It is currently 24 Jun 2017, 00:21

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# plz post your reasoning can we use decimal numbers to

Author Message
Manager
Joined: 05 Oct 2005
Posts: 200

### Show Tags

08 Aug 2006, 15:48
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

can we use decimal numbers to disprove statement

Is A+B+C EVEN ?

ST1 A -C - B IS EVEN

ST2 ( A- C) B IS ODD

thanks
_________________

when there is a will there is a way

best regards

VP
Joined: 14 May 2006
Posts: 1403

### Show Tags

08 Aug 2006, 16:08
e+e+e=e
o+o+o=o
e+o+o=e
o+e+e=o

(1) a-c-b=even, so either all even or there are 2 odds and in either case the subtraction is same as addition, so addition must be even
SUFF

(2) (a-c)b=odd, for odd result in multiplication, both factors must be odd..., so here (a-c)=odd and b must be odd, because there is no evens in multiplication with odd oucome... however, (a-c)...either a or c could be odd, so we will get o+o+e, which makes outcome even...
SUFF

so I say D... let's see if I screwed up
Manager
Joined: 26 Jun 2006
Posts: 152

### Show Tags

08 Aug 2006, 16:26
mandy,
If it does not say numbers must be integers, then yes, you can try using decimals or fractions.

Anyway,

I am getting (C)

ST1. A-C-B is even, or A-C-B=2k

If we choose A,B,C all integers and even, then A+B+C is even
However, if we use fractions, it does A+B+C does not have to be even

Let's see: A=2.2, B=0.1, C=0.1 and A-B-C=2 even. But A+B+C is not

ST2. (A-C)*B is odd

It follows that either
(1) A-C is even and B is odd
or (2) A-C is odd and B is even
or (3) A-C is a fraction and B is integer such that (A-C)*B=odd
or (4) A-C is integer and B is a fraction such that (A-C)*B=odd

Not sufficient.

Now, combine ST1 and ST2.

Neither A-C nor B can be fractions, otherwise ST1 is not valid.

So, all numbers -- A, B, and C -- must be integers.

Moreover, either 1 or all 3 must be odd (otherwise ST2 is violated.)

Therefore, A+B+C is even

(C) is sufficient
VP
Joined: 29 Dec 2005
Posts: 1341

### Show Tags

08 Aug 2006, 18:35
mand-y wrote:
can we use decimal numbers to disprove statement?
Is A+B+C EVEN ?

ST1. A -C - B IS EVEN
ST2. ( A- C) B IS ODD

E. good question. if A, B and C are not whole numbers/integers or are fractions, we cannot say.

initially i also thought B but after going through your question, should be E.

this is indeed another good question............................
Manager
Joined: 26 Jun 2006
Posts: 152

### Show Tags

08 Aug 2006, 19:51
Prof,

Can you give me two examples where both ST1 and ST2 are satisfied and we get 2 different answers for the main question?

I am still convinced that it should be (C)...
VP
Joined: 29 Dec 2005
Posts: 1341

### Show Tags

08 Aug 2006, 20:05
v1rok wrote:
Prof, Can you give me two examples where both ST1 and ST2 are satisfied and we get 2 different answers for the main question?

I am still convinced that it should be (C)...

if a = 4.5, b = 1 and c = 1.5
st 1. a - c - b = 4.5 - 1.5 - 1 = 2.........even
st 2. (a - c) b = (4.5 - 1.5) 1 = 3........odd

a + b + c = 4.5 + 1.5 + 1 = 7 ...odd.

if a = 4, b = 1 and c = 1
st 1. a - c - b = 4 - 1 - 1 = 2.. even
st 2. (a - c) b = (4 - 1) 1 = 3 .. odd

a + b + c = 4 + 1 + 1 = 6 ... even.

Quote:
Is A+B+C EVEN ?

ST1. A -C - B IS EVEN
ST2. ( A- C) B IS ODD
CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

### Show Tags

08 Aug 2006, 23:53
This is E.

NEVER assume integers until explicitly stated.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Manager
Joined: 05 Oct 2005
Posts: 200

### Show Tags

09 Aug 2006, 18:39
THANKS FOR yours answers but i thought that even number have to be integer
becaause even is an integer when divided by 2
oa will come
_________________

when there is a will there is a way

best regards

Manager
Joined: 05 Oct 2005
Posts: 200

### Show Tags

09 Aug 2006, 18:41
ps_dahiya wrote:
This is E.

NEVER assume integers until explicitly stated.

plz can you explain me why
thanks
_________________

when there is a will there is a way

best regards

Manager
Joined: 22 May 2006
Posts: 71

### Show Tags

09 Aug 2006, 19:15
A---
let A be 5.1,C be 2.1,A be 1
A-B-C=5.1-1-2.1=2(even)
but A+B+C=8.2 not even...
but if A is 6,B is 1 C is 3 both stmts are even hence A and D goes out
now consider B,
noe (5.1-2.1)1=3,odd
consider the same above combinations above,we will get yes for one
combinaton and no for another combination.hence B goes out
similarly for c we will get yes for 6,1,3 while no for 5.1,2.1,1
hence C also goes out
and giving us the answer E
CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

### Show Tags

09 Aug 2006, 21:14
mand-y wrote:
ps_dahiya wrote:
This is E.

NEVER assume integers until explicitly stated.

plz can you explain me why
thanks

A+B+C is even does not mean that A,B and C are also integers. A = 1.5 B = 1.75 and C = 0.75 yields A+B+C = 4 which is even. These are simpler values. more complex would be A = 1.564325465879256 and B and C also like that........ this way the answer will be certainly E.

Moral of the Story: ALL VARIABLES ARE REAL NUMBERS UNLESS EXPLICITLY STATED
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Manager
Joined: 05 Oct 2005
Posts: 200

### Show Tags

10 Aug 2006, 08:15
ps_dahiya wrote:
mand-y wrote:
ps_dahiya wrote:
This is E.

NEVER assume integers until explicitly stated.

plz can you explain me why
thanks

A+B+C is even does not mean that A,B and C are also integers. A = 1.5 B = 1.75 and C = 0.75 yields A+B+C = 4 which is even. These are simpler values. more complex would be A = 1.564325465879256 and B and C also like that........ this way the answer will be certainly E.

Moral of the Story: ALL VARIABLES ARE REAL NUMBERS UNLESS EXPLICITLY STATED

so I understood well the whole sum is an even integer it does not mean
that each part -each - has to be an integer
_________________

when there is a will there is a way

best regards

10 Aug 2006, 08:15
Display posts from previous: Sort by