbles will be picked up both of them.
Ultimately, we left with 150-8x i.e.6 marbles on the table.
Since after a person's turn if there are less than 2 marbles on the table, then the game ends. If Rinesh picks five marbles, then there will be one marble on the table, the game ends, Rinesh will be the winner.
If Rinesh picks six marbles, there he will try to make the number of marbles after his turn to be a multiple of 8, and he will be the winner.

Since we have to find the minimum number of marbles, Rinesh will pick five marbles.
B is the correct answer.

Since both of them play intelligently, and the person who picks the last marble is the winner, If it is Rinesh's turn to pick the coin, he will
ensure that after he picks the marbles, the number of marbles left on the table is a multiple of 8.
In a single round, 8 marbles will be picked up both of them.
Ultimately, we left with 150-8x i.e.6 marbles on the table.
Since after a person's turn if there are less than 2 marbles on the table,then he has to pick all the marbles. If Rinsha picks five marbles, then there will be one marble on the table,Rinsha has to pick that marble as well. Rinesh will be the winner.
If Rinsha picks four marbles, then there will be 2 marbles which Rinesh has to pick.
If Rinsha picks three marbles, then even if Rinesh picks two marbles in his turn since there are less than 2 marbles on the table, Rinesh has to pick the third leftover marble aswell.

Since we have to find the maximum number of marbles which Rinsha has to pick, she will pick four marbles.

It is given that the annual income of one among Augusta, Boliv and Centy is the average of the annual incomes of the other two. So, their incomes can be (40, 45, 50) lakhs or (40, 50, 60) lakhs or (45, 60, 75) lakhs in any order. Also, the absolute difference Augusta's income and Elin's income is 15 lakhs. So, their incomes can be (45, 60) lakhs or (60, 75) lakhs.
If Augusta's, Boliv,s and Centy's incomes are (45, 60, 75) lakhs in any order, we will have three cases:

In case I, if Augusta's income is 45 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case I is not possible.
In case II, if Augusta's income is 60 lakhs, Elin's income must be 75 lakhs or 45 lakhs but Boliv's and Centy's income are 75 lakhs and 45 lakhs. Case II is not possible.
In case III, if Augusta's income is 75 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case III is not possible.
So, Augusta's, Boliv's and Centy's income must be either (40, 45, 50) lakhs or (40, 50, 60) lakhs. Augusta's income can either be 45 lakhs or 60 lakhs. We get the following cases:

Diaz does not have the highest income. So, case I is not valid. Elin is the youngest male and Boliv is the older female. It is also given that the older female has more income. So, Boliv's income cannot be 40 lakhs because it is the least income. Thus, Boliv's income is 50 lakhs and Centy's income is 40 lakhs.

Centy is older that Boliv and Elin. So, Centy cannot be a female. Also, Augusta cannot be a female as Augusta's income is more than Boliv's. So, Diaz must be female.

Also, Centy is older than Boliv and Elin and Centy is having the least income. So, he can not be the oldest. Also, Elin has the highest income. So, he must be younger than one of the females and older than the other. Since, Elin is younger than Boliv, he must be older than the other female, Diaz. Thus, Augusta must be the oldest. The arrangement in the decreasing order of age is :

From the arrangement we can see that Augusta is the oldest and his salary is 60 lakhs.
Hence, option A is the correct answer.

It is given that the annual income of one among Augusta, Boliv and Centy is the average of the annual incomes of the other two. So, their incomes can be (40, 45, 50) lakhs or (40, 50, 60) lakhs or (45, 60, 75) lakhs in any order. Also, the absolute difference Augusta's income and Elin's income is 15 lakhs. So, their incomes can be (45, 60) lakhs or (60, 75) lakhs. If Augusta's, Boliv,s and Centy's incomes are (45, 60, 75) lakhs in any order, we will have three cases:

In case I, if Augusta's income is 45 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case I is not possible. In case II, if Augusta's income is 60 lakhs, Elin's income must be 75 lakhs or 45 lakhs but Boliv's and Centy's income are 75 lakhs and 45 lakhs. Case II is not possible. In case III, if Augusta's income is 75 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case III is not possible. So, Augusta's, Boliv's and Centy's income must be either (40, 45, 50) lakhs or (40, 50, 60) lakhs. Augusta's income can either be 45 lakhs or 60 lakhs. We get the following cases:

Diaz does not have the highest income. So, case I is not valid. Elin is the youngest male and Boliv is the older female. It is also given that the older female has more income. So, Boliv's income cannot be 40 lakhs because it is the least income. Thus, Boliv's income is 50 lakhs and Centy's income is 40 lakhs.

Centy is older that Boliv and Elin. So, Centy cannot be a female. Also, Augusta cannot be a female as Augusta's income is more than Boliv's. So, Diaz must be female.

Also, Centy is older than Boliv and Elin and Centy is having the least income. So, he can not be the oldest. Also, Elin has the highest income. So, he must be younger than one of the females and older than the other. Since, Elin is younger than Boliv, he must be older than the other female, Diaz. Thus, Augusta must be the oldest. The arrangement in the decreasing order of age is :

From the arrangement we can see that sum of incomes of Boliv and Diaz is (50 + 45) lakhs = 95 lakhs.
Hence, option C is the correct answer.

It is given that the annual income of one among Augusta, Boliv and Centy is the average of the annual incomes of the other two. So, their incomes can be (40, 45, 50) lakhs or (40, 50, 60) lakhs or (45, 60, 75) lakhs in any order. Also, the absolute difference Augusta's income and Elin's income is 15 lakhs. So, their incomes can be (45, 60) lakhs or (60, 75) lakhs. If Augusta's, Boliv,s and Centy's incomes are (45, 60, 75) lakhs in any order, we will have three cases:

In case I, if Augusta's income is 45 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case I is not possible. In case II, if Augusta's income is 60 lakhs, Elin's income must be 75 lakhs or 45 lakhs but Boliv's and Centy's income are 75 lakhs and 45 lakhs. Case II is not possible. In case III, if Augusta's income is 75 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case III is not possible. So, Augusta's, Boliv's and Centy's income must be either (40, 45, 50) lakhs or (40, 50, 60) lakhs. Augusta's income can either be 45 lakhs or 60 lakhs. We get the following cases:

Diaz does not have the highest income. So, case I is not valid. Elin is the youngest male and Boliv is the older female. It is also given that the older female has more income. So, Boliv's income cannot be 40 lakhs because it is the least income. Thus, Boliv's income is 50 lakhs and Centy's income is 40 lakhs.

Centy is older that Boliv and Elin. So, Centy cannot be a female. Also, Augusta cannot be a female as Augusta's income is more than Boliv's. So, Diaz must be female.

Also, Centy is older than Boliv and Elin and Centy is having the least income. So, he can not be the oldest. Also, Elin has the highest income. So, he must be younger than one of the females and older than the other. Since, Elin is younger than Boliv, he must be older than the other female, Diaz. Thus, Augusta must be the oldest. The arrangement in the decreasing order of age is :

From the arrangement we can see that in both the cases, only the position of Boliv is same.
Hence, 1 is the correct answer.

It is given that the annual income of one among Augusta, Boliv and Centy is the average of the annual incomes of the other two. So, their incomes can be (40, 45, 50) lakhs or (40, 50, 60) lakhs or (45, 60, 75) lakhs in any order. Also, the absolute difference Augusta's income and Elin's income is 15 lakhs. So, their incomes can be (45, 60) lakhs or (60, 75) lakhs. If Augusta's, Boliv,s and Centy's incomes are (45, 60, 75) lakhs in any order, we will have three cases:

In case I, if Augusta's income is 45 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case I is not possible. In case II, if Augusta's income is 60 lakhs, Elin's income must be 75 lakhs or 45 lakhs but Boliv's and Centy's income are 75 lakhs and 45 lakhs. Case II is not possible. In case III, if Augusta's income is 75 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case III is not possible. So, Augusta's, Boliv's and Centy's income must be either (40, 45, 50) lakhs or (40, 50, 60) lakhs. Augusta's income can either be 45 lakhs or 60 lakhs. We get the following cases:

Diaz does not have the highest income. So, case I is not valid. Elin is the youngest male and Boliv is the older female. It is also given that the older female has more income. So, Boliv's income cannot be 40 lakhs because it is the least income. Thus, Boliv's income is 50 lakhs and Centy's income is 40 lakhs.

Centy is older that Boliv and Elin. So, Centy cannot be a female. Also, Augusta cannot be a female as Augusta's income is more than Boliv's. So, Diaz must be female.

Also, Centy is older than Boliv and Elin and Centy is having the least income. So, he can not be the oldest. Also, Elin has the highest income. So, he must be younger than one of the females and older than the other. Since, Elin is younger than Boliv, he must be older than the other female, Diaz. Thus, Augusta must be the oldest. The arrangement in the decreasing order of age is :

From the arrangement we can see that the sum of the incomes of Elin and Diaz is (75 + 45) lakhs = 120 lakhs. Average = 60 lakhs
Hence, 60 is the correct answer.

It is given that the annual income of one among Augusta, Boliv and Centy is the average of the annual incomes of the other two. So, their incomes can be (40, 45, 50) lakhs or (40, 50, 60) lakhs or (45, 60, 75) lakhs in any order. Also, the absolute difference Augusta's income and Elin's income is 15 lakhs. So, their incomes can be (45, 60) lakhs or (60, 75) lakhs. If Augusta's, Boliv,s and Centy's incomes are (45, 60, 75) lakhs in any order, we will have three cases:

In case I, if Augusta's income is 45 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case I is not possible. In case II, if Augusta's income is 60 lakhs, Elin's income must be 75 lakhs or 45 lakhs but Boliv's and Centy's income are 75 lakhs and 45 lakhs. Case II is not possible. In case III, if Augusta's income is 75 lakhs, Elin's income must be 60 lakhs but either Boliv's or Centy's income is 60 lakhs. Case III is not possible. So, Augusta's, Boliv's and Centy's income must be either (40, 45, 50) lakhs or (40, 50, 60) lakhs. Augusta's income can either be 45 lakhs or 60 lakhs. We get the following cases:

Diaz does not have the highest income. So, case I is not valid. Elin is the youngest male and Boliv is the older female. It is also given that the older female has more income. So, Boliv's income cannot be 40 lakhs because it is the least income. Thus, Boliv's income is 50 lakhs and Centy's income is 40 lakhs.

Centy is older that Boliv and Elin. So, Centy cannot be a female. Also, Augusta cannot be a female as Augusta's income is more than Boliv's. So, Diaz must be female.

Here, option A is not true, since the salary of Elin is 75 L. Option B is not true since Augusta is Male. option D is not true since the salary of Boliv is 50 L .

In such a tournament, the total number of points scored by all the teams together would be a constant in all the arrangements.

For n teams, if there are no draws, the total number of wins would be [n(n-1)/2 ] * points per win. Hence the total number of 
points in this case is given by [5*4/2]*1 = 10 points. 

The total number of points scored among all the teams together would be 10. 
For each arrangement we try to distribute the points such that they would add up to 10. We find that there are 9 such arrangements and the arrangements are given by: 

If we fix Aces, Beasts, Cavaliers, Dashers and Enigmas in the place of Team 1 , Team 2 , Team 3 , Team 4 , Team 5,  there are 9 arrangements possible.

All the possible point distributions are as follows.

The only arrangement in which the third team scores 3 points is in Ar_5.

Hence, only one arrangement of team scores is possible. Now we need to arrange team names in these 5 slots. As the top 3 teams are tied, the order would be decided by alphabetical order. Hence, if we select 3 teams for the top slot, only one arrangement of teams is possible. Hence, number of arrangements of top 3 teams = 5C3 * 1 = 10 ways.

The bottom 2 teams can be arranged in 2! ways.

The number of possible arrangements are, 5C3 * 2 = 20 ways.

Enigmas cannot be tied with any team at the top of the table, as by virtue of their rank in the alphabetic order they would be pushed down. Hence the total arrangements that are possible with Enigmas at the top of the table is as follows.

Now different arrangements of teams 2,3,4,5 has to be made for each points distribution.

Arr_1: Each of the remaining 4 teams can come in any of the four spots. Hence, the number of ways of arrangement would be 4! = 24

Arr_2 : One of the teams must be 3 and the remaining three teams would be 1. The teams that score 1 point would arrange themselves alphabetically. So the number of ways of arrangement would be the same as picking the team which scored 3 points. This is done in 4C1 = 4 ways.

Arr_3: This is arranged similar to Arr_2. The team that scored 0 points is chosen from 4 teams. This can be done in 4C1 = 4ways.

Arr_4: Two teams score 2 points and two teams score one point. Hence, the number of ways of arrangement would be the same as choosing two teams that scored 2 points from the four teams. They will arrange themselves alphabetically. The teams that are not chosen score 1 point and they would arrange themselves alphabetically as well. Therefore the number of ways is 4C2 = 6 ways

Arr_8 : This is similar to Arr_2 and Arr_3.  The team that scores 1 point must be chosen from the 4 teams. The number of ways is 4C1 = 4 ways.

Total combinations = 42

The different distribution of points in which atleast 3 teams have equal number of points is as follows.

Number of ways teams can be arranged in Arr_2 : The three teams that score 1 point can be selected in 5C3 ways. They would be arranged alphabetically automatically. The remaining two teams can be arranged in 2 ways. Hence total arrangement is 5C3*2=20

Number of ways teams can be arranged in Arr_3 =5C3*2=20

Number of ways teams can be arranged in Arr_5 =5C3*2=20

Number of ways teams can be arranged in Arr_8 =5C3*2=20

Number of ways teams can be arranged in Arr_9 =1

Total number of arrangements = 81

The different arrangements where each team win at least one game are coloured below.

In Arr_2, the three teams which scored equal points will be arranged in one way (i.e. alphabetical order).

Selection of 3 teams out of 5 teams = $$^5C_3=\frac{5!}{3!\times\ 2!}=10\ ways$$. The rest two teams can be arranged in 2! ways.

Hence total ways = 10 x 2! = 10 x 2 = 20 ways.

Similarly

For Arr_4:- Number of ways = $$^5C_2\times\ ^3C_2\times\ 1!=\frac{5!}{3!\times\ 2!}\times\ \frac{3!}{2!\times\ 1!}=10\times\ 3=30\ ways$$

For Arr_7:- Number of ways = $$^5C_2\times\ ^3C_2\times\ 1!=\frac{5!}{3!\times\ 2!}\times\ \frac{3!}{2!\times\ 1!}=10\times\ 3=30\ ways$$

For Arr_8:- Number of ways = $$^5C_3\times\ 2!=\frac{5!\times\ 2!}{3!\times\ 2!}=20\ ways.$$

For Arr_9:- All teams have an equal score. Hence they will be arranged in one way. 

Total possible arrangements = 20 + 30 + 30 + 20 + 1 = 101 ways.

The answer is Option (B) 

Using 2, 3 and 4, 

From 1, $$\ \dfrac{\ 4}{3}=\ \dfrac{\ 20+2y+x}{20+y+x}$$
=> 80+4y+4x=60+6y+3x
=> 20+x=2y.......(i)
From 5, x+y+40 = 65
=> x+y =25 ......(ii)
From (i) and (ii), x =10, y=15
The number of arcade games= x+20+2y = 10+20+30 = 60

Using 2, 3 and 4, 

From 1, $$\ \dfrac{\ 4}{3}=\ \dfrac{\ 20+2y+x}{20+y+x}$$
=> 80+4y+4x=60+6y+3x
=> 20+x=2y.......(i)
From 5, x+y+40 = 65
=> x+y =25 ......(ii)
From (i) and (ii), x =10, y=15
The number of arcade shooting games= 20+2y = 20+2*15 =50
The number of strategy shooting games = 20+x = 30
Hence, the ratio = $$\ \frac{\ 50}{30}=\ \frac{\ 5}{3}$$

Using 2, 3 and 4,

From 1, $$\ \dfrac{\ 4}{3}=\ \dfrac{\ 20+2y+x}{20+y+x}$$
=> 80+4y+4x=60+6y+3x
=> 20+x=2y.......(i)
From 5, x+y+40 = 65
=> x+y =25 ......(ii)
From (i) and (ii), x =10, y=15
The number of games which are both racing and shooting = 20+20 =40
Total number of games produced = 40+2x+3y = 40+20+3*15 = 105
Required percentage = $$\ \frac{\ 40\times\ 100}{105}$$ = 38

Using 2, 3 and 4,

From 1, $$\ \dfrac{\ 4}{3}=\ \dfrac{\ 20+2y+x}{20+y+x}$$
=> 80+4y+4x=60+6y+3x
=> 20+x=2y.......(i)
From 5, x+y+40 = 65
=> x+y =25 ......(ii)
From (i) and (ii), x =10, y=15
In arcade, ratio of racing games to shooting games = $$\ \frac{\ 10+20}{20+30}$$ = $$\ \frac{\ 3}{5}$$

Using 2, 3 and 4,

From 1, 43= 20+2�+�20+�+� 3 4​= 20+y+x 20+2y+x​
=> 80+4y+4x=60+6y+3x
=> 20+x=2y.......(i)
From 5, x+y+40 = 65
=> x+y =25 ......(ii)
From (i) and (ii), x =10, y=15

Hence number of shooting games = 40
Hence number of racing games = 25

Hence number of combined games = 40 

hence ratio of racing , shooting and both games = 5:8:8

After multiplying this ratio with respective sales price ratio we get - 55:72:104