Problem solving for elementary school but without algebra

Question

 

Joe had a herd of cattle. Then he bought more cattle so that his total number was three times the original number. He was given six more cattle. Cattle prices were up so he sold half the herd. After the summer drought he felt lucky because he only lost 4 cattle. When fall arrived, Joe had 41 cattle left. How many cattle did he have at the start?

 

I can't use algebra but what kind of strategy can i come up with for this problem?

Answer 1

Work backward.

 

41 + 4 = 45

 

The 45 was half the cattle, so he had 90.

 

90 - 6 = 84 cattle before he was given 6.

 

If 84 is 3 times what he had, then he had 84/3 = 28.