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

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VP
Joined: 21 Jul 2006
Posts: 1452
If a, b, and c are integers such that b>a, is b+c>a?  [#permalink]

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22 Jul 2008, 08:32
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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

I know that this can be solved by experimenting with different numbers, but I honestly hate this approach because my mind just turns nuts and even more confused. More importantly, experimenting will make you waste time by doing the same approach multiple times until you reach your desired answer. I prefer the algebraic approach because you have to do the approach only once and then there you have your result. So would anyone please show me how to approach this problem using strictly algebra?? Here is what I did so far:

b+c>a
- b>a
c>0, so what the question is really asking is whether C is positive

(1) we don't know whether both are positive or negative

(2) either all the variables are positive or only 2 of the variables are negative, which C can be one of them

(1&2) Together, when we look at the worst case scenario that 2 variables are negative for abc>0, and when taking b>a and c>a into consideration, since a is smaller than both b and c, we know that a must be at least negative. However, what algebraic approach can we do to determine whether c or b is positive?

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SVP
Joined: 30 Apr 2008
Posts: 1841
Location: Oklahoma City
Schools: Hard Knocks

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22 Jul 2008, 08:44
Rather than subtracting out c>a, I would add c>a to b>a

this gives you b+c>2a

#1 should be sufficient. We already know that b>a. If c>a, then 2 numbers larger than a must have a sum larger than a, even if c is negative. Try picking some numbers.

b = 3, a = -2 c = -1
3 + -1 > -2 True.

b = 1, a = -1 c = 0
1 + 0 > -1

I don't believe #2 is sufficient.

tarek99 wrote:
If a, b, and c are integers such that b>a, is b+c>a?

(1) c>a

(2) abc>0

I know that this can be solved by experimenting with different numbers, but I honestly hate this approach because my mind just turns nuts and even more confused. More importantly, experimenting will make you waste time by doing the same approach multiple times until you reach your desired answer. I prefer the algebraic approach because you have to do the approach only once and then there you have your result. So would anyone please show me how to approach this problem using strictly algebra?? Here is what I did so far:

b+c>a
- b>a
c>0, so what the question is really asking is whether C is positive

(1) we don't know whether both are positive or negative

(2) either all the variables are positive or only 2 of the variables are negative, which C can be one of them

(1&2) Together, when we look at the worst case scenario that 2 variables are negative for abc>0, and when taking b>a and c>a into consideration, since a is smaller than both b and c, we know that a must be at least negative. However, what algebraic approach can we do to determine whether c or b is positive?

_________________

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a\$\$.

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Senior Manager
Joined: 06 Apr 2008
Posts: 393

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22 Jul 2008, 08:57
1
If you pick numbers b=c=-2 and a=-3 then although b>a and c>a but b+c<a

Combining statement 1) and 2)

b>a and c>a , b + c > 2a and abc>0

This means either a>0, b>0, c>0 or a<0, b<0 and c>0 or a<0, c<0 and b>0

If all are +ve and b + c > 2a then b + c > a is true as well

If a<0, b<0 and c>0 , b > a and b + c > 2a then b + c > a is true as well

If a<0, c<0 and b>0 , c > a and b + c > 2a then b + c > a is true as well

jallenmorris wrote:
Rather than subtracting out c>a, I would add c>a to b>a

this gives you b+c>2a

#1 should be sufficient. We already know that b>a. If c>a, then 2 numbers larger than a must have a sum larger than a, even if c is negative. Try picking some numbers.

b = 3, a = -2 c = -1
3 + -1 > -2 True.

b = 1, a = -1 c = 0
1 + 0 > -1

I don't believe #2 is sufficient.

tarek99 wrote:
If a, b, and c are integers such that b>a, is b+c>a?

(1) c>a

(2) abc>0

I know that this can be solved by experimenting with different numbers, but I honestly hate this approach because my mind just turns nuts and even more confused. More importantly, experimenting will make you waste time by doing the same approach multiple times until you reach your desired answer. I prefer the algebraic approach because you have to do the approach only once and then there you have your result. So would anyone please show me how to approach this problem using strictly algebra?? Here is what I did so far:

b+c>a
- b>a
c>0, so what the question is really asking is whether C is positive

(1) we don't know whether both are positive or negative

(2) either all the variables are positive or only 2 of the variables are negative, which C can be one of them

(1&2) Together, when we look at the worst case scenario that 2 variables are negative for abc>0, and when taking b>a and c>a into consideration, since a is smaller than both b and c, we know that a must be at least negative. However, what algebraic approach can we do to determine whether c or b is positive?
Senior Manager
Joined: 07 Jan 2008
Posts: 279

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22 Jul 2008, 09:13
1) b>a; c>a; Is b+c>a?
{a,b,c} = {-2,1,-1} works
= {-3,-2,-2} does not work

2) {a,b,c}={-2,1,-5} does not work
{1,2,3} works
No, I don't think they are just asking us to check whether C is +ve or -ve!
VP
Joined: 21 Jul 2006
Posts: 1452

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22 Jul 2008, 09:43
guys, I know that "if I pick numbers..." then I can do this and that.....I'm asking for the algebraic approach rather than the picking number approach.

OA will be displaced once the algebraic approach is displayed. This will be better for all of us.
VP
Joined: 21 Jul 2006
Posts: 1452

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23 Jul 2008, 08:35
jallenmorris wrote:
Rather than subtracting out c>a, I would add c>a to b>a
you mean subtracting out the b>c, right??
this gives you b+c>2a

#1 should be sufficient. We already know that b>a. If c>a, then 2 numbers larger than a must have a sum larger than a, even if c is negative. Try picking some numbers.

b = 3, a = -2 c = -1
3 + -1 > -2 True.

b = 1, a = -1 c = 0
1 + 0 > -1

I don't believe #2 is sufficient.

tarek99 wrote:
If a, b, and c are integers such that b>a, is b+c>a?

(1) c>a

(2) abc>0

I know that this can be solved by experimenting with different numbers, but I honestly hate this approach because my mind just turns nuts and even more confused. More importantly, experimenting will make you waste time by doing the same approach multiple times until you reach your desired answer. I prefer the algebraic approach because you have to do the approach only once and then there you have your result. So would anyone please show me how to approach this problem using strictly algebra?? Here is what I did so far:

b+c>a
- b>a
c>0, so what the question is really asking is whether C is positive

(1) we don't know whether both are positive or negative

(2) either all the variables are positive or only 2 of the variables are negative, which C can be one of them

(1&2) Together, when we look at the worst case scenario that 2 variables are negative for abc>0, and when taking b>a and c>a into consideration, since a is smaller than both b and c, we know that a must be at least negative. However, what algebraic approach can we do to determine whether c or b is positive?

we do have to subtract the b>a from b+c>a because these are the information given in the question. We have to first manipulate what we already have and then use is to justify every statement.
VP
Joined: 21 Jul 2006
Posts: 1452

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23 Jul 2008, 10:13
no input? The OA is C.
Senior Manager
Joined: 19 Mar 2008
Posts: 344

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23 Jul 2008, 10:35
If a, b, and c are integers such that b>a, is b+c>a?

(1) c>a

(2) abc>0

(1) Given b > a and c > a
so, b + c > a + a = 2a > a
So (1) is sufficient.

(2) only tells me that the sign of the three integter is either 3 + , or 1 + and 2 -

Not sufficient

So (A)
Director
Joined: 20 Sep 2006
Posts: 632

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23 Jul 2008, 11:18
tarek99 wrote:
OA is C

I am still not convinced. Do you agree A is suff by airthmatic?
VP
Joined: 21 Jul 2006
Posts: 1452

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23 Jul 2008, 11:43
No because we don't know whether both c and a are positives. It is also possible that both c and a are negatives, for example:

c= -2

a= -3

so, c>a is the same as -2>-3

I'm trying to avoid using plug ins, but I had to use number here just to explain to you why statement 1 is not sufficient, because we don't know whether both c & a are positives, or both negatives, or even whether c is positive and a is negative. see what I mean?? c can be bigger than a in all these 3 different scenarios.
Retired Moderator
Joined: 18 Jul 2008
Posts: 897

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23 Jul 2008, 15:23
Is it me? but it looks like {-3,-2,-2} does work indeed?

mbawaters wrote:
1) b>a; c>a; Is b+c>a?
{a,b,c} = {-2,1,-1} works
= {-3,-2,-2} does not work

2) {a,b,c}={-2,1,-5} does not work
{1,2,3} works
No, I don't think they are just asking us to check whether C is +ve or -ve!
Intern
Joined: 20 Jul 2008
Posts: 24

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23 Jul 2008, 21:04
so the algebraic approach is not going to work with this type of problems?
Current Student
Joined: 28 Dec 2004
Posts: 3287
Location: New York City
Schools: Wharton'11 HBS'12

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24 Jul 2008, 05:13
i am getting C we need to know if C>a

we are told abc>0

suppose a<0 C<0 b>0

c=-4 a-5 b=1

1-4=-3>-5

suppose b<0 C>0 a<0
b=-2 a=-3 c=1

-2+1=-1>-3
Senior Manager
Joined: 19 Mar 2008
Posts: 344

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05 Sep 2008, 21:54
Guys, I am going through my previous post and found this

try to explain instead on try numbers.
Given b > a, we have to prove whether b + c > a, which is always true if c > or = 0.

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

Three possible situations:
Case(1):
a, b , c are all + ve
the statement b + c > a is always true, because c > 0

Case (2):
b is +ve 0, c and a are -ve
since c > a, lcl < lal
then b - lcl > a (for all b > 0 and c > a and c, a are -ve)
So, the statement is true.

Case (3):
c is +ve 0, b and a are -ve
the statement b + c > a is always true, because c > 0

So, ans is C
VP
Joined: 21 Jul 2006
Posts: 1452

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06 Sep 2008, 07:26
judokan wrote:
Guys, I am going through my previous post and found this

try to explain instead on try numbers.
Given b > a, we have to prove whether b + c > a, which is always true if c > or = 0.

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

Three possible situations:
Case(1):
a, b , c are all + ve
the statement b + c > a is always true, because c > 0

Case (2):
b is +ve 0, c and a are -ve
since c > a, lcl < lal
then b - lcl > a (for all b > 0 and c > a and c, a are -ve)
So, the statement is true.

Case (3):
c is +ve 0, b and a are -ve
the statement b + c > a is always true, because c > 0

So, ans is C

nice approach. but thanks to your approach, i just figured out a really cool way to approach this. Check this out:

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

(1) c>a

(2) abc>0

First of all, the question says that b>a, so (b-a) > 0, which is positive. The inequality b+c>a can be turned into (b-a)+c>0. The question is asking whether the sum of these variables are positive. For this to be true, we have to see whether c is positive or zero.

(1) both the variables could be positive or negatives
(2) c could be positive or negative

(1&2) because both b and c are greater than a, a will definitely be at least negative because you can't expect a to be positive and then consider the possibility for it to be bigger than either b or c, both of which could be negative....agree? We know that if all of the variables are positive, the answer will be true to our given inequality, but what if only 2 of the variables are negative? so:

if b is positive:

from statement 1, we know that c>a, which means (c-a) > 0, which is positive. So the inequality b+c>a can be rearranged to: b + (c-a) >0, which is positive because we've consider b to be positive ----->true

if c is positive (which we already know that it make the inequality hold true anyways, but for the sake of practice, will illustrate it anyways):

we know from the question that b>a, which means (b-a) > 0, which is positive. So the inequality b+c>a can be rearranged to: (b-a) + c > 0, which is positive because we've considered c to be positive----->true.

Note: when you sit for the real test, you can actually skip the scenario of c being positive because we already know from the given question that if c is positive, then the overall inequality will hold true . so you're required to check ONLY the scenario of b being positive

what can we learn from this problem? a difficult question such as this one would try to confuse you by dragging you to consider the different negative possibilities which can be time consuming. This question can be A LOT easier when you isolate only one of the three variables as negative, and then consider the different scenarios of the remaining 2 variables as POSITIVE because that would automatically take care of the negative possibilities by themselves. However, the question has indirectly given us the clue that the inequality will hold true for one of the remaining variables being positive, and i'm talking about c. So you only have to check what will happen if ONLY b is positive.
Intern
Joined: 03 Sep 2008
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Updated on: 06 Sep 2008, 14:57
You should try to think about these problems visually using a number line. Doing that I came up with some numbers that seem to indicate the answer is E

a= -10, b= -9, c = -8. b+c>a? No

a= -10, b= -9, c= 2. b+c>a? Yes

Edit: oops, as someone pointed out my first set of numbers doesn't work out. Don't know what I was thinking there. It's difficult to do this problem algebraically because you really have to see visually what's going on. Sometimes plugging in numbers makes a problem more concrete so you can then proceed to the more abstract algebraic approach but it's best to use as little math as possible when doing GMAT problems so if you can do it visually that's the best approach. On the other hand sometimes a little plugging in and algebra is what it takes to visualize the problem.

Originally posted by lsmv479 on 06 Sep 2008, 11:53.
Last edited by lsmv479 on 06 Sep 2008, 14:57, edited 1 time in total.
VP
Joined: 05 Jul 2008
Posts: 1332

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06 Sep 2008, 11:59
lsmv479 wrote:
You should try to think about these problems visually using a number line. Doing that I came up with some numbers that seem to indicate the answer is E

a= -10, b= -9, c = -8. b+c>a? No

a= -10, b= -9, c= 2. b+c>a? Yes

abc > 0 . Your first set of numbers does not satisfy that ( if you were posting that C is incorrect )

IMO, Its never easy to pick numbers or algebra for any problem. I believe it comes down to personal preference

I try to solve it algebraically in the first 30 sec or so. If I dont get any where I look for numbers. But this approach has been a little bit time consuming. I guess over a period of time practice will pay off
VP
Joined: 21 Jul 2006
Posts: 1452

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07 Sep 2008, 10:45
lsmv479 wrote:
You should try to think about these problems visually using a number line. Doing that I came up with some numbers that seem to indicate the answer is E

a= -10, b= -9, c = -8. b+c>a? No

a= -10, b= -9, c= 2. b+c>a? Yes

Edit: oops, as someone pointed out my first set of numbers doesn't work out. Don't know what I was thinking there. It's difficult to do this problem algebraically because you really have to see visually what's going on. Sometimes plugging in numbers makes a problem more concrete so you can then proceed to the more abstract algebraic approach but it's best to use as little math as possible when doing GMAT problems so if you can do it visually that's the best approach. On the other hand sometimes a little plugging in and algebra is what it takes to visualize the problem.

well, take a look at my last post, which is right before yours. I made a long post, but it's a very detailed one. I think that should make things a lot clearer for you.
have a look at it.

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Re: DS: is b+c>a? &nbs [#permalink] 07 Sep 2008, 10:45
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