中一數學題

1.108 sweets and 88 chocolates are to be shared among some children. If each child gets the same number of sweets and chocolates, how many sweets and chocolates each child gets at least? (Hint: Find the largest number of children first.)

Ans:

the largest number of children is 88 ,because we just have 88 chocolates

so they at least get 1 sweets and 1 chocolates.....

2.There is a bottle of soft drink with a capacity of 1.5 L. Amy drinks 1/3 of it. Betty drinks 0.12 L. Then Cindy drinks 2/5 of the remaining soft drink. How much soft drink does Cindy drink?

Amy drinks 1/3 , it left 1.5x(1-1/3)=1 L

Betty drinks 0.12 L it left 1-0.12=0.88

Then Cindy drinks 2/5 of the remaining soft drink

that mean 0.88 x(2/5)=0.352 L=352 mL//

1.The HCF between 108 sweets and 88 chocolates is:
2[108 88=2[54 44=27 22
So, it is 4.
Because of the HCF between 108 sweets and 88 chocolates is 4 ,
So, each children can get 108/4=27sweets and 88/4=22 chocolates.
2.Amy drinks: 1.5*(1/3)=0.5L
the remaining soft drink before Cindy drinks:
1.5L(The total capacityof the soft drink)-0.5L(Amy drinks)-0.12L(Betty drinks)
=0.88L
Cindy drinks:0.88*2/5=0.352L
So, Cindy drinks 0.352L.
有d混亂,唔好意思.
希望可以幫到你啦,加油!!=)