Concept: Average

CONTENTS

INTRODUCTION

The concept of averages is important since questions based on averages regularly appear in Mathematics and Data-interpretation sections of various MBA entrance examinations. Average is a very simple but an effective way of representing an entire group by a single value.

The sum of all observations divided by the number of observations is referred to as average of those observations.

Average = $\frac{Sum of all quantities}{Number of quantities}$

It is also referred to as the Arithmetic Mean.

AVERAGE

Let the n quantities be x₁, x₂, x₃…x_n, then the formula for the average a is $a = \frac{x_{1} + x_{2} + x_{3} + \dots + x_{n}}{n} = \frac{\sum x}{n}$

If out of n quantities, x₁ occurs f₁ times, x₂ occurs f₂ times, …, x_k occurs f_k times, then their average a is given by $a = \frac{{f_{1} x}_{1} + {f_{2} x}_{2} + \dots + f_{k} x_{k}}{f_{1} + f_{2} + \dots + f_{n}} = \frac{\sum fx}{\sum f}$

For example, if the wages of five workers are 80, 120, 70, 100, 110, then the average wage per worker is

a = $\frac{Total wages}{Number of workers}$ = $\frac{80 + 70 + 120 + 100 + 110}{5}$ = $\frac{480}{5}$ = Rs.96

If a is the average of n quantities, then the sum of all the quantities S is given by S = a × n

If the average and Sum of certain quantities is a and S respectively, then the number of quantities is given by n = $\frac{S}{a}$

Remember:

If the value of each item is increased by the same value p, then the average of the group of items will also get increased by p.
If the value of each item is decreased by the same value p, then the average of the group of items will also get decreased by p.
If the value of each item is multiplied by the same value p, then the average of the group of items will also get multiplied by p.
If the value of each item is divided by the same value p (p ≠ 0), then the average of group of items will also be divided by p.
The average of a group of item will always lie between the smallest value in the group and largest value in the group.
If all the observations form an Arithmetic Progression, then the middle term (Median) will be the average of the set of observations.

Example : Find the average of the following numbers: 351, 244, 479, 588, 105, 78, 483, 536.
(a) 479
(b) 356
(c) 424
(d) 358
(e) None of these

Solution :
Average = $\frac{351 + 244 + 479 + 588 + 105 + 78 + 483 + 536}{8}$ = 2864/8 = 358

Example : There are six numbers 30, 72, 53, 68, X and 87, out of which X is unknown. The average of the numbers is 60. What is the value of X?
(a) 40
(b) 60
(c) 70
(d) 30
(e) None of these

Solution :
Sum of all 6 numbers = 60 × 6 = 360

So, 30 + 72 + 53 + 68 + X + 87 = 360

⇒ 310 + X = 360

⇒ x = 50

Example : The average age of a woman and her daughter is 16 years. The ratio of their ages is 7 : 1 respectively. What is the woman’s age?
(a) 4 years
(b) 28 years
(c) 32 years
(d) 6 years
(e) None of these

Solution :
Let the woman’s age be 7x years and her daughter’s age be x years.

According to the question, 7x + x = 2 × 16

⇒ 8x = 32 ⇒ x = 4

⇒ Woman’s age = 7x = 7 × 4 = 28 years.

Example : The average of 4 consecutive even numbers is 103. What is the product of the smallest and the largest number?
(a) 10400
(b) 10504
(c) 10605
(d) 10600
(e) None of these

Solution :
Sum of the four consecutive even numbers = 4 × 103 = 412.

⇒ x + (x + 2) + (x + 4) + (x + 6) = 412

4x + 12 = 412 ⇒ 4x = 400 ⇒ x = 100

Thus, product of the smallest and the largest number = 100 × (100 + 6) = 10600.

Example : The average of 6 consecutive odd numbers is 24. What is the largest number?
(a) 25
(b) 27
(c) 29
(d) 31
(e) None of these

Solution :
Let the first odd number be x.

As per the question, x + (x + 2) + (x + 4) + (x + 6) + (x + 8) + (x + 10) = 24 × 6 = 144

⇒ 6x + 30 = 144 ⇒ 6x = 114 ⇒ x = 19

Thus, the largest number = 19 + 10 = 29

Example : If 36a + 36b = 576, then what is the average of a and b?
(a) 16
(b) 8
(c) 12
(d) 6
(e) 10

Solution :
36a + 36b = 576

⇒ 36(a + b) = 576 ⇒ (a + b) = 16

Thus, average of a and b = (a + b)/2 = 8

Example : The average monthly expenditure of a family was Rs. 2,200 during first 3 months, Rs. 2,550 during next 4 months and Rs. 3,120 during last 5 months of the year. If the total saving during the year was Rs. 1,260, find average monthly income.

Solution :
Total yearly income

= yearly expenditure + yearly saving
= [2200 × 3 + 2550 × 4 + 3120 × 5] + [1260]
= Rs. 33,660

∴ Average monthly income = 33660/12 = Rs.2805

Example : The average age of 5 children is 8 years. If the age of the father be included, the average is increased by 7 years, find the age of the father.

Solution :
The total age of 5 children = 5 × 8 = 40 years

Total ages of (5 children + the father) = 6 × (8 + 7) = 90 years

∴ Age of the father = 90 years – 40 years = 50 years

Example : The average weight of 6 men is 65 kg. Two of them weigh 70 and 48 kg respectively, what is the average weight of the remaining four?

Solution :
Total weight of six = 6 × 65 = 390 kg

The weight of the remaining four = 390 – 70 – 48 = 272

∴ Average weight of the remaining four = 272/4 = 68 kgs

Example : The average temperature for Sunday, Monday & Tuesday is 37.3° C, whereas the average temperature for Monday, Tuesday & Wednesday is 38.7°C. If Sunday’s temperature recorded is 40°C, find Wednesday’s temperature.

Solution :
Total temperature of Sunday, Monday & Tuesday = 37.3 × 3 = 111.9° C

∴ Monday + Tuesday = 111.9 – 40 = 71.9° C

Total temperature of Monday, Tuesday & Wednesday = 38.7 × 3 = 116.1° C

Wednesday’s temperature = 116.1° C – 71.9° C = 44.2° C

Example : There are three numbers. The second is half of the first and the third is half of the second. If the average of the three numbers is 28, find the first number.

Solution :
Sum of the three numbers = 3 × 28 = 84

Let, the first number be 4x, so the second number is 2x and the third number is x.

∴ 4x + 2x + x = 84 ⇒ x = 12

∴ First number is 4x = 48

Example : 30 horses were purchased for Rs. 12000. The average cost of 12 horses out of them is Rs. 250. Find the average cost of the remaining horses.

Solution :
Cost of 30 horses = Rs. 12000

Cost of 12 horses out of them = 12 × 250 = 3000

∴ Average cost of remaining horses =(12000-3000)/18 = Rs. 500

ASSUMED AVERAGE METHOD

In some cases, we can avoid tedious calculations of adding all the numbers and dividing by the number of observations. The calculation of averages can be simplified using the assumed average method. Take some arbitrary average (A), which lies between the smallest and the largest value. Then, calculate the deviations (differences) of the given values from A and find the average of all these deviations. Add this to A to find the average.

∴ Required average = Assumed average + Average of deviations

Example : Find the average of 101, 98, 105, 96, 103.

Solution :
Here, let’s take the assumed average as 100.

Now, the required average = assumed average + average of deviations

∴ Average = 100 + $\frac{(101 - 100) + (98 - 100) + (105 - 100) + (96 - 100) + (103 - 100)}{5}$ = 100 + $\frac{1 - 2 + 5 - 4 + 3}{5}$ = 100 + $\frac{3}{5}$ = 100.6

Remember:

If the average of ‘n’ numbers is ‘a‘ and

a new number equal to ‘a’ is added to the group, the average of the new group will remain same.
a new number greater than ‘a’ is added to the group, the average of the new group will increase.
a new number less than ‘a’ is added to the group, the average of the new group will decrease.

a new number equal to ‘a’ is removed from the group, the average of the new group will remain same.
a number greater than ‘a’ is removed from the group, the average of the new group will decrease.
a number less than ‘a’ is removed from the group, the average of the new group will increase.

a existing number is replaced with a greater number, the average of the group will increase.
a existing number is replaced with a smaller number, the average of the group will decrease.

Let’s discuss these in detail.

A PERSON JOINING THE GROUP

When a number is added/removed from a group of numbers, we can calculate how much more/less, the number being added/removed, is from the earlier average by multiplying the remaining no. of numbers with the change in average.

Example : The average weight of a class of 24 students is 36 kg. When the weight of the teacher is also included, the average weight increases by 1kg. What is the weight of the teacher?
(a) 60 kg
(b) 61 kg
(c) 37 kg
(d) 39 kg

Solution :
The average weight of a class of 24 students = 36 kgs.

Therefore, the total weight of the class = 24 × 36 = 864 kgs

When the weight of the teacher is included, the average weight increases by 1 kg i.e. the new average weight = 37 kgs.

Therefore, the total weight of the 24 students plus the teacher = 25 × 37 = 925 kgs

∴ 864 + Teacher’s weight = 925

∴ Weight of the teacher = 925 - 864 = 61 kgs.

Now, the same question can also be solved is a direct manner

Example : The average weight of a class of 24 students is 36 kg. When the weight of the teacher is also included, the average weight increases by 1 kg. What is the weight of the teacher?
(a) 60 kg
(b) 61 kg
(c) 37 kg
(d) 39 kg

Solution :
The average weight of a class of 24 students = 36 kgs.

Had Teacher’s weight been 36 kgs, the average of the new group would not change. But since, the average of new group increases, it means Teacher weights more than 36 kgs.

Now, since the average of now 25 people increases by 1 kg, that means Teacher brought in 25 × 1 = 25 kg weight more than the earlier average of the group.

i.e. Teacher’s weight – 36 = 25

⇒ Weight of the Teacher = 36 + 25 = 61 kgs.

Example : The average weight of a class of 24 students is 36 kg. When a new student joins the class, the average weight decreases by 0.2 kg. What is the weight of the new student?
(a) 30 kg
(b) 31 kg
(c) 37 kg
(d) 29 kg

Solution :
The average weight of a class of 24 students = 36 kgs.

Had new student’s weight been 36 kgs, the average of the new group would not change. But since, the average of new group decreases, it means new student weights less than 36 kgs.

Now, since the average of now 25 people decreases by 1 kg, that means new student brought in 25 × 0.2 = 5 kg weight less than the earlier average of the group.

i.e. 36 – New student’s weight = 5

⇒ Weight of the new student = 36 - 5 = 31 kgs.

Example : The average weight of class of 20 students is 30 kgs. If a new student whose weight is 51 kgs is added to the group, what will be the new average of the group.

Solution :
Had the new student weighed 30 kgs, the new average would not change.

But, the new student brought in 51 – 30 = 21 kgs weight more than the average.

To calculate the average, this extra weight will be divided equally among the 21 students.

∴ Each student will gain 21/21 = 1 kg.

Hence, the new average of the class will be 30 + 1 = 31 kgs

A PERSON LEAVING THE GROUP

Example : The average weight of a class of 24 students is 36 kg. When one of the students leaves the group, the average weight decreases by 1 kg. What is the weight of the student leaving?
(a) 60 kg
(b) 61 kg
(c) 57 kg
(d) 59 kg

Solution :
The average weight of a class of 24 students = 36 kgs.

Therefore, the total weight of the class = 24 × 36 = 864 kgs

When the student leaves the group, the average weight decreases by 1 kg i.e. the new average weight = 35 kgs.

Therefore, the total weight of the remaining 23 students = 23 × 35 = 805 kgs

∴ 805 + Leaving student’s weight = 864

∴ Weight of the student leaving = 964 - 805 = 59 kgs.

Alternately,
The average weight of a class of 24 students = 36 kgs.

If the weight of the student leaving was 36 kgs, the average of the new group would not change. But since, the average of new group decreases, it means the student leaving weights more than 36 kgs.

Now, since the average of now 23 people decreases by 1 kg, that means the student leaving took 23 × 1 = 23 kgs weight more than the earlier average of the group.

i.e. Weight of the student leaving – 36 = 23

⇒ Weight of the student leaving = 36 + 23 = 59 kgs.

Example : The average weight of a class of 24 students is 36 kg. When one of the students leaves the group, the average weight increases by 0.2 kg. What is the weight of the student leaving?
(a) 30.5 kg
(b) 31.4 kg
(c) 32 kg
(d) 31.6 kg

Solution :
The average weight of a class of 24 students = 36 kgs.

If the weight of the student leaving was 36 kgs, the average of the new group would not change. But since, the average of new group increases, it means the student leaving weights less than 36 kgs.

Now, since the average of now 23 people increases by 0.2 kg, that means the student leaving took 23 × 0.2 = 4.6 kgs weight less than the earlier average of the group.

i.e. 36 - Weight of the student leaving = 4.6

⇒ Weight of the student leaving = 36 – 4.6 = 31.4 kgs.

Example : The average weight of class of 21 students is 50 kgs. If a student whose weight is 30 kgs is removed to the group, what will be the new average of the group.

Solution :
Had the student leaving weighed 50 kgs, the new average would not change.

But, the student leaving took 50 – 30 = 21 kgs weight less than the average.

To calculate the average, this extra weight will be divided equally among the remaining 20 students.

∴ Each student will gain 20/20 = 1 kg.

Hence, the new average of the class will be 50 + 1 = 51 kgs.

Example : When a student weighing 45 kg left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the original group of students?
(a) 57
(b) 56.8
(c) 58.2
(d) 52.2

Solution :
Let the average weight of the original group be a kg.

∴ new average weight of the class will be = (a + 0.2) kg

Now, the average weight of 59 students increases by 0.2 kg. It means the student leaving the group took 59 × 0.2 = 11.8 kgs less than the earlier average.

⇒ Weight of the student leaving = (a - 11.8) kgs

∴ a – 11.8 = 45

⇒ a = 56.8 kgs

A PERSON BEING REPLACED IN THE GROUP

Example : The average weight of a class of 25 students is 36 kg. When one of the students whose weight is 30 kgs is replaced by another in the group, the average weight of the group increases by 0.2 kg. What is the weight of the student joining?
(a) 35 kg
(b) 31 kg
(c) 33.2 kg
(d) 35 kg

Solution :
The average weight of a class of 25 students = 36 kgs.

Therefore, the total weight of the class = 25 × 36 = 900 kgs

When the student weighing 30 kgs leaves the group and another students weighing x kgs joins the group, the average weight increases by 0.2 kg i.e. the new average weight = 36.2 kgs.

Therefore, the total weight of the group = 25 × 36.2 = 905 kgs

∴ 900 – 30 + x = 905

∴ Weight of the student joining = 905 – 900 + 30 = 35 kgs.

Alternately,
The average weight of a class of 25 students = 36 kgs.

If the weight of the student leaving and joining were same, the average of the new group would not change. But since, the average of new group increases, it means the student leaving weighs less than the student coming in.

Now, since the average of total 25 people increases by 0.2 kg, that means the student leaving weighs 25 × 0.2 = 5 kgs less than the students coming in.

i.e. Weight of the student joining - Weight of the student leaving = 5

⇒ Weight of the student joining = 30 + 5 = 35 kgs.

Example : The average age of 10 men in a play is increased by 1.5 years when two men aged 31 years and 49 years are replaced by two females. Find the average age of females.

Solution :
Overall increase in the total age = 10 × 1.5 years = 15 years

∴ Total age of two females = 31 + 49 + 15 = 95 years.

Average age of two females = 95/2 = 47.5 years

WEIGHTED AVERAGE

Let W₁, W₂, … ,Wn be the weights assigned to the quantities X₁, X₂, … , X_n respectively, then their weighted average (X_W ) is defined as :

$\bar{X_{W}} = \frac{\sum WX}{\sum W} = \frac{W_{1} X_{1} + W_{2} X_{2} + \dots + W_{n} x_{n}}{W_{1} + W_{2} + \dots + W_{n}}$

Weighted averages are used in connection with average speed, average price, average consumption etc.

Combined Average

If the number of items and their averages are given for two or more groups, then the formula for their combined average is

Combined average $\bar{X} = \frac{n_{1} \bar{X_{1}} + n_{2} \bar{X_{2}} + \dots \dots n_{k} \bar{X_{k}}}{n_{1} + n_{2} + \dots \dots n_{k}}$

⇒ $\bar{X} = \frac{\sum n_{k} \bar{X_{k}}}{\sum n_{k}}$

Example : The average age of the students in section A of 40 students is 40 years and the average of students in section B of 30 students is 12 years. Find the average age of students in both sections taken together.

Solution :
Average = $\frac{40 \times 40 + 30 \times 12}{40 + 30}$ = $\frac{1960}{70}$ = 28 years
∴ Average age of all the students is 28 years.

Example : A man bought 13 shirts of Rs. 50 each, 15 pairs of trousers of Rs. 60 each and 12 pairs of shoes at Rs. 65 each. Find the average value of each article.

Solution :
Average = $\frac{13 \times 50 + 15 \times 60 + 12 \times 65}{13 + 15 + 12}$ = Rs. 58.25

Example : The average of marks obtained by 60 students in a certain examination is 35. If the average marks of students who passed are 39 and that of the students who failed are 15, how many students passed in the examination?

Solution :
Let the number of passing students be x, then 39x + 15 (60 - x) = 60 × 35
⇒ 24x = 1200 ⇒ x = 50
∴ Number of passing students = 50

Example : The average monthly salary of 8 workers and 1 supervisor in a factory was Rs. 195. The supervisor,who used to get a monthly salary of Rs. 300, retired and a new supervisor was appointed in his place as a result of which the average monthly salary of the factory workers reduced to Rs. 185. Calculate the monthly salary of new supervisor.

Solution :
Average monthly salary of 8 workers and 1 supervisor = Rs. 195
Total salary of these 9 persons = 9 × 195 = Rs. 1,755
Total salary after the retirement of the supervisor = 1,755 – 300 = Rs. 1,455
Let the salary of new supervisor be Rs. x
∴ Total salary after the entry of new supervisor = Rs. (1,455 + x)
Average salary after the new supervisor joined = Rs. 185and total salary = 9 × 185 = Rs. 1,665
Hence, 1,455 + x = 1,665
⇒ x = 1,665 – 1,455 = Rs. 210

Example : 30 persons live in a hotel. On an increase of 10 persons, expenses of food increased by Rs. 60 but average expenses are reduced by Rs. 1. Find out what was the earlier total monthly expenditure of hotel for 30 persons?

Solution :
Let the monthly average expenditure of 30 persons = $\bar{x}$
∴ Total monthly expenditure of 30 persons = 30 $\bar{x}$
Average monthly expenditure of 40 (= 30 + 10) persons = $\bar{x}$ – 1
∴ Total monthly expenditure of these 40 persons = 40 ( $\bar{x}$ – 1)
According to the question, 40( $\bar{x}$ - 1) - 30 $\bar{x}$ = 60
⇒ 40 $\bar{x}$ - 40 - 30 $\bar{x}$ = 60
⇒ 10 $\bar{x}$ = 100

∴ x ̅ = 10
Hence, Earlier total expenditure = 30 × 10 = Rs. 300

Example : Visitors to a show were charged Rs. 15 each on the first day, Rs. 7.5 each on the second, Rs. 2.5 each on the third and the total attendance on three days were in the ratio 2 : 5 : 13 respectively. Find the average charge per person for the whole show.

Solution :
Suppose the attendance on the first, second and third day were 2x, 5x and 13x respectively
∴ Total attendance = 2x + 5x + 13x = 20x
Money collected on the first day = 15 × 2x = 30x
Money collected on the second day = 7.5 × 5x = 37.5x
Money collected on the third day = 2.5 × 13x = 32.5 x
∴ Total money collected on all the three days = Rs. (30x + 37.50x + 32.50x) = Rs. 100x
∴ The average charge per person for the whole show = $\frac{100 x}{20 x}$ = Rs. 5

Example : The average weight of boys in a class is 43 kg. Later, four boys, whose weights are respectively 42 kg, 36.5 kg, 39 kg and 42.5 kg, join them. The average now becomes 42.5 kg. Find the initial number of boys in the class.

Solution :
Suppose initial number of boys in the class be x.
As their average weight = 43 kg
Their total weight = 43x kg
∴ Total weight of (x + 4) boys
= (43 x + 42 + 36.5 + 39 + 42.5) kg
= (43x + 160) kg
∴ Average weight of (x + 4) boys = $\frac{43 x + 160}{x + 4}$ kg
According to the question, $\frac{43 x + 160}{x + 4}$ = 42.5
⇒ 43x + 160 = 42.5x + 170
⇒ 43x – 42.5 x = 170 – 160
⇒ 0.5x = 10 ⇒ x = 10/0.5 = 20
Hence, initial number of boys in class = 20

Example : The average of 8 results is 20. The average of first two results is 15 1/2, The average of next three results is 21 1/3. The sixth result is 4 less than seventh and 7 less than eighth. Find the result.

Solution :
As the average of results is 20, the total of 8 results = 20 × 8 = 160.
Also, average of first two results =15 1/2
⇒ Total of first two result = $15 \frac{1}{2}$ × 2 = 31
And, average of next three results = $21 \frac{1}{3}$
⇒ Total of next three results = $21 \frac{1}{3}$ × 3 = 64
∴ Total of the remaining three results = 160 – (31 + 64) = 160 – 95 = 65

Suppose 6th result = x, then 7th result = x + 4 and 8th result = x + 7
∴ x + (x + 4) + (x + 7) = 65
⇒ 3x + 11 = 65
⇒ 3x = 54 ⇒ x = 18
∴ Eighth result = 18 + 7 = 25

Example : In each row of a hand written book of 180 pages, there are 12 words on an average and each page consists of 16 rows on an average. After printing, in each row of the book there are 18 words on average and each page contains 20 rows on average. Find the number of pages of the printed book.

Solution :
No. of words in hand-written book = 180 × 12 × 16
Let the number of printed pages be x. Then, number of words in printed book = 18 × 20 × x
∴ 18 × 20 × x = 180 × 12 × 16
⇒ x = $\frac{180 \times 12 \times 16}{18 \times 20}$ = 96
∴ Required number of pages = 96

Example : There were 35 students in a hostel. If the number of students increased by 7, the expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs. 1. Find the initial expenditure of the mess.

Solution :
Let the initial expenditure be x. Then the initial average expenditure =x/35
New average expenditure = $\frac{x + 42}{42}$
⇒ $\frac{x}{35} - \frac{x + 42}{42}$ = 1 ⇒ x = 420
∴ Initial expenditure = Rs. 420

Example : Three years ago, the average of a family of 5 members was 17 years. A baby having been born, the average of the family is the same today, find the age of the baby.

Solution :
Present age of 5 members of the family + new born baby = 17 × 6 = 102
Present age of 5 members of the family = 17 × 5 + 3 × 5 = 100 years
Age of the baby = (102 - 100) years i.e., 2 years

Example : 5 years ago, the average age of Ram and Shyam was 20 years. At present, the average age of Ram, Shyam and Mohan is 30 years. What will be Mohan’s age 10 years hence?

Solution :
Total age of Ram & Shyam 5 years ago = (2×20) = 40
∴ Total age of Ram and Shyam now = (40 + 5 + 5) = 50
Total age of Ram, Shyam and Mohan now = (3×30) = 90
Mohan’ age now = 90 – 50 = 40
Mohan’s age 10 years hence = 40 + 10 = 50 years

Example : A batsman has a certain average runs for 16 innings. In the 17th inning, he made a score of 87 runs thereby increasing his average by 3. What is his average after 17th inning?

Solution :
Let the average for 16 innings be x runs
Total runs in 16 innings = 16x
Total runs in 17 innings = 16x + 87
Average of 17 innings = $\frac{16 x + 87}{17}$
∴ $\frac{16 x + 87}{17}$ = x + 3
⇒ x = 36
Thus, average of 17 innings = 36 + 3 = 39

Example : The average age of the husband and wife who got married 7 years ago was 25 years then.The average age of the family, including a child born afterwards, is 22 years now. How old is the child now?

Solution :
Total age of Husband and wife, 7 years ago = (2 × 25) = 50
Total age of husband & wife now = (50 + 7 + 7) = 64
Total age of husband, wife & child now = (3 × 22) = 66
Age of child now = 66 – 64 = 2 years

Example : Out of 24 students of a class, 6 are 1 m 15 cm in height; 8 are 1 m 5 cm and the rest are 1 m 11 cm. What is the average height of all the students?

Solution :
Sum of the heights of all the 24 students = (6 × 115 + 8 × 105 + 10 × 111) cm = 2640 cm.
∴ Average height =2640/24=110 cm = 1 m 10 cm.

Example : The average age of a family of 6 members is 22 years. If the age of the youngest member be 7 years, then what was the average age of the family just before the birth of the youngest member?

Solution :
Total age of family = 6 × 22 = 132 years
7 years ago, total sum of ages = 132 – (6 × 7) = 90 years
But at that time there are 5 members in the family
∴ Average at that time =90/5=18 years

Example : 10 years ago the average age of a family of 4 members was 24 years. 2 children having been born, the average age of the family is same today. Find the present age of the two children, assuming that the children’s ages differ by 2 years.

Solution :
10 years ago, the total age of 4 members = 4 × 24 = 96 years
Total age of these 4 members at present = 96 + 10 × 4 = 136 years
∵ The average age of family is same after the birth of 2 children.
∴ Total age of 6 (= 4 + 2) members
= 24 × 6 = 144 years
∴ Total age of the 2 children = 144 – 136 = 8 years
Let the age of children be ‘x’ years and ‘x + 2’ years
Hence, x + (x + 2) = 8
⇒ 2x = 8 – 2 = 6 ⇒ x = 6/2 = 3 years
Hence, ages of the children would be 3 years and 5 years.

Concept: Average

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