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If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is it

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If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is it  [#permalink]

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New post 16 Oct 2018, 01:03
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A
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C
D
E

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88% (01:16) correct 12% (01:15) wrong based on 31 sessions

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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?

\(A. 1\)
\(B. a+b\)
\(C. ab\)
\(D. \frac{1}{(a+b)}\)
\(E. \frac{1}{ab}\)

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Re: If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is it  [#permalink]

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New post 16 Oct 2018, 01:09
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

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Re: If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is it  [#permalink]

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New post 16 Oct 2018, 01:34
MathRevolution wrote:
[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?

\(A. 1\)
\(B. a+b\)
\(C. ab\)
\(D. \frac{1}{(a+b)}\)
\(E. \frac{1}{ab}\)


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
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Re: If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is it  [#permalink]

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New post 17 Oct 2018, 06:49
MathRevolution wrote:
[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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Re: If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is it  [#permalink]

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New post 18 Oct 2018, 01:12
=>

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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Re: If a rectangle’s length is a+b and its area is (1/a)+(1/b), what is it   [#permalink] 18 Oct 2018, 01:12
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