Multiple choice

Three solid lead spheres of diameters 6 cm, 8 cm and 10 cm are melted together and recast as a solid sphere. What is the percentage diminution of the surface area as compared to the sum of the surface areas of the three spheres?

  1. 25%

  2. 26%

  3. 27%

  4. 28%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Sum of surface areas = 4*pi*(3^2 + 4^2 + 5^2) = 4*pi*(9+16+25) = 200*pi. New radius R^3 = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216, so R = 6. New area = 4*pi*(6^2) = 144*pi. Diminution = (200-144)/200 = 56/200 = 28%.

AI explanation

Let the radii of the three spheres be 3, 4, and 5 centimeters. The sum of their surface areas is 4 times pi times (3 squared plus 4 squared plus 5 squared), which equals 4 times pi times 50, or 200 pi. The combined volume is 4 divided by 3 times pi times (3 cubed plus 4 cubed plus 5 cubed), which equals 4 divided by 3 times pi times 216; setting this equal to the volume of the new sphere gives a new radius of 6 centimeters. The surface area of the new sphere is 4 times pi times 6 squared, which is 144 pi. The percentage diminution is calculated as ((200 pi minus 144 pi) divided by 200 pi) times 100, resulting in a 28 percent reduction.

Keep practicing — related questions