Logical Reasoning & DI (LRDI) Test - 38

Starting Point:

Start with fixing A as A attends one of the days before Wednesday i.e., A either attends on Monday or Tuesday.

Clues: C attends the class immediately before the one who attends the Computer class. Three persons attend the class between A and the one who attends the class of Computer.

Inference: Now, there are possibilities that either there is no holiday between the one who attends the Computer classes and A or there is a holiday between them.

Case I: When A attend on Tuesday but there is no holiday between A and the one who attends the class of Computer:

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II: When A attends the class on Monday but there is no holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: There is a gap of three days between the classes of C and the one who attends the class of Marketing. Number of days after the holiday is the same as the number of days before the one who attends the Current affairs class.

Inference: Now, case I and II are invalid as we can't fix the one who attends the class of Marketing.

Now, in case I(a), Friday must be the holiday and in case II(a), Wednesday must be the holiday so, that we can fix the one who attends the classes of Current affairs.

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: B attends the class immediately after the Current affairs class. One person attends the class between D and the one who attends Quant classes.

Inference: Now, in case II(a), we can't fix the one who attends the class of Quant as only one person attends between the one who attends the class of Quant and D so, case II(a), is invalid.

Clues: Number of days before D attends the class is same as the number of persons attend the class between F and the one who attend Reasoning classes.

Inference: Now, there are two days before D which means there must be two persons between F and the one who attends the class in Reasoning. So, F must attend the class on Monday.

The final arrangement is as follows:

No one attends the class on the day which is immediately before the day on which C attends the class.

Hence, option 1 is correct.

Starting Point:

Start with fixing A as A attends one of the days before Wednesday i.e., A either attends on Monday or Tuesday.

Clues: C attends the class immediately before the one who attends the Computer class. Three persons attend the class between A and the one who attends the class of Computer.

Inference: Now, there are possibilities that either there is no holiday between the one who attends the Computer classes and A or there is a holiday between them.

Case I: When A attend on Tuesday but there is no holiday between A and the one who attends the class of Computer:

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II: When A attends the class on Monday but there is no holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: There is a gap of three days between the classes of C and the one who attends the class of Marketing. Number of days after the holiday is the same as the number of days before the one who attends the Current affairs class.

Inference: Now, case I and II are invalid as we can't fix the one who attends the class of Marketing.

Now, in case I(a), Friday must be the holiday and in case II(a), Wednesday must be the holiday so, that we can fix the one who attends the classes of Current affairs.

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: B attends the class immediately after the Current affairs class. One person attends the class between D and the one who attends Quant classes.

Inference: Now, in case II(a), we can't fix the one who attends the class of Quant as only one person attends between the one who attends the class of Quant and D so, case II(a), is invalid.

Clues: Number of days before D attends the class is same as the number of persons attend the class between F and the one who attend Reasoning classes.

Inference: Now, there are two days before D which means there must be two persons between F and the one who attends the class in Reasoning. So, F must attend the class on Monday.

The final arrangement is as follows:

E attends the class of Computer.

Hence, option 4 is correct.

Starting Point:

Start with fixing A as A attends one of the days before Wednesday i.e., A either attends on Monday or Tuesday.

Clues: C attends the class immediately before the one who attends the Computer class. Three persons attend the class between A and the one who attends the class of Computer.

Inference: Now, there are possibilities that either there is no holiday between the one who attends the Computer classes and A or there is a holiday between them.

Case I: When A attend on Tuesday but there is no holiday between A and the one who attends the class of Computer:

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II: When A attends the class on Monday but there is no holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: There is a gap of three days between the classes of C and the one who attends the class of Marketing. Number of days after the holiday is the same as the number of days before the one who attends the Current affairs class.

Inference: Now, case I and II are invalid as we can't fix the one who attends the class of Marketing.

Now, in case I(a), Friday must be the holiday and in case II(a), Wednesday must be the holiday so, that we can fix the one who attends the classes of Current affairs.

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: B attends the class immediately after the Current affairs class. One person attends the class between D and the one who attends Quant classes.

Inference: Now, in case II(a), we can't fix the one who attends the class of Quant as only one person attends between the one who attends the class of Quant and D so, case II(a), is invalid.

Clues: Number of days before D attends the class is same as the number of persons attend the class between F and the one who attend Reasoning classes.

Inference: Now, there are two days before D which means there must be two persons between F and the one who attends the class in Reasoning. So, F must attend the class on Monday.

The final arrangement is as follows:

The one who attends the class of Quant attends the class on Saturday.

Hence, option 2 is correct.

Starting Point:

Start with fixing A as A attends one of the days before Wednesday i.e., A either attends on Monday or Tuesday.

Clues: C attends the class immediately before the one who attends the Computer class. Three persons attend the class between A and the one who attends the class of Computer.

Inference: Now, there are possibilities that either there is no holiday between the one who attends the Computer classes and A or there is a holiday between them.

Case I: When A attend on Tuesday but there is no holiday between A and the one who attends the class of Computer:

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II: When A attends the class on Monday but there is no holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: There is a gap of three days between the classes of C and the one who attends the class of Marketing. Number of days after the holiday is the same as the number of days before the one who attends the Current affairs class.

Inference: Now, case I and II are invalid as we can't fix the one who attends the class of Marketing.

Now, in case I(a), Friday must be the holiday and in case II(a), Wednesday must be the holiday so, that we can fix the one who attends the classes of Current affairs.

Case I(a): When there is a holiday between A and the one who attends the class of Computer:

Case II(a): When A attends the class on Monday but there is a holiday between A and the one who attends the class of Computer:

Clues: B attends the class immediately after the Current affairs class. One person attends the class between D and the one who attends Quant classes.

Inference: Now, in case II(a), we can't fix the one who attends the class of Quant as only one person attends between the one who attends the class of Quant and D so, case II(a), is invalid.

Clues: Number of days before D attends the class is same as the number of persons attend the class between F and the one who attend Reasoning classes.

Inference: Now, there are two days before D which means there must be two persons between F and the one who attends the class in Reasoning. So, F must attend the class on Monday.

The final arrangement is as follows:

F attends the class on Monday.

Hence, option 3 is correct.

According to the question,

Number of water bodies in region 'P' is least, therefore, number of water bodies in region 'P' will be only 1.

E doesn't lie in region 'P', therefore, 'E' lies either in region 'Q' or region 'R'.

'C' lie in region 'Q'

The water body that lie in region 'R' doesn't have least area, therefore, 'B' doesn't lie in region 'R'.

The water body in region 'P' and region 'R' have same area, therefore, 'A' and 'D' lie in either of regions 'P' of 'R'

'D' lies in cold climatic condition, therefore, 'D' lie in region 'R' (because region 'R' is in cold climatic condition)

Therefore, 'A' lie in region 'P'

Since, the number of water bodies in regions 'Q' and 'R' is equal, therefore, 'B' lie in region 'Q' and hence 'E' will lie in region 'R'.

Therefore,

Also,

Quantity I:

Required difference = (25000 + 20000) - 15000 = 30000 sq. km

Quantity II:

Required area = 950 X (20 + 12) = 30400 sq. km

Hence, option 2 is correct.

According to the question,

Number of water bodies in region 'P' is least, therefore, number of water bodies in region 'P' will be only 1.

E doesn't lie in region 'P', therefore, 'E' lies either in region 'Q' or region 'R'.

'C' lie in region 'Q'

The water body that lie in region 'R' doesn't have least area, therefore, 'B' doesn't lie in region 'R'.

The water body in region 'P' and region 'R' have same area, therefore, 'A' and 'D' lie in either of regions 'P' of 'R'

'D' lies in cold climatic condition, therefore, 'D' lie in region 'R' (because region 'R' is in cold climatic condition)

Therefore, 'A' lie in region 'P'

Since, the number of water bodies in regions 'Q' and 'R' is equal, therefore, 'B' lie in region 'Q' and hence 'E' will lie in region 'R'.

Therefore,

Also,

Required percentage = {(48000 - 19200)/48000} X 100 = 60%

Hence, option 3 is correct.

According to the question,

Number of water bodies in region 'P' is least, therefore, number of water bodies in region 'P' will be only 1.

E doesn't lie in region 'P', therefore, 'E' lies either in region 'Q' or region 'R'.

'C' lie in region 'Q'

The water body that lie in region 'R' doesn't have least area, therefore, 'B' doesn't lie in region 'R'.

The water body in region 'P' and region 'R' have same area, therefore, 'A' and 'D' lie in either of regions 'P' of 'R'

'D' lies in cold climatic condition, therefore, 'D' lie in region 'R' (because region 'R' is in cold climatic condition)

Therefore, 'A' lie in region 'P'

Since, the number of water bodies in regions 'Q' and 'R' is equal, therefore, 'B' lie in region 'Q' and hence 'E' will lie in region 'R'.

Therefore,

Also,

Required ratio = 36000:4800 = 15:2

Hence, option 1 is correct.

According to the question,

Number of water bodies in region 'P' is least, therefore, number of water bodies in region 'P' will be only 1.

E doesn't lie in region 'P', therefore, 'E' lies either in region 'Q' or region 'R'.

'C' lie in region 'Q'

The water body that lie in region 'R' doesn't have least area, therefore, 'B' doesn't lie in region 'R'.

The water body in region 'P' and region 'R' have same area, therefore, 'A' and 'D' lie in either of regions 'P' of 'R'

'D' lies in cold climatic condition, therefore, 'D' lie in region 'R' (because region 'R' is in cold climatic condition)

Therefore, 'A' lie in region 'P'

Since, the number of water bodies in regions 'Q' and 'R' is equal, therefore, 'B' lie in region 'Q' and hence 'E' will lie in region 'R'.

Therefore,

Also,

The number of water bodies in brackish category is least.

Hence, option 3 is correct.

According to the question,

(x + y) = 320.... (1)

And, (3x - y) = 0.15(x + y + 2y - x + 3x - y + x + 2y)

Or, 3x = 2y.... (2)

On solving equation (1) and (2), we get

x = 128 and y = 192

Therefore, number of article 'P' sold = x + y = 320

Number of article 'Q' sold = 2y - x = 256

Number of article 'R' sold = 3x - y = 192

Number of article 'S' sold = 128

Number of article 'T' sold = 2y = 384

Also,

Total selling price of article 'P' = (0.75 X 320 X 2 X 150) + (0.25 X 320 X 4 X 20) = INR 78400

Total selling price of article 'S' = (0.25 X 128 X 4 X 150) + (0.75 X 128 X 2 X 20) = INR 23040

Required ratio = 78400:23040 = 245:72

Hence, option 3 is correct.

According to the question,

(x + y) = 320.... (1)

And, (3x - y) = 0.15(x + y + 2y - x + 3x - y + x + 2y)

Or, 3x = 2y.... (2)

On solving equation (1) and (2), we get

x = 128 and y = 192

Therefore, number of article 'P' sold = x + y = 320

Number of article 'Q' sold = 2y - x = 256

Number of article 'R' sold = 3x - y = 192

Number of article 'S' sold = 128

Number of article 'T' sold = 2y = 384

Also,

Total selling price of article 'R' = 192 X 5 X 25 = INR 24000

Amount returned by him to the customers = (0.25 X 192 X 6 X 5) = INR 1440

Therefore, actual revenue earned by the seller = 24000 - 1440 = INR 22560

Hence, option 4 is correct.