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

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A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 06 Aug 2008, 17:44
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72% (03:14) correct 28% (50:46) wrong based on 68 sessions
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?

A. 3
B. 5
C. 6
D. 7
E. 8
[Reveal] Spoiler: OA
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Re: A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 06 Aug 2008, 18:08
Assuming that there are no 2 headed cats/sailors (like in recent news).

6 cats and 7 sailors

take out the captain and the cook, you have 13 heads and 38 legs

sailor + cat = 13 --> S = 13 -C

2 legs for each sailor and 4 legs for each cat.

2S + 4C = 38

2(13 - C) + 4C = 38
2C = 12
C = 6
S = 7
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Re: A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 06 Aug 2008, 18:31
GMAT_700 wrote:
If you know how to solve the following question and you know how to explain your answer, please let me know. By the way, I already found some explanations for this question, but I did not understand their explanations.

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?
a)3
b)5
c)6
d)7
e)8

The correct answer is 6.


Q + cats + Sailors + a cook + a captain = 15
1 + cats + Sailors + 1 + 1 = 15
cats + Sailors = 12

2Q + 4(cats) + 2(Sailors) + 2 (a cook) + a captain = 41
2 + 4(cats) + 2(Sailors) + 2 + 1 = 41
4(cats) + 2(Sailors) = 36
2(cats) + (cats) + (Sailors) + (cats) + (Sailors) = 36
2(cats) + 12 + 12 = 36
cats + = (36 - 24)/2 = 12

So Cats = 6
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Re: A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 07 Aug 2008, 04:03
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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
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Re: A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 07 Aug 2008, 06:31
GMAT_700 wrote:
If you know how to solve the following question and you know how to explain your answer, please let me know. By the way, I already found some explanations for this question, but I did not understand their explanations.

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?
a)3
b)5
c)6
d)7
e)8

The correct answer is 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
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Re: A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 07 Aug 2008, 16:48
x2suresh wrote:
GMAT_700 wrote:
If you know how to solve the following question and you know how to explain your answer, please let me know. By the way, I already found some explanations for this question, but I did not understand their explanations.

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?
a)3
b)5
c)6
d)7
e)8

The correct answer is 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
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Re: A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 25 Dec 2014, 11:53
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Re: A luxury liner, Queen Marry II, is transporting several cats as well [#permalink] New post 25 Dec 2014, 21:38
Expert's post
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:
[Reveal] Spoiler:
C


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Re: A luxury liner, Queen Marry II, is transporting several cats as well   [#permalink] 25 Dec 2014, 21:38
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