Question
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 :
Answer: Option A
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 % :$$\eqalign{
& = \left( {\frac{{9lb}}{{200}} \times \frac{1}{{lb}} \times 100} \right)\% \cr
& = 4.5\% \cr} $$
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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 % :$$\eqalign{
& = \left( {\frac{{9lb}}{{200}} \times \frac{1}{{lb}} \times 100} \right)\% \cr
& = 4.5\% \cr} $$
Was this answer helpful ?
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