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(B) Let $X = ABC$. Taking the natural logarithm on both sides,we get $\ln X = \ln A + \ln B + \ln C$. Differentiating both sides,we get $\frac{dX}{X}

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$a + b + c$

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(B) Let $X = ABC$. Taking the natural logarithm on both sides,we get $\ln X = \ln A + \ln B + \ln C$. Differentiating both sides,we get $\frac{dX}{X} = \frac{dA}{A} + \frac{dB}{B} + \frac{dC}{C}$. For small errors,this can be written as $\frac{\Delta X}{X} = \frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{\Delta C}{C}$. Given that $a, b, c$ are the percentage errors,we have $a = \frac{\Delta A}{A} 	imes 100$,$b = \frac{\Delta B}{B} 	imes 100$,and $c = \frac{\Delta C}{C} 	imes 100$. Therefore,the percentage error in $X = ABC$ is $\frac{\Delta X}{X} 	imes 100 = (\frac{\Delta A}{A} 	imes 100) + (\frac{\Delta B}{B} 	imes 100) + (\frac{\Delta C}{C} 	imes 100) = a + b + c$.

If $a, b, c$ are the percentage errors in the measurement of $A, B$ and $C$,then the percentage error in $ABC$ would be approximately

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