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Re: Direct and inverse proportionality [#permalink]

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06 Jun 2011, 01:34

According to expression Q= 5w/4xz² now, w'=4w, x'=2x z'=3z Put these values in a same order, we get Q'= 5(4w)/4(2x)(3z)² Q'=5w/18xz² Q'=2/9*(5w/4xz²) So, Q'= 2/9(Q),q will be a multiple of factor 2/9. This is the most effective way to solve such question.

Would anyone be able please to elaborate on it? I would really appreciate it.

Serge.

If the function Q is defined by the formula Q = 5w/(4x(z^2)), by what factor will Q be multiplied if w is quadrupled, x is doubled, and z is tripled? A. 1/9 B. 2/9 C. 4/9 D. 3/9 E. 2/27

Given: \(Q=\frac{5w}{4x*z^2}\).

Now, quadruple \(w\), so make it \(4w\); double \(x\) so make it \(2x\); triple \(z\) and substitute these values instead of \(x\), \(y\), and \(z\) in the original equation:

\(\frac{5(4w)}{4(2x)*(3z)^2}=\frac{4*5w}{4*2x*9*z^2}=\frac{4*5w}{18*(4x*z^2)}=\frac{4}{18}*\frac{5w}{4x*z^2}=\frac{2}{9}*\frac{5w}{4x*z^2}\). Thus Q is multiplied by \(\frac{2}{9}\).

Answer: B.

Else plug-in values for \(x\), \(y\), and \(z\). Let \(x=y=z=1\) --> \(Q=\frac{5w}{4x*z^2}=\frac{5}{4}\).

\(4w=4\), \(2x=2\) and \(3z=3\) --> \(\frac{5*4}{4*2*3^2}=\frac{4}{18}*\frac{5}{4}=\frac{2}{9}*\frac{5}{4}\). Thus Q is multiplied by \(\frac{2}{9}\).

Re: If the function Q is defined by the formula Q = 5w/(4x(z^2)) [#permalink]

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20 Sep 2016, 00:17

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