If one Buffalo give one litre milk and one cow give 2litre milk and one goat give 0.25litre milk Then how can we take 20 litre milk by taking 20 animals from all of them?
Answers
Answer:
Short answer: You milk them.
Long answer:
Let B = number of buffalo, C = number of cows, G = number of goats
Total number of animals = 100, so B + C + G = 100 (equation 1)
Total milk yield = 100 litres, so 5B + 0.50C + 0.25G = 100 (equation 2)
But we need 3 simultaneous equations to solve for 3 variables. Let’s at least eliminate one of the variables.
Multiplying equation 2 by 4 gives: 20B + 2C + G = 400
Subtract equation 1: 19B + C = 300
B and C are both positive integers so B <= 300/19 so B <= 15.79
C = 300 - 19B and B + C <= 100
so 300 - 19B + B <= 100, so 18B >= 200, so B >= 200/18, so B >= 11.11
Finding C and G for all integral values of B between 11.11 and 15.79,where C = 300 - 19B and G = 100 - B - C:
Solution i) B = 12; C= 300 - (19 × 12) = 72; G = 100 - 12 - 72 = 16
Solution ii) B = 13; C = 300 - (19 × 13) = 53; G = 100 - 13 - 53 = 34
Solution iii) B = 14; C = 300 - (19 × 14) = 34; G = 100 - 14 - 34 = 52
Solution iv) B = 15; C = 300 - (19 x 15) = 15; G = 100 - 15 - 15 = 70
Checking milk yields, 5B + 0.50C + 0.25G:
Solution i) (5 × 12) + (0.50 × 72) + (0.25 × 16) = 60 + 36 + 4 = 100
Solution ii) (5 × 13) + (0.50 × 53) + (0.25 × 34) = 65 + 26.5 + 8.5 = 100
Solution iii) (5 x 14) + (0.50 × 34) + (0.25 × 52) = 70 + 17 + 13 = 100
Solution iv) (5 × 15) + (0.50 x 15) + (0.25 × 70) = 75 + 7.5 + 17.5 = 100