**Quantitative Aptitude Average Tutorial (Study Material)**

Contents

The term average is used frequently in everyday life to express an amount that is typical for a group of people or things. For example, you may read in a newspaper that on average people watch 3 hours of television per day. We understand from the use of the term average that not everybody watches 3 hours of television each day, but that some watch more and some less. However , we realize from the use of the term average that the figure of 3 hours per day is a good indicator of the amount of TV watched in general.

** Quantitative Aptitude Average Tutorial (Download PDF)**

**Introduction**

The idea of average is not new to us. We all are familiar with the following types of statements.

The average runs scored by Sachin Tendulkar in a series is 72.

The average marks secured by Kana is 78%.

If a man earns Rs 40. Rs 50. Rs 56. Rs 46 and Rs 48 on five consecutive days of a week, then he earns a total of Rs (40 + 50 + 56 + 46 + 48) = Rs 240.

To find his average earning per day, his total earning is divided by the number of days, i.e…

Average = 240/5 = Rs 48

Average earning does not mean that he earned Rs 48 everyday . But had he earned Rs 48 everyday, then his total earnings would have also been Rs 240 in 5 days.

Hence, to find the average of given quantities:

**Step 1** The given quantities arc added to get a Sum

**Step 2** The Sum is divided by the Number of items to get the Average.

Sum of all the items /Number of items = Average—- (1)

Note: The average is also called the Mean.

The quantities, whose average is to be determined, should he in the same unit.

Hence, Sum of all the items = Average x no. of items (2)

**Average of Different Groups**

Sometimes, the average of two different groups are known and the average of a third group (made by combining these two groups) is to be found out.

Let

(Group l) + | (Group 2) | makes (Combined Group (1+2)) | |

No. of items = | m | n | m + n |

Average = | a | h | A |

Sum of all items = | ma | nh | ma + nb |

Therefore, average of combined group = Sum of all Items/ No. of Items

A= ma+nb/m+n

This formula is also applicable for more than two groups forming the combined group.

**Addition or Removal of Items and Change in Average**

Since Average = Sum of Items / Number of Items

So, the original average may change (increase/decrease), if number of items change. The number of items may change in the following two cases,

**Case I**

**When one or more than one New items are added**

Let the average of N items = A

Now n New items are added and the average increases or decreases by x. then

Average of New items added = A ± (1+ N/n)

Use (-). when average decreases (+), when average increases.

**Case II**

**When one or more than one items are removed**

In this case, items arc removed, so on placing-N/n in place of +N/n in formula it becomes.

Average of items removed = A ± (1- N/n)

Use (-), when average decreases (+), when average increases

When only One item is removed, put n = 1, then

Value of the item removed = A ± (1 – N)x

**Replacement of some of the Items**

Sometimes, when a number of items of a group are removed and these arc replaced with equal number of different items, then the average of the group changes, (increases or decreases) by x.

Let there are N items in the group, then

Sum of New items added – Sum of removed items = ± Nx

Use (-). when average decreases (+), when average increases.

**Example**: When a man weighing 80 kg is replaced by anolher man in a group of five persons, the average weight decreases by 3 kg. What is the weight of new man?

Solution: Using formula (6).

Weight of new man – Weight of removed man = -Nx (-ve, average decrease)

Weight of new man – 80 = -5 x 3

Weight of new man = 80 – 15

= 65 kg.

**Some Problem-Specific Formulae**

Before t years, the average age of ‘Ar members of a family was *T” years. If the average remains same even after one more member joins the family, then present age of new member = T – Nt.

Example: Four years ago, the average age of six members of a family was 26 years. On the birth of a child in the family, the average remains the same. Find the present age of the child.

Solution: Present age of the child = 26-6 x 4 = 2 years.

Out of the given numbers, if the average of frst n numbers is x and that of last n numbers is y, then

First number – last number = n(x – y).

**Example**: The average temperature of June, July and August was 3I°C. The average temperature of July. August and September was 30 °C. If the temperature of June was 29 °C. find the temperature of September.

**Solution**: In the given problem, four months have been indicated, i.e.

June, July, August. September

(first) 2 3 4 (last)

Out of these, the average temperature of first three (n = 3) months = 31°C°= x

and the average temperature of last three (n = 3) months = 30 °C = y Then, by the formula

Temperature of first month – temperature of last month = n(x – y)

temperature of June – temperature of September = 3(31 – 30).

29 – temp, of September = 3

Temperature of September = 26 °C

** Quantitative Aptitude Average Tutorial (Download PDF)**

## Leave a Reply