If the base radius and the height of a right circular cone are increas

1.	If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately :A. 60 B. 68 C. 73 D. 78
Answer» Option : (C) Let base radius of cone = x Let height of cone = y Volume of cone = \(\frac{1}{3}\)πx²y New radius of cone = x + x\(\frac{20}{100}\) = \(\frac{6x}{5}\) New height of cone = y + y\(\frac{20}{100}\) = \(\frac{6y}{5}\) New volume = \(\frac{1}{3}\)π\(\frac{36x^2}{25}\) x \(\frac{6y}{5}\) = \(\frac{1}{3}\)π\(\frac{216x^2y}{125}\) Increase in volume = \(\frac{216x^2y}{125}\) - x²y = \(\frac{91x^2y}{125}\) Percentage increase in volume = \(\frac{\frac{91x^2y}{125}}{x^2y}\) x 100 = 72.8 % Approx 73%

If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately :A. 60 B. 68 C. 73 D. 78

Answer»

Option : (C)

Let base radius of cone = x

Let height of cone = y

Volume of cone = \(\frac{1}{3}\)πx²y

New radius of cone = x + x\(\frac{20}{100}\) = \(\frac{6x}{5}\)

New height of cone = y + y\(\frac{20}{100}\) = \(\frac{6y}{5}\)

New volume = \(\frac{1}{3}\)π\(\frac{36x^2}{25}\) x \(\frac{6y}{5}\)

= \(\frac{1}{3}\)π\(\frac{216x^2y}{125}\)

Increase in volume = \(\frac{216x^2y}{125}\) - x²y

= \(\frac{91x^2y}{125}\)

Percentage increase in volume = \(\frac{\frac{91x^2y}{125}}{x^2y}\) x 100

= 72.8 %

Approx 73%