How far does the plane fly in total?

Question

A certain flight from LaGuardia airport has two legs, with layover in between.

The first leg of the flight is 200 miles shorter than the second leg, and the total time in the air is 4.2 hours. 

If the plane has an average speed of 400 mph during the first leg, and 500 mph during the second leg, how far does the plane fly in total?

Answer 1

FIRST, DEFINE SOME VARIABLES

L1 = LEG1

L2 = LEG2

T1 = TIME TO FLY LEG1 (HRS)

T2 = TIME TO FLY LEG2 (HRS) 

 

WE KNOW THAT LEG1 IS 200MI SHORTER THAN LEG2, SO

 

#1)   L1 = L2 - 200  (MILES)

 

#2)   T1 + T2 = 4.2 HOURS    (T1 = TIME TO FLY L1,  T2 = TIME TO FLY L2)

#3)   T2 = 4.2 - T1                (SUBTRACT T1 EACH SIDE OF EQ 2)

 

NOW, WE KNOW THAT RATE (SPEED) X TIME = DISTANCE

 

#4)   L1 = 400*T1

        L2 = 500*T2     OR

 

#5)   L2 = 500*(4.2 - T1)   (FROM #3, SUBSTITUTE FOR T2)

 

SUBSTITUE L1 AND L2 INTO #1

 

400*T1  = 500*(4.2 - T1) - 200

400*T1  = 2100  -  500*T1 - 200    OR,  

 

400*T1  = 1900  -  500*T1

 

COMBINING T1 TERMS AND SOLVE FOR T1

 

900*T1 = 1900       SO T1 = 1900/900  OR

 

T1 = 2.11111 HRS 

 

SINCE THE TOTAL TIME IS 4.2 HRS,  T2 = 4.2 HRS - 2.11111   OR  

 

T2 = 2.08888 HRS

 

 

NOW, GOING BACK TO OUR RATE(SPEED) X TIME = DISTANCE EQUATIONS

 

SO LEG1 = 400*2.11111 = 844.44444   

 

     LEG2 = 500*2.08888 = 1044.4444

 

TOTAL DISTANCE = LEG1 + LEG2 = 1888.8888 MILES

 

 

 

AS A CHECK, LEG2 - LEG1 = 1044.4444 - 844.4444 = 200 MI DIFFERENCE

 

 

 

 

 

 

Answer 2

Let x = distance of first leg

 

Then, x+200 = distance of second leg

 

Distance = (Rate)(Time), so Time = Distance/Rate

 

Time of first leg + Time of second leg = 4.2

 

So, x/400 + (x+200)/500 = 4.2

 

Multiply by (400)(500) to get:

 

500x + 400(x+200) = 840000

 

900x + 80000 = 840000

 

900x = 760000

 

x = 844.4 miles

x+200 = 1044.4 miles

 

Total distance = 1888.8 miles (approximately)