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a, b, and c are three consecutive odd integers such that a <

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a, b, and c are three consecutive odd integers such that a < [#permalink] New post 19 Aug 2018, 04:08
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83% (03:28) correct 16% (01:10) wrong based on 6 sessions
a, b, and c are three consecutive odd integers such that a < b < c. If a is halved to become m, b is doubled to become n, c is tripled to become p, and k = mnp, which of the following is equal to k in terms of a?

(A) \(3a^3 + 18a^2 + 24a\)
(B) \(3a^3 + 9a^2 + 6a\)
(C) \(\frac{11}{2}a + 16\)
(D) \(6a^2 + 36a + 24\)
(E) \(a^3 + 6a^2 + 4a\)
[Reveal] Spoiler: OA

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Re: a, b, and c are three consecutive odd integers such that a < [#permalink] New post 01 Feb 2019, 09:02
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Solution,

Since the question asked about the value of k in terms of a,

let us suppose three consecutive odd integers a, b and c be a, a + 2 and a + 4 such that a<b<c

Given that,
m = 1/2 a
n = 2( a + 2 )
p = 3 ( a + 4 )

k = mnp
= 1/2 a * 2 ( a+2 ) * 3 ( a+4 )
= 6a (a +2) (a + 4)/2
= 3a ( a^2 + 6a + 8)
= 3a ^ 3 + 18 a ^ 2 + 24a

Thus, the answer is A.
Re: a, b, and c are three consecutive odd integers such that a <   [#permalink] 01 Feb 2019, 09:02
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a, b, and c are three consecutive odd integers such that a <

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