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Standard VII

Mathematics

Converting Percentages to Fractions

An experiment...

Question

An experiment measures quantities a, b and c, and X is calculated from $X = a b^{2} / c^{3}$ . If the percentage error in a, b and c are $\pm 1 %$ , $\pm 3 %$ and $\pm 2 %$ respectively, the percentage error in X will be:

\pm 13

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\pm 7

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\pm 4

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\pm 1

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Solution

The correct option is A $\pm 13$ %
$x = a b^{2} / c^{3}$
Thus, $\frac{△ x}{x} = \frac{△ a}{a} + 2 \frac{△ b}{b} + 3 \frac{△ c}{c}$
$= \pm 1 + 2 (\pm 3) + 3 (\pm 2)$
$= \pm 1 \pm 6 \pm 6 = \pm 13 %$

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An experiment measures quantities a, b and c, and X is calculated from X=ab2/c3. If the percentage error in a, b and c are ±1% , ±3% and ±2% respectively, the percentage error in X will be:

The correct option is A ±13% x=ab2/c3Thus, △xx=△aa+2△bb+3△cc =±1+2(±3)+3(±2) =±1±6±6=±13%

An experiment measures quantities a, b and c, and X is calculated from $X = a b^{2} / c^{3}$ . If the percentage error in a, b and c are $\pm 1 %$ , $\pm 3 %$ and $\pm 2 %$ respectively, the percentage error in X will be:

The correct option is A $\pm 13$ %
$x = a b^{2} / c^{3}$
Thus, $\frac{△ x}{x} = \frac{△ a}{a} + 2 \frac{△ b}{b} + 3 \frac{△ c}{c}$
$= \pm 1 + 2 (\pm 3) + 3 (\pm 2)$
$= \pm 1 \pm 6 \pm 6 = \pm 13 %$