Question

# For a party, three solid cheese balls with a diameter of 2 inches, 4 inches and 6 inches, respectively, were combined to form a single cheese ball. What was the approximate diameter, in inches, of the new cheese ball? (The volume of a sphere is $\dfrac{4}{3}\pi {r^3}$, where r is the radius.)A. 12B. 6C. $\sqrt[3]{{16}}$ D. $3\sqrt[3]{8}$E. $2\sqrt[3]{{36}}$

Hint: In this question, we are supposed to combine three cheese balls with diameter 2 inches, 4 inches and 6 inches, when we combine the volume of the bigger cheese ball will be the sum of the volume of the three smaller cheese balls.

Volume of the bigger cheese ball = Sum of volume of the three smaller cheese balls.
Volume of new cheese ball = $\dfrac{4}{3}\pi \left( {{r_1}^3 + {r_2}^3 + {r_3}^3} \right)$
Putting the values of r in the above equation,
Volume of bigger cheese ball = $\dfrac{4}{3}\pi \left( {{2^3} + {4^3} + {6^3}} \right)$
Since the cheese ball is in spherical shape the volume can be taken as
Volume of bigger cheese ball =$\dfrac{4}{3}\pi {R^3}$
On equating both the equations,
$\dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi \left( {8 + 64 + 216} \right)$
Cancelling the common terms, we get,
${R^3} = \left( {8 + 64 + 216} \right)$
${R^3} = 288$
On solving it further, we get,
$R = \sqrt[3]{{288}}$
On factoring, we get,
$R = \sqrt[3]{{8 \times 36}}$
$R = 2\sqrt[3]{{36}}$ inches

Note: Do not make the mistake of equating the sum of areas or perimeter because with the change of structure, the value of these entities will also change, but the value of volume will remain the same.