A milk man has 175 litres of cow’s milk and 105 litres of buffalow�

1.	A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following (i) Capacity of a can (ii) Number of cans of cow’s milk (iii) Number of cans of buffalow’s milk.
Answer» 175 litres of cow’s milk. 105 litres of goat’s milk. H.C.F of 175 & 105 by using Euclid’s division algorithm. 175 = 105 x 1 + 70, the remainder 70 ≠ 0 Again using division algorithm, 105 = 70 x 1 + 35, the remainder 35 ≠ 0 Again using division algorithm. 70 = 35 x 2 + 0, the remainder is 0. ∴ 35 is the H.C.F of 175 & 105. (i) ∴ The milk man’s milk can’s capacity is 35 litres. (ii) No. of cow’s milk obtained = 175/35 = 5 cans (iii) No. of buffalow’s milk obtained = 105/35 = 3 cans

A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following (i) Capacity of a can (ii) Number of cans of cow’s milk (iii) Number of cans of buffalow’s milk.

Answer»

175 litres of cow’s milk.

105 litres of goat’s milk.

H.C.F of 175 & 105 by using Euclid’s division algorithm.

175 = 105 x 1 + 70, the remainder 70 ≠ 0

Again using division algorithm,

105 = 70 x 1 + 35, the remainder 35 ≠ 0

Again using division algorithm.

70 = 35 x 2 + 0, the remainder is 0.

∴ 35 is the H.C.F of 175 & 105.

(i) ∴ The milk man’s milk can’s capacity is 35 litres.

(ii) No. of cow’s milk obtained = 175/35 = 5 cans

(iii) No. of buffalow’s milk obtained = 105/35 = 3 cans