A) 12 | B) 13 |

C) 15 | D) Cannot be determined |

Explanation:

In a usual dice, the sum of the numbers on any two opposite faces is always 7. Thus, 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4.

Consequently, when 4, 3, 1 and 5 are the numbers on the top faces, then 3, 4, 6 and 2 respectively are the numbers on the face touching the ground. The total of these numbers = 3 + 4 + 6 + 2 = 15.

A) 56 | B) 60 |

C) 64 | D) 72 |

Explanation:

(Total numbers of cubes in a line x Number of stack / tower) + ...

= (6x1)+(5x2)+(4x3)+(3x4)+(5x2)+(6x1)

= 6+10+12+12+10+6 = 56

A) 56 | B) 48 |

C) 32 | D) 64 |

Explanation:

We know that Cubes with no surface painted can be find using ${\left(x-2\right)}^{3}$, where x is number of cuttings. Here x=6.

$\therefore {\left(6-2\right)}^{3}=64$

A) Option 1,2 and 3 | B) Option 1,3 and 4 |

C) Option 2 and 3 | D) Option 2,3 and 4 |

Explanation:

When the sheet in fig. (A) is folded, then one of the faces of the cube formed will be of the form and this face will lie opposite the face bearing a square. Also, one of the blank faces lies opposite another blank face and the third blank face lies opposite the face bearing an '=' sign. Clearly, all the three blank faces cannot appear adjacent to each other. So, the cube shown in fig. (2) which has all the three blank faces adjacent to each other cannot be formed. Hence, only the cubes shown in figures A, C and D can be formed.

A) Option 1,2 and 3 | B) Option 1,2 and 4 |

C) Option 2 and 3 | D) Option 2,3 and 4 |

Explanation:

when a cube is formed by folding the sheet shown in fig. (A), then is one of the faces of the cube and this face lies opposite to a blank face. Also, a face bearing a square lies opposite to another blank face. The remaining two blank faces lie opposite to each other. Clearly, in the cube shown in fig. (1), the face consisting of the four symbols is not the same as that formed (as shown above). Hence, the cube in fig. (1) cannot be formed

A) 130 | B) 132 |

C) 138 | D) 140 |

Explanation:

In the figure, there are

5 columns containing 4 cubes each;

33 columns containing 3 cubes each;

9 columns containing 2 cubes each and 3 columns containing 1 cube each.

$\therefore $ Total Number of cubes = ( 5 x 4) + (33 x 3) + (9 x 2) + (3 x 1) = 20 + 99 + 18 + 3 = 140

A) 45 | B) 46 |

C) 48 | D) 49 |

Explanation:

In the figure, there are 21 columns containing 2 cubes each and 4 columns containing 1 cube each.

$\therefore $Total number of cubes = (21 x 2) + (4 x 1) = 42 + 4 = 46.

A) 80 | B) 87 |

C) 89 | D) 90 |

Explanation:

In the figure, there are 9 columns containing 5 cubes each, 7 columns containing 4 cubes each, 5 columns containing 3 cubes each and 1 column containing 1 cube.

$\therefore $Total number of cubes = (9 x 5) + (7 x 4) + (5 x 3) + (1 x 1) = 89

A) 64 | B) 66 |

C) 68 | D) 70 |

Explanation:

In the figure, there are 34 columns containing 2 cubes each.

$\therefore $ Total number cubes = (34 x 2) = 68