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

Question

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?

A) 5 miles
B) 20 miles
C) 55 miles
D) 95 miles
E) Both trucks arrive at Betaville simultaneously

A) 5 miles
B) 20 miles
C) 55 miles
D) 95 miles
E) Both trucks arrive at Betaville simultaneously

chetan2u · Accepted Answer

One of the easiest way would be to equate time..

In 12 minutes, S covers = 12/60*50=10 miles..

Let the two catch up at x miles in third stretch
1) S covers 20 miles at 50mph, then 10 and 40mph and say x at 50mph ----- Time taken  \frac{20}{50}+\frac{10}{40}+\frac{x}{50} 
2) F covers 30 miles at 60mph, then 10 and 40mph and x at 55mph ----- Time taken  \frac{30}{60}+\frac{10}{40}+\frac{x}{55}

Both should be equal..
So, ----- Time taken  \frac{20}{50}+\frac{10}{40}+\frac{x}{50} = \frac{30}{60}+\frac{10}{40}+\frac{x}{55} .......... \frac{30}{60}-\frac{20}{50}=\frac{x}{50}-\frac{x}{55} ..
 \frac{1}{10}=\frac{5x}{55*50}......x=55 .....

So they meet when they have covered 55 miles in last 60 miles. Hence 5 miles short of final destination.

A

zhanbo · Answer

Wow. I got the right answer  (A) , but it took me a whopping 6 minutes! I can only afford to spend that much time on a single question throughout the whole Quant section!

(1) First, I establish that in the first 30 miles (50 m/h for S, 60 m/h for F), it took 36 minutes for S and 30 minutes for F. So, S is still ahead of F by 6 minutes' drive at 50 m/h, which is 5 miles.
(2) Second, both trucks run at 40 m/h for the next 10 miles. The distance between the 2 remains 5 miles.
(3) For the remaining 60 miles, let x is the hours it takes for F (55m/h) to overtake S(50m/h), we have
50x + 5 = 55x => x = 1. 
That is exactly one hour. In one hour, F travels 55 miles. So the answer is 60 - 55 = 5.

chetan2u · Answer

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.

Maria240895chile · Answer

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

gmatchile1 · Answer

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

white_wiz · Answer

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

anrocha · Answer

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?

A) 5 miles
B) 20 miles
C) 55 miles
D) 95 miles
E) Both trucks arrive at Betaville simultaneously

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.

Gmathub · Answer

Check video solution here:

https://www.youtube.com/watch?v=40jhTQuHkIw

Thanks
Gmathub

awesome_burst · Answer

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

Kinshook · Answer

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

quosrepudiandae · Answer

the point is time difference
for S, t1=0.6 t2=0.25 t3=1.2
for F, t1=0.5 t2=0.25 t3=1+1/11
Given S leave 0.2h earlier
we can infer S arrive the final 60 miles 0.1h earlier
so it needs 1h (0.1*50)/(55-50)=1h for F catching up with S
thus there are 5 miles left (60-55*1h=5)

Shivam2027 · Answer

is this an official question?

Bunuel · Answer

Please check the source tag: https://gmatclub.com/forum/download/file.php?id=88181 GMAT-Club-Forum-4da37kww.png

Adit_ · Answer

Key point to note is that the distances travelled by F is always 12 mins behind S. So considering that we get S traveling 45 miles in 57 mins and F traveling 40 miles in 45 mins.
The gap is 5 miles and speed gap is also 5mph.
Thus it will take 1 more hour to catch up.

Thus they both will meet at 95 miles or 5 miles short of the endline.

Great question that tests patience and dealing with multi variables. Took quite a bit of time.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards