### Brain Teasers

# Camel and Watermelons

In city A on desert, there is a guy with his camel and 1000 watermelons. He wants to sell the watermelons in city B where a fair takes place. The distance between A and B is 99km.

However, the camel can only carry a maximum of 100 watermelons at a time. In addition to that, the camel eats one watermelon per each kilometer it travels.

The guy wants to bring to the fair as many watermelons as possible. How can he do it? And what is the maximum number?

However, the camel can only carry a maximum of 100 watermelons at a time. In addition to that, the camel eats one watermelon per each kilometer it travels.

The guy wants to bring to the fair as many watermelons as possible. How can he do it? And what is the maximum number?

### Hint

What if the distance was just 1km? What if it was 2? What if it was 5?### Answer

If the distance was 1, the obvious maximum would be 981 (as the camel needs to travel 19 times). If the distance was 2, the result is same as if the man had 981 watermelons and to go only 1km, so the result would be 962. When using this method, we can calculate that the final number of watermelons is 140. (Note: The result for distance of 6 is 886 but from this moment the camel needs to travel only 17 times to get all the watermelons 1km further, so the result for distance of 7 is 869)Hide Hint Show Hint Hide Answer Show Answer

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## Comments

Nov 21, 2001

The answer seems to be missing.

I think it is 1

(after all with a 99 km journey he has not got enough to get back for a second trip

I think it is 1

(after all with a 99 km journey he has not got enough to get back for a second trip

I fully agree with the previous comment

so he can sell 10 watermelons TOTAL, since he needs to make 10 trips.

Excellent! I really enjoyed this one. I believe however, that the answer is 139 and not 140 though.

Hey bluetwo, how did he get back to make the secon trip? the camel didn't have enough watermelons to get home. We have been imagining the letter home: Dear mum, camel dead, Watermelons gone, send money. Is there a fair closer to home for next year??

I had a lot of fun working this one out and the answer is indeed 140. When you have reached kilometer 11, you will have 801 watermelons. Making an extra trip back for watermelon #801 uses 2 watermelons, so leave it behind.

I don't follow the answer on this one...

I don't get it

I got 140 too after working it out. My procedure was a bit tedious, maybe there is a simpler method. Here's how I got the answer.. Initially, for every km, the camel uses up 19 watermelons. So after 6km, the camel has 886 left. Then it needs only 17 per km, so at the end of 11km, it has 801 left. Now it makes 8 trips(forgetting the 1 extra melon) thus requiring 15 mlons per km. So it has 695 melons after 18 km. Now its 13 melons per km, so after 26 km, its 591 melons. Then 11 melons per km, so 492 left after 35.Then 9 per km, leaving 393 after 46, then 7per km, leaving 295 after 60. Then 5 per km, leaving 200 after 79 kms. For the remaining 20 km, we use up 3 per km leaving a final answer of 140 watermelons to sell at the fair!

Very good teaser. I did not consider leaving any melons behind, so I ended up with 139.

whoosh....over my head

I suppose the guy will then sell his camel at the fair, since he's got no watermellons left to make the trip home with the camel.

No,no,no! The answer is ONE-HUNDRED-FORTY-TWO (142) melons.

While 19 crossings are initially needed to advance the mound a single km, there is a better move. Make the initial advance 100km/19. The guy then has exactly 900 melons and is, to the nearest meter, 93.737 kilometers from City B.

Because the second advance involves 100 fewer melons, it involves two fewer, but longer, crossings. Each will be 100km/17.

The eighth advance involves 5 crossings, each 20km, and puts 200 melons 19.008km from the goal.

On the final 3 crossings, the camel gets 19.008 X 3 melons, i.e. just over 57 melons. This leaves 142 intact for the fair.

For those using calculators instead of spreadsheet programs, the advances can be rounded down to a multiple of 50km and the guy will still arrive with 142 intact melons.

While 19 crossings are initially needed to advance the mound a single km, there is a better move. Make the initial advance 100km/19. The guy then has exactly 900 melons and is, to the nearest meter, 93.737 kilometers from City B.

Because the second advance involves 100 fewer melons, it involves two fewer, but longer, crossings. Each will be 100km/17.

The eighth advance involves 5 crossings, each 20km, and puts 200 melons 19.008km from the goal.

On the final 3 crossings, the camel gets 19.008 X 3 melons, i.e. just over 57 melons. This leaves 142 intact for the fair.

For those using calculators instead of spreadsheet programs, the advances can be rounded down to a multiple of 50km and the guy will still arrive with 142 intact melons.

Here is the how-to for people concerned about fractional melons.

If the crossing is 4 3/4 km, the return trip is made with 5 1/4 melons. (I like to avoid having 3 pieces of melon.) 1/2 km from return's end, the excess 1/2 melon is unloaded to be picked up on the next outward crossing. (While it can't be closer, it only needs to be this close because I like to minimize the time a melon is split.)

If the crossing is 4 3/4 km, the return trip is made with 5 1/4 melons. (I like to avoid having 3 pieces of melon.) 1/2 km from return's end, the excess 1/2 melon is unloaded to be picked up on the next outward crossing. (While it can't be closer, it only needs to be this close because I like to minimize the time a melon is split.)

MY ERROR-- in my first comment I intended to write that you could round donw to multiple of 50 meters, not km.

Oct 04, 2006

1) lets start!! 1000 melons at home.

2) now let the camel carry 200 melons.. the camel says NO..

3) so carry only 100 melons instead... camel says alright!

4) camel travels the 10 kms, not suprisingly the camel ate 10 melons as well, so 90 melons left right??

5) lets finish 99th Km. that means 99 melon eaten as well..

6) ok then sell all the melons left.. oh boy..that is only 1 pc of ***king melon.

7) ok then ill go back home and get some more..

camel started to travel back home..

9) camel stops. after 1st km. waiting for a wetermelon to eat.

10) unfortunately no melon.

11) camel died

12) the man is crying!

13) went back to fair

14) buys a skateboard.

15) heads home!

16) 900 melons left was stolen.

17) man died!

2) now let the camel carry 200 melons.. the camel says NO..

3) so carry only 100 melons instead... camel says alright!

4) camel travels the 10 kms, not suprisingly the camel ate 10 melons as well, so 90 melons left right??

5) lets finish 99th Km. that means 99 melon eaten as well..

6) ok then sell all the melons left.. oh boy..that is only 1 pc of ***king melon.

7) ok then ill go back home and get some more..

camel started to travel back home..

9) camel stops. after 1st km. waiting for a wetermelon to eat.

10) unfortunately no melon.

11) camel died

12) the man is crying!

13) went back to fair

14) buys a skateboard.

15) heads home!

16) 900 melons left was stolen.

17) man died!

Stil is pretty much correct! Here is how to think about it. The inefficiency arrives when the camel has to pick up a load which is less than 100. So we should aim to move the total load a distance that will allow the camel to consume 100 watermelons at a time. Therefore our first dump should be t a place where there will be 900 watermelons remaining. Let x be the distance travelled. Watermelons remaining = 9(100 - 2x) + (100 - x) = 900. x = 100/19

There is no need to drop any watermelons or pieces. Say the camel eats a watermelon 100 metres from base. It continues to base, picks up a load and continues another 900 metres until it eats the next. Final answer is 142.9766 or almost 143 watermelons. I reckon he just abandons the camel a few hundred metres from the end of the final trip and carries

the last watermelon home himself. That would make it a round 143 melons!

Good teaser - bad answer.

There is no need to drop any watermelons or pieces. Say the camel eats a watermelon 100 metres from base. It continues to base, picks up a load and continues another 900 metres until it eats the next. Final answer is 142.9766 or almost 143 watermelons. I reckon he just abandons the camel a few hundred metres from the end of the final trip and carries

the last watermelon home himself. That would make it a round 143 melons!

Good teaser - bad answer.

I based my answers on the concept that the camel would be nibbling her watermelon as she traveled. (If she downs a whole melon before starting, what's to say she can't down 2 or 3 or the 198 it would take to make it to the market and back?) The comment about stashing pieces was for anyone who didn't like the idea of starting the last leg of the first advance with pieces from 9 melons.

Actually the last watermelon would be due to be consumed only 3 metres from home but there would still be 43 melons on the camel so I guess he'd have to make several trips of 3 metres to get the last 43 melons home. If watermelons sold for $100 each then I reckon it'd be worth going back and forth a few times to save that last watermelon. I still think that 143 is a reasonable answer.

Jimbo's math was okay when he calculated 142.9766 melons: that means there are 976.6 meters of camel power available in the last melon started. That, however, means that melon would be started 23.4 meters from market, not 3 meters. This close to the market, the guy couldn't cache any melons secretly. By the time he got back from lugging in the few he could carry, some of the others would have be stolen.

Yeah - you're right Stil that hard stuff like 2.4% of 1000 metres is too difficult for me. Okay he gets to within 24 metres of the town and he has 143 melons right? So near and yet so far. There's no way man I'm feedin' one more melon or part thereof to this gluttenous camel. I wave a melon coloured hanky in front of the camel and he chases me for the last 24 metres!

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