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

I think its an elimination of the options work much better here.

There is a 12 min gap; which is not covered in first 40 kms. The only realistic options are A and B.

Posted from my mobile device

Two trucks travel from Alphaburg to Betaville along the same route. The speed limit for the first 30 miles is 60 miles per hour. The speed limit for the next 10 miles is 40 miles per hour, and the limit for the final 60 miles is 55 miles per hour. Truck F has a maximum speed of 70 miles per hour, but Truck S has a speed-limiting governor installed to cap its maximum speed at 50 miles per hour. Truck S departs Alphaburg 12 minutes before Truck F. If each truck travels at the lesser of the speed limit or the maximum speed of the truck, how far from Betaville is the point where Truck F catches up with Truck S?

Speed Limit:
30 miles is 60 miles per hour
10 miles is 40 miles per hour
60 miles is 55 miles per hour

Maximum speed:
F - 70 miles per hour
S - 50 miles per hour

So speed for each truck:
F:
30 miles is 60 miles per hour
10 miles is 40 miles per hour
60 miles is 55 miles per hour

S:
30 miles is 50 miles per hour
10 miles is 40 miles per hour
60 miles is 50 miles per hour

Truck S 12 minutes before Truck F.

S=D/T --> T=D/S
1st part of the road:
F: 30/60 = 1/2 h = 30 min
S: 30/50 = 3/5 h = 36 min

Then S arrives 6 min first because it had 12 min in advance.

2nd part of the road:
Equal speed, so need to compute

3rd part of the road:
F: 60/55 = 12/11 h = ~65min
S: 60/50 = 6/5 h = 72min

So, the difference is bigger than 6 min S got in advance. They will meet before the end.

If the distance they go is the same (because they meet in the same point), and we know each truck's speed and truck S had 6 min in advance (6/60 h = 1/10 h):
55 * t = 50 * (t+1/10)
5 * t = 5
t = 1

As t=1h, they met at 55 miles. Point B was at 60 miles, then there were 5 miles left.

Check video solution here:



Thanks
Gmathub

This question seems impossible to answer in 2 minutes. Even by elimination, it seems A,B,E how to solve this in 2 mins?

Two trucks travel from Alphaburg to Betaville along the same route. The speed limit for the first 30 miles is 60 miles per hour. The speed limit for the next 10 miles is 40 miles per hour, and the limit for the final 60 miles is 55 miles per hour. Truck F has a maximum speed of 70 miles per hour, but Truck S has a speed-limiting governor installed to cap its maximum speed at 50 miles per hour. Truck S departs Alphaburg 12 minutes before Truck F.

If each truck travels at the lesser of the speed limit or the maximum speed of the truck, how far from Betaville is the point where Truck F catches up with Truck S?

Let's assume that Truck S depart at 00:00 hrs and Truck F departs at 00:12 hrs

First 30 miles (max speed limit = 60 miles per hour) : -
Truck S cross in 30/50 = .6 hrs = 36 minutes
Truck F cross in 30/60 = .5 hrs = 30 minutes
Truck F gains 6 minutes w.r.t. Truck S
Time left to make up = 12 - 6 = 6 minutes = .1 hours

Next 10 miles (max speed limit = 40 miles per hour) : -
Truck S cross in 10/40 = .25 hrs = 15 minutes
Truck F cross in 10/40 = .25 hrs = 15 minutes
Truck F gains 0 minutes w.r.t. Truck S

Final x miles (max speed limit = 55 miles per hour) : -
Truck S cross in x/50 hrs
Truck F cross in x/55 hrs
Truck F gains (x/50 - x/55 = x/550) hrs w.r.t. Truck S
x/550 = .1
x = 55 miles

The point where Truck F catches up with Truck S = 30+10+55 = 95 miles away from Alphaburg
The point where Truck F catches up with Truck S = 30+10+60 - 95 = 5 miles away from Betaville

IMO A