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arithmetic – A camel transporting bananas

Its fairly positive that, If the camel takes all of the 1000 bananas without delay to the top level,he can be left with 0 bananas on the finish.
So, there must some Intermediate Points which are in between O km and 1000 km.

So, to show my level, Let us assume 4 factors.

let’s referred to as level A——>Beginning Point
level B——> End Point
level Q——>First Intermediate Point
level W——>Second Intermediate Point

3000 2000 1001 533
<——5X———-> <—-3Y—–> <————Z———————>

If we take 1000 bananas without delay to 1km,then there shall be 3 journeys ahead, and a pair of journeys backward, so , there shall be a complete of 5 journeys.

so, When we take 1000 bananas, we’re solely left with 600 bananas and equally for the opposite one, however for the final one, there shall be only one ahead journey and so, its a complete of 5 journeys.
Let’s show that with a easy equation,

So, Point Q is 200km away from the begining.

Now, we’re solely left with 2000 bananas, and taking it to W,
As we’re solely left with 2000 bananas, we will make a most of two journeys ahead, and 1 journey backward.

It may be understood as now we have solely 2000 bananas left, and the camel can take 1000 bananas a time.

Let’s discover the gap with the Equation,
So, Y=333. But because the Camel is consuming 1 banana per km,so, it may possibly’t be in fraction,
thus, Y=333
As we diminished the time period,
so, will improve 1 banana,i.e, 1000+1=1001.

Again,there’ll solely be 1 final journey, with the remaining distance,
So, Remaining Distance=467.
And 1000-467=533

Thus, complete Bananas reached at vacation spot=533.




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