The absolute errors in four questions A, B, C and D are known to be $ΔA, ΔB, ΔC and ΔD$ respectively. Find the relative error in Z if $Z = \frac{A^{4} B^{1 / 3}}{{CD}^{3 / 2}}$ .

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$\frac{ΔZ}{Z} = 4 \frac{ΔA}{A} + \frac{1}{3} \frac{ΔB}{B} + \frac{ΔC}{C} + \frac{3}{2} \frac{ΔD}{D}$

$\frac{ΔZ}{Z} = 4 \frac{ΔA}{A} + \frac{1}{3} \frac{ΔB}{B} - \frac{ΔC}{C} + \frac{3}{2} \frac{ΔD}{D}$

$\frac{ΔZ}{Z} = 4 \frac{ΔA}{A} + \frac{1}{3} \frac{ΔB}{B} - \frac{ΔC}{C} - \frac{3}{2} \frac{ΔD}{D}$

$\frac{ΔZ}{Z} = 4 \frac{ΔA}{A} + \frac{1}{3} \frac{ΔB}{B} + \frac{ΔC}{C} - \frac{3}{2} \frac{ΔD}{D}$

answer is A.

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Detailed Solution

$Z = \frac{A^{4} B^{1 / 3}}{{CD}^{3 / 2}}$ The relative error in Z is given by

$\frac{Δ Z}{Z} = 4 \frac{Δ A}{A} + \frac{1}{3} \frac{Δ B}{B} + \frac{Δ C}{C} + \frac{3}{2} \frac{Δ D}{D}$

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