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

Given: a + (1/b) = 400
Subtract 1/b from both sides to get: a = 400 - 1/b

Also given: a < -300
Replace a with 400 - 1/b to get: 400 - 1/b < -300
Add 1/b to both sides: 400 < -300 + 1/b
Add 300 to both sides: 700 < 1/b

IMPORTANT: 700 < 1/b, then b must be POSITIVE
Since b is POSITIVE, we can safely multiply both sides of the inequality by b to get: 700b < 1
Finally, divide both sides by 700 to get: b < 1/700

Answer: E

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a + (1/b) = 400
1/b = 400-a
let a= -300
1/b= 400+300
so 1/b = 700
b =1/700
now since a < -300
b < 1/700
E

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
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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.
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