Question

111 cows, 185 sheep, 296 goats are to be taken across a river. There is only one boat and the boatman says that he will take the same number and same kind of animals in each trip. Find the largest number of animals in each trip and number of trips he will make.

Answer 1

To calculate the highest number which can divide all these three numbers, we will have to find their  HCF . 
the factors of 111 are -  3 and 37
of 185 are -- 5 and 37 
and of 296 are -- 2 ,2,2 and 37 
so, the hcf is 37. 
therefore, maximum 37 animals can be taken in the boat at a time. 
total no. of animals = 111+185+296 = 592 
so, no. of trips = 592 divided by 37 = 16 

Answer 2

Answer:

16.

Step-by-step explanation:

H.C.F of 111 cows = 3 and 37

H.C.F of 185 sheep = 5 and 37

H.C.F of 296 goats = 2,2,2 and 37

now,

H.C.F=37

Therefore, maximum 37 animals can be taken in the boat at a time.  

Total number of animals = 185+111+296 = 592  

So, number of trips = 592 divided by 37 = 16