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16 Oct 2018, 01:03
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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[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?
\(A. 1\)
\(B. a+b\)
\(C. ab\)
\(D. \frac{1}{(a+b)}\)
\(E. \frac{1}{ab}\)
If a rectangle’s length is \(a+b\) and its area is \((\frac{1}{a})+(\frac{1}{b})\), what is its width?
\(A. 1\)
\(B. a+b\)
\(C. ab\)
\(D. \frac{1}{(a+b)}\)
\(E. \frac{1}{ab}\)
16 Oct 2018, 01:09
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Length given is (a+b)
Area = (1/a+1/b)
Area of rectangle = length * width
(1/a+1/b) = (a+b) * width
Width = 1/ab
Option E is the correct answer
Posted from my mobile device
Area = (1/a+1/b)
Area of rectangle = length * width
(1/a+1/b) = (a+b) * width
Width = 1/ab
Option E is the correct answer
Posted from my mobile device
16 Oct 2018, 01:34
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MathRevolution
The area is given as a sum of two fractions - let us find this sum.
\((\frac{1}{a})+(\frac{1}{b}) = (\frac{b}{ab} + \frac{a}{ab}) = \frac{(a+b)}{ab}\)
since the length is \((a+b)\) width must be Area/length ...
\(\frac{(a+b)}{ab} * \frac{1}{(a+b)}\)
Thus width is = \(\frac{1}{ab}\)
Hence Option (E) is our choice.
Best,
Gladi
17 Oct 2018, 06:49
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MathRevolution
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.
