Let the numbers be x, x+1, x+2

Then x + x+1 + x+2 = 42

=> 3x + 3 = 42

=> 3x = 39

=> x = 13

The numbers would be 13, 14, 15

Sum of the digits of middle number = 1+4 = 5

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.

A) 83 | B) 89 |

C) 85 | D) 79 |

Explanation:

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F of 551 and 1073 = 29;

First number = 551/29 = 19

Third number = 1073/29 = 37.

Required sum = 19 + 29 + 37 = 85.

A) 28 | B) 19 |

C) 37 | D) 17 |

Explanation:

Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.

P + Q = 8 ----- (1)

(10Q + P) - (10P + Q) = 54

9(Q - P) = 54

(Q - P) = 6 ----- (2)

Solve (1) and (2) P = 1 and Q = 7

The required number is = 17

A) 87 | B) 81 |

C) 83 | D) 85 |

Explanation:

Cp1 = p2 - p1 ,

Cp2 = p3 - p2

..

..

..

Cp23 = p24 - p23

Sum of series = (p2-p1) + (p3-p2) + .....(p23-p22) + (p24-p23)

All terms get cancelled, except p1 = 2 and p24 = 89

So Sum = -p1 + p24

Sum of series = -2 + 89 = 87