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# M01 Q07

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21 Feb 2009, 23:35
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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?

(A) 3
(B) 5
(C) 6
(D) 7
(E) 8

[Reveal] Spoiler: OA
C

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.
$$\left{ \begin{eqnarray*} 4C + 2S &=& 38\\ C + S &=& 13\\ \end{eqnarray*}$$
$$\begin{eqnarray*} 4C + 2(13-C) &=& 38\\ 2C = 38 - 26 &=& 12\\ C &=& 6\\ \end{eqnarray*}$$

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?

Thanks
JF
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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:

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

Solving equations 1 and 2 for C gives C=6.
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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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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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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

C +S + 1 (Captain) + 1 (cook) = 15 Heads

C+S = 13......eqn 2

From Eqn 1 and 2...

C=6...

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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 = ( 41-3) = 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...
Manager
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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.
Manager
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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

Posted from my mobile device
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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 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.)

Then,
p = 15-c

Thus,
4c + 2(15-c) =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 = 38-26 = 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 7-1=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!
Manager
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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 (15-1captain = 14)
4c+2r=41 (41-1 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) = 15
x + y = 13 ...(2)

Solving equation 1 and 2 we get x = 6(no. of cats head) and y = 7
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perfect plan executed next week.’

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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.

Re: M01 Q07   [#permalink] 22 Feb 2012, 11:50

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# M01 Q07

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