1) Basic Problems

2) Weighted average problems

3) Average Speed problems  

Concept 1: 

Basic Models:

In this, the candidate will learn about the all basic problems in Averages. Entire averages topic is depends on one standard formulae i.e; 

Average = Sum of Obs. / No. of Obs.

Along with the above formulae, the candidate must learn some more additional formulae in order to solve the problems in Averages.

Important Formulae:

1. The average of first ‘n’ consecutive odd numbers is ‘n’.

2. The average of first ‘n’ consecutive even numbers is ‘n+1’.

3. Average of odd numbers from 1 to n is 

Average = (Last odd number + 1) / 2

4. Average of even numbers from 1 to n is 

Average = (Last even number + 2) / 2

5. If the difference between any two consecutive numbers is same, then

Average = (First number + Last number) / 2

6. The average of ‘n’ numbers is A.

a) if each number is multiplied by ‘k’, then new average = A × k

b) if each number is added by ‘k’, then new average = A + k

c) if each number is subtracted by ‘k’, then new average = A – k

d) if each number is divided by ‘k’, then new average = A ÷ k

Eg: Find the average of 35, 43, 45, 48, 56, 67, 70.

Sol: Average =  Sum of Obs. / No. of Obs.

Average = (35 + 43 + 45 + 48 + 56 + 67 + 70) / 7 = 364 / 7 = 52.

Concept 2: 

Weighted Average Problems:

Eg: There are 40 boys and 50 girls in a class. If the average weight of boys is 54 kg and the average weight of girls is 45 kg. What is the average weight of entire class?

Sol: Average weight of entire class = (Sum of boys + Sum of girls) / Total Strength 

Sum of boys = Average of boys × No. of boys = 54 × 40 = 2160 kg

Sum of girls = Average of girls × No. of girls = 45 × 50 = 2250 kg           Average weight of entire class = (2160 + 2250) / (40 + 50) = 49 kg.

Concept 3:

Average Speed:

In general, average speed depends on two conditions.

1) Distances are equal 2) Distances are not equal

1) Distances are equal: 

If there are two equal distances, then the 

Average Speed = 2S₁S₂ / (S₁ + S₂)

If there are three equal distances, then the

Average Speed = 3S₁S₂S₃ / (S₁S₂ + S₂S₃ + S₃S₁)

2) Distances are not equal:

If distances are not equal, then the

Average Speed = Total Distance / Total Time

Eg: A boy is going to school from his house at a speed of 30 kmph and return home at a speed of 45 kmph. Find the average speed.

Sol: In the given problem, distances are equal in both the cases.

Average speed = 2S₁S₂ / (S₁ + S₂)

             ⸫ Average speed = 2 × 30 × 45 / (30 + 45) = 36 kmph.

Eg: A man covers three equal distances at a speeds of 15 kmph, 20 kmph and 30 kmph respectively, then find the average speed during the entire journey.

Sol: According to question, man covers 3 equal distances. 

Then, Average speed = 3S₁S₂S₃ / (S₁S₂ + S₂S₃ + S₃S₁) 

⸫ Average speed = 3 × 15 × 20 × 30 / (15 × 20 + 20 × 30 + 30 × 15) = 20 kmph.

Eg: An aeroplane travels distances 1200 km, 1500 km and 2000 km with a speeds of 600 kmph, 500 kmph and 400 kmph respectively. What is the average speed of entire journey?

What is the average of all numbers 78, 64, 92, 69, 88, 97, 114?

a) 81                             b) 85                            c) 86                             d) 89   

Solution:

Average = (Sum of Obs./No. of Obs.) = (78+64+92+69+88+97+114)/7 = 86

Alternate method:

By observing all the given numbers, take one approximate average round figure number which is in between lowest and highest numbers.

Consider 80 is the approximate average round figure number. Now, compare every number with approximate average. If the number is less take –ve sign, if the number is more take +ve sign.

⸫ Average = 80 + (-2 -16 + 12 -11 + 8 +17 + 34)/7 = 80 + (42/7) = 86.

Answer: c

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