A luxury liner, Queen Marry II, is transporting several cats as well

Easier approach

head count...
cats + humans without captain = 14 (since only one captain is there)
let c= #of cats, h=#humans (non captain)

c+h=14

leg count.
4(c) + 2(h) = 40. Captain had only one leg
so 2c+h=20

solve to get c=6

heads 1 (cook) 1(captain) x(sailors) y(cats)
legs 2 (cook) 1(captain) 2x(sailors) 4 y(cats)

1+1+x+y =15 (total heads)
2+1+2x+4y= 41

solve above y=6

solved and got the same

Hi All,

The other explanations properly explain the algebra approach to this question, but the question can also be solved by TESTing THE ANSWERS.

**Note: the question probably means for us to assume that the cats each have 4 legs and each of the sailors and cook have 2 legs each**

We're told that there are 15 heads and 41 legs, so we're looking for an answer that makes THAT result happen.

Let's TEST...

Answer B: 5 cats

5 cats
1 captain
1 cook
8 sailors (because there are total heads)

5 cats = 20 legs
1 captain = 1 leg
1 cook = 2 legs
8 sailors = 16 legs
Total legs = 39 legs
This is TOO SMALL (there are supposed to be 41 legs).

Since we need 2 more legs, we can get that by replacing 1 sailor with 1 cat.

Answer C: 6 cats

6 cats = 24 legs
1 captain = 1 leg
1 cook = 2 legs
7 sailors = 14 legs
Total legs = 41 legs
This is a MATCH for what we're looking for, so this MUST be the answer!

Final Answer:

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Rich

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