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16 Sep 2014, 01:18
Kudos
Bookmarks
D
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Difficulty:
35%
(medium)
Question Stats:
77% (02:37) correct
23%
(02:56)
wrong
based on 188
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A plane flying against the wind covers a 900-kilometer distance between two airports in 2 hours. When flying with the wind, the same plane covers this distance in 1 hour and 48 minutes. If the speed of the plane in still air is constant and the speed of the wind also remains constant, what is the speed of the wind?
A. 18 kmh
B. 20 kmh
C. 22 kmh
D. 25 kmh
E. 27 kmh
A. 18 kmh
B. 20 kmh
C. 22 kmh
D. 25 kmh
E. 27 kmh
10 Jun 2024, 17:38
Kudos
Bookmarks
Hi, I would like to know if this method is also correct to solve -
Calculate Speed -
1. 900/2 - 450km/h
2. 900/1 hour 48 min - 500 km/h
with wind -
speed of plane +speed of wind = 450
speed of plane - speed of wind = 500
Calculate and speed of wind - 25 km/s
Is this method logically correct?
Calculate Speed -
1. 900/2 - 450km/h
2. 900/1 hour 48 min - 500 km/h
with wind -
speed of plane +speed of wind = 450
speed of plane - speed of wind = 500
Calculate and speed of wind - 25 km/s
Is this method logically correct?
General Discussion
16 Sep 2014, 01:18
Kudos
Bookmarks
Official Solution:
A plane flying against the wind covers a 900-kilometer distance between two airports in 2 hours. When flying with the wind, the same plane covers this distance in 1 hour and 48 minutes. If the speed of the plane in still air is constant and the speed of the wind also remains constant, what is the speed of the wind?
A. 18 kmh
B. 20 kmh
C. 22 kmh
D. 25 kmh
E. 27 kmh
Let \(p\) represent the speed of the plane and \(w\) represent the speed of the wind.
From the given information, we can set up the equations: \(\frac{900}{p - w} = 2\) and \(\frac{900}{p + w} = 1.8\) (since 1 hour and 48 minutes is equal to 1.8 hours).
From the first equation, we have \(900=2(p-w)\), and from the second, we have \(900=1.8(p+w)\). Hence, \(2(p-w) = 1.8(p+w)\), which leads to \(p=19w\).
By substituting this into the first equation, we obtain \(\frac{900}{19w - w} = 2\), which leads to \(36w=900\), resulting in \(w=25\).
Answer: D
A plane flying against the wind covers a 900-kilometer distance between two airports in 2 hours. When flying with the wind, the same plane covers this distance in 1 hour and 48 minutes. If the speed of the plane in still air is constant and the speed of the wind also remains constant, what is the speed of the wind?
A. 18 kmh
B. 20 kmh
C. 22 kmh
D. 25 kmh
E. 27 kmh
Let \(p\) represent the speed of the plane and \(w\) represent the speed of the wind.
From the given information, we can set up the equations: \(\frac{900}{p - w} = 2\) and \(\frac{900}{p + w} = 1.8\) (since 1 hour and 48 minutes is equal to 1.8 hours).
From the first equation, we have \(900=2(p-w)\), and from the second, we have \(900=1.8(p+w)\). Hence, \(2(p-w) = 1.8(p+w)\), which leads to \(p=19w\).
By substituting this into the first equation, we obtain \(\frac{900}{19w - w} = 2\), which leads to \(36w=900\), resulting in \(w=25\).
Answer: D
