RRB Group D Exam Maths 03 Aug All Shift Exam Review By Shubham Sir

RRB Group D Exam Maths 03 Aug All Shift Exam Review By Shubham Sir helps students understand the exam pattern and prioritise their preparation. The Mathematics section, in particular, shows consistent patterns, making Previous Year Questions (PYQs) essential for success. Insights into other subjects and effective study strategies are also covered for comprehensive preparation.

Also Read:

• RRB Group D Exam Science 03 Aug All Shift Exam Review By Yogesh Sir
• RRB Group D Exam Analysis 2026: Reasoning Review (03 August All Shifts) By Prashant Sir
• RRB Group D Exam GK GS 03 Aug All Shift Exam Review By Raja Gupta Sir

RRB Group D Exam Analysis: 3rd August 2026 (All Shifts)

This session reviews the types of questions and their distribution from the Group D exam held on August 3rd, 2026, across all shifts. This analysis aims to help students understand the exam pattern and prioritize their preparation.

Mathematics Section Analysis

An overview of question distribution from the Mathematics section:

• High-Weightage Chapters:

• Percentage, Profit & Loss, Discount: 5-6 questions. This combined topic is highlighted as most important and carries the highest weightage in the exam.

• Other Chapters:

• Trigonometry: 1 question.

• Algebra: 0 questions (specifically in the first and second shifts).

• Coordinate Geometry: 2 questions. This chapter is important and consistently features questions.

• Time & Work (including Pipes & Cisterns): 2-3 questions.

• Simplification: 2 questions, primarily based on the BODMAS rule. These questions are generally easy.

• Speed, Time & Distance: 2 questions. Questions on Boat & Stream were not observed.

• Mensuration: 2 questions.

• Interest (Simple & Compound): 2 questions. A 3-year Compound Interest (CI) question was seen. This topic is considered important.

• Average & Age: Questions were present.

• Ratio & Proportion: 1 question.

• Number System: 1-2 questions.

• Geometry: 1-2 questions.

• Statistics: 1 question (typically covering Mean, Median, Mode).

• LCM & HCF: No questions were observed in the morning and afternoon shifts.

Strategic Advice for Mathematics:

Students preparing for upcoming shifts should focus heavily on the Percentage, Profit & Loss, and Discount chapters due to their high frequency.

RRB Group D 2026 Overall Exam Pattern and Strategy

• Mathematics Exam Pattern:

• The overall pattern for Mathematics remains unchanged.

• The exam primarily consists of Previous Year Questions (PYQs). The Railway board has reproduced PYQs without introducing any new patterns.

• Out of 25 questions, 21-22 questions are easy to moderate and are doable for students who have practiced PYQs.

• Strategic Tip: If a question is highly calculative, it is advisable to skip it to save time.

• Other Subjects Analysis:

• General Knowledge (GK) / General Science (GS):

• History, Polity, Geography questions were minimal or absent.

• Current Affairs dominates, with 15-16 questions out of 20.

• Key Focus: Current affairs from the last six months of 2026 are highly relevant; 2025 current affairs are not seen.

• Reasoning:

• The Reasoning section is generally very easy, with many students capable of attempting all 30 questions.

• High-Frequency Topic: Direction-based questions were frequently observed.

• Science:

• Questions are drawn from all three segments: Biology, Chemistry, and Physics.

• Traditionally, Physics has a higher weightage in Group D, but this trend was not observed in the initial shifts. In the morning shift, Chemistry questions were more prominent than Physics or Biology. The distribution tends to be balanced (e.g., 7-8 questions each).

• Overall Preparation Conclusion:

• Mathematics, Reasoning, and Science are highly scoring if students have thoroughly practiced Previous Year Questions (PYQs).

• Crucial Focus: In addition to PYQs for Math, Reasoning, and Science, students must also thoroughly prepare Current Affairs.

RRB Group D Maths Analysis: Problem Solving

Percentage Increase

Problem: The price of an article is successively increased by 9% and 13%. Find the total percentage increase in the price of the article.

Solution:

Use the formula for successive percentage changes: A + B + (AB / 100).

Here, A = 9% and B = 13%.

Total increase = 9 + 13 + (9 * 13 / 100) = 22 + (117 / 100) = 22 + 1.17 = 23.17%.

(Memory Tip: PYQ stands for Previous Year Question Paper. It is crucial to study these for the exam.)

Number Theory: Finding the Greatest Divisor

Problem: A positive integer n divides 228, leaving a remainder of 18. What is the greatest two-digit value of n?

Solution:

If 228 divided by n leaves a remainder of 18, then (228 - 18) = 210 must be perfectly divisible by n.

We need the greatest two-digit divisor of 210.

Divisors of 210 include 70 (210/70 = 3) and 42 (210/42 = 5).

The greatest two-digit value of n is 70.

Time and Work: Pipes and Cisterns

Problem: Pipes A and B together fill a tank in 10 hours. Pipes B and C together fill it in 20 hours. Pipes C and A together fill it in 30 hours. In how much time will pipes A, B, and C together fill the tank?

Solution:

1. Total Work (LCM): LCM (10, 20, 30) = 60 units.

2. Individual Efficiencies:

• (A+B) = 60 / 10 = 6 units/hour

• (B+C) = 60 / 20 = 3 units/hour

• (C+A) = 60 / 30 = 2 units/hour

3. Combined Efficiency of (A+B+C):
   2(A+B+C) = 6 + 3 + 2 = 11 units/hour
   (A+B+C) = 11 / 2 units/hour

4. Time Taken:
   Time = Total Work / Combined Efficiency = 60 / (11/2) = 120 / 11 hours = 10 and 10/11 hours.

Simplification (Order of Operations)

Problem: Simplify the expression as derived during the lecture: (10/21) ÷ (29/21) + (10/29).

Solution:

1. Perform division: (10/21) ÷ (29/21) = (10/21) * (21/29) = 10/29.

2. Perform addition: 10/29 + 10/29 = 20/29.
   The simplified value is 20/29.

Speed, Time, Distance: Train Problems

Problem: Two trains, 130m and 120m long, are traveling in the same direction at speeds of 62 km/h and 44 km/h respectively. How much time will the faster train take to completely cross the slower train?

Solution:

1. Total Distance: Sum of lengths = 130 m + 120 m = 250 m.

2. Relative Speed (Same Direction): 62 km/h - 44 km/h = 18 km/h.

3. Unit Conversion (km/h to m/s): 18 km/h * (5/18) m/s = 5 m/s.
   (Memory Tip: To convert km/h to m/s, multiply by 5/18; to convert m/s to km/h, multiply by 18/5.)

4. Time Taken: Time = Distance / Speed = 250 m / 5 m/s = 50 seconds.

Average

Problem: The average of four numbers is 15. If three of the numbers are 10, 20, and 25, find the fourth number.

Solution:

1. Total Sum: Average * Number of items = 15 * 4 = 60.

2. Sum of three numbers: 10 + 20 + 25 = 55.

3. Fourth Number: Total Sum - Sum of three numbers = 60 - 55 = 5.

Geometry: Angles in a Triangle

Problem: In a triangle ABC, angle A = (3x - 7)°, angle B is twice angle A, and angle C = (4x + 6)°. Find the value of angle A + angle C.

Solution:

1. Angle B: 2 * Angle A = 2 * (3x - 7) = (6x - 14)°.

2. Triangle Angle Sum Theorem: Angle A + Angle B + Angle C = 180°.
   (3x - 7) + (6x - 14) + (4x + 6) = 180
   13x - 15 = 180
   13x = 195 => x = 15.

3. Angle A + Angle C: (3x - 7) + (4x + 6) = 7x - 1.
   Substitute x = 15: 7(15) - 1 = 105 - 1 = 104°.

Time and Work: Remaining Work

Problem: A and B can complete a piece of work in 12 days and 6 days respectively. They work together for 2 days, and then A leaves. In how many days will the whole work be completed?

Solution:

1. Total Work (LCM) and Efficiencies:

• LCM (12, 6) = 12 units.

• Efficiency of A = 12 / 12 = 1 unit/day.

• Efficiency of B = 12 / 6 = 2 units/day.

2. Work done together (2 days):

• Combined Efficiency (A+B) = 1 + 2 = 3 units/day.

• Work done = 3 units/day * 2 days = 6 units.

3. Remaining Work: 12 - 6 = 6 units.

4. Time for B to complete remaining work: 6 units / 2 units/day = 3 days.

5. Total Time: Time worked together + Time B worked alone = 2 days + 3 days = 5 days.

Profit and Loss: Unit Price

Problem: Surjit buys 20 kg of sugar for ₹900. He then sells 18 kg of sugar for ₹918. What is Surjit's profit percentage?

Solution:

1. Cost Price (CP) per kg: ₹900 / 20 kg = ₹45/kg.

2. Selling Price (SP) per kg: ₹918 / 18 kg = ₹51/kg.

3. Profit per kg: ₹51 - ₹45 = ₹6/kg.

4. Profit Percentage: (Profit per kg / CP per kg) * 100 = (₹6 / ₹45) * 100 = (2 / 15) * 100 = 200 / 15 = 13 and 1/3%.

Profit & Loss: Discount, CP, MP

Problem: Ravi buys an article at a 25% discount on the marked price. He then sells it for ₹1320, making a profit of 10%. What was the marked price of the article?

Solution:

1. Calculate Cost Price (CP) for Ravi:
   SP = ₹1320, Profit = 10%. So, ₹1320 represents 110% of CP.
   CP = ₹1320 / 1.10 = ₹1200.

2. Calculate Marked Price (MP):
   Ravi bought at 25% discount on MP, so his CP (₹1200) is 75% of MP.
   MP = ₹1200 / 0.75 = ₹1200 / (3/4) = ₹1200 * (4/3) = ₹1600.

Mensuration: Ratio of Sphere Volumes

Problem: The ratio of the radii of two spheres is 2:3. If the radius of the first sphere is increased by 50% and the radius of the second sphere is increased by 20%, find the ratio of their new volumes.

Solution:

1. Initial Radii Ratio: R1_initial : R2_initial = 2 : 3.

2. New Radii:

• R1_new = 2 * (1 + 0.50) = 2 * 1.5 = 3.

• R2_new = 3 * (1 + 0.20) = 3 * 1.2 = 3.6.

3. Ratio of New Radii: 3 : 3.6 = 30 : 36 = 5 : 6.

4. Volume Ratio (V ∝ r³):
   V1_new : V2_new = (R1_new)³ : (R2_new)³ = 5³ : 6³ = 125 : 216.

Third Proportion

Problem: Find the third proportion to 8 and 48.2.

Solution:

For three numbers a, b, and c in proportion (a:b = b:c), the third proportion c is given by c = b² / a.

Here, a = 8 and b = 48.2.

Third Proportion = (48.2)² / 8 = (48.2 * 48.2) / 8 = 290.49.

(Memory Tip: Digital Sum can verify complex calculations. Add digits of each number until a single digit remains (treating 9 as 0). Perform the operation on these sums. The digital sum of the result should match. This helps eliminate incorrect options quickly.)

Partnership: Profit Sharing

Problem: Two friends, P and Q, start a business. P invests ₹3000 and Q invests ₹600. At the end of the year, the business earns a total profit of ₹54,846. What is P's share of the profit?

Solution:

1. Investment Ratio (P:Q): ₹3000 : ₹600 = 5 : 1.

2. Total Ratio Parts: 5 + 1 = 6.

3. Value of 1 Part: Total Profit / Total Parts = ₹54,846 / 6 = ₹9,141.

4. P's Share: P's ratio part * Value of 1 part = 5 * ₹9,141 = ₹45,705.

Number Theory: LCM, HCF, and Product of Numbers

Problem: Two positive numbers are in the ratio 2:3. If the product of their LCM and HCF is 294, find the sum of the two numbers.

Solution:

1. Property: For two numbers N1 and N2, N1 * N2 = LCM * HCF.

2. Represent Numbers: Let the numbers be 2x and 3x.

3. Apply Property: (2x) * (3x) = 294
   6x² = 294
   x² = 49 => x = 7.

4. Sum of Numbers: 2x + 3x = 5x = 5 * 7 = 35.

Review of Decimal Division Problem

Problem: Given that 324 * 24 = 7776. Now, calculate 7.776 / 2.4.

Solution:

1. Base Division: From the given, 7776 / 24 = 324.

2. Adjust for Decimals: To calculate 7.776 / 2.4, multiply numerator and denominator by 10 to remove the decimal from the divisor:
   (7.776 * 10) / (2.4 * 10) = 77.76 / 24.

3. Perform Division: Since 7776 / 24 = 324, then 77.76 / 24 will have the decimal point shifted two places from the right.
   Therefore, 77.76 / 24 = 3.24.

Homework Problem - Rectangle Area Calculation

Problem Statement: If the diagonal of a rectangle is 10 cm long and its perimeter is 28 cm, what is the area of the rectangle?

Solution:

1. Given diagonal = 10 cm, perimeter = 28 cm.

2. For a diagonal of 10 cm, common Pythagorean triples (3:4:5 scaled by 2) suggest sides of 6 cm and 8 cm (6² + 8² = 36 + 64 = 100 = 10²).

3. Verify perimeter: 2 * (6 cm + 8 cm) = 2 * 14 cm = 28 cm. This matches.

4. Area of the rectangle = length × width = 6 cm × 8 cm = 48 cm².