UPSC ONLINE ACADEMY

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Q.1 A hollow cube of size 5 cm is taken with a thickness of 1 cm. It is made of smaller cubes of size 1 cm. If only 2 faces of the outer surface of the cube are painted, totally how many faces of the smaller cubes remain unpainted ?

A) 485

B) 486

C) 487

D) 488

Ans. D

The big cube is completely hollow.. But, the thickness of its side is 1cm & this is made up of smaller cubes.

 

Just like the walls of a room. The room is hollow, but, its thickness is due to smaller bricks. The 4 walls of this room is painted from the outside. And, you are asked totally how many sides of all the bricks are now unpainted.

 

First, you need to know how many bricks (small cubes) are there.

 

Size of big cube = 5cm

Total volume of the big cube = 5*5*5 = 125cm^3

 

Size of the hollow cube inside the big cube = 3cm

Volume of the hollow space inside the big cube = 3*3*3 = 27cm^3

 

Therefore, volume occupied by small cubes (or volume of thickness) = 125 – 27 = 98cm^3

 

Size of each small cube = 1cm

Volume of each small cube = 1*1*1 = 1cm^3

Total number of small cubes in wall = 98 / 1 = 98

 

In short, 98 small cubes make up the wall of the big cube.

Each cube has 6 faces, so 98 cubes have = 98*6 faces = 588

 

The four sides of the big cube have 100 painted faces.

Because each big side has 25 faces of the small cubes.

 

Therefore, total unpainted faces = 588 – 100 = 488

 

Q.2 A large cube is painted green on all six faces & then cut into a certain number of smaller but identical cubes. It was found that amongst the smaller cubes, there were 8 cubes which had no face painted at all. What is the maximum number of cubes that were cut from large original cube ?

 

A) 46

B) 56

C) 64

D) 54

Ans. C

When a cube is painted & cut into smaller n pieces along each side then the total number of smaller cubes obtained will be (n*n*n). From these, remove the outer layer of cubes, that is 2 from each face, then there will be [(n-2) * (n-2) * (n-2)] cubes left. Now. Number of cubes that do not have any face painted is (2*2*2)

As, (n-2) = 2

i.e. n=4

Therefore, the maximum number of cubes that were cut from original cube are 4*4*4= 64 Cube

 

Q.3 Pointing towards a boy, Veena said, “He is the son of the only son of my grandfather”. How is that boy related to Veena?

 

A) Brother

B) Nephew

C) Father

D) None of these

Ans. A

 

EXPLANATION:

 

 

 

Q.4 Pointing to Kapil, Shilpa said, “His mother’s brother is the father of my son Ashish”. How is Kapil related to Shilpa?

 

A) Sister–in–law

B) Nephew

C) Niece

D) Aunt

Ans. B

 

 

 

 

Kapil is the nephew of Shilpa.

 

Q.5 A cube is painted green on all the sides. Then it is cut into 512 cubes of equal size. How many of smaller cubes are painted on 1 side ?

A) 18

B) 216

C) 12

D) 14

Ans. B

Cube root of 512 is 8. Si, cube is 8*8*8. Each side of the cube  has 64 visible small cube faces, 4 of which are corners (3 green sides) & 24 more are edges (2 coloured sides), so  there are only 36 small cubes with one green face on each of the six sides of the large cube.

So, 36*6= 216

Shortcut:

Learn this: Vertices are: 8

Edges are: 12

Faces are :6

*To Find 1 face painted:

(N/n-2)2 *6

= (8/1-2)2 *6 = 36 *6

=216

*Similarly if it asks you to find 2 faces painted, then use this formula,

(N/n-2) * 12

*For 3 faces painted, use this formula,

just find how many vertices are there & these are 8 which is answer

*For finding no. of smaller cubes: (N/n)3

 

Q.6 Find the number of triangles in the given figure.

 

 

A) 22

 

B) 24

 

C) 26

 

D) 28

Ans. D

The figure may be labelled as shown.

 

The simplest triangles are AGH, GFO, LFO, DJK, EKP, PEL and IMN i.e. 7

in number.

The triangles having two components each are GFL, KEL, AMO, NDP, BHN,

 CMJ, NEJ and HFM i.e. 8 in number.

The triangles having three components each are IOE, IFP, BIF and CEI i.e. 4 in

number.

The triangles having four components each are ANE and DMF i.e. 2 in number.

The triangles having five components each are FCK, BGE and ADL i.e. 3 in

Number.

The triangles having six components each are BPF, COE, DHF and AJE i.e. 4 in

Number

Total number of triangles in the figure = 7 + 8 + 4 + 2 + 3 + 4 = 28.