Answer : Option B

Explanation :

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Solution 1
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Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

$MF#%\begin{align}&\text{Lost 20% of length}\\
&\Rightarrow\text{New length = Original length}\times \dfrac{(100 - 20)}{100} = 100 \times \dfrac{80}{100} = 80\\\\\\
&\text{Lost 10% of breadth}\\
&\Rightarrow \text{New breadth= Original breadth}\times \dfrac{(100 - 10)}{100} = 100 \times \dfrac{90}{100} = 90\\\\
\end{align} $MF#%

$MF#%\text{Percentage of decrease in area = }\dfrac{\text{Decrease in Area}}{\text{Original Area}} \times 100 = \dfrac{2800}{10000} \times 100 = 28\% $MF#%

$MF#%\begin{align}&\text{Lost 20% of length}\\
&\Rightarrow\text{New length = Original length}\times \dfrac{(100 - 20)}{100} = l\times \dfrac{80}{100} = \dfrac{80l}{100}\\\\\\\\\\
&\text{Lost 10% of breadth}\\
&\Rightarrow \text{New breadth= Original breadth}\times \dfrac{(100 - 10)}{100} = b \times \dfrac{90}{100} = \dfrac{90b}{100}\\\\\\\\\\
&\text{New area =}\dfrac{80l}{100} \times \dfrac{90b}{100} = \dfrac{7200lb}{10000}= \dfrac{72lb}{100}\\\\
&\text{Decrease in area = Original Area - New Area = }lb - \dfrac{72lb}{100} = \dfrac{28lb}{100}\\\\\\
&\text{Percentage of decrease in area = }\dfrac{\text{Decrease in Area}}{\text{Original Area}} \times 100 \\
&= \dfrac{\left(\dfrac{28lb}{100}\right)}{lb} \times 100 = \dfrac{28lb \times 100}{100lb} = 28\%
\end{align} $MF#%