Question

dare to do ☺
1 buffalo gives 5 litre milk, 1 cow gives 1 litre milk , 1 goat gives 250 ml milk so what you will do so that the sum of animals become 20 and the milk become 20 litres

Answer 1

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

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