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If a, b and c are integers such that b > a, is b+c > a [#permalink]
 26 May 2009, 02:46
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68% (01:21) wrong based on 3 sessions
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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Last edited by Bunuel on 10 Aug 2012, 00:22, edited 1 time in total.
Edited the question.
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sondenso wrote: If a, b and c are integers such that b > a, is b+c > a ?
(1) c > a
(2) abc > 0 E a = -4 or 3 b = -3 or 4 The above satifies b>a, using - or + as examples. (1) c>a c = -2 or 4 Insuff (2) abc>0 This means either 2 of the intergers are negative or all of the intergers are positive. Insuff
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sondenso wrote: If a, b and c are integers such that b > a, is b+c > a ?
(1) c > a
(2) abc > 0 b>a <==> 0>a-b <==> a-b is negative Question:(b+c>a)? Question:(c>a-b)? Question:(c >= 0)? (1) Insufficient, as a-b<0 could mean that a<0 or a>0, which means c<0 or c>=0. (2) We have an even # of negatives, ie 0/2. If we have 0 negatives, then sufficient. Again if we have 2 insufficient. I can see the pattern, the answer is E. Here are the Yes/No cases: Yes: 4 5 7 No: -9 3 -10 Final Answer, E.
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For me its straight A)
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why cant it be C?
1. b>a and c>a and abc>0:
If a<0, b can be less than or greater than 0.
If a<0, b<0 then c>0 (a=-6, b=-4, c=2) => b+c >a
If a<0, b>0, c<0 (a=-3, b=5, c=-1) => b+c >a
If all are positive, then also b+c >a.
Hence, C.
Are there any assumptions which do not satisfy both conditions simultaneously?
What is the OA?
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The answer is C
statement 1 does work for positives, but not for negatives. a= -2 b= -1 c= -1.
statement 2 doesnt work on its own because c could be way less than a, negating b,
a= -3 b=2 c= -5
however, both together mean that there can only be two negatives, and c must be larger than A.
for any value A, negative or not, two values that are greater than it with one positive are going to together be greater than A.
if A is positive = a=2, c=3, b=3, a=1 b=2 c=2
if A is negative = a= -3, c= -2, b=1, a= -2 b=-1 c=1
There is no solution with (1) and (2) true that doesn't end up with B + C > A
gmatprep is right
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Has anybody considered taking both the conditions 1 and 2 together? What is the OA for this? I still think its C.
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b>a
stmt 1 :
c>a so b+c>2a
we cannot tell whether b+c >a may or may not. So insufficient.
stmt 2 :
abc>0 and b >a
a = +ve b = +ve c =+ve so definetly b+c > a
a = -ve b = -ve c = +ve so definetly b+c > a. ( since c =+ve and b>a)
a = -ve b = +ve c = -ve ( here it is just the opposite u know b =+ve but do not know c>a).so u cant tell whether b+c > a
Combing : yes u know you have got what u wanted so C.
Thanks gmatprep09 and dk94588.
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I agree C is the answer. If a,b and c and positive,it is possible.
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tkarthik wrote -
c>a so b+c>2a
we cannot tell whether b+c >a may or may not. So insufficient
If b+c>2a how in the world can it not be > a !!!! I m confused.Can you pls. elaborate or give example numbers.
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Hi mdfrahim
you need to take some negative value.
Say a,b,c are +ve then b+c > a and b+c>2a
say b = -2 c = -3 a =-4 ( c> a and b > a)
b + c = -5 which less than a but will be greater than 2a.
I meant to say this only. Hope i have made it clear.
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Thx. TKathik, you made it clear.
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Whoops the answer is C If you look at both together, we have that b>a & c>a. from (1) we get b+c>2a and we're almost sufficient, but only if 2a>a. Well if a<0, no, but if a>0, yes. (2) tells us that 0 or 2 are negative. If all 3 are positive, sufficient. Now if 2 are negative, then a has to be negative as well. Suppose a were positive-- then b/c would be negative, and b>a & c>a would be false. Hence a has to be positive. (C) Very tricky
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Re: Property of Integers, Inequalities [#permalink]
23 Oct 2009, 00:02
(1) because b and c both can be negative, thus (1) is insufficient
(2) insufficient for example, a = -2, b = 1, c = -4
Both are sufficient because either b or c is always greater than a.
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Re: If a, b and c are integers such that b > a, is b+c > a [#permalink]
09 Aug 2012, 23:40
Bunuel can you please provide your method
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Re: If a, b and c are integers such that b > a, is b+c > a [#permalink]
10 Aug 2012, 01:15
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If a, b and c are integers such that b > a, is b+c > a ?Question: is b+c > a ? --> or: is b+c-a > 0? (1) c > a. If a=1, b=2 and c=3, then the answer is clearly YES but if a=-3, b=-2 and c=-1, then the answer is NO. Not sufficient. (2) abc > 0. Either all three unknowns are positive (answer YES) or two unknowns are negative and the third one is positive. Notice that in the second case one of the unknowns that is negative must be a (because if a is not negative, then b is also not negative so we won't have two negative unknowns). To get a NO answer for the second case consider a=-3, b=1 and c=-4 and . Not sufficient. (1)+(2) We have that b > a, c > a ( c-a>0) and that ether all three unknowns are positive or a and any from b and c is negative. Again for the first case the answer is obviously YES. As for the second case: say a and c are negative and b is positive, then b+(c-a)=positive+positive>0 (you can apply the same reasoning if a and b are negative and c is positive). So, we have that in both cases we have an YES answer. Sufficient. Answer: C.
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Re: If a, b and c are integers such that b > a, is b+c > a [#permalink]
26 Feb 2013, 06:52
Bunuel wrote: If a, b and c are integers such that b > a, is b+c > a ?
Question: is b+c > a ? --> or: is b+c-a > 0?
(1) c > a. If a=1, b=2 and c=3, then the answer is clearly YES but if a=-3, b=-2 and c=-1, then the answer is NO. Not sufficient.
(2) abc > 0. Either all three unknowns are positive (answer YES) or two unknowns are negative and the third one is positive. Notice that in the second case one of the unknowns that is negative must be a (because if a is not negative, then b is also not negative so we won't have two negative unknowns). To get a NO answer for the second case consider a=-3, b=-4 and c=1. Not sufficient.
(1)+(2) We have that b > a, c > a (c-a>0) and that ether all three unknowns are positive or a and any from b and c is negative. Again for the first case the answer is obviously YES. As for the second case: say a and c are negative and b is positive, then b+(c-a)=positive+positive>0 (you can apply the same reasoning if a and b are negative and c is positive). So, we have that in both cases we have an YES answer. Sufficient.
Answer: C. In 2nd case you have taken a=-3,b=-4,c=1. But the problem clearly stating that b must be grater than a ( b>a). But you have taken in a different way which might result in wrong answer.Correct me if i am wrong?
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Re: If a, b and c are integers such that b > a, is b+c > a [#permalink]
26 Feb 2013, 07:42
mannava189 wrote: Bunuel wrote: If a, b and c are integers such that b > a, is b+c > a ?
Question: is b+c > a ? --> or: is b+c-a > 0?
(1) c > a. If a=1, b=2 and c=3, then the answer is clearly YES but if a=-3, b=-2 and c=-1, then the answer is NO. Not sufficient.
(2) abc > 0. Either all three unknowns are positive (answer YES) or two unknowns are negative and the third one is positive. Notice that in the second case one of the unknowns that is negative must be a (because if a is not negative, then b is also not negative so we won't have two negative unknowns). To get a NO answer for the second case consider a=-3, b=-4 and c=1. Not sufficient.
(1)+(2) We have that b > a, c > a (c-a>0) and that ether all three unknowns are positive or a and any from b and c is negative. Again for the first case the answer is obviously YES. As for the second case: say a and c are negative and b is positive, then b+(c-a)=positive+positive>0 (you can apply the same reasoning if a and b are negative and c is positive). So, we have that in both cases we have an YES answer. Sufficient.
Answer: C. In 2nd case you have taken a=-3,b=-4,c=1. But the problem clearly stating that b must be grater than a ( b>a). But you have taken in a different way which might result in wrong answer.Correct me if i am wrong? The values of b and c should be switched. It should be a=-3, b=1 and c=-4. Typo edited thank you.
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Re: If a, b and c are integers such that b > a, is b+c > a
[#permalink]
26 Feb 2013, 07:42
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