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

Question

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

Chethan92 · Answer

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.

PKN · Answer

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)

hudacse6 · Answer

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.

jahongirnorimov · Answer

One algebraic solution involves defining all three terms in terms of a. Since the terms are consecutive odd
integers, they are 2 apart from each other, as such:

a
b= a + 2
c = a + 4

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

1/2*a=b
2b = n
2(a + 2) = n
2a + 4 = n

3c = p
3(a + 4) = p
3a + 12 = p

Since k = mnp, multiply the values for m, n, and p:

k = ( 1/2*a) (2a+4) (3a+12)
k = ( 1/2*a) (6a2 + 24a + 12a +48)
k = ( 1/2*a) (6a2 + 36a + 48)
k = 3a3 + 18a2 + 24a

This is a match with answer choice (A).

luisdicampo · Answer

Deconstructing the Question  
 ,   are consecutive odd integers with  a<b<c , so:
 b=a+2,\quad c=a+4 
Define:
 m=a/2,\ n=2b,\ p=3c 
and  k=mnp . Find  k  in terms of  a .

Step-by-step  
Substitute  b=a+2 ,  c=a+4 :
 m=a/2 
 n=2b=2(a+2)=2a+4 
 p=3c=3(a+4)=3a+12

Compute:
 k=\frac{a}{2}(2a+4)(3a+12)

Factor:
 2a+4=2(a+2),\quad 3a+12=3(a+4)

So:
 k=\frac{a}{2}\cdot 2(a+2)\cdot 3(a+4)=3a(a+2)(a+4)

Expand:
 (a+2)(a+4)=a^2+6a+8 
 k=3a(a^2+6a+8)=3a^3+18a^2+24a

Answer: A

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

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