Two cyclists leave towns 112 kilometers apart at the same time and travel toward each other. One cyclist travels 4 kph faster than the other. If they meet in 2 hours, what is the rate of each cyclist?

解

Tw o cyc l i s t s l e a v e t o w n s 112 ki l o m e t ers a p a r t a t t h e s am e t im e an d t r a v e l t o w a r d e a c h o t h er . O n e cyc l i s t t r a v e l s 4 k p h f a s t er t han t h e o t h er . I f t h ey m ee t in 2 h o u rs, w ha t i s t h e r a t e o f e a c h cyc l i s t ?

first cyclis t^{'} s speed : 30

second cyclis t^{'} s speed : 26

解答ステップ

Translate the problem into an equation:

x \cdot 2 + (x - 4) \cdot 2 = 112

x

を解く

x \cdot 2 + (x - 4) \cdot 2 = 112 : x = 30

発見する speed

first cyclis t^{'} s speed : 30

second cyclis t^{'} s speed : 26