Category	Code	Composition per 100g of food grains				Total
Category	Code	Carbs	Protein	Fat	Others	Total
Cereal	C1			0	12	100
Cereal	C2			3	10	100
Millet	M1	62	10			100
	M2			7	16	100
	M3	56		12		100
Pseudo-cereal	P1	66			10	100
Pseudo-cereal	P2		14		8	100

Understanding the set:

In this question, the very first and most important step will be to understand that the sum of every row will be 100. The table given below shows the amount, in grams, of carbohydrate, protein, fat and all other nutrients per 100 grams. Therefore, in a food grain, the total amount of nutrients will sum up to 100 grams. —--

Step 1: Add Multiplication Indicators

Category	Code	Composition per 100g of food grains				Total
Category	Code	Carbs (5x)	Protein (4x)	Fat (4x)	Others (4x)	Total
Cereal	C1			0	12	100
Cereal	C2			3	10	100
Millet	M1	62	10			100
	M2			7	16	100
	M3	56		12		100
Pseudo-cereal	P1	66			10	100
Pseudo-cereal	P2		14		8	100

Clue 3 and clue 4 say: All the missing values of carbohydrate amounts (in grams) for all the food grains are non-zero multiples of 5. All the missing values of protein, fat and other nutrients amounts (in grams) for all the food grains are non-zero multiples of 4. —--

Step 2: Solving for C1

Category	Code	Composition per 100g of food grains				Total
Category	Code	Carbs (5x)	Protein (4x)	Fat (4x)	Others (4x)	Total
Cereal	C1	80	8	0	12	100
Cereal	C2			3	10	100
Millet	M1	62	10			100
	M2			7	16	100
	M3	56		12		100
Pseudo-cereal	P1	66			10	100
Pseudo-cereal	P2		14		8	100

Using clue 2: Both the cereals had higher amounts of carbohydrate than any pseudo-cereal. This implies that carbs in C1 & C2 > carbs in P1 & P2

We can see that P1 = 66. Therefore, C1 & C2 will be a multiple of 5 greater than 66.

Calculating for C1 first: We know that the sum of all nutrients in a food grain = 100. carbs + protein + fats + other = 100 In C1: carbs + protein + 0 + 12 = 100 carbs + protein = 88 Therefore, the possibilities are: 70, 75, 80, 85.

In C1, we know protein is a multiple of 4. If we put carbs = 70: protein = 18 (not a multiple of 4) If we put carbs = 75: protein = 13 (not a multiple of 4) If we put carbs = 85: protein = 3 (not a multiple of 4) If we put carbs = 80: protein = 8 (a multiple of 4) Therefore, the only possibility left for C1 is carbs = 80 and protein = 8. —--

Step 3: Solving for C2

Category	Code	Composition per 100g of food grains				Total (should be 100)
Category	Code	Carbs (5x)	Protein (4x)	Fat (4x)	Others (4x)	Total (should be 100)
Cereal	C1	80	8	0	12	100
Cereal	C2	75	12	3	10	100
Millet	M1	62	10			72
	M2			7	16	23
	M3	56		12		68
Pseudo-cereal	P1	66			10	76
Pseudo-cereal	P2		14		8	22

Calculating the same for C2: We know P1 = 66. Therefore, C2 will be a multiple of 5 greater than 66. We know that the sum of all nutrients in a food grain = 100. carbs + protein + fats + other = 100 In C2: carbs + protein + 3 + 10 = 100 carbs + protein = 87 Therefore, the possibilities are: 70, 75, 80, 85.

But in C2, we know protein is a multiple of 4. If we put carbs = 70: protein = 17 (not a multiple of 4) If we put carbs = 75: protein = 12 (a multiple of 4) If we put carbs = 85: protein = 2 (not a multiple of 4) If we put carbs = 80: protein = 7 (not a multiple of 4) Therefore, the only possibility left for C2 is carbs = 75 and protein = 12. —--

Step 4: Solving for P2

Category	Code	Composition per 100g of food grains				Total (should be 100)
Category	Code	Carbs (5x)	Protein (4x)	Fat (4x)	Others (4x)	Total (should be 100)
Cereal	C1	80	8	0	12	100
Cereal	C2	75	12	3	10	100
Millet	M1	62	10			72
	M2			7	16	23
	M3	56		12		68
Pseudo-cereal	P1	66			10	76
Pseudo-cereal	P2	70	14	8	8	100

Now continuing with the 2nd clue and using clue 1 alongside, which says: Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet. We know carbs in P1 & P2 < carbs in C1 & C2. And carbs in P1 & P2 > carbs in M1, M2 & M3 (according to clue 2). Therefore, for P2: We know P2 is a multiple of 5 and will be less than 75 but greater than 62. The only possibilities are: 65 & 70.

But we know that all missing values of fat are multiples of 4. If carbs in P2 = 65, then fats = 13 (not a multiple of 4). If carbs in P2 = 70, then fats = 8 (a multiple of 4). Therefore, carbs in P2 = 70 and fats = 8. —--

Step 5: Solving for M2

Category	Code	Composition per 100g of food grains				Total (should be 100)
Category	Code	Carbs (5x)	Protein (4x)	Fat (4x)	Others (4x)	Total (should be 100)
Cereal	C1	80	8	0	12	100
Cereal	C2	75	12	3	10	100
Millet	M1	62	10			72
	M2	65	12	7	16	100
	M3	56		12		68
Pseudo-cereal	P1	66			10	76
Pseudo-cereal	P2	70	14	8	8	100

Continuing with clue 1: We know protein in P1 & P2 > protein in M1, M2 & M3. Therefore, protein in M2 & M3 will be less than 14 (protein in P2). The possibilities are: 0, 4, 8, 12.

We know: carbs + protein + fats + other = 100. For M2: carbs + protein + 7 + 16 = 100. carbs + proteins = 77 Now, trying all possibilities from the previous slide, we get: If we put protein = 0: carbs = 77 (not a multiple of 5) If we put protein = 4: carbs = 73 (not a multiple of 5) If we put protein = 8: carbs = 69 (not a multiple of 5) If we put protein = 12: carbs = 65 (a multiple of 5) Therefore, in M2, carbs = 65 and protein = 12. —--

Final Solution: Complete Table

Category	Code	Composition per 100g of food grains
Category	Code	Carbs (5x)	Protein (4x)	Fat (4x)	Others (4x)
Cereal	C1	80	8	0	12
Cereal	C2	75	12	3	10
Millet	M1	62	10	20	8
	M2	65	12	7	16
	M3	56	8	12	24
Pseudo-cereal	P1	66	16	8	10
Pseudo-cereal	P2	70	14	8	8

We can do the same for M3 as well. The protein in M3 can be 0, 4, 8 or 12. Clue 5 says: P1 contained double the amount of protein that M3 contains. 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 cannot be 0 or 8. This leaves 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. This gives us 8 grams of protein in M3 and 24 grams of others in M3.

For M1: carbs + protein + fats + other = 100 62 + 10 + fats + other = 100 fats + other = 28 Since both must be multiples of 4, the combinations could be: (4,24), (8,20), (12,16), (16,12), (20,8), (24,4). Given the constraints and checking consistency, fats = 20 and other = 8 works for M1.

CAT (DILR) PYQs

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