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

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a, b, and c are three consecutive odd integers such that a < b < c. If  [#permalink]

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New post 19 Aug 2018, 09:35
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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) \(a^3 + 6a^2 + 4a\)

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Re: a, b, and c are three consecutive odd integers such that a < b < c. If  [#permalink]

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New post 19 Aug 2018, 09:43
let's consider 3 odd integers. a=1, b=3, c=5.
Now a=0.5=m
b=6=n
c=15=p
k=mnp = 0.5*6*15 = 45.
Only answer A satisfies the requirement.
A is the answer.
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a, b, and c are three consecutive odd integers such that a < b < c. If  [#permalink]

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New post 19 Aug 2018, 12:30
Bunuel wrote:
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) \(a^3 + 6a^2 + 4a\)


Another approach:

We know, consecutive odd numbers differ by 2.
so, we can say, the three consecutive odd numbers are:
a, b=a+2, c=a+4
Given, a/2=m, 2(a+2)=n and 3(a+4)=p
Given, \(k=mnp=\frac{a}{2}*2(a+2)*3(a+4)\)=\(3a(a+2)(a+4)=3a(a^2+6a+8)=3a^3+18a^2+24a\)

Ans. (A)
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a, b, and c are three consecutive odd integers such that a < b < c. If   [#permalink] 19 Aug 2018, 12:30
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