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If a, b and c are integers such that b > a, is b+c > a

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If a, b and c are integers such that b > a, is b+c > a [#permalink] New post 25 Sep 2007, 06:33
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If a, b and c are integers such that b > a, is b+c > a ?

(1) c > a
(2) abc > 0
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 [#permalink] New post 25 Sep 2007, 07:21
From stem, all variables are integers, but no indication whether they are negative, non-negative, etc. so negative, positive and zero must be tried. From statement 1 - c>a. That's fine, but what if a = -2, c is -1, and b is -1? Then b+c not bigger than a. Could plug in numbers to make statement true as well - therefore insufficient.
From statement 2 - abc>0. So all integers are positive. but no indication that they are different integers. If a = 1, b = 2, and c = 1, then answer is no. But would be easy to plug in other #'s to make it true. So statement 2 also insufficient.
Together - c >a, b>a, all numbers are positive...therefore b+c>a. Sufficient.
C it is.
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 [#permalink] New post 25 Sep 2007, 14:12
uphillclimb wrote:
From stem, all variables are integers, but no indication whether they are negative, non-negative, etc. so negative, positive and zero must be tried. From statement 1 - c>a. That's fine, but what if a = -2, c is -1, and b is -1? Then b+c not bigger than a. Could plug in numbers to make statement true as well - therefore insufficient.
From statement 2 - abc>0. So all integers are positive. but no indication that they are different integers. If a = 1, b = 2, and c = 1, then answer is no. But would be easy to plug in other #'s to make it true. So statement 2 also insufficient.
Together - c >a, b>a, all numbers are positive...therefore b+c>a. Sufficient.
C it is.


Not necessarily true, but otherwise perfect!
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 [#permalink] New post 25 Sep 2007, 16:55
kevincan wrote:
uphillclimb wrote:
From stem, all variables are integers, but no indication whether they are negative, non-negative, etc. so negative, positive and zero must be tried. From statement 1 - c>a. That's fine, but what if a = -2, c is -1, and b is -1? Then b+c not bigger than a. Could plug in numbers to make statement true as well - therefore insufficient.
From statement 2 - abc>0. So all integers are positive. but no indication that they are different integers. If a = 1, b = 2, and c = 1, then answer is no. But would be easy to plug in other #'s to make it true. So statement 2 also insufficient.
Together - c >a, b>a, all numbers are positive...therefore b+c>a. Sufficient.
C it is.


Not necessarily true, but otherwise perfect!


I wud say it is B.

Here's an example:

Given:
1. b > a ( From Question Stem)
2. abc > 0 ( From Statement II)

From 2: either all are positive or two of them are negative.

eg 1: a=-3;b=-2;c=1 ==> abc>0(holds); b>a(holds) b+c=(1) > a(-3)(holds)

eg 2: a=-3;b=2;c=-1 ==> abc>0(holds); b>a(holds); b+c(1) > a(-3) holds

eg: a=1;b=2;c=1 ==> abc>0(holds);b>a(holds); b+c(3) > a(1) (holds)

Statement II alone is sufficient.

Any sample of numbers for which this fails?
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 [#permalink] New post 25 Sep 2007, 17:35
St1:
c > a
If c = -1, a = -2, and b = 0, then b>a and b+c > a
If c = -2, a = -3, and b = -1, then b>a but b+c = a
Insufficient.

St2:
abc > 0
Could be all positive, or two negatives and one positive.

All positive:
b = 2, a = 1, c = 3, then b+c > a
b+c > a --> for b positive, a and c negative
b+c > a --> for c positive, a and b negative

(no need to test a positive and b & c negative as b > a and c > a)
Sufficient.

Ans C

Last edited by ywilfred on 25 Sep 2007, 18:06, edited 1 time in total.
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 [#permalink] New post 25 Sep 2007, 17:44
This sample set in your inputs fails to hold the condition b > a provided in the question stem.

b = 1, c = 2, a = 6, then b+c < a
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 [#permalink] New post 25 Sep 2007, 18:05
done_gre_now_gmat wrote:
This sample set in your inputs fails to hold the condition b > a provided in the question stem.

b = 1, c = 2, a = 6, then b+c < a


well, take it out and the answer is still C =)
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Re: DS Three variable inequality [#permalink] New post 25 Sep 2007, 18:08
kevincan wrote:
If a, b and c are integers such that b > a, is b+c > a ?

(1) c > a
(2) abc > 0



C.

S1: b>a and c>a. u can think of many + values that will make this work. to make this not work try negs. a>b: -2>-3 c>a: -2>-3

-4 not greater than -3. so S1 insuff.

S2: abc>0. could all be + or could be two negatives

Insuff.


1 and 2. b/c c>a and b>a. and abc>0. only positives work here.
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Re: DS Three variable inequality [#permalink] New post 25 Sep 2007, 18:11
1 and 2. b/c c>a and b>a. and abc>0. only positives work here.[/

should all necessarily be positive: c=3, b=-1 and a=-2 still holds true,
Re: DS Three variable inequality   [#permalink] 25 Sep 2007, 18:11
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