GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
a, b, and c are three consecutive odd integers such that a <
[#permalink]
19 Aug 2018, 04:08
19 Aug 2018, 04:08
Expert Reply
3
Bookmarks
00:00
Question Stats:
65% (02:30) correct
34% (02:13) wrong based on 44 sessions
Hide Show timer Statistics
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) \(3a^3 + 18a^2 + 24a\)
(B) \(3a^3 + 9a^2 + 6a\)
(C) \(\frac{11}{2}a + 16\)
(D) \(6a^2 + 36a + 24\)
(E) \(a^3 + 6a^2 + 4a\)
Re: a, b, and c are three consecutive odd integers such that a <
[#permalink]
01 Feb 2019, 09:02
01 Feb 2019, 09:02
1
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.
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]
13 Oct 2019, 18:05
13 Oct 2019, 18:05
2
sandy 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) \(3a^3 + 18a^2 + 24a\)
(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.
