Logical Reasoning & DI (LRDI) Test

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Logical Reasoning & DI (LRDI) Test - 12

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Question 1
3 / -1

Directions For Questions

A cell phone assembling plant has 7 processes to assemble a cell phone. The seven processes are - P, Q, R, S, T, U and V. The process can only be started with S or T. S takes 12 days, while T takes 7 days. Q can be started only after S and T. Q takes 5 days. R takes 8 days and can be started only after T. R and S are prerequisites for U and it can be completed in 5 days. Q and R are perquisites for V, which takes 2 days. P, which is the final assembly process, can only start if all the other processes are completed. P takes 6 days. Further, it is known that processes can be carried out simultaneously.

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If the process starts on 1st March, the process can be completed at the earliest by:

A
21 March

B
22 March

C
26 March

D
27 March

Solution

Let’s first tabulate the time required and prerequisites for each task.

Since tasks can be started simultaneously, S and T should start first. T takes 7 days to complete. Since, R has only T as prerequisite, R can be started at the end of 7th day. Q has S and T as prerequisites. Hence, Q can be started at the end of 12th day. U can be started after 15 days. V can be started after 17 days. All processes except P will be completed in 20 days. P takes 6 more days to complete. Hence, a total of 26 days are required.

26 days from 1st March is 26 March. Hence, option C is the correct answer.
Question 2
3 / -1

Directions For Questions

A cell phone assembling plant has 7 processes to assemble a cell phone. The seven processes are - P, Q, R, S, T, U and V. The process can only be started with S or T. S takes 12 days, while T takes 7 days. Q can be started only after S and T. Q takes 5 days. R takes 8 days and can be started only after T. R and S are prerequisites for U and it can be completed in 5 days. Q and R are perquisites for V, which takes 2 days. P, which is the final assembly process, can only start if all the other processes are completed. P takes 6 days. Further, it is known that processes can be carried out simultaneously.

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A new machine is bought which halves the time for process R. If the whole process is completed in the minimum possible time, what percent of the time is spent doing exactly one of the processes ( Percentage of time during which multiple processes are not run in parallel)?

A
36

B
364

C
367

D
369

Solution

Let’s first tabulate the time required and prerequisites for each task.

Since tasks can be started simultaneously, S and T should start first. T takes 7 days to complete. Since, R has only T as prerequisite, R can be started at the end of 7^th day. Q has S and T as prerequisites. Hence, Q can be started at the end of 12^th day. U can be started after 12 days. V can be started after 17 days. All processes except P will be completed in 19 days. P takes 6 more days to complete. Hence, a total of 25 days are required.

We see that there 2 days between completion of U and start of P where only one process runs. There is one day between completion of R and start of U. There are 6 days when only P runs.

Thus, a total of 9 days exists such that only one process runs.

Percentage = (9/25) * 100 = 36%
Question 3
3 / -1

Directions For Questions

A cell phone assembling plant has 7 processes to assemble a cell phone. The seven processes are - P, Q, R, S, T, U and V. The process can only be started with S or T. S takes 12 days, while T takes 7 days. Q can be started only after S and T. Q takes 5 days. R takes 8 days and can be started only after T. R and S are prerequisites for U and it can be completed in 5 days. Q and R are perquisites for V, which takes 2 days. P, which is the final assembly process, can only start if all the other processes are completed. P takes 6 days. Further, it is known that processes can be carried out simultaneously.

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A new machine is bought, which reduces the time for S by half. If the process is started on 2nd April, the earliest when the whole process will be completed is:

A
20 April

B
23 April

C
27 April

D
28 April

Solution

Let’s first tabulate the time required and prerequisites for each task.

Since tasks can be started simultaneously, S and T should start first. T takes 7 days to complete. Since, R has only T as prerequisite, R can be started at the end of 7^th day. Q has S and T as prerequisites. Hence, Q can be started at the end of 7^th day. U can be started after 15 days. V can be started after 12 days. All processes except P will be completed in 20 days. P takes 6 more days to complete. Hence, a total of 26 days are required.

`Thus, work will be completed on 27^th April.
Question 4
3 / -1

Directions For Questions

A cell phone assembling plant has 7 processes to assemble a cell phone. The seven processes are - P, Q, R, S, T, U and V. The process can only be started with S or T. S takes 12 days, while T takes 7 days. Q can be started only after S and T. Q takes 5 days. R takes 8 days and can be started only after T. R and S are prerequisites for U and it can be completed in 5 days. Q and R are perquisites for V, which takes 2 days. P, which is the final assembly process, can only start if all the other processes are completed. P takes 6 days. Further, it is known that processes can be carried out simultaneously.

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The management is planning to invest in a new machine which will reduce the process time for one of the processes among P, Q, R, S, T, U and V by half. The management decides in such a way that there is maximum reduction in completion time as compared to the original time. What is the percentage reduction in time that the management can expect?

A
3.84%

B
23.07%

C
11.53%

D
7.69%

Solution

Let’s first tabulate the time required and prerequisites for each task.

On observation, we see that since process P is dependent on all other processes and no other process takes place simultaneously with it, reducing P will have the maximum impact.

Since tasks can be started simultaneously, S and T should start first. T takes 7 days to complete. Since, R has only T as prerequisite, R can be started at the end of 7^th day. Q has S and T as prerequisites. Hence, Q can be started at the end of 12^th day. U can be started after 15 days. V can be started after 17 days. All processes except P will be completed in 20 days. P takes 3 more days to complete. Hence, a total of 23 days are required.

There is a reduction of 3 days.

Thus, percentage reduction = (3/26) * 100 = 11.53%
Question 5
3 / -1

Directions For Questions

A cell phone assembling plant has 7 processes to assemble a cell phone. The seven processes are - P, Q, R, S, T, U and V. The process can only be started with S or T. S takes 12 days, while T takes 7 days. Q can be started only after S and T. Q takes 5 days. R takes 8 days and can be started only after T. R and S are prerequisites for U and it can be completed in 5 days. Q and R are perquisites for V, which takes 2 days. P, which is the final assembly process, can only start if all the other processes are completed. P takes 6 days. Further, it is known that processes can be carried out simultaneously.

...view full instructions

A new machine is bought, which reduces the processing time of activity 'S' by 3/4th. Then what percentage of the total time are two activities running parallel if all the 7 processes are completed in the minimum time possible?

A
42.39

B
44.6

C
33.52

D
38.46

Solution

Let’s first tabulate the time required and prerequisites for each task.

But the process S takes only $$\frac{1}{4}th$$ of the previous time. Hence the time taken by process s = $$\frac{1}{4}\times\ 12=3\ Days$$

Since tasks can be started simultaneously, S and T should start first. T takes 7 days to complete. Since R has only T as a prerequisite, R can be started at the end of 7^th day. Q has S and T as prerequisites. Hence, Q can be started at the end of the 7^th day. U can be started after 15 days. V can be started after 15 days. All processes except P will be completed in 20 days. P takes 6 more days to complete. Hence, a total of 26 days are required.

Total time = 26 days.

The number of days when two activities run parallely = 3 + 5 + 2 = 10 days.

The percentage of the total time, when two activities are running parallel = $$\frac{10}{26}\ \times\ 100=38.46\%\ $$

The answer is Option (D).
Question 6
3 / -1

Directions For Questions

Ronaldo, Messi, and Neymar participate in a football tournament consisting of Penalty shootouts. They try to score one after another in that order without any breaks. In every round, every player gets exactly 1 chance to score. The tournament has ‘n’ rounds. For any player 'P' the player taking the penalty just before him (not necessarily in the same round) is called as the preceding player for that round.
The following observations were made:-
1. If P scores the penalty in the previous round and the preceding player also scores the penalty, then P will miss the penalty in the current round.
2. If P scores the penalty in the previous round and the preceding player misses the penalty, then P will score the penalty in the current round.
3. If P misses the penalty in the previous round and the preceding player also misses the penalty, then P will miss the penalty in the current round.
4. If P misses the penalty in the previous round and the preceding player scores the penalty, then P will score the penalty in the current round.
5. At least 1 player scored the penalty in every round.

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What is the number of penalties that Neymar scores in the first 21 rounds if Ronaldo is the only player to score the penalty in Round 1?

A
15

B
10

C
9

D
12

Solution

Let 1 denote penalty scored and 0 denote penalty missed.
Let only Ronaldo scores the penalty in Round 1 thus, in round 1 the table will be:-

In round 2, Ronaldo will score the penalty as he scored the penalty in round 1 and Neymar(preceding player) missed the penalty in round 1.
Messi will score the penalty as he missed the penalty in round 1 and Ronaldo(preceding player) scored the penalty in round 2.
Neymar will score the penalty as he missed the penalty in round 1 and Messi(preceding player) scored the penalty in round 2.

Let A represents the penalty missed or scored by a Player in the previous round, B represents the penalty missed or scored by the preceding player and C represents the penalty missed or scored by the Player in the next round.
Thus, as per the given conditions, we get

From this, the table for the first 9 rounds will be:-

As we can see that the pattern repeats after every 7 rounds.
In the first 7 rounds, Neymar will score the penalty 4 times.
Thus, in the first 21 rounds, Neymar will score the penalty 12 times.
Hence, option D is the correct answer.
Question 7
3 / -1

Directions For Questions

Ronaldo, Messi, and Neymar participate in a football tournament consisting of Penalty shootouts. They try to score one after another in that order without any breaks. In every round, every player gets exactly 1 chance to score. The tournament has ‘n’ rounds. For any player 'P' the player taking the penalty just before him (not necessarily in the same round) is called as the preceding player for that round.
The following observations were made:-
1. If P scores the penalty in the previous round and the preceding player also scores the penalty, then P will miss the penalty in the current round.
2. If P scores the penalty in the previous round and the preceding player misses the penalty, then P will score the penalty in the current round.
3. If P misses the penalty in the previous round and the preceding player also misses the penalty, then P will miss the penalty in the current round.
4. If P misses the penalty in the previous round and the preceding player scores the penalty, then P will score the penalty in the current round.
5. At least 1 player scored the penalty in every round.

...view full instructions

What is the maximum number of penalties scored in any 8 consecutive rounds?

A
15

B
12

C
16

D
14

Solution

Let 1 denote penalty scored and 0 denote penalty missed.
Let only Ronaldo scores the penalty in Round 1 thus, in round 1 the table will be:-

In round 2, Ronaldo will score the penalty as he scored the penalty in round 1 and Neymar(preceding player) missed the penalty in round 1.
Messi will score the penalty as he missed the penalty in round 1 and Ronaldo(preceding player) scored the penalty in round 2.
Neymar will score the penalty as he missed the penalty in round 1 and Messi(preceding player) scored the penalty in round 2.

Let A represents the penalty missed or scored by a Player in the previous round, B represents the penalty missed or scored by the preceding player and C represents the penalty missed or scored by the Player in the next round.
Thus, as per the given conditions, we get

From this, the table for the first 9 rounds will be:-

As we can see that the pattern repeats after every 7 rounds
Thus, the maximum number of penalties scored in any 8 consecutive rounds will be when in the first and last round all the 3 players score the penalty.
In any 7 consecutive rounds total 12 penalties are scored.
If in the 8th round all the 3 scores then the total number of penalties will be 12+3 = 15.
Hence, option A is the correct answer.
Question 8
3 / -1

Directions For Questions

Ronaldo, Messi, and Neymar participate in a football tournament consisting of Penalty shootouts. They try to score one after another in that order without any breaks. In every round, every player gets exactly 1 chance to score. The tournament has ‘n’ rounds. For any player 'P' the player taking the penalty just before him (not necessarily in the same round) is called as the preceding player for that round.
The following observations were made:-
1. If P scores the penalty in the previous round and the preceding player also scores the penalty, then P will miss the penalty in the current round.
2. If P scores the penalty in the previous round and the preceding player misses the penalty, then P will score the penalty in the current round.
3. If P misses the penalty in the previous round and the preceding player also misses the penalty, then P will miss the penalty in the current round.
4. If P misses the penalty in the previous round and the preceding player scores the penalty, then P will score the penalty in the current round.
5. At least 1 player scored the penalty in every round.

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If Messi missed the penalty in Round 1 then what is the minimum number of penalties he can score in the first 13 rounds?

A
5

B
8

C
6

D
7

Solution

Let 1 denote penalty scored and 0 denote penalty missed.
Let only Ronaldo scores the penalty in Round 1 thus, in round 1 the table will be:-

In round 2, Ronaldo will score the penalty as he scored the penalty in round 1 and Neymar(preceding player) missed the penalty in round 1.
Messi will score the penalty as he missed the penalty in round 1 and Ronaldo(preceding player) scored the penalty in round 2.
Neymar will score the penalty as he missed the penalty in round 1 and Messi(preceding player) scored the penalty in round 2.

Let A represents the penalty missed or scored by a Player in the previous round, B represents the penalty missed or scored by the preceding player and C represents the penalty missed or scored by the Player in the next round.
Thus, as per the given conditions, we get

From this, the table for the first 9 rounds will be:-

As we can see that the pattern repeats after every 7 rounds
In any 7 consecutive rounds Messi will score 4 penalties. In next 6 rounds, he will score a minimum of 3 goals.
Thus, if Messi missed the penalty in Round 1, then 7 is the minimum number of penalties he can score in the next 13 rounds.
Hence, option D is the correct answer.
Question 9
3 / -1

Directions For Questions

Ronaldo, Messi, and Neymar participate in a football tournament consisting of Penalty shootouts. They try to score one after another in that order without any breaks. In every round, every player gets exactly 1 chance to score. The tournament has ‘n’ rounds. For any player 'P' the player taking the penalty just before him (not necessarily in the same round) is called as the preceding player for that round.
The following observations were made:-
1. If P scores the penalty in the previous round and the preceding player also scores the penalty, then P will miss the penalty in the current round.
2. If P scores the penalty in the previous round and the preceding player misses the penalty, then P will score the penalty in the current round.
3. If P misses the penalty in the previous round and the preceding player also misses the penalty, then P will miss the penalty in the current round.
4. If P misses the penalty in the previous round and the preceding player scores the penalty, then P will score the penalty in the current round.
5. At least 1 player scored the penalty in every round.

...view full instructions

All the 3 players scored the penalties in a round. What is the total number of penalties scored in the next 51 rounds?

A
78

B
92

C
63

D
87

Solution

Let 1 denote penalty scored and 0 denote penalty missed.
Let only Ronaldo scores the penalty in Round 1 thus, in round 1 the table will be:-

In round 2, Ronaldo will score the penalty as he scored the penalty in round 1 and Neymar(preceding player) missed the penalty in round 1.
Messi will score the penalty as he missed the penalty in round 1 and Ronaldo(preceding player) scored the penalty in round 2.
Neymar will score the penalty as he missed the penalty in round 1 and Messi(preceding player) scored the penalty in round 2.

Let A represents the penalty missed or scored by a Player in the previous round, B represents the penalty missed or scored by the preceding player and C represents the penalty missed or scored by the Player in the next round.
Thus, as per the given conditions, we get

From this, the table for the first 9 rounds will be:-

As we can see that the pattern repeats after every 7 rounds
In any 7 consecutive rounds total 12 penalties are scored.
Thus, in the first 49 rounds, the total number of penalties scored will be 7*12 = 84.
In 50th and 51st round they would have scored 1 and 2 penalties respectively.
Thus, the required sum = 84+3 = 87.
hence, option D is the correct answer
Question 10
3 / -1

Directions For Questions

Ronaldo, Messi, and Neymar participate in a football tournament consisting of Penalty shootouts. They try to score one after another in that order without any breaks. In every round, every player gets exactly 1 chance to score. The tournament has ‘n’ rounds. For any player 'P' the player taking the penalty just before him (not necessarily in the same round) is called as the preceding player for that round.
The following observations were made:-
1. If P scores the penalty in the previous round and the preceding player also scores the penalty, then P will miss the penalty in the current round.
2. If P scores the penalty in the previous round and the preceding player misses the penalty, then P will score the penalty in the current round.
3. If P misses the penalty in the previous round and the preceding player also misses the penalty, then P will miss the penalty in the current round.
4. If P misses the penalty in the previous round and the preceding player scores the penalty, then P will score the penalty in the current round.
5. At least 1 player scored the penalty in every round.

...view full instructions

In round 1, only Messi and Neymar scored penalties. In which round will Ronaldo score his 3rd penalty?

A
6

B
5

C
4

D
7

Solution

Let A represents the penalty missed or scored by a Player in the previous round, B represents the penalty missed or scored by the preceding player, and C represents the penalty missed or scored by the Player in the next round.
Thus, as per the given conditions, we get

In round 1, Ronaldo didn't score, but his preceding player scored. Hence in round 2, he will score. Messi will not score in round 2, because he scored in round 1 and Ronaldo scored in round 2. likewise, we get the following table.

Hence the answer is Option (B)
Question 11
3 / -1

Directions For Questions

There are 3 sections in the test and the tables shown below show the score in all 3 sections among 100 students. A score is shown in the left column and the number of test-takers who have scored less then that marks is shown in the right column. The overall score is an integral value and the individual score is also an integer. Assume that if the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in other two sections also. For example, if student A has rank 3 in section 1, then he would have the same rank 3 in section 2 and 3 also.

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What is the total score of the test taker who scored the highest?

A
26

B
27

C
28

D
29

Solution

Take a look at the exact score and the number of test takers who scored that score in the table shown below.

The highest scorer scored = 10+9+8 = 27
Question 12
3 / -1

Directions For Questions

There are 3 sections in the test and the tables shown below show the score in all 3 sections among 100 students. A score is shown in the left column and the number of test-takers who have scored less then that marks is shown in the right column. The overall score is an integral value and the individual score is also an integer. Assume that if the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in other two sections also. For example, if student A has rank 3 in section 1, then he would have the same rank 3 in section 2 and 3 also.

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What is the number of candidates who scored 5 or more than 5 in each of the three sections?

A
25

B
30

C
20

D
15

Solution

Number of candidates who scored 5 or more than 5 in each of the three sections = 30 (18+7+2+1+1+1), 25 (13+5+4+2+1) and 20 (12+4+3+1).

It is given that the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in other two sections also. This means that the student who is ranked 51st in section 1 will also be ranked 51st for the other two sections also and this will be true for all the other students.

So, the number of students who scored 5 or more in ALL the three sections is 20 (which is the smallest out of the three numbers 30, 25 and 20)
Question 13
3 / -1

Directions For Questions

There are 3 sections in the test and the tables shown below show the score in all 3 sections among 100 students. A score is shown in the left column and the number of test-takers who have scored less then that marks is shown in the right column. The overall score is an integral value and the individual score is also an integer. Assume that if the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in other two sections also. For example, if student A has rank 3 in section 1, then he would have the same rank 3 in section 2 and 3 also.

...view full instructions

If 50% of the test-takers cleared the overall cutoff, what is the minimum overall score of a person who cleared the cutoff?

A
6

B
7

C
8

D
9

Solution

A score is shown in the left column and the number of test-takers who have scored less then that marks is shown in the right column and individual score of each column is an integer.

Hence we can draw a table for the number of test-takers who have scored exact marks.

In section 1 test-takers who scored exact 0=18

Similarly test-takers who scored exact 1 = 32-18=14

We can create a following table on the basis of the given data.

It is given that the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in the other two sections also. This means that the student who is ranked 51st in section 1 will also be ranked 51st for the other two sections also
and this will be true for all the other students.

If we rank the students according to their marks then we can say that

Students with ranks from 83 to 100 have scored 0 in all three sections so their overall score is also 0.

Students with ranks from 76 to 82 have scored 1 in section 1 and 0 in the rest of the sections so their overall score is 1.

Students with ranks from 70 to 75 have scored 1 in both section 1 and section 2 and 0 in section 3 so their overall score is 2.

Students with rank 69 have scored 1 in all three sections so their overall score is 3.

Students with ranks from 61 to 68 have scored 2 in section 1 and 1 in the other two sections so their overall score is 4.

Students with ranks from 53 to 60 have scored 2 in sections 1 and 2 and 1 in section 3 so their overall score is 5.

Students with ranks from 51 to 52 have scored 2 in all three sections so their overall score is 6.

Student ranked 50 scored a total of 4 in section 1, 3 in section 2, and 2 in section 1 => overall score is 9. Students whose ranks are 1 - 49 will have scored greater than or equal to 9. => => This is the minimum score of someone who cleared the cutoff.
Question 14
3 / -1

Directions For Questions

There are 3 sections in the test and the tables shown below show the score in all 3 sections among 100 students. A score is shown in the left column and the number of test-takers who have scored less then that marks is shown in the right column. The overall score is an integral value and the individual score is also an integer. Assume that if the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in other two sections also. For example, if student A has rank 3 in section 1, then he would have the same rank 3 in section 2 and 3 also.

...view full instructions

What is the number of candidates who scored 4 or more than 4 in each of the three sections?

A
25

B
30

C
35

D
50

Solution

Number of candidates who scored 4 or more than 4 in each of the three sections = 50 (20+18+7+2+1+1+1), 35 (10+13+5+4+2+1) and 30 (10+12+4+3+1).

It is given that the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in the other two sections also. This means that the student who is ranked 51st in section 1 will also be ranked 51st for the other two sections also
and this will be true for all the other students.

So the number of candidates who scored 4 or more than 4 in all the three sections will be 30.
Question 15
3 / -1

Directions For Questions

There are 3 sections in the test and the tables shown below show the score in all 3 sections among 100 students. A score is shown in the left column and the number of test-takers who have scored less then that marks is shown in the right column. The overall score is an integral value and the individual score is also an integer. Assume that if the test-takers are arranged in descending order of their scores in one section, then the same arrangement would occur in other two sections also. For example, if student A has rank 3 in section 1, then he would have the same rank 3 in section 2 and 3 also.

...view full instructions

What is the total score of the person with rank 18?

A
15

B
18

C
12

D
17

Solution

The exact score and the number of test takers who scored that score are shown below.

The score of the 18th rank holder in Section 1 is 5, as 17 people above him scored 5 or more in section 1.

Similarly,

In section 2, the 18th rank holder scored 5.

In section 3, the 18th rank holder scored 5.

Total marks of Overall 18th ranker = 5 + 5 + 5 = 15 Marks.

The answer is Option (A).
Question 16
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Directions For Questions

There are 850 households in a colony. Exactly three newspapers - Hindu, TOI and ET are circulated in this colony. 370 households subscribed to Hindu, 336 households subscribed to TOI and 298 households subscribed to ET. 96 households subscribed to all the three newspapers and 128 households have not subscribed to any of the newspapers.

...view full instructions

How many households subscribed to at least two newspapers?

A
186

B
1864

C
1867

D
1869

Solution

The given data is represented in the below figure.

a + d + f = 370 - 96 = 274
b + e + d = 336 - 96 = 240
c + f + e = 298 - 96 = 202
=> a + b + c + 2(d + e + f) = 716 - Eqn(1)
a + b + c + d + e + f + 96 + 128 = 850
=> a + b + c + d + e + f = 626 - Eqn(2)
Eqn(1) - Eqn(2)
=> d + e + f = 90
At least 2 newspapers => 90 + 96 = 186
Question 17
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Directions For Questions

There are 850 households in a colony. Exactly three newspapers - Hindu, TOI and ET are circulated in this colony. 370 households subscribed to Hindu, 336 households subscribed to TOI and 298 households subscribed to ET. 96 households subscribed to all the three newspapers and 128 households have not subscribed to any of the newspapers.

...view full instructions

How many households subscribed to exactly one newspaper?

A
536

B
5364

C
5367

D
5369

Solution

The given data is represented in the below figure.

a + d + f = 370 - 96 = 274
b + e + d = 336 - 96 = 240
c + f + e = 298 - 96 = 202
=> a + b + c + 2(d + e + f) = 716 - Eqn(1)
a + b + c + d + e + f + 96 + 128 = 850
=> a + b + c + d + e + f = 626 - Eqn(2)
Eqn(1) - Eqn(2)
=> d + e + f = 90
=> a + b + c = 626 - 90 = 536
Question 18
3 / -1

Directions For Questions

There are 850 households in a colony. Exactly three newspapers - Hindu, TOI and ET are circulated in this colony. 370 households subscribed to Hindu, 336 households subscribed to TOI and 298 households subscribed to ET. 96 households subscribed to all the three newspapers and 128 households have not subscribed to any of the newspapers.

...view full instructions

How many households have not subscribed to TOI or ET, given that exactly 234 households subscribed for only Hindu?

A
362

B
3624

C
3627

D
3629

Solution

The given data is represented in the below figure.

a + d + f = 370 - 96 = 274
b + e + d = 336 - 96 = 240
c + f + e = 298 - 96 = 202
=> a + b + c + 2(d + e + f) = 716 - Eqn(1)
a + b + c + d + e + f + 96 + 128 = 850
=> a + b + c + d + e + f = 626 - Eqn(2)
Eqn(1) - Eqn(2)
=> d + e + f = 90
=> a + b + c = 626 - 90 = 536
Not subscribed to at least one newspaper between TOI and ET => Households who subscribed for only Hindu + Households who have not subscribed for any of the newspapers
= 234 + 128
= 362
Question 19
3 / -1

Directions For Questions

There are 850 households in a colony. Exactly three newspapers - Hindu, TOI and ET are circulated in this colony. 370 households subscribed to Hindu, 336 households subscribed to TOI and 298 households subscribed to ET. 96 households subscribed to all the three newspapers and 128 households have not subscribed to any of the newspapers.

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If exactly 84 households are subscribed to only TOI and Hindu, how many households subscribed to only ET?

A
196

B
1964

C
1967

D
1969

Solution

The given data is represented in the below figure.

a + d + f = 370 - 96 = 274
b + e + d = 336 - 96 = 240
c + f + e = 298 - 96 = 202
=> a + b + c + 2(d + e + f) = 716 - Eqn(1)
a + b + c + d + e + f + 96 + 128 = 850
=> a + b + c + d + e + f = 626 - Eqn(2)
Eqn(1) - Eqn(2)
=> d + e + f = 90
=> a + b + c = 626 - 90 = 536
It is given that d = 84 => e + f = 6
Number of households subscribed to only ET => c = 298 - 96 - (e + f) = 298 - 96 - 6 = 196
Question 20
3 / -1

Directions For Questions

There are 850 households in a colony. Exactly three newspapers - Hindu, TOI and ET are circulated in this colony. 370 households subscribed to Hindu, 336 households subscribed to TOI and 298 households subscribed to ET. 96 households subscribed to all the three newspapers and 128 households have not subscribed to any of the newspapers.

...view full instructions

If exactly 202 people subscribed to only ET, Find how many people subscribed to both Hindu and TOI.

A
186

B
196

C
90

D
96

Solution

The given data is represented in the below figure.

a + d + f = 370 - 96 = 274
b + e + d = 336 - 96 = 240
c + f + e = 298 - 96 = 202
=> a + b + c + 2(d + e + f) = 716 - Eqn(1)
a + b + c + d + e + f + 96 + 128 = 850
=> a + b + c + d + e + f = 626 - Eqn(2)
Eqn(1) - Eqn(2)
=> d + e + f = 90
=> a + b + c = 626 - 90 = 536

It is given that c = 202

c + e + f = 202 => e + f = 0

Then d + e + f = 90 => d = 90.

The number of households subscribed to both Hindu & TOI = d + 96 = 90 + 96 = 186.

The answer is Option (A).