Two jets leave an air base at the same time and travel in opposite directions. One jet travels 74 mph faster than the other. If the two jets are 9492 miles apart after 7 hours, what is the rate of each jet?

Solution

Tw o j e t s l e a v e an ai r ba se a t t h e s am e t im e an d t r a v e l in o pp os i t e d i rec t i o n s . O n e j e t t r a v e l s 74 m p h f a s t er t han t h e o t h er . I f t h e tw o j e t s a re 9492 mi l es a p a r t a f t er 7 h o u rs, w ha t i s t h e r a t e o f e a c h j e t ?

first je t^{'} s speed : 715

second je t^{'} s speed : 641

Solution steps

Translate the problem into an equation:

x \cdot 7 + (x - 74) \cdot 7 = 9492

Solve for

x, x \cdot 7 + (x - 74) \cdot 7 = 9492 : x = 715

Find speed

first je t^{'} s speed : 715

second je t^{'} s speed : 641