# Numbers Questions

A) 4 | B) 6 |

C) 9 | D) 0 |

Explanation:

Last digit for the power of 6 is 6 (always)

Power cycle of 7 is 7, 9, 3, 1.

Now 467/4 gives a remainder of 3

Then the last digit is ${7}^{3}$ = 3

Last digit is 6 + 3 = 9.

A) 8475 | B) 8500 |

C) 8550 | D) 8525 |

Explanation:

On dividing we get

75)8485(113

75

------

98

75

------

235

225

------

10

------

Required number = (8485-10)=8475.

A) 8 | B) 9 |

C) 10 | D) 11 |

Explanation:

The maximum number of bows will be 4 yards (= 4 x 36 inches) divided by 15 inches.

This gives 9.6. But as a fraction of a bow is no use, we can only make 9 bows.

A) 6 | B) 7 |

C) 8 | D) 9 |

Explanation:

The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3

If the sum is 24, we need to solve the equation 3n + 3 = 24;

=> 3n = 21;

=> n = 7

The greatest of the three numbers is therefore 7 + 2 = 9

A) 4 | B) 8 |

C) 12 | D) None |

Explanation:

we can do this by trial and error method.

Putting x = 2,we get 2²(2² - 1) = 12.

checking with other integers, the above equation always gives a value which is a multiple of 12,

So, x²(x² - 1) is always divisible by 12.

A) 3 | B) 2 |

C) 1 | D) 0 |

Explanation:

Let n=4*q + 3. Then, 2*n = 8*q + 6 = 4(2*q + 1) + 2.

Thus when 2*n is divided by 4, the reminder is 2.

A) 6 only | B) 6 and 12 both |

C) 12 only | D) by 18 only |

Explanation:

$\left(6{n}^{2}+6n\right)=6n\left(n+1\right)$, which is always divisible by 6 and 12 both, since n(n+1) is always even.

A) 1 | B) 0 |

C) -1 | D) Infinity |

Explanation:

The mutiplicative inverse of a number is nothing but a reciprocal of a number.

Now, the product of a number and its multiplicative inverse is always equal to **1**.

**For example :**

Let the number be 15

Multiplicative inverse of 15 = 1/15

The product of a number and its multiplicative inverse is = **15 x 1/15 = 1.**