Daily Practice Questions Questions for CAT with Answers PDF

Number of smaller cubes with no face coloured = (n – 2)³ = (4 – 2)³ = 8

Number of smaller cubes with two surfaces painted red = (n – 2) × 12 = (4 – 2) × 12 = 24

Since out of 125 total number of cubes, we removed 4 columns of 5 cubes each, the remaining number of cubes = 125 – (4 × 5) = 125 – 20 = 105

There are 9 + (4 × 4) = 25 such cubes

Since the surfaces of the big cube are painted with some colours, so the number of smaller cubes with only one surface painted is (n – 2) × 12 = (4 – 2) × 12 = 2 × 12 = 24.

From figures (i) and (ii), we conclude that numbers 1, 5, 6 and 3 appear adjacent to number 2.
Therefore, number 2 must appear opposite to number 4.
Thus, if number 2 is on the top, then number 4 is at the bottom.

From figures (i) and (ii), we conclude that 5, 4, 6 and 2 dots appear adjacent to 3 dots.
Therefore, 1 dot must appear opposite to 3 dots.
Thus, if the face with 1 dot is at the bottom, then the face with 3 dots will appear at the top.

From figures (i) and (ii), we conclude that the numbers 1, 2, 3 and 4 appear adjacent to 6.
Thus, the number 5 will appear opposite 6.
Therefore, when 6 is at the bottom, then 5 will be at the top.

Clearly, the six faces of the dice are labelled as shown:
Face I → a, Face II → b, Face III → c, Face IV → d, Face V → e, Face VI → f
Therefore, ‘a’ appears opposite ‘c’.
Hence, when ‘c’ is at the top, then ‘a’ will be at the bottom.

According to the given figures,
Top face in (i) dice contain 3 dots;
Top face in (ii) dice contain 1 dots;
Top face in (iii) dice contain 3 dots;
Therefore,
Total dots = 3 + 1 + 3 = 7
Correct Answer: 7