Question 212474: I need help with quadratic word problems!
Flight 725 left New York at 10.28 am flying north. Flight 245 left from the same airport at 11.18 am flying south at 120 kph less than three times the speed of flight 725. At 1.06 pm they are 1,030 kilometers apart. What is the average rate of speed for Flight 245 and for Flight 725?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! flight 725 flight time is 1:06 pm minus 10:28 am.
I'm assuming this is hours and minutes.
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Flight time for flight 725 is 2 hours and 38 minutes.
10:28 to 12:28 is 2 hours.
Another 32 minutes gets you to 1 pm.
another 6 minutes gets you to 1:06 pm.
total time is 2 hours plus 32 minutes plus 6 minutes equals 2 hours and 38 minutes.
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Flight time for flight 245 is 1 hour 48 minutes.
11:18 to 12:18 is 1 hour.
12:18 to 1:00 pm is 42 minutes.
1:00 pm to 1:06 pm is another 6 minutes.
total time is 1 hour plus 42 minutes plus 6 minutes equals 1 hour and 48 minutes.
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We need to convert these times to hours.
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2 hours 38 minutes is equivalent to 2.63333... hours.
1 hour 48 minutes is equivalent to 1.8 hours
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you take the minutes and divide them by 60 and add them to the hours to convert from hours and minutes to hours.
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the rate of speed for flight 725 is x (we assigned it).
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the rate of speed for flight 245 is 3x - 120 (it was given that flight 245 traveled 3 times as fast as flight 725 - 120 kph. since speed of flight 725 is x kph, then speed of flight 245 is 3x - 120 kph).
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the basic equation to use is rate * time = distance.
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The distance between them is 1030 kilometers.
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In order to get this distance, flight 725 went north for a distance of y kilometers (unknown distance so we assign a variable to it).
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flight 245 went south for a distance of z kilometers (another unknown distance so we assign another variable to it.).
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We know that the distance of 1030 kilometers is the sum of the distance that flight 725 traveled and the distance that flight 245 traveled.
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We have an equation that says:
y + z = 1030
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This means that the distance flight 725 traveled plus the distance that flight 245 traveled equals 1030 kilometers.
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We know that flight 725 traveled y kilometers.
We also know that flight 725 traveled at x kilometers per hour.
We also know that flight 725 traveled for 2.63333... hours.
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This tells us that flight 725 traveled y kilometers and that y = 2.63333... * x
because rate * time = distance and:
distance = y
time = 2.6333... hours
rate = x kilometers per hour.
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We know that flight 245 traveled z kilometers.
We also know that flight 245 traveled at (3x-120) kilometers per hour.
We also know that flight 245 traveled for 1.8 hours.
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This tells us that flight 245 traveled z kilometers and that z = 1.8 * (3x-120) because rate *& time = distance and:
distance = z
time = 1.8 hours
rate = (3x-120) kilometers per hour.
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We started off with an equation that says:
y + z = 1030
We know that:
y = 2.63333... * x
We also know that:
z = 1.8 * (3x-120)
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Our equation:
y + z = 1030
becomes:
2.63333... * x + 1.8 * (3x-120) = 1030
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Now we have one equation in one unknown and we can solve for x.
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expanding the equation by removing the parentheses, we get:
(2.6333... * x) + (1.8 * 3x) - (1.8 * 120) = 1030
this becomes:
(2.6333... * x) + (5.4 * x) - 216 = 1030
combine like terms to get:
8.0333... * x = 1030 + 216
this becomes:
8.0333... * x = 1246
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divide both sides of this equation by 8.0333... to get:
x = 155.1037344
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if x = 155.1037344 kph, then:
3x - 120 = 345.3112033 kph.
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We have enough now to figure out the kilometers.
flight 725 traveled for 2.6333... hours at 155.1037344 kph to make a distance of 408.439834 kilometers.
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flight 245 traveled for 1.8 hours at 345.3112033 kph to make a distance of 621.560166 kilometers.
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sum of these kilometers is 408.439834 + 621.560166 = 1030 so we have the right rates of travel.
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answer to the question is:
rate of speed of flight 725 is 155.1037344 kph and rate of speed of flight 245 is 345.3112033 kph.
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