Arithmetic Word Problems Cheat Sheet - IBPS PO Prelims

1. Overview & Exam Analysis

Arithmetic word problems form the core of the Quantitative Aptitude section in the IBPS PO (Probationary Officer) Prelims examination. Based on recent PYQ Analysis and updated NTA Pattern trends, arithmetic concepts carry a substantial weightage of around 10 to 15 questions out of 35 in the Prelims stage. Mastering these topics is essential not only for direct word problems but also for Data Interpretation (DI) sets that rely on arithmetic concepts.

2. The 4-Step Problem Strategy

When tackling complex word problems under tight time constraints, follow this structured four-step strategy to ensure both speed and accuracy:

1. Understand: Read the question carefully to identify the given parameters and the exact unknown value to calculate.
2. Translate: Convert word statements into algebraic equations using appropriate variables.
3. Solve: Perform algebraic manipulations step-by-step to arrive at the solution.
4. Verify: Double-check the answer against the problem context to rule out calculation errors.

3. Translation Dictionary

Converting English statements into mathematical operations is a critical step in solving arithmetic problems. Refer to the table below for quick translation mappings:

English Keywords	Mathematical Operation
Sum, total, more than	\(+\) (Addition)
Difference, less than	\(-\) (Subtraction)
Product, of, times	\(\times\) (Multiplication)
Quotient, per, ratio	\(\div\) (Division)
Is, was, will be, equals	\(=\) (Equality)
What, a number, unknown	\(x, y, z\) (Variables)

4. Percentage & Ratio Formulas

Percentages and interest calculations frequently appear in IBPS PO Prelims arithmetic sets.

Percentage Increase/Decrease

To compute relative change:

\[\text{Percentage Change} = \frac{\text{New Value} - \text{Old Value}}{\text{Old Value}} \times 100\]

Simple Interest (SI)

Calculated directly on the principal amount:

\[SI = \frac{P \times R \times T}{100}\]

where \(P\) is Principal, \(R\) is Rate of interest per annum, and \(T\) is Time in years.

Compound Interest (CI)

For interest compounded annually:

\[A = P \bigl(1 + \frac{R}{100}\bigr)^T\] \[CI = A - P\]

where \(A\) represents the total amount accumulated after time \(T\).

5. Speed, Distance & Time

Speed, Distance, and Time problems test fundamental kinetic relationships and average speed principles.

Core Relations

\[\text{Distance} = \text{Speed} \times \text{Time}\] \[\text{Speed} = \frac{\text{Distance}}{\text{Time}}, \quad \text{Time} = \frac{\text{Distance}}{\text{Speed}}\]

Unit Conversion

To convert speed from kilometers per hour (km/hr) to meters per second (m/s):

\[\text{km/hr} \times \frac{5}{18} = \text{m/s}\]

Average Speed (Equal Distances)

When an object covers equal distances at two distinct speeds \(x\) and \(y\):

\[\text{Avg Speed} = \frac{2xy}{x + y}\]

6. Work & Time Problems

Time and work efficiency concepts rely on unit work rates.

If a person completes a total job in \(n\) days, their 1-day work contribution is \(\frac{1}{n}\).

Combined Work

If Person A completes a task in \(a\) days and Person B completes it in \(b\) days, their combined completion time working together is:

\[\text{Time} = \frac{ab}{a + b} \text{ days}\]

7. Mixtures & Alligations

The rule of alligation is an efficient technique to determine the ratio in which two ingredients are combined.

To mix two ingredients of cost prices \(C_1\) (cheaper) and \(C_2\) (dearer) to obtain a mixture of mean cost \(C_m\):

\[\frac{\text{Quantity of Cheaper}}{\text{Quantity of Dearer}} = \frac{C_2 - C_m}{C_m - C_1}\]

8. Frequently Asked Questions (People Also Ask)