# 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

2
by aishu1

2014-08-20T05:26:06+05:30

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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.

i guess monkey can carry 533.33 bananas at max..... not 666.67 bananas
2014-08-20T20:09:02+05:30
Lets say, A is starting point and D is stoppage.
At max you can carry 1000 bananas.

(1) First part of journey
we have 3000 bananas so,
A -----> B (1000 bananas)
B -----> A (come back to take rest 2000 bananas)
A -----> B (1000 bananas)
B -----> A (come back to take last 1000 bananas)
A -----> B (1000 bananas)
Now lets make some optimization and assume that A ----> B lap is 1 km in distance, So we need to give monkey 5 bananas to eat according to question.
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, monkey covers 200 km distance.

(2) Second part of journey,
we have 2000 bananas so,
B -----> C (1000 bananas)
C -----> B (come back to take last 1000 bananas)
B -----> C (1000 bananas)
Now lets make some optimization and assume that B ----> C lap is 1 km in distance, So we need to give monkey 3 bananas to eat according to question.
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, monkey covers 1000/3 km distance.

(3) Third part of journey
we have 1000 bananas so,
C -----> D (1000 bananas)
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
(1) 200 km for first 1000 bananas
(2) 1000/3 km for next 1000 bananas
(3) 1400/3 km for last 1000 bananas

Also journey (1) and (2) consumed 1000, 1000 bananas each, so we have only 1000 bananas and distance to be covered is 1400/3 km So, total no. of bananas eaten by monkey in journey (3) = 1400/3

So at last bananas transferred at point D = 1000 - 1400/3 = 1600/3 = 533.33