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dsaqwert

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3000 kg of bananas are need to be moved from city A to city B which are 1000 km apart from each other. An elephant will do the task. However he has some shortcomings; he can lift/carry 1000 kg at most and he need to eat 1 kg of bananas for moving 1 km (with or without load). Is it possible to transport any bananas to city B in spite of this voracious elephant? If so, what is the maximum value?
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I think he can move only 1 kg from city A to City B.

 

When he starts feed it one kg and load it with 1000kg. As it starts with 1 kg to complete the distance it will need 999kg ( from 2nd km onwards).

 

so when he reaches town B only 1 kg will be left.

 

Think shd choose an alternative mode of transport.

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I can manage 500Kg

 

0-------250-------500--------------1000

 

1) Carry 1000 to 250 mark, dump 500, return to 0 mark

2) Carry 1000 to 250 mark, dump 500, return to 0 mark

3) Carry 1000 to 500 mark, dump 250, return to 250 mark

4) Carry 1000 from 250 mark, pick up 250 from the 500 mark, go to 1000

 

You're left with 500 bananas. Kill and cook the elephant and you have a feast.

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i've thot of a similar approach. This is just an atempt to elaborate arjmen's post.

 

250Km 250Km 1000Km

-----!-----------!-------------------

500 500 500

500 500

750

 

We can transfer a minimum of 500 Kg from A to B if we follow the steps below.

 

1. Elephant takes 1000Kg. Reaches the 250 Km mark with 750 Kg of bananas.Let it download 750Kg of bananas there and come back. During the return journey let it consume the 250 Kg of bananas remaining.

2. After the third iteration, there will be 500 + 500 + 750 Kgs of bananas at the 250Km mark. 750 Kg because it need not return to the starting pt. during the 3rd iteration.

3. From the 250th KM mark, take 1000 kg and put 500Kg in the 500Km mark and come back.

4. Carry the remaining 750 Kg to the 500 Km mark.

5. Now we have 500 + 500 Kg at the 500 Km mark. Carry it to the end pt and you still have 500 Kg of bananas.

 

HTH:)

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Another try

 

0-------P1-------P2--------------1000

 

The optimum condition would be to have a multiple of 2000 bananas at P1 and 1000 bananas at P2 since the lousy creature can carry only 1000kg at a time

 

0 to P1: 2 round trips and 1 one-way trip leaving us with 2000 bananas at P1. i.e. 5*P1=1000 => P1=200 Km

 

P1 to P2: 1 round trip and 1 one-way trip leaving us with 1000 bananas at P2.

i.e. 3*(P2-P1) = 1000 => P2-P1 = 1000/3 => P2 = 1600/3

 

At P2 we have 1000 bananas and (3000-1600)/3 = 1400/3 km to go

 

So at the 1000Km mark we are left with 1600/3 bananas.

 

I still say kill and cook the elephant and have a feast.

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