Let required fraction be then

............. (i)

and ............ (ii)

Solving (i) and (ii) we get x=3, y=5

Required fraction =

A) 476190476 | B) 48617 |

C) 47619 | D) 4587962 |

Explanation:

By hit and trail, we find that

47619 x 7 = 333333.

7) 333333 (47619

333333

----------------

0

And 476190476 x 7 = 3333333333 but smallest number is 47619.

A) 6 | B) 4 |

C) 3 | D) 12 |

A) 42 | B) 54 |

C) 46 | D) 58 |

Explanation:

Let x and y be the two parts of 96.

x/7 = y/9 => x:y = 7:9

=> The smallest part is = 7/16 x 96 = 42.

A) 1107 | B) 1080 |

C) 1100 | D) 1208 |

Explanation:

LCM of 4, 6, 8 and 10 = 120

120) 1000 (8

960

------

40

The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080.

A) 45 | B) 48 |

C) 360 | D) 36 |

Explanation:

16200 =

A perfect cube has a property of having the indices of all its prime factors divisible by 3.

Required number = = 9x5 = 45.