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# 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. $2\sqrt[3]{{36}}$

Last updated date: 16th Sep 2024
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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