Question 103110: marylin drove her car from amarillo to dallas at an average rate of 55 miles per hour and returned over icy roads averaging only 30 miles per hour.find the time going and returning if the time returning was 4 hours more than the time going
p.s this is a work sheet
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! We always start by listing what we know to be equal. In this case, it is the distance driven to/from Amarillo. Let's call it d.
The standard distance equation is d=rt, where d=distance, r=rate (e.g., velocity), and t=time.
We do not know how long it took to get to Amarillo, but we know it took 4 hours longer to get back. So, let x = time driving to Amarillo and x+4 = time driving back.
We know r=55 going to Amarillo, and r=30 returning from Amarillo.
dto = 55x
dfrom = 30(x+4)
dto = dfrom (by definition, distance to Amarillo is same as from Amarillo)
So,
55x = 30(x+4)
55x = 30x + 120
Subtracting 30x from both sides
25x = 120
Dividing by 25
x = 4.8 = time spent driving to Amarillo from home
x+4 = 4.8+4 = 8.8 = time spent driving from Amarillo to home
ALWAYS check!
We know dto = dfrom (discussed above), so check the truth of this assertion.
dto=55x=55*4.8=264 miles
dfrom=30(8.8)=264 miles
Check!
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