Question

Three cubes of edges 20 cms, 12 cms and 16 cms are meted without loss of metal into a sir
edge of the new cube will be:
A)18 cms B)24 cms C)16 cms D)20 ​

Answer 1

Given :-

▪ Three cubes edges 20 cm, 12 cm, and 16 cm are melted without loss of metal into a big cube.

To Find :-

▪ Edge of the new cube.

Solution :-

Since the cubes are melted without any loss of metal, so the total volume of the three small cubes will be equal to the volume of the big cube.

Cube 1 ( Edge = 20 cm )

We know,

⇒ Volume of cube = (Edge)³

⇒ V₁ = (20)³

⇒ V₁ = 400 × 20

⇒ V₁ = 8000 cm³

Cube 2 ( Edge = 12 cm )

We know,

⇒ Volume of cube = (Edge)³

⇒ V₂ = (12)³

⇒ V₂ = 144 × 12

⇒ V₂ = 1728 cm³

Cube 3 ( Edge = 16 cm )

We know,

⇒ Volume of cube = (Edge)³

⇒ V₃ = (16)³

⇒ V₃ = 256 × 16

⇒ V₃ = 4096 cm³

Now, Let the edge of the big cube be x , So as discussed before

⇒ Volume of New cube = V₁ + V₂ + V₃

⇒ (x)³ = 8000 + 1728 + 4096

⇒ x³ = 13824

⇒ x = 24 cm

Hence, The edge of the new cube formed by melting other cubes is 24 cm.

Answer 2

$\mathcal{\green{\underline{\red{GIVEN:}}}}$

There are three cubes:

• Side of cube 1 = 20 cm.
• Side of cube 2 = 12 cm.
• Side of cube 3 = 16 cm.

$\mathcal{\green{\underline{\red{TO \:\: FIND:}}}}$

We need to find the edge of the new cube formed when combining the three smaller cubes.

$\mathcal{\green{\underline{\red{SOLUTION:}}}}$

Find the volume of the three cubes.

$\boxed{\sf{\blue{Volume = side ^3}}}$

Volume of cube 1:

$\sf V_1=20^3$

$\sf V_1=8000 cm^3.$

Volume of cube 2:

$\sf V_2=12^3.$

$\sf V_2=1728 cm^3.$

Volume of cube 3:

$\sf V_3=16^3$

$\sf V_3=4096cm^3.$

Now, the volumes of the smaller cubes are combined to form a bigger cube. Thus, assuming volume of bigger cube as V₄:

$\sf V_4=V_1+V_2+V_3$

$\sf V_4=8000+1728+4096.$

$\sf V_4=13824 cm^3.$

To find out side, we have a formula:

$\boxed{\sf{\blue{side=\sqrt[3]{Volume} }}}$

Substituting the values:

$\sf Side = \sqrt[3]{13824}$

$\sf Side = 24 cm.$

OPTION B IS CORRECT  ✓✓✓.