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On a certain transatlantic crossing, 20 percent of a ship's passengers
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09 Mar 2014, 14:17
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers
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09 Mar 2014, 14:17
SOLUTIONOn a certain transatlantic crossing, 20 percent of a ship's passengers held roundtrip tickets and also took their cars aboard the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship, what percent of the ship's passengers held roundtrip 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 roundtrip tickets AND cars. As 60% of the passengers with roundtrip tickets did not take their cars then 40% of the passengers with roundtrip tickets did take their cars, so # of passengers with roundtrip tickets AND cars is 40% of the passengers with roundtrip 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
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09 Mar 2014, 17:18
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
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09 Mar 2014, 18:59
Answer = (C) 50% Lets say total passengers = 100 Passengers with round ticket & cars = 20 If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship which means 40% with round trip tickets took cars40x/100 = 20 x = 50 Answer = 50%
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers
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09 Mar 2014, 19:14
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
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09 Mar 2014, 20:09
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
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09 Mar 2014, 20:37
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) Time Taken 2:00



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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers
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09 Mar 2014, 23:28
Option C. If 60% of passengers with roundtrip do not have cars,40% of passengers with roundtrip 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
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15 Jun 2015, 10:07
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
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05 Jun 2016, 12:59
pepo wrote: On a certain boat, 20 % of the passengers have roundtrip tickets and are trasporting cars. if 60% of the passengers who have roundtrip tickets are not transporting cars, what percent of all the passengers have roundtrip ticketsOn a certain boat, 20 % of the passengers have roundtrip tickets and are trasporting cars. if 60% of the passengers who have roundtrip tickets are not transporting cars, what percent of all the passengers have roundtrip tickets?
A 40% B 50% C 60% D 70% E 80% Let's suppose total customers are 100 and customers who have roundtrip tickets are r if 60% of the passengers who have roundtrip 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 roundtrip 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
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10 Aug 2016, 11:37
I have a slight difficulty in understanding the question: 1.On a certain transatlantic crossing, 20 percent of a ship's passengers held roundtrip tickets and also took their cars aboard the ship. 2.If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship, what percent of the ship's passengers held roundtrip 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
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04 Apr 2018, 03:55
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionOn a certain transatlantic crossing, 20 percent of a ship's passengers held roundtrip tickets and also took their cars aboard the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship, what percent of the ship's passengers held roundtrip 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 ProjectEach 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 x20 into 20x
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers
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04 Apr 2018, 05:55
dave13 wrote: Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionOn a certain transatlantic crossing, 20 percent of a ship's passengers held roundtrip tickets and also took their cars aboard the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship, what percent of the ship's passengers held roundtrip 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 ProjectEach 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 x20 into 20x Hi dave13Let the ship's passengers who hold roundtrip tickets be x P(Hold roundtrip tickets) = x We are told that 60% of this number did not take their cars P(Only hold roundtrip tickets) = 0.6x Also, P(Both) = 20%. If total number of people is 100, P(Both) = 20 P(Hold roundtrip ticket) = P(Only hold roundtrip 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
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04 Apr 2018, 11:29
pushpitkc wrote: dave13 wrote: Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionOn a certain transatlantic crossing, 20 percent of a ship's passengers held roundtrip tickets and also took their cars aboard the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship, what percent of the ship's passengers held roundtrip 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 ProjectEach 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 x20 into 20x Hi dave13Let the ship's passengers who hold roundtrip tickets be x P(Hold roundtrip tickets) = x We are told that 60% of this number did not take their cars P(Only hold roundtrip tickets) = 0.6x Also, P(Both) = 20%. If total number of people is 100, P(Both) = 20 P(Hold roundtrip ticket) = P(Only hold roundtrip 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
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04 Apr 2018, 22:35
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 dave13Please find the corrected diagram Attachment:
Diagram.png [ 4.61 KiB  Viewed 2906 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 roundtrip tickets(which we are assuming to be x) 60% of the passengers who hold roundtrip tickets do not take their cars. Therefore, the total number of passengers who only have roundtrip tickets are 60% of x or \(\frac{60}{100}*x = 0.6x\) In the diagram, it is clear why P(Hold roundtrip ticket) = P(Only hold roundtrip 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
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06 Apr 2018, 07:46
Bunuel wrote: On a certain transatlantic crossing, 20 percent of a ship's passengers held roundtrip tickets and also took their cars aboard the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship, what percent of the ship's passengers held roundtrip 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 roundtrip 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 roundtrip tickets. If n = the number of passengers with roundtrip tickets, we have: 20 = 0.4n n = 20/0.4 = 200/4 = 50 Thus, 50/100 = 50% of the ship’s passengers held roundtrip tickets. Answer: C
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