How Many Ladies Had All Four Items at the Church Fete?

[ilvpat]

Q:
Out of 100 ladies attending the church fete, 85 had a white handbag; 75 had black shoes; 60 carried an umbrella; 90 wore a ring, How many ladies must have had all four items?


I feel like the problem is under-determined. But if anybody can explain it to me it would be great. Thank you very much.

 

[arildno]

It certainly looks that way, but we can reinterpret "must have had" as a lower bound, that is, we are interested in finding out what is the least number of ladies that had to have all for items!

Now, consider the ladies with rings, and those with handbags.
At least 75 of them must have had both (why?)

Now, furthermore consider the R&H ladies (those with rings AND handbg), and those with black shoes (S). Now, of the S population, only 25 max could not be wearing both R and H, since the 100 ladies can be divided into those having both R and H (at least 75) and those not having both (at most 25).

Thus, of the S ladies, at least 50 of them are R&H&S ladies.

Finally, only max 50 of the umbrella ladies U could not be R&H&S ladies; hence, at least 10 U's are also R&H&S ladies.

Thus, at least 10 ladies had all four objects, up to a maximum of 60.

 

[Edgardo]

Welcome to physicsforums ilvpat,

my advice is to try a simplified version of the problem.
Out of 100 ladies, 85 have a white handbag, 75 have black shoes.
How many ladies must have both items?

I think if you can solve this simplified problem, you can also
solve the original one.

 

[ilvpat]

I see now. The way I interpreted the problem leads me to the wrong direction. I was looking for a unique solution rather than a bound. Thanks guys!

 

[jd2718]

It might be simpler to look at complements:
15 are w/o a white handbag
25 are w/o black shoes
40 w/o umbrella
10 w/o ring

That's 90 without something, leaving 10 with everything.

(what happened? We maximized the number with 3 of the 4)