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Let \(\frac{x}{y} = a\) and \(\frac{w}{z} = b\)

=> a + b = 2

We need to find the value of \(\frac{1}{a} + \frac{1}{b}\) = \(\frac{a+b}{ab}\)

We know the value of a + b but we don't know the value of a*b

Hence option E
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Is the answer not A.1/2 because \(\frac{x}{y}\) + \(\frac{w}{z}\) is not reciprocal of \(\frac{y}{x}\) + \(\frac{z}{w}\)?
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Is the answer not A.1/2 because \(\frac{x}{y}\) + \(\frac{w}{z}\) is not reciprocal of \(\frac{y}{x}\) + \(\frac{z}{w}\)?

\(\frac{x}{y}\) + \(\frac{w}{z}\) is NOT reciprocal of \(\frac{y}{x}\) + \(\frac{z}{w}\)

\(\frac{x}{y}\) + \(\frac{w}{z}\) = \(\frac{xz+wy}{yz}\) ------- 1

\(\frac{y}{x}\) + \(\frac{z}{w}\) = \(\frac{wy+xz}{xw}\) -------- 2

1 and 2 are not reciprocal to each other.
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