Hey I'm new here and i dont want just the answer, i'd like to understand the problem as well.

The small plane could go 6 times as fast as the small car. Thus, the plane could go 1200 miles in only 1 hour less than it took the car to go 250 miles. Find the rate and the time of the small plane and the rate and the time of the small car.

Thankz for your help.

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Originally posted by Atorius:
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The small plane could go 6 times as fast as the small car. Thus, the plane could go 1200 miles in only 1 hour less than it took the car to go 250 miles. Find the rate and the time of the small plane and the rate and the time of the small car.

Welcome, Atorius! You have 4 unknowns here: the speed (rate) of the car (v_c), the speed of the plane (v_p), the time it takes the car to travel 250 miles (t_c), and the time it takes the plane to travel 1200 miles (t_p).You must first know the general equation that relates speed, distance, and time, which is
distance = average speed*time, that is,
x = vt
Knowing that, you are also given 4 equations, which are

(1.)v_p=6v_c
(2.)t_p= t_c - 1
(3.)250 = v_ct_c(from the general equation I just mentioned), and
(4.)1200 = v_pt_p(also from general equation)
Thus, you have 4 equations to use to solve for the 4 unknowns.
Plugging equation (1.)and equation (2.)into equation (4.),
1200=6v_c(t_c - 1),or
1200=6v_ct_c - 6v_c
and plugging equation (3.) into this last equation we get
1200=6(250)-6v_c, or
1200=1500-6v_c
solving this last equation, v_c=50 mph,and then the plane's speed is 6 times that or
v of plane= 300mph
and from equation (3.), the time of the car is 5 hours, and the plane made its 1200 mile trip in an hour less, or 4 hours.
This is the 'substitution' method of solving this problem. There are other ways as well. Thanks for choosing Answerpool!