Logical Reasoning & DI (LRDI) Test - 9

Let the child who took the minimum chocolates have x chocolates. Which implies Mohit would have taken x + 8 chocolates.
If x = 1, number of chocolates with Mohit = 9. The others should have a different number among 2 to 8. The sum of numbers from 1 to 9 is 45, out of which 5 can be omitted to get a sum of 40.

If x = 2, number of chocolates with Mohit = 10. The sum of consecutive numbers from 2 to 10 is 54. The sum of 40 cannot be achieved by omitting any one number. Hence, x > 1 is not possible.
So, x = 1, Mohit has the maximum chocolates - 9.
Others have one among 1, 2, 3, 4, 6, 7 and 8.
Asha took the second lowest number of chocolates; hence she took 2 chocolates.
If everyone took an equal number of chocolates, they would have taken 5 chocolates each.
Therefore, Eshan + Govind = 10. The possibilities are (6, 4) or (3, 7).
But, Aman took one chocolate more than Deepa - The possibilities are (3,4), (6,7) or (7, 8)
From the above statements, it is clear that Aman and Deepa took 8 and 7 chocolates respectively.
Eshan and Govind took 6 and 4 chocolates in any order.
The remaining chocolates 1 and 3 could have been taken by anyone among Tarun and Arun.
Tabulating the results:

Asha took 2 chocolates.

Let the child who took the minimum chocolates have x chocolates. Which implies Mohit would have taken x + 8 chocolates.
If x = 1, number of chocolates with Mohit = 9. The others should have a different number among 2 to 8. The sum of numbers from 1 to 9 is 45, out of which 5 can be omitted to get a sum of 40.

If x = 2, number of chocolates with Mohit = 10. The sum of consecutive numbers from 2 to 10 is 54. The sum of 40 cannot be achieved by omitting any one number. Hence, x > 1 is not possible.
So, x = 1, Mohit has the maximum chocolates - 9.
Others have one among 1, 2, 3, 4, 6, 7 and 8.
Asha took the second lowest number of chocolates; hence she took 2 chocolates.
If everyone took an equal number of chocolates, they would have taken 5 chocolates each.
Therefore, Eshan + Govind = 10. The possibilities are (6, 4) or (3, 7).
But, Aman took one chocolate more than Deepa - The possibilities are (3,4), (6,7) or (7, 8)
From the above statements, it is clear that Aman and Deepa took 8 and 7 chocolates respectively.
Eshan and Govind took 6 and 4 chocolates in any order.
The remaining chocolates 1 and 3 could have been taken by anyone among Tarun and Arun.
Tabulating the results:

The least number of chocolates is 1, which could have been taken either by Tarun or Arun. Hence, it cannot be determined.

Let the child who took the minimum chocolates have x chocolates. Which implies Mohit would have taken x + 8 chocolates.
If x = 1, number of chocolates with Mohit = 9. The others should have a different number among 2 to 8. The sum of numbers from 1 to 9 is 45, out of which 5 can be omitted to get a sum of 40.

If x = 2, number of chocolates with Mohit = 10. The sum of consecutive numbers from 2 to 10 is 54. The sum of 40 cannot be achieved by omitting any one number. Hence, x > 1 is not possible.
So, x = 1, Mohit has the maximum chocolates - 9.
Others have one among 1, 2, 3, 4, 6, 7 and 8.
Asha took the second lowest number of chocolates; hence she took 2 chocolates.
If everyone took an equal number of chocolates, they would have taken 5 chocolates each.
Therefore, Eshan + Govind = 10. The possibilities are (6, 4) or (3, 7).
But, Aman took one chocolate more than Deepa - The possibilities are (3,4), (6,7) or (7, 8)
From the above statements, it is clear that Aman and Deepa took 8 and 7 chocolates respectively.
Eshan and Govind took 6 and 4 chocolates in any order.
The remaining chocolates 1 and 3 could have been taken by anyone among Tarun and Arun.
Tabulating the results:

Except for Arun, Tarun, Esha and Govind, the number of chocolates is fixed.
The following combinations are possible:

Therefore, 4 ways.

Let the child who took the minimum chocolates have x chocolates. Which implies Mohit would have taken x + 8 chocolates.
If x = 1, number of chocolates with Mohit = 9. The others should have a different number among 2 to 8. The sum of numbers from 1 to 9 is 45, out of which 5 can be omitted to get a sum of 40.

If x = 2, number of chocolates with Mohit = 10. The sum of consecutive numbers from 2 to 10 is 54. The sum of 40 cannot be achieved by omitting any one number. Hence, x > 1 is not possible.
So, x = 1, Mohit has the maximum chocolates - 9.
Others have one among 1, 2, 3, 4, 6, 7 and 8.
Asha took the second lowest number of chocolates; hence she took 2 chocolates.
If everyone took an equal number of chocolates, they would have taken 5 chocolates each.
Therefore, Eshan + Govind = 10. The possibilities are (6, 4) or (3, 7).
But, Aman took one chocolate more than Deepa - The possibilities are (3,4), (6,7) or (7, 8)
From the above statements, it is clear that Aman and Deepa took 8 and 7 chocolates respectively.
Eshan and Govind took 6 and 4 chocolates in any order.
The remaining chocolates 1 and 3 could have been taken by anyone among Tarun and Arun.
Tabulating the results:

Govind can take 4 chocolates. Other statements are definitely false.

Let the child who took the minimum chocolates have x chocolates. This implies Mohit would have taken x + 8 chocolates.
If x = 1, the number of chocolates with Mohit = 9. The others should have a different number from 2 to 8. The sum of numbers from 1 to 9 is 45, out of which 5 can be omitted to get a sum of 40.

If x = 2, the number of chocolates with Mohit = 10. The sum of consecutive numbers from 2 to 10 is 54. The sum of 40 cannot be achieved by omitting any one number. Hence, x > 1 is not possible.
So, x = 1, Mohit has the maximum chocolates - 9.
Others have one among 1, 2, 3, 4, 6, 7 and 8.
Asha took the second lowest number of chocolates; hence she took 2 chocolates.
If everyone took an equal number of chocolates, they would have taken 5 chocolates each.
Therefore, Eshan + Govind = 10. The possibilities are (6, 4) or (3, 7).
But, Aman took one chocolate more than Deepa - The possibilities are (3,4), (6,7) or (7, 8)
From the above statements, it is clear that Aman and Deepa took 8 and 7 chocolates, respectively.
Eshan and Govind took 6 and 4 chocolates in any order.
The remaining chocolates 1 and 3 could have been taken by anyone among Tarun and Arun.
Tabulating the results:

Hence Tarun can have either 1 chocolate or 3 chocolates.

The answer is Option (B). 

Let A, B, C, and D denote the players with initial ratings 0, 200, 400, and 600 respectively. Then there are 3 ways in which the matches on 1st day could happen,

1. A plays B and C plays D

2. A plays C and B plays D

3. A plays D and B plays C.

Since each match on the first day is upset, the following ratings are possible:

A has a rating of 0 and B has a rating of 200 initially. The value of a will be 50 in this case. Hence, the rating of A increases by 50 and that of B decreases by 50. A similar process happens in every case mentioned in the table.

It has been given that the values of ratings on the first two days are multiples of 10. This means that the value of a, which we subtract/add to ratings, should be a multiple of 10 too, as the possible ratings after day 1 are multiples of 10. $$\frac{\left(X-Y\right)}{4}=10n$$ or X-Y = 40n.

Hence, the people who play on Day 2 must have a rating difference which is of the form 40n.

In Case 1, where A-B and C-D, the matches on Day 2 could be A-C and B-D or A-D and B-C, as matches, A-B and C-D have already happened. But A-D and B-C case is not possible on Day 2, as their rating differences after Day 1 are 500 and 300 respectively. Hence, on Day 2, the matches for Case 1 would be A-C and B-D.

Similarly, for Case 2, Day 2 matches are: A-B and C-D. A match between BC is not possible here as the value of a becomes 0 in that case, and it has been mentioned that during the course of the tournament, the value of a was a natural number in each case.

Similarly, for Case 3, Day 2 matches are: A-C and B-D. 

Since all the matches on Day 2 are also upsets, the ratings for each case:

For Day 3, each player has only 1 player left to play with. Hence for each case, we have the matches:

1. A-D and B-C

2. A-D and B-C

3, A-B and C-D

On Day 3, A won all the matches. But since the initial rating of A is 0, A denotes Tal. Hence, Anand can be B or C. Since player D lost all matches, and had the highest absolute decrease in his rating, we can see that Garry cannot be D. Thus Anand is B and his final rating is 275.

Let A, B, C, and D denote the players with initial ratings 0, 200, 400, and 600 respectively. Then there are 3 ways in which the matches on 1st day could happen,

1. A plays B and C plays D

2. A plays C and B plays D

3. A plays D and B plays C.

Since each match on the first day is upset, the following ratings are possible:

A has a rating of 0 and B has a rating of 200 initially. The value of A will be 50 in this case. Hence, the rating of A increases by 50 and that of B decreases by 50. A similar process happens in every case mentioned in the table.

It has been given that the values of ratings on the first two days are multiples of 10. This means that the value of a, which we subtract/add to ratings, should be a multiple of 10 too, as the possible ratings after day 1 are multiples of 10. $$\frac{\left(X-Y\right)}{4}=10n$$ or X-Y = 40n.

Hence, the people who play on Day 2 must have a rating difference which is of the form 40n.

In Case 1, where A-B and C-D, the matches on Day 2 could be A-C and B-D or A-D and B-C, as matches, A-B and C-D have already happened. But A-D and B-C case is not possible on Day 2, as their rating differences after Day 1 are 500 and 300 respectively. Hence, on Day 2, the matches for Case 1 would be A-C and B-D.

Similarly, for Case 2, Day 2 matches are: A-B and C-D. 

A match between BC is not possible here as the value of a becomes 0 in that case, and it has been mentioned that during the course of the tournament, the value of a was a natural number in each case.

Similarly, for Case 3, Day 2 matches are: A-C and B-D. 

Since all the matches on Day 2 are also upsets, the ratings for each case:

For Day 3, each player has only 1 player left to play with. Hence for each case, we have the matches:

1. A-D and B-C

2. A-D and B-C

3, A-B and C-D

Since each match was an upset, Tal must have won all the matches, as his rating is lowest each day. Since A denotes the player whose initial rating was 0, hence A denotes Tal. Thus, the rating of Tal after Day 3 is 225.

Let A, B, C, and D denote the players with initial ratings 0, 200, 400, and 600 respectively. Then there are 3 ways in which the matches on 1st day could happen,

1. A plays B and C plays D

2. A plays C and B plays D

3. A plays D and B plays C.

Since each match on the first day is upset, the following ratings are possible:

A has a rating of 0 and B has a rating of 200 initially. The value of A will be 50 in this case. Hence, the rating of A increases by 50 and that of B decreases by 50. The same happens in every case mentioned in the table.

It has been given that the values of ratings on the first two days are multiples of 10. This means that the value of a, which we subtract/add to ratings, should be a multiple of 10 too, as the possible ratings after day 1 are multiples of 10. $$\frac{\left(X-Y\right)}{4}=10n$$ or X-Y = 40n.

Hence, the people who play on Day 2 must have a rating difference which is of the form 40n.

In Case 1, where A-B and C-D, the matches on Day 2 could be A-C and B-D or A-D and B-C, as matches, A-B and C-D have already happened. But A-D and B-C case is not possible on Day 2, as their rating differences after Day 1 are 500 and 300 respectively. Hence, on Day 2, the matches for Case 1 would be A-C and B-D.

Similarly, for Case 2, Day 2 matches are: A-B and C-D. A match between BC is not possible here as the value of a becomes 0 in that case, and it has been mentioned that during the course of the tournament, the value of a was a natural number in each case.

Similarly, for Case 3, Day 2 matches are: A-C and B-D.

Since all the matches on Day 2 are also upsets, the ratings for each case:

For Day 3, each player has only 1 player left to play with. Hence for each case, we have the matches:

1. A-D and B-C

2. A-D and B-C

3, A-B and C-D

Only in case 3 do we see that the rating of two players is the same after Day 2. In that case, the highest rating with any player is 400 after day 2.

Let A, B, C, and D denote the players with initial ratings 0, 200, 400, and 600 respectively. Then there are 3 ways in which the matches on 1st day could happen,

1. A plays B and C plays D

2. A plays C and B plays D

3. A plays D and B plays C.

Since each match on the first day is upset, the following ratings are possible:

A has a rating of 0 and B has a rating of 200 initially. The value of A will be 50 in this case. Hence, the rating of A increases by 50 and that of B decreases by 50. The same happens in every case mentioned in the table.

It has been given that the values of ratings on the first two days are multiples of 10. This means that the value of a, which we subtract/add to ratings, should be a multiple of 10 too, as the possible ratings after day 1 are multiples of 10. $$\frac{\left(X-Y\right)}{4}=10n$$ or X-Y = 40n.

Hence, the people who play on Day 2 must have a rating difference which is of the form 40n.

In Case 1, where A-B and C-D, the matches on Day 2 could be A-C and B-D or A-D and B-C, as matches, A-B and C-D have already happened. But A-D and B-C case is not possible on Day 2, as their rating differences after Day 1 are 500 and 300 respectively. Hence, on Day 2, the matches for Case 1 would be A-C and B-D.

Similarly, for Case 2, Day 2 matches are: A-B and C-D. A match between BC is not possible here as the value of a becomes 0 in that case, and it has been mentioned that during the course of the tournament, the value of a was a natural number in each case.

Similarly, for Case 3, Day 2 matches are: A-C and B-D. 

Since all the matches on Day 2 are also upsets, the ratings for each case:

For Day 3, each player has only 1 player left to play with. Hence for each case, we have the matches:

1. A-D and B-C

2. A-D and B-C

3, A-B and C-D

We can see that in every case, the difference between the highest-rated player and the lowest-rated player is 150. 

Let A, B, C, and D denote the players with initial ratings 0, 200, 400, and 600 respectively. Then there are 3 ways in which the matches on 1st day could happen,

1. A plays B and C plays D

2. A plays C and B plays D

3. A plays D and B plays C.

Since each match on the first day is upset, the following ratings are possible:

A has a rating of 0 and B has a rating of 200 initially. The value of A will be 50 in this case. Hence, the rating of A increases by 50 and that of B decreases by 50. The same happens in every case mentioned in the table.

It has been given that the values of ratings on the first two days are multiples of 10. This means that the value of a, which we subtract/add to ratings, should be a multiple of 10 too, as the possible ratings after day 1 are multiples of 10. $$\frac{\left(X-Y\right)}{4}=10n$$ or X-Y = 40n.

Hence, the people who play on Day 2 must have a rating difference which is of the form 40n.

In Case 1, where A-B and C-D, the matches on Day 2 could be A-C and B-D or A-D and B-C, as matches, A-B and C-D have already happened. But A-D and B-C case is not possible on Day 2, as their rating differences after Day 1 are 500 and 300 respectively. Hence, on Day 2, the matches for Case 1 would be A-C and B-D.

Similarly, for Case 2, Day 2 matches are A-B and C-D. A match between BC is not possible here as the value of a becomes 0 in that case, and it has been mentioned that during the course of the tournament, the value of a was a natural number in each case.

Similarly, for Case 3, Day 2 matches are A-C and B-D.

Since all the matches on Day 2 are also upsets, the ratings for each case:

For Day 3, each player has only 1 player left to play with. Hence for each case, we have the matches:

1. A-D and B-C

2. A-D and B-C

3, A-B and C-D

From the 1st and 2nd tables, the fall in ratings of C and D in case 1 is high among all players in all cases. 

Hence, the answer is 450 - 350 = 550 - 450 = 100.

The answer is Option (B).

CDEH -> Girls

ABFG -> Boys

CEF -> cooks

At least 2 cooks should be selected in a team of 5

CASE 1: only C and E

From 1, A cannot be selected

From 3 and 4, G and F should be selected which contradicts the Case of selecting only C and E as cooks.

Hence this case is rejected

CASE 2: only C and F

From 4, G should be selected.

H cannot be selected because if H is selected then B should be selected => D should be selected and a team of 6 will be formed

Possible teams -> CFGAD and CFGBD

CASE 3: only E and F

Possible teams -> none

CASE 4: C E F

Possible team -> CEFGD

Possible combinations: CFGAD, CFGBD and CEFGD

Hence, 4 people, CDFG are always selected

CDEH -> Girls

ABFG -> Boys

CEF -> cooks

At least 2 cooks should be selected in a team of 5

CASE 1: only C and E

From 1, A cannot be selected

From 3 and 4, G and F should be selected which contradicts the Case of selecting only C and E as cooks.

Hence this case is rejected

CASE 2: only C and F

From 4, G should be selected.

H cannot be selected because if H is selected then B should be selected => D should be selected and a team of 6 will be formed

Possible teams -> CFGAD and CFGBD

CASE 3: only E and F

Possible teams -> none

CASE 4: C E F

Possible team -> CEFGD

Possible combinations: CFGAD, CFGBD and CEFGD

CDEH -> Girls

ABFG -> Boys

CEF -> cooks

At least 2 cooks should be selected in a team of 5

CASE 1: only C and E

From 1, A cannot be selected

From 3 and 4, G and F should be selected which contradicts the Case of selecting only C and E as cooks.

Hence this case is rejected

CASE 2: only C and F

From 4, G should be selected.

H cannot be selected because if H is selected then B should be selected => D should be selected and a team of 6 will be formed

Possible teams -> CFGAD and CFGBD

CASE 3: only E and F

Possible teams -> none

CASE 4: C E F

Possible team -> CEFGD

Possible combinations: CFGAD, CFGBD and CEFGD

CDEH -> Girls

ABFG -> Boys

CEF -> cooks

At least 2 cooks should be selected in a team of 5

CASE 1: only C and E

From 1, A cannot be selected

From 3 and 4, G and F should be selected which contradicts the Case of selecting only C and E as cooks.

Hence this case is rejected

CASE 2: only C and F

From 4, G should be selected.

H cannot be selected because if H is selected then B should be selected => D should be selected and a team of 6 will be formed

Possible teams -> CFGAD and CFGBD

CASE 3: only E and F

Possible teams -> none

CASE 4: C E F

Possible team -> CEFGD

Possible combinations: CFGAD, CFGBD and CEFGD

Team CEFGD has 3 girls

CDEH -> Girls

ABFG -> Boys

CEF -> cooks

At least 2 cooks should be selected in a team of 5

CASE 1: only C and E

From 1, A cannot be selected

From 3 and 4, G and F should be selected which contradicts the Case of selecting only C and E as cooks.

Hence this case is rejected

CASE 2: only C and F

From 4, G should be selected.

H cannot be selected because if H is selected then B should be selected => D should be selected and a team of 6 will be formed

Possible teams -> CFGAD and CFGBD

CASE 3: only E and F

Possible teams -> none

CASE 4: C E F

Possible team -> CEFGD

Possible combinations: CFGAD, CFGBD and CEFGD

There can be only one combination with both C and E.

The answer is option(B)

First, we will tabulate the given data in the scatter plot and then proceed.

Now, we will calculate the percentage difference of BMI with respect to Body fat % using the formula, % difference= $$\frac{BMI-Body\ Fat\%}{Body\ Fat\%}\times100\ $$

We will get the following table:

When arranged in ascending order, we get P9, P2, P8, P14, P1, P10, P7, P4, P3, P12, P13, P5, P11, P6

Group 1 has: P9, P2, P8, P14, P1, P10, P7

Group 2 has: P4, P3, P12, P13 and

Group 3 has: P5, P11, P6.

Now, the point 3/ clue 3 says that the data of children were at the extremes. So, P9 andP7 from G1; P4 and P13 from G2; and one or both of P5 and P11 along with P6 are children.

In G1, the data of P14 forms the median. Hence, he is a male and rest all are the females.

Therefore, in G1: Children= P9 and P7. Male= P14. Female= P2, P8, P1, P10.

In G2: Children= P4 and P13. It has no male (Clue 4). Females= P3 and P12.

In G3:

Case 1- P5 and P6 and children. P11 is a male.

Case 2- P6 and P11 are children. P5 is a male.

Considering all the cases, we get the same number of females, which is equal to 6.

First, we will tabulate the given data in the scatter plot and then proceed.

Now, we will calculate the percentage difference of BMI with respect to Body fat % using the formula, % difference= $$\frac{BMI-Body\ Fat\%}{Body\ Fat\%}\times100\ $$

We will get the following table:

When arranged in ascending order, we get P9, P2, P8, P14, P1, P10, P7, P4, P3, P12, P13, P5, P6, P11

'.' There are 14 members in the team, the median % change of BMI w.r.t. Body Fat % will be the average of P7 and P4.
.'. Median % change= $$\frac{\left(26.32+31.58\right)}{2}=\frac{57.9}{2}=28.95$$

First, we will tabulate the given data in the scatter plot and then proceed.

Now, we will calculate the percentage difference of BMI with respect to Body fat % using the formula, % difference= $$\frac{BMI-Body\ Fat\%}{Body\ Fat\%}\times100\ $$

We will get the following table:

When arranged in ascending order, we get P9, P2, P8, P14, P1, P10, P7, P4, P3, P12, P13, P5, P6, P11

Group 1 has: P9, P2, P8, P14, P1, P10, P7

Group 2 has: P4, P3, P12, P13 and

Group 3 has: P5, P6, P11.

Now, the point 3/ clue 3 says that the data of children were at the extremes. So, P9 andP7 from G1; P4 and P13 from G2; and one or both of P5 and P6 along with P11 are children.

In G1, the data of P14 forms the median. Hence, he is a male and rest all are the females.

Therefore, in G1: Children= P9 and P7. Male= P14. Female= P2, P8, P1, P10.

In G2: Children= P4 and P13. It has no male (Clue 4). Females= P3 and P12.

In G3:

Case 1- P5 and P6 and children. P11 is a male.

Case 2- P6 and P11 are children. P5 is a male.

In Case-1 => The two males selected are P14 and P11, and in Case-2 => The two males selected are P14 and P5. Therefore, P14 is definitely present in both cases, as there are only two males, and both need to be selected.

First, we will tabulate the given data in the scatter plot and then proceed.

Now, we will calculate the percentage difference of BMI with respect to Body fat % using the formula, % difference= $$\frac{BMI-Body\ Fat\%}{Body\ Fat\%}\times100\ $$

We will get the following table:

When arranged in ascending order, we get P9, P2, P8, P14, P1, P10, P7, P4, P3, P12, P13, P5, P6, P11

Group 1 has: P9, P2, P8, P14, P1, P10, P7

Group 2 has: P4, P3, P12, P13 and

Group 3 has: P5, P6, P11.

Now, the point 3/ clue 3 says that the data of children were at the extremes. So, P9 andP7 from G1; P4 and P13 from G2; and one or both of P5 and P6 along with P11 are children.

In G1, the data of P14 forms the median. Hence, he is a male and rest all are the females.

Therefore, in G1: Children= P9 and P7. Male= P14. Female= P2, P8, P1, P10.

In G2: Children= P4 and P13. It has no male (Clue 4). Females= P3 and P12.

In G3:

Case 1- P5 and P6 and children. P11 is a male.

Case 2- P6 and P11 are children. P5 is a male.

In total, there are 6 children in the team of 14 members- P9, P7, P4, P13, P6 and P11/P5. [P11 and P5 both have 'k' value of 66.67]

Average % change of BMI w.r.t. Body fat%= $$\dfrac{\left(11.54+26.32+31.58+41.18+144.44+66.67\right)}{6}=53.62$$

First, we will tabulate the given data in the scatter plot and then proceed.

Now, we will calculate the percentage difference of BMI with respect to Body fat % using the formula, % difference= $$\frac{BMI-Body\ Fat\%}{Body\ Fat\%}\times100\ $$

We will get the following table:

When arranged in ascending order, we get P9, P2, P8, P14, P1, P10, P7, P4, P3, P12, P13, P5, P11, P6

Group 1 has: P9, P2, P8, P14, P1, P10, P7

Group 2 has: P4, P3, P12, P13 and

Group 3 has: P5, P11, P6.

Now, the point 3/ clue 3 says that the data of children were at the extremes. So, P9 andP7 from G1; P4 and P13 from G2; and one or both of P5 and P11 along with P6 are children.

In G1, the data of P14 forms the median. Hence, he is a male and rest all are the females.

Therefore, in G1: Children= P9 and P7. Male= P14. Female= P2, P8, P1, P10.

In G2: Children= P4 and P13. It has no male (Clue 4). Females= P3 and P12.

In G3:

Case 1- P5 and P6 and children. P11 is a male.

Case 2- P6 and P11 are children. P5 is a male.

Considering all the cases, we get the same number of females, which is equal to 6.

After tabulating the BMI of females

The average BMI = $$\frac{29+25+20+24+25+21}{6}=\frac{144}{6}=24$$