105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatsman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally take the largest possible number each time.Can you tell me how many animals went in each trip?
Answers
Given that:
- 105 goats, 140 donkeys and 175 cows have to be taken across a river.
- He insists that he will take the same number of animals in every trip and they have to be of the same kind.
- He will naturally take the largest possible number each types.
To Find:
- How many animals went in each trip?
Finding the number of animals went in each trip:
Finding HCF using prime factorisation.
- Factors of 105 = 3 × 5 × 7
- Factors of 140 = 2² × 5 × 7
- Factors of 175 = 5² × 7
HCF (105, 140 & 175) = 5 × 7 = 35
HCF of 105, 140 and 175 is 35.
Hence,
- 35 animals of same kind went in each trip.
To find the largest possible number of animals ,we will find the H.C.F of 105,140 and 175.
To find the largest possible number of animals ,we will find the H.C.F of 105,140 and 175.105=3×5×7
To find the largest possible number of animals ,we will find the H.C.F of 105,140 and 175.105=3×5×7140=2×2×5×7
To find the largest possible number of animals ,we will find the H.C.F of 105,140 and 175.105=3×5×7140=2×2×5×7175=5×5×7
To find the largest possible number of animals ,we will find the H.C.F of 105,140 and 175.105=3×5×7140=2×2×5×7175=5×5×7HCF of 105,140 and 175=5×7=35
To find the largest possible number of animals ,we will find the H.C.F of 105,140 and 175.105=3×5×7140=2×2×5×7175=5×5×7HCF of 105,140 and 175=5×7=35Hence The number of animals went in each trip is 35
answer is = 35