"> Word Problem 2 | Job Exam Rare Mathematics (JEMrare)

Word Problem 2

Problem 2: If there are three metallic cubes with sides of 3 cm, 4 cm and5 cm respectively, and if these cubes are melted together to make one cube, what will be the side of the new cube?



Solution:

There are three metallic cube boxes and suppose there volumes are V1, V2, and V3

Now let us calculate volumes of each cube box;

We use formula for calculation of volume;

Volume = Side1 \times Side2 \times Side3

For cube Box with each side of 3cm;

Volume1 = 3 \times 3 \times 3 = 27 cm^{3}For cube Box with each side of 4cm;

Volume2 = 4 \times 4 \times 4 = 64 cm^{3}

For cube Box with each side of 5cm;

Volume2 = 5 \times 5 \times 5 = 125 cm^{3}

Now suppose when all these three  boxes are melted and mixed to form a single box, the each side of that single cube box is ‘y’

So, then for that single cube box ;

Volume = V = y \times y \times y = y cm^{3}

But this volume of single cube box should be equal to sum of volumes of all three above cube boxes !

i-e

V = Volume1 + Volume2+ Volume3

 \Rightarrow y^{3}= 27+64+125

 \Rightarrow y^{3}= 216

 \Rightarrow y^{3}= 8^{3}

Cancel out powers on both sides;

 \Rightarrow y}= 8

So, each side of coalesced  cube is 8 cm length

So, Volume of new coalesced cube ;

Volume=V= 8\times 8\times 8 =512 cm^{3}

(basic mathematics for general public)

 

 

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