If t ≠0, is r > 0?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
(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.
