# A monkey has to carry 3000 bananas over 1000kms at a time 1000 bananas it can carry..provided monkey eat 1 banana per km..find maximum no of bananas which can reach the other end

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by aishu1

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by aishu1

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Observe the diagram enclosed.

First, we see that if monkey carries the maximum, the there is possibility of maximum bananas remaining. But he/she finally ends up eating all bananas, if he carries in the regular way. So there has to be some other way.

Let us suppose monkey starts from A carrying 1000 bananas (maximum possible). He/she halts in between at B, which is X km from the start (A). He /she deposits (1000 - X) the bananas remaining after eating X bananas, over there at B. Now he/she goes back to A and fetches 1000 more bananas. Again deposits 1000-X at B. Finally goes back and fetches the last lot of 1000 and deposits 1000 - X at B. She/He does not any on the way return, as there no bananas to carry or to eat.

Now at B we have (3000 - 3 X) bananas. There are two possibilities.

Monkey makes one trip from B to C or two trips fro m B to C.

__ONE TRIP from B to C:__

** **

That means we have a maximum 1000 bananas.

So, 3000 - 3 X <= 1000 => X >= 2000/3 km = 666.67 km

Then the number bananas carried to the other end C are:

(3000 - 3 X) - (1000 - X) = 2000 - 2 X as X >= 666.67 km, this number is

<= 2000 - 1333.34 =__666.66 bananas left __ __ (this is the maximum.__)

If the halting point is farther than 666.67 km, then number of bananas reaching C is less.

**TWO TRIPS FROM B to C**

Bananas remaining at B: 1000 <= (3000 - 3 X) <= 2000

so 333.34 km <= X <= 666.67 km

Number of Bananas reaching the other end is = (3000 - 3 X) - 2 (1000 - X)

= 1000 - X

Substituting the above range of X, we get 666.67 to 333.33 bananas.

*So, when X = 333.34 km, monkey carries over 666.67 bananas - maximum possible.*

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Answer 666.67 bananas, with an intermediate halt at 333.33 km or 666.67 km from the start.

First, we see that if monkey carries the maximum, the there is possibility of maximum bananas remaining. But he/she finally ends up eating all bananas, if he carries in the regular way. So there has to be some other way.

Let us suppose monkey starts from A carrying 1000 bananas (maximum possible). He/she halts in between at B, which is X km from the start (A). He /she deposits (1000 - X) the bananas remaining after eating X bananas, over there at B. Now he/she goes back to A and fetches 1000 more bananas. Again deposits 1000-X at B. Finally goes back and fetches the last lot of 1000 and deposits 1000 - X at B. She/He does not any on the way return, as there no bananas to carry or to eat.

Now at B we have (3000 - 3 X) bananas. There are two possibilities.

Monkey makes one trip from B to C or two trips fro m B to C.

That means we have a maximum 1000 bananas.

So, 3000 - 3 X <= 1000 => X >= 2000/3 km = 666.67 km

Then the number bananas carried to the other end C are:

(3000 - 3 X) - (1000 - X) = 2000 - 2 X as X >= 666.67 km, this number is

<= 2000 - 1333.34 =

If the halting point is farther than 666.67 km, then number of bananas reaching C is less.

Bananas remaining at B: 1000 <= (3000 - 3 X) <= 2000

so 333.34 km <= X <= 666.67 km

Number of Bananas reaching the other end is = (3000 - 3 X) - 2 (1000 - X)

= 1000 - X

Substituting the above range of X, we get 666.67 to 333.33 bananas.

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Answer 666.67 bananas, with an intermediate halt at 333.33 km or 666.67 km from the start.

At max you can carry 1000 bananas.

(1) First part of journey

we have 3000 bananas so,

Now lets make some optimization and assume that A ----> B lap is 1 km in distance, So we need to

But what we need to have 2000 bananas left so, say the monkey can travel to d km, 3000 - 5d1 = 2000,

So, d1 = 200

So,

(2) Second part of journey,

we have 2000 bananas so,

Now lets make some optimization and assume that B ----> C lap is 1 km in distance, So we need to

But what we need to have 1000 bananas left so, say the monkey can travel to d km, 2000 - 3d1 = 1000,

So, d1 = 1000/3

So,

(3) Third part of journey

we have 1000 bananas so,

Also, the total distance covered before starting third part is (200) + (1000/3) = 1600/3 km. So the rest distance that is covered in third part of journey is (1000) - (1600/3) = (1400/3) km.

Finally, we have

Also

So at last