CAT 2024 Slot 3 DILR Set 5: Tables (4 Questions) with Solutions

Row totals give: M2 carbohydrate + protein = 100 − 7 − 16 = 77; P2 carbohydrate + fat = 100 − 14 − 8 = 78; C1 carbohydrate + protein = 100 − 0 − 12 = 88; C2 carbohydrate + protein = 100 − 3 − 10 = 87.

P1: protein + fat = 100 − 66 − 10 = 24. Fact 5 makes P1 protein double M3 protein. M3 protein + other = 100 − 56 − 12 = 32. M3 protein is 4 or 8, so P1 protein is 8 or 16. Fact 1 says pseudo-cereal protein exceeds M1's 10, so P1 protein is 16 and M3 protein is 8.

M2: protein must be below both pseudo-cereal proteins (16 and 14) and be a multiple of 4, so it is at most 12. Carbohydrate $= 77 - \text{protein}$ must be a multiple of 5, which only protein 12 allows. M2 carbohydrate is 65.

P2: carbohydrate $= 78 - \text{fat}$ with fat a multiple of 4. Carbohydrate must be a multiple of 5 and exceed every millet (fact 1), so it is 70.

Cereals must exceed P1's 66 and P2's 70 (fact 2). C1: 88 splits as 80 + 8, the only multiple-of-5 option above 70 with protein a multiple of 4. C2: 87 splits as 75 + 12.

Carbohydrate values: C1 80, C2 75, M1 62, M2 65, M3 56, P1 66, P2 70.

Grains above M1's 62: C1, C2, M2, P1, P2, which is 5 grains.

M2 has fat 7 and other nutrients 16, so carbohydrate + protein = 100 − 7 − 16 = 77.

Fact 1 says every pseudo-cereal has more protein than any millet. The pseudo-cereals' proteins are P2 = 14 and P1 (at least 12, as shown next). P1 protein is double M3's, and M3 protein + other = 100 − 56 − 12 = 32 with both multiples of 4. P1 protein must also exceed M1's 10, so P1 protein is 16 and M3's is 8. M2 protein is therefore a multiple of 4 below 14, so at most 12.

Fact 3 requires M2 carbohydrate $= 77 - \text{protein}$ to be a multiple of 5. Protein 4 gives 73 and protein 8 gives 69, so both fail. Protein 12 gives 65, which works.

M2 protein is 12 g.

M3 has carbohydrate 56 and fat 12, so protein + other = 100 − 56 − 12 = 32.

P1 has carbohydrate 66 and other 10, so protein + fat = 24. Both are non-zero multiples of 4, so P1 protein is at most 20.

Fact 5 makes P1 protein double M3 protein, so M3 protein is at most 10. As a multiple of 4, it is 4 or 8, and P1 protein is 8 or 16.

Fact 1 says P1 must have more protein than the millet M1, which has 10. P1 protein 8 fails, so P1 protein is 16 and M3 protein is 8.

M3 other nutrients $= 32 - 8 = 24$.

M3 and P1: P1 protein must exceed M1's 10 (fact 1) and equals 2 × M3 protein (fact 5). M3 protein + other = 32 and P1 protein + fat = 24 with all values multiples of 4, so P1 protein is 16 and M3 protein is 8.

M2: carbohydrate + protein = 77. Protein must be a multiple of 4 below 14, and carbohydrate a multiple of 5. Only protein 12 with carbohydrate 65 works.

P2: carbohydrate $= 78 - \text{fat}$ must be a multiple of 5 and above every millet's carbohydrate (62, 65, 56). That gives 70, so the cereals' carbohydrate must exceed 70 (fact 2).

C1: carbohydrate + protein = 88, so carbohydrate is 80 and protein is 8. C2: carbohydrate + protein = 87, so carbohydrate is 75 and protein is 12.

Proteins: C1 8, C2 12, M1 10, M2 12, M3 8, P1 16, P2 14.

Sorted: 8, 8, 10, 12, 12, 14, 16. The fourth of seven values is 12.