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An odd number of stones lie along a straight path, the

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Manager
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An odd number of stones lie along a straight path, the [#permalink] New post 28 Apr 2008, 03:17
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An odd number of stones lie along a straight path, the distance between consecutive stones being 8m. the stones are to be collected at the place where the middle stone lies. Aman carry only one stone at a time. He starts carrying the stones beginning at the extreme. If he covers a path of 2.4km, how many stones are there?
1) 13
2) 25
3) 21
4) 39
5) 33

Please explain.
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Re: PS-Sequence [#permalink] New post 28 Apr 2008, 05:48
Value wrote:
An odd number of stones lie along a straight path, the distance between consecutive stones being 8m. the stones are to be collected at the place where the middle stone lies. Aman carry only one stone at a time. He starts carrying the stones beginning at the extreme. If he covers a path of 2.4km, how many stones are there?
1) 13
2) 25
3) 21
4) 39
5) 33

Please explain.


B. 25

I have developed some weird formula :)
I shall post it soon...need to think over the explanation

for now, do [32*(sum(n/2-1...1) + 3* n/2 * 8 )]
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Re: PS-Sequence [#permalink] New post 28 Apr 2008, 07:44
i get 33..i am not sure though..
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Re: PS-Sequence [#permalink] New post 29 Apr 2008, 06:09
Value wrote:
An odd number of stones lie along a straight path, the distance between consecutive stones being 8m. the stones are to be collected at the place where the middle stone lies. Aman carry only one stone at a time. He starts carrying the stones beginning at the extreme. If he covers a path of 2.4km, how many stones are there?
1) 13
2) 25
3) 21
4) 39
5) 33

Please explain.


(B) We have N-1 stones that should be transfered to the middle. To make things easier, I divided the whole set into 2 parts - left and right, a man will have to bring "n" stones from the left and "n" stones from the right, where n=(N-1)/2, N=2n+1.
So, we have an arithmetical progression with n stones that should be brought to the "zero" point. To bring the first one a man will have to travel 8+8=16 meters, the second one - 16+16=32 meters, the 3rd one - 24+24=64m .. The formula of progression is: Mn=16+16(n-1), where M - number of meters to bring a stone number n to the zero point. the sum of all units:
(16 + 16+16(n-1))/2 * n. It gives us quantity of meters that a man will travel to collect half of all stones.
But this value doesn't equal 2400/2 meters, because "He starts carrying the stones beginning at the extreme". So, if we want this man to travel 2 times for each stone, we should add 8*n meters - a way he should travel from the middle to one of the extreme stones. Then
(16 + 16+16(n-1))/2 * n = (2400 + 8n)/2
(16 + 16n)/2 * n = 1200 + 4n
(8 + 8n) * n = 1200 + 4n
8n^2 + 4n = 1200
2n^2 + n - 300 = 0
n1=12; n2=-12.5 <0 - leave out
so, N= 2n+1 = 25..
Ufff.... But, actually, I don't think we are supposed to solve this problem like that and it's better to:
1. divide the whole set into 2 equal parts (left and right) with the abovementioned formula An=16+16(n-1) for each item (number of meters that should be traveled to collect a stone) and the formula (16 + 16+16(n-1))/2 *n for the sum of all "items" of progression.
2. take the sum of all items of progression as 2400/2=1200 meters and don't mind that "He starts carrying the stones beginning at the extreme"
3. Try to plug different numbers from the offered variants, keeping in mind that n=(N-1)/2, i.e. that from the variant (B) you take not 25, but (25-1)/2=12.

Hope not to meet such questions in the test. It's a real time killer..
Re: PS-Sequence   [#permalink] 29 Apr 2008, 06:09
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