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On a certain train, children’s tickets cost $6 and adult tic
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Updated on: 20 Jul 2018, 08:36
Updated on: 20 Jul 2018, 08:36
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Question Stats:
67% (01:36) correct
32% (01:45) wrong based on 180 sessions
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On a certain train, tickets cost $6 each for children and $9 each for adults. The total train ticket cost for a certain group of six passengers was between $44 and $50.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The number of children in the group |
The number of adults in the group |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: On a certain train, children’s tickets cost $6 and adult tic
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10 Oct 2018, 03:53
10 Oct 2018, 03:53
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Let the number of children = c and the number of adults = a
6*c + 9*a = (44 to 50)
c + a = 6
c = 6 - a
6*(6 - a) + 9*a = (44 to 50)
3*a = (8 to 14) then a can be 3 or 4 so there are two cases [one where a = 3 equal c = 3, while the other (a = 4) > (c = 2)]
So the correct answer is D)The relationship cannot be determined from the information given.
6*c + 9*a = (44 to 50)
c + a = 6
c = 6 - a
6*(6 - a) + 9*a = (44 to 50)
3*a = (8 to 14) then a can be 3 or 4 so there are two cases [one where a = 3 equal c = 3, while the other (a = 4) > (c = 2)]
So the correct answer is D)The relationship cannot be determined from the information given.
General Discussion
Re: On a certain train, children’s tickets cost $6 and adult tic
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28 May 2018, 13:47
28 May 2018, 13:47
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Explanation
Even though the range of costs ($44 to $50) is fairly small, there is still more than one possibility. A good way to work this out is to start with the simplest scenario: 3 adults and 3 children. Their tickets would cost 3(9) + 3(6) = $45. That’s in the range, so it’s one possibility.
Since children’s tickets are cheaper, you don’t want to add more children to the mix (4 children, 2 adults will give you too small a total), but try switching out 1 child for 1 adult.
For 4 adults and 2 children, tickets would cost 4(9) + 2(6) = $48. Thus, Quantity A and Quantity B could be equal, or Quantity B could be greater, so the relationship cannot be determined from the information given.
Hence option D is correct!
Even though the range of costs ($44 to $50) is fairly small, there is still more than one possibility. A good way to work this out is to start with the simplest scenario: 3 adults and 3 children. Their tickets would cost 3(9) + 3(6) = $45. That’s in the range, so it’s one possibility.
Since children’s tickets are cheaper, you don’t want to add more children to the mix (4 children, 2 adults will give you too small a total), but try switching out 1 child for 1 adult.
For 4 adults and 2 children, tickets would cost 4(9) + 2(6) = $48. Thus, Quantity A and Quantity B could be equal, or Quantity B could be greater, so the relationship cannot be determined from the information given.
Hence option D is correct!
