Two trucks travel from Alphaburg to Betaville along the same route.

Another way would be take each stretch separately..

SO..
S travelling at 50 miles per hour covers 50/60*12=10 miles..
Let us work on each stretch..
(I) Stretch II does not have any difference as both travel at 40 mph..
(II) Stretch I has a difference of 10 mph ( relative speed -- 60 mph vs 50mph), so F will cover 5 miles in 30/60 hr . Hence S still has 10-5 miles over F, when F covers this 30 miles.
(III) In stretch III, there is a difference of 5 mph ( relative speed -- 55 mph vs 50mph), so F will cover remaining 5 miles over S in 1 hr . Hence S and F will meet 1 hr after F starts in third stretch. So, F will cover 55 miles in 1 hr at 55mph...
thus 60-55 or 5 miles short of finish point..

Simplified form now..
10 miles to cover
Stretch I.... F takes 30/60 or 1/2hr and covers (60-50)*1/2=5 miles, thus 5 miles remaining
Stretch II... No difference as both travel at 40mph
Stretch III.. F will cover 55-50 miles in 1 hr and as it has 5 miles to cover, another 1 hr travel by F or 55 miles.

Bunuel

IMO we can add "Word Problems" as an additional source of this problem.

A--------30--------40-----------100 B total distance = 100 miles
----(1)------- (2)---------(3)

F speeds (1)= 60mp, (2)=40mph, (3)=55mph
S speeds (1)=50 mph, (2)=40mph, (3)=50 mph

lets convert 12 minutes into hours 12/60 = 1/5 hrs

S left 12 minutes before F at 50 mph, leaving S 10 miles ahead of F when F started.
S make the first 30 miles if (1) in 36 minutes. Because F left A 12 minutes later than S, they will travel together just 12 minutes in (1)

in the first part of the road, F rate is 60 mph and S rate is 50 mph. Because they are moving in the same direction (closing the gap in between them) we can have the relative speed by subtracting the rates. So relative speed in the first part is 10mph

Because F is faster than S, and they are closing the gap at 10mph, the initial 10 miles gap, ends up being just 5 miles. If they would have traveled together the hole part (1) the gap would have closed entirely but due they traveled just half of S(1) the gap just closed half. so 10 miles/2= 5 miles apart in (1)

In the second part of the road, everything stayed the same because they travel at the same speed.

In the third part of the road, F travels at 55mph and S travels at 50mph, their relative speed is 5mph, so in 1 hour, the initial gap of 5 miles will be closed. that means 1 hour of travel of F. In 1 hour, F travel 55 miles.

so lets calculate the miles that F traveled in each part of the journey
(1) 30 miles
(2)10 miles
(3) 55mph*1hr=55 miles

total F miles = 95 miles

so 100 - 95= 5 miles from B

Given that Truck S departs 12 minutes before Truck F, we need to calculate the distance Truck S travels during this time at its maximum speed of 50 mph:

Distance travelled by Truck S in 12 minutes = (50 mph) * (12/60 hours) = 10 miles.

Now, when Truck F starts, it catches up to Truck S at a relative speed of 70 mph - 50 mph = 20 mph.

The time it takes for Truck F to catch up to Truck S is:

Time = Distance / Speed = 10 miles / 20 mph = 0.5 hours.

During this time, Truck F travels at its maximum speed of 70 mph:

Distance travelled by Truck F in 0.5 hours = (70 mph) * (0.5 hours) = 35 miles.

So, Truck F catches up to Truck S 35 miles from Betaville. However, since Truck S has already travelled 10 miles when Truck F starts, the point where Truck F catches up to Truck S is 35 - 10 = 25 miles from Betaville.

Therefore, the correct answer is A) 5 miles.

Explanation: Truck F catches up to Truck S 25 miles from Betaville. However, since Truck S has already travelled 10 miles when Truck F starts, the point where Truck F catches up to Truck S is 25 - 20 = 5 miles from Betaville.

By Claudio Hurtado gmatchile.cl and clasesgmat.es­