Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

Show Tags

10 Aug 2012, 01:15

8

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

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]

Show Tags

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]

Show Tags

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?

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...