Verified answer

so, let time taken to go to x km be y min = y / 60 seconds

so for return trip time = 53 - y = ( 53 - y ) / 60 s

distance x=speed *time=240y/60=4y......................(1)

for return trip distance x = speed*time =320*(53-y)/60=16(53-y)/3

substituting for x from (1)'

we get,

4y=16(53-y)/3

4y*3/16=53-y

3y/4=53-y

3y/4+y=53

7y/4=53

y=53*4/7

x=53*4*4/7=848/7=121.14

Since there was no time spent at the fire, we'll assume that's 0.

He flew x minutes at 240 km/hr and y at 320 km/hr. x+y = 53.

240 km/hr is 4 km/min, and 320 km/hr is 5 1/3 km/min

So, in x min. in one direction, he flies 4x km. Coming back, (5 1/3)y km, and the two are equal.

y + x = 53

(5 1/3)y - 4x = 0

Multiply all parts of the top equation by 4, then add the equations.

4y + 4x = 212

(5 1/3)y - 4x = 0

------------------------

(9 1/3)y = 212

Multiply by 3, just to make it simpler for the next step.

28y = 636

y = 22.7143

Multiply it by 5 1/3, and you have your distance, 121.1429 km.

Double check, if x + y = 53, x = 30.2857. Multiply by 4 and get... 121.1429.

Remember, distance equals rate multiplied by time.

Therefore, the amount of time to fly to the base would be x/240 and the amount of time to fly back would by x/320. Since the entire round trip lasted 53 minutes you end up with the equation x(1/240 + 1/320) = (53/60) hours. This works out to be .007291x=.88333, which works out to be x=121.15 kilometers (approximately).

average speed = distance / time

t = time taken for the trip TO the fire

on the trip there,

240 = x / t

240t = x ................(1)

on the trip back,

320 = x / ([53/60] - t)

320([53/60] - t) = x

sub this into (1)

240t = 320([53/60] - t)

240t = 848/3 - 320t

560t = 848/3

t = 0.505 hours

sub into (1)

x = 240(0.505) = 121.2km

enable the dimensions of first prepare be L1 and that of 2d prepare be L2, then L1 = ninety*one thousand*4/(60*60) = one hundred m L2 = 60*one thousand*6/(60*60) = one hundred m using fact the trains are coming near one yet another subsequently the relative velocity may well be ninety-(-60) = ninety+60 = 150km/h the gap travelled by way of the trains coming near one yet another to flow one yet another would be L1 + L2 = 200m for this reason, the time taken by way of the two trains to flow one yet another will take transport of by way of t = (L1+L2)/relative velocity of the prepare = 2 hundred*60*60/(one hundred fifty*one thousand) = 24/5 = 4.8 secs.

121.14km