Aptitude - Volume and Surface Area - Discussion
Discussion Forum : Volume and Surface Area - General Questions (Q.No. 14)
14.
A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?
Answer: Option
Explanation:
Volume of the large cube = (33 + 43 + 53) = 216 cm3.
Let the edge of the large cube be a.
So, a3 = 216 
a = 6 cm.
 Required ratio = |
 |
6 x (32 + 42 + 52) |  |
= | 50 | = 25 : 18. |
| 6 x 62 | 36 |
Discussion:
24 comments Page 3 of 3.
Ramesh said:
9 years ago
Can anyone explain it in any other ways?
Prasanna said:
10 years ago
Cube total surface area = 6Xa^2.
So the small cubes total surface area = 6(3^2+4^2+5^2).
Lager Cube former from this cube would be = 6X(12^2).
Because a = 3+4+5.
So, the required ratio small cubes : large cube = 50:144.
So the small cubes total surface area = 6(3^2+4^2+5^2).
Lager Cube former from this cube would be = 6X(12^2).
Because a = 3+4+5.
So, the required ratio small cubes : large cube = 50:144.
MANVI MEHTA said:
1 decade ago
@Hary.
Can you please explain the required ratio step more clearly?
Can you please explain the required ratio step more clearly?
Harsha said:
1 decade ago
@Hary.
This shows the area of each side of individual small cubes. For the cube with 3 cm side area of one wall is 3*3 and for 6 wall its 6(3*3).
Likewise for all the 3 smaller cubes. And final is the sum of S.A of all three smaller cubes.
This shows the area of each side of individual small cubes. For the cube with 3 cm side area of one wall is 3*3 and for 6 wall its 6(3*3).
Likewise for all the 3 smaller cubes. And final is the sum of S.A of all three smaller cubes.
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