CAT 2024 | DILR Set 1
Direction Q.1 to 5:
The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.
The following additional information is known.
- The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
- The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is12 km.
1. Which of the following statements is correct?
- The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.
- The cash requirement of the ATM placed at the (R-C, V2) intersection cannot be uniquely determined.
- The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 8 Lakhs.
- There is no ATM placed at the (R-C, V2) intersection.
By keeping all conditions in mind, We can see There should be an ATM placed at intersection of V3 and R-C.
Also, By condition (2), we can judge there is only one possibility to place ATM with second highest cash requirement at intersection of V2 and R-B.
Because this is the only place which is 12 km away.
Also, we cannot use distinct integers ranging from 7 to 15.
So, By hit and trial, there are only two cases to arrange all 6 ATM’s in following manner:
Option (A) is correct as The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakh.
2. How many ATMs have cash requirements of Rs. 10 Lakhs or more?
By keeping all conditions in mind, We can see There should be an ATM placed at intersection of V3 and R-C.
Also, By condition (2), we can judge there is only one possibility to place ATM with second highest cash requirement at intersection of V2 and R-B.
Because this is the only place which is 12 km away.
Also, we cannot use distinct integers ranging from 7 to 15.
So, By hit and trial, there are only two cases to arrange all 6 ATM’s in following manner:
In Case 1, ATMs with cash requirement of 15 L, 12 L, 11L are more than 10 L.
In case (2), ATMs with cash requirement of 15 L, 12 L, 11L are more than 10 L.
Hence there are 3 ATMs with cash requirements of Rs. 10 Lakh or more.
3. Which of the following two statements is/are DEFINITELY true? Statement A: Each of R-A, R-B, and R-C has two ATMs. Statement B: Each of V1, V2, and V3 has two ATMs.
- Only Statement B
- Neither Statement A nor Statement B
- Both Statement A and Statement B
- Only Statement A
By keeping all conditions in mind, We can see There should be an ATM placed at intersection of V3 and R-C.
Also, By condition (2), we can judge there is only one possibility to place ATM with second highest cash requirement at intersection of V2 and R-B.
Because this is the only place which is 12 km away.
Also, we cannot use distinct integers ranging from 7 to 15.
So, By hit and trial, there are only two cases to arrange all 6 ATM’s in following manner:
As we can see in both cases only statement (1) is right but statement (2) is wrong in case (1). Hence option (D) is the answer.
4. What is the number of ATMs whose locations and cash requirements can both be uniquely determined?
By keeping all conditions in mind, We can see There should be an ATM placed at intersection of V3 and R-C.
Also, By condition (2), we can judge there is only one possibility to place ATM with second highest cash requirement at intersection of V2 and R-B.
Because this is the only place which is 12 km away.
Also, we cannot use distinct integers ranging from 7 to 15.
So, By hit and trial, there are only two cases to arrange all 6 ATM’s in following manner:
We can determine only 3 ATMs’ locations uniquely and they are (V2, R-B), (V2, R-C) and (V3, R-C).
5. What best can be said about the road distance (in km) between the ATMs having the second highest and the second lowest cash requirements?
- 4 km
- 5 km
- Either 4 km or 7 km
- 7 km
By keeping all conditions in mind, We can see There should be an ATM placed at intersection of V3 and R-C.
Also, By condition (2), we can judge there is only one possibility to place ATM with second highest cash requirement at intersection of V2 and R-B.
Because this is the only place which is 12 km away.
Also, we cannot use distinct integers ranging from 7 to 15.
So, By hit and trial, there are only two cases to arrange all 6 ATM’s in following manner:
In case (1), ATMs (V2, R-B) and (V3, R-B) are the ATMs with second highest and second lowest requirements.
Hence distance between them is 7 KM.
In case (2), ATMs ((V1, R-B) and (V2, R-B) are the ATMs with second lowest and second highest requirements.
Hence distance between them is 4 KM.
So ans is either D or 7 KM.
CAT 2024 | DILR Set 2
Direction Q.6 to 9:
Over the top (OTT) subscribers of a platform are segregated into three categories:
i) Kid, ii) Elder, and iii) Others.
Some of the subscribers used one app and the others used multiple apps to access the platform. The figure below shows the percentage of the total number of subscribers in 2023 and 2024 who belong to the ‘Kid’ and ‘Elder’ categories.
The following additional facts are known about the numbers of subscribers.
- The total number of subscribers increased by 10% from 2023 to 2024.
- In 2024, 1/2 of the subscribers from the ‘Kid’ category and 2/3 of the subscribers from the ‘Elder’ category subscribers use one app.
- In 2023, the number of subscribers from the ‘Kid’ category who used multiple apps was the same as the number of subscribers from the ‘Elder’ category who used one app.
- 10,000 subscribers from the ‘Kid’ category used one app and 15,000 subscribers from the ‘Elder’ category used multiple apps in 2023.
6. How many subscribers belonged to the ‘Others’ category in 2024?
- 45000
- 55000
- Cannot be determined
- 65000
Let the total number of subscribers in 2023 = 100x
From point 1, the total number of subscribers in 2024 = 100x + 10% of 100x = 110x
In 2023,
Total number of subscribers from Kid category = 15% of 100x = 15x
From point 3 and 4, in 2023
Number of subscribers from Kid category using one app = 10000.
So, number of subscribers from Kid category using multiple apps = 15x – 10000 = number of subscribers from Elder category using one app.
Also, number of subscribers from Elder category using multiple apps = 15000
So, total number of subscribers from Elder category = 15x – 10000 + 15000 = 20% of 100x = 20x
Solving, x = 1000
So, the total number of subscribers in 2023 = 100000 and in 2024 = 110000
The rest of the information can be gathered as follows-
Number of subscribers belonged to the Others category in 2024 = 55000
7. What percentage of subscribers in the ‘Kid’ category used multiple apps in 2023?
- 50.00%
- 5.00%
- 33.33%
- 25.50%
Let the total number of subscribers in 2023 = 100x
From point 1, the total number of subscribers in 2024 = 100x + 10% of 100x = 110x
In 2023,
Total number of subscribers from Kid category = 15% of 100x = 15x
From point 3 and 4, in 2023
Number of subscribers from Kid category using one app = 10000.
So, number of subscribers from Kid category using multiple apps = 15x – 10000 = number of subscribers from Elder category using one app.
Also, number of subscribers from Elder category using multiple apps = 15000
So, total number of subscribers from Elder category = 15x – 10000 + 15000 = 20% of 100x = 20x
Solving, x = 1000
So, the total number of subscribers in 2023 = 100000 and in 2024 = 110000
8. What was the percentage increase in the number of subscribers in the ‘Elder’ category from 2023 to 2024?
- 50%
- 65%
- 60%
- 40%
Let the total number of subscribers in 2023 = 100x
From point 1, the total number of subscribers in 2024 = 100x + 10% of 100x = 110x
In 2023,
Total number of subscribers from Kid category = 15% of 100x = 15x
From point 3 and 4, in 2023
Number of subscribers from Kid category using one app = 10000.
So, number of subscribers from Kid category using multiple apps = 15x – 10000 = number of subscribers from Elder category using one app.
Also, number of subscribers from Elder category using multiple apps = 15000
So, total number of subscribers from Elder category = 15x – 10000 + 15000 = 20% of 100x = 20x
Solving, x = 1000
So, the total number of subscribers in 2023 = 100000 and in 2024 = 110000
9. What could be the minimum percentage of subscribers who used multiple apps in 2024?
- 20.0%
- 10.0%
- 22.00%
- 16.5%
Let the total number of subscribers in 2023 = 100x
From point 1, the total number of subscribers in 2024 = 100x + 10% of 100x = 110x
In 2023,
Total number of subscribers from Kid category = 15% of 100x = 15x
From point 3 and 4, in 2023
Number of subscribers from Kid category using one app = 10000.
So, number of subscribers from Kid category using multiple apps = 15x – 10000 = number of subscribers from Elder category using one app.
Also, number of subscribers from Elder category using multiple apps = 15000
So, total number of subscribers from Elder category = 15x – 10000 + 15000 = 20% of 100x = 20x
Solving, x = 1000
So, the total number of subscribers in 2023 = 100000 and in 2024 = 110000
CAT 2024 | DILR Set 3
Direction Q.10 to 12:
Out of 10 countries — Country 1 through Country 10 — Country 9 has the highest gross domestic product (GDP), and Country 10 has the highest GDP per capita. GDP per capita is the GDP of a country divided by its population. The table below provides the following data about Country 1 through Country 8 for the year 2024.
- Column 1 gives the country’s identity.
- Column 2 gives the country’s GDP as a fraction of the GDP of Country 9.
- Column 3 gives the country’s GDP per capita as a fraction of the GDP per capita of Country10.
- Column 4 gives the country’s annual GDP growth rate.
- Column 5 gives the country’s annual population growth rate.
Assume that the GDP growth rates and population growth rates of the countries will remain constant for the next three years.
10. Which one among the countries 1 through 8, has the smallest population in 2024?
- Country 8
- Country 5
- Country 3
- Country 7
11. The ratio of Country 4’s GDP to Country 5’s GDP in 2026 will be closest to
- 1.032
- 1.195
- 1.314
- 0.963
12. Which one among the countries 1, 4, 5, and 7 will have the largest population in 2027?
- Country 7
- Country 4
- Country 5
- Country 1
CAT 2024 | DILR Set 4
Direction Q.13
The air-conditioner (AC) in a large room can be operated either in REGULAR mode or in POWER mode to reduce the temperature.
If the AC operates in REGULAR mode, then it brings down the temperature inside the room (called inside temperature) at a constant rate to the set temperature in 1 hour. If it operates in POWER mode, then this is achieved in 30 minutes.
If the AC is switched off, then the inside temperature rises at a constant rate so as to reach the temperature outside at the time of switching off in 1 hour. The temperature outside has been falling at a constant rate from 7 pm onward until 3 am on a particular night. The following graph shows the inside temperature between 11 pm (23:00) and 2 am (2:00) that night.
The following facts are known about the AC operation that night.
- The AC was turned on for the first time that night at 11 pm (23:00).
- The AC setting was changed (including turning it on/off, and/or setting different temperatures) only at the beginning of the hour or at 30 minutes after the hour.
- The AC was used in POWER mode for longer duration than in REGULAR mode during this3-hour period.
13. How many times the AC must have been turned off between 11:01 pm and 1:59 am?
- 0
- 2
- 1
- cannot be determined
As per the graph, the AC must have been turned off 2 times (12 am and 1 pm, when the inside temperature started rising) between 11:01 pm and 1:59 am
CAT 2024 | DILR Set 5
DIRECTIONS for the question: Study the table below and answer the following question.
Out of 10 countries — Country 1 through Country 10 — Country 9 has the highest gross domestic product (GDP), and Country 10 has the highest GDP per capita. GDP per capita is the GDP of a country divided by its population. The table below provides the following data about Country 1 through Country 8 for the year 2024.
- Column 1 gives the country’s identity.
- Column 2 gives the country’s GDP as a fraction of the GDP of Country 9.
- Column 3 gives the country’s GDP per capita as a fraction of the GDP per capita of Country10.
- Column 4 gives the country’s annual GDP growth rate.
- Column 5 gives the country’s annual population growth rate.
Assume that the GDP growth rates and population growth rates of the countries will remain constant for the next three years.
14. For how many countries among Country 1 through Country 8 will the GDP per capita in 2027 be lower than that in 2024?
Since all the GDP growth rates are positive i.e., GDP is increasing. So, there is no country among C1 through C8 whose GDP per capita in 2027 be lower than that in 2024.
CAT 2024 | DILR Set 6
DIRECTIONS for the question: Analyse the graph/s given below and answer the question that follows.
The air-conditioner (AC) in a large room can be operated either in REGULAR mode or in POWER mode to reduce the temperature.
If the AC operates in REGULAR mode, then it brings down the temperature inside the room (called inside temperature) at a constant rate to the set temperature in 1 hour. If it operates in POWER mode, then this is achieved in 30 minutes.
If the AC is switched off, then the inside temperature rises at a constant rate so as to reach the temperature outside at the time of switching off in 1 hour. The temperature outside has been falling at a constant rate from 7 pm onward until 3 am on a particular night. The following graph shows the inside temperature between 11 pm (23:00) and 2 am (2:00) that night.
The following facts are known about the AC operation that night.
- The AC was turned on for the first time that night at 11 pm (23:00).
- The AC setting was changed (including turning it on/off, and/or setting different temperatures) only at the beginning of the hour or at 30 minutes after the hour.
- The AC was used in POWER mode for longer duration than in REGULAR mode during this3-hour period.
15. What was the temperature outside, in degree Celsius, at 1 am?
Given that the AC was turned on for the first time that night at 11 pm.
So the temperature inside the room = 38 °C = the temperature outside the room.
Let the falling rate of temperature outside the room per hour = x °C
So, at 1 am (after 2 hours), the outside temperature = (38 – 2x) °C
Also given that it takes 1 hour to reach the temperature outside while AC was switched off.
Now, from 12 am to 12:30 pm, the temperature reached from 26 °C to 31 °C while AC was switched off, so it should have been reached from 31 °C to 36 °C from 12:30 pm to 1 pm (as per the constant rise in the rate), but along with that outside temperature is also falling at x °C per hour.
So, at 1 pm, the outside temperature should be = (36 – x) °C
Equating, (38 – 2x) °C = (36 – x) °C => x = 2 °C per hour
So, the temperature outside at 1 pm = 36 – 2 = 34 °C
16. What was the temperature outside, in degree Celsius, at 9 pm?
Given that the AC was turned on for the first time that night at 11 pm.
So the temperature inside the room = 38 °C = the temperature outside the room.
Let the falling rate of temperature outside the room per hour = x °C
So, at 1 am (after 2 hours), the outside temperature = (38 – 2x) °C
Also given that it takes 1 hour to reach the temperature outside while AC was switched off.
Now, from 12 am to 12:30 pm, the temperature reached from 26 °C to 31 °C while AC was switched off, so it should have been reached from 31 °C to 36 °C from 12:30 pm to 1 pm (as per the constant rise in the rate), but along with that outside temperature is also falling at x °C per hour.
So, at 1 pm, the outside temperature should be = (36 – x) °C
Equating, (38 – 2x) °C = (36 – x) °C => x = 2 °C per hour
At 9 pm (2 hours prior to 11 pm), the temperature outside = 38 + 2x = 42 °C
17. What best can be concluded about the number of times the AC must have either been turned on or the AC temperature setting been altered between 11:01 pm and 1:59 am?
- Exactly 2
- Either 2 or 3
- More than 3
- Exactly 3
Between 11:01 pm and 1:59 pm, the AC have been turned on at 12:30 am and 1:30 am, when the inside temperature starts falling. Along with that it is given that the AC used in POWER mode is for longer duration that in REGULAR mode during 11 pm to 2 am and this alteration can be done after every 30 minutes. This is possible as follows- From 11 pm to 11:30 pm – REGULAR mode From 11:30 pm to 12 pm – POWER mode From 12:30 pm to 1 pm – POWER mode From 1:30 pm to 2 pm – POWER mode So, there must be an alteration at 11:30 pm Hence, the number of time the AC must have been turned on or settings altered = 2 + 1 = 3 (exactly)
18. What was the maximum difference between temperature outside and inside temperature, in degree Celsius, between 11:01 pm and 1:59 am?
Between 11:01 pm and 11:59 am, the maximum difference between temperature outside and inside was at 12 am = 36 – 26 = 10 °C
CAT 2024 | DILR Set 7
Direction Q.19 to 22:
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 foodgrain category and the second column its codename. The table has some missing values.
The following additional facts are known.
- Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
- Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
- All the missing values of carbohydrate amounts (in grams) for all the foodgrains are non-zero multiples of 5.
- All the missing values of protein, fat and other nutrients amounts (in grams) for all the foodgrains are non-zero multiples of 4.
- P1 contained double the amount of protein that M3 contains.
19. How many foodgrains had a higher amount of carbohydrate per 100 grams of nutrients than M1?
From point 1, 2 and 3, for all categories, Carbohydrate (Cereal) > Carbohydrate (Pseudo-cereal) > Carbohydrate (Millet) and all of these missing values are non-zero multiples of 5.
Let C1 (carbohydrate) = A, C2 (carbohydrate) = B, M2 (carbohydrate) = C and P2 (carbohydrate) = D such that A, B > 66, D > 62, C, 56
From point 4 and 5, all missing values of protein, fat and other nutrients are non-zero multiples of 4 such that P1 (protein) = 2 × M3 (protein)
Let C1 (protein) = P, C2 (protein) Q, M2 (protein) = R, M3 (protein) = S and P1 (protein = 2S such that 2S, 14 > 10, R, S (point 1)
Let M1 (fat) = E, P1 (fat) = F and P2 (fat) = G
Let M1 (other nutrients) = X and M3 (other nutrients) = Y
Solving, A + P + 0 + 12 = 100 => A + P = 88
Possible values of A (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering P is a multiple of 4, so A = 80 and P = 8 is the only possibility
Solving, B + Q + 3 + 10 = 100 => B + Q = 87
Possible values of B (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering Q is a multiple of 4, so B = 75 and Q = 12 is the only possibility
Solving, 62 + 10 + E + X = 100 => E + X = 28
Solving, C + R + 7 + 16 = 100 => C + R = 77
Possible values of R (< 14) = 12, 8 or 4, but considering C is a multiple of 5, only possible value of C = 65 and R = 12
Solving, 56 + S + 12 + Y = 100 => S + Y = 32
Solving, 66 + 2S + F + 10 = 100 => 2S + F = 24
Considering, S is a multiple of 4, 2S must be a multiple of 8 and > 10
The only possible values of 2S = 16 and F = 8
So, S = 8 and Y = 24
Solving, D + 14 + G + 8 = 100 => D + G = 78
Possible values of D (> 62 and multiple of 5) = 65, 70, 75
But considering G is a multiple of 4, so D = 70 and G = 8
The rest of the information can be gathered as follows-
The number of food grains having a higher amount of carbohydrate per 100 grams of nutrients that M1 = 5 (C1, C2, M2, P1 and P2)
20. How many grams of protein were there in 100 grams of nutrients in M2?
From point 1, 2 and 3, for all categories, Carbohydrate (Cereal) > Carbohydrate (Pseudo-cereal) > Carbohydrate (Millet) and all of these missing values are non-zero multiples of 5.
Let C1 (carbohydrate) = A, C2 (carbohydrate) = B, M2 (carbohydrate) = C and P2 (carbohydrate) = D such that A, B > 66, D > 62, C, 56
From point 4 and 5, all missing values of protein, fat and other nutrients are non-zero multiples of 4 such that P1 (protein) = 2 × M3 (protein)
Let C1 (protein) = P, C2 (protein) Q, M2 (protein) = R, M3 (protein) = S and P1 (protein = 2S such that 2S, 14 > 10, R, S (point 1)
Let M1 (fat) = E, P1 (fat) = F and P2 (fat) = G
Let M1 (other nutrients) = X and M3 (other nutrients) = Y
Solving, A + P + 0 + 12 = 100 => A + P = 88
Possible values of A (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering P is a multiple of 4, so A = 80 and P = 8 is the only possibility
Solving, B + Q + 3 + 10 = 100 => B + Q = 87
Possible values of B (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering Q is a multiple of 4, so B = 75 and Q = 12 is the only possibility
Solving, 62 + 10 + E + X = 100 => E + X = 28
Solving, C + R + 7 + 16 = 100 => C + R = 77
Possible values of R (< 14) = 12, 8 or 4, but considering C is a multiple of 5, only possible value of C = 65 and R = 12
Solving, 56 + S + 12 + Y = 100 => S + Y = 32
Solving, 66 + 2S + F + 10 = 100 => 2S + F = 24
Considering, S is a multiple of 4, 2S must be a multiple of 8 and > 10
The only possible values of 2S = 16 and F = 8
So, S = 8 and Y = 24
Solving, D + 14 + G + 8 = 100 => D + G = 78
Possible values of D (> 62 and multiple of 5) = 65, 70, 75
But considering G is a multiple of 4, so D = 70 and G = 8
The rest of the information can be gathered as follows-
The amount of protein in 100 grams of nutrients in M2 = 12 grams
21. How many grams of other nutrients were there in 100 grams of nutrients in M3?
From point 1, 2 and 3, for all categories, Carbohydrate (Cereal) > Carbohydrate (Pseudo-cereal) > Carbohydrate (Millet) and all of these missing values are non-zero multiples of 5.
Let C1 (carbohydrate) = A, C2 (carbohydrate) = B, M2 (carbohydrate) = C and P2 (carbohydrate) = D such that A, B > 66, D > 62, C, 56
From point 4 and 5, all missing values of protein, fat and other nutrients are non-zero multiples of 4 such that P1 (protein) = 2 × M3 (protein)
Let C1 (protein) = P, C2 (protein) Q, M2 (protein) = R, M3 (protein) = S and P1 (protein = 2S such that 2S, 14 > 10, R, S (point 1)
Let M1 (fat) = E, P1 (fat) = F and P2 (fat) = G
Let M1 (other nutrients) = X and M3 (other nutrients) = Y
Solving, A + P + 0 + 12 = 100 => A + P = 88
Possible values of A (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering P is a multiple of 4, so A = 80 and P = 8 is the only possibility
Solving, B + Q + 3 + 10 = 100 => B + Q = 87
Possible values of B (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering Q is a multiple of 4, so B = 75 and Q = 12 is the only possibility
Solving, 62 + 10 + E + X = 100 => E + X = 28
Solving, C + R + 7 + 16 = 100 => C + R = 77
Possible values of R (< 14) = 12, 8 or 4, but considering C is a multiple of 5, only possible value of C = 65 and R = 12
Solving, 56 + S + 12 + Y = 100 => S + Y = 32
Solving, 66 + 2S + F + 10 = 100 => 2S + F = 24
Considering, S is a multiple of 4, 2S must be a multiple of 8 and > 10
The only possible values of 2S = 16 and F = 8
So, S = 8 and Y = 24
Solving, D + 14 + G + 8 = 100 => D + G = 78
Possible values of D (> 62 and multiple of 5) = 65, 70, 75
But considering G is a multiple of 4, so D = 70 and G = 8
The rest of the information can be gathered as follows-
The amount of other nutrients in 100 grams of nutrients in M3 = 24 grams
22. What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?
From point 1, 2 and 3, for all categories, Carbohydrate (Cereal) > Carbohydrate (Pseudo-cereal) > Carbohydrate (Millet) and all of these missing values are non-zero multiples of 5.
Let C1 (carbohydrate) = A, C2 (carbohydrate) = B, M2 (carbohydrate) = C and P2 (carbohydrate) = D such that A, B > 66, D > 62, C, 56
From point 4 and 5, all missing values of protein, fat and other nutrients are non-zero multiples of 4 such that P1 (protein) = 2 × M3 (protein)
Let C1 (protein) = P, C2 (protein) Q, M2 (protein) = R, M3 (protein) = S and P1 (protein = 2S such that 2S, 14 > 10, R, S (point 1)
Let M1 (fat) = E, P1 (fat) = F and P2 (fat) = G
Let M1 (other nutrients) = X and M3 (other nutrients) = Y
Solving, A + P + 0 + 12 = 100 => A + P = 88
Possible values of A (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering P is a multiple of 4, so A = 80 and P = 8 is the only possibility
Solving, B + Q + 3 + 10 = 100 => B + Q = 87
Possible values of B (> 66 and multiple of 5) = 70, 75, 80 or 85
But considering Q is a multiple of 4, so B = 75 and Q = 12 is the only possibility
Solving, 62 + 10 + E + X = 100 => E + X = 28
Solving, C + R + 7 + 16 = 100 => C + R = 77
Possible values of R (< 14) = 12, 8 or 4, but considering C is a multiple of 5, only possible value of C = 65 and R = 12
Solving, 56 + S + 12 + Y = 100 => S + Y = 32
Solving, 66 + 2S + F + 10 = 100 => 2S + F = 24
Considering, S is a multiple of 4, 2S must be a multiple of 8 and > 10
The only possible values of 2S = 16 and F = 8
So, S = 8 and Y = 24
Solving, D + 14 + G + 8 = 100 => D + G = 78
Possible values of D (> 62 and multiple of 5) = 65, 70, 75
But considering G is a multiple of 4, so D = 70 and G = 8
The rest of the information can be gathered as follows-
The numbers of grams of proteins in 100 grams of nutrients among given food grains in increasing order are 8, 8, 10, 12, 12, 14 and 16. The median value = 12
