Mensuration Shape Transformation Type Problem

mensuration

Dear Reader, this type is very common in government exams. Let us see an example and learn how to solve problems of this type.

Example Question:
Ramu is a blacksmith. He has a cone-shaped steel of base radius 20 cm and height 30 cm. A client gives him a task of creating spherical steel balls of each 2 cm in radius. Ramu decides to melt the cone and create steel balls. How many such balls he can make?

Here’s How To Solve This Problem:
Ramu is going to melt cone into spherical balls. First, you have to find the volume of steel he will get from the cone. Second, you have to find the volume of material needed for each new spherical ball. If you divide the volume of the cone by volume of the spherical ball, you will get the answer.

Solution:
The volume of material Ramu will get by melting the cone
= volume of the cone = 1/3 Π r2 h
But we know, radius of cone = r = 20 cm and height = h = 30 cm
Therefore, volume of the cone = 1/3 Π (20 x 20 x 20) x 30
= Π x 20 x 20 x 20 x 10
= Π x 80000 cm cube

Tip: To solve such problems, you should remember the formulas for volumes of shapes.

The volume of material required for each spherical ball
= volume of the sphere = Π r3
But we know, radius of the sphere r = 2 cm
Volume of each new sphere = Π x 2 x 2 x 2 = Π x 8 cm cube
Total number of balls he can create = volume of the cone / volume of each sphere = Π x 80000 / Π x 8 = 10000

Hope you find this problem and solution useful. If you want to prepare aptitude for your exams fast in 4 weeks, check this Quantitative Aptitude Fast Preparation Book by MirchiExperts.

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