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

A) 26 | B) 23 |

C) 27 | D) 21 |

Explanation:

Total number of exams written by 100 students = 48 + 45 + 38 = 131

Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.

Therefore, x + 2y + 3z = 131 also x + y + z = 100.

Given that z = 5. So x + 2y = 116 and x + y = 95.

Solving we get y = 21.

So 21 members are writing exactly 2 exams.