Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?
A) 90
B) 130
C) 150
D) 320
E) 450
I'm confused how to set up the formulas here. Can any experts help?
Two heavily loaded sixteen-wheeler transport trucks are 770
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In the first hour, A travels on its own at a rate of 90 kph, with the result that the 770-kilometer distance between A and B is reduced by 90 miles to 680 miles.ardz24 wrote:Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?
A) 90
B) 130
C) 150
D) 320
E) 450
After the first hour, A and B are both travel TOWARD each other, implying that they WORK TOGETHER to cover the remaining 680 kilometers.
When elements work together, ADD THEIR RATES.
Since A's rate is 90 kph and B's rate is 80 kph, the combined rate for A and B working together = 90+80 = 170 kph.
Of the 170 kilometers covered every hour by A and B working together, B travels 80 kilometers.
Thus, B will travel 80/170 of the remaining 680 kilometers between A and B:
(80/170)(680) = 320 kilometers.
Since B travels 320 kilometers, the balance of the 770 kilometers between the two rest stops must be traveled by A:
770-320 = 450 kilometers.
Thus:
(A's distance) - (B's distance) = 450 - 320 = 130 kilometers.
The correct answer is B.
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Let t = the driving time of driver B, and so (t + 1) = the driving time of driver A. We can create the equation:BTGmoderatorAT wrote:Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?
A) 90
B) 130
C) 150
D) 320
E) 450
I'm confused how to set up the formulas here. Can any experts help?
90(t + 1) + 80t = 770
90t + 90 + 80t = 770
170t = 680
t = 4
When the two meet, driver A will have driven 90 x 5 = 450, and driver B will have driven 80 x 4 = 320 miles. Thus, driver A will have driven 450 - 320 = 130 miles more.
Answer: B
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