Cachers and Whackos Riddle - Can you solve it?

91. Cachers and Whackos Riddle

I think all sane people are cachers

and one third of all cachers are sane

but half of all whackos are cachers

with only one whacko that is sane

If eight whackos are cachers

and ninety are attending my ball

how many cachers are neither sane nor whacko at all?

Added 8 July 2010 · Updated 8 July 2026

Show Solution

Solution:

53.
Working it through:

"Ninety are attending my ball" → the cachers' ball, so C = 90.
"One third of all cachers are sane" → sane cachers = 90/3 = 30 (and since all sane are cachers, that's all sane people).
"Eight whackos are cachers" → whacko cachers = 8.
"Only one whacko that is sane" → the sane‑and‑whacko overlap = 1 (and since that person is sane, they're already a cacher).

Inclusion‑exclusion on the cachers:
cachers that are sane or whacko = 30 + 8 − 1 = 37
cachers that are neither = 90 − 37 = 53

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