M20-04 : GMAT Club Tests

sanket1991

Manager

Joined: 14 Sep 2014

Posts: 74

Own Kudos [?]: 95 [1]

Given Kudos: 51

WE:Engineering (Consulting)

Send PM

Re: M20-04 [#permalink]

08 Jan 2015, 10:41

1

Kudos

I think this question is good and helpful.

passportsizephoto

Intern

Joined: 03 Nov 2014

Posts: 7

Own Kudos [?]: 22 [22]

Given Kudos: 13

Location: India

Schools: Kellogg '19 CBS '20 (S)

GMAT 1: 740 Q47 V45

GPA: 3

Send PM

Re: M20-04 [#permalink]

15 Apr 2015, 09:06

22

Kudos

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 A-C+B and D+B-A, even and odd respectively.

On subtracting the two, you get, A-C+B-D-B+A = 2A-C-D (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.

B

Abhigyanashakuntalam

Current Student

Joined: 04 Nov 2014

Status:Remember to Always Think Twice

Posts: 51

Own Kudos [?]: 23 [4]

Given Kudos: 16

Location: India

GMAT 1: 740 Q49 V40 Verified score

GPA: 3.51

Send PM

Re: M20-04 [#permalink]

19 Jan 2016, 06:25

3

Kudos

1

Bookmarks

There is an easier way this question can be solved.
Lets say A-C+B = 2n (even)
and after rearranging the second equation, -A+D+B=2m+1 (odd)

Adding the two equations we get, 2B+D-C=2(m+n)+1

From this, 2B is even and 2(m+n) is even, hence we conclude,
D-C=odd

When difference of two integers is odd, there sum will also be odd;

Hence D+C = odd

P

mvictor

Board of Directors

Joined: 17 Jul 2014

Posts: 2163

Own Kudos [?]: 1180 [17]

Given Kudos: 236

Location: United States (IL)

Concentration: Finance, Economics

Schools: Stanford '20 (WD)

GMAT 1: 650 Q49 V30 Verified score

GPA: 3.92

WE:General Management (Transportation)

Send PM

Re: M20-04 [#permalink]

20 Feb 2016, 13:58

14

Kudos

2

Bookmarks

I solved it in a different way:
add both expressions
A-C+B+D+B-A=odd.
ok, rewrite everything as:
2B+D-C=odd
2B is always even, it must mean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

B

Raunak_Bhatia

Intern

Joined: 31 Jul 2016

Posts: 26

Own Kudos [?]: 50 [3]

Given Kudos: 13

Location: Singapore

Schools: NTU PMBA"21 (I) NTU '20 (II) NUS PT MBA '21 (S)

Send PM

Re: M20-04 [#permalink]

27 Nov 2016, 06:40

2

Kudos

1

Bookmarks

Add or Substract the two equations-

A-C+B=Even
D+B-A=Odd
=2A+C-D (Even-Odd is always Odd) and we know 2A is always even

Therefore C-D should be Odd and thus also C+D

B

joondez

Intern

Joined: 01 Nov 2016

Posts: 40

Own Kudos [?]: 65 [1]

Given Kudos: 70

Concentration: Technology, Operations

Send PM

Re: M20-04 [#permalink]

06 May 2017, 14:56

1

Kudos

y4nkee wrote:

mvictor wrote:

I solved it in a different way:
add both expressions
A-C+B+D+B-A=odd.
ok, rewrite everything as:
2B+D-C=odd
2B is always even, it must bean only 1 thing: D-C is always odd. if D-C 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.

A-C+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 1-2+3 = 2 which is even.
D+B-A is odd. Re-use 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

B

supra2411

Intern

Joined: 22 Mar 2017

Posts: 2

Own Kudos [?]: 3 [1]

Given Kudos: 1

Send PM

Re: M20-04 [#permalink]

08 Jun 2017, 08:06

1

Kudos

Given A-C+B =Even and D+B-A=Odd
Subtract 1) and 2)
2A-C-D=Even-odd=Odd=> 2A=Odd- (C+D)
Left side of the equation is even hence C+D has to be Odd.

B

nitsrj

Current Student

Joined: 01 Aug 2017

Posts: 6

Own Kudos [?]: 5 [1]

Given Kudos: 87

Location: India

Schools: ISB '23 (A)

GMAT 1: 700 Q49 V35

GPA: 3.7

Send PM

Re: M20-04 [#permalink]

31 May 2018, 08:29

1

Kudos

The basic Rule to solve this question is if
A+B = even then A-B will also be even and vice versa
Same way if A+B is odd the A-B will also be odd and vice versa
Now
A-C+B = Even ---->1
D+B-A = Odd ---->2
adding 1 & 2 we have
D-C +2B = Odd
D-C = Odd- 2B
D-C = Odd

hence D+C is odd.

B

Richin

Intern

Joined: 19 Jun 2017

Posts: 2

Own Kudos [?]: 3 [1]

Given Kudos: 20

Send PM

Re: M20-04 [#permalink]

11 Jun 2018, 07:53

1

Kudos

A - C + B = even
-A+D +B = ODD
---------------------
D - C + 2B = ODD --> 2B is always odd.

so D-C = ODD
Therefore, D+C = ODD

Please let me know if the approach is correct and can be followed

Thanks & Warm Regards,
Richin

P

energetics

Senior Manager

Joined: 05 Feb 2018

Posts: 312

Own Kudos [?]: 794 [1]

Given Kudos: 325

Send PM

Re: M20-04 [#permalink]

02 Jul 2019, 16:51

1

Kudos

I solved by making a small chart to list possibilities:

If A-C+B = E E E
-A+D+B = E O E

If A-C+B = O E O
-A+D+B = O O O

By checking the answers against this, you can see that only C+D will be E+O = Odd number both ways.

B

AashishGautam

Intern

Joined: 21 Apr 2018

Posts: 47

Own Kudos [?]: 10 [1]

Given Kudos: 14

Location: India

Schools: ISB '24 (A)

GMAT 1: 710 Q50 V36

GPA: 3.3

Send PM

Re: M20-04 [#permalink]

19 May 2020, 01:25

1

Bookmarks

A comprehensive solution :
To explain all options ;
E : even, O: Odd

A B C D
Options E E E O
O O E O
O E O E
E O O E

As we can see C, D are always of opposite parity.

B

ishitaz

Intern

Joined: 25 Aug 2020

Posts: 10

Own Kudos [?]: 3 [0]

Given Kudos: 120

Send PM

Re: M20-04 [#permalink]

10 Sep 2020, 06:14

Hi,
we know that Odd - Even = Odd
we are given, D+B - A is odd
(Odd) (Even)

Which suggests D+B is odd, right? Not sure why this can't be the answer.

Any help appreciated, thanks in advance,
Ishita

L

Bunuel

Math Expert

Joined: 02 Sep 2009

Posts: 92912

Own Kudos [?]: 618926 [0]

Given Kudos: 81595

Send PM

Re: M20-04 [#permalink]

10 Sep 2020, 07:54

Expert Reply

ishitaz wrote:

Hi,
we know that Odd - Even = Odd
we are given, D+B - A is odd
(Odd) (Even)

Which suggests D+B is odd, right? Not sure why this can't be the answer.

Any help appreciated, thanks in advance,
Ishita

We are given that D + B - A = odd but it's possible that A is odd, B is odd, and D is odd. In this case D + B = even.

D

happypuppy

Manager

Joined: 29 Mar 2020

Posts: 220

Own Kudos [?]: 331 [0]

Given Kudos: 238

Location: India

Concentration: Entrepreneurship, Technology

Schools: Stanford '25 (S) Booth '25 (S) Wharton '25 (S) ISB '24 (A) HBS '25 (D)

GMAT 1: 720 Q50 V38 (Online) Verified score

GPA: 3.5

Send PM

Re: M20-04 [#permalink]

01 Feb 2022, 11:10

bartone89 wrote:

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

How can we say, 2k and 2k+1, or (2) - (1) = 1. We don't know about the value of A, B, C or D.
Rearrangement does not seem correct using 2k or 2k + 1 values.

Can you explain if I am wrong or missing something?

L

IanStewart

GMAT Tutor

Joined: 24 Jun 2008

Posts: 4128

Own Kudos [?]: 9243 [1]

Given Kudos: 91

Q51 V47

Send PM

Re: M20-04 [#permalink]

01 Feb 2022, 12:01

1

Kudos

Expert Reply

happypuppy wrote:

bartone89 wrote:

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

How can we say, 2k and 2k+1, or (2) - (1) = 1. We don't know about the value of A, B, C or D.
Rearrangement does not seem correct using 2k or 2k + 1 values.

Can you explain if I am wrong or missing something?

No, you're not missing anything

That solution isn't correct -- it could become correct, if the first value was "2k" and the second value was "2m + 1", but using "k" both times isn't right. That solution is a bit inefficient though. We know D + B - A is odd and A - C + B is even. So if we subtract that second expression from the first one, we must get something odd, so D + B - A - (A - C + B) is odd, and simplifying, D + C - 2A is odd. Subtracting the even number 2A won't change evenness or oddness at all, so D + C is odd.

B

joe123x

Manager

Joined: 03 Oct 2022

Posts: 90

Own Kudos [?]: 12 [1]

Given Kudos: 53

GMAT 1: 610 Q40 V34

Re: M20-04 [#permalink]

13 Mar 2023, 04:38

1

Kudos

I took the second equation and added the first equation to it.
the result is: 2B+D-C = odd
from this we can say that 2B is even so (D-C) has to be odd.
that means that either D = odd or C = even
or
D = even or C = odd
D and C cannot be odd at the same time or even at the same time.

is this logic good?

L

Bunuel

Math Expert

Joined: 02 Sep 2009

Posts: 92912

Own Kudos [?]: 618926 [0]

Given Kudos: 81595

Send PM

Re: M20-04 [#permalink]

13 Mar 2023, 05:26

Expert Reply

joe123x wrote:

I took the second equation and added the first equation to it.
the result is: 2B+D-C = odd
from this we can say that 2B is even so (D-C) has to be odd.
that means that either D = odd or C = even
or
D = even or C = odd
D and C cannot be odd at the same time or even at the same time.

is this logic good?

Yes. D - C = odd means that D + C = odd.

gmatclubot

Re: M20-04 [#permalink]

13 Mar 2023, 05:26