IBPS PO 2026 Quant Preparation Strategy - Topic-Wise Plan for Prelims & Mains

Mastering the IBPS PO 2026 Quant Preparation Strategy is the defining factor for candidates aiming to clear the cutoff in both Prelims and Mains. The Quantitative Aptitude section is often perceived as the most challenging, yet with a structured approach to high-weightage topics like Data Interpretation, Speed Maths, and Quadratic Equations, it can become your highest-scoring area. Here is the exact topic-wise roadmap you need to follow, providing expert shortcuts for approximation and step-by-step methods for decoding complex DI sets. If you are a beginner or looking to sharpen your calculation speed, this plan is designed to help you tackle the 2026 exam pattern with confidence and precision.

Data Interpretation: Population in Five Cities Introduction

This section focuses on solving a Data Interpretation (DI) problem from a pie chart, a common type in competitive exams. The problem requires calculating population distributions across five cities based on given percentages and conditional notes, which is crucial for accurately answering subsequent questions.

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Data Interpretation Set 1: Population in Five Cities

This problem uses a pie chart to show population distribution across five cities. Extracting and structuring the data correctly is essential.

Initial Data & Conditions

• Pie Chart Percentage Distribution:

• Jaipur: 20%

• Kanpur: 25%

• Bharatpur: 15%

• Surat: 10%

• Goa: 30%

• Note 1: The total number of people in Jaipur and Goa is x. The value of x is the cube of the smallest two-digit even number.

• The smallest two-digit even number is 10.

• Therefore, x = 10³ = 1000.

• Note 2: In all cities except for Kanpur and Surat, the number of females is greater than the number of males.

• For Jaipur, Bharatpur, and Goa: Females > Males.

• For Kanpur and Surat: Males > Females.

Framing the Data Table

The first step involves calculating absolute population values for each city, followed by determining male and female populations.

Step 1: Calculate Total Population

1. Combined percentage for Jaipur and Goa from the pie chart: 20% + 30% = 50%.

2. From Note 1, this 50% equals 1000.

3. If 50% = 1000, then the total population (100%) across all five cities is 2000.

Step 2: Calculate Population for Each City

Using the total population of 2000:

• Jaipur (20%): 0.20 * 2000 = 400

• Kanpur (25%): 0.25 * 2000 = 500

• Bharatpur (15%): 0.15 * 2000 = 300

• Surat (10%): 0.10 * 2000 = 200

• Goa (30%): 0.30 * 2000 = 600

Step 3: Calculate Male and Female Populations

The problem provides the difference between male and female populations for each city.

(To quickly calculate males and females when their sum and difference are known: First, assign the 'difference' value to the larger group. Subtract this difference from the total, then divide the remainder equally between the two groups. Add this equal share to the initial amounts.)

Calculations:

• Jaipur: Total = 400, Difference = 80 (Females > Males). Females = 240, Males = 160.

• Kanpur: Total = 500, Difference = 120 (Males > Females). Males = 310, Females = 190.

• Bharatpur: Total = 300, Difference = 60 (Females > Males). Females = 180, Males = 120.

• Surat: Total = 200, Difference = 40 (Males > Females). Males = 120, Females = 80.

• Goa: Total = 600, Difference = 70 (Females > Males). Females = 335, Males = 265.

Solved Questions from DI Set 1

Question 1: Population in Agra

• Condition: Ratio of Bharatpur to Agra is 5:8. Females in Agra are 20% more than females in Goa. Males in Agra increase by 28. Find new total males.

• Calculation:

• Population of Bharatpur = 300. Agra population = (8/5) * 300 = 480.

• Females in Goa = 335. Females in Agra = 335 * 1.20 = 402.

• Initial Males in Agra = 480 - 402 = 78.

• New Males in Agra = 78 + 28 = 106.

Question 2: Educated vs. Uneducated in Jaipur

• Condition: In Jaipur (Total=400), 60% are uneducated. Uneducated males are half the males in Surat. Find the difference between Educated Males and Uneducated Females.

• Calculation:

• Uneducated in Jaipur = 400 * 0.60 = 240.

• Males in Surat = 120. Uneducated Males in Jaipur = 120 / 2 = 60.

• Uneducated Females in Jaipur = 240 - 60 = 180.

• Total Males in Jaipur = 160. Educated Males = 160 - 60 = 100.

• Difference (Uneducated Females - Educated Males) = 180 - 100 = 80.

Question 3: Population Transfer from Bharatpur to Surat

• Condition: 45% of people from Bharatpur transfer to Surat. Remaining ratio of Males to Females in Bharatpur is 6:5. Find new Male:Female ratio in Surat.

• Calculation:

• People transferred = 300 * 0.45 = 135. Remaining in Bharatpur = 300 - 135 = 165.

• Remaining Males in Bharatpur = (6/11) * 165 = 90. Remaining Females = (5/11) * 165 = 75.

• Males transferred to Surat = 120 (Initial) - 90 (Remaining) = 30.

• Females transferred to Surat = 180 (Initial) - 75 (Remaining) = 105.

• New Males in Surat = 120 (Initial) + 30 (Transferred) = 150.

• New Females in Surat = 80 (Initial) + 105 (Transferred) = 185.

• New Ratio (Male: Female) = 150: 185 = 30: 37.

Question 4: Population in Mohali

• Condition: Total people in Mohali = √(x - 100). The difference between males and females ('a') is the square of the third-largest single-digit even number. Males > Females. Find females in Mohali.

• Calculation:

• x = 1000. Total people = √(1000 - 100) = √900 = 30.

• Single-digit even numbers: 2, 4, 6, 8. The third-largest is 4.

• Difference 'a' = 4² = 16.

• Male + Female = 30, Male - Female = 16.

• 2 * Female = 30 - 16 = 14. Females = 7.

Speed Maths: Approximation

Approximation questions test your ability to quickly estimate values to simplify complex calculations.

Question 1

√4624.2 * 13.98 - 13.01 * 14.1 = x + 489.9

• Approximation: √4624 * 14 - 13 * 14 = x + 490

• (Note: √4624 = 68, as 60²=3600, 70²=4900, and it ends in 4).

• Calculation: 68 * 14 - 182 = x + 490 => 952 - 182 - 490 = x => 770 - 490 = x => x = 280

Question 2

559.98 ÷ (480.01 ÷ 11.99) + 18.02 * 1.98 = x

• Approximation: 560 ÷ (480 ÷ 12) + 18 * 2 = x

• Calculation: 560 ÷ 40 + 36 = x => 14 + 36 = x => x = 50

Question 3

89.89% of 40.01 + √576.02 * 5.99 = x + 120% of 179.98

• Approximation: 90% of 40 + √576 * 6 = x + 120% of 180

• Calculation: 36 + 24 * 6 = x + (6/5) * 180 => 36 + 144 = x + 216 => 180 = x + 216 => x = -36 (or x = 36 if the options suggest positive difference).

Question 4

112.01 + (982.02 ÷ 8.01) - (377.9 ÷ 5.98) + x = 191.99

• Approximation: 112 + (982 ÷ 8) - (378 ÷ 6) + x = 192

• 982 ÷ 8 ≈ 123

• 378 ÷ 6 = 63

• Calculation: 112 + 123 - 63 + x = 192 => 112 + 60 + x = 192 => 172 + x = 192 => x = 20

Question 5

(8.01³ + 22.01²) % of 79.99 + 62.5% of 1240.1 - 194.01 = x

• Approximation: (8³ + 22²) % of 80 + (5/8) * 1240 - 194 = x

• 8³ = 512, 22² = 484. Sum = 996. Round 996% to 1000% for approximation.

• Calculation: 1000% of 80 + (5/8) * 1240 - 194 = x => 800 + 5 * 155 - 194 = x => 800 + 775 - 194 = x => 1575 - 194 = x => x ≈ 1381

Data Interpretation Set 2: Pen Sales in Five Shops

This DI problem involves a table with total pens sold and the percentage of Ball Pens, requiring a critical value 'a' to be derived first.

• Core Condition: The number of Gel Pens sold in Shop D is equal to the number of Ball Pens sold in Shop C.

Derivation of 'a'

1. Gel Pens in Shop D: Total Pens = 800. Ball Pens = 70% of 800. Gel Pens = 30% of 800 = 240.

2. Ball Pens in Shop C: Total Pens = 4a. Ball Pens = 60% of 4a.

3. Equating: 60% of 4a = 240 => 0.60 * 4a = 240 => 2.4a = 240 => a = 100.

With a=100, the complete data table can be formed to solve further questions related to pen sales.

Quadratic Equations

Solving quadratic equations and comparing their roots is a frequent topic in competitive exams.

Question 1

• w² + 7w + 12 = 0 => (+, +) roots are (-, -). Factors are 4, 3. Roots: -4, -3.

• x² + 3x - 4 = 0 => (+, -) roots are (-, +). Factors are 4, 1. Roots: -4, +1.

• Comparison: -3 is greater than -4 but less than +1. Relationship cannot be established (CND).

Question 2

• x² - x - 72 = 0

• y² + 5y - 66 = 0

• (If the constant term 'c' is negative in both ax² + bx - c = 0 equations, the relationship between x and y cannot be established as roots will always be (+, -) leading to overlap.)

• Answer: Relationship cannot be established (CND).

Question 3

• 2x² + 26x + 84 = 0 (Divide by 2: x² + 13x + 42 = 0) => (+, +) roots are (-, -). Factors are 7, 6. Roots: -7, -6.

• y² + 15y + 56 = 0 => (+, +) roots are (-, -). Factors are 8, 7. Roots: -8, -7.

• Comparison: -6 is greater than both -8 and -7. -7 is equal to -7 and greater than -8. Therefore, x ≥ y.

Question 4

• x² + 3x - 10 = 0 => (+, -) roots are (-, +). Factors 5, 2. Roots: -5, +2.

• y² + 5y + 6 = 0 => (+, +) roots are (-, -). Factors 3, 2. Roots: -3, -2.

• Comparison: +2 is greater than both -3 and -2. -5 is less than both -3 and -2. Relationship cannot be established (CND).

Question 5

• x² - 12x + 27 = 0 => (-, +) roots are (+, +). Factors 9, 3. Roots: +9, +3.

• 2y² - 15y + 28 = 0 => (-, +) roots are (+, +). Factors 8, 7. Roots: +8, +7.

• (To compare, either divide y-roots by 2: +4, +3.5, or multiply x-roots by 2: +18, +6. Multiplying is often easier to avoid decimals.)

• New comparison: x-roots are 18, 6. y-roots are 8, 7.

• Comparison: 18 is greater than both 8 and 7. 6 is less than both 8 and 7. Relationship cannot be established (CND).

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