+936 Because of a traffic jam, Alana's 18-mile commute to work took 4 minutes longer than usual, and her average speed was decreased by 9 mph. How many minutes did it take her to get to work that day?
+128475 Let her normal time be T
Let her normal rate be R
So
RT = D
RT = 18 ⇒ R = 18/T (1)
We need to convert 4 minutes to hours = 4/60 = 1/15 hr
So....when she is slowed we have
(R - 9) ( T + 1/15) = 18 (2)
Sub (1) into (2) and we have
(18/T - 9) (T + 1/15) = 18 simplify
18 - 9T + (18)/ (15T) - 9/15 = 18 subtract 18 from both sides
- 9T + (6)/(5T) - 3/5 = 0 multiply through by -5T
45T^2 - 6 + 3T = 0
45T^2 + 3T - 6 = 0 divide through by 3
15T^2 + T- 2 = 0
(5T + 2) ( 3T - 1) = 0
The first factor set to 0 and solved for T produces a negative anser
The second factor set to 0 and solved for T produces T =1/3
This is in hrs
So (1/3) hr = 20 min = her normal time to get to work
So....now...at the slowed rate it takes her 4 minutes longer = 24min
And 24 minutes = 24/60 hrs = 2/5 hrs
Check
R = 18 / (1/3) = 54mph
And 54mph * 1/3 hr = 18 miles
And at the slower rate
(54 - 9) ( 1/3 + 1/15)
(45) ( 18/ 45) = 18 miles
+36916 ANother perspective;
the situation in the question is described by the following equation:
18/(x-9) = 18/x +4/60 where x = her usual speed in mph (x-9) = new (slower speed) & 4 minutes = 4/60 hr
solve for x
18 = (18x-162)/x +4/60x - 4/60(9)
18x = 18x-162 + 4/60 x^2 - 36/60 x
0 = .06667 x^2 -.6x - 162 Use Quadratic Formula to find x = 54 mph
Her new (slower) speed = 54-9 = 45 mph so the trip will take her 18 m / 45m/hr = .6hr= 24 minutes
