Question 249444: Mrs. Smith drove at a rate of 45 mph from her home to her sisters house. she spent 1.5 hrs having lunch with her sister. She then drove back home at a rate of 55 mph. The entire trip, including lunch, took 4 hrs. How far does Mrs. Smith live from her sister?
Found 2 solutions by oberobic, checkley77:
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! The elapsed time is 4 hr of which 2.5 hr was driving time.
The distance formula is d = rt, where d=distance, r=rate, t=time
We assume the distance to her sister's is the same as it is home (or she drives the same path both ways without any detours).
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We can define the equation for driving to her sister's house:
d = 45(t)
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Then the equation for driving home from her sister's house is:
d = 55(2.5-t)
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In this way, we have defined the two unknown times in terms of a single unknown variable 't' and '2.5-t'.
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Since d = d, we can say:
45t = 55(2.5-t)
45t = 137.5 - 55t
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Adding 55t to both sides
100t = 137.5
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Dividing both sides by 100
t = 1.375
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So the time spend driving to her sister's house at 45 mph = 1.375 hr
The time spend driving home at 55 mph = 2.5 - 1.375 = 1.125 hr
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How far does she travel at 45 mph?
d = 45 mph * 1.375 hr = 61.875 miles.
How far does she travel at 55 mph?
d = 55 mph * 1.125 hr = 61.875 miles.
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So we've checked our work and are confident the distance is 61.875 = 61 7/8 miles.
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We could be asked what her average speed was, too.
Was it simply (45+55)/2?
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Well, we see that 2d = 2*61.875 = 123.75 miles
And it took 2.5 hr, so
t = d/r = 123.75/2.5 = 49.5.
So her average speed was 49.5.
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Done.
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! D=RT
D=45T THE TRIP TO HER SISTER'S HOME.
D=55(4-T-1.5) THE RETURN TRIP
45T=55(2.5-T)
45T=137.5-55T
45T+55T=137.5
100T=137.5
T=137.5/100
T=1.375 HOURS FOR THE TO TO HER SISTER'S HOUSE.
4-1.5-1.375=1.125HOURS FOR THE RETURN TRIP.
PROOF:
45*1.375=55(4-1.375-1.5)
61.875=55*1.125
61.875=61.875
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