## The Truck and the Plane

A delivery truck from the post office is sent to the airport to meet a cargo plane at its planned arrival time. The plane lands ahead of schedule and its contents are brought toward the post office by bicycle. After a half hour, the bicycle meets the truck and the mail is transferred.

The truck returns from the post office 20 minutes early. How early did the plane arrive? (Assume all transactions are instantaneous)

The delivery truck arrived back 20 minutes early, so it would have taken 20 minutes to go from where it met the bicycle to get to the airport and back. Therefore, the bicycle and truck met when the truck was 10 minutes from the airport. Adding those 10 minutes to the 30 minutes the truck had already driven to meet the bicycle means **the plane arrived 40 minutes ahead of schedule**.

15 Comments on "The Truck and the Plane"Ben says

July 16, 2014 @ 08:43

This doesn’t make sense, why would you add, the time the truck was travelling for to the the ten minutes?

Dan says

July 18, 2014 @ 13:24

What is your solution? You’ll see why you have to add the truck’s travel time if you work it out.

Jeff says

March 24, 2015 @ 19:17

Hi Dan,

Here’s a different solution:

Truck leaves post office at 9:00am to meet plan upon arrival. It takes 40 mins to reach the airport therefore the plane normally lands at 9:40am. The truck usually returns to the post office by 10:20 (80mins). Because the plane was early the truck returns at 10:00am (20 mins early). If the plane landed 20 mins early at 9:20am and a car drove 10mins toward the truck it would have met the truck at 9:30.

What do you think?

Jeff

William says

April 30, 2015 @ 05:08

Because the speed of the truck and car aren’t mentioned it could be close to 10 minutes early at minimum and an almost infinte maximum value.

example 1:

A delivery truck from the post office is sent to the airport at 50 km/h to meet a cargo plane. The plane lands ahead of schedule by 8 hours and 30 minutes, and its contents are brought toward the post office by car moving at 1 km/h. After a half hour (from the time the truck was sent), the car meets the truck and they exchange the contents.

The truck returns from the post office 20 minutes early.

example 2:

A delivery truck from the post office is sent to the airport at 3 km/h to meet a cargo plane. The plane lands ahead of schedule by 11 minutes, and its contents are brought toward the post office by car moving at 30 km/h. After a half hour (from the time the truck was sent), the car meets the truck and they exchange the contents.

The truck returns from the post office 20 minutes early.

William says

April 30, 2015 @ 05:18

Also, as a comment on the “answer” provided. The correct way to calculate it if they are moving at the same speed is to subtract the time not add it leading to 30-10=20 minutes not 30+10=40 minutes. acording to your solution the car was moving at a third of the speed the truck moves, wich could be totally correct if we were told that the car moves at 1/3 of the trucks speed.

Dan says

April 30, 2015 @ 10:52

All you have to go on is the information given. You can’t assume the car and truck travel at the same speed since it’s not stated. The car is in fact traveling 1/3 of the truck’s speed, but it’s a consequence of the time durations, not a determining factor.

I’ve changed car to bicycle to make it a little less confusing.

If the plane only arrived 10 minutes early, how could the truck have arrived back at the post office 20 minutes early? That wouldn’t make sense.

30 minutes pass after the plane lands and before the car (or bicycle) and the truck meet. You don’t subtract from the 30 minutes because they’ve already passed. In a normal on-time scenario, those 30 minutes would have passed as well. The truck would have needed to travel another 10 minutes to get to the airport, so you

addthe 30 minutes that already elapsed for a normal on-time mail pickup to calculate how early the plane was.Leon says

September 28, 2015 @ 13:41

Still doesn’t make sense. You can’t calculate the answer using only the original information, you need to know how fast the bike and the truck are travelling relative to each other.

Let’s say Superman is riding the bicycle and it’s travelling 10x faster than the truck. The plane could have landed 1 minute ahead of schedule and the truck would still be back 20 minutes early. In that 1 minute, the bike would have carried the parcel the same distance as the truck would have in 10 minutes, saving it a journey time of 20 minutes in total.

Your solution only makes sense once you tell us that the bike is travelling 1/3 of the speed of the lorry. Without that information, it’s impossible to determine how early the plane lands.

Dan says

September 28, 2015 @ 16:04

Leon, you need to forget about speed. It’s confusing you (and others) and completely orthogonal to the problem. All you information you need to solve it is provided in the problem. Maybe it’s not clear to you that the post office truck left at the usual time to meet the plane when it landed?

Both the truck and the bicycle traveled 30 minutes toward one another. The truck returned 20 minutes early. This means if the plane hadn’t arrived early, the truck would have had to travel another 10 minutes to get to the airport and 10 minutes to return to the spot where they met (10 minutes + 10 minutes = 20 minutes early). That’s the 20 minutes the truck saved.

I highly recommend you (or anyone else who is working on this) sketch a diagram of the problem. It will make it easier to understand the solution.

It doesn’t matter if Superman is riding the bicycle. All you know is the truck met the bicycle 10 minutes from the airport. How do we know this? Because the truck would have driven the remaining 10 minutes to the airport, picked up the items, then driven 10 minutes back to the spot where they met, and returned to the Post Office at his normal time.

In other words,

the bicycle met the truck at a point that saved the truck 20 minutes of driving time. The speeds are irrelevant (other than being constant).I hope that makes it clearer.

Jon says

January 23, 2016 @ 17:11

@Dan, the confusion from your answer is you keep suggesting that the problem is stating that the truck drove for 30 min before meeting the bike. That may be true but has no impact on the answer. The critical info is that the bike drove for 30 min before reaching the truck, which is stated in the question but not very clearly as to which drove 30 min.

If you don’t consider that the bike drove for 30 min, then the plane could have arrived a month early and it took the poor biker that distance to drive what would have taken the truck it’s remaining 10 minutes.

So the solution is: the truck saved 20 min so it met the bike 10 min short of the usual arrival time. Then add 30 min that the BIKE already drove and you get 40. It’s coincidence that the truck also already drove 30 min.

Me says

December 18, 2016 @ 11:47

I understand the answer.

BUT! You need to say transactions are instantaneous.

Raj says

April 9, 2017 @ 10:47

Nowhere does the problem say that Bicycle traveled for 30 minutes …It’s the truck that travelled 30 minutes …so your statement …”Both the truck and the bicycle traveled 30 minutes toward one another” . …Is also an assumption …Neither is cycle speed mentioned ….So it could be a fraction of a second more than 10 minutes till endless time ….Depending on your imagination

Abhishek Patkar says

April 29, 2017 @ 04:53

Well,

Nowhere in the problem is it stated that the truck and the bicycle left from their respective locations at the same time. So I don’t understand why are we assuming that the truck had travelled for 30 mins.

The bicycle left as soon as the plane landed (instantaneous transactions). By the time bicycle reached the truck it is clear that it had been 30 mins. since the plane had arrived at the airport as the bicycle took 30 mins. to reach the truck. Had the bicycle been not there then it would have taken the truck further 10 mins

to reach the airport which adds to the 30 mins for which the plane had been at the airport. So assuming that the truck was scheduled to leave the post office so that it could reach the airport exactly at the time of the plane’s arrival, the plane had arrived 40 mins. before schedule.

Bob says

May 1, 2017 @ 12:51

The grammar of the problem statement says truck had travelled 30 mins before meeting the bike,

at which point you cannot solve the problem since you need to know how long the bike travelled.

If you know how long the bike travelled, add it to the 10 mins to get your solution. For example,

let’s say the plane landed at 9:00am, bike travelled 30 mins and met truck at 9:30am, and truck

was expecting to meet the plane at 9:40am. So in that case, the plane arrived 40 mins early.

The wording of the problem needs to be fixed to not imply the truck was sent out 30 mins before

meeting with the bike.

Doug says

May 28, 2017 @ 20:30

40 minutes is the correct answer but it is not explained correctly in the published answer.

The published answer spuriously claims the truck had already driven 30 minutes to reach the bicycle which may or may not be correct but in any case is irrelevant.

The bicycle had been ridden for 30 minutes to meet the truck at a point when the truck was 10 minutes from the plane. Assuming instantaneous transactions, the plane landed 30 minutes before the bicycle met the truck, saving the truck 10 minutes of travel to the plane. There’s your 40 minutes. Why make it more difficult than that?

Arjun says

August 10, 2018 @ 06:34

The Plane arrived 10 minutes early .

It took 30 minutes for the truck to meet the bicycle and therefore will take 30 minutes back to the post office ( assuming the perfect system)

So the truck say left at 9 am and was back at the post office at 10 am when it should have been back at 10:20 am

The truck was scheduled to reach the airport at 9:40 am and be back at 10:20 — 40 minutes to go and 40 minutes to come in a perfect system

Plane was schedule to land at 9:40 am

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