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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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19 Aug 2018, 09:35
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45% (medium)

Question Stats:

70% (02:45) correct 30% (02:33) wrong based on 81 sessions

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

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

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13 Oct 2019, 19:07
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) $$a^3 + 6a^2 + 4a$$

A smart numbers solution would be to pick three consecutive odd integers for a, b, and c. When picking numbers for a Variables-in-the-Choices problem, avoid picking 0, 1, or any of the numbers in the problem (this can sometimes cause more than one answer to appear to be correct, thus necessitating starting over with another set of numbers).

So:
a = 3
b = 5
c = 7

Then, a is halved to become m, b is doubled to become n, and c is tripled to become p,

so:
1.5 = m
10 = n
21 = p

Since k = mnp, multiply the values for m, n, and p:
k = (1.5)(10)(21)
k = 315

Now, plug a = 3 (the value originally selected) into the answer choices to see which choice equals 315. Only (A) works.

Because the correct answer is a mathematical way of writing the situation described in the problem, this will work for any value you pick for a, provided that a, b, and c are consecutive odd integers and you calculate k correctly.
Re: a, b, and c are three consecutive odd integers such that a < b < c. If   [#permalink] 13 Oct 2019, 19:07
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