# Formula - Average

**AVERAGE**

‘**Average**’ is a very simple but effective way of representing an entire group by a single value.

Average or Mean = \(\frac{Sum \; of \; given \; quantities}{Number \; of \; quantities}\)

To calculate the sum of quantities, they should be in the same unit.

**Example 1**

**1. The average of the first nine prime numbers is:**

(a) 9

(b) 11

(c) 11(1/9)

(d) 11(2/9)

**Sol.(c) Average**

**Example 2**

**2. In three numbers, the first is twice the second and thrice the third. If the average of these three numbers is 44,then the first number is :**

**Sol.** (a) Let the three numbers be x, y and z

Therefore, x = 2y = 3z,

y = x/2 and z = x/3.

Now, \(\frac{x + \frac{x}{2} + \frac{x}{3}}{3} = 44\)

or 11x/18 = 44 or x = 72.

**Example 3**

**3. The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?**

(a) 2

(b) 5

(c) 8

(d) Cannot be determined

**Sol.(c)** Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.

Then, \(\frac{x + \left ( x + 2 \right ) + \left ( x + 4 \right ) + \left ( x + 6 \right ) + \left ( x + 8 \right )}{5} = 11\)

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

So, required difference = (57 + 8) – 57 = 8.

**Average of a group consisting two different groups when their averages are known**

Let Group A contains m quantities and their average is a and Group B contains n quantities and their average is b, then **average of group C** containing a + b quantities = \(\frac{ma + mb}{m + n}\).

**Example 4**

**4. There are 30 student in a class. The average age of the first 10 students is 12.5 years. The average age of the next 20 students is 13.1 years. The average age of the whole class is :**

(a) 12.5 years

(b) 12.7 years

(c) 12.8 years

(d) 12.9 years

**Sol.(d)** Total age of 10 students = 12.5 × 10 = 125 years

Total age of 20 students = 13.1 × 20 = 262 years

Average age of 30 students = \(\frac{125 + 262}{30} = 12.9 \; years\)

**Example 5**

**5. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is**

(a) 1 : 2

(b) 2 : 3

(c) 3 : 4

(d) 3 : 5

**Sol.(b)** Let the number of boys in a class be x.

Let the number of girls in a class be y.

∴ Sum of the ages of the boys = 16.4 x

Sum of the ages of the girls = 15.4 y

∴ 15.8 (x + y) = 16.4 x + 15.4 y

=> 0.6 x = 0.4 y => x/y = 2/3

∴ Required ratio = 2 : 3

**Example 6**

**6. The average monthly expenditure of a family was 2200 during the first 3 months; 2250 during the next 4 months and 3120 during the last 5 months of a year.If the total saving during the year were 1260, then the average monthly income was**

(a) 2605

(b) 2805

(c) 2705

(d) 2905

**Sol.(c)** Total annual income

= 3 × 2200 + 4 × 2250 + 5 × 3120 + 1260

= 6600 + 9000 + 15600 + 1260 = 32460

∴ Average monthly income = 32460/12 = 2705

If, in a group, one or more new quantities are added or excluded, then the new quantity or sum of added or excluded quantities = [Change in no. of quantities × original average] ± [change in average × final no. of quantities]
Take +ve sign if quantities added and

take –ve sign if quantities removed.

**Example 7**

**7. The average weight of 29 students in a class is 48 kg. If the weight of the teacher is included, the average weight rises by 500 g. Find the weight of the teacher.**

**Sol.** Here, weight of the teacher is added and final average of the group increases.

∴ Change in average is (+)ve, using the formula

Sum of the quantities

= [Change in no. of quantities × original average] ± [change in average × final no. of quantities]

=> weight of teacher = (1 × 48) + (0.5 × 30) = 63 kg.

∴ weight of teacher is 63 kg.

**Example 8**

**8. The average age of 40 students in a class is 15 years. When 10 new students are admitted, the average is increased by 0.2 year. Find the average age of the new students.**

**Sol.** Here, 10 new students are admitted.

∴ change in average is +ve. Using the formula

Sum of the quantities added

= [Change in no. of quantities × original average] ± [change in average × final no. of quantities]

=> Sum of the weight of 10 new students admitted

= (10 × 15) + (0.2 × 50) = 160 kg.

∴ Average age of 10 new students = \(\frac{s_{a}}{n_{a}} = \frac{160}{10} = 16\)

∴ Average age of 10 new students is 16 years. If a certain distance is covered at a kmph and an equal distance at b kmph, then the average speed during whole journey = 2ab/a + b kmph.

**Example 9**

**9. A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is :**

(a) 35

(b) 37

(c) 37.5

(d) 40

**Sol.(c)** Average speed

= 37.5 km/hr.

**Example 10**

**10. The mean of the marks secured by 25 students of section A of class X is 47, that of 35 students of section B is 51 and that of 30 students of section C is 53. Find the combined mean of the marks of students of three sections of class X.**

**Sol.** Mean of the marks of 25 students of XA = 47

∴ Sum of the marks of 25 students = 25 × 47 = 1175 ........(i)

Mean of the marks of 35 students of XB = 51

∴ Sum of the marks of 35 students = 35 × 51 = 1785 .......(ii)

Mean of the marks of 30 students of XC = 53

∴ Sum of the marks of 30 students = 30 × 53 = 1590 .......(iii)

Adding (i), (ii) and (iii)

Sum of the marks of (25 + 35 + 30) i.e., 90 students

= 1175 + 1785 + 1590 = 4550

Thus the combined mean of the marks of students of three sections = 4550/90 = 50.56

**Example 11**

**11. Find the A.M. of the sequence 1, 2, 3,....., 100.**

**Sol.** We have sum of first n natural numbers = n/2(n + 1)

here n = 100

=> Sum = 100/2 × 101 = 101 × 50

=> AM = Sum/100 = (101 × 50)/100 = 50.5

**Example 12**

**12. A sequence of seven consecutive integers is given. The average of the first five given integers is n. Find the average of all the seven integers.**

**Sol.** Let the seven consecutive integers be

x, x + 1, x + 2, ......., x + 6

The sum of the first five is

x + x + 1 + x + 2 + x + 3 + x + 4 = 5x + 10

The average of these five is (5x + 10)/5 = x + 2 = n

The average of the seven will be

\(\frac{5x + 10 + x + 5 + x + 6}{7} = \frac{7x + 21}{7} = x + 3\)As x + 2 = n, so x + 3 = x + 2 + 1 = n + 1

**Example 13**

**13. The average of 11 results is 50. If the average of first six result is 49 and that of last six results is 52, find the sixth result.**

**Sol.** Average of 11 results

1 2 3 4 5 6 7 8 9 10 11

Average of last 6 results = 52

Average of first 6 results = 49

It is quite obvious that the sixth result is included twice,once in the first six results and second in the last six results.

∴ Value of the sixth result = (Sum of first six results) + (Sum of last six results) – Sum of 11 results

= 6 × 49 + 6 × 11 × 50 = 56

**Example 14**

**14.Typist A can type a sheet in 5 minutes, typist B in 6 minutes and typist C in 8 minutes. The average number of sheets typed per hour per typist is ......... sheets.**

**Sol.** A types 12 sheets in 1 hour

B types 10 sheets in 1 hour

C types 7.5 sheets in 1 hour

Average number of sheets types per hour per typist