Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 17 Jul 2019, 09:47

### GMAT 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56275
a, b, and c are three consecutive odd integers such that a < b < c. If  [#permalink]

### Show Tags

19 Aug 2018, 09:35
00:00

Difficulty:

45% (medium)

Question Stats:

65% (02:34) correct 35% (02:35) wrong based on 41 sessions

### HideShow 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) $$a^3 + 6a^2 + 4a$$

_________________
NUS School Moderator
Joined: 18 Jul 2018
Posts: 978
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: a, b, and c are three consecutive odd integers such that a < b < c. If  [#permalink]

### Show Tags

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.
_________________
Press +1 Kudos If my post helps!
VP
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1028
WE: Supply Chain Management (Energy and Utilities)
a, b, and c are three consecutive odd integers such that a < b < c. If  [#permalink]

### Show Tags

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)
_________________
Regards,

PKN

Rise above the storm, you will find the sunshine
a, b, and c are three consecutive odd integers such that a < b < c. If   [#permalink] 19 Aug 2018, 12:30
Display posts from previous: Sort by