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A luxury liner, Queen Marry II, is transporting several cats as well as the crew (sailors, a cook, and onelegged 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 Source: GMAT Club Tests  hardest GMAT questions 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. \(\left{ \begin{eqnarray*} 4C + 2S &=& 38\\ C + S &=& 13\\ \end{eqnarray*}\) \(\begin{eqnarray*} 4C + 2(13C) &=& 38\\ 2C = 38  26 &=& 12\\ C &=& 6\\ \end{eqnarray*}\) 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? Thanks JF



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Re: M01. Q.7 [#permalink]
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22 Feb 2009, 10:28
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As cat has 4 legs , and sailor has 2 legs. Total legs = 4C +2S = ( 413) = 38.
Hope this helps



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Re: M01. Q.7 [#permalink]
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23 Feb 2009, 08:40
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...



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Re: M01 Q07 [#permalink]
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30 Aug 2009, 11:15
Number of Cats = C Number of Sailors = S
Since the cook and the captain together have 2 heads and 3 legs
Eq 1) Total heads = 15 C + S = 15  2 = 13
Eq 2) Total Legs = 41 4C + 2S = 41  3 = 38
Solving Eq 1 and Eq 2, we get, C = 6.
Hence No. of Cats = 6.



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Re: M01 Q07 [#permalink]
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01 Oct 2009, 20:58
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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:
Onelegged 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
Solving equations 1 and 2 for C gives C=6.



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Re: M01 Q07 [#permalink]
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10 Feb 2010, 06:33
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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.



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Re: M01 Q07 [#permalink]
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11 Feb 2010, 11:56
It can be solved by forming equation as explained above and the ans is 6
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Re: M01 Q07 [#permalink]
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29 Jul 2010, 19:38
"A ship is transporting several cats and a crew (sailors, a cook, and a onelegged 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.)
Then, p = 15c
Thus, 4c + 2(15c) =42 ======> 4c + 30 2c=42 =====>, 2c=12, and c=6
Would that kind of thinking get me into trouble on test day?



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Re: M01 Q07 [#permalink]
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15 Feb 2011, 06:37
c*4 + 2 * S + 2 + 1 = 41 (S > # of sailors) C + S + 2 = 15 4c + 2s = 38 c + s = 13 2c = 3826 = 12 => c = 12/2 = 6 So the answer is C.
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Re: M01 Q07 [#permalink]
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15 Feb 2011, 08:41
lets take the middle number from the answer i.e 6 cats. then cats leg = 24. 41  24 =17 captain has one leg= 1 total persons = 16/2 =8 persons
so it is 8+1+6=15.
so the answer is 6 cats.



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Re: M01 Q07 [#permalink]
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16 Feb 2011, 01:29
ans:c cat+crew=14 4cat+2crew=40 by solving this two equation No of cat=6
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Re: M01 Q07 [#permalink]
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21 Feb 2011, 06:49
I tried to do it by myself before even looking at the answer. I am not sure if my way is right too:
The total number of heads and legs is: 56
A cat comprises 1 head and 4 legs equals 5 A human being comprises 1 head and 2 legs equal 3
Total number of heads and legs of one cat and one human being is 8.
8*?=56 8*7=56
7. As the question says that there is one legged captain so deduct 71=6 is the answer.
Plz let me know if this way also right without forming an equation.



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Re: M01 Q07 [#permalink]
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17 Feb 2012, 06:37
Finally an easy one!



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Re: M01 Q07 [#permalink]
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17 Feb 2012, 21:06
What level is this question considered? I found the two equations pretty quickly and was able to solve within the time limit.



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Re: M01 Q07 [#permalink]
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17 Feb 2012, 22:04
very easy.
just subtract the captain from the first equation c+r=15 (151captain = 14) 4c+2r=41 (411 leg = 40)
c = 6.



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Re: M01 Q07 [#permalink]
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19 Feb 2012, 23:57
Ans is C. Lets say x = no of cats and y = no of sailors
4x + 2y + 2(Cook's legs) + 1 (Captain's legs) = 41 4x + 2y = 38 ...(1)x + y + 1 (Cook's head) + 1 (Captain's head) = 15x + y = 13 ...(2)Solving equation 1 and 2 we get x = 6(no. of cats head) and y = 7
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Re: M01 Q07 [#permalink]
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21 Feb 2012, 05:23
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



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Re: M01 Q07 [#permalink]
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22 Feb 2012, 11:50
Thanks for the alternate approach. I was totally confused.



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Re: M01 Q07 [#permalink]
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26 Feb 2013, 22:32
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One Captain (1 legged), 1 cook, S sailors and C cats.
For legs: 4 (legs) * C + 2 * S + 1 (Captain) + 2 (cook) = 41 legs
4C + 2S = 38......eqn 1
For heads:
C +S + 1 (Captain) + 1 (cook) = 15 Heads
C+S = 13......eqn 2
From Eqn 1 and 2...
C=6...
Answer: C



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Re: M01 Q07 [#permalink]
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12 Mar 2013, 05:01
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Let no. of cats = x Let no. of crew except one captain = y
The equations can be set up as
4x+2y+1=41 and x+y+1=15
solving for x we get x=6 hence correct ans = c







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