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MathRevolution
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MathRevolution
[Math Revolution GMAT math practice question]

If a rectangle’s length is \(a+b\) and its area is \((\frac{1}{a})+(\frac{1}{b})\), what is its width?

\(A. 1\)
\(B. a+b\)
\(C. ab\)
\(D. \frac{1}{(a+b)}\)
\(E. \frac{1}{ab}\)
Just a glance at the alternative choices makes us realize that all of them get different values when a=b=3, hence...

Let´s explore this PARTICULAR CASE!

\({\rm{rectangle}}\,\,\left\{ \matrix{\\
\,{\rm{length}} = \,\,6 \hfill \cr \\
\,{\rm{area}} = \,\,{2 \over 3} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,?\,\,\left( {{\rm{target}}} \right)\,\,\,\, = \,\,\,\,{{\,\,{2 \over 3}\,\,} \over 6}\,\, = \,\,{1 \over 9}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( E \right)\)

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

Regards,
Fabio.
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=>

Let \(x\) be the width of the rectangle. Then
\(x(a+b) = \frac{1}{a} + \frac{1}{b} = \frac{(a+b)}{ab}.\)
Thus, \(x = \frac{1}{ab}.\)

Therefore, the answer is E.
Answer: E
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