islands80 wrote:If t ≠ 0, is r greater that zero?

(1) rt = 12
(2) r + t = 7


I apologize if this has been posted before. I couldn't find this problem on the forum.

OA C

If t ≠ 0, is r > 0?

(1) rt = 12
If r = 3, t = 4, then rt = 3 * 4 = 12. Here r > 0.
If r = -3, t = -4, then rt = -3 * -4 = 12. Here r < 0.
No definite answer; NOT sufficient.

(2) r + t = 7
If r = 3, t = 4, then r+ t = 3 + 4 = 7. Here r > 0.
If r = -5, t = 12, then r + t = -5 + 12 = 7. Here r < 0.
No definite answer; NOT sufficient.

Combining (1) and (2), rt = 12 and r + t = 7
Substitute the value of t from r + t = 7 implies t = 7 - r, in rt = 12, we get
r(7 - r) = 12
7r - r² = 12
r² - 7r + 12 = 0
r² - 4r - 3r + 12 = 0
r(r - 4) - 3(r - 4) = 0
(r - 3)(r - 4) = 0
r = 3, 4, both values are > 0; SUFFICIENT.

The correct answer is C.

Shadow88 wrote:Guys/Girls what do you say about my reasoning to the right answer.
When putting (1) and (2) together, rt=12 and r+t=12.
We know that for rt to equal 12, both r and t need to have the same sign, but since r+t=12, both t and r must be positive, otherwise we would have r+t=-12 if r and t were negative. And if r or t were negative in r+t=12, then rt would equal -12, so r and t must be positive. r>0, am i thinking wrong here?

Just trying to get to a conclusion that take less time. Please correct me if i am wrong.

Excellent logic, Shadow88

Since we aren't trying to find the exact value of r here, the method you used to conclude that r must be positive is nice and quick.

Cheers,
Brent