Let required fraction be then

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

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

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

Required fraction =

A) 57 | B) 75 |

C) 48 | D) 39 |

Explanation:

Let the two-digit number be 10a + b

a + b = 12 --- (1)

If a>b, a - b = 6

If b>a, b - a = 6

If a - b = 6, adding it to equation (1), we get

2a = 18 => a =9

so b = 12 - a = 3

Number would be 93.

if b - a = 6, adding it to the equation (1), we get

2b = 18 => b = 9

a = 12 - b = 3.

Number would be 39.

There fore, Number would be 39 or 93.

A) 1020 | B) 1040 |

C) 1060 | D) 1022 |

Explanation:

Lowest 4-digit number is 1000.

LCM of 3, 4 and 5 is 60.

Dividing 1000 by 60, we get the remainder 40. Thus, the lowest 4-digit number that exactly divisible by 3, 4 and 5 is 1000 + (60 - 40) = 1020.

Now, add the remainder 2 that's required. Thus, the answer is 1022.

A) 270 | B) 1270 |

C) 350 | D) 720 |

Explanation:

Let the smallest number be x.

Then larger number = (x + 1365)

x + 1365 = 6x + 15

= 5x = 1350

x = 270

Smaller number = 270.

A) 17 | B) 19 |

C) 13 | D) 11 |

Explanation:

(385/1001) = 5/13

First Number is 5

=> Fourth number is 13

A) 87 | B) 59 |

C) 92 | D) 81 |

Explanation:

Number = 271 x 96 + 0 = 26016

=> 95) 26016 (273

25935

--------

81

Required number = 81.