CAT Table Based DI Sets Questions
Master CAT Table Based DI Sets Questions with practice questions and detailed solutions.
Instructions
The table given below shows the amount, in grams, of carbohydrate, protein, fat and all other nutrients, per 100 grams of nutrients in seven food grains. The first column shows the food grain category and the second column its codename. The table has some missing values.
The following additional facts are known.
1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the food grains are non-zero multiples of 5.
4. All the missing values of protein, fat and other nutrients amounts (in grams) for all the food grains are non-zero multiples of 4.
5. P1 contained double the amount of protein that M3 contains.
The table given below shows the amount, in grams, of carbohydrate, protein, fat and all other nutrients, per 100 grams of nutrients in seven food grains. The first column shows the food grain category and the second column its codename. The table has some missing values.
The following additional facts are known.
1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the food grains are non-zero multiples of 5.
4. All the missing values of protein, fat and other nutrients amounts (in grams) for all the food grains are non-zero multiples of 4.
5. P1 contained double the amount of protein that M3 contains.
Question 1.
How many grams of protein were there in 100 grams of nutrients in M2?
How many grams of protein were there in 100 grams of nutrients in M2?
Text Explanation:
The set's starting point is that the sum of each row must be 100
Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that the missing elements of the other three columns are multiples of 4.
Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo cereal.
We have the reference point of P1 at 66
Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95.
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a multiple of 4.
Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated.
Leaving only 80 grams of carb in C1 and 8 grams of protein.
Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are eliminated.
Taking 85 carbs would give 2 grams of protein, which can be eliminated.
80 and 70 would also not work for the same reason.
Leavin has only 75 grams of carbs, leaving 12 grams of protein.
Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.
Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0
The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24 grams of others in M3.
We can see that there were 12 grams of protein in M2.
Question 2.
How many foodgrains had a higher amount of carbohydrate per 100 grams of nutrients than M1?
How many foodgrains had a higher amount of carbohydrate per 100 grams of nutrients than M1?
Text Explanation:
The set's starting point is that the sum of each row must be 100
Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that the missing elements of the other three columns are multiples of 4.
Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo cereal.
We have the reference point of P1 at 66
Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95.
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a multiple of 4.
Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated.
Leaving only 80 grams of carb in C1 and 8 grams of protein.
Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are eliminated.
Taking 85 carbs would give 2 grams of protein, which can be eliminated.
80 and 70 would also not work for the same reason.
Leavin has only 75 grams of carbs, leaving 12 grams of protein.
Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.
Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0
The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24 grams of others in M3.
We can see that all other foodgrains except M3 had more carbohydrates than M1.
Therefore, 5 is the correct answer.
Question 3.
How many grams of other nutrients were there in 100 grams of nutrients in M3?
How many grams of other nutrients were there in 100 grams of nutrients in M3?
Text Explanation:
The set's starting point is that the sum of each row must be 100
Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that the missing elements of the other three columns are multiples of 4.
Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo cereal.
We have the reference point of P1 at 66
Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95.
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a multiple of 4.
Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated.
Leaving only 80 grams of carb in C1 and 8 grams of protein.
Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are eliminated.
Taking 85 carbs would give 2 grams of protein, which can be eliminated.
80 and 70 would also not work for the same reason.
Leavin has only 75 grams of carbs, leaving 12 grams of protein.
Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.
Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0
The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24 grams of others in M3.
There were 24 grams of other nutrients in M3.
Question 4.
What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?
What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?
Text Explanation:
The set's starting point is that the sum of each row must be 100
Clue 3 tells us that all the missing elements in the Carbs column are multiple of 5. Similarly, clue 4 tells us that the missing elements of the other three columns are multiples of 4.
Note that we look at clue 1, which says that the carbs in C1 and C2 must be more than any carbs in any pseudo cereal.
We have the reference point of P1 at 66
Trying to fill in for C1:
The possible values are 75, 80, 85, 90, 95
Since 12 grams is of other nutrients, we can eliminate 90 and 95.
If we take 85, 85+12 = 97, which would leave 3 grams of protein, but from clue 4, we know this has to be a multiple of 4.
Taking 75, 75+12 = 87 would leave 13 grams of protein, which, too, is eliminated.
Leaving only 80 grams of carb in C1 and 8 grams of protein.
Similar logic is to be applied for C2; we have 13 grams already present, so the values of 90 and 95 are eliminated.
Taking 85 carbs would give 2 grams of protein, which can be eliminated.
80 and 70 would also not work for the same reason.
Leavin has only 75 grams of carbs, leaving 12 grams of protein.
Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.
Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0
The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24 grams of others in M3.
The proteins in the foodgrains, when arranged in ascending order, are: 8, 8, 10, 12, 12,14, 16
Giving the median as 12.
Instructions
Five restaurants, coded R1, R2, R3, R4 and R5 gave integer ratings to five gig workers -
Ullas, Vasu, Waman, Xavier and Yusuf, on a scale of 1 to 5.
The mean of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively.
The summary statistics of these ratings for the five workers is given below.
* Range of ratings is defined as the difference between the maximum and minimum ratings awarded to a worker.
The following is partial information about ratings of 1 and 5 awarded by the restaurants to the workers.
(a) R1 awarded a rating of 5 to Waman, as did R2 to Xavier, R3 to Waman and Xavier, and R5 to Vasu.
(b) R1 awarded a rating of 1 to Ullas, as did R2 to Waman and Yusuf, and R3 to Yusuf.
Five restaurants, coded R1, R2, R3, R4 and R5 gave integer ratings to five gig workers -
Ullas, Vasu, Waman, Xavier and Yusuf, on a scale of 1 to 5.
The mean of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively.
The summary statistics of these ratings for the five workers is given below.
* Range of ratings is defined as the difference between the maximum and minimum ratings awarded to a worker.
The following is partial information about ratings of 1 and 5 awarded by the restaurants to the workers.
(a) R1 awarded a rating of 5 to Waman, as did R2 to Xavier, R3 to Waman and Xavier, and R5 to Vasu.
(b) R1 awarded a rating of 1 to Ullas, as did R2 to Waman and Yusuf, and R3 to Yusuf.
Question 5.
To how many workers did R2 give a rating of 4?
To how many workers did R2 give a rating of 4?
Text Explanation:
Given that the means of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively.
=> The sum of ratings given by R1, R2, R3 R4, R5 are 5*means = 17, 11, 19, 14, and 17 respectively.
Similarly the sum of ratings received by U, V, W, X and Y are 5*means = 11, 19, 17, 18, and 13 respectively.
Also capturing the absolute data given in the partial information (a) and (b) and representing as a table, we get:
Now,
Consider U
Given median = 2, mode = 2 and range = 3
=> His ratings should be of the form 1, a , 2, b, 4 => 1 + 2 + 4 + a + b = 11 => a + b = 4. For mode = 2 => a = b = 2
=> U's ratings are 1, 2, 2, 2, 4.
Consider V
Given median = 4, mode = 4 and range = 3
=> His ratings should be of the form 2, a, 4, b, 5 => 2 + 4 + 5 + a + b = 19 => a + b = 8 => For mode = 4 => a = b = 4
=> V's ratings are 2, 4, 4, 4, 5.
Consider W
Given median = 4, mode = 5 and range = 4
=> His ratings should be of the form 1, a, 4, 5, 5 => 1 + a + 4 + 5 + 5 = 17 => a = 2
=> W's ratings are 1, 2, 4, 5, 5.
Consider X
Given median = 4, mode = 5 and range = 4
=> His ratings should be of the form 1, a, 4, 5, 5 => a + 1 + 4 + 5 + 5 = 18 => a = 3
=> X's ratings are 1, 3, 4, 5, 5
Consider Y
Given median = 3, mode = 1 & 4, Range = 3
=> His ratings are 1, 1, 3, 4, 4.
Capturing this data in the table, we get:
Now, consider column R3 => The two missing entries should add up to 19 - 1 - 5 - 5 = 8, (only possibility is 4 + 4) => We can fill the row "U" and 4 in the row "V"
Now, consider column R2 => Missing entry should be 11 - 2 - 1 - 5 - 1 = 2
Consider column R1, the missing elements should add up to 17 - 5 - 4 - 1 = 7 (3 + 4 or 4 + 3) ----(1)
Consider R5, the missing elements should add up to 10 => 2 + 4 + 4 or 4 + 3 + 3 (not possible) as (1) requires a 3.
Now, we can fill column R1 as 3 + 4 and the remaining in column R4 and we can get the complete table
R2 gave ratings of 1, 1, 2, 2, 5 => He gave 4 to 0 workers => 0 is the answer.
Question 6.
How many individual ratings cannot be determined from the above information?
How many individual ratings cannot be determined from the above information?
Text Explanation:
Given that the means of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively.
=> The sum of ratings given by R1, R2, R3 R4, R5 are 5*means = 17, 11, 19, 14, and 17 respectively.
Similarly the sum of ratings received by U, V, W, X and Y are 5*means = 11, 19, 17, 18, and 13 respectively.
Also capturing the absolute data given in the partial information (a) and (b) and representing as a table, we get:
Now,
Consider U
Given median = 2, mode = 2 and range = 3
=> His ratings should be of the form 1, a , 2, b, 4 => 1 + 2 + 4 + a + b = 11 => a + b = 4. For mode = 2 => a = b = 2
=> U's ratings are 1, 2, 2, 2, 4.
Consider V
Given median = 4, mode = 4 and range = 3
=> His ratings should be of the form 2, a, 4, b, 5 => 2 + 4 + 5 + a + b = 19 => a + b = 8 => For mode = 4 => a = b = 4
=> V's ratings are 2, 4, 4, 4, 5.
Consider W
Given median = 4, mode = 5 and range = 4
=> His ratings should be of the form 1, a, 4, 5, 5 => 1 + a + 4 + 5 + 5 = 17 => a = 2
=> W's ratings are 1, 2, 4, 5, 5.
Consider X
Given median = 4, mode = 5 and range = 4
=> His ratings should be of the form 1, a, 4, 5, 5 => a + 1 + 4 + 5 + 5 = 18 => a = 3
=> X's ratings are 1, 3, 4, 5, 5
Consider Y
Given median = 3, mode = 1 & 4, Range = 3
=> His ratings are 1, 1, 3, 4, 4.
Capturing this data in the table, we get:
Now, consider column R3 => The two missing entries should add up to 19 - 1 - 5 - 5 = 8, (only possibility is 4 + 4) => We can fill the row "U" and 4 in the row "V"
Now, consider column R2 => Missing entry should be 11 - 2 - 1 - 5 - 1 = 2
Consider column R1, the missing elements should add up to 17 - 5 - 4 - 1 = 7 (3 + 4 or 4 + 3) ----(1)
Consider R5, the missing elements should add up to 10 => 2 + 4 + 4 or 4 + 3 + 3 (not possible) as (1) requires a 3.
Now, we can fill column R1 as 3 + 4 and the remaining in column R4 and we can get the complete table
=> All ratings can be determined uniquely => 0.
Instructions
In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months - January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known.
1. In every month, both online and offline registration numbers were multiples of 10.
2. In January, the number of offline registrations was twice that of online registrations.
3. In April, the number of online registrations was twice that of offline registrations.
4. The number of online registrations in March was the same as the number of offline registrations in February.
5. The number of online registrations was the largest in May.
In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months - January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known.
1. In every month, both online and offline registration numbers were multiples of 10.
2. In January, the number of offline registrations was twice that of online registrations.
3. In April, the number of online registrations was twice that of offline registrations.
4. The number of online registrations in March was the same as the number of offline registrations in February.
5. The number of online registrations was the largest in May.
Question 7.
What was the total number of registrations in April?
What was the total number of registrations in April?
Text Explanation:
Given that in every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was twice that of online registrations.
=> If x is number of online registrations => 2x is the number of offline registrations => 3x is the total number of registrations.
According to the data given in the table => 3x should lie between the minimum and maximum total number of registrations. => x = 40 (as x should also be a multiple of 10)
=> In Jan => (40,80) are the online and offline registrations respectively.
Similarly from (3) => In Apr (80,40) are the online and offline registrations respectively.
From-5, the number of online registrations is highest in may => In may there are 100 online registrations. The lowest possible number of offline registrations is 30 and maximum possible total registrations is 130 => In May (100,30) are the online and offline registrations respectively.
Let us assume, 'x' to be the number of offline registrations in May = number of online registrations in March.
Let us capture all this data in a table:
From the table given in the question, 50 is the median for Offline data
=> x should lie between 50 and 80 (included)
For 80 to be the median for the online data => y lie between 80 and 100 (included).
Now, consider Feb => Minimum value of y + x = 80 + 50 = 130 (which is the maximum value possible of the total possible registrations)
=> x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March => 50 + z = 110 => z = 60.
Now, filling the complete table we get,
The total number of registrations in April is 120.
Question 8.
What was the number of online registrations in January?
What was the number of online registrations in January?
Text Explanation:
Given that in every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was twice that of online registrations.
=> If x is number of online registrations => 2x is the number of offline registrations => 3x is the total number of registrations.
According to the data given in the table => 3x should lie between the minimum and maximum total number of registrations. => x = 40 (as x should also be a multiple of 10)
=> In Jan => (40,80) are the online and offline registrations respectively.
Similarly from (3) => In Apr (80,40) are the online and offline registrations respectively.
From-5, the number of online registrations is highest in may => In may there are 100 online registrations. The lowest possible number of offline registrations is 30 and maximum possible total registrations is 130 => In May (100,30) are the online and offline registrations respectively.
Let us assume, 'x' to be the number of offline registrations in May = number of online registrations in March.
Let us capture all this data in a table:
From the table given in the question, 50 is the median for Offline data
=> x should lie between 50 and 80 (included)
For 80 to be the median for the online data => y lie between 80 and 100 (included).
Now, consider Feb => Minimum value of y + x = 80 + 50 = 130 (which is the maximum value possible of the total possible registrations)
=> x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March => 50 + z = 110 => z = 60.
Now, filling the complete table we get,
The number of online registrations in Jan is 40.
Question 9.
Which of the following statements can be true?
I. The number of offline registrations was the smallest in May.
II. The total number of registrations was the smallest in February.
Which of the following statements can be true?
I. The number of offline registrations was the smallest in May.
II. The total number of registrations was the smallest in February.
Text Explanation:
Given that in every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was twice that of online registrations.
=> If x is number of online registrations => 2x is the number of offline registrations => 3x is the total number of registrations.
According to the data given in the table => 3x should lie between the minimum and maximum total number of registrations. => x = 40 (as x should also be a multiple of 10)
=> In Jan => (40,80) are the online and offline registrations respectively.
Similarly from (3) => In Apr (80,40) are the online and offline registrations respectively.
From-5, the number of online registrations is highest in may => In may there are 100 online registrations. The lowest possible number of offline registrations is 30 and maximum possible total registrations is 130 => In May (100,30) are the online and offline registrations respectively.
Let us assume, 'x' to be the number of offline registrations in May = number of online registrations in March.
Let us capture all this data in a table:
From the table given in the question, 50 is the median for Offline data
=> x should lie between 50 and 80 (included)
For 80 to be the median for the online data => y lie between 80 and 100 (included).
Now, consider Feb => Minimum value of y + x = 80 + 50 = 130 (which is the maximum value possible of the total possible registrations)
=> x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March => 50 + z = 110 => z = 60.
Now, filling the complete table we get,
1) In May, there are 30 offline registrations (smallest) => True
2) In Mar, we have smallest number of total registrations => False.
Question 10.
What best can be concluded about the number of offline registrations in February?
What best can be concluded about the number of offline registrations in February?
Text Explanation:
Given that in every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was twice that of online registrations.
=> If x is number of online registrations => 2x is the number of offline registrations => 3x is the total number of registrations.
According to the data given in the table => 3x should lie between the minimum and maximum total number of registrations. => x = 40 (as x should also be a multiple of 10)
=> In Jan => (40,80) are the online and offline registrations respectively.
Similarly from (3) => In Apr (80,40) are the online and offline registrations respectively.
From-5, the number of online registrations is highest in may => In may there are 100 online registrations. The lowest possible number of offline registrations is 30 and maximum possible total registrations is 130 => In May (100,30) are the online and offline registrations respectively.
Let us assume, 'x' to be the number of offline registrations in May = number of online registrations in March.
Let us capture all this data in a table:
From the table given in the question, 50 is the median for Offline data
=> x should lie between 50 and 80 (included)
For 80 to be the median for the online data => y lie between 80 and 100 (included).
Now, consider Feb => Minimum value of y + x = 80 + 50 = 130 (which is the maximum value possible of the total possible registrations)
=> x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March => 50 + z = 110 => z = 60.
Now, filling the complete table we get,
The number of offline registrations in Feb is 50.
Question 11.
Which pair of months definitely had the same total number of registrations?
I. January and April
II. February and May
Which pair of months definitely had the same total number of registrations?
I. January and April
II. February and May
Text Explanation:
Given that in every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was twice that of online registrations.
=> If x is number of online registrations => 2x is the number of offline registrations => 3x is the total number of registrations.
According to the data given in the table => 3x should lie between the minimum and maximum total number of registrations. => x = 40 (as x should also be a multiple of 10)
=> In Jan => (40,80) are the online and offline registrations respectively.
Similarly from (3) => In Apr (80,40) are the online and offline registrations respectively.
From-5, the number of online registrations is highest in may => In may there are 100 online registrations. The lowest possible number of offline registrations is 30 and maximum possible total registrations is 130 => In May (100,30) are the online and offline registrations respectively.
Let us assume, 'x' to be the number of offline registrations in May = number of online registrations in March.
Let us capture all this data in a table:
From the table given in the question, 50 is the median for Offline data
=> x should lie between 50 and 80 (included)
For 80 to be the median for the online data => y lie between 80 and 100 (included).
Now, consider Feb => Minimum value of y + x = 80 + 50 = 130 (which is the maximum value possible of the total possible registrations)
=> x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March => 50 + z = 110 => z = 60.
Now, filling the complete table we get,
Total registrations in Jan = Apr = 120 and Feb = May = 130.
Instructions
There are only three female students - Amala, Koli and Rini - and only three male students - Biman, Mathew and Shyamal - in a course. The course has two evaluation components, a project and a test. The aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.
The projects are done in groups of two, with each group consisting of a female and a male student. Both the group members obtain the same score in the project.
The following additional facts are known about the scores in the project and the test.
1. The minimum, maximum and the average of both project and test scores were identical - 40, 80 and 60, respectively.
2. The test scores of the students were all multiples of 10; four of them were distinct and the remaining two were equal to the average test scores.
3. Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Yet Amala had the highest aggregate score.
4. Shyamal scored the second highest in the test. He scored two more than Koli, but two less than Amala in the aggregate.
5. Biman scored the second lowest in the test and the lowest in the aggregate.
6. Mathew scored more than Rini in the project, but less than her in the test.
There are only three female students - Amala, Koli and Rini - and only three male students - Biman, Mathew and Shyamal - in a course. The course has two evaluation components, a project and a test. The aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.
The projects are done in groups of two, with each group consisting of a female and a male student. Both the group members obtain the same score in the project.
The following additional facts are known about the scores in the project and the test.
1. The minimum, maximum and the average of both project and test scores were identical - 40, 80 and 60, respectively.
2. The test scores of the students were all multiples of 10; four of them were distinct and the remaining two were equal to the average test scores.
3. Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Yet Amala had the highest aggregate score.
4. Shyamal scored the second highest in the test. He scored two more than Koli, but two less than Amala in the aggregate.
5. Biman scored the second lowest in the test and the lowest in the aggregate.
6. Mathew scored more than Rini in the project, but less than her in the test.
Question 12.
What was Rini’s score in the project?
What was Rini’s score in the project?
Text Explanation:
It is given that there are only three female students - Amala, Koli, and Rini - and only three male students - Biman, Mathew, and Shyamal - in a course.
It is also known that the aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.
Let the project score component be x, which implies the test score component will be (1-x). The projects are done in groups of two, with each group consisting of a female and a male student, which implies there are three groups for the project. It is also known that both the group members obtain the same score in the project. The score obtained in the project is 40, 60, and 80, respectively.
Therefore, we can say that each female student will consist of a different group, and no two male students or female students will be in the same group.
For the test scores, there are six scores given for six students among which four are distinct and the remaining two are average scores, which is 60. It is also known that the maximum score possible is 80, and the minimum score is 40.
Hence, the distinct scores are 80, 70, 50, and 40 (since all the test scores are multiple of 10), and the remaining two scores are 60, and 60, respectively.
From point 3, we know that Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Hence, we can say the score obtained by Amala in the project is 80, and the score obtained by Koli is 40, which implies the score obtained by Rini in the project is 60. Now, Koli scored 20 more than Amala in the test, which implies the score obtained by Koli can be either 80, 70, or 60.
The score obtained by them is given below:
It is known that Amala had the highest aggregate score, and Shyamal scored the second highest on the test. He scored two more than Koli, but two less than Amala in the aggregate.
Hence, the score obtained by Shyamal in the test is 70, which implies Koli can't score 70 in the test => Amala can't score 50 in the test.
It is given that Shyamal scored two more than Koli, but two less than Amala in the aggregate. Hence, the aggregate score of Amala is 4 more than Koli. It is also known that Amala had the highest aggregate score.
Case 1: The test score of Amala is 40
Therefore, 40(1-x)+80x = 60(1-x)+40x+4
=> 60x = 24
=> x = 0.4
Hence, the aggregate score obtained by Amala is 40(1-0.4)+80*4 = 56
The minimum aggregate score of Shyamal is 70(1-0.4)+ 40*0.4 = 58, which is greater than Amala.
Hence, Case 1 is not possible.
Hence, the table is given below:
Therefore, 60(1-x)+80x = 80(1-x)+40x+4
=> 60+20x = 84-40x
=> 60x = 24 => x = 0.4
Hence, the aggregate score of Amala is 60(1-0.4)+80*0.4 = 68, which implies the aggregate score of Shyamal is (68-2) = 66
Hence, the score obtained by Shyamal in Project is {66-70*(0.6)}/0.4 = 60.
It is also known that Biman scored second lowest in the test, which implies the score of Biman in the test is 50, and he scored the lowest in the aggregate. It is also known that Mathew scored more than Rini in the project, but less than her in the test. Hence, Mathew scored 80 in the project (since Rini scored 60 in the project), and Biman scored 40 in the project.
Similarly, Rini Scored more than Mathew on the test, which implies the score obtained by Rini is 60, and the score obtained by Mathew is 40 in the test.
Hence, the final table will look like this:
Hence, the score obtained by Rini in the project is 60
Question 13.
What was the weight of the test component?
What was the weight of the test component?
Text Explanation:
It is given that there are only three female students - Amala, Koli, and Rini - and only three male students - Biman, Mathew, and Shyamal - in a course.
It is also known that the aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.
Let the project score component be x, which implies the test score component will be (1-x). The projects are done in groups of two, with each group consisting of a female and a male student, which implies there are three groups for the project. It is also known that both the group members obtain the same score in the project. The score obtained in the project is 40, 60, and 80, respectively.
Therefore, we can say that each female student will consist of a different group, and no two male students or female students will be in the same group.
For the test scores, there are six scores given for six students among which four are distinct and the remaining two are average scores, which is 60. It is also known that the maximum score possible is 80, and the minimum score is 40.
Hence, the distinct scores are 80, 70, 50, and 40 (since all the test scores are multiple of 10), and the remaining two scores are 60, and 60, respectively.
From point 3, we know that Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Hence, we can say the score obtained by Amala in the project is 80, and the score obtained by Koli is 40, which implies the score obtained by Rini in the project is 60. Now, Koli scored 20 more than Amala in the test, which implies the score obtained by Koli can be either 80, 70, or 60.
The score obtained by them is given below:
It is known that Amala had the highest aggregate score, and Shyamal scored the second highest on the test. He scored two more than Koli, but two less than Amala in the aggregate.
Hence, the score obtained by Shyamal in the test is 70, which implies Koli can't score 70 in the test => Amala can't score 50 in the test.
It is given that Shyamal scored two more than Koli, but two less than Amala in the aggregate. Hence, the aggregate score of Amala is 4 more than Koli. It is also known that Amala had the highest aggregate score.
Case 1: The test score of Amala is 40
Therefore, 40(1-x)+80x = 60(1-x)+40x+4
=> 60x = 24
=> x = 0.4
Hence, the aggregate score obtained by Amala is 40(1-0.4)+80*4 = 56
The minimum aggregate score of Shyamal is 70(1-0.4)+ 40*0.4 = 58, which is greater than Amala.
Hence, Case 1 is not possible.
Hence, the table is given below:
Therefore, 60(1-x)+80x = 80(1-x)+40x+4
=> 60+20x = 84-40x
=> 60x = 24 => x = 0.4
Hence, the aggregate score of Amala is 60(1-0.4)+80*0.4 = 68, which implies the aggregate score of Shyamal is (68-2) = 66
Hence, the score obtained by Shyamal in Project is {66-70*(0.6)}/0.4 = 60.
It is also known that Biman scored second lowest in the test, which implies the score of Biman in the test is 50, and he scored the lowest in the aggregate. It is also known that Mathew scored more than Rini in the project, but less than her in the test. Hence, Mathew scored 80 in the project (since Rini scored 60 in the project), and Biman scored 40 in the project.
Similarly, Rini Scored more than Mathew on the test, which implies the score obtained by Rini is 60, and the score obtained by Mathew is 40 in the test.
Hence, the final table will look like this:
Hence, the weight of the test component is 0.6
The correct option is A
Question 14.
What was the maximum aggregate score obtained by the students?
What was the maximum aggregate score obtained by the students?
Text Explanation:
It is given that there are only three female students - Amala, Koli, and Rini - and only three male students - Biman, Mathew, and Shyamal - in a course.
It is also known that the aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.
Let the project score component be x, which implies the test score component will be (1-x). The projects are done in groups of two, with each group consisting of a female and a male student, which implies there are three groups for the project. It is also known that both the group members obtain the same score in the project. The score obtained in the project is 40, 60, and 80, respectively.
Therefore, we can say that each female student will consist of a different group, and no two male students or female students will be in the same group.
For the test scores, there are six scores given for six students among which four are distinct and the remaining two are average scores, which is 60. It is also known that the maximum score possible is 80, and the minimum score is 40.
Hence, the distinct scores are 80, 70, 50, and 40 (since all the test scores are multiple of 10), and the remaining two scores are 60, and 60, respectively.
From point 3, we know that Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Hence, we can say the score obtained by Amala in the project is 80, and the score obtained by Koli is 40, which implies the score obtained by Rini in the project is 60. Now, Koli scored 20 more than Amala in the test, which implies the score obtained by Koli can be either 80, 70, or 60.
The score obtained by them is given below:
It is known that Amala had the highest aggregate score, and Shyamal scored the second highest on the test. He scored two more than Koli, but two less than Amala in the aggregate.
Hence, the score obtained by Shyamal in the test is 70, which implies Koli can't score 70 in the test => Amala can't score 50 in the test.
It is given that Shyamal scored two more than Koli, but two less than Amala in the aggregate. Hence, the aggregate score of Amala is 4 more than Koli. It is also known that Amala had the highest aggregate score.
Case 1: The test score of Amala is 40
Therefore, 40(1-x)+80x = 60(1-x)+40x+4
=> 60x = 24
=> x = 0.4
Hence, the aggregate score obtained by Amala is 40(1-0.4)+80*4 = 56
The minimum aggregate score of Shyamal is 70(1-0.4)+ 40*0.4 = 58, which is greater than Amala.
Hence, Case 1 is not possible.
Hence, the table is given below:
Therefore, 60(1-x)+80x = 80(1-x)+40x+4
=> 60+20x = 84-40x
=> 60x = 24 => x = 0.4
Hence, the aggregate score of Amala is 60(1-0.4)+80*0.4 = 68, which implies the aggregate score of Shyamal is (68-2) = 66
Hence, the score obtained by Shyamal in Project is {66-70*(0.6)}/0.4 = 60.
It is also known that Biman scored second lowest in the test, which implies the score of Biman in the test is 50, and he scored the lowest in the aggregate. It is also known that Mathew scored more than Rini in the project, but less than her in the test. Hence, Mathew scored 80 in the project (since Rini scored 60 in the project), and Biman scored 40 in the project.
Similarly, Rini Scored more than Mathew on the test, which implies the score obtained by Rini is 60, and the score obtained by Mathew is 40 in the test.
Hence, the final table will look like this:
Hence, the maximum aggregate score obtained is 68. The correct option is A
Question 15.
What was Mathew’s score in the test?
What was Mathew’s score in the test?
Text Explanation:
It is given that there are only three female students - Amala, Koli, and Rini - and only three male students - Biman, Mathew, and Shyamal - in a course.
It is also known that the aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.
Let the project score component be x, which implies the test score component will be (1-x). The projects are done in groups of two, with each group consisting of a female and a male student, which implies there are three groups for the project. It is also known that both the group members obtain the same score in the project. The score obtained in the project is 40, 60, and 80, respectively.
Therefore, we can say that each female student will consist of a different group, and no two male students or female students will be in the same group.
For the test scores, there are six scores given for six students among which four are distinct and the remaining two are average scores, which is 60. It is also known that the maximum score possible is 80, and the minimum score is 40.
Hence, the distinct scores are 80, 70, 50, and 40 (since all the test scores are multiple of 10), and the remaining two scores are 60, and 60, respectively.
From point 3, we know that Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Hence, we can say the score obtained by Amala in the project is 80, and the score obtained by Koli is 40, which implies the score obtained by Rini in the project is 60. Now, Koli scored 20 more than Amala in the test, which implies the score obtained by Koli can be either 80, 70, or 60.
The score obtained by them is given below:
It is known that Amala had the highest aggregate score, and Shyamal scored the second highest on the test. He scored two more than Koli, but two less than Amala in the aggregate.
Hence, the score obtained by Shyamal in the test is 70, which implies Koli can't score 70 in the test => Amala can't score 50 in the test.
It is given that Shyamal scored two more than Koli, but two less than Amala in the aggregate. Hence, the aggregate score of Amala is 4 more than Koli. It is also known that Amala had the highest aggregate score.
Case 1: The test score of Amala is 40
Therefore, 40(1-x)+80x = 60(1-x)+40x+4
=> 60x = 24
=> x = 0.4
Hence, the aggregate score obtained by Amala is 40(1-0.4)+80*4 = 56
The minimum aggregate score of Shyamal is 70(1-0.4)+ 40*0.4 = 58, which is greater than Amala.
Hence, Case 1 is not possible.
Hence, the table is given below:
Therefore, 60(1-x)+80x = 80(1-x)+40x+4
=> 60+20x = 84-40x
=> 60x = 24 => x = 0.4
Hence, the aggregate score of Amala is 60(1-0.4)+80*0.4 = 68, which implies the aggregate score of Shyamal is (68-2) = 66
Hence, the score obtained by Shyamal in Project is {66-70*(0.6)}/0.4 = 60.
It is also known that Biman scored second lowest in the test, which implies the score of Biman in the test is 50, and he scored the lowest in the aggregate. It is also known that Mathew scored more than Rini in the project, but less than her in the test. Hence, Mathew scored 80 in the project (since Rini scored 60 in the project), and Biman scored 40 in the project.
Similarly, Rini Scored more than Mathew on the test, which implies the score obtained by Rini is 60, and the score obtained by Mathew is 40 in the test.
Hence, the final table will look like this:
Hence, the score obtained by Mathew in the test is 40
Question 16.
Which of the following pairs of students were part of the same project team?
i) Amala and Biman
ii) Koli and Mathew
Which of the following pairs of students were part of the same project team?
i) Amala and Biman
ii) Koli and Mathew
Text Explanation:
It is given that there are only three female students - Amala, Koli, and Rini - and only three male students - Biman, Mathew, and Shyamal - in a course.
It is also known that the aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.
Let the project score component be x, which implies the test score component will be (1-x). The projects are done in groups of two, with each group consisting of a female and a male student, which implies there are three groups for the project. It is also known that both the group members obtain the same score in the project. The score obtained in the project is 40, 60, and 80, respectively.
Therefore, we can say that each female student will consist of a different group, and no two male students or female students will be in the same group.
For the test scores, there are six scores given for six students among which four are distinct and the remaining two are average scores, which is 60. It is also known that the maximum score possible is 80, and the minimum score is 40.
Hence, the distinct scores are 80, 70, 50, and 40 (since all the test scores are multiple of 10), and the remaining two scores are 60, and 60, respectively.
From point 3, we know that Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Hence, we can say the score obtained by Amala in the project is 80, and the score obtained by Koli is 40, which implies the score obtained by Rini in the project is 60. Now, Koli scored 20 more than Amala in the test, which implies the score obtained by Koli can be either 80, 70, or 60.
The score obtained by them is given below:
It is known that Amala had the highest aggregate score, and Shyamal scored the second highest on the test. He scored two more than Koli, but two less than Amala in the aggregate.
Hence, the score obtained by Shyamal in the test is 70, which implies Koli can't score 70 in the test => Amala can't score 50 in the test.
It is given that Shyamal scored two more than Koli, but two less than Amala in the aggregate. Hence, the aggregate score of Amala is 4 more than Koli. It is also known that Amala had the highest aggregate score.
Case 1: The test score of Amala is 40
Therefore, 40(1-x)+80x = 60(1-x)+40x+4
=> 60x = 24
=> x = 0.4
Hence, the aggregate score obtained by Amala is 40(1-0.4)+80*4 = 56
The minimum aggregate score of Shyamal is 70(1-0.4)+ 40*0.4 = 58, which is greater than Amala.
Hence, Case 1 is not possible.
Hence, the table is given below:
Therefore, 60(1-x)+80x = 80(1-x)+40x+4
=> 60+20x = 84-40x
=> 60x = 24 => x = 0.4
Hence, the aggregate score of Amala is 60(1-0.4)+80*0.4 = 68, which implies the aggregate score of Shyamal is (68-2) = 66
Hence, the score obtained by Shyamal in Project is {66-70*(0.6)}/0.4 = 60.
It is also known that Biman scored second lowest in the test, which implies the score of Biman in the test is 50, and he scored the lowest in the aggregate. It is also known that Mathew scored more than Rini in the project, but less than her in the test. Hence, Mathew scored 80 in the project (since Rini scored 60 in the project), and Biman scored 40 in the project.
Similarly, Rini Scored more than Mathew on the test, which implies the score obtained by Rini is 60, and the score obtained by Mathew is 40 in the test.
Hence, the final table will look like this:
From the table, we can see that (Amala, Mathew), (Koli, Biman), and (Shyama, Rini) are the three groups for the project.
Hence, the correct option is C
Instructions
Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing.
It is known that the duration of calls from Faruq to Eshan was 200 minutes.
Also, there were no calls from:
i. Bijay to Eshan,
ii. Chetan to Anu and Chetan to Deepak,
iii. Deepak to Bijay and Deepak to Faruq,
iv. Eshan to Chetan and Eshan to Deepak.
Anu, Bijay, Chetan, Deepak, Eshan, and Faruq are six friends. Each of them uses a mobile number from exactly one of the two mobile operators - Xitel and Yocel. During the last month, the six friends made several calls to each other. Each call was made by one of these six friends to another. The table below summarizes the number of minutes of calls that each of the six made to (outgoing minutes) and received from (incoming minutes) these friends, grouped by the operators. Some of the entries are missing.
It is known that the duration of calls from Faruq to Eshan was 200 minutes.
Also, there were no calls from:
i. Bijay to Eshan,
ii. Chetan to Anu and Chetan to Deepak,
iii. Deepak to Bijay and Deepak to Faruq,
iv. Eshan to Chetan and Eshan to Deepak.
Question 17.
What was the duration of calls (in minutes) from Bijay to Anu?
What was the duration of calls (in minutes) from Bijay to Anu?
Text Explanation:
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The duration of calls from Bijay to Ajay is 50 minutes, which is the outgoing minutes to operator Xitel of Bijay.Hence, the correct answer is 50.
Question 18.
What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?
What was the total duration of calls (in minutes) made by Anu to friends having mobile numbers from Operator Yocel?
Text Explanation:
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The total duration of calls made by Anu to friends having mobile numbers from Operator Yocel is Anu's outgoing minutes to Operator Yocel, which is 525 minutes.Hence, the correct answer is 525.
Question 19.
What was the total duration of calls (in minutes) made by Faruq to friends having mobile numbers from Operator Yocel?
What was the total duration of calls (in minutes) made by Faruq to friends having mobile numbers from Operator Yocel?
Text Explanation:
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
The total duration of calls made by Faruq to friends having mobile numbers from Operator Yocel is the outgoing minutes from Faruq to Operator Yocel, which is 350 minutes.Hence, the correct answer is 350.
Question 20.
What was the duration of calls (in minutes) from Deepak to Chetan?
What was the duration of calls (in minutes) from Deepak to Chetan?
Text Explanation:
We are told that Anu and Bijay use the Xitel operator, and Chetan, Deepak, Esthan and Faruq use the Yocel operator.The outgoing call from Anu to the Xitel operator has to be the incoming call from Xitel to Bijay because there are only two friends using Xitel, and if one of the Xitel users is calling another Xitel user, then those two have to be Anu and Bijay.
The outgoing to Xitel by Anu = 100
The incoming from Xitel for Bijay = c = The outgoing to Xitel by Anu = 100
So, the value of c in the above table is 100.
The outgoing to Xitel by Bijay = b
The incoming from Xitel for Anu = 50 = The outgoing to Xitel by Bijay = b
So, the value of b in the above table is 50.
The sum of Outgoing from Anu and Bijay to operator Yocel must be equal to the sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq.
The sum of Outgoing from Anu and Bijay to operator Yocel = a + 200
The sum of Incoming minutes from Xitel for Chetan, Deepak, Esthan and Faruq = 250 + 275 + 100 + 100 = 725
Equating them, we get,
a + 200 = 725
a = 525
So, the value of a in the above table is 525.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel must be equal to the sum of Incoming minutes from Yocel for Anu and Bijay.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Xitel = 50 + 100 + d + 0 = 150 + d
The sum of Incoming minutes from Yocel for Anu and Bijay = 225 + 125 = 350
Equating them, we get,
150 + d = 350
d = 200
So, the value of d in the above table is 200.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel must be equal to the sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq.
The sum of outgoing minutes from Chetan, Deepak, Esthan, and Faruq to operator Yocel = 175 + 150 + 100 + e = 425 + e
The sum of Incoming minutes from Yocel for Chetan, Deepak, Esthan, and Faruq = 150 + 100 + 375 + 150 = 775
Equating them, we get,
425 + e = 775
e = 350
So, the value of e in the above table is 350.
The final table is,
Label the unknown value we need as$x=\text{minutes from Deepak to Chetan}$
Use the operator groups from the table:
Xitel users: Anu, Bijay.
Yocel users: Chetan, Deepak, Eshan, Faruq.
Chetan’s incoming from Yocel = 150 minutes.
Yocel$\to$Chetan can come only from Deepak, Eshan, or Faruq. But the problem states Eshan $\to$ Chetan = 0, so
$(\text{Deepak}\to\text{Chetan}) + (\text{Faruq}\to\text{Chetan}) = 150$
Hence
$x + (\text{Faruq}\to\text{Chetan}) = 150 \qquad(1)$
Faruq’s outgoing to Yocel = 350 minutes (table). Those 350 minutes are split among Yocel recipients Chetan, Deepak and Eshan. We are told Faruq$\to$Eshan = 200 (given). So
$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) + 200 = 350,$
hence
$(\text{Faruq}\to\text{Chetan}) + (\text{Faruq}\to\text{Deepak}) = 150 \qquad(2)$
Consider Deepak’s incoming from Yocel (table) = 100 minutes. The ones in Yocel that can call Deepak are Chetan, Eshan, and Faruq. But we are told Chetan $\to$ Deepak = 0 and Eshan $\to$ Deepak = 0, so the only Yocel caller to Deepak is Faruq. Therefore
$(\text{Faruq}\to\text{Deepak}) = 100 \qquad(3)$
Substitute (3) into (2):
$(\text{Faruq}\to\text{Chetan}) + 100 = 150 \quad\Rightarrow\quad (\text{Faruq}\to\text{Chetan}) = 50.$
Now use (1):
$x + (\text{Faruq}\to\text{Chetan}) = 150$.
With $\text{Faruq}\to\text{Chetan} = 50$, we get,
$x + 50 = 150$
$x = 100$
Thus Deepak $\to$ Chetan = 100 minutes.
Hence, the correct answer is option A.
Free CAT DILR Questions
