(solved)Question 20 SSC-CGL 2018 June 4 Shift 1 | A solid cube with a volume of 13824 cu. cm is cut into eight cubes of the same volume. The ratio of the surface area of the original cube and the total sum of the surface area of three smaller cubes will be?

Parveen, Feb 11, 2020

Question 20
SSC-CGL 2018 June 4 Shift 1

A solid cube with a volume of 13824 cu. cm is cut into eight cubes of the same volume. The ratio of the surface area of the original cube and the total sum of the surface area of three smaller cubes will be?

2 : 3
2 : 1
4 : 3
8 : 3

Solution 1

See NCERT class XI math chapter solid geometry concept of octants, first exercise.

When a cube of side $\displaystyle 2s$ is cut into eight equal octants, the side of each resulting cube would be $\displaystyle s$. Hence the required ratio of the surface areas of the original cube to the three smaller ones is $\displaystyle \frac{6 \times (2s)^2}{3 \times 6 s^2}$, which gives 4 : 3 as the answer!

Solution 2

Side of the original cube from the given volume is $\displaystyle \sqrt [3]{13824} = 24$ cm.

Volume of each new cube is $\displaystyle \frac{13824}{8} = 1728$ cc. The side is $\displaystyle \sqrt [3]{1728} = 12$ cm.

The required ratio is $\displaystyle \frac{6 \times (24)^2}{3 \times 6 \times (12)^2}$, which again gives 4 : 3 as the answer!

Solution 3

Standard Shortcut Formula: If a cube is split into $\displaystyle n$ equal cubes, the ratio of surface area of the original to the surface area of one cube is $\displaystyle n^{\frac 23}$

In the present case the ratio of surface areas of the larger to one smaller would be $\displaystyle 8^{\frac 23} = 4$

Hence, the ratio of larger to three smaller would be 4 : 3 answer!