1. **Problem:** A student approximated 0.03621 as 0.0269 instead of rounding to 2 significant figures. Find the percentage error. 2. **Problem:** A boy measures the length and breadth of a cricket field as 58.6 m and 30.3 m with errors of 6 cm and 40 cm respectively. Find the percentage error in the calculation of the perimeter. --- 1. 1. Find actual value: $0.03621$ 2. Approximate value: $0.0269$ 3. Calculate absolute error: $$|0.03621 - 0.0269| = 0.00931$$ 4. Calculate percentage error: $$\frac{0.00931}{0.03621} \times 100 = 25.7\%$$ 5. Answer: Percentage error = $25.7\%$ to 2 significant figures. --- 2. 1. Convert errors to meters: Length error = $0.06$ m, Breadth error = $0.40$ m 2. Actual length = $58.6$ m, breadth = $30.3$ m 3. Actual perimeter: $$P = 2(L + B) = 2(58.6 + 30.3) = 2(88.9) = 177.8\text{ m}$$ 4. Maximum error in length and breadth: Length $\pm 0.06$ m, breadth $\pm 0.40$ m 5. Maximum possible error in perimeter: $$2(0.06 + 0.40) = 2(0.46) = 0.92\text{ m}$$ 6. Percentage error in perimeter: $$\frac{0.92}{177.8} \times 100 = 0.517\%$$ 7. Answer: Percentage error in perimeter calculation = $0.52\%$ to 2 significant figures.