Sets. Help Needed.?

23. (a)

Friday. Since everyone is in F circle, then Survey must have occurred on Friday.

Everyone who is there on Friday must shop on Fridays (they are there now taking the survey.

of the 100 people all either come on only day or only two days or only 3 days

so 1 dayer + 2 dayer + 3 dayer = 100

100 - 1 dayer only -2 dayer only 100- 48 - 37 = 15

23. (b) (i) I agree with the answer.

so 15 people shops on all three days.

48 = (F and S) + (F and D)

but F and S = 28

The 37 who shop on one day

only shop on Friday. Since the S and D circle are enclosed by the F circle . Therefore is

no one who shopped on both Saturday and Sunday, but not Friday.

48-28 = 20 people shop on Friday and Sunday.

Checking

Friday alone = 37

F and S only= 28

F and D only = 20

15 on all three

37 + 20 + 28 + 15 = 100

so Sunday = All 3 days + (D and F only) = 20 + 15 = 35 people shop on Sunday

So I don't see anyone can get an answer of 50.

Outer area Friday only = 37

Friday and Saturday area = 28

Friday and Sunday = 20

Friday , Saturday and Sunday = 15

23 (b) (ii) I disagree with the answer and I believe it is not possible. 37 do only one day and

that must be Friday. (28 are there only on Friday and Saturday) + 37 +28 = 65 so

you have accounted for 65 people, there aren't 50 people let to shop on Sunday. It cannot be true.

Sunday and other days = 35 ( I don't agree with answer of 50)

Saturday and other days y = 28+ 15 = 43

Friday and other days= 100

100+35 + 43 = 178

178 - 1* double counts - 2*triple counts should equal 100 If lucky and right

178 - 48 - 2*15 = 100 (it checks )