SOLUTION: During the day, a truck driver encounters light, moderate, and heavy traffic. If he travels 1/3 hour in light traffic, 1/2 hour in moderate traffic, and 1/5 hour in heavy traffic,
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Question 171283: During the day, a truck driver encounters light, moderate, and heavy traffic. If he travels 1/3 hour in light traffic, 1/2 hour in moderate traffic, and 1/5 hour in heavy traffic, he can cover 21 miles. If he travels 1/5 hour in light, 1/4 hour in moderate, and 1 hour in heavy traffic, he can cover 16 miles. The sum of his speeds in moderate and heavy traffic is 5 miles per hour less than his speed in light traffic. Find his speed in each type of traffic. Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website! d=rt.....this is our basic equation. In this instance we have to keep in mind that rates and times are different under different conditions
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we are looking for the rates under different conditions. lets call the rates under light, moderate, and heavy conditions:L,M,and H respectively. We are given 2 different distances along with there times. We will take this information and write two equations. We are also give the fact that the sum of the speeds M+H equals 5 less than the speed in Light traffic L-5.
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M+H=L-5--->L=M+H+5.........eq 1
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(1/3)L+(1/2)M+(1/5)H=21....eq 2---->this is the rates and times of each added
....................................together to equal the distance...rt=d
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(1/5)L+(1/4)M+(1)H=16......eq 3
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Lets get rid of the fractional parts in eq 2 and 3 by multiplying eq 2 by 30 and eq 3 by 20....Least common multiples.
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10L+15M+6H=630....revised eq 2
4L+5M+20H=320.....revised eq 3
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Now lets take L's value from eq 1 and plug it into revised eq 2 and eq 3.
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10(M+H+5)+15M+6H=630-->distribute-->10M+10H+50+15M+6H=630
4(M+H+5)+5M+20H=320--->distribute--->4M+4H+20+5M+20M=320
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after combining the terms we arrive at:
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25M+16H=580.(4).mult this eq by 3..we are doing this to eliminate the H terms
9M+24H=300..(5)..multiply this eq by -2
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What folllows is what we have left as the H terms cancel out 48H-48H=0
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3(25M)+(-2)9M=3(580)+(-2)300
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75M-18M=1140
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57M=1140
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now we plug M's found value into either equation marked (4) or (5)...I chose (5)
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9(20)+24H=300---->24H=300-180--->24H=120
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Now plug M's and H's found values into eq 1
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speeds for ight, oderate and eavy traffic
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