Infosys Logical Reasoning Questions and Answers pdf

1. Step 1: Analyzing the Argument

   Look for differences across the answer options to identify what is being tested in the sentence.

   • Linda’s argument has a cause-effect relationship and a comparison element as well.
     • The causal argument is that the safety of women (or low mortality among women) is dependent on the hygiene awareness among the doctors.
     • The comparison element is that women today are better off than the women in the 1800s.
   • If Fiona is to question Linda’s argument, she has to establish that one of these two (the causal and the comparison) relationships do not actually hold true. This is the reason that Fiona’s argument tries to establish that awareness is not enough. Mortality is also affected by the actual practices of the doctors.
   • To further weaken Linda’s argument, the correct answer Option must extend this discussion or question whether women today are actually better off than the women from the 1800s.

Step 2: Eliminating Options

• Option (B) can be eliminated because we do not know whether these hospital-acquired infections were because of poor practices of the doctors or because of other factors.
• Option (C) can be eliminated because the option adds further support to Linda’s argument, when in fact the objective is to weaken her argument.
• Option (D) can be eliminated because the option again adds support to Linda’s argument. Some people wonder about the usage of the words “self-reported”, which might imply that the 90% figure may not be accurate. However, it cannot be assumed that the doctors are lying or that the actual rate is much lower than 90%.
• Option (E) seems to imply that the nurses do not have proper practices in place. At first glance, the option seems to work. However, the option assumes that doctors do not adhere to the schedules, implying that nurses are as bad. Nurses and doctors could both be GOOD at sticking to the cleansing schedules. Moreover, the focus of the discussion is on the practice of doctors and not of nurses.
• Option (A) works because it establishes another reason for carelessness among doctors when it comes to hygiene. If doctors believe they do not carry pathogens, they are more likely to be careless about cleansing schedules

Solution:
In questions, where the seating arrangement is a circle, then the following points should be kept in mind.
→Always start filling the position from the bottom-most position
→If phrase to the right is used then the position should be filled in anti clock-wise direction

→If the phrase to the left is used then clockwise direction.

Clue: Abhijeet focus the centre – sits 3rd to the right of Rakesh

→Always the 1st person in the 1st clue is filled in the 1st position of the circle
→Here, Abhijeet focus centre – so filled in position 1 facing centre
→Now, he sits 3rd to the right to Rakesh (facing/not facing centre)
→So tentative position of Rakesh
· If facing centre, 6
· If facing outside, 4

This is an almost impossibly tough question if you do not have a grasp on finding prime numbers. Fortunately, it will be well worth our time to find the three digit prime numbers under 300, since the whole question is based on this. Since we only have to check up to 300, we have to bother checking all multiples of primes 17 and under only (since 17 square is 289). Let\'s create a table where we list down all odd numbers under 300, and eliminating all multiples of 3 and 5 (these are the easiest odd numbers to eliminate since divisibility test for 3 is straightforward.)

We get this table:

We can list all odd multiples of 7, 11, 13 and 17 above 100 and under 300 and cross this off our table.

Multiples of 7: 105, 119, 133, 147, 161, 175, 189, 203, 217, 231, 245, 259, 273, 287 (We can generate this list quickly by step counting 14).

Multiples of 11:121, 143, 165, 187, 209, 231, 253, 275, 287 (We can generate this list quickly by step counting 22).

Multiples of 13:117, 143, 169, 195, 221, 247, 273, 299 (We can generate this list by step counting 26). Multiples of 17:119, 153, 187, 221, 255, 289 (We can generate this by step counting 34).

Now, if we cross out any common numbers in our list and the table that we drew earlier, we will have a list of all prime numbers under 300. This whole process is just an application of the Sieve of Eratosthenes (more interesting stuff to read about!)

Now that we have all the primes, we\'ve got to do a little thinking. We\'ve got two lists of five consecutive numbers to work with. Also, the puzzle tells us this list behaves like this: If the first floor had numbers 101, 102, 103, 104, 105; the second floor has 201, 202, 203, 204, 205.

In any list of 5 consecutive numbers, we can have a maximum of three odd numbers. Our required prime numbers will definitely be odd. Also, no three consecutive odd numbers can be concurrently prime. Think about this. The numbers can be of the form \'n\', \'n+2\', and \'n+4\'. One of these three odd numbers will definitely be a multiple of 3 (try to prove this, it\'s quite fun to establish).

This essentially means that one floor cannot have 3 primes. This in turn implies that each floor should have exactly 2 primes. So any set of two primes that are within the range of 5 consecutive numbers needs to be looked at. This also means that one pair of primes (at least) will be have common ten\'s and unit\'s digits (that is 1ab and 2ab. Let\'s refine our table a little bit to show only the primes:

Let\'s look at the primes in the 200 series (There are fewer to work with). From this point it is trial and error. We notice that 227 and 229 are a possible pair, and so are 127 and 131. A possible sequence of numbers can be 127, 128, 129, 120, 131, and 227, 228, 229, 230 and 231.

Another possible pair is 239 and 241. We notice that 137 and 139 are also close together. So another sequence is 137, 138, 139, 140, 141, and 237, 238, 239, 240, 241.

Another pair is 277 and 281. We notice that 179 and 181 are also close together. So another sequence is 177, 178, 179, 180, 181 and 277, 278, 279, 280 and 281. The final pair is 281 and 283. The sequence is 179, 180, 181, 182, 183 and 279, 280, 281, 282, 283. The possible room numbers are listed below, with the primes in red. The average of all room numbers in each sequence should also be a prime number

We see that there there are two possibilities. Both Sequence 1 and 3 are possible solutions at this stage. But Xuomi said he wanted the first room on the first floor to also be a prime number. This is possible only in Sequence 1. So this is the answer.

The average of all rooms in the first floor is just the middle number (since the numbers are in AP, the middle term is the average).

General Solution

This is an almost impossibly tough question if you do not have a grasp on finding prime numbers. Fortunately, it will be well worth our time to find the three digit prime numbers under 300, since the whole question is based on this. Since we only have to check up to 300, we have to bother checking all multiples of primes 17 and under only (since 17 square is 289). Let\'s create a table where we list down all odd numbers under 300, and eliminating all multiples of 3 and 5 (these are the easiest odd numbers to eliminate since divisibility test for 3 is straightforward.)

We get this table:

We can list all odd multiples of 7, 11, 13 and 17 above 100 and under 300 and cross this off our table.

Multiples of 7: 105, 119, 133, 147, 161, 175, 189, 203, 217, 231, 245, 259, 273, 287 (We can generate this list quickly by step counting 14).

Multiples of 11:121, 143, 165, 187, 209, 231, 253, 275, 287 (We can generate this list quickly by step counting 22).

Multiples of 13:117, 143, 169, 195, 221, 247, 273, 299 (We can generate this list by step counting 26). Multiples of 17:119, 153, 187, 221, 255, 289 (We can generate this by step counting 34).

Now, if we cross out any common numbers in our list and the table that we drew earlier, we will have a list of all prime numbers under 300. This whole process is just an application of the Sieve of Eratosthenes (more interesting stuff to read about!)

Now that we have all the primes, we\'ve got to do a little thinking. We\'ve got two lists of five consecutive numbers to work with. Also, the puzzle tells us this list behaves like this: If the first floor had numbers 101, 102, 103, 104, 105; the second floor has 201, 202, 203, 204, 205.

In any list of 5 consecutive numbers, we can have a maximum of three odd numbers. Our required prime numbers will definitely be odd. Also, no three consecutive odd numbers can be concurrently prime. Think about this. The numbers can be of the form \'n\', \'n+2\', and \'n+4\'. One of these three odd numbers will definitely be a multiple of 3 (try to prove this, it\'s quite fun to establish).

This essentially means that one floor cannot have 3 primes. This in turn implies that each floor should have exactly 2 primes. So any set of two primes that are within the range of 5 consecutive numbers needs to be looked at. This also means that one pair of primes (at least) will be have common ten\'s and unit\'s digits (that is 1ab and 2ab. Let\'s refine our table a little bit to show only the primes:

Let\'s look at the primes in the 200 series (There are fewer to work with). From this point it is trial and error. We notice that 227 and 229 are a possible pair, and so are 127 and 131. A possible sequence of numbers can be 127, 128, 129, 120, 131, and 227, 228, 229, 230 and 231.

Another possible pair is 239 and 241. We notice that 137 and 139 are also close together. So another sequence is 137, 138, 139, 140, 141, and 237, 238, 239, 240, 241.

Another pair is 277 and 281. We notice that 179 and 181 are also close together. So another sequence is 177, 178, 179, 180, 181 and 277, 278, 279, 280 and 281. The final pair is 281 and 283. The sequence is 179, 180, 181, 182, 183 and 279, 280, 281, 282, 283. The possible room numbers are listed below, with the primes in red. The average of all room numbers in each sequence should also be a prime number

We see that there there are two possibilities. Both Sequence 1 and 3 are possible solutions at this stage. But Xuomi said he wanted the first room on the first floor to also be a prime number. This is possible only in Sequence 1. So this is the answer.

The largest number is 231.

General Solution

This is an almost impossibly tough question if you do not have a grasp on finding prime numbers. Fortunately, it will be well worth our time to find the three digit prime numbers under 300, since the whole question is based on this. Since we only have to check up to 300, we have to bother checking all multiples of primes 17 and under only (since 17 square is 289). Let\\\'s create a table where we list down all odd numbers under 300, and eliminating all multiples of 3 and 5 (these are the easiest odd numbers to eliminate since divisibility test for 3 is straightforward.)

We get this table:

We can list all odd multiples of 7, 11, 13 and 17 above 100 and under 300 and cross this off our table.

Multiples of 7: 105, 119, 133, 147, 161, 175, 189, 203, 217, 231, 245, 259, 273, 287 (We can generate this list quickly by step counting 14).

Multiples of 11:121, 143, 165, 187, 209, 231, 253, 275, 287 (We can generate this list quickly by step counting 22).

Multiples of 13:117, 143, 169, 195, 221, 247, 273, 299 (We can generate this list by step counting 26). Multiples of 17:119, 153, 187, 221, 255, 289 (We can generate this by step counting 34).

Now, if we cross out any common numbers in our list and the table that we drew earlier, we will have a list of all prime numbers under 300. This whole process is just an application of the Sieve of Eratosthenes (more interesting stuff to read about!)

Now that we have all the primes, we\\\'ve got to do a little thinking. We\\\'ve got two lists of five consecutive numbers to work with. Also, the puzzle tells us this list behaves like this: If the first floor had numbers 101, 102, 103, 104, 105; the second floor has 201, 202, 203, 204, 205.

In any list of 5 consecutive numbers, we can have a maximum of three odd numbers. Our required prime numbers will definitely be odd. Also, no three consecutive odd numbers can be concurrently prime. Think about this. The numbers can be of the form \\\'n\\\', \\\'n+2\\\', and \\\'n+4\\\'. One of these three odd numbers will definitely be a multiple of 3 (try to prove this, it\\\'s quite fun to establish).

This essentially means that one floor cannot have 3 primes. This in turn implies that each floor should have exactly 2 primes. So any set of two primes that are within the range of 5 consecutive numbers needs to be looked at. This also means that one pair of primes (at least) will be have common ten\\\'s and unit\\\'s digits (that is 1ab and 2ab. Let\\\'s refine our table a little bit to show only the primes:

Let\\\'s look at the primes in the 200 series (There are fewer to work with). From this point it is trial and error. We notice that 227 and 229 are a possible pair, and so are 127 and 131. A possible sequence of numbers can be 127, 128, 129, 120, 131, and 227, 228, 229, 230 and 231.

Another possible pair is 239 and 241. We notice that 137 and 139 are also close together. So another sequence is 137, 138, 139, 140, 141, and 237, 238, 239, 240, 241.

Another pair is 277 and 281. We notice that 179 and 181 are also close together. So another sequence is 177, 178, 179, 180, 181 and 277, 278, 279, 280 and 281. The final pair is 281 and 283. The sequence is 179, 180, 181, 182, 183 and 279, 280, 281, 282, 283. The possible room numbers are listed below, with the primes in red. The average of all room numbers in each sequence should also be a prime number

We see that there there are two possibilities. Both Sequence 1 and 3 are possible solutions at this stage. But Xuomi said he wanted the first room on the first floor to also be a prime number. This is possible only in Sequence 1. So this is the answer.

The smallest number is 127.

General Solution

This is an almost impossibly tough question if you do not have a grasp on finding prime numbers. Fortunately, it will be well worth our time to find the three digit prime numbers under 300, since the whole question is based on this. Since we only have to check up to 300, we have to bother checking all multiples of primes 17 and under only (since 17 square is 289). Let\'s create a table where we list down all odd numbers under 300, and eliminating all multiples of 3 and 5 (these are the easiest odd numbers to eliminate since divisibility test for 3 is straightforward.)

We get this table:

We can list all odd multiples of 7, 11, 13 and 17 above 100 and under 300 and cross this off our table.

Multiples of 7: 105, 119, 133, 147, 161, 175, 189, 203, 217, 231, 245, 259, 273, 287 (We can generate this list quickly by step counting 14).

Multiples of 11:121, 143, 165, 187, 209, 231, 253, 275, 287 (We can generate this list quickly by step counting 22).

Multiples of 13:117, 143, 169, 195, 221, 247, 273, 299 (We can generate this list by step counting 26). Multiples of 17:119, 153, 187, 221, 255, 289 (We can generate this by step counting 34).

Now, if we cross out any common numbers in our list and the table that we drew earlier, we will have a list of all prime numbers under 300. This whole process is just an application of the Sieve of Eratosthenes (more interesting stuff to read about!)

Now that we have all the primes, we\'ve got to do a little thinking. We\'ve got two lists of five consecutive numbers to work with. Also, the puzzle tells us this list behaves like this: If the first floor had numbers 101, 102, 103, 104, 105; the second floor has 201, 202, 203, 204, 205.

In any list of 5 consecutive numbers, we can have a maximum of three odd numbers. Our required prime numbers will definitely be odd. Also, no three consecutive odd numbers can be concurrently prime. Think about this. The numbers can be of the form \'n\', \'n+2\', and \'n+4\'. One of these three odd numbers will definitely be a multiple of 3 (try to prove this, it\'s quite fun to establish).

This essentially means that one floor cannot have 3 primes. This in turn implies that each floor should have exactly 2 primes. So any set of two primes that are within the range of 5 consecutive numbers needs to be looked at. This also means that one pair of primes (at least) will be have common ten\'s and unit\'s digits (that is 1ab and 2ab. Let\'s refine our table a little bit to show only the primes:

Let\'s look at the primes in the 200 series (There are fewer to work with). From this point it is trial and error. We notice that 227 and 229 are a possible pair, and so are 127 and 131. A possible sequence of numbers can be 127, 128, 129, 120, 131, and 227, 228, 229, 230 and 231.

Another possible pair is 239 and 241. We notice that 137 and 139 are also close together. So another sequence is 137, 138, 139, 140, 141, and 237, 238, 239, 240, 241.

Another pair is 277 and 281. We notice that 179 and 181 are also close together. So another sequence is 177, 178, 179, 180, 181 and 277, 278, 279, 280 and 281. The final pair is 281 and 283. The sequence is 179, 180, 181, 182, 183 and 279, 280, 281, 282, 283. The possible room numbers are listed below, with the primes in red. The average of all room numbers in each sequence should also be a prime number

We see that there there are two possibilities. Both Sequence 1 and 3 are possible solutions at this stage. But Xuomi said he wanted the first room on the first floor to also be a prime number. This is possible only in Sequence 1. So this is the answer.

The overall average is 179 that has already been calculated. We did this to satisfy one of the conditions (refer to the table above).

General Solution

This is an almost impossibly tough question if you do not have a grasp on finding prime numbers. Fortunately, it will be well worth our time to find the three digit prime numbers under 300, since the whole question is based on this. Since we only have to check up to 300, we have to bother checking all multiples of primes 17 and under only (since 17 square is 289). Let\'s create a table where we list down all odd numbers under 300, and eliminating all multiples of 3 and 5 (these are the easiest odd numbers to eliminate since divisibility test for 3 is straightforward.)

We get this table:

We can list all odd multiples of 7, 11, 13 and 17 above 100 and under 300 and cross this off our table.

Multiples of 7: 105, 119, 133, 147, 161, 175, 189, 203, 217, 231, 245, 259, 273, 287 (We can generate this list quickly by step counting 14).

Multiples of 11:121, 143, 165, 187, 209, 231, 253, 275, 287 (We can generate this list quickly by step counting 22).

Multiples of 13:117, 143, 169, 195, 221, 247, 273, 299 (We can generate this list by step counting 26). Multiples of 17:119, 153, 187, 221, 255, 289 (We can generate this by step counting 34).

Now, if we cross out any common numbers in our list and the table that we drew earlier, we will have a list of all prime numbers under 300. This whole process is just an application of the Sieve of Eratosthenes (more interesting stuff to read about!)

Now that we have all the primes, we\'ve got to do a little thinking. We\'ve got two lists of five consecutive numbers to work with. Also, the puzzle tells us this list behaves like this: If the first floor had numbers 101, 102, 103, 104, 105; the second floor has 201, 202, 203, 204, 205.

In any list of 5 consecutive numbers, we can have a maximum of three odd numbers. Our required prime numbers will definitely be odd. Also, no three consecutive odd numbers can be concurrently prime. Think about this. The numbers can be of the form \'n\', \'n+2\', and \'n+4\'. One of these three odd numbers will definitely be a multiple of 3 (try to prove this, it\'s quite fun to establish).

This essentially means that one floor cannot have 3 primes. This in turn implies that each floor should have exactly 2 primes. So any set of two primes that are within the range of 5 consecutive numbers needs to be looked at. This also means that one pair of primes (at least) will be have common ten\'s and unit\'s digits (that is 1ab and 2ab. Let\'s refine our table a little bit to show only the primes:

Let\'s look at the primes in the 200 series (There are fewer to work with). From this point it is trial and error. We notice that 227 and 229 are a possible pair, and so are 127 and 131. A possible sequence of numbers can be 127, 128, 129, 120, 131, and 227, 228, 229, 230 and 231.

Another possible pair is 239 and 241. We notice that 137 and 139 are also close together. So another sequence is 137, 138, 139, 140, 141, and 237, 238, 239, 240, 241.

Another pair is 277 and 281. We notice that 179 and 181 are also close together. So another sequence is 177, 178, 179, 180, 181 and 277, 278, 279, 280 and 281. The final pair is 281 and 283. The sequence is 179, 180, 181, 182, 183 and 279, 280, 281, 282, 283. The possible room numbers are listed below, with the primes in red. The average of all room numbers in each sequence should also be a prime number

We see that there there are two possibilities. Both Sequence 1 and 3 are possible solutions at this stage. But Xuomi said he wanted the first room on the first floor to also be a prime number. This is possible only in Sequence 1. So this is the answer.

We have to check which sequence when extended to the 300s also has two primes.
Only sequence 4 has two primes in the 300s. In this case, the average of all numbers is: 281 (which ironically is a prime number).

General Solution

Total revenues are Rs. 1800 crores. Overall profit margin is 10%. Overall profit is Rs. 180 crores. Exactly one division has the same profit margin as the overall Company. So, for this division, share of revenues should be equal to share of profits. Or, one of Company A or D has a profit margin of 10%.

Let us also compute absolute profit numbers – the 5 numbers are Rs. 35 crores, Rs. 40 crores, Rs. 25 crores, Rs. 30 crores and Rs. 50 crores
The division with the largest revenue has the least profit margin but not the least profits.
So, E has the least profit margin but not the least profits. Company D generates more profits than Company E, so Company E does not generate the least profit either.
So, Company E should generate profits of Rs. 35 crores or Rs. 40 crores.
Now, this is the critical statement – “The division with the profit margin higher than all others generates the least profit.” Company C has the least revenue, and by a distance. So, chances are division C should have the highest profit margin. The least profit margin for division C should be 16.67%. Now, if the Rs. 25 crore profit were for any other division, the profit margin of that division would be less than 16.67%. This tells us that the Rs. 25 crore profit is made by division C at a profit margin of 16.67%.
So, this is where we stand now.

The three missing profit numbers are Rs. 50 crores, Rs. 40 crores and Rs. 35 crores.
Company E did not generate profit of Rs. 50 crores. So, the Rs. 50 crore profit number should have been seen by either B or D. If it had been D, then D would also have had a profit margin of 16.67%. But this is not possible as the question says – “The division with the profit margin higher than all others generates the least profit.”. The above statement tells us that no company aprt from C had a profit margin of 16.67% or more.
So, Company B should have seen a profit of Rs. 50 crores.

Company D generates more profit than E, so E should have seen a profit of Rs. 35 crores and D a profit of Rs. 40 crores.

Company A saw profits of Rs. 30 crores. Choice C.
We do not even need to solve the whole grid for this. The statement “Exactly one division has the same profit margin as the overall Company.” alone is sufficient.

General Solution

Total revenues are Rs. 1800 crores. Overall profit margin is 10%. Overall profit is Rs. 180 crores. Exactly one division has the same profit margin as the overall Company. So, for this division, share of revenues should be equal to share of profits. Or, one of Company A or D has a profit margin of 10%.

Let us also compute absolute profit numbers – the 5 numbers are Rs. 35 crores, Rs. 40 crores, Rs. 25 crores, Rs. 30 crores and Rs. 50 crores
The division with the largest revenue has the least profit margin but not the least profits.
So, E has the least profit margin but not the least profits. Company D generates more profits than Company E, so Company E does not generate the least profit either.
So, Company E should generate profits of Rs. 35 crores or Rs. 40 crores.
Now, this is the critical statement – “The division with the profit margin higher than all others generates the least profit.” Company C has the least revenue, and by a distance. So, chances are division C should have the highest profit margin. The least profit margin for division C should be 16.67%. Now, if the Rs. 25 crore profit were for any other division, the profit margin of that division would be less than 16.67%. This tells us that the Rs. 25 crore profit is made by division C at a profit margin of 16.67%.
So, this is where we stand now.

The three missing profit numbers are Rs. 50 crores, Rs. 40 crores and Rs. 35 crores.
Company E did not generate profit of Rs. 50 crores. So, the Rs. 50 crore profit number should have been seen by either B or D. If it had been D, then D would also have had a profit margin of 16.67%. But this is not possible as the question says – “The division with the profit margin higher than all others generates the least profit.”. The above statement tells us that no company aprt from C had a profit margin of 16.67% or more.
So, Company B should have seen a profit of Rs. 50 crores.

Company D generates more profit than E, so E should have seen a profit of Rs. 35 crores and D a profit of Rs. 40 crores.

General Solution

Total revenues are Rs. 1800 crores. Overall profit margin is 10%. Overall profit is Rs. 180 crores. Exactly one division has the same profit margin as the overall Company. So, for this division, share of revenues should be equal to share of profits. Or, one of Company A or D has a profit margin of 10%.

Let us also compute absolute profit numbers – the 5 numbers are Rs. 35 crores, Rs. 40 crores, Rs. 25 crores, Rs. 30 crores and Rs. 50 crores
The division with the largest revenue has the least profit margin but not the least profits.
So, E has the least profit margin but not the least profits. Company D generates more profits than Company E, so Company E does not generate the least profit either.
So, Company E should generate profits of Rs. 35 crores or Rs. 40 crores.
Now, this is the critical statement – “The division with the profit margin higher than all others generates the least profit.” Company C has the least revenue, and by a distance. So, chances are division C should have the highest profit margin. The least profit margin for division C should be 16.67%. Now, if the Rs. 25 crore profit were for any other division, the profit margin of that division would be less than 16.67%. This tells us that the Rs. 25 crore profit is made by division C at a profit margin of 16.67%.
So, this is where we stand now.

The three missing profit numbers are Rs. 50 crores, Rs. 40 crores and Rs. 35 crores.
Company E did not generate profit of Rs. 50 crores. So, the Rs. 50 crore profit number should have been seen by either B or D. If it had been D, then D would also have had a profit margin of 16.67%. But this is not possible as the question says – “The division with the profit margin higher than all others generates the least profit.”. The above statement tells us that no company aprt from C had a profit margin of 16.67% or more.
So, Company B should have seen a profit of Rs. 50 crores.

Company D generates more profit than E, so E should have seen a profit of Rs. 35 crores and D a profit of Rs. 40 crores.