Arithmetic Word Problems Concept with Examples for LIC AAO

Arithmetic Word Problems are one of the most asked and high-weightage topics in the Quantitative Aptitude section of LIC AAO exam. It covers 13 to 15 questions in the LIC AAO Prelims exam from various arithmetic topics like Profit and loss, Percentages, Ages, Time and Distance, Time and Work, Partnership, Mensuration, Mixture and Alligation, Trains, Boats, Pipes, Averages, and SI & CI. A complete understanding of these topics can fetch you quick marks in this exam. In this blog, we will explain each Arithmetic Word Problems Concept with Examples for LIC AAO exam. Continue reading to understand basic concepts of Arithmetic and boost your score in the Quant section. 

What are Arithmetic Word Problems in LIC AAO?

Quantitative Aptitude section in LIC AAO Prelims exam covers questions based on Arithmetic word problems. They test the application of mathematical concepts in real-life situations. They cover various topics like percentages, ratios, time & work, time, speed, and distance, profit & loss, averages, etc. To solve these word problems, they are written in simple forms for easy calculations. It is suggested by experts to learn squares, cubes, multiplication tables, componendo-dividendo, factors, and percentages, and basic calculations to easily solve word problems. Let’s understand the basic concept and formula of each type of arithmetic word problem with the help of a solved example to get an idea of these type of questions asked in the exam. 

Arithmetic Word Problems Basic Concepts with a Solved Example

Arithmetic word problems also called word problems covers important topics as mentioned earlier in this article. They are easy to solve if you remember the formula. They fetch you quick marks in the exam. If you learn the basic concepts of these topics, and remember their formula,  you will be able to easily solve them when you practice questions based on these topics. They also form the base to solve most Data Interpretation questions. Let’s discuss and understand each one of them using a solved example. 

1. Percentages

Percentages are such word problems which involve increase/decrease, discounts, or marks scored.

Example:
A shop applies two successive discounts of 10% and 20% on the marked price of a laptop. After both discounts, the customer pays ₹7,200.

Solution:

Let the marked price be M.

1. After the first discount of 10% the price becomes:

M×(1−0.10)= 0.90M

2. After the second discount of 20% on the reduced price the final price becomes:

0.90M×(1−0.20)= 0.90M×0.80= 0.72M

3. We are told the final price is ₹7,200:

0.72M= 7200 ⇒ M= 7200/ 0.72

Therefore, M=10,000

So the marked price = ₹10,000.

4. Overall discount percentage:
   The effective price paid is 0.72M, so the overall discount fraction is:

1−0.72= 0.28= 28%

So the overall discount = 28%.

2. Profit and Loss

These questions usually involve successive percentage changes, discounts, or profit-loss with marked price.

The selling price of 40 articles is same as cost price of 24 articles. Find the gain/loss percentage on selling each article.

A) 60% loss

B) 40% gain

C) 50% gain

D) 60% gain

E) 40% loss

Solution:

Let the cost price of each article be Rs. ‘x’.

So, cost price of 24 articles = Rs. ’24x’

So, selling price of 40 articles = 24x ÷ 40 = Rs. ‘0.6x’

So, loss percentage = {(x – 0.6x)/x} X 100 = 40%

Hence, option e.

3. Simple Interest (SI) & Compound Interest (CI)

Based on principal, rate, time.

Example (SI):
Find the compound interest earned on investing Rs. 18,000 at compound interest of 24% p.a. for 2 years given that interest is compounded once a year.

A) Rs. 7,786.80

B) Rs. 9,676.80

C) Rs. 9,236.20

D) Rs. 7,436.20

E) None of these

Solution:

Amount at the end of two years = 18000 X (1.24)² = Rs. 27,676.80

So, interest earned = 27676.8 – 18000 = Rs. 9,676.80

Hence, option b.

4. Time and Work

Based on work capacity.

Example:
A contractor hired 25 men to finish some work in 40 days. If 9 men left after working for 24 days, then find the time required to finish the remaining work.

A) 30 days

B) 20 days

C) 18 days

D) 24 days

E) 25 days

Solution:
Let the efficiency of each man be ‘x’ units/day

So, total work = 25 X 40 X x = ‘1000x’ units/day

So, work done in 24 days = 24 X x X 25 = ‘600x’ units/day

So, work remaining = 1000x – 600x = ‘400x’ units

So, required time = 400x ÷ {x X (25 – 9)} = 25 days

Hence, option e.

5. Time, Speed, and Distance

Covers average speed, relative speed, trains, boats.

Example:
A boat takes 270 minutes to cover 162 km in downstream. If the speed of the boat in still water is 25% less than the downstream speed of the boat, then find the speed of the stream.

A) 12 km/h

B) 8 km/h

C) 9 km/h

D) 6 km/h

E) 10 km/h

Solution:

Downstream speed of the boat = (162/270) X 60 = 36 km/h

So, speed of the boat in still water = 36 X 0.75 = 27 km/h

So, speed of the stream = 36 – 27 = 9 km/h

Hence, option c.

6. Ratio and Proportion

Used in age problems, partnership, and mixtures.

Example:
Three numbers i.e. ‘A’, ‘B’ and ‘C’ are in the ratio 29:39:79, respectively. If each number is reduced by 27, then ‘A’, ‘B’ and ‘C’ would be in the ratio 2:3:7, respectively. Find the sum of three original numbers.

A) 524

B) 441

C) 481

D) 216

E) 36

Solution:

Let ‘A’, ‘B’ and ‘C’ be ’29x’, ’39x’ and ’79x’, respectively.

Required sum = 29x + 39x + 79x = ‘147x’

ATQ;

{(29x – 27)/(39x – 27)} = (2/3)

Or, 87x – 81 = 78x – 54

Or, 9x = 27

So, x = 3

So, required sum = 147x = 147 X 3 = 441

Hence, option b.

7. Mixtures & Alligation

Problems with milk-water or cost mixtures.

Example:
A 400 litres mixture of milk and honey contains 36% honey. If 150 litres of the mixture is replaced with same quantity of water, then find the ratio of quantity of honey to that of water, in the resultant mixture.

A) 9:16

B) 16:9

C) 16:25

D) 25:16

E) 3:5

Solution:
Quantity of honey in the original mixture = 400 X 0.36 = 144 litres

Quantity of honey removed = 150 X 0.36 = 54 litres

So, quantity of honey left = 144 – 54 = 90 litres

Quantity of water in the mixture = 150 litres

So, required ratio = 90:150 = 3:5

Hence, option e.

8. Income & Expense

Finding mean value.

Example:
Rohan spent 36% of his monthly income on food and 25% of the rest on rent. He spent some amount on amenities and saved 24% of the total monthly income. If every month, he spent Rs. 10,800 on amenities, then find the annual income of Rohan.

A) Rs. 4,50,000

B) Rs. 45,000

C) Rs. 37,500

D) Rs. 40,000

E) Rs. 5,40,000

Solution:
Let the monthly income of Rohan be Rs. ‘100x’.

Amount spent on food = 100x X 0.36 = Rs. ’36x’

Amount spent on rent = (100x – 36x) X 0.25 = Rs. ’16x’

Amount saved = 100x X 0.24 = Rs. ’24x’

Amount spent on amenities = 100x – 36x – 16x – 24x = Rs. ’24x’

ATQ;

24x = 10,800

Or, x = 450

So, monthly income of Rohan = 450 X 100 = Rs. 45,000

Annual income of Rohan = 45000 X 12 = Rs. 5,40,000

Hence, option e.

9. Partnership

Business investment sharing.

Example:
‘A’ and ‘B’ started a business by investing Rs. 45,000 and Rs. 75,000, respectively. Few months later, ‘B’ withdrew his investment such that at the end of the year profits were divided among ‘A’ and ‘B’ in ratio 9:5, respectively. After how many months did ‘B’ withdraw his investment?

A) 4

B) 8

C) 6

D) 7

E) 5

Solution:
Let the number of months for which ‘B’ kept his investment be ‘X’.

So, ratio of profits of ‘A’ and ‘B’ at the end of the year = (45000 X 12):(75000 X X) = (36/5X)

ATQ;

(36/5X) = (9/5)

Or, X = 4

So, ‘B’ withdrew his investment after 4 months.

Hence, option a.

10. Mensuration (Volume & Area Problems)

Practical geometry-based word problems.

Example:
The height of a cone is 16.8 cm. If the radius of the cone is 7 cm, then find the lateral surface area of the cone. (Take π = 22/7)

A) 384.6 cm²

B) 332.6 cm²

C) 432.4 cm²

D) 400.4 cm²

E) 456.8 cm²

Solution:

Lateral surface area of cone = πrl {Where ‘r’ is radius and ‘l’ is slant height}

Slant height = √(16.8² + 7²) = 18.2 cm {Since, slant height cannot be negative}

So, required area = (22/7) X 18.2 X 7 = 400.4 cm²

Hence, option d.

11. Age Based Problem

These involve problems based on the calculation of the age of a particular person.

Example:

Difference between present ages of ‘A’ and ‘B’ is 8 years. Present ages of ‘C’ is 25% more than that of ‘B’ whereas present age of ‘A’ is 25% less than that of ‘B’. Find the present age of ‘C’.

A) 36 years

B) 30 years

C) 32 years

D) 40 years

E) 24 years

Solution:

Let the present age of ‘C’ be ’10x’ years.

So, present age of ‘B’ = 10x ÷ 1.25 = ‘8x’ years

And present age of ‘A’ = 8x X 0.75 = ‘6x’ years

ATQ;

8x – 6x = 8

Or, 2x = 8

So, x = 4

So, present age of ‘C’ = 4 X 10 = 40 years

Hence, option d.

Conclusion

Almost all types of word problems asked in the LIC AAO exam have been covered in the blog. The difficulty level of the questions will vary in the exam. For your better understanding of basic concepts, a solved example is given. Solve previous year papers and attempt mocks of LIC AAO based on the latest exam pattern. Now you know the basic concepts, and you can start practicing questions from these topics through topic tests available on the PracticeMock platform. With consistent practice of word problems, you will be able to secure good marks in the Quantitative Aptitude section.

Know What Our Successful Aspirant Says…

Join our exclusive Telegram group, where our experts are ready to answer all your queries, guide you in banking exam preparation, and give personalised tips to boost your success. Get access to real-time solutions, expert advice, and valuable resources to improve your study journey.

Also read other related blogs:

LIC AAO Syllabus and Exam Pattern	LIC AAO Previous Year Cut-offs
LIC AAO Eligibility Criteria	LIC AAO Salary
LIC AAO Study Plan	LIC AAO Previous Year Papers

Our Banking Preparation Package includes topic tests, sectional tests, rank boosters for prelims, previous year paper tests, e-books, CA tests, Quizzes, live tests, PDF Course, and more. Discover our banking and insurance packages in detail from the link provided below!

FAQs