Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
23 Nov 2010, 11:38

2

This post received KUDOS

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

72% (02:27) correct
28% (02:20) wrong based on 169 sessions

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?

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%

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.

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
23 Nov 2011, 20:18

1

This post received KUDOS

Expert's post

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. _________________

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
02 Jan 2012, 12:13

Fairly easy question this. Easy to get to 50%, the correct answer. 60% of the people who have not taken the car also gives us 40% of the people who have taken the car (people with round trip tickets btw). From there, it is easy to get to 50%. _________________

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
09 Jan 2012, 14:15

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% _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
01 May 2013, 02:37

Another method?

Take the total number of passengers to be 100. therefore those with both round trip tickets and cars is 20, leaving 80 passengers with either round trip ticket only and cars only (the question is silent on the number of passengers who have neither round trip tickets nor cars, so assuming that to be zero). let number of passengers with round trip tickets be - x therefore number of passengers with cars is - 0.6x hence 80= x + 0.6x x=50%

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
25 May 2013, 11:20

can anyone please clarify this through a Venn diagram. I understand that this is a percent of a percent problem. I just need to picture it. Thanks _________________

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
27 May 2013, 06:24

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.

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
27 May 2013, 08:20

Expert's post

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. _________________

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
27 Sep 2013, 10:07

Let total no of passengers be T Let the no of passengers with round trip tickets be X

Given: 20% of total passengers have round trip tickets and took their cars aboard = 0.2*T

60% of passengers with round trip tickets did not take their cars => 40% of passengers with round trip tickets took their cars => 0.4*X=0.2*T => X=1/2*T => X=50% of T

Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
31 Jul 2014, 00:41

azule45 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%

a = total passenger b = passenger owning R ticket ( including car (C) and no car (NC))

passenger owning R ticket and C = 20%*a passenger owning R ticket and NC = 60%*b

so, b = 20%a + 60%b -> 40%b = 20%a -> b/a = 1/2 -> b = 50% _________________

......................................................................... +1 Kudos please, if you like my post