# SOLUTION: Two cyclists, 90 mi apart, start riding toward each other at the same time. One cycles twice as fast as the other. If they meet 2h later, at what average speed is each cyclist trav

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 Question 199039This question is from textbook algebra and trigonometry : Two cyclists, 90 mi apart, start riding toward each other at the same time. One cycles twice as fast as the other. If they meet 2h later, at what average speed is each cyclist traveling?This question is from textbook algebra and trigonometry Found 2 solutions by checkley75, jojo14344:Answer by checkley75(3666)   (Show Source): You can put this solution on YOUR website!D=RT 90=(X+2X)2 90=3X*2 90=6X X=90/6 X=15 MPH FOR THE SLOWER CYCLIST. 2*15=30 MPH FOR THE FASTER CYCLIST. PROOF; 90=(15+30)2 90=45*2 90=90 Answer by jojo14344(1512)   (Show Source): You can put this solution on YOUR website! Just to show a small illustration here where the two Cyclists are: We know Distance Formula = Speed x time Both covered the Distance of 90miles in 2 hours: *note: , Cyclist[1] is twice as fast Cyclist[2] ---> , Speed of Also, , Speed of Check, Thank you, Jojo