In measuring the length and breadth of a rectangle, errors of $5 %$ and $3 %$ in excess, respectively, are made. The error percent in the calculated area is

$7.15 %$

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$6.25 %$

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$8.15 %$

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$8.35 %$

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Solution

The correct option is C
$8.15 %$

Let actual length and breadth of the rectangle be 'l' and 'b' respectively.

The measured sides are $(l + 0.05 l)$ and $(b + 0.03 b)$ i.e. $1.05 l$ and $1.03 b$
$∴$ Measured area $= (1.05 l) (1.03 b)$

Actual area $= l b$

$E r r o r i n a r e a = 1.0815 l b - l b = 0.0815 l b$

$∴ % e r r o r = \frac{E r r o r i n a r e a}{A c t u a l a r e a} \times 100$

$\Rightarrow % e r r o r = \frac{0.0815 l b}{l b} \times 100 % = 8.15 %$

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