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m04,Q-13

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m04,Q-13 [#permalink] New post 24 Jan 2011, 12:10
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Hello TimeTrader,

I see no know discussion for M#04 ,Q- 13.
If a and b are integers, is x even?

1. x = a/b, where the division leaves no remainder and b is odd.
2. x= a/b, where the division leaves no remainder and both a and b are odd


I think the answer should be E.

We dont know the values of a and b. All we know is they are odd

x = a/b

if a = 9, b=3 ->x=3 odd
a= 3,b=9 ->x=1/3 Not odd

Therefor B is not suff

1 and 2 are not suff together => answer is E
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Re: m04,Q-13 [#permalink] New post 24 Jan 2011, 14:08
goalsnr wrote:
Hello TimeTrader,

I see no know discussion for M#04 ,Q- 13.
If a and b are integers, is x even?

1. x = a/b, where the division leaves no remainder and b is odd.
2. x= a/b, where the division leaves no remainder and both a and b are odd


I think the answer should be E.

We dont know the values of a and b. All we know is they are odd

x = a/b

if a = 9, b=3 ->x=3 odd
a= 3,b=9 ->x=1/3 Not odd

Therefor B is not suff

1 and 2 are not suff together => answer is E


Hi,

when you pick numbers, you have to follow the rules that you've been given. The second set of numbers you chose for statement (2) contradict the statement, so those numbers are "illegal".

In terms of quotients and remainders, 3/9 has a quotient of 0 and a remainder of 3; since (2) tells us that a/b leaves no remainder, you cannot choose those numbers.

The only way to get no remainder is for a to be a multiple of b; if a and b are both odd, then we have an odd divided by an odd which, if it's an integer, must be odd (and we know that it has to be an integer, since there's no remainder). Accordingly, (2) tells us that x is definitely NOT even and is therefore sufficient.

As an aside, even with the illegal numbers you chose you should have arrived at (b) as the correct answer. The question is "is x even?" and if you plug in x = 1/3 the question becomes "is 1/3 even?", which also has a "no" answer. A statement is only insufficient if it can generate both a "yes" and a "no" answer to the original question.
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Re: m04,Q-13 [#permalink] New post 24 Jan 2011, 14:28
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Couple of words in addition.

Statement (2): "x=a/b, where the division leaves no remainder ..." simply means that a is divisible by b so x must be an integer.

Generally an integer a is a multiple of an integer b (integer a is a divisible by an integer b) means that \frac{a}{b}=integer.

Also on GMAT when we are told that a is divisible by b (or which is the same: "a is multiple of b", or "b is a factor of a"), we can say that:
1. a is an integer;
2. b is an integer;
3. \frac{a}{b}=integer.

Hope it helps.
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Re: m04,Q-13 [#permalink] New post 25 Jan 2011, 11:52
Bunuel wrote:
Couple of words in addition.

Statement (2): "x=a/b, where the division leaves no remainder ..." simply means that a is divisible by b so x must be an integer.

Generally an integer a is a multiple of an integer b (integer a is a divisible by an integer b) means that \frac{a}{b}=integer.

Also on GMAT when we are told that a is divisible by b (or which is the same: "a is multiple of b", or "b is a factor of a"), we can say that:
1. a is an integer;
2. b is an integer;
3. \frac{a}{b}=integer.

Hope it helps.


I dont know if I can agree with that (with point 3 yes, but with 1 and 2 no)

Lets say a=4/9 and b=1/9.

a is divisible by b (or which is the same: "a is multiple of b", or "b is a factor of a"), and neither a nor b are integers.
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Re: m04,Q-13 [#permalink] New post 25 Jan 2011, 12:01
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noboru wrote:
Bunuel wrote:
Couple of words in addition.

Statement (2): "x=a/b, where the division leaves no remainder ..." simply means that a is divisible by b so x must be an integer.

Generally an integer a is a multiple of an integer b (integer a is a divisible by an integer b) means that \frac{a}{b}=integer.

Also on GMAT when we are told that a is divisible by b (or which is the same: "a is multiple of b", or "b is a factor of a"), we can say that:
1. a is an integer;
2. b is an integer;
3. \frac{a}{b}=integer.

Hope it helps.


I dont know if I can agree with that (with point 3 yes, but with 1 and 2 no)

Lets say a=4/9 and b=1/9.

a is divisible by b (or which is the same: "a is multiple of b", or "b is a factor of a"), and neither a nor b are integers.


All 3 points are true, at least for the GMAT. Generally every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So "4/9 is divisible by 1/9" doesn't make any sense.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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.


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Re: m04,Q-13   [#permalink] 25 Jan 2011, 12:01
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