Eberlecaleidn
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Miguel can drive 4 times as fast as Raul can ride his bicycle. If it takes Raul 3 hours longer than Miguel to travel 68 miles, how fast (in mph) can Raul ride his bicycle? Round your answer to two decimal places, if needed.

Sagot :

Mhanifaidn

Answer:

  • 17 mph

Step-by-step explanation:

  • Miguel's speed = m, Raul's speed = r

Given

  • m = 4r
  • Distance = 68 miles
  • Time difference = 3 hours

Solution

  • time = distance/speed

Substitute values and solve for r the equation of the time difference:

  • 68/m + 3 = 68/r
  • 68/4r + 3 = 68/r
  • 17/r + 3 = 68/r
  • 17 + 3r = 68
  • 3r = 68 - 17
  • 3r = 51
  • r = 17

Raul's speed is 17 mph

Answer:

[tex] \huge \boxed{ \boxed{ \red{ \tt17 \: mph}}}[/tex]

Step-by-step explanation:

to understand this

you need to know about:

  • equation
  • equation word problems
  • PEMDAS

given:

  • Miguel can drive 4 times as fast as Raul can ride his bicycle. If it takes Raul 3 hours longer than Miguel to travel 68 miles, how fast (in mph) can Raul ride his bicycle

tips and formulas:

  • t=d/s

to find

  • how fast can Raul ride his bicycle

let's solve:

let's the speed of Raul be x

likewise

Miguel can drive 4 times as fast as Raul can ride his bicycle

therefore

4x

according to the question

[tex] \sf \frac{68}{4x}+3=\frac{68}{x}[/tex]

[tex] \implies \sf \frac{68 + 12x}{4x} = \frac{68}{x} [/tex]

[tex] \sf factor \: out \: 4 \: and \: reduce \: it : \\ \sf \implies \frac{4(17 + 3x)}{4x} = \frac{68}{x} \\ \\ \implies\sf \frac{17 + 3x}{x} = \frac{68}{x}[/tex]

[tex] \sf multipy \: both \: sides \: by \: x : \\ \sf \implies (\frac{17 +3 x}{x} )x = (\frac{68}{x} )x \\ \implies \sf 17 + 3x = 68[/tex]

[tex] \sf cancel \: 17 \: from \: both \: sides : \\ \implies \sf 17 - 17 + 3x = 68 - 17 \\ \implies \sf 3x = 51[/tex]

[tex] \sf divide \: both \: side s \: by \: 3 : \\ \sf \implies \frac{3x}{3} = \frac{51}{3} \\ \therefore x = 17[/tex]

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