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 boatman 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 like to take the largest possible number each time. Can you tell how many animals went on each trip?

Given:

Number of goats = 105

Number of donkeys = 140

Number of cows = 175

To find: Here we have to find the largest possible number of animals the boatman can transport each time.

Solution:

To find the largest possible number of animals the boatman can transport each time we need to calculate the HCF of 105, 140 and 175.

First, let's find HCF of 105 and 140 using Euclid's division algorithm:

Using Euclid’s lemma to get: 
  • $140\ =\ 105\ \times\ 1\ +\ 35$

Now, consider the divisor 105 and the remainder 35, and apply the division lemma to get:
  • $105\ =\ 35\ \times\ 3\ +\ 0$

The remainder has become zero, and we cannot proceed any further. 

Therefore the HCF of 105 and 140 is the divisor at this stage, i.e., 35.


Now, let's find HCF of 35 and 175 using Euclid's division algorithm:

Using Euclid’s lemma to get: 
  • $175\ =\ 35\ \times\ 5\ +\ 0$

The remainder has become zero, and we cannot proceed any further. 

Therefore the HCF of 35 and 175 is the divisor at this stage, i.e., 35.


So, the largest possible number of animals the boatman can transport each time is 35.

Updated on: 10-Oct-2022

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