Logical Reasoning & DI (LRDI) Test - 11

Consider the second statement of B - I do not always speak the truth.

A truth-teller won't say this statement. Even a liar won't say this statement. Hence, B must be an alternator. And this statement is true. Hence, B's first and third statements must be false.

B's first statement is that A is a liar. Hence, A is not a liar. He must be a truth-teller or an alternator.

B's third statement is that he knows French, which is false.

A's second statement is that B knows French, which is false, hence, A must be an alternator. Hence, his first and third statements must be true.

Hence, C is from Australia.

Also, D is a liar or an alternator.

Let us analyse D's statements.

The second statement of D is: 

There is only one liar.

We know that A and B are not liars. Hence, if this statement is true, then C is the only liar and D is the alternator. If this statement is false, since there must be at least one liar, in this case, both C and D are liars.

Since we know that in both cases, C is a liar, we can analyze his statements.

D is from Europe since C is a liar.

Also, A is from Asia and A speaks Spanish.

B must be from Africa.

A who is an alternator says that B knows French as his second statement, which is false.

We know that the first and third statements of D are false since he is a liar or an alternator(with the second statement true). Hence, D does not speak French.

B must be from Africa.

Hence, we get the following 2 arrangements.

C, who is from Australia, speaks French.

Consider the second statement of B - I do not always speak the truth.

A truth-teller won't say this statement. Even a liar won't say this statement. Hence, B must be an alternator. And this statement is true. Hence, B's first and third statements must be false.

B's first statement is that A is a liar. Hence, A is not a liar. He must be a truth-teller or an alternator.

B's third statement is that he knows French, which is false.

A's second statement is that B knows French, which is false, hence, A must be an alternator. Hence, his first and third statements must be true.

Hence, C is from Australia.

Also, D is a liar or an alternator.

Let us analyse D's statements.

The second statement of D is:

There is only one liar.

We know that A and B are not liars. Hence, if this statement is true, then C is the only liar and D is the alternator. If this statement is false, since there must be at least one liar, in this case, both C and D are liars.

Since we know that in both cases, C is a liar, we can analyze his statements.

D is from Europe since C is a liar.

Also, A is from Asia and A speaks Spanish.

B must be from Africa.

A who is an alternator says that B knows French as his second statement, which is false.

We know that the first and third statements of D are false since he is a liar or an alternator(with the second statement true). Hence, D does not speak French.

B must be from Africa.

Hence, we get the following 2 arrangements.

B is from Africa.

Consider the second statement of B - I do not always speak the truth.

A truth-teller won't say this statement. Even a liar won't say this statement. Hence, B must be an alternator. And this statement is true. Hence, B's first and third statements must be false.

B's first statement is that A is a liar. Hence, A is not a liar. He must be a truth-teller or an alternator.

B's third statement is that he knows French, which is false.

A's second statement is that B knows French, which is false, hence, A must be an alternator. Hence, his first and third statements must be true.

Hence, C is from Australia.

Also, D is a liar or an alternator.

Let us analyse D's statements.

The second statement of D is:

There is only one liar.

We know that A and B are not liars. Hence, if this statement is true, then C is the only liar and D is the alternator. If this statement is false, since there must be at least one liar, in this case, both C and D are liars.

Since we know that in both cases, C is a liar, we can analyze his statements.

D is from Europe since C is a liar.

Also, A is from Asia and A speaks Spanish.

B must be from Africa.

A who is an alternator says that B knows French as his second statement, which is false.

We know that the first and third statements of D are false since he is a liar or an alternator(with the second statement true). Hence, D does not speak French.

B must be from Africa.

Hence, we get the following 2 arrangements.

If German is spoken by a liar, it corresponds to the first case, we can simplify the table as:

Hence, 2 statements of A are true and one statement of B is true. So, a total of 3 true statements are spoken.

Consider the second statement of B - I do not always speak the truth.

A truth-teller won't say this statement. Even a liar won't say this statement. Hence, B must be an alternator. And this statement is true. Hence, B's first and third statements must be false.

B's first statement is that A is a liar. Hence, A is not a liar. He must be a truth-teller or an alternator.

B's third statement is that he knows French, which is false.

A's second statement is that B knows French, which is false, hence, A must be an alternator. Hence, his first and third statements must be true.

Hence, C is from Australia.

Also, D is a liar or an alternator.

Let us analyse D's statements.

The second statement of D is:

There is only one liar.

We know that A and B are not liars. Hence, if this statement is true, then C is the only liar and D is the alternator. If this statement is false, since there must be at least one liar, in this case, both C and D are liars.

Since we know that in both cases, C is a liar, we can analyze his statements.

D is from Europe since C is a liar.

Also, A is from Asia and A speaks Spanish.

B must be from Africa.

A who is an alternator says that B knows French as his second statement, which is false.

We know that the first and third statements of D are false since he is a liar or an alternator(with the second statement true). Hence, D does not speak French.

B must be from Africa.

Hence, we get the following 2 arrangements.

B - Alternator - Africa - German is one of the possibilities.

Consider the second statement of B - I do not always speak the truth.

A truth-teller won't say this statement. Even a liar won't say this statement. Hence, B must be an alternator. And this statement is true. Hence, B's first and third statements must be false.

B's first statement is that A is a liar. Hence, A is not a liar. He must be a truth-teller or an alternator.

B's third statement is that he knows French, which is false.

A's second statement is that B knows French, which is false, hence, A must be an alternator. Hence, his first and third statements must be true.

Hence, C is from Australia.

Also, D is a liar or an alternator.

Let us analyse D's statements.

The second statement of D is:

There is only one liar.

We know that A and B are not liars. Hence, if this statement is true, then C is the only liar and D is the alternator. If this statement is false, since there must be at least one liar, in this case, both C and D are liars.

Since we know that in both cases, C is a liar, we can analyze his statements.

D is from Europe since C is a liar.

Also, A is from Asia and A speaks Spanish.

B must be from Africa.

A who is an alternator says that B knows French as his second statement, which is false.

We know that the first and third statements of D are false since he is a liar or an alternator(with the second statement true). Hence, D does not speak French.

B must be from Africa.

Hence, we get the following 2 arrangements.

Overall, there are four possible arrangements of attributes.

The answer is option A.

Projects completed before E = A, C and Projects completed before F=B, D

Minimum time required to complete E = max (time required to complete A, time required to complete C) + 37   = max (A, B-C) + 37 = (35, 18+16) +37 = 71

Minimum day to complete F = max(B,D)+19

Time taken to complete B = 18

Time taken to complete D = Time taken to complete A and time taken to complete D = 35+23=58

Minimum day to complete F = max(B,D)+19 = 58+19=77

Minimum days to complete I = max (B,C,F) +51

Time taken to complete C = B+C = 18+16=34

Minimum days to complete I = max (B,C,F) +51 = 77+51=128

Minimum time taken to complete H = F+H=77+13=90

Minimum time taken to complete J =max(A,E,H)+21=90+21=111

Total time taken to complete = max(A,B,C,D,E,F,G,H,I,J) =128 

We can represent the processes in the following diagram.

Projects completed before E = A, C and Projects completed before F=B, D

Minimum time required to complete E = max (time required to complete A, time required to complete C) + 37   = max (A, B-C) + 37 = (35, 18+16) +37 = 72

Minimum day to complete F = max(B,D)+19

Time taken to complete B = 18

Time taken to complete D = Time taken to complete A and time taken to complete D = 35+23=58

Minimum day to complete F = max(B,D)+19 = 58+19=77

Minimum days to complete I = max (B,C,F) +51

Time taken to complete C = B+C = 18+16=34

Minimum days to complete I = max (B,C,F) +51 = 77+51=128

Minimum time taken to complete H = F+H=77+13=90

Minimum time taken to complete J =max(A,E,H)+21=90+21=111

Except I, every project can be completed in 125 days.

Projects completed before G = E, F

Projects completed before E = A, C and Projects completed before F=B, D

Minimum time required to complete E = max (time required to complete A, time required to complete C) + 37   = max (A, B-C) + 37 = (35, 18+16) +37 = 72

Minimum time required to complete F = max (time required to complete B, time required to complete D) + 19 = max (B, A-D) +19 = (18, 35+23) +19 = 77

Minimum time required to complete G = max (E, F) +43 = 77+43=120

Minimum day to complete F = max(B,D)+19

Time taken to complete B = 18

Time taken to complete D = Time taken to complete A and time taken to complete D = 35+23=58

Minimum day to complete F = max(B,D)+19 = 58+19=77

Minimum days to complete I = max (B,C,F) +51

Time taken to complete C = B+C = 18+16=34

Minimum days to complete I = max (B,C,F) +51 = 77+51=128

The percentage by which the minimum number of days required to complete F is greater or less than the minimum number of days required to complete I = ((128-77)*100)/128 = 5100/128 = 39.84

Minimum time to complete E = 37 + max(A,C) 

Time taken to complete A = 35

Time taken to complete C = Time taken to complete B and time taken to complete C = 18 + 16 = 34

Minimum time to complete E = 37 + max(A,C) = 37 + max(35,34) = 37 + 35 = 72

Minimum days to complete J = max (A,E,H) + 21

Time taken to complete A = 35

Minimum time taken to complete E = 72

Minimum time taken to complete H = 13 + Minimum time taken to complete F = 13 + 77 = 90

Minimum days to complete J = max (A,E,H) + 21 = max(35,72,90) + 21 = 90 + 21 = 111

Required percentage = $$\ \frac{111-72}{111}\times100=35.13\%$$

The answer is option D.

The number of people who stayed in the mall for more than 5 hours will be maximum when the maximum people entering between 10 and 11 exits after 3 pm. The number of people entering at 10 and not exiting between 11-12 = 154-101 =53.
For maximizing the number of people who stayed in the mall for more than 5 hours, we will try to keep these 53 in the mall only as far as possible.
The total number of people in the mall at 12 pm = 53+121 = 174 out of these 89 leaves the mall and leaving a total of =174-89 = 85 people.
Out of these 53 have entered the mall between 10-11 and 32 have entered the mall between 11-12.
The total number of people at 1 pm = 85+119 = 204 out of this 129 leave the mall leaving 75 people at the mall.
Out of these 75, 53 have entered the mall between 10-11 and 22 have entered the mall between 11-12.
The total number of people at the mall at 2 pm = 91+75 = 166 out of these 148 people leave the mall leaving a total of 166-148 = 18 people at the mall.
These 18 people have entered the mall between 10 and 11.
Thus, the number of people who stayed in the mall for more than 5 hours is at most 18.

Number of people entered is equal to number of people exit. So 154+121+119+91+66-(101+89+129+148)=84

Thus we can make the following table

The 101 people leaving between 11-12 must have left within 2 hours. Thus there is an excess of 53 people at the mall. To minimize the number of people leaving the mall in 2 hours we will have to assume that these 53 people will leave the mall in between 12-1.
Thus, the number of people entering the mall in between 11-12 and leaving with 2 hours = 89-53 = 36
The excess of people entering the mall in between 11-12 = 121-36 = 85.
They will leave the mall in between 1-2.
Thus, the number of people entering the mall in between 12-1 and leaving with 2 hours = 129-85 = 44.
The excess of people entering the mall in between 12-1 = 119-44 =75
They will leave the mall in between 1-2.
Thus, the number of people entering the mall in between 1-2 and leaving with 2 hours = 148-75 = 73.

Thus, the number of people entering the mall in between 1-2 and leaving with 2 hours = 84-18 = 66
All the 66 people entering the mall in between 2-3 must have left within 2 hours
Thus, the required number = 101+36+44+73+66 = 320

Out of 154 people entered 101 exited the mall, but the remaining 53 could have exited the mall in between 1 pm and 2 pm
Thus, the required number = 53.

The 101 people leaving between 11-12 must have left within 2 hours. Thus there is an excess of 53 people at the mall. To minimize the number of people leaving the mall in 2 hours we will have to assume that these 53 people will leave the mall in between 12-1.
Thus, the number of people entering the mall in between 11-12 and leaving with 2 hours = 89-53 = 36
The excess of people entering the mall in between 11-12 = 121-36 = 85.
They will leave the mall in between 1-2.
Thus, the number of people entering the mall in between 12-1 and leaving with 2 hours = 129-85 = 44.
The excess of people entering the mall in between 12-1 = 119-44 =75
They will leave the mall in between 1-2.
Thus, the number of people entering the mall in between 1-2 and leaving with 2 hours = 148-75 = 73.
All the 66 people entering the mall in between 2-3 must have left within 2 hours
Thus, the minimum number of people who left the temple within 2 hrs of entering =101+36+44+73+66= 320.
The total number of people entering the mall = 154+121+119+91+66 = 551
The total number of people leaving the mall before 3 pm = 467
Thus, the number of people leaving the mall after 3 pm = 551-467 = 84
Thus, the required percentage = 84*100/320 = 26.25%

Out of 154 who entered between 10 - 11, 101 people left between 11-12. So 53 are remaining.

It is given that all those who entered between 10 - 11 left before 1 pm. Hence, out of 89 who left between 12 pm - 1 pm, 53 (remaining people from 10 - 11) will be from the 10 - 11 batch. So rest 36 (89 - 53 = 36) will be from the 11 am - 12 pm batch.

The remaining people from the 11 am - 12 pm batch will be 121 - 36 = 85.

We need to find the maximum number of people who entered between 11 am & 12 pm and left between 2 pm - 3 pm. Hence, out of 85, we have to maximise the number of people who leaves between 2 pm - 3 pm.

The total number of people present between 1 pm - 2 pm is = 119 + 85 = 204. Out of 204, 129 will leave the mall. To maximise number people among 85 who stay in mall, 119 people who entered between 12 pm - 1 pm will leave. It implies, 129 - 119 = 10, 10 people (from 85 people) will also leave. The remaining 75 (85-10=75) will leave between 2 pm - 3 pm. Hence maximum number of people who enters  leaves between $$11<x\le\ 12$$ and left between $$2<x\le\ 3$$ is 75.

The answer is Option (A)

We can draw a Venn diagram as shown which indicates the number of students who have passed in each of the three subjects.
The numbers of the Venn diagram are obtained as explained below:
There are a total of 180 students in 11th class.
It is given that the passing percentage is 40%, that is, 72 students.
So the number of students who passed all three subjects is 72.
It is also given that 16 students have failed.
So the number of students who did not pass in any subject are 16.
It is given that the number of students who did not clear only Physics is 13 which is present in the Venn diagram in the area which indicates students who have passed in both Maths and Chemistry but not Physics.
It is given that the number of students who did not clear exactly two subjects Maths and Physics is 9. It is shown in the Venn diagram by the area indicated by students who have passed only in Chemistry.
Now it is given that the number of students who did not clear Maths is 44. This includes the 16 students who did not clear any of the subjects. So 44-16 = 28 students cleared atleast one subject but not Maths. Among them we have 9 students who cleared only Chemistry. Excluding them we have 28-9 = 19.
These are the students who either cleared only Physics or both Physics and Chemistry. They are divided into ‘x’ and ’19-x’ as shown in the Venn diagram.
Now the remaining number of students who have not been accounted for are 180-16-13-9-72-19 = 51.
These students have been divided into ‘y’ and ’51-y’ as shown in the Venn diagram.
Now the number of students who cleared Maths but not Chemistry are 51 as can be seen from the Venn diagram.

We can draw a Venn diagram as shown which indicates the number of students who have passed in each of the three subjects.
The numbers of the Venn diagram are obtained as explained below:
There are a total of 180 students in 11th class.
It is given that the passing percentage is 40%, that is, 72 students.
So the number of students who passed all three subjects is 72.
It is also given that 16 students have failed.
So the number of students who did not pass in any subject are 16.
It is given that the number of students who did not clear only Physics is 13 which is present in the Venn diagram in the area which indicates students who have passed in both Maths and Chemistry but not Physics.
It is given that the number of students who did not clear exactly two subjects Maths and Physics is 9. It is shown in the Venn diagram by the area indicated by students who have passed only in Chemistry.
Now it is given that the number of students who did not clear Maths is 44. This includes the 16 students who did not clear any of the subjects. So 44-16 = 28 students cleared atleast one subject but not Maths. Among them we have 9 students who cleared only Chemistry. Excluding them we have 28-9 = 19.
These are the students who either cleared only Physics or both Physics and Chemistry. They are divided into ‘x’ and ’19-x’ as shown in the Venn diagram.
Now the remaining number of students who have not been accounted for are 180-16-13-9-72-19 = 51.
These students have been divided into ‘y’ and ’51-y’ as shown in the Venn diagram.
Now the number of students who cleared Maths = 13+51+72 = 136
Among these the number of students who have cleared at most one other subject = 51-y+y+13 = 64.
Thus the percentage of those students is $$ \frac{64}{136}\times100=47$$%

We can draw a Venn diagram as shown which indicates the number of students who have passed in each of the three subjects.
The numbers of the Venn diagram are obtained as explained below:
There are a total of 180 students in 11th class.
It is given that the passing percentage is 40%, that is, 72 students.
So the number of students who passed all three subjects is 72.
It is also given that 16 students have failed.
So the number of students who did not pass in any subject are 16.
It is given that the number of students who did not clear only Physics is 13 which is present in the Venn diagram in the area which indicates students who have passed in both Maths and Chemistry but not Physics.
It is given that the number of students who did not clear exactly two subjects Maths and Physics is 9. It is shown in the Venn diagram by the area indicated by students who have passed only in Chemistry.
Now it is given that the number of students who did not clear Maths is 44. This includes the 16 students who did not clear any of the subjects. So 44-16 = 28 students cleared atleast one subject but not Maths. Among them we have 9 students who cleared only Chemistry. Excluding them we have 28-9 = 19.
These are the students who either cleared only Physics or both Physics and Chemistry. They are divided into ‘x’ and ’19-x’ as shown in the Venn diagram.
Now the remaining number of students who have not been accounted for are 180-16-13-9-72-19 = 51.
These students have been divided into ‘y’ and ’51-y’ as shown in the Venn diagram.
Now the number of students who cleared Physics and Chemistry is 72+x.
But we do not know the value of x.
Thus the percentage of students who cleared Physics also cleared Chemistry cannot be determined.

We can draw a Venn diagram as shown which indicates the number of students who have passed in each of the three subjects.
The numbers of the Venn diagram are obtained as explained below:
There are a total of 180 students in 11th class.
It is given that the passing percentage is 40%, that is, 72 students.
So the number of students who passed all three subjects is 72.
It is also given that 16 students have failed.
So the number of students who did not pass in any subject are 16.
It is given that the number of students who did not clear only Physics is 13 which is present in the Venn diagram in the area which indicates students who have passed in both Maths and Chemistry but not Physics.
It is given that the number of students who did not clear exactly two subjects Maths and Physics is 9. It is shown in the Venn diagram by the area indicated by students who have passed only in Chemistry.
Now it is given that the number of students who did not clear Maths is 44. This includes the 16 students who did not clear any of the subjects. So 44-16 = 28 students cleared atleast one subject but not Maths. Among them we have 9 students who cleared only Chemistry. Excluding them we have 28-9 = 19.
These are the students who either cleared only Physics or both Physics and Chemistry. They are divided into ‘x’ and ’19-x’ as shown in the Venn diagram.
Now the remaining number of students who have not been accounted for are 180-16-13-9-72-19 = 51.
These students have been divided into ‘y’ and ’51-y’ as shown in the Venn diagram.
The number of students who have cleared Physics is 72+19+y which is given as 104.
So y = 13.
The number of students who cleared exactly two subjects such that one of them is Maths = y+13 = 13+13 = 26.
The answer is option C.

We can draw a Venn diagram as shown which indicates the number of students who have passed in each of the three subjects.
The numbers of the Venn diagram are obtained as explained below:
There are a total of 180 students in 11th class.
It is given that the passing percentage is 40%, that is, 72 students.
So the number of students who passed all three subjects is 72.
It is also given that 16 students have failed.
So the number of students who did not pass in any subject are 16.
It is given that the number of students who did not clear only Physics is 13 which is present in the Venn diagram in the area which indicates students who have passed in both Maths and Chemistry but not Physics.
It is given that the number of students who did not clear exactly two subjects Maths and Physics is 9. It is shown in the Venn diagram by the area indicated by students who have passed only in Chemistry.
Now it is given that the number of students who did not clear Maths is 44. This includes the 16 students who did not clear any of the subjects. So 44-16 = 28 students cleared atleast one subject but not Maths. Among them we have 9 students who cleared only Chemistry. Excluding them we have 28-9 = 19.
These are the students who either cleared only Physics or both Physics and Chemistry. They are divided into ‘x’ and ’19-x’ as shown in the Venn diagram.
Now the remaining number of students who have not been accounted for are 180-16-13-9-72-19 = 51.
These students have been divided into ‘y’ and ’51-y’ as shown in the Venn diagram.

In the question, it is given that the number of students who failed in physics and are classified as tolerated is 52, i.e.

13 + 9 + 51 - y = 52

y = 21

The number of students who failed only in chemistry is the number of students who passed only maths and physics, i.e. y = 21

Therefore, the answer is option D.