Is r more than -12?

I solved this algebraically:

Question is asking if \(r > -12\)

1
\(q+r>-12\)
or \(r>-12-q\)
no info on q, so insufficient

2
\(q>-12\)
no info on r, so insufficient

1+2,
Since we can add or subtract inequalities with the same sign,
\(q+r>-12\) (from 1)
\(q>-12\) (from 2)
Subtract 2 from 1,
\(r+q (-) q >-12 (-) -12\)
Therefore,
\(r > 0\)
Hence, C.

Where am I going wrong?

Note from Bunuel:

ADDING/SUBTRACTING INEQUALITIES:

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

RAISING INEQUALITIES TO EVEN/ODD POWER:

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Coming back to the question

So, we have q+r >-12 and q > -12
Adding the 2 we get

\(q+r+q> 0\) or\(2q+r >0\)

So, we can have multiple option where r may be greater than or less than -12.

hope it helps.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.