A) 17 | B) 18 |

C) `19 | D) 20 |

Explanation:

The fence will consist of one more post than there are chains. (e.g. P-c-P-c-P).

Therefore, a total length has to be a multiple of the length of the chain plus one post (5.5) plus one post extra.We have length = (5.5n + 0.5), where n can be any positive whole number. If n= 3, length =17

The Multiples of 4 are **4, 8, 12, 16, 20, 24, 28, 32, 36, 40 **upto 40. Mutiples of 4 means which can be divided by 4 leaving remainder '0'.

**Common Multiples of 4 & 6** are **12, 24, 36, 48, 60 **upto 60.

A) 31 | B) 33 |

C) 29 | D) 27 |

Explanation:

Let the three consecutive odd numbers be **x, x+2, x+4**

**Then,**

**x + x + 2 + x + 4 = 93**

**=> **3x + 6 = 93

=> 3x = 87

=> x = 29 => **29, 31, 33 are three consecutive odd numbers.**

Therefore, the middle number is **31.**

A) 7 | B) 9 |

C) 11 | D) 5 |

Explanation:

LCM of 2, 3, 7 is 42.

=> (700 – 300)/42 = 9 22/42 => 9 Numbers.

A) 138 to 164 | B) 125 to 147 |

C) 148 to 158 | D) 152 to 164 |

Explanation:

A+7B=112

it is clear that A = 14, then it becomes A>=B, but A is smallest angle) given

so, range of A is 0.0001 to 13.9999 ( I am taking upto 4 decimal places)

so, range of B becomes 14 to 16 ( after rounding off to 4 decimal places)

so, range of C becomes 152 to 164 ( after rounding off to 4 decimal places)

A) 788 | B) 786 |

C) 784 | D) 792 |

Explanation:

(kx22)/100 = 340 - 166.64 = 173.36

k = (173.36 x 100)/22

k = 788

A) 28 | B) 19 |

C) 37 | D) 46 |

Explanation:

Let the original number is 82 i.e 8 + 2 = 10 and when the digits are reversed i.e 28

The difference is 82 - 28 = 54 i.e, the number is decreased by 54.

A) 40 | B) 45 |

C) 32 | D) 30 |

Explanation:

Probability of left handed = 4/5

total student = 40

total left handed = 40 x(4/5)

= 32

A) 3 | B) 2 |

C) 1 | D) 0 |

Explanation:

Let assume the value of a,b &c that will satisfy the equation.

so a=1,b=1,c= -1 or a=1,b= -1 &c=1 or a= -1,b=1,c=1

here all three cases satisfy the equation

assume any one a=1,b=1,c= -1

Hence (a+b)(b+c)(c+a)=(1+1)(1-1)(-1+1) = 0