monkeygirl said:
No. I know how to set up, but not clear how to do this one because
there are more than four numbers 9,10,1,6,300
I dont get if I use the 9of 10 as a fraction or the 1 of 6 as a fraction
and since it is it is asking how many of the 300 exersize regularly
do I just use 1,6,300?-1over6, x over 300?
because 9out of 10 think its good so do I leave that out.
basically Im confused because its a word problem and dosent make sense .because its not asking how many of the 300 think to exersize
so how does the 9 of ten who think its a good idea fit in , the question is asking for how many of 300 employees exersise regularly
9 of 10 think to exersise
of the 9 only 1 of 6 actually exersise
300 not sure how many of these exersise regularly
so either I do them as fractions or I have to elemanate two numbers?
How about doing the problem in two stages (which is basically what the previously demonstrated solution involved)?
Stage 1:
9 out of 10 think exercise is a good idea.
how many out of 300 think it is a good idea, then?
9/10 = x/300
10x = 2700
x = 270
So, now we know that there are 270 people who think that exercise is a good idea.
Stage 2:
1 out of 6 who think it is a good idea
actually exercise. So, how many of this group of 270 actually exercise?
1/6 = y/270
Solve this proportion for y, and you will have the answer to the question "How many of the 300 people exercise regularly?"
Or, you can use this approach.
We know that 9/10 of the people say exercise is a good idea, but of these, only 1/6 actually exercise. So, 1/6 of 9/10 of the people actually exercise. 1/6 of 9/10 is
(1/6)*(9/10) = 3/20
So, 3 out of 20 responders actually exercise. And you can use this proportion:
3/20 = z/300
to determine the number of actual exercisers in the group of 300. You should get the same results as in the previously-described two-stage process.