A) 11 | B) 18 |

C) 20 | D) 21 |

Explanation:

Clearly, From 1 to 100, there are ten numbers with 3 as the unit's digit - 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.

So, required number = 10 + 10 = 20.

A) 0 | B) 1 |

C) 2 | D) 3 |

Explanation:

From the given information in the question

24 ÷ 4 + 6 - 3 x 4

6 + 6 - 12 = 0

A) 3 | B) 5 |

C) 7 | D) 9 |

Explanation:

There are total 5 ways.

we have only one kg of weight so,

2kg = 1kg + 1kg

4kg = 1 + 2 + 1

8kg = 1 + 2 + 4 + 1

16kg = 1 + 2 + 4 + 8 + 1

31kg = 1 + 2 + 4 + 8 + 16

So 5 weights are needed.

A) 5 | B) 2 |

C) 7 | D) 4 |

Explanation:

Given number is 14278359

As given 1st and 5th digits, 2nd & 6th, 3rd & 7th and 4th & 8th digits are to to be interchanged.

Therfore, the rearranged number is 83591427

Then, the second digit from the right end is 2.

A) 125 | B) 96 |

C) 35 | D) 25 |

Explanation:

We know that, to be in the form that the number of rows and the number of columns to be equal the number should be a perfect square ().

Given number is 1200

The perfect number which is above and near to 1200 is 1225 which is 35 35.

The minimum number of coins he need is 25.

A) 60 | B) 30 |

C) 70 | D) 90 |

Explanation:

Let 'x' be the number of new Rs.2000 notes.

Given for one new 2000 note there are three old 500 notes.

3x

Given 10 more new Rs.2000 notes are added to the collection

and the ratio of new Rs.2000 notes to old Rs.500 notes

Therefore, number of new Rs.2000 notes x+10 = 30.

old Rs.500 notes 3x = 60.

Thus, the total number of notes in the collection 30 + 60 = 90.