Practice Questions: Averages | CSAT Preparation - UPSC PDF Download

The arithmetic mean, often simply called the average, is the sum of all observations divided by the number of observations. It represents the central value of a data set.

Formula:

Arithmetic Mean = (Sum of Observations) ÷ (Number of Observations)

This document explores calculation methods and practical applications of the arithmetic mean across different types of problems.

Easy Level

Example 1: The average of the first ten whole numbers is 

(a) 4.5 

(b) 5 

(c) 5.5 

(d) 4

Ans: (a)

Sol: Required average = (0 + 1 + 2 + … + 9)/10 = 45/10 = 4.5

Example 2: The average of the first ten prime numbers is

(a) 15.5

(b) 12.5

(c) 10

(d) 12.9

Ans: (d)

Sol: Required average = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10
= 129/10 = 12.9

Example 3: There are three fractions A, B, and C. If A = 1/5 and B = 1/8, the average of A, B, and C is 1/10. What is the value of C?

(a) –1/20

(b) – 1/60

(c) –1/30

(d) – 1/40 

Ans: (d)

Sol:

(1/5 + 1/8 + C) ÷ 3 = 1/10

= 1/5 + 1/8 + C = 3 × 1/10

= 1/5 + 1/8 + C = 3/10

= 3/10 − 1/5 − 1/8

 = 3/10 − 2/10 − 1/8

 = 1/10 − 1/8

 = (4 − 5) / 40

 = −1/40

Example 4: The average age of 10 men is increased by 3 years when one of them, whose age is 54 years is replaced by a woman. What is the age of the woman?

(a) 68 years

(b) 82 years

(c) 72 years

(d) 84 years

Ans: (d)

Sol: 

Increase in average = 3 years for 10 people

 Total increase in sum of ages = 10 × 3 = 30 years

 Age of woman = Age of replaced man + total increase = 54 + 30

= 84 years

Example 5: The average height of 13 people reduces by 2 cm if a person of height 184 cm is replaced by a new person. Find the height of the new person.

(a) 154 cm 

(b) 159 cm 

(c) 197 cm 

(d) 158 cm

Ans: (d)

Sol: 

Let the average height of the 13 people initially be A cm.

Total height of 13 people initially = 13 × A cm.

A person with height 184 cm is replaced, and the new average height becomes A − 2 cm.

Step 1: Calculate the New Total Height

New total height = 13 × (A − 2) = 13A − 26 cm.

Step 2: Find the Height of the New Person

The new total height is equal to the original total height minus the replaced person’s height plus the new person’s height:

13A − 184 + New Person’s Height = 13A − 26.

Step 3: Solve for the Height of the New Person

New Person’s Height = 184 − 26 = 158 cm.

Thus, the height of the new person is 158 cm.

Example 6: The monthly salaries of two persons A and B are in the ratio of 3: 5 respectively. If both of them received an increment of Rs. 250, then the ratio becomes 2 : 3. What were their respective salaries before the increment? 

(a) Rs. 850 & Rs. 1,275 

(b) Rs. 700 & Rs. 1,050 

(c) Rs.750 & Rs. 1,250 

(d) Rs. 650 & Rs. 975

Ans: (c)

Sol: 

Let the original monthly salaries of A and B be 3x and 5x respectively.

After an increment of ₹250, their new salaries become:

• A's new salary = 3x + 250
• B's new salary = 5x + 250

According to the problem, the new ratio becomes:

(3x + 250)/(5x + 250) = 2/3

Step 1: Cross multiply
3(3x + 250) = 2(5x + 250)

9x + 750 = 10x + 500

9x - 10x = 500 - 750
-x = -250 ⟶ x = 250

• A's original salary = 3 × 250 = ₹750
• B's original salary = 5 × 250 = ₹1250

Medium Level

Almost 70% of questions in exam are Medium-level questions. Though conceptually they seem easier, the trick is to solve the calculations faster & we curated problems for you to help you do problems more easily.

Example 1: With an average speed of 25 km/h, a train reaches its destination in time. If it goes with an average speed of 20 km/h, it is late by 1 hour. The length of the total journey is:

(a) 90 km 

(b) 100 km 

(c) 120 km 

(d) 80 km

Ans: (b)

Sol: 

Let the total length of the journey be D kilometers, and the time taken by the train at 25 km/h (on time) be T hours.

Using the formula Distance = Speed × Time, we can write:

D = 25 × T (when the train is on time)

If the train travels at 20 km/h, it is late by 1 hour. So, the new time taken is T + 1 hours, giving:

D = 20 × (T + 1)

Since both equations represent the same total distance D, we equate them:

25T = 20(T + 1)

Expand the equation:

25T = 20T + 20

5T = 20 ⟶ T = 4 hours

Now that we know T = 4 hours, substitute it into the equation D = 25 × T:

D = 25 × 4 = 100 km

Example 2: In Hotel Clarks, the rooms are numbered from 101 to 150 on the first floor, 201 to 240 on the second floor, and 316 to 355 on the third floor. In May 2018, the room occupancy was 50% on the first floor, 50% on the second floor, and 30% on the third floor. If it is also known that the room charges are ₹ 2000, ₹1000, and ₹1500 on each of the floors, then find the average income per room (in ₹) for May 2018. 

(a) 676.92 

(b) 880.18 

(c) 783.3 

(d) 650.7

Ans: (a)

Sol:

Step 1: Number of rooms on each floor

 First floor: rooms 101–150 ⟶ 50 rooms

 Second floor: rooms 201–240 ⟶ 40 rooms

Third floor: rooms 316–355 ⟶ 40 rooms

Step 2: Occupied rooms in May

First floor occupied = 50% of 50 = 25 rooms

 Second floor occupied = 50% of 40 = 20 rooms

 Third floor occupied = 30% of 40 = 12 rooms

Step 3: Revenue from each floor

 First floor revenue = 25 × ₹2,000 = ₹50,000

 Second floor revenue = 20 × ₹1,000 = ₹20,000

Third floor revenue = 12 × ₹1,500 = ₹18,000

 Total revenue = 50,000 + 20,000 + 18,000 = ₹88,000

Step 4: Average income per room

 Total number of rooms = 50 + 40 + 40 = 130

Average income per room = 88,000 ÷ 130 = 676.923…

 ≈ ₹676.92 

Example 3: The average weight of 10 men is decreased by 5 kg when one of them weighing 100 kg is replaced by another person. This new person is again replaced by another person, whose weight is 10 kg lower than the person he replaced. What is the overall change in the average due to this dual change?

(a) 5 kg 

(b) 6 kg 

(c) 12 kg 

(d) 15 kg 

Ans: (b)

Sol: 

Step 1: First replacement effect

 Average drops by 5 kg for 10 people ⟶ total drop = 10 × 5 = 50 kg

 So the new person's weight = 100 − 50 = 50 kg

Step 2: Second replacement effect

Third person is 10 kg lighter than the second ⟶ third person's weight = 50 − 10 = 40 kg

Step 3: Net change in total weight

Original person 100 replaced by 40 ⟶ net drop = 100 − 40 = 60 kg

Change in average = 60 ÷ 10 = 6 kg (decrease)

Example 4: The average price of 3 precious diamond-studded platinum thrones is ₹ 97610498312. If their prices are in the ratio 4:7:9. The price of the cheapest is 

(a) 5, 65, 66, 298.972 

(b) 5, 85, 66, 29, 8987.2 

(c) 58, 56, 62, 889.72 

(d) None of these 

Ans: (b)

Sol: 

Average price = 97,610,498,312

 Total price of three = 3 × 97,610,498,312 = 292,831,494,936

 Ratio 4 : 7 : 9 ⟶ sum of ratio parts = 4 + 7 + 9 = 20

Cheapest share = (4/20) × 292,831,494,936

 Cheapest share = (1/5) × 292,831,494,936

Cheapest share = 58,566,298,987.2

Example 5: The average age of a group of 15 persons is 25 years and 5 months. Two persons, each 40 years old, left the group. What will be the average age of the remaining persons in the group?

(a) 23.17 years

(b) 24.25 years

(c) 25.35 years

(d) 25 years

Ans: (a)

Sol:

Step 1: Convert ages to months

Average = 25 years 5 months = 25 × 12 + 5 = 305 months

 Total age of 15 persons = 15 × 305 = 4,575 months

Step 2: Two persons of 40 years leave

 Each 40 years = 480 months, so two persons = 2 × 480 = 960 months

Remaining total = 4,575 − 960 = 3,615 months

 Average for remaining 13 persons = 3,615 ÷ 13 = 278.0769 months

 Convert back to years = 278.0769 ÷ 12 = 23.1731 years

 ≈ 23.17 years

Example 6: There are 24 students in a class whose average marks in a subject, the maximum marks of which is 100, is 89. If 3 students leave the class, then what is the maximum by which the average could go up? 

(a) 10.7 

(b) 10.5 

(c) 11.2 

(d) 11

Ans: (d)

Sol: Sum of marks for 24 students = 24 × 89 = 2136
Average marks of a student cannot increase beyond 100.
So, total marks for 21 students cannot exceed 2100.
So, maximum increase in average = 100 – 89 = 11.

Example 7: Ajay started a firm with a capital of Rs. 28,000. After 5 months, Boman joined him and invested Rs. 40,000 in the firm. Chirag was also added as a new partner with an individual investment of Rs. 56,000 after 7 months of commencement. If at the end of the year, the profit of the firm is Rs. 32,000, what is the share of Boman? 

(a) Rs. 12,000 

(b) Rs. 8,000 

(c) Rs. 14,000 

(d) Rs. 10,000

Ans: (d)

Sol: 

Step 1: Compute time-weighted investments (capital × months)

 Ajay: ₹28,000 for 12 months = 28,000 × 12 = 336,000

 Boman: ₹40,000 for (12 − 5) = 7 months = 40,000 × 7 = 280,000

 Chirag: ₹56,000 for (12 − 7) = 5 months = 56,000 × 5 = 280,000

Step 2: Ratio of shares

Ajay : Boman : Chirag = 336,000 : 280,000 : 280,000 = 6 : 5 : 5

 Boman's share = (5 / (6 + 5 + 5)) × 32,000 = (5 / 16) × 32,000

= ₹10,000

Hard Level

Around 25% of these types of questions come in exam - If your target is above 95%ile, we recommend you to solve these questions as well.

Example 1: A salesman gets a bonus according to the following structure: If he sells articles worth ₹ x then he gets a bonus of ₹(x/10 – 1000). In the month of January, the value of his sales was ₹10000, in February it was ₹12000, from March to November it was ₹30000 for every month and in December it was ₹12000. Apart from this, he also receives a basic salary of ₹3000 per month from his employer. Find his average income per month (in ₹) during the year. 

(a) 4533 

(b) 4517 

(c) 4532 

(d) 4668

Ans: (a)

Sol: 

Step 1: Compute monthly bonus using Bonus = x/10 − 1000

January (x = 10,000): bonus = 1,000 − 1,000 = 0

February (x = 12,000): bonus = 1,200 − 1,000 = 200

 March–November (9 months, x = 30,000 each): bonus per month = 3,000 − 1,000 = 2,000

December (x = 12,000): bonus = 200

Step 2: Total bonus for the year

 Total bonus = 0 + 200 + (9 × 2,000) + 200

= 0 + 200 + 18,000 + 200

= ₹18,400

Step 3: Total salary for the year

 Basic salary = 3,000 × 12 = ₹36,000

 Total annual income = 36,000 + 18,400 = ₹54,400

Step 4: Average monthly income

 54,400 ÷ 12 = 4,533.33

 Closest option: (a) 4533

Example 2: Ramu appears in six different papers in his semester examination, where the maximum marks were 50 for each paper. His marks in these papers are in the proportion of 8 : 9 : 10 : 13 : 14 : 15. Considering his aggregate in all the papers together, he fails to obtain 50% of the total marks. What is the minimum possible additional marks Ramu should get to obtain 50% of the total marks, given that he got integral marks in each paper? 

(a) 81 

(b) 57 

(c) 12 

(d) 18

Ans: (b)

Sol: 

Step 1: Let k be the scaling factor (integer) so marks are: 8k, 9k, 10k, 13k, 14k, 15k.

= Maximum mark per paper is 50 ⟶ 15k ≤ 50 ⟶ k ≤ 3

Step 2: Compute totals for allowed k

= For k = 1: total = 8 + 9 + 10 + 13 + 14 + 15 = 69 ⟶ needs 150 − 69 = 81 more marks

= For k = 2: total = 2 × 69 = 138 ⟶ needs 150 − 138 = 12 more marks

= For k = 3: total = 3 × 69 = 207 ⟶ already above 150 (so he would have passed)

Step 3: Since he fails to obtain 50% initially, valid k are 1 or 2; the minimum additional marks required is the smaller of 81 and 12.

= Minimum additional marks = 12 (option c)