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Akron and Philadelphia are 400 miles apart. Brad's average speed is 10 miles per hour faster than Gail's. Find Gail's average speed, to the nearest tenth, if she travels from Akron to Philadelphia in 1.5 hours more time than Brad. (A) 36.9 mph (B) 46.9 mph (C) 56.9 mph (D) 66.9 mph

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rvkacademic

Answer:

(B) 46.9 mph

Step-by-step explanation:

  • The distance traveled by both Brad and Gail is the same = 400 miles
  • Let Gail's speed be x mph
  • Brad's speed is 10 mph more than x = x + 10
  • Time taken for travel is given by the formula T = D/S
    where
    T = time taken
    D = distance traveled
    S = average speed

  • Time taken for Gail to travel 400 miles = [tex]\dfrac{400}{x} \;hours[/tex]
  • Time taken for Brad to travel 400 miles [tex]=\dfrac{400}{x + 10}\;hours[/tex]
  • Since Gail takes 1.5 hours more than Brad to travel the same distance we get the following equation"
    [tex]\dfrac{400}{x+10} - \dfrac{400}{x} = 1.5\\[/tex]
  • To get rid of the denominator, multiply throughout by the LCM which is [tex]x(x+10)[/tex]   to get

    [tex]\dfrac{400}{x + 10}\cdot x(x+10) - \dfrac{400}{x}\cdot x(x+10) = 1.5x(x+10)[/tex]
    [tex]\longrightarrow 400x - 400(x + 10) = 1.5x^2 + 15x \\\\\longrightarrow 400x - 400x - 4000 = 1.5x^2 + 15x\\\\\text{400x cancels out from both sides}\\\\-4000 = 1.5x^2 + 15x\\\\\text{Switch sides:}\\1.5x^2 +15x = - 4000\\\\\text{Add 4000 to both sides:}1.5x^2 +15x - 4000=0\\[/tex]
  • This is a quadratic equation which can be solved using the quadratic formula and with the help of a quadratic solution calculator
  • The solution set for the above quadratic equation is
    [tex]x = 46.8813, \; x = -56.8813[/tex]
  • We can obviously discard the negative value and state that
    [tex]x = 46.8813\;mph[/tex]
  • Rounded to the nearest tenth that would be
    [tex]x = 46.9 \;mph[/tex]
  • Hence Gail's speed is 46.9 mph -> Option B
anbu40

Answer:

B)46.9 mph

Step-by-step explanation:

Algebraic equation:

To find Gail's average speed, we can frame and algebraic equation and then solve it.

[tex]\boxed{Time = \dfrac{Distance}{speed}}[/tex]

Let Gail's average speed be 'x mph'. Brad's average speed is 10 mph faster than Gail's average speed.

Brad's average speed = (x + 10) mph

[tex]\sf Time \ taken \ by \ Gail =\dfrac{400}{x}\\\\\\Time \ taken \ by \ Brad = \dfrac{400}{x + 10}[/tex]

As Brad is traveling fast, time taken by Brad will be less.

Time difference of the journey = 1.5 hour

[tex]\sf 1.5 = 1\dfrac{5}{10} =1\dfrac{1}{2}=\dfrac{3}{2}[/tex]

Time taken by Gail - Time taken by Brad =  (3/2)

[tex]\sf ~~~~~~~~~~~~ ~~~~~~~~\dfrac{400}{x}-\dfrac{400}{x + 10}=\dfrac{3}{2}\\\\\\\dfrac{400*(x +10)}{x(x+10)}-\dfrac{400*x}{(x +10)*x}=\dfrac{3}{2}\\\\\\~~~~~~~~~~~~~\dfrac{400(x + 10)-400x}{x(x +10)}=\dfrac{3}{2}\\\\\\~~~~~~~~~~~~ \dfrac{400x + 4000 - 400x}{x^2+10x}=\dfrac{3}{2}\\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~ \dfrac{4000}{x^2+10x}=\dfrac{3}{2}[/tex]

Cross multiply,

                           4000*2 = 3*(x² + 10x)

                              8000 = 3x² + 30x

          3x² + 30x - 8000 = 0    

We can use Quadratic formula to find the value of x,

a = 3 ; b = 30 ; c = -8000

D = b² - 4ac

   = 30² - 4*3*(-8000)

   = 900 + 96000

   = 96900

[tex]\sf x = \dfrac{-b \± \sqrt{D}}{2a}\\\\\\x =\dfrac{-30 \± \sqrt{96900}}{2*3}\\\\x = \dfrac{-30\±311.29}{6}[/tex]

[tex]\sf x = \dfrac{-30-311.29}{6} \ \text{This is ignored as speed will not be negative}[/tex]

[tex]\sf x = \dfrac{-30+311.29}{6}\\\\\\x = \dfrac{281.29}{6}\\\\\\x= 46.88\\\\x = 46.9 \ mph[/tex]

Gail's average speed = 46.9 mph

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