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

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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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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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Re: 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 09 Mar 2014, 18:18
Assume total passengers = 100,

So people with round ticket and having cars = 20

These 20 people = 40% of total round ticket holders. So total round ticket holders = 60
% of Round Ticket holders = 60/100 = 60%
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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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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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

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New post 09 Mar 2014, 21:09
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Total Passengers = 100
Both RT and Cars = 20

RT*(1-.6) = 20
RT = 50
RT/Passengers = 50/100

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 09 Mar 2014, 21:37
As there are only percents involve here and no numbers, we can assume some comfortable numbers.

Total = 100
Passenger with round trip+ cars = 20% of Total = 20
Passenger with round trip+no car = 60% of round trip

Hence,
20 = 100% of round trip - 60% of round trip = 40% of round trip
Therefore,
roundtrip = 50
which is 50% of Total

Answer C
Difficulty level 650 (It took me complete 2 mins which I think makes this problem lengthy)
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers  [#permalink]

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New post 10 Mar 2014, 00:28
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Option C.
If 60% of passengers with round-trip do not have cars,40% of passengers with round-trip tickets will have their cars on board.
40% of total=20
=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 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 05 Jun 2016, 13:59
pepo wrote:
On a certain boat, 20 % of the passengers have round-trip tickets and are trasporting cars. if 60% of the passengers who have round-trip tickets are not transporting cars, what percent of all the passengers have round-trip ticketsOn a certain boat, 20 % of the passengers have round-trip tickets and are trasporting cars. if 60% of the passengers who have round-trip tickets are not transporting cars, what percent of all the passengers have round-trip tickets?

A 40%
B 50%
C 60%
D 70%
E 80%


Let's suppose total customers are 100 and customers who have round-trip tickets are r


if 60% of the passengers who have round-trip tickets are not transporting cars or 40% who have round trips are transporting cars.

i.e. .6r are not transporting cars and .4r are transporting cars.

Its given that 20 % of the passengers have round-trip tickets and are transporting cars (20 out of 100)

.4r=20
r=50%

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

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New post 10 Aug 2016, 12:37
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I have a slight difficulty in understanding the question:
1.On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship.
2.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?

>> So 2nd statement means that 40% have Round trip ticket and also took car.

But in the first sentence it said that percentage of such people who have round trip ticket and brought their cars is 20% ?

How did you guys understand this question ?
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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
1
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
1
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 1716 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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