## Farm messengers: Hard Math Puzzle

One dairy farmer sent off his boy with a message to the melon farmer down the road, and at the same time the melon farmer sent his girl off with a message to the dairy farmer.

One ran faster than the other, and they were seen to cross at a spot 720 yards from the melon farmer’s house. Each stopped 10 minutes at their destination and then started on the return journey, when it was found that they passed each other at a spot 400 yards from the melon farmer’s house.

**How far apart are the 2 farmer’s houses?** Of course each child went at a uniform speed throughout.

Not many have been able to solve this. So far, we have:

hex| PUZZLE MASTER | Profile August 12th, 2009 - 2:46 am1280 yards

aaronlau| Profile August 12th, 2009 - 6:12 amspeed of boy = a

speed of girl = b

distance between = d

at first intercept:

Time taken = 720/b = (d – 720)/a

at second intercept:

Time taken = (2d – 400)/b = (d + 400)/a

(As 10 mins is taken for both, it is negated.)

Therefore, b/a = 720/(d – 720) = (2d – 400)/(d + 400)

Solving, d = 1280 yards

jason| Profile August 12th, 2009 - 10:16 amless than a mile as the crow flies! =)

makada| Profile August 12th, 2009 - 11:09 pmThe answer is 1280 yards

makada| Profile August 12th, 2009 - 11:23 pmHow did I reach 1280 yards.

Variables: Distance from dairy to melon farm = s

Speed of boy = b

Speed of girl = g

The boy and girl cross each other twice.

Time taken by girl to reach first crossing point = 720 / g

Time taken by boy to reach first crossing point = (s- 720)/ b

These two times should be the same as they started at the same time.

So equating 720/g and (s-720)/b we get

b/g = (s-720)/720 ……. Equation (1)

Now coming to the second crossing point.

Time taken by the boy to reach the second crossing point = s/b + 10 + 400/b

(Where s/b is the time taken by the boy to reach the melon farmer, 10 is the time he spent there and 400/b is the time he took to cover 400 yards to meet the girl again)

Similarly for the girl we get,

Time taken by the girl to reach the second crossing point = s/g + 10 + (s-400)/g

These two times have to be same as they started at the same time and are meeting in real time.

Therefore equating these two expressions we get

(s+400)/b = (2s-400)/g [ The 10’s got canceled ]

From which we get,

b/g = (s+400)/(2s-400) ……………Equation (2)

Since equation 1 and equation 2 are equal, as they are both equal to b/g, we can equate them and get:

(s+400)/(2s-400) = (s-720)/720

From which we derive

s^2 – 1280s = 0

Solving which we get

s = 1280

rachel| Profile August 13th, 2009 - 12:18 ami took a shot at it not sure if it right though..

farms are 1280 yards apart

bobg| Profile August 13th, 2009 - 4:51 amA____720________–?–B

A__400__—(720!)—-B

The distance from B->A on the way back we KNOW is 720 (that is the distance travelled from A->B) so whatever is left is the remaining distance home.

720+400=1120=distance between houses.

JosephBach| Profile August 13th, 2009 - 12:15 pmThe distance between houses is 1280 yards.

bobg| Profile August 14th, 2009 - 1:40 amoops, my last comment was wrong, I forgot to account for the extra travel time of the slower walker.

bobg| Profile August 14th, 2009 - 2:14 am1120 is the distance travelled by the slower child between meetings.

2Y+320 is the distance travelled by the faster child between meetings.

720 is the distance travelled by the faster child before meeting

Y is the distance travelled by the slower child before meeting.

720+Y= Distance

1120+2Y+320= Distance

Solve for Y )

poopsicle| Profile August 14th, 2009 - 11:03 amVb = speed of boy (yd/time)

Vg = speed of girl (yd/time)

D = total distance between houses (yd)

x=Vb/Vg

time for boy to go (D-720)yd = time for girl to go 720 yd

(D-720)/Vb=720/Vg

D=720x+720

time for boy to go 720 yd, wait 10 min, go 400 yd= time for girl to go (D-720)yd, wait 10 min, go (D-400) yd

(720+400)/Vb=((D-720)+(D-400))/Vg

1120/x=2D-1120

D=560/x+560

720x+720=560/x+560

x=-1 or 7/9

x=7/9

D=720(7/9)+720

D=1280

1280 yards

I think there also might be a solution where the boy goes so slow that he doesn’t make it to the melon farmer’s before the second time the girl passes him, but in this situation the distance depends on the girl’s speed, which isn’t given

In this case,

D=160/x-5Vg+560

where x=Vb/Vg=((Vg^2+64Vg+19456)^(1/2)-Vg-32)/288

this yields

D=46080/((Vg^2+64Vg+19456)^(1/2)-Vg-32)-5Vg+560

D decreases asymptotically to 720 with increasing Vg.

Minimum distance for this scenario is 988.7 yards

poopsicle| Profile August 14th, 2009 - 3:15 pmoops, I meant ‘maximum’ distance in the last line of my previous post

MFox| Profile August 14th, 2009 - 6:43 pmI think that the farm houses are 1280 yards apart, and that the dairy farmer’s son was running seven ninths as fast as the melon farmer’s daughter.

I didn’t use any clever math to work this out, but I hope that my inelegant approach still led me there in the end.

I simply drew several simulations on paper, in which I established an arbitrary distance between the houses and varied the speeds of the two children. After a few repetitions, it became clear that the distance the girl traveled before the first meeting with the boy is precisely half the distance that she travels between the first and second meeting, irrespective of their relative speeds. So if she traveled 720 yards before the first meeting, then she traveled and additional 1440 yards before the second meeting, and only 400 more yards remained until she made it back home.

So if (d) is the total distance between the farmhouses, then:

(3 x 720)+ 400 = 2560 = 2d

d = 1280

michaelc| Profile August 14th, 2009 - 6:59 pmAs the crow flies or in country miles?

Let the distance equal X.

So the girl goes 720 yds, and the boy goes X-720 yds in the same amount of time.

Since they both met on the trail the second time around, means they both waited 10 minutes. So this is useful because we can assume the ratio of their speed is equal for both distances traveled.

So the 2nd time around the girl travels X-720 yds plus X-400 yds. The boy travels 720 yds plus 400 yds.

So 720 / (X-720) = (2x- 1,120)/1,120.

Solving for X you see X= 0, 1,280 yds. 0 is impossible, so we know it must be the other one.

Bobo The Bear| PUZZLE MASTER | Profile August 17th, 2009 - 2:36 pmThey are 1280 yds apart. The key idea is recognizing that at any time when they’re both traveling, the ratio of their speeds (G:B) equals the ratio of the distance they’ve traveled (for example, if the girl goes twice as fast, then she goes twice as far). Because the speed ratio is the same at both times they cross, the distance ratios will be the same as well.

At the first time that they cross, the distances they traveled are 720 for the girl and (D-720) for the boy. The second time they cross, the distances traveled are (2D-400) for the girl and (D+400) for the boy. Set these as ratios and make the ratios equal to each other, and you have:

720D-720) = (2D-400)D+400)

Solving this gives you D = 1280 yds.

(The 10 minute layover is a bit of a red herring – as long as it’s the same time for both, it doesn’t matter how long the layover is)

RK| Founder | Profile August 17th, 2009 - 8:13 pmThis one was very hard, and as you can see very few got the answer.

Aaronlau, MichaelC, MFox, Makada, Poopsicle, Bobo The Bear provide nice explanations.

makada| Profile August 18th, 2009 - 9:53 amI was just pondering….

Had they stayed at their respective destinations for different periods of time ( instead of the identical 10 mins wait ) , then would this problem be solvable.

For Example had the boy stayed at the melon farmer’s house for 5 mins and the girl for 10 mins at the dairy farmer’s house, then is it possible to solve this problem ? ( Keeping all other problem characteristics the same )

JosephBach| Profile August 18th, 2009 - 1:26 pmIn answer to makada’s pondering…

I believe that in order to solve the problem with differing wait times, you would need additional information as to how fast one of the messengers was going. As it stands, the problem allows you to find their relative speeds, but the 10 minute wait is the only invariable time measurement given.

If the girl was walking 47 yards per minute and the boy was walking 40, with wait times of 10 minutes and 5 minutes respectively, the distance between houses would become 1221 yards. However, with the girl walking roughly 42 yards per minute and the boy walking 30, using the same wait times as before, the distance between houses becomes 1236 yards.

makada| Profile August 19th, 2009 - 9:02 amSo I deduce that with different wait times , we cannot reach any single figure as the distance between the two houses. But is it possible to obtain a range ( minimum and maximum ) distance between the two farms. For example keeping wait time of 5 and 10 mins for boy and girl and other variables in the problem the same ?

michaelc| Profile August 22nd, 2009 - 9:29 pmHello makada,

Perhaps I can help you with your question?

In the example as given, we still do not know the speed of the boy or the girl. We only know the ratio between the 2, and this it turned out we could use to find the distance between the houses.

Let’s suppose the girl was traveling 720 yds/min. (12yds/sec kind of moving on!) Since we know the ratio of their speeds (or we could use the distance the boy traveled in this case), we can calculate the boy’s speed to be 560 yds /min. If we assume the girl is traveling at a lesser speed, the boy’s speed must lessen as well by the ratio which we know to be correct. Same thing if we assume the girl is traveling at a greater speed.

Now that we understand the first problem better, let’s move on to your question. Suppose the girl’s speed was near the speed of light, and the boy’s speed was that of a normal boy. The speed of the girl would make them meet almost simultaneously from the time they started. So the minimum distance would be ever so slighty above 720 yds. The girl of course would have to wait 10 minutes, and let the boy get to the other side, wait 5 minutes, and start again. The boy’s speed would have to be roughly 1,120 yds/5 min as this would be his approximate distance traveled in that time frame.

I myself found the absolute maximum distance much more harder to calculate. I’ll post that answer hopefully in a little bit.

michaelc| Profile August 22nd, 2009 - 10:57 pmWith the girl waiting 10 minutes, and the boy waiting 5 minutes, the distance must be greater than 720 yds, but less than 1,440 yds.

If the speeds were the same, they would meet half way. Yet the girl has 1,440-400 yds to travel and 10 minutes to wait to the boy’s 5 minute wait and 400 yds to travel (they both had to travel 720 yds before the start of the wait). So we know this isn’t possible. The girl has to go much faster than the boy, to wait the extra time and go the distance.

I didn’t calculate the exact number as it’s getting pretty late, but it looks like the distance could be no more than about 1,200 yds or so with the new waiting times.

The short answer is yes it is possible to obtain a min and max distance between the 2 farms.

Hope this helped!

MichaelC

makada| Profile August 23rd, 2009 - 7:59 pmThat’s a nice explanation michaelc

Thnx

Satya| Profile October 9th, 2009 - 12:58 pmIt must be 720 + 400, As both moved at their constant speeds.

1120 is the distance

RogHyde| Profile October 11th, 2009 - 5:45 pm1280 Yards

It took me a long time to work out but eventually I realised that between them they travelled a combined total of three times the distance between the places. The boy travelled 720 yards when they first met and as they had then travelled in total the actual distance between the places this must have been one-third of his total distance travelled, i.e. 2160 yards. He was then 400 yards from his destination (on his return trip) so (2160 + 400) = 2560 yards/2 = 21280 yards.

Roger

thesmarticlekid| Profile November 11th, 2009 - 8:59 pmsimple algebra

speed of boy = a

speed of girl = b

distance between = d

at first intercept:

Time taken = 720/b = (d – 720)/a

at second intercept:

Time taken = (2d – 400)/b = (d + 400)/a

(As 10 mins is taken for both, it is negated.)

Therefore, b/a = 720/(d – 720) = (2d – 400)/(d + 400)

Solving, d = 1280 yards

Triquetra| Profile December 6th, 2009 - 4:43 amGreat answers! I think I did this the hard way.

Rg = Rate of the girl

Rb = Rate of the boy

D = distance from 720yd mark to dairy farm

The girl travels the 720 yards at the rate Rg to the first meeting. The boy travels the unknown distance D at his rate, or D*Rb. So D*Rb = 720*Rg

The girl travels the unknown distance D twice plus 320yrds (720 – 400) from the time they meet at 720 yards to the time they meet at 400yds. Or 2*D*Rg + 320*Rg. From the same meeting place the boy travels 720yards to the melon farm plus 400 yards back to the second meeting. That is 1120*Rb. Set these equal and you have 2*D*Rg + 320*Rg = 1120*Rb. Solve D*Rb = 720*Rg for Rb by dividing both sides by D to get Rb = (720*Rg)/D. Substitute this in for Rb and get 2*D*Rg + 320*Rg = 1120*(720*Rg)/D. Cancel a Rg in each term to get 2*D + 320 = 1120*(720)/D. Multiply by D on both sides to get 2*D^2 + 320D = 806400.

Solve the quadratic equation to D = 560 (or -720). So the total distance from farm to farm is 720 + 560 = 1280.