Two sides of a rectangle were measured. One of the sides (length) was measured 10% more than its actual length and the other side (width) was measured 5% less than its actual length. The percentage error in measure obtained for the area of the rectangle is :

Solution(By Examveda Team)

Let the actual length and width of the rectangle be $$l$$ and b respectively.
Then, measured length :
$$ = 100\% {\text{ of }}l = \frac{{11l}}{{10}}$$
Measured width :
$$ = 95\% {\text{ of }}b = \frac{{19b}}{{20}}$$
Actual area = $$lb$$
Measured area :
$$\eqalign{ & = \left( {\frac{{11l}}{{10}} \times \frac{{19b}}{{20}}} \right) \cr & = \frac{{209lb}}{{200}} \cr} $$
Error in measurement :
$$\eqalign{ & = \left( {\frac{{209lb}}{{200}} - lb} \right) \cr & = \frac{{9lb}}{{200}} \cr} $$
∴ Error % :
$$\eqalign{ & = \left( {\frac{{9lb}}{{200}} \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 4.5\% \cr} $$