If a + (1/b) = 400, and a < -300, then

Since a = 400 - 1/b, we have:

400 - 1/b < -300

700 < 1/b

We note at this point that b is positive (which is because a is negative and a + (1/b) is positive; therefore 1/b must be positive, therefore b must be positive). Let’s multiply each side of this inequality by the positive number b:

700b < 1

b < 1/700

Answer: E

The winning triad asks for DATA connected to FOCUS... but it is the FOCUS the fundamental leg of the triad!

\(?\,\,\,:\,\,\,b\,\,{\rm{inequality}}\)

Hence:

\(a + {1 \over b} = 400\,\,\,\, \Rightarrow \,\,\,{1 \over b} = 400 - a = 400 + \left( { - a} \right)\mathop > \limits^{\left( * \right)} 700\)

\(\left( * \right)\,\,\,a < - 300\,\,\,\, \Rightarrow \,\,\, - a > 300\)

\({1 \over b} > 700\,\,\,\,\mathop \Rightarrow \limits^{{\rm{positives}}!} \,\,\,\,\,?\,\,:\,\,b < {1 \over {700}}\)

Do not forget the two "Golden rules": when numbers have the same signs, the greater has the lower reciprocal... Mathematically speaking:

\(x > y > 0\,\,\,\mathop \Rightarrow \limits^{:\,\,xy\, > \,0} \,\,\,{1 \over x} < {1 \over y}\)

\(q < w < 0\,\,\,\mathop \Rightarrow \limits^{:\,\,qw\, > \,0} \,\,\,{1 \over w} < {1 \over q}\)

\(\left( {x = {1 \over b}\,\,\,{\rm{and}}\,\,y = 700\,\,{\rm{in}}\,\,{\rm{our}}\,\,{\rm{case}}} \right)\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

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.