Question 508764: I wrote this question earlier but I forgot to show what I tried but am not sure if it's right
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Exactly 30 minutes after the Andersons' head south on the highway, the Quentins' set out from the same point and follow the same route. The Andersons' travel at 50 mi/h. The Quentins' travel at 65 mi/h. How long does it take the Quentins' to overtake the Andersons'?
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~Chart(This is what I did & I converted 30 minutes to 1/2)
_____________|Rate x Time = Distance |
Andersons____| 50 x t = 50t |
Quentins_____| 65 x t-1/2 = 65(t-1/2)|
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~Equation(When I solved it I felt the answer didn't make sense please tell me if I got this right or wrong)
50t= 65(t-1/2)
50t= 65t-65/2
-65t=-65t-65/2
-15t=-65/2
t= 2.1666666 repeats
please tell me what I did wrong or if I got it right and tell me the right equation and answer so I can try and work out the problem. Thank you and please get back to me before 8:00 today 10/3/2011
Found 3 solutions by stanbon, bucky, MathTherapy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Exactly 30 minutes after the Andersons' head south on the highway, the Quentins' set out from the same point and follow the same route.
The Andersons' travel at 50 mi/h.
The Quentins' travel at 65 mi/h.
How long does it take the Quentins' to overtake the Andersons'?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Anderson DATA:
rate = 50 mph ; time = t hrs ; distance = r*t = 50t miles
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Quentin DATA:
rate = 65 mph ; time = t-(1/2) hrs ; distance = r*t = 65t-65/2 miles
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Equation:
distance = distance
50t = 65t-(65/2)
15t = (65/2)
time = 2 hrs 10 minutes (time for the Quentins to overtake the Andersons)
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Cheers,
Stan H.
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Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! You said:
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~Equation(When I solved it I felt the answer didn't make sense please tell me if I got this right or wrong)
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50t= 65(t-1/2) <-- this is OK
50t= 65t-65/2 <--this is OK
-65t=-65t-65/2 <-- this is confusing. You want to subtract 65t from both sides. Do not list the -65/2 for subtraction
-15t=-65/2 <-- this is OK after the subtraction of 65t from both sides.
t= 2.1666666 repeats <-- this is OK
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Your answer is correct if you start the clock running at the time the Andersons depart. 2.166666 hours after the Andersons depart the Quentins catch up to them.
(Note 2.166666 hours is 2 hours and ten minutes.)
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However, the problem asks you how long it takes the Quentins to catch up. That implies that the question is asking how long the Quentins will be on the road. It would be 0.5 hours (or 30 minutes) less than 2.166666 hours (2 hours and 10 minutes). So when the Quentins depart at 65 mph they will drive 1 hour and 40 minutes to catch the Andersons.
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An equation you can use is to start the clock when the Quentins depart. They will travel a distance of 65t. Meanwhile, the Andersons have (at 50 mph) a 1/2 hour head start so they have gone 25 miles before the Andersons start out. From the time that the Quentins start, the Andersons will travel an additional distance of 50T. So you could write the equation:
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65T = 25 + 50T
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Subtract 50T from both sides to get:
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15T = 25
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Solve for T by dividing both sides by 15 to get a time:
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T = 25/15 = 1.6666666 = 1 hr and 40 minutes.
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This tells you that from the time that the Quentins start, they will drive for 1 hour and 40 minutes to catch the Andersons.
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A mileage versus time in table form might show:
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Andersons ... Quentins .... Elapsed Minutes
...0 .........................0 ..................... 0
..25 .........................0 ................... 0
..50 ......................32.5 .................. 30
..100 .....................97.5 .................. 90
..108.3333 ......108.3333 .............. 100
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Note that the clock for the elapsed time started when the Quentins started, not when the Andersons started. Also note that the distance traveled by both parties is 108.3333 miles from the starting point.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
What you did was calculate the time it took the Andersons to cover the distance
What you need, as the question asks is to determine the time it took the Quentins to overtake the Andersons, which is the same as the time it took the Quentins to cover the same distance that the Andersons covered.
You got the Andersons' time as  , or  , or  hours, or 2 hours 10 mins.
Now, in order to find the time it took the Quentins' to catch up to the Andersons, we simply subtract  hour, or 30 mins from the Andersons' time to get the Quentins' time, which becomes  hours, or 1 hour and 40 mins.
Remember that the Andersons will take a longer time to cover the distance that the Quentins will eventually cover, 30 minutes after the Andersons left. With the Andersons traveling at a slower speed, and having left before the Quentins, the Quentins higher speed and lesser hours will allow them to catch up to and eventually pass the Andersons. The other thing you need to note is that both will have covered the same distance when the Quentins catch up to the Andersons. I hope this makes it a little clearer for you.
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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