If a , b, and c are integers and
ab^2/c is a positive even integer, which of the following must
be true?
I. ab is even
II. ab > 0
III. c is even
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
If a , b, and c are integers
This topic has expert replies
-
earth@work
- Master | Next Rank: 500 Posts
- Posts: 248
- Joined: Mon Aug 11, 2008 9:51 am
- Thanked: 13 times
IMO: I only
ab^2/c is a positive even integer
I. ab is even : since ab^2 is even integer the either a or b or both have to be even implies ab is even, both a & b cannot be odd
either case ab will be even...true always
II. ab > 0 not necessary true, ab^2/c positive implies a,b^2 & c are all positive but we cannot say about b...it cud be positive or negative
III. c is even...not necessarily true, c cud be odd if a is odd as ab^2 is an integer.
if b is even : a cud be even leads to c even; OR a is odd leads to c odd or even
if b is odd, then a is even & c cud be even or odd
ab^2/c is a positive even integer
I. ab is even : since ab^2 is even integer the either a or b or both have to be even implies ab is even, both a & b cannot be odd
either case ab will be even...true always
II. ab > 0 not necessary true, ab^2/c positive implies a,b^2 & c are all positive but we cannot say about b...it cud be positive or negative
III. c is even...not necessarily true, c cud be odd if a is odd as ab^2 is an integer.
if b is even : a cud be even leads to c even; OR a is odd leads to c odd or even
if b is odd, then a is even & c cud be even or odd
you just answered yourself.pbanavara wrote:well .. why not I and II ?
ab has to be positive because .. if b is negative .. a can be negative and c can be negative. Am I missing something ?
- pradeep
The question is what MUST be true.
if A was negative and B was positive, then ab < 0,
but if C was negative, the statement is true again that the value is positive.
Therefore, it does NOT have to be true at all times.
-
pbanavara
- Master | Next Rank: 500 Posts
- Posts: 279
- Joined: Wed Sep 24, 2008 8:26 am
- Location: Portland, OR
- Thanked: 6 times
Oh man .. missed that MUST clause .. dankahwiya320 wrote:you just answered yourself.pbanavara wrote:well .. why not I and II ?
ab has to be positive because .. if b is negative .. a can be negative and c can be negative. Am I missing something ?
- pradeep
The question is what MUST be true.
if A was negative and B was positive, then ab < 0,
but if C was negative, the statement is true again that the value is positive.
Therefore, it does NOT have to be true at all times.
-
niraj_a
- Legendary Member
- Posts: 708
- Joined: Sun Jun 01, 2008 4:59 am
- Location: USA
- Thanked: 13 times
- Followed by:1 members
I only
i did the MGMAT may (dunno why it clicked)
ab^2 = c * even
(note how doing this doesn't violate the rule that c could be 0, because the question doesn't test for that here).
so for RHS to be correct, LHS must be even as well. II is out since b's sign can be hidden coz of the square. c doesn't have to be even for RHS to be even, because anything multiplied by even will be even.
i did the MGMAT may (dunno why it clicked)
ab^2 = c * even
(note how doing this doesn't violate the rule that c could be 0, because the question doesn't test for that here).
so for RHS to be correct, LHS must be even as well. II is out since b's sign can be hidden coz of the square. c doesn't have to be even for RHS to be even, because anything multiplied by even will be even.
GMAT/MBA Expert
-
Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7274
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We are given that (a*b^2)/c is a positive even integer. Therefore, a*b^2 must be even. (If a*b^2 is odd, (a*b^2)/c can't ever be even.)logitech wrote:If a, b, and c are integers and a*b^2/c is a positive even integer, which of the following must be true?
I. ab is even
II. ab > 0
III. c is even
A. I only
B. II only
C. I and II
D. I and III
E. I, II, and III
Now recall that the product of an even number and any integer is even, so either a or b, or both, must be even. Thus we see that ab must be an even integer. However, ab DOES NOT have to be greater than zero, since a could be -2 and b could be 1. Finally, we see that c does not have to be even, since a could be -2, b could be 1, and c = -1. Thus, only Roman numeral I must be true.
Answer: A
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
