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A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]
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04 Nov 2014, 09:37
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Tough and Tricky questions: Sequences. A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A? A. 24(5^12) B. 23(5^12) C. 24(5^6) D. 23(5^12) E. 24(5^12) Kudos for a correct solution.
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A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]
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10 Nov 2014, 02:14
\(H = 5^{13}\) \(G = 5^{12}\) \(H  G = 5^{13}  5^{12} = 4 * 5^{12}\) A ...... 1....... B ........ 2....... C ........ 3......... D .......... 4........... E ...... 5....... F ........ 6......... G ................... HFrom point A, \(4 * 5^{12}\) has to be added 6 times to reach point \(G = 5^{12}\) \(A + 6* 4 * 5^{12} = 5^{12}\) \(A = 23 * 5^{12}\) Answer = B
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]
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04 Nov 2014, 10:29
Bunuel wrote: Tough and Tricky questions: Sequences. A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A? A. 24(5^12) B. 23(5^12) C. 24(5^6) D. 23(5^12) E. 24(5^12) Kudos for a correct solution.Ans : B Assume that the numbers appear as shown below on the number line ABCDEFGH   (5^12) (5^13) As the values for G and H are given , we can calculate the difference between any two terms of the series . Common Difference ,d = (5^13)  (5^12) = (5^12) *[ 51 ] = (5^12)*(4) Also F + d = G as the terms are in equidistant and in increasing order. So F + (5^12)*(4) = (5^12). That is , F = (5^12)  (5^12)*(4) = (5^12)[ 14] = (5^12) (3) Similarly , E = F  d = (5^12)[34] = (5^12)*(7) You can see a 4 getting added to the nonexponent part of the values . That is , according to the pattern , D SHOULD BE (5^12)*(74)= (5^12)*(11) Following this pattern , A = (5^12)*(23) Ans : B



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A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]
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17 Nov 2015, 17:49
Bunuel wrote: Tough and Tricky questions: Sequences. A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A? A. 24(5^12) B. 23(5^12) C. 24(5^6) D. 23(5^12) E. 24(5^12) H = 5^13 G = 5^12 Distance between H and G = 5^13  5^12 = 5^12(5^1  1) = (5^12)(4) = (4)(5^12)So, F = 5^12  (4)(5^12)E = 5^12  (4)(5^12)  (4)(5^12) D = 5^12  (4)(5^12)  (4)(5^12)  (4)(5^12)C = 5^12  (4)(5^12)  (4)(5^12)  (4)(5^12)  (4)(5^12)B = 5^12  (4)(5^12)  (4)(5^12)  (4)(5^12)  (4)(5^12)  (4)(5^12)A = 5^12  (4)(5^12)  (4)(5^12)  (4)(5^12)  (4)(5^12)  (4)(5^12)  (4)(5^12)So, A = 5^12  [ (4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12) + (4)(5^12)] = 5^12  (24)(5^12)= (1)(5^12)  (24)(5^12)= (23)(5^12) = Answer choice B Cheers, Brent
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]
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16 Jan 2017, 18:12
Bunuel wrote: Tough and Tricky questions: Sequences. A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^13, respectively, what is the value of A? A. 24(5^12) B. 23(5^12) C. 24(5^6) D. 23(5^12) E. 24(5^12) We are given that A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size, and that the distance between any two consecutive numbers, which we can denote by the variable n, is constant. Since G and H are equal to 5^12 and 5^13, respectively, then the distance between them is: n = 5^13  5^12 n = 5^12(5  1) n = 5^12(4) Since A is 6 numbers ahead of G, A is 6n less than G, and therefore: A = G  6n A = 5^12  6[5^12(4)] A = 5^12[1  6(4)] A = 5^12[23] = 23(5^12) Answer: B
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Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa [#permalink]
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31 Aug 2017, 05:39
How I solved it is: A, B, C, D, E, F, G, H (Total of 8 numbers = n) Since its a sequence and each term is a certain constant more than the other. H  G = 5^13  5^12 = (5^12) (51) = 4 x (5^12) = Our common difference, d.
Also unrelated, G = A + (n1) d 5^12 = A + (71) 4 (5^12) As n=7, because its the 7th term in the sequence
Therefore, A = (5^12)  24 (5^12) = (5^12) (124) Taking 5^12 common =  23 (5^12) = Answer choice B




Re: A, B, C, D, E, F, G, and H are all integers, listed in order of increa
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31 Aug 2017, 05:39






