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On a certain transatlantic crossing, 20 percent of a ship's passengers

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, what percent of the ship's passengers held round-trip tickets?

(A) 33 1/3%
(B) 40%
(C) 50%
(D) 60%
(E) 66 2/3%

Problem Solving
Question: 146
Category: Arithmetic Percents
Page: 81
Difficulty: 600

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Originally posted by Bunuel on 20 Jun 2008, 00:16.
Last edited by Bunuel on 01 Feb 2019, 01:17, edited 2 times in total.
Edited the question.
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On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 09 Mar 2014, 15:17
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SOLUTION

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, what percent of the ship's passengers held round-trip tickets?

(A) 33 1/3%
(B) 40%
(C) 50%
(D) 60%
(E) 66 2/3%

Let the total # of passengers be 100.

Now, 20 passengers held round-trip tickets AND cars.

As 60% of the passengers with round-trip tickets did not take their cars then 40% of the passengers with round-trip tickets did take their cars, so # of passengers with round-trip tickets AND cars is 40% of the passengers with round-trip tickets.

If we take # of the passengers with round trip tickets to be \(x\) then we'll have \(0.4x=20\) --> \(x=50\).

Answer: C.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 20 Jun 2008, 13:39
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ritula wrote:
On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip tickets?
A. 33 1/3%
B. 40%
C. 50%
D. 60%
E. 66 2/3%


C
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 20 Jun 2008, 00:28
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On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip tickets?

Solution: Let total number of passengers be 100
According to Q stem 40% of passengers who had round-trip tics have taken cars - let number of passengers with round trip be X then

40% of X = 20 => X= 50.

Thus 50% of passengers have round-trip tics with them!!
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On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 23 Nov 2010, 13:04
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 21 Nov 2011, 09:56
Let round ticket = x
so,
0.60x+0.20 = x
x = 50%
Ans. C

Please help if i am not correct.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 23 Nov 2011, 21:18
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Baten80 wrote:
Let round ticket = x
so,
0.60x+0.20 = x
x = 50%
Ans. C

Please help if i am not correct.


If the thought process was that x is the fraction of total passengers who have the round ticket and total passengers = 1 (theoretically since we need percentages), then the logic is sound.

Otherwise, note that 0.20 on its own doesn't mean anything. It needs to be 0.20p where p is the total number of passengers.
0.60x + 0.20p = x
x/p = .50 = 50%

or you can say that since 60% people with round trip ticket did not take their car, 40% people with round trip ticket did take their car.
40% people with round trip ticket = 20% total people
So people with round trip ticket are 1/2 (i.e.50%) of total people.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 09 Jan 2012, 15:15
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You can use a table approach as recommended by MGMAT.

-----R---nR---Total
C---20-------------
nC--0.6x-----------
Total x---------100

We get, 0.4x = 20 -> x = 50%
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 27 May 2013, 07:24
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i might sound a little silly with my query but if one looks at the question and tries to translate it seems some what like this:

1)20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship which means that 20% held roundtrip tickets and took their cars aboard the ship
2) 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship which means that another 60% held round trip tickets but didn't take their cars abroad. so aren't they like 2 different groups
i dunno but i am so confused.
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New post 27 May 2013, 09:20
mohnish104 wrote:
i might sound a little silly with my query but if one looks at the question and tries to translate it seems some what like this:

1)20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship which means that 20% held roundtrip tickets and took their cars aboard the ship
2) 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship which means that another 60% held round trip tickets but didn't take their cars abroad. so aren't they like 2 different groups
i dunno but i am so confused.


Read the stem carefully: "60 percent of the passengers with round-trip tickets did not take their cars abroad the ship..." So, 60% of some particular group did not take their cars abroad the ship.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 09 Mar 2014, 19:59
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Answer = (C) 50%

Lets say total passengers = 100

Passengers with round ticket & cars = 20

If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship which means 40% with round trip tickets took cars

40x/100 = 20

x = 50

Answer = 50%
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New post 09 Mar 2014, 20:14
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20% of the passengers were RT passengers with cars aboard. 60% of RT passengers did not bring cars. Thus, 20% of all passengers corresponds to 40% of RT passengers.

r = RT
t = total

2r/5 = t/5

r = 5t/10

= t/2

= 50%

OR 40% = 2(20%). 10% of all passengers are 20% of RT, so 100% of RT are 50% of passengers.
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New post 20 May 2014, 09:38
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I think this is a poorly worded problem:
They should make the distinctions in the second sentence more clearly that the 60% is only of the passangers WITH the round-trip tickets would have been more obvious. Upon reading the problem, I ended up with 80%.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 15 Jun 2015, 11:07
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We can solve this problem using Double matrix. Unknown NUmber is here the number of pasengers with round trip tickets - let it be X, then 60% of X = 0,6X
Let the Total be 100, then people with Round Trip Tickets and Cars = 20% of 100 = 20

See the attached screenshot and then solve.

20 + 0,6X = X
X = 50
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 04 Apr 2018, 04:55
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, what percent of the ship's passengers held round-trip tickets?

(A) 33 1/3%
(B) 40%
(C) 50%
(D) 60%
(E) 66 2/3%

Problem Solving
Question: 146
Category: Arithmetic Percents
Page: 81
Difficulty: 600


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generis, pushpitkc,

do you guys have any tips for overlapping sets ? i need some methodologic approach to solve these type of questions :)

i dont understand whats wrong with my solution :-)

the only thing I would change in my diagram x-20 into 20-x :)
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 04 Apr 2018, 06:55
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dave13 wrote:
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, what percent of the ship's passengers held round-trip tickets?

(A) 33 1/3%
(B) 40%
(C) 50%
(D) 60%
(E) 66 2/3%

Problem Solving
Question: 146
Category: Arithmetic Percents
Page: 81
Difficulty: 600


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!


generis, pushpitkc,

do you guys have any tips for overlapping sets ? i need some methodologic approach to solve these type of questions :)

i dont understand whats wrong with my solution :-)

the only thing I would change in my diagram x-20 into 20-x :)


Hi dave13

Let the ship's passengers who hold round-trip tickets be x
P(Hold round-trip tickets) = x

We are told that 60% of this number did not take their cars
P(Only hold round-trip tickets) = 0.6x
Also, P(Both) = 20%.

If total number of people is 100, P(Both) = 20
P(Hold round-trip ticket) = P(Only hold round-trip tickets) + P(Both)

\(x = 0.6x + 20\) -> \(x - 0.6x = 20\) -> \(0.4x = 20\) -> \(x = \frac{20}{0.4} = 50\)
Therefore, the percentage of ship's passengers who hold round trip tickets is \(\frac{50}{100} = 50\)% (Option C)

Hope this helps you.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 04 Apr 2018, 12:29
pushpitkc wrote:
dave13 wrote:
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, what percent of the ship's passengers held round-trip tickets?

(A) 33 1/3%
(B) 40%
(C) 50%
(D) 60%
(E) 66 2/3%

Problem Solving
Question: 146
Category: Arithmetic Percents
Page: 81
Difficulty: 600


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!


generis, pushpitkc,

do you guys have any tips for overlapping sets ? i need some methodologic approach to solve these type of questions :)

i dont understand whats wrong with my solution :-)

the only thing I would change in my diagram x-20 into 20-x :)


Hi dave13

Let the ship's passengers who hold round-trip tickets be x
P(Hold round-trip tickets) = x

We are told that 60% of this number did not take their cars
P(Only hold round-trip tickets) = 0.6x
Also, P(Both) = 20%.

If total number of people is 100, P(Both) = 20
P(Hold round-trip ticket) = P(Only hold round-trip tickets) + P(Both)

\(x = 0.6x + 20\) -> \(x - 0.6x = 20\) -> \(0.4x = 20\) -> \(x = \frac{20}{0.4} = 50\)
Therefore, the percentage of ship's passengers who hold round trip tickets is \(\frac{50}{100} = 50\)% (Option C)

Hope this helps you.


hi pushpitkc, thanks!

here \(x = 0.6x + 20\) why do you denote 20 as number and not percent ? 60 and 20 are both percentages ?

do you mind correcting my diagram so as i can see visually what i did wrong ? :)
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 04 Apr 2018, 23:35
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dave13 wrote:

hi pushpitkc, thanks!

here \(x = 0.6x + 20\) why do you denote 20 as number and not percent ? 60 and 20 are both percentages ?

do you mind correcting my diagram so as i can see visually what i did wrong ? :)


Hey dave13

Please find the corrected diagram
Attachment:
Diagram.png
Diagram.png [ 4.61 KiB | Viewed 8793 times ]


Here, we are assuming the total number of passengers to be 100.
However, we have no clarity about the total number of passengers with round-trip tickets(which we are assuming to be x)

60% of the passengers who hold round-trip tickets do not take their cars.
Therefore, the total number of passengers who only have round-trip tickets are 60% of x or \(\frac{60}{100}*x = 0.6x\)

In the diagram, it is clear why P(Hold round-trip ticket) = P(Only hold round-trip tickets) + P(Both)

So, we arrive at the equation x=0.6x+20

Hope that it's clear now!
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 06 Apr 2018, 08:46
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Bunuel wrote:

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, what percent of the ship's passengers held round-trip tickets?

(A) 33 1/3%
(B) 40%
(C) 50%
(D) 60%
(E) 66 2/3%


We can let the number of passengers = 100.

Thus, 20 passengers held round trip tickets and took their cars aboard the ship. Since we are given that 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, the 20 passengers who held round trip tickets and took their cars aboard the ship represent 40 percent of the passengers with round-trip tickets. If n = the number of passengers with round-trip tickets, we have:

20 = 0.4n

n = 20/0.4 = 200/4 = 50

Thus, 50/100 = 50% of the ship’s passengers held round-trip tickets.

Answer: C
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers   [#permalink] 06 Apr 2018, 08:46

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