# Verbal Reasoning - Cube and Cuboid - Discussion

Discussion Forum : Cube and Cuboid - Cube and Cuboid 8 (Q.No. 2)
Directions to Solve

There are 128 cubes with me which are coloured according to two schemes viz.

1. 64 cubes each having two red adjacent faces and one yellow and other blue on their opposite faces while green on the rest.
2. 64 cubes each having two adjacent blue faces and one red and other green on their opposite faces, while red on the rest. They are then mixed up.

2.
What is the total number of red faces ?
0
64
320
128
Explanation:

No. of red faces among first 64 cubes = 128

No. of red faces among second 64 cubes = 192

Therefore, total number of red faces = 128 + 192 = 320

Discussion:
7 comments Page 1 of 1.

How do the second 64 cubes contain (64*3=192). How it is 3 faces with red colour?

Ans should be 256. (64*2+64*2)

No. of red faces among first cube = 64*2 = 128.
No. of red faces among Second cube = 64*2 = 128.
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Total no. of red faces = 256.
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Shekhar Ghate said:   1 decade ago
The second cube have three faces red. Two opposite to two adjacent blue (rest in question) and one opposite to green. Therefore 320 is correct answer.

DON said:   10 years ago
No.of red faces in 1st cube(64*2) = 128 because 2 red colours are painted.
No.of red faces in 2nd cube(64*3) = 192 because 3 red colours are painted.

Total no.of red faces = 320.

Sanyam said:   8 years ago
According to me,

There are 2 faces colored red in the 1st cube. So 32 red faces are there in the first cube.

In the 2nd cube, there are 3 red colored faces. So red faces in the second cube = 48.
So total no of red faces = 48 + 32 = 80.
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