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An odd number of stones lie along a straight path, the
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28 Apr 2008, 04:17
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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: An odd number of stones lie along a straight path, the
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28 Apr 2008, 06:48
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
Re: An odd number of stones lie along a straight path, the
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29 Apr 2008, 07:09
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..
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gmatclubot
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