first one IMO A?

zaarathelab wrote:1. If a, b, c are integers such that b > a, is b + c > a?

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

Experts, what is the shortest possible to solve the above? I took 4 minutes on the first one, especially because of the 2nd statement.

The question boils down to finding out whether c > 0 or c <= 0. In my opinion, the best way would be to prove that the statement holds true in both the cases; prove insufficiency and move on. It may not make sense to prove using inequality fundaes in a situation like this.

For eg.

Let a = 0.5, b = 1 and c = -1000

Here, abc > c while c is a -ve number

Let a = 3, b = 4 and c = +1000

Here, abc > c while c is a +ve number.

Hence, 2nd statement is insufficient by itself.

I hope this helps. Thanks.

cans wrote:first one IMO A?

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

from (1) c > a: let a = -3, b = -2, and c = -1 ; => -2-1 > -3 (FALSE)
let a = 4, C = 5, and b = 6; => 11> 4 ( True)
i think it's not sufficient!

from (2), it's not sufficient as well! ( already proved by @4GMAT_Mumbai)

Using both the statements 1 and 2, we have c>a and abc>C

i.e. Let a = -4, b = 3, c = -1 ; => 2 > -4 ( True)
one more case: a = -4, b = -3, c = 1 ; => -2 > -4 ( True)
Let a = 3, b = 4 and c = +1000 ; => 1004 > 3 (True)

Hence Answer : C!

what abt the second qsn?
Is it A for that?

parul9 wrote:what abt the second qsn?
Is it A for that?

i think even it's C!

from 2nd stmt: abc>0, a = -3, b= -2, c = 4 ; => -2+4 > -3 ( True)
a = 8, b = 10, c = 4; => 10 +4 > 8 ( True)
a = -8, b = 10, c = -20; => 10-20 > -8 ( False)

Not sufficient!

Using both the statements 1 and 2, we have c>a and abc>0

i.e. Let a = -4, b = 3, c = -1 ; => 2 > -4 ( True)
one more case: a = -4, b = -3, c = 1 ; => -2 > -4 ( True)
Let a = 3, b = 4 and c = +1000 ; => 1004 > 3 (True)

Hence Answer : C

4GMAT_Mumbai wrote:
For eg.

Let a = 0.5, b = 1 and c = -1000

There is a problem with the number selection here - the variable 'a' cannot be 0.5 because it is an integer. But your overall approach is good..

Can an expert please explain how this can be solved algebraically or within 2 mins? Thanks

zaarathelab wrote:2. If a, b, c are integers such that b > a, is b + c > a?
(1) c > a
(2) abc > 0

Given:
b>a

Ques:
Is b+c>a

Statement 1: c > a

True when all are positve
but might not be true if all are negative

This statement is not sufficient

Statement 2: abc > 0

This means either two of the variables are negative or none of them are negative.
Case 1 [none of the variables negative]
b > a
c => 0

b+c > a [value of c does not matter as long as it is non negative]

Case 2 [two of the vaeiables are negative]
a and b are negative , c is positve
a+b > c

b and c cannot be negative as b has to be greater than a

a and c are negative , b is positive
here if |c| > b-a,
b+c < a

if |c| < b-a
b+c > a

if |c| = b-a
b+c = a

Therefore Statement 2 alone is not suffiicient

Combining statement 1 and 2
c > a

if a and c are negative and b is positve
|c| < b-a [because c is greater than a]
hence b+c > a

|c| != b-a [!= means not equal to]

and all three can never be negative

Hence, sol can be found using both the statements
C is the correct option

1) E
2) C

zaarathelab wrote:1. If a, b, c are integers such that b > a, is b + c > a?

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

2. If a, b, c are integers such that b > a, is b + c > a?
(1) c > a
(2) abc > 0

Experts, what is the shortest possible to solve the above? I took 4 minutes on the first one, especially because of the 2nd statement.

IMO: A for both.

What is the OA?

Thanks.

This is a hard question, here is one explanation.

Statement 1 Alone: Given b>a and c>a, add the two inequalities, this means b+c>2a, now we need to determine if b+c>a?. If a is positive, then 2a is to the right of a on the number line, and b+c>a(Yes). However, if a is negative, then 2a is to the left of a and b+c could be less than a.

If the above approach is not clear then consider the following examples:
a = 5, b = 7, c = 9, these values satisfy b>a and c>a, and the answer to the question is b+c>a, is a Yes.
a = -4, b = -2, c = -3, these values again satisfy b>a, and c>a, however the answer to the question b+c>a, is now No.
Therefore, Insufficient.

Statement 2 Alone: abc>c, rewrite abc-c>0, factor c, c(ab-1)>0.
Reuse the first example above:
a = 5, b = 7, c = 9, these values satisfy statement 2 and b>a, the answer to the question is b+c>a is Yes.
Now change the numbers to c=-8, a = -3, and b=2, we do satisfy statement 2, -8[(-3)(2)-1] = 56>0, and b>a, however, b+c=-6 which is less than a=-3. Answer to the question is No.
Insufficient.

Statements 1 and 2 Together: c(ab-1)>0 and b>a and c>a.
Case 1: If a>0, then both b and c are positive because b>a and c>a. Under this scenario we also satisfy c(ab-1)>0. Just as we did in Statment 1 adding b>a and c>a yields b+c>2a, and because 2a is to the right of a(a is positive here), therefore b+c must be greater than a. This is a definite Yes scenario.
Case 2: If a<0, then either c>0 and b<0 or c<0 and b>0, to satisfy c(ab-1)>0. In both scenarios b and c are to the right of a on the number line, because b>a and c>a, and the answer to the question is b+c>a, will always be Yes. Writing the two cases on the number line makes it easier to visualize:
a<0, c>0, and b<0
--------a-------b------0--------c-------
Here b+c>a, because adding a positive number(c) to a number b that is already greater than a.
a<0, c<0, and b>0
-------a------c------0------b----------
Again, b+c>a, because adding a positive number(b) to a number c that is already greater than a.

Therefore 1 and 2 together are sufficient, and answer is C. Might be better if I made a video.

Dabral

zaarathelab wrote:1. If a, b, c are integers such that b > a, is b + c > a?

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

Experts, what is the shortest possible to solve the above? I took 4 minutes on the first one, especially because of the 2nd statement.