# Average Questions

**FACTS AND FORMULAE FOR AVERAGE QUESTIONS**

**1**. Average =$\frac{Sumofobservations}{No.ofobservations}$

**2**. Suppose a man covers a certain distance **x** kmph and an equal distance at** y** kmph. Then, the average speed during the whole journey is$\frac{2xy}{x+y}$Kmph

A) 76 | B) 79 |

C) 85 | D) 87 |

Explanation:

Average = total runs / no.of innings = 32

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs

Total runs = new avg x new no. of innings = 36 x 11 = 396

Runs made in the 11th inning = 396 - 320 = 76

A) 70 kg | B) 75 kg |

C) 80 kg | D) 85 kg |

Explanation:

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

A) 4991 | B) 5467 |

C) 5987 | D) 6453 |

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs.[(6500 x 6) - 34009]

= Rs. (39000 - 34009)

= Rs. 4991.

A) 4 | B) 8 |

C) 12 | D) 16 |

Explanation:

Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.

Then [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.

or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) - 57 = 8

A) 0 | B) -1 |

C) 1 | D) none of these |

Explanation:

Average of 20 numbers = 0.

Sum of 20 numbers = (0 * 20) = 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

A) 21 | B) 22 |

C) 23 | D) 24 |

Explanation:

Let the numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6,

Then (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.

or 7x + 21 = 140 or 7x = 119 or x =17.

Latest number = x + 6 = 23.

A) 45 kg | B) 46 kg |

C) 47 kg | D) 48 kg |

Explanation:

Weight of the teacher = (35.4 x 25 - 35 x 24) kg = 45 kg.