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A train takes 15 seconds to cross a bridge at 50 mph, and at
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30 Apr 2019, 01:26
30 Apr 2019, 01:26
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Question Stats:
45% (01:43) correct
54% (03:36) wrong based on 70 sessions
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A train takes 15 seconds to cross a bridge at 50 mph, and at the same speed takes 10 seconds to cross the same bridge when the train's length is halved.
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 |
Length of the bridge |
Original length of the train |
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: A train takes 15 seconds to cross a bridge at 50 mph, and at
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02 May 2019, 11:08
02 May 2019, 11:08
2
Solutions for this problem please?
Re: A train takes 15 seconds to cross a bridge at 50 mph, and at
[#permalink]
03 May 2019, 01:42
03 May 2019, 01:42
2
Expert Reply
Recall the key formula
Distance = speed * time
Now, we do have actually two distances.
Let the bridge being B and the train is L
The common factor here is the speed = 50
1) \(\frac{B+L}{15} = 50\)
\(\frac{B+\frac{L}{2}}{10} = 50\)
Equate the two \(\frac{B+L}{15} = \frac{B+\frac{L}{2}}{10}\)
\(2(B + L) = 3(B + \frac{L}{2})\)
\(2B + 2L = 3B + \frac{3L}{2}\)
\(\frac{L}{2} = B\)
\(L = 2B\)
The length of the train is twice the length of the bridge.
Regards
Distance = speed * time
Now, we do have actually two distances.
Let the bridge being B and the train is L
The common factor here is the speed = 50
1) \(\frac{B+L}{15} = 50\)
\(\frac{B+\frac{L}{2}}{10} = 50\)
Equate the two \(\frac{B+L}{15} = \frac{B+\frac{L}{2}}{10}\)
\(2(B + L) = 3(B + \frac{L}{2})\)
\(2B + 2L = 3B + \frac{3L}{2}\)
\(\frac{L}{2} = B\)
\(L = 2B\)
The length of the train is twice the length of the bridge.
Regards
