In measuring the side of a rectangle, there is an excess of 5% on one side and 2% deficit on the other. Then the error percent in the area is

Let the length and breadth of the rectangle be 20 cm and 50 cm respectively.

=> Area = $$A=20\times50=1000$$ $$cm^2$$

Length is taken 5% in excess, => New length = $$20+(\frac{5}{100}\times20)=21$$ cm

Similarly, new breadth = $$50-(\frac{2}{100}\times50)=49$$ cm

=> New area = $$A'=21\times49=1029$$ $$cm^2$$

$$\therefore$$ Error percent in the area = $$\frac{(1029-1000)}{1000}\times100=2.9\%$$

=> Ans - (C)