1.(A) 2.(A) 3.(A) 4.(C) 5.(B)

1. In all these kinds of questions, the number of cubes which would be painted on three sides would always be the corner cubes. There would be 8 corner cubes.

2. The two faces colored cubes would be the cubes along each edge (except for the corner cubes). It can be visualized that the top edges of the cube would have 7+7+3+3=20 cubes and the bottom edges would also have the same number ( i.e. 20 cubes). The vertical edges would have a further 7+7+7+7=28 cubes.

Out of the total of 68 cubes which are on the edges, 8 cubes would be corner cubes. Hence, the number of cubes painted on two sides would be 68-8=60

3. The number of one face colored cubes would be : On the front and back surfaces: 7 x 5 + 7 x 5= 70 cubes( You can remember this as (m-2)x(n-2) where m and n are the number of rows and columns on the surface in question). On the top and bottom surfaces : 5 x 3 + 5 x 3=30 cubes.

On the lateral surfaces : 7 x 3 + 7 x 3=42 cubes. Thus a total of 142 cubes would have one side painted.

4. The number of cubes with no face colored would be given by 7 x 5 x 3=105 [which can again be remembered as (m-2)(n-2)(p-2) where m, n and p are the number of parts in which the cube surfaces have been cut- in this case m=9, n=7 and p=5]

5. The cubes along the vertical edges would be painted both red and green. There are 5 such cubes ( with 2 faces coloured) on each of the 4 edges of the cuboid. Thus, the total number of cubes which would be painted on two sides( in red and green) would be 5 x 4=20.

Ans

1) A

2) A

3) A

4) C

5) B

.

Solution: 1) 8 Corner (With 3 Clr)

2) (7+5+3)*4=60

(L+B+H)*Sides

L=9-2Clr

B=7-2clr

H=5-2clr

Cube Side=4

3) (15+35+21)*2=142

First Draw Cube Sketch then Solve

4) 7*5*3=105

l*b*h

L=9-2Clr

B=7-2clr

H=5-2clr