Asked by sylvia
1. A man completing a 40km trip finds that by travelling one more km per hour, he could have made the journey in 2 hours less time. At what speed did he actually travel?
2. One pipe alone can fill a tub in 6 minutes less than it takes a second pipe alo e to fill the same tub. Both pipes together can fill tne tub in 4 minutes. How long does it take each pipe to fill tne tub?
2. One pipe alone can fill a tub in 6 minutes less than it takes a second pipe alo e to fill the same tub. Both pipes together can fill tne tub in 4 minutes. How long does it take each pipe to fill tne tub?
Answers
Answered by
Reiny
1.
let his actual speed by x km/hr
time to do his 40 km trip = 40/x
let his supposed speed be x+1 km/h
time taken = 40/(x+1)
40/x - 40(x+1) = 2
times x(x+1) , the LCD
40(x+1) - 40x = 2x^2 + 2x
2x^2 + 2x - 40 = 0
x^2 + x - 20 = 0
(x+5)(x-4) = 0
x = 4 or a negative
he went at 4 km/h
2. time to fill tub by slow pipe = x min
rate of that pipe = 1/x
time to fill tub by other pipe = x-6
rate of other pipe = 1/(x-6)
combined rate = 1/x + 1/(x-6)
= (x-6 + x)/(x(x-6))
= (2x - 6)/(x^2 - 6x)
but 1 / (2x - 6)/( (x^2 - 6x) ) = 4
(x^2 - 6x)/(2x - 6) = 4
x^2 - 6x = 8x - 24
x^2 - 14x + 24 = 0
check my arithmetic, and solve this quadratic
let his actual speed by x km/hr
time to do his 40 km trip = 40/x
let his supposed speed be x+1 km/h
time taken = 40/(x+1)
40/x - 40(x+1) = 2
times x(x+1) , the LCD
40(x+1) - 40x = 2x^2 + 2x
2x^2 + 2x - 40 = 0
x^2 + x - 20 = 0
(x+5)(x-4) = 0
x = 4 or a negative
he went at 4 km/h
2. time to fill tub by slow pipe = x min
rate of that pipe = 1/x
time to fill tub by other pipe = x-6
rate of other pipe = 1/(x-6)
combined rate = 1/x + 1/(x-6)
= (x-6 + x)/(x(x-6))
= (2x - 6)/(x^2 - 6x)
but 1 / (2x - 6)/( (x^2 - 6x) ) = 4
(x^2 - 6x)/(2x - 6) = 4
x^2 - 6x = 8x - 24
x^2 - 14x + 24 = 0
check my arithmetic, and solve this quadratic
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