Please help with this math problem?
Question by Looney_: Please help with this math problem?
MIke and Adam left a bus terminal at the same time and traveled in opposite directions. Mike’s bus was in heavy traffic and had to travel 20 mi/h slower than Adam’s bus. After 3 hours, their buses were 270 miles apart. HOw fast was each bus going?
Explain the steps please i gota show work on my hw i just dont get this
Best answer:
Answer by Jane
So let’s pretend that the distance Adam travels per hour = x. So that means Mike’s speed is x-20.
If the bus traveled 3 hours, and were 270 apart, that means Adam’s bus traveled x distance the first hour, x the second hour, and x the third.
Mike’s bus traveled x-20 the first hour, x-20 the second hour, and x-20 the third hour.
So x+x+x+(x-20)+(x-20)+(x-20) = 270
6x – 60 = 270
6x = 330
x = 55
So Adam’s bus = 55 mph and Mike’s bus = 35 mph
Give your answer to this question below!
You must first make an equation
M= speed of Mike’s bus
A= speed of Adam’s Bus
M=A-20
after 3 hours,
M traveled 3M miles
A traveled 3A miles
3M+3A= 270 miles
3(A-20) + 3A = 270
6A -60 = 270
6A = 330
A=55
so M= 55-20=35
You check it here:
3 x 55 + 3 x 35 = 270
Hope this helps
Victor
Lets assume that the speed of Mike’s bus was x mi/h. Since it’s 20 mi/h slower than Adam’s bus, that makes Adam’s bus 20 mi/h faster than Mike’s bus. Thus, Adam’s bus has a speed of x + 20 mi/h.
Since they are travelling in the opposite direction, the rate at which the distance between them is increasing is the sum of their speeds. That rate therefore, is x + x + 20 or 2x + 20 mi/h.
At this rate, the distance they will be apart after 3 hours will be 3 * (2x + 20) or 6x + 60 miles.
Since that distance is given as 270 miles, we have the equation:
6x + 60 = 270
6x = 210
x = 35.
Thus, Mike’s bus was travelling at 35 mi/h while the speed of Adam’s bus was 35 + 20 or 55 mi/h.