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A metal spherical ball was melted and formed into three spherical balls, such that the radii of the two balls was 50% and 80% of the radius of the original ball. If 2% of the metal remained, then the radius of the third spherical ball is what percentage of the radius of the original ball?

A metal spherical ball was melted and formed into three spherical balls, such that the radii of the two balls was 50% and 80% of the radius of the original ball. If 2% of the metal remained, then the radius of the third spherical ball is what percentage of the radius of the original ball? Correct Answer 70%

Given:

Radii of the two balls was 50% and 80% of the radius of the original ball

2% of the metal remained while forming

Formula used:

Volume of sphere = (4/3)π × (radius)3

Calculation:

Let the radius of the original ball be R units and that of the third ball be r cm

⇒ Radii of two spherical balls formed is 0.5R and 0.8R

∵ 2% of the metal remained,

⇒ Volume of three spherical balls = 98% of Volume of original ball

Now, Volume of sphere = (4/3)π × (radius)3

⇒ + + = 98% of

⇒ (5R/10)3 + (8R/10)3 + (r)3 = 98% of (R)3

⇒ 125R3/1000 + 512R3/1000 + r3 = 98R3/100

⇒ 637R3/1000 + r3 = 980R3/1000

⇒ r3 = (980 – 637)R3/1000

⇒ r3 = 343R3/1000

⇒ r = 7R/10 = 0.7R

⇒ r = 70% of R

∴ The radius of the third spherical ball is 70% of the radius of the original ball

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