Asked by Summer
A man drove his automobile to New York, a distance of 550 miles, at a certain average rate. He returned from New York at the rate of 11 mph more. The entire trip required 22.5 hours. What were his rates returning and going?
Answers
Answered by
Reiny
first rate ---- x mph
return rate --- x+11
sum of times = 22.5
550/x + 550/(x+11) = 22.5
times x(x+11)
550x + 6050 + 550x = 22.5x(x+11)
1100x + 6050 = 22.5x^2 + 247.5x
22.5x^2 - 852.5x - 6050 = 0
using the formula
x = 44 or a negative
he went at 44 mph going there, and 55 mph coming back
check:
550/44 + 550/55 = 22.5 , YEAHH
return rate --- x+11
sum of times = 22.5
550/x + 550/(x+11) = 22.5
times x(x+11)
550x + 6050 + 550x = 22.5x(x+11)
1100x + 6050 = 22.5x^2 + 247.5x
22.5x^2 - 852.5x - 6050 = 0
using the formula
x = 44 or a negative
he went at 44 mph going there, and 55 mph coming back
check:
550/44 + 550/55 = 22.5 , YEAHH
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