Math Question on speed?

Question

A pilot flew a jet from Montreal to Los Angeles, a distance of 2600 miles. On the return trip the average speed was 20% faster than the outbound speed. The round trip took 9 hours 45 minutes.

What was the speed from Montreal to Los Angeles?

Could you also please show how this question is solved? thanks

Answer 1

d=distance Montreal to Los Angeles=2600 miles
v1=Velocity flight
v2=Velocity return
t1= Time flight [h]
t2=Time return [h]

From problem Statement:
T=Total time=9.75 h (9 hours 45 minutes)
v2=1.2*v1

Total time:
T=t1+t2
Velocity equations:
v1=d/t1 => t1=d/v1
v2=d/t2 => t2=d/v2

Substituting:
T=d/v1+d/v2
T=d/v1+d/(1.2*v1)
Taking comon factor:
v1=d/T*(1+1/1.2)

The las expression allows to evaluate the velocity:

v1=2600 miles/9.75 hours*(1+1/1.2)

v1=2600/9.75*(1+1/1.2) miles/hour
v1=266.67*(1.8333) miles/hour

v1=488.9 miles/hour

I tried to be as clear as I could.

Answer 2

Let
r = speed from Montreal to Los Angeles
t = time from Montreal to Los Angeles
(5/6)t = time from Los Angeles to Montreal

t + (5/6)t = 9.75
11t/6 = 9.75
t = 9.75*6/11 = 58.5/11

rate = distance/time

r = 2600/(58.5/11) = 488 8/9 miles/hour

Answer 3

d = 2600 miles
s1 = x
s2 = x - 20%x = x - 0.2x = 0.8x (we subtract 20% since the speed is faster hence less)
t = 9 + 45/60 = 9.75

d = st
d = (s1 + s2)t
2600 = (x + 0.8x)9.75 = (1.8x)9.75
1.8x = 2600/9.75
x = 266.667/1.8 = 148.148 miles/hr

Answer 4

2600(1/v + 1/1.2v) = 9.75 hr
2600(1.2 + 1)/1.2v = 9.75 hr
2600(2.2)/1.2v = 9.75 hr
v = 2600(2.2)/(1.2*9.75) mph
v = 489 mph