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?

dgfgrafgdfge111 Nov 21, 2018

#1**+2 **

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

CPhill Nov 22, 2018

#2**0 **

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

ElectricPavlov Nov 22, 2018