HS Thomas

Ranch Hand

Posts: 3404

posted 13 years ago

A: The Chelsea Pensioners (Lewis Carroll)

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If 70% have lost an eye, 75% an ear, 80% an arm and 85% a leg, what percentage at least must have lost all four?

B: What mathematical symbol can be put between 2 and 3 to make a number greater than two but less than three?

regards

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If 70% have lost an eye, 75% an ear, 80% an arm and 85% a leg, what percentage at least must have lost all four?

B: What mathematical symbol can be put between 2 and 3 to make a number greater than two but less than three?

regards

Mark Herschberg

Sheriff

Posts: 6037

posted 13 years ago

I think the answer to the first (morbid) problem is 10%.

Suppose there are 100 people and 70 have lost an eye. Then we consider the 75 who have lost an ear. To get the minimum set, we consider that all 25 who did not lose an ear come out of the set of people who lost an eye. Hence, the minimum set who lost both an eye and an ear is 45 people.

We use similar logic for the 80 folks who lost an arm. We assume that the 20 who were spared all come from the set of 45, reducing the minimum number to get hit with all three tragedies to 25.

Finally, we assume the 15 who did not lose a leg belong to the unfortunate 25. This leaves 10 people who suffered all four losses.

--Mark

Suppose there are 100 people and 70 have lost an eye. Then we consider the 75 who have lost an ear. To get the minimum set, we consider that all 25 who did not lose an ear come out of the set of people who lost an eye. Hence, the minimum set who lost both an eye and an ear is 45 people.

We use similar logic for the 80 folks who lost an arm. We assume that the 20 who were spared all come from the set of 45, reducing the minimum number to get hit with all three tragedies to 25.

Finally, we assume the 15 who did not lose a leg belong to the unfortunate 25. This leaves 10 people who suffered all four losses.

--Mark

HS Thomas

Ranch Hand

Posts: 3404

Vivek Kumar

Greenhorn

Posts: 9

posted 13 years ago

A simpler analogy would be, let the 4 losses are termed as A, B, C and D.

A+B+C+D = 310

To minimize number of peple getting all 4, distribute 3 losses randomly among all, so if originally there are 100 persons, even after all 100 get 3 losses each, we have 10 losses to account for. So atleast 10 people have all 4 losses.

A+B+C+D = 310

To minimize number of peple getting all 4, distribute 3 losses randomly among all, so if originally there are 100 persons, even after all 100 get 3 losses each, we have 10 losses to account for. So atleast 10 people have all 4 losses.

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