Question 87389
1. Thomas left Miami and drove at a speed of 41 mph. Katherine left 1 hour and 28 minutes later and drove at a speed of 57 mph. How long will it take Katherine to catch up with Thomas?
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Let t = amt of time for K to catch up with T:
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Convert 28 min to hrs: 28/60 = .467, making K left 1.467 hrs after T
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When K catches up we know they will have traveled the same distance, write a simple equation from this fact: Dist = Speed * time
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K's dist = T's dist
57t = 41(t+1.467)
57t = 41t + 60.147
57t - 41t = 60.147
16t = 60.147
t = 60.147/16
t = 3.76 hrs which is 3 hrs 45.6 min, when K catches up with T
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Check solutions by finding the distance both K & T have traveled
K: 57 * 3.76 = 214.3 mi
T: 41 * (3.76 + 1.467) = 214.3 mi also
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2. EdHelper Airlines flight H3488 from Boston to Dallas averaged a speed of 90 mph more than the return trip back to Boston. The return trip took 1 109/308 times longer than the trip going to Dallas. The distance between Boston and Dallas is one thousand, seven hundred fifty miles. What was the average speed from Dallas to Boston?
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Convert 1 109/308 to a decimal: 417/308 = 1.3539
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Let s = Average speed from Dallas to Boston
Then:
(s+90) = Average speed from Boston to Dallas
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Write a time equation Time = Dist/speed:
1.3539*{{{1750/((s+90))}}} = {{{1750/s}}}
{{{2369.32/((s+90))}}} = {{{1750/s}}}
Cross multiply:
2369.32s = 1750(s+90)
2369.32s - 1750x + 157500
619.32s = 157500
s = 157500/619.32
s = 254.3 mph Dallas to Boston
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Check solution using speed and time
254.3 + 90 = 344.3 mph Boston to Dallas
 1750/254.3 = 6.88 hrs
 1750/344.3 = 5.08 hr
 6.88/5.08 = 1.354 ~ 1 & 109/308