Detailed Notes: Averages | CSAT Preparation - UPSC PDF Download

Introduction

This chapter forms the backbone concept of most questions in the Quantitative Aptitude & Data Interpretation sections. This is a crucial chapter, & quick-solving methods in this concept will help you save time - which is an essential factor for your success.

Definition

Simple Average (or Mean) is defined as the ratio of the sum of the quantities to the number of quantities.

Average Formula

Example (illustration): Take the first five natural numbers 1, 2, 3, 4 & 5.

Now add two more 3's to these five numbers.

Here are a couple more worked examples to clarify the idea of simple average.

Example 1: If a person with age 45 joins a group of 5 persons with an average age of 39 years. What will be the new average age of the group?

Sol:

Total age of the original 5 persons = 5 × 39.

Total age after the new person joins = 5 × 39 + 45.

Number of persons now = 6.

New average = (5 × 39 + 45) ÷ 6.

New average = (195 + 45) ÷ 6.

New average = 240 ÷ 6.

New average = 40.

Alternate quick method:

Difference of new person's age from old average = 45 − 39 = 6.

Increase in average = 6 ÷ 6 = 1.

New average = 39 + 1 = 40.

Example 2: Two students with marks 50 and 54 leave class VIII A and move to class VIII B. As a result, the average marks of class VIII A fall from 48 to 46. How many students were there initially in class VIII A?

Sol:

Let the original number of students in class VIII A be N.

Original total marks = N × 48.

After two students of marks 50 and 54 leave, new average = 46 and new number of students = N − 2.

New total marks = (N − 2) × 46.

Decrease in total marks due to movers = N × 48 − (N − 2) × 46.

Compute the decrease:

N × 48 − (N − 2) × 46 = 48N − (46N − 92) = 2N + 92.

But the two leaving students carried away marks equal to 50 + 54 = 104.

Therefore, 2N + 92 = 104.

Thus, 2N = 12.

N = 6.

So, there were 6 students initially in class VIII A.

Weighted Average (or Weighted Mean)

• If we want the combined average of two or more groups, and those groups have different sizes, we must use the weighted average.
• The weight attached to each value is usually the number of items in that group (for example, number of students in a section).
• Let N₁, N₂, ..., N_n be the weights attached to values X₁, X₂, ..., X_n respectively. The weighted arithmetic mean is:

• When two quantities are taken in different ratios, the weighted average shifts toward the value having greater weight - like a see-saw: the larger the weight, the greater the pull towards that value.

Example 4: The average marks of 30 students in a section of class X are 20, while that of 20 students of the second section is 30. Find the average marks for the entire class X.

Sol:

Total marks of first section = 30 × 20 = 600.

Total marks of second section = 20 × 30 = 600.

Total marks for the entire class = 600 + 600 = 1,200.

Total number of students = 30 + 20 = 50.

Combined average = 1,200 ÷ 50.

Combined average = 24.

Important Facts about Average

1. If each number is increased (or decreased) by a constant n, the mean also increases (or decreases) by n.
2. If each number is multiplied (or divided) by a constant n, the mean is multiplied (or divided) by n.
3. If the same value is added to half the quantities and the same value is subtracted from the other half, the overall mean does not change.

Average Speed

• Average speed for a journey is defined as the total distance travelled divided by the total time taken.

• If distances d₁ and d₂ are covered at speeds v₁ and v₂ respectively, and times taken are t₁ and t₂ respectively, then the average speed over the entire distance is:

Tip: Average speed is always between the minimum and maximum of the individual speeds. It can never be double or more than double of either speed if the other speed is finite and positive.

• If both distances are equal (d₁ = d₂), average speed =which is the harmonic mean of two velocities.
• If both time intervals are equal (t₁ = t₂), average speed =which is the arithmetic mean of the two velocities.

Example 5: The average of 10 consecutive numbers starting from 21 is:

Sol:

For any set of consecutive integers, the arithmetic mean equals the average of the middle two terms when the number of terms is even.

The 5th and 6th numbers in this list are 25 and 26 respectively.

Average = (25 + 26) ÷ 2.

Average = 51 ÷ 2.

Average = 25.5.

Solved Examples

Example 1: A man travels 120 km at 60 km per hour and returns the same distance at 40 km per hour. What is his average speed for the entire journey?

Sol:

For a two-part journey where distances are equal, average speed = (2 × v₁ × v₂) ÷ (v₁ + v₂).

Here, v₁ = 60 km/h and v₂ = 40 km/h.

Average speed = (2 × 60 × 40) ÷ (60 + 40).

Average speed = 4,800 ÷ 100.

Average speed = 48 km/h.

Therefore, the average speed for the entire journey is 48 km per hour.

Example 2: A grocer mixes two types of sugar. Type A costs ₹30 per kg, and Type B costs ₹50 per kg. How many kilograms of Type B sugar must he mix with 10 kg of Type A to get a mixture worth ₹40 per kg?

Sol:

Let x be the kilograms of Type B to be mixed.

Total cost of mixture = 30 × 10 + 50 × x.

Total quantity = 10 + x.

Required average price: (30 × 10 + 50 × x) ÷ (10 + x) = 40.

Equation: 300 + 50x = 40(10 + x).

300 + 50x = 400 + 40x.

50x − 40x = 400 − 300.

10x = 100.

x = 10.

So, 10 kg of Type B must be mixed with 10 kg of Type A.

Example 3: A company has three departments:

• Department A has 10 employees with an average salary of ₹60,000.
• Department B has 15 employees with an average salary of ₹80,000.
• Department C has 5 employees with an average salary of ₹50,000.
• What is the overall average salary of all employees in the company?

Sol:

Total salary of Department A = 10 × 60,000 = 600,000.

Total salary of Department B = 15 × 80,000 = 1,200,000.

Total salary of Department C = 5 × 50,000 = 250,000.

Total salary of company = 600,000 + 1,200,000 + 250,000 = 2,050,000.

Total number of employees = 10 + 15 + 5 = 30.

Overall average salary = 2,050,000 ÷ 30.

Overall average salary = 68,333.33.

Answer: ₹68,333.33.

Example 4

The average mark of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is

(a) 20

(b) 19

(c) 21

(d) 22

Sol:

Let the marks of each girl be g and of each boy be b.

Given, (4g + 6b) ÷ 10 = 24.

Therefore, 4g + 6b = 240.

Divide both sides by 2: 2g + 3b = 120. (1)

Given constraint: b ≤ g ≤ 2b.

From (1), g = (120 − 3b) ÷ 2.

Substitute into the inequality b ≤ g ≤ 2b to get bounds on b:

b ≤ (120 − 3b) ÷ 2 ≤ 2b.

Left inequality gives:

2b ≤ 120 − 3b.

5b ≤ 120.

b ≤ 24.

Right inequality gives:

(120 − 3b) ÷ 2 ≤ 2b.

120 − 3b ≤ 4b.

120 ≤ 7b.

b ≥ 120 ÷ 7 = 17 1/7.

Thus b ranges over real numbers satisfying 120/7 ≤ b ≤ 24.

We are to find the number of possible distinct integer values of the total marks of 2 girls and 6 boys, i.e., of T = 2g + 6b.

Use (1) to express T in terms of b:

From (1), 2g + 3b = 120.

Therefore, T = 2g + 6b = (2g + 3b) + 3b = 120 + 3b.

We require T to be an integer. Thus 3b must be an integer.

Let 3b = m where m is an integer.

Then b = m ÷ 3 and the allowable range for m follows from b's bounds:

120/7 ≤ m ÷ 3 ≤ 24.

Multiply through by 3: 360/7 ≤ m ≤ 72.

Compute 360/7 ≈ 51.428... so the smallest integer m ≥ 52.

Largest integer m ≤ 72 is 72.

Hence m can take integer values 52, 53, ..., 72.

Number of integers from 52 to 72 inclusive = 72 − 52 + 1 = 21.

Therefore, T = 120 + m can take 21 distinct integer values.

Hence, option (c) is correct.

Example 5: The average marks of 3 classes are as follows:

• Class A: 50 students with an average of 75.
• Class B: 30 students with an average of 80.
• Class C: 20 students with an average of 85.
• What is the overall average of all the students?

Sol:

Total marks Class A = 50 × 75 = 3,750.

Total marks Class B = 30 × 80 = 2,400.

Total marks Class C = 20 × 85 = 1,700.

Total marks overall = 3,750 + 2,400 + 1,700 = 7,850.

Total students = 50 + 30 + 20 = 100.

Overall average = 7,850 ÷ 100 = 78.5.

Example 6: The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is:

(a) 5

(b) 1

(c) 3

(d) 4

Sol:

Sum of the original three integers = 3 × 13 = 39.

When n is included, average of four becomes an odd integer. Let that odd integer be O.

Then (39 + n) ÷ 4 = O, where O is odd.

Write O = 2k + 1 for some integer k.

Therefore, 39 + n = 4(2k + 1) = 8k + 4.

Rearrange: n = 8k + 4 − 39 = 8k − 35.

We need the smallest natural number n > 0 of the form 8k − 35.

Try successive integer k values until n > 0.

For k = 5, n = 8×5 − 35 = 40 − 35 = 5, which is positive.

For smaller k, n is non‐positive.

Hence the minimum possible value of n is 5.

Therefore, option (a) is correct.