Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 46142

Question Stats:
67% (01:48) correct 33% (02:06) wrong based on 168 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 46142

Re M2004 [#permalink]
Show Tags
16 Sep 2014, 01:07
Official Solution:If \(A\), \(B\), \(C\), and \(D\) are integers such that \(A  C + B\) is even and \(D + B  A\) is odd, which of the following expressions is always odd? A. \(A + D\) B. \(B + D\) C. \(C + D\) D. \(A + B\) E. \(A + C\) Rewrite \(A  C + B\) as \((A + B)  C\) and \(D + B  A\) as \(D + (B  A)\). If \(A + B\) is even, \(B  A\) is even too. If \(A + B\) is odd, \(B  A\) is odd too. Now it follows from the stem that:  if \(A + B\) is even, \(C\) is even and \(D\) is odd
 if \(A + B\) is odd, \(C\) is odd and \(D\) is even
Thus, \(C + D\) is always odd. Answer: C
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 07 Sep 2014
Posts: 4

Re: M2004 [#permalink]
Show Tags
20 Dec 2014, 19:54
I think I may have found a much more simple methodology to answer this question:
(1) A  C + B = even; thus, A  C + B = 2k (where k is an integer)
(2) D + B  A = odd; thus, D + B  A = 2k + 1 (where k is an integer)
rearrange (2): D + B  A  1 = 2k
Sub: rearranged (2) into (1): A  C + B = D + B  A  1 > 2A + 1 = C + D
thus, per my method above, if C + D = 2A + 1 (where A is an integer), C + D is ALWAYS ODD



Manager
Joined: 14 Sep 2014
Posts: 91
WE: Engineering (Consulting)

Re M2004 [#permalink]
Show Tags
08 Jan 2015, 10:41
I think this question is good and helpful.



Intern
Joined: 03 Nov 2014
Posts: 7
Location: India
GPA: 3

Re: M2004 [#permalink]
Show Tags
15 Apr 2015, 09:06
bartone89 I believe your method is wrong because it assumes the values of the two equations are consecutive. ie 2k and 2k+1. I hope I havent misunderstood you solution. Anyway, I think I may have found a better method. The two equations are AC+B and D+BA, even and odd respectively. On subtracting the two, you get, AC+BDB+A = 2ACD (which is odd, since even  odd = odd) 2A(C+D) is odd. Which means C+D is odd since we know that 2A has to be even. C+D is one of the options. Hope this makes sense.



Manager
Status: Remember to Always Think Twice
Joined: 04 Nov 2014
Posts: 54
Location: India
GPA: 3.51

Re: M2004 [#permalink]
Show Tags
19 Jan 2016, 06:25
There is an easier way this question can be solved. Lets say AC+B = 2n (even) and after rearranging the second equation, A+D+B=2m+1 (odd) Adding the two equations we get, 2B+DC=2(m+n)+1 From this, 2B is even and 2(m+n) is even, hence we conclude, DC=odd When difference of two integers is odd, there sum will also be odd; Hence D+C = odd
_________________
breathe in.. and breathe out!



Board of Directors
Joined: 17 Jul 2014
Posts: 2730
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: M2004 [#permalink]
Show Tags
20 Feb 2016, 13:58
I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd.



Manager
Joined: 16 Feb 2016
Posts: 53
Concentration: Other, Other

Re: M2004 [#permalink]
Show Tags
26 Apr 2016, 18:25
mvictor wrote: I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd. Very good solution mvictor!



Intern
Joined: 31 Jul 2016
Posts: 19

Re: M2004 I solved it with a different approach! Just Sharing [#permalink]
Show Tags
27 Nov 2016, 06:40
Add or Substract the two equations
AC+B=Even D+BA=Odd =2A+CD (EvenOdd is always Odd) and we know 2A is always even
Therefore CD should be Odd and thus also C+D



Manager
Joined: 01 Nov 2016
Posts: 69
Concentration: Technology, Operations

Re: M2004 [#permalink]
Show Tags
06 May 2017, 14:56
y4nkee wrote: mvictor wrote: I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd. Very good solution mvictor! I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic. AC+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 12+3 = 2 which is even. D+BA is odd. Reuse the same variables from before. D + 3  1 is D + 2. Set D to an easy odd number which is 5. So D = 5. Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers. A. A+D = 1 + 5 = 6 B. B+D = 3 + 5 = 8 C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C. D. A+B E. A+C



Intern
Joined: 22 Mar 2017
Posts: 7

Re: M2004 [#permalink]
Show Tags
08 Jun 2017, 08:06
Given AC+B =Even and D+BA=Odd Subtract 1) and 2) 2ACD=Evenodd=Odd=> 2A=Odd (C+D) Left side of the equation is even hence C+D has to be Odd.



Intern
Joined: 15 Oct 2014
Posts: 10
GPA: 4
WE: Project Management (Energy and Utilities)

Re: M2004 [#permalink]
Show Tags
25 Jun 2017, 04:01
Given AC+B=Even1 & D+BA = Odd2 Add 1 & 2 even + odd = odd hence CD= 2B  Odd= Odd if CD is odd, then C+D is also Odd. (eg 43=1, 4+3=7)



Intern
Status: Don't watch the clock,Do what it does, Keep Going.
Joined: 10 Jan 2017
Posts: 45

Re: M2004 [#permalink]
Show Tags
18 Sep 2017, 19:15
+Ac+B= even D+BA = odd Now equation 1  equation2 which gives 2A(D+C)=odd 2A is always even so (D+C) is always odd Hence C



Board of Directors
Joined: 17 Jul 2014
Posts: 2730
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: M2004 [#permalink]
Show Tags
20 Nov 2017, 16:41
joondez wrote: y4nkee wrote: mvictor wrote: I solved it in a different way: add both expressions AC+B+D+BA=odd. ok, rewrite everything as: 2B+DC=odd 2B is always even, it must bean only 1 thing: DC is always odd. if DC is odd, then D+C is always odd. Very good solution mvictor! I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic. AC+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 12+3 = 2 which is even. D+BA is odd. Reuse the same variables from before. D + 3  1 is D + 2. Set D to an easy odd number which is 5. So D = 5. Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers. A. A+D = 1 + 5 = 6 B. B+D = 3 + 5 = 8 C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C. D. A+B E. A+C the thing is  when presented with such expressions, you have to either add or subtract one from another. the thing is, you have 3 variables in each expression, therefore, you might lose some extra time to test all the possible options... my approach is very direct and fast though...and I always have the same approach to solving similar questions.



Intern
Joined: 01 Aug 2017
Posts: 6

Re: M2004 [#permalink]
Show Tags
31 May 2018, 08:29
The basic Rule to solve this question is if A+B = even then AB will also be even and vice versa Same way if A+B is odd the AB will also be odd and vice versa Now AC+B = Even >1 D+BA = Odd >2 adding 1 & 2 we have DC +2B = Odd DC = Odd 2B DC = Odd
hence D+C is odd.



Intern
Joined: 09 Oct 2016
Posts: 5

Re M2004 [#permalink]
Show Tags
09 Jun 2018, 01:02
How to approach question in this way. There can be multiple other combinations to start with.



Intern
Joined: 19 Jun 2017
Posts: 6

Re: M2004 [#permalink]
Show Tags
11 Jun 2018, 07:53
A  C + B = even A+D +B = ODD  D  C + 2B = ODD > 2B is always odd.
so DC = ODD Therefore, D+C = ODD
Please let me know if the approach is correct and can be followed
Thanks & Warm Regards, Richin










