posted by Emily on .
Ha! Mom said if I could find help online to go for it so here I am asking for your help. Here is one problem she gave me from a page called Superstars.
Mr. Jackson is preparing bags of treats to give trick or treaters. He has 48 pieces of candy and 60 pieces of gum. He uses all the candy and gum, and he puts the same ratio of candy to gum in each bag. What is the largest number of bags he could have made?
Yes, the answer would be great, but I love math and am trying to figure out what method I should use to figure this out.
PS: i may be back now that I found you!
What factors divides both numbers evenly? For instance 2,
Each of two bags has 30gum, and 24 candies.
Maybe the common factor needs to be larger.
'k .. I think my brain is not working .. I don't think other 8 year olds have to do this .. My mother is creating a short circuit. I am confused. I should have gone higher to 12. Thank you for that tip. So would the answer be 4 bags since gum would be too many or would it be 5 bags ... Okay,now that I am thinking, I think 5 bags because I see it says ALL CANDY AND GUM ARE USED, so it must be the higher number. Please tell me I am right and if not then tell my mom this is too hard. She thinks I can do it but I think she thinks too much! lol! Again, thank you for your help! emily
Thank you very much. I think I got it. Can you tell me if the answer would be 8 with 2 gums left over? If not I will try again. My mom by the way is laughing. She didn't think I could find help so quickly! lol ..Thanks for helping me.
to get 8 candies, the divisor is six, which goes into 60 evenly, so no gums will be left over.
I wonder if 12 goes into each evenly?
Okay I will try again. I was thinking that 6 was the largest number and since the question said what would be the largest number of bags I thought 8 because 6 goes into 48, 8 times and into 60 10 times, which would be 8 bags possible to make, keeping the bags even, leaving you 2 pieces of gum leftover .... but I will try again ... :-( .. thank you. Emily.