Question

111 cows,185sheeps and 296 giats are to be taken acroos river.there is only one boat and the boat mam says that he will take the same number and same kind of animals in each trip .find the greatest number of animaks in each trip and number of trips he will make

Answer 1

Answer:

Step-by-step explanation:

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, 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:56

Step-by-step explanation:

111 de 78 ye 89ud