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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.

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.

If a, b and c are integers such that b > a, is b+c > a [#permalink]

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21 Oct 2014, 22:12

sondenso wrote:

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

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

The question stem says that b > a. There are following possibilities for this a. Both Positive Lets say \(b = 5 and a = 2\)

b. Both Negative, lets say \(b = -3 and a = -5\)

c. One positive one negative, lets say \(b = 3 and a = -1\)

d. Either of the two zero

1. c > a

C can be positive or negative and can fit in any of the above mentioned scenarios.

Insufficient

2. abc > 0

This means either all are positive or any two are negative. Since b > a, there are numerous possibilities for c.

So, Insufficient.

Adding the two

c > a and abc > 0

This would mean that either all are positive - b + c > a

or two negatives and one one positive - a has to be negative, either of b and c has to be negative and the other one positive. This would also mean that c + b > a

Re: If a, b and c are integers such that b > a, is b+c > a [#permalink]

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07 Mar 2016, 14:51

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 \(a=-3\), \(b=1\) 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.

Answer: C.

Bunuel could you please explain how you can solve the problem with the approach above in 2 min?

In this problem can we subtract the following inequalities?

b>a b+c>a

and arrive at 'is c>0??'

Let me try to answer.

The answer is NO you can not.

b>a and b+c>a can mean both c<0 and c>0 by taking the following cases:

b=5, c=-3 and a=1 but the given conditions are also true for b=5, c=3 and a=1. So you have both the cases possible.

With,

b>a and b+c>a, you can not directly subtract the 2 inequalities as the sign for b>a MUST be reversed. You can only ADD in inequalities without thinking about the signs.

If I give you,

b>a and b+c>a ---> then the only MUST be true statement will be ---> b+b+c > 2a ---> 2b+c > 2a

BUT,

If you are given, b<a and b+c > a, as you can not add 2 inequalities with dissimilar inequality signs ---> you must do an additional operation (b<a ---> -b>-a) before you add the 2.

After the manipulation you get,

-b > -a and this added to b+c > a --> -b+b+c > 0 ---> c>0