In measuring the sides of a rectangle, errors of $5 %$ and $3 %$ in excess 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 sides of the rectangle are 'l' and 'b' respectively.

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

Actual area $= l b$
Error in area $= 1.0815 l b - l b = 0.0815 l b$
$∴ % e r r o r = \frac{0.0815 l b}{l b} \times 100 % = 8.15 %$

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

The correct option is C 8.15%Let actual sides of the rectangle are 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b Therefore, Measured area =(1.05l)(1.03b) Actual area =lb Error in area =1.0815lb−lb=0.0815lb ∴% error=0.0815lblb×100%=8.15%

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