CAT 2024 Slot 3 · DILR · Set 2: Nutrients in Seven Food Grains

The full set and all 4 questions (Q6–Q9) from CAT 2024 Slot 3, with answers and solutions.

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.

Composition per hundred grams of nutrients in the food grains

Food grain category	Codename of the food grain	Carbohydrate	Protein	Fat	Other nutrients
Cereal	C1			0	12
Cereal	C2			3	10
Millet	M1	62	10
Millet	M2			7	16
Millet	M3	56		12
Pseudo-cereal	P1	66			10
Pseudo-cereal	P2		14		8

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 6 · Tables

How many foodgrains had a higher amount of carbohydrate per 100 grams of nutrients than M1?

Answer & solution

Answer: 5

Solving the set

Step 1 – Each row adds up to 100 g, so the missing values in each row must add up to a known number.

Grain	Known values	Missing values must add up to
C1	Fat 0, Other 12	Carb + Protein = 88
C2	Fat 3, Other 10	Carb + Protein = 87
M1	Carb 62, Protein 10	Fat + Other = 28
M2	Fat 7, Other 16	Carb + Protein = 77
M3	Carb 56, Fat 12	Protein + Other = 32
P1	Carb 66, Other 10	Protein + Fat = 24
P2	Protein 14, Other 8	Carb + Fat = 78

Step 2 – M2: carbohydrate (multiple of 5) + protein (multiple of 4) = 77. Possible pairs: (65, 12), (45, 32), (25, 52), (5, 72). P2's protein (14) must exceed M2's protein (fact 1), so M2 = (65, 12).

Step 3 – P2: carbohydrate + fat = 78 gives (70, 8), (50, 28), (30, 48) or (10, 68). P2's carbohydrate must exceed M2's 65 (fact 1), so P2 carbohydrate = 70, fat = 8.

Step 4 – P1 and M3: P1's protein + fat = 24, and P1's protein must exceed 12 (M2's protein), so it is 16 or 20. P1's protein = 2 × M3's protein (fact 5), and M3's protein is a multiple of 4, so P1 protein = 16 (M3 protein = 8). Then P1 fat = 8 and M3 other nutrients = 32 − 8 = 24.

Step 5 – Cereals: carbohydrate must exceed both pseudo-cereals (66 and 70), so it is more than 70.

• C1: carbohydrate + protein = 88. For the protein to be a multiple of 4, the carbohydrate must be a multiple of 20 → 80, protein 8.

• C2: carbohydrate + protein = 87. Carbohydrate above 70 and a multiple of 5: 75 (protein 12), 80 (7 ✗), 85 (2 ✗). So 75 and 12.

Step 6 – M1: fat + other = 28 can be split in several ways (4 + 24, 8 + 20, …, 24 + 4), so M1's fat and other nutrients cannot be fixed.

Grain	Carbohydrate	Protein	Fat	Other nutrients	Total
C1	80	8	0	12	100
C2	75	12	3	10	100
M1	62	10	f (4 to 24)	28 − f	100
M2	65	12	7	16	100
M3	56	8	12	24	100
P1	66	16	8	10	100
P2	70	14	8	8	100

This question

From the final table, the carbohydrate values above M1's 62 g are C1 (80), C2 (75), M2 (65), P1 (66) and P2 (70). M3 (56) is lower.

So 5 food grains have more carbohydrate than M1.

Hence, the answer is 5.

Question 7 · Tables

How many grams of protein were there in 100 grams of nutrients in M2?

Answer & solution

Answer: 12

Solving the set

Step 1 – Each row adds up to 100 g, so the missing values in each row must add up to a known number.

Grain	Known values	Missing values must add up to
C1	Fat 0, Other 12	Carb + Protein = 88
C2	Fat 3, Other 10	Carb + Protein = 87
M1	Carb 62, Protein 10	Fat + Other = 28
M2	Fat 7, Other 16	Carb + Protein = 77
M3	Carb 56, Fat 12	Protein + Other = 32
P1	Carb 66, Other 10	Protein + Fat = 24
P2	Protein 14, Other 8	Carb + Fat = 78

Step 3 – P2: carbohydrate + fat = 78 gives (70, 8), (50, 28), (30, 48) or (10, 68). P2's carbohydrate must exceed M2's 65 (fact 1), so P2 carbohydrate = 70, fat = 8.

Step 5 – Cereals: carbohydrate must exceed both pseudo-cereals (66 and 70), so it is more than 70.

• C1: carbohydrate + protein = 88. For the protein to be a multiple of 4, the carbohydrate must be a multiple of 20 → 80, protein 8.

• C2: carbohydrate + protein = 87. Carbohydrate above 70 and a multiple of 5: 75 (protein 12), 80 (7 ✗), 85 (2 ✗). So 75 and 12.

Step 6 – M1: fat + other = 28 can be split in several ways (4 + 24, 8 + 20, …, 24 + 4), so M1's fat and other nutrients cannot be fixed.

Grain	Carbohydrate	Protein	Fat	Other nutrients	Total
C1	80	8	0	12	100
C2	75	12	3	10	100
M1	62	10	f (4 to 24)	28 − f	100
M2	65	12	7	16	100
M3	56	8	12	24	100
P1	66	16	8	10	100
P2	70	14	8	8	100

This question

From Step 2, M2 has 12 g of protein.

Hence, the answer is 12.

Question 8 · Tables

How many grams of other nutrients were there in 100 grams of nutrients in M3?

Answer & solution

Answer: 24

Solving the set

Step 1 – Each row adds up to 100 g, so the missing values in each row must add up to a known number.

Grain	Known values	Missing values must add up to
C1	Fat 0, Other 12	Carb + Protein = 88
C2	Fat 3, Other 10	Carb + Protein = 87
M1	Carb 62, Protein 10	Fat + Other = 28
M2	Fat 7, Other 16	Carb + Protein = 77
M3	Carb 56, Fat 12	Protein + Other = 32
P1	Carb 66, Other 10	Protein + Fat = 24
P2	Protein 14, Other 8	Carb + Fat = 78

Step 3 – P2: carbohydrate + fat = 78 gives (70, 8), (50, 28), (30, 48) or (10, 68). P2's carbohydrate must exceed M2's 65 (fact 1), so P2 carbohydrate = 70, fat = 8.

Step 5 – Cereals: carbohydrate must exceed both pseudo-cereals (66 and 70), so it is more than 70.

• C1: carbohydrate + protein = 88. For the protein to be a multiple of 4, the carbohydrate must be a multiple of 20 → 80, protein 8.

• C2: carbohydrate + protein = 87. Carbohydrate above 70 and a multiple of 5: 75 (protein 12), 80 (7 ✗), 85 (2 ✗). So 75 and 12.

Step 6 – M1: fat + other = 28 can be split in several ways (4 + 24, 8 + 20, …, 24 + 4), so M1's fat and other nutrients cannot be fixed.

Grain	Carbohydrate	Protein	Fat	Other nutrients	Total
C1	80	8	0	12	100
C2	75	12	3	10	100
M1	62	10	f (4 to 24)	28 − f	100
M2	65	12	7	16	100
M3	56	8	12	24	100
P1	66	16	8	10	100
P2	70	14	8	8	100

This question

From Step 4, M3 has 8 g protein, so its other nutrients = 100 − 56 − 8 − 12 = 24 g.

Hence, the answer is 24.

Question 9 · Tables

What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?

Answer & solution

Answer: 12

Solving the set

Step 1 – Each row adds up to 100 g, so the missing values in each row must add up to a known number.

Grain	Known values	Missing values must add up to
C1	Fat 0, Other 12	Carb + Protein = 88
C2	Fat 3, Other 10	Carb + Protein = 87
M1	Carb 62, Protein 10	Fat + Other = 28
M2	Fat 7, Other 16	Carb + Protein = 77
M3	Carb 56, Fat 12	Protein + Other = 32
P1	Carb 66, Other 10	Protein + Fat = 24
P2	Protein 14, Other 8	Carb + Fat = 78

Step 3 – P2: carbohydrate + fat = 78 gives (70, 8), (50, 28), (30, 48) or (10, 68). P2's carbohydrate must exceed M2's 65 (fact 1), so P2 carbohydrate = 70, fat = 8.

Step 5 – Cereals: carbohydrate must exceed both pseudo-cereals (66 and 70), so it is more than 70.

• C1: carbohydrate + protein = 88. For the protein to be a multiple of 4, the carbohydrate must be a multiple of 20 → 80, protein 8.

• C2: carbohydrate + protein = 87. Carbohydrate above 70 and a multiple of 5: 75 (protein 12), 80 (7 ✗), 85 (2 ✗). So 75 and 12.

Step 6 – M1: fat + other = 28 can be split in several ways (4 + 24, 8 + 20, …, 24 + 4), so M1's fat and other nutrients cannot be fixed.

Grain	Carbohydrate	Protein	Fat	Other nutrients	Total
C1	80	8	0	12	100
C2	75	12	3	10	100
M1	62	10	f (4 to 24)	28 − f	100
M2	65	12	7	16	100
M3	56	8	12	24	100
P1	66	16	8	10	100
P2	70	14	8	8	100

This question

Protein values: C1 8, C2 12, M1 10, M2 12, M3 8, P1 16, P2 14. Arranged in order: 8, 8, 10, 12, 12, 14, 16. With 7 values, the median is the 4th value = 12.

Hence, the answer is 12.

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