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Re: If a, b and c are integers such that b > a, is b+c > a [#permalink]
10 Aug 2012, 00:15
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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 either 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.
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
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.
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.