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

Question

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%

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

Bunuel · Accepted Answer

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%

(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.

brokerbevo · Answer

C

Bunuel · Answer

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iamcartic · Answer

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%

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!!

MHIKER · Answer

Let round ticket = x
so,
0.60x+0.20 = x
x = 50% 
Ans. C

Please help if i am not correct.

KarishmaB · Answer

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.

shinbhu · Answer

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%

mohnish104 · Answer

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.

Bunuel · Answer

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.

PareshGmat · Answer

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%

CCMBA · Answer

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.

tedchou12 · Answer

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%.

Bunuel · Answer

The question clearly says "60 percent of the passengers   with round-trip tickets  ..."

BrainLab · Answer

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

JeffTargetTestPrep · Answer

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

ManishaPrepMinds · Answer

https://drive.google.com/file/d/17G_dASjQrje5KiP3kO6h0W88usiZq2fa/view?usp=sharing  https://drive.google.com/file/d/17G_dASjQrje5KiP3kO6h0W88usiZq2fa/view?usp=sharing Let's solve the problem using a grid or table method to clearly visualize the relationships between the different groups of passengers.
 Given Information Recap: 
 
 20% of the ship's passengers had round-trip tickets and took their cars aboard the ship. 
 60% of the passengers with round-trip tickets did not take their cars aboard.  
 Using the Grid Method: 
Let's use a grid to organize this information. We'll divide the passengers into groups based on:
 
 Whether they have round-trip tickets. 
 Whether they took their cars aboard.  
 Let's Define: 
 
 Let the  total number of passengers  be 100. 
 Let the  number of passengers with round-trip tickets  be R.  
 Step-by-Step Solution: 
 
  Passengers with Round-Trip Tickets and Cars: 
 
 20% of the total passengers took their cars aboard and held round-trip tickets. 
 This is given by 20   
  Passengers with Round-Trip Tickets but No Cars: 
 
 60% of the passengers with round-trip tickets did not take their cars aboard. 
 Therefore, the remaining 40% (100% - 60%) of the passengers with round-trip tickets did take their cars aboard. 
 This means that 0.4R=20 because 20% of the total passengers took their cars aboard and held round-trip tickets.   
 Solve for R:
0.4R=20 => R =50

50% of the total number of passengers.   
  Conclusion: 
50% of the ship's passengers held round-trip tickets.

Correct Answer:  (C) 50% .

.

totaltestprepNick · Answer

C https://www.youtube.com/watch?v=GzF9rbpJXl4 Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience tutoring@totaltestprep.net

Usernamevisible · Answer

Take 100 passengers.

Round-trip + car = 20

60% of round-trip passengers did NOT take cars → 40% DID take cars.

So, 40% of all round-trip passengers = 20.

Total round-trip passengers = 20 / 0.40 = 50

Therefore, 50% of passengers held round-trip tickets.

Answer: (C) 50%

Important: The 20% represents only the 40% of round-trip passengers who took cars, not all round-trip passengers.

PGTLrowanhand · Answer

Here's my video solution:  https://www.youtube.com/watch?v=On-a1o1W-wI

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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