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A luxury liner, Queen Marry II, is transporting several cats as well as the crew (sailors, a cook, and one-legged captain) to a nearby port. Altogether, these passengers have 15 heads and 41 legs. How many cats does the ship host?

Total number of heads and legs on the ships is 15 and 41 respectively. Since we know that captain and cook together have 2 heads and 3 legs, then cats and sailors would have together 13 heads and 38 legs. From this information we can construct system of equalities, where \(S\) is number of sailors and \(C\) - cats. \(\left{ \begin{eqnarray*} 4C + 2S &=& 38\\ C + S &=& 13\\ \end{eqnarray*}\) \(\begin{eqnarray*} 4C + 2(13-C) &=& 38\\ 2C = 38 - 26 &=& 12\\ C &=& 6\\ \end{eqnarray*}\)

Therefore, the ship has 6 cats on board.

On actual test it might be easier to construct the system of equations and plug the answer choices into it. The correct answer is C.

-------------------------------------------------------------------------------------- How do we know that there are 4C + 2S?? Like where is the 4 cats coming from?

4C is not 4 cats here... it means 4 * No of cats because each Cat will have 4 legs and similarly 2S means 2 * No of sailors because each sailor will have 2 legs... I hope this cleared your doubt...

A faster way of tackling this question would be to consider the cook as a sailor, since no feature quantitatively distinguishes a cook from a sailor. If the question asked for number of sailors, that would be a different matter, but because it only asks for cats, the following solution is easier:

One-legged captain= 1 leg and 1 head

Remaining cats (C) and people (P) have 14 heads and 40 legs.

Equation 1 (heads): C + P = 14 Equation 2 (legs): 4C + 2P = 40

Let all x cats stand on only two legs and count captain's wooden leg as a leg. Then total number of legs on the floor will be 15x2 =30. The remaining (41+1) – 30 =12 legs are in the air and belong to the cats (2 legs per each). 2x = 12 and x=6. Answer. is C.

"A ship is transporting several cats and a crew (sailors, a cook, and a one-legged captain) to a nearby port. If these passengers combined have 15 heads and 41 legs, then how many cats is the ship transporting?"

Could this problem -- or a problem like it -- be solved effectively in this way:

c= cats, p= people

Two equations: First: c + p = 15

Second: 4c + 2P = 42 (42, because if Captain Barbossa had had his other leg, the Black Pearl's passengers would have had a total of 15 heads and 42 legs.)

I think a quicker way to do this is to use a most likely answer from the options given so that it satisfies the no. of heads and legs. i.e 13 H and 38L