Can you answer this?

Question

GOING BANANAS!

Cleopatra ("Cleo") the Camel works for the owner of a small, remote banana plantation. This year's harvest consists of three thousand bananas. Cleo can carry up to one thousand bananas at a time. The market place where the bananas are sold is one thousand miles away. Unfortunately, Cleo eats one banana each and every mile she walks.

Your Task:

Of the three thousand bananas harvested, what is the largest number of bananas Cleo can get to market?

and the answer is not 0 or 3. She can stop and drop of bananas and get more, but she even has to eat a banana for those miles she walked back.

Answer 1

533 and i know it cuz i saw the anser

Answer 2

I don't think Cleo can get any bananas at all to the market. Just think about it for a minute or two. If Cleo can carry 1000 bananas at a time and eats one per mile of a 1000 mile long trip, that will leave zero bananas left when she reaches the market.

1000 bananas carried at a time - 1000 (one eaten per mile) = 0

Answer 3

If Cleo makes three trips to a spot X miles from the plantation, she will accumulate there a total of Y bananas:

2.5 round trips (i.e. 2 full round trips + 1 half trip) x [1000 bananas max load per trip - (2 x Xmiles)] = Y bananas. (There are 2.5 round trips because Cleo will carry the maximum number of bananas, 1000, for each load.)

Note that Y <= 1000 because this is the maximum number of bananas Cleo can carry on the second leg of her journey, which is the remaining distance to the market.

From here the total carried to market = Y bananas (max 1000) - (1000 - X miles)

If we optimize the first equation so that it equals 1000:

2.5(1000 - 2X) = 1000
2500 -5X = 1000
1500 = 5X
300 = X

we find that the optimal distance for the first stop is 300 miles. Plugging this into the second equation we find that the maximum number of bananas that Cleo can get to market is 300.

Answer 4

Answer: 600 bananas.
b= bananas
m = miles
S= start point
M1= interm. point 1
M2= interm. point 2
M3= interm. point 3
M4 = interm. point 4
A= arrival
1. Cleo walks 100 m from S carrying 1.000 b; eats 100 b. Leaves 800 b in M1(in storage) and walks back the 100 m with 100 b to eat; arrives at S without b.
2. Cleo repeats this once more. There are now 1.600 b in M1 because she took again 100 b for her last trip to S.
3. Arrives at the M1 with 900 b. (Doesn´t have to go back; all b in S were either carried or eaten).
4. She has now 2.500 b and more 900 m to go.
5. Cleo takes 1.000 b to M2, leaving 1.500 b in point M1.
6. At M2 leaves 800 b, taking 100 b to eat in her way back to M1.
7. At M1 takes 1.000 b to M2, leaving 500 in M1.
8. Leaves 800 b in M2, goes back to M1, eating 100b.
9. Takes the 500 b left in M1. Arrives at M2 with 400b. There are now 2.000 b in M2.
10. Cleo takes 1.000 b from M2 to M3. As usual, leaves there 800 b at M3, and goes back with her 100 b provision to reach M2.
11. In M2 she carries her 1.000 b to M3. Leaves there 800 b and goes back to M2, placidly chewing her share of 100 b.
12. Arrives there empty handed so she can carry all that was left, i. e., 1.000 b to M3. Arrives in M3 with 900 b. Now, she has at M3 1.700 b.
13. Carries 1.000 b to M4; leaves 800 b. Goes back to M3, almost fed up with b.
14. Takes the 700 b that were in M3. Arrives in M4 with 600 b.
She has now 1.400 b in M4.
15. Takes 1.000 b to M5; leaves there 800 b.
16. Back to M4 carries all the 400 b to M5. Arrives with 300.
17. More 1.000 b from M5 to M6. Leaves 800 b in M6.
18. Goes back to M5 eating her part of 100 b. Takes all 400 b to M6 and arrives there with 300 b.
19. Gets some rest, drinks some water. Gets up early in the morning and carries her last 1.000 b, leaving 100 b behind to be eaten by the vultures.
20. She goes very slowly the last 400 m, while eating 400 b.
21. Arrives in A with 600 b, swearing she will never eat a b again til the next week.
Uff! That was fun!

Answer 5

1000 bannanas