RRB Group D 2026 Paper 7 August: All Shift Exam Review Maths By Shubham Difficulty Level

RRB Group D 07 Aug All Shift Exam Review Maths By Shubham Sir: The Railway Recruitment Board (RRB) is organising the RRB Group D CBT exam from 3rd August to 25th August 2026. It is for candidates who applied for the Group D exam in the Railways. Most candidates struggle with the complex calculations involved in the Maths subject of the CBT exam.

RRB Group D 07 Aug All Shift Exam Review Maths by Shubham Sir can help candidates to get insights into the important topics and questions discussed based on the previous year paper trends. Practice and mentor guidance are useful for candidates to improve their scores in the Maths paper. 

Also Read: 

• RRB Group D Admit Card 2026 Out

• RRB Group D City Intimation Slip 2026

Topics Discussed In RRB Group D 7 August 2026 Maths Review

Shubham Sir has conducted a session on the RRB Group D Maths review. All the important topics with the questions were discussed with the candidates. It can help them revise the formulas for the examination. 

• Profit and Loss: Impact of Simultaneous Percentage Reduction on CP and SP

• LCM of Decimal Numbers

• Discounts and Delivery Charges

• Work and Time: Individual and Combined Work

• Age Problems: Using Ratios

• Weighted Average: Deviation Method for Simplification

• Compound Interest for Fractional Time Periods

• Income, Expenditure, and Savings Calculations

• Height and Distance: Angle of Elevation Change

• BODMAS: Order of Operations

• Simple Interest: Calculating Principal

• Trapezium Properties: Midpoints of Diagonals and Sides

• Speed, Time, and Distance: "Late/Early" Problems

Questions Covered In RRB Group D Maths Session

1. Profit and Loss: Impact of Simultaneous Percentage Reduction on CP and SP

Problem Statement: An article is sold at a profit of 20%. If both the Cost Price (CP) and Selling Price (SP) are reduced by 10%, what is the new Profit Per cent?

Correct Principle and Efficient Approach:

When both the Cost Price and Selling Price are reduced by the same percentage, the profit percentage remains unchanged. This principle holds true because the ratio between profit and cost price is preserved when both components are scaled down by the same factor. Therefore, if the initial profit was 20%, the profit percentage will still be 20% after both CP and SP are reduced by 10%. This method allows for a quick answer during an exam (Memory Tip: When CP and SP change by the same percentage, profit per cent is constant).

2. LCM of Decimal Numbers

Problem Statement: Find the Least Common Multiple (LCM) of the numbers 3, 7.2, and 0.64.

Methodology for LCM of Decimals:

1. Convert to Fractions: Express all numbers as fractions with a common denominator.

• 3 = 300/100

• 7.2 = 72/10 = 720/100

• 0.64 = 64/100

2. Calculate LCM: To find the LCM of decimal numbers, calculate the LCM of the numerators and divide it by the HCF (Highest Common Factor) of the common denominators.

• LCM (300, 720, 64) / HCF (100, 100, 100) = LCM (300, 720, 64) / 100.

• Prime factorization:

• 300 = 2² * 3 * 5²

• 720 = 2⁴ * 3² * 5

• 64 = 2⁶

• LCM (300, 720, 64) = 2⁶ * 3² * 5² = 64 * 9 * 25 = 14400.

• Final LCM = 14400 / 100 = 144.

3. Discounts and Delivery Charges

Problem Statement: A customer orders food worth ₹1000. They receive a 20% discount from the food delivery app, followed by an additional 5% discount using a discount card. An ₹80 delivery charge is then applied to the final bill. What is the final amount paid?

Solution:

1. First Discount (20%):

• Amount after 20% discount = ₹1000 * (1 - 0.20) = ₹1000 * 0.80 = ₹800.

2. Second Discount (Additional 5%):

• Amount after additional 5% discount = ₹800 * (1 - 0.05) = ₹800 * 0.95 = ₹760.

3. Add Delivery Charge:

• Final amount paid = ₹760 + ₹80 = ₹840.
  This problem demonstrates the straightforward application of successive discounts and simple addition, often appearing lengthy but being direct in calculation.

4. Work and Time: Individual and Combined Work

Problem Statement: A and B can complete a task in 12 and 20 days respectively. They work together for 3 days. After 3 days, A leaves. How many days will B take to complete the remaining work alone?

Methodology (LCM Method):

1. Calculate Total Work: Find the LCM of the individual days.

• LCM (12, 20) = 60 units.

2. Calculate Individual Efficiencies:

• A's efficiency = 60 units / 12 days = 5 units/day.

• B's efficiency = 60 units / 20 days = 3 units/day.

3. Calculate Combined Work:

• Combined efficiency = 5 + 3 = 8 units/day.

• Work done in 3 days = 8 units/day * 3 days = 24 units.

4. Calculate Remaining Work:

• Remaining work = 60 units - 24 units = 36 units.

5. Calculate Time for B:

• Time for B = 36 units / 3 units/day = 12 days.

5. Age Problems: Using Ratios

Problem Statement: Four years ago, Ravi was three times as old as Kaviraj. At present, Ravi is twice as old as Kaviraj. What is Ravi's current age?

Methodology (Ratio Method with Age Gap Equalization):

1. Write down Ratios:

• 4 years ago (Ravi : Kaviraj) = 3 : 1 (Difference = 2)

• Present (Ravi : Kaviraj) = 2 : 1 (Difference = 1)

2. Equalize Age Gaps: Multiply the present ratio by 2 to match the difference.

• Present (Ravi : Kaviraj) = 2 * (2 : 1) = 4 : 2.

3. Compare Ratios Across Time:

• Ravi's age ratio changed from 3 (4 years ago) to 4 (Present), a change of 1 unit.

• Kaviraj's age ratio changed from 1 (4 years ago) to 2 (Present), also a change of 1 unit.

4. Relate Ratio Units to Actual Time: The change from "4 years ago" to "Present" is 4 years.

• 1 unit in the ratio corresponds to 4 years.

5. Calculate Ravi's Present Age: Ravi's present age in the ratio is 4 units.

• Ravi's present age = 4 units * 4 years/unit = 16 years.

6. Weighted Average: Deviation Method for Simplification

Problem Statement: A factory produces three types of widgets (A, B, C).

• Type A: 12 units, Average Weight 47 kg

• Type B: 8 units, Average Weight 62 kg

• Type C: 15 units, Average Weight 53 kg
  What is the average weight of the widgets produced in that week?

Efficient Approach: Deviation Method

The Deviation Method simplifies average calculations by working with smaller numbers.

1. Assume a Mean: Let the assumed average be 55 kg.

2. Calculate Deviations:

• For A: 47 - 55 = -8

• For B: 62 - 55 = +7

• For C: 53 - 55 = -2

3. Calculate Weighted Deviations:

• For A: 12 * (-8) = -96

• For B: 8 * (+7) = +56

• For C: 15 * (-2) = -30

4. Sum Weighted Deviations: -96 + 56 - 30 = -70

5. Calculate Mean Deviation: Divide by total units (12 + 8 + 15 = 35).

• Mean Deviation = -70 / 35 = -2

6. Calculate Correct Average:

• Correct Average = Assumed Mean + Mean Deviation = 55 + (-2) = 53 kg.
  This method significantly reduces calculation complexity compared to direct computation.

7. Compound Interest for Fractional Time Periods

Problem Statement: Find the Compound Interest (CI) on a principal of ₹25,600 at an annual rate of 12.5% for a period of 2 years and 3 months.

Methodology (Ratio Method):

1. Convert Annual Rate to Fraction: 12.5% = 1/8.

2. Handle Full Years: For each full year, the principal-to-amount ratio is 8:9.

• Year 1: 8:9

• Year 2: 8:9

3. Handle Fractional Period (3 Months):

• Rate for 3 months = (1/8) * (3/12) = (1/8) * (1/4) = 1/32.

• For the 3-month period, the ratio is 32:33.

4. Calculate Total Principal-to-Amount Ratio:

• Multiply ratios: (8 * 8 * 32) : (9 * 9 * 33) = 2048 : 2673.

5. Relate Ratio to Principal:

• 2048 units = ₹25,600

• 1 unit = ₹25,600 / 2048 = ₹12.50.

6. Calculate Compound Interest (CI):

• CI in units = 2673 - 2048 = 625 units.

• CI in rupees = 625 * ₹12.50 = ₹7812.50.

8. Income, Expenditure, and Savings Calculations

Problem Statement: A family's monthly income was ₹45,000 and monthly expenditure was ₹36,000. If their monthly income increases by 10% and monthly expenditure increases by 20%, what are their new savings?

Calculations:

1. Original Savings: ₹45,000 - ₹36,000 = ₹9,000.

2. New Income:

• Increase = 10% of ₹45,000 = ₹4,500.

• New Income = ₹45,000 + ₹4,500 = ₹49,500.

3. New Expenditure:

• Increase = 20% of ₹36,000 = ₹7,200.

• New Expenditure = ₹36,000 + ₹7,200 = ₹43,200.

4. New Savings:

• New Income - New Expenditure = ₹49,500 - ₹43,200 = ₹6,300.

9. Height and Distance: Angle of Elevation Change

Problem Statement: The angle of elevation of the sun changes from 60° to 45°, causing the shadow of a tower to increase by 20 meters. Find the height of the tower.

Trigonometric Approach:

Let h be the height of the tower and x be the initial length of the shadow.

1. At 60° elevation:

• tan 60° = h / x => √3 = h / x => x = h / √3.

2. At 45° elevation: The shadow becomes x + 20.

• tan 45° = h / (x + 20) => 1 = h / (x + 20) => h = x + 20.

3. Substitute and Solve for h:

• Substitute x in the second equation: h = (h / √3) + 20.

• h - (h / √3) = 20

• h (1 - 1/√3) = 20

• h ( (√3 - 1) / √3 ) = 20

• h = 20√3 / (√3 - 1)

4. Rationalise the Denominator:

• h = [20√3 / (√3 - 1)] * [(√3 + 1) / (√3 + 1)]

• h = 20√3(√3 + 1) / (3 - 1)

• h = 20(3 + √3) / 2

• h = 10(3 + √3)

5. Substitute the value of √3 (≈ 1.732):

• h = 10(3 + 1.732) = 10(4.732) = **47.32 meters**.

10. BODMAS: Order of Operations

Problem Statement: Simplify the following expression:

(7/13 * 13/33) ÷ ( (16/8 * 1/3) + 5/11 ) + (7/3) ÷ ( (3/20) of (20/37) )

Step-by-Step Simplification:

This problem requires careful application of the BODMAS/PEMDAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction). "Of" acts as multiplication and has precedence over division.

1. First Main Term: (7/13 * 13/33) ÷ ( (16/8 * 1/3) + 5/11 )

• (7/13 * 13/33) simplifies to 7/33.

• (16/8 * 1/3) simplifies to 2 * 1/3 = 2/3.

• The divisor term (2/3 + 5/11) becomes (22 + 15)/33 = 37/33.

• So, the first main term is (7/33) ÷ (37/33) = (7/33) * (33/37) = 7/37.

2. Second Main Term: (7/3) ÷ ( (3/20) of (20/37) )

• (3/20 of 20/37) simplifies to 3/37.

• Following the lecture's simplification, this entire term evaluates to 7/37.

3. Final Addition:

• Add the results of the two main terms: 7/37 + 7/37 = **14/37**.

(Memory Tip: In BODMAS problems, checking options based on common denominators or prime factors can sometimes aid in elimination).

11. Simple Interest: Calculating Principal

Problem Statement: If a sum is invested at an 8% annual simple interest rate for 9 months, and it yields an interest of ₹18,000, find the principal amount invested.

Methodology (Effective Rate Method):

1. Calculate Effective Rate (RT%): For simple interest, the total percentage interest is R * T.

• Annual Rate (R) = 8%

• Time (T) = 9 months = 9/12 years = 3/4 years.

• Effective Rate = 8% * (3/4) = 6%.

2. Relate Effective Rate to Interest Earned: The 6% effective rate corresponds to the interest earned.

• If 6% = ₹18,000

• Then 1% = ₹18,000 / 6 = ₹3,000.

3. Calculate Principal (100%):

• Principal = 100% = 100 * ₹3,000 = ₹3,00,000.
  This method allows for quick calculation without using the full SI formula.

12. Trapezium Properties: Midpoints of Diagonals and Sides

Problem Statement: ABCD is a quadrilateral where AB is parallel to CD. AC and BD are its diagonals. E and F are the midpoints of the diagonals AC and BD respectively. If AB = 23 cm and CD = 35 cm, find the length of EF.

Definition: A quadrilateral with exactly one pair of parallel sides (AB || CD) is a Trapezium (or Trapezoid).

Properties and Formulas for Trapezium:

Solution for EF:

• Given parallel sides: AB = 23 cm, CD = 35 cm.

• Using the formula for midpoints of diagonals:

• EF = |35 - 23| / 2 = 12 / 2 = 6 cm.

13. Speed, Time, and Distance: "Let/Early" Problems

Problem Statement: A train covers a certain distance. If its speed were 14 km/h more, it would take 35 minutes less. If its speed were 10 km/h less, it would take 30 minutes more. Find the distance covered by the train.

Methodology (Using "Let/Early Problems" Distance Formula):

The general formula for distance when speed changes result in time differences is:

Distance = (S₁ * S₂) / |S₁ - S₂| * ΔT

Let original speed be S km/h.

1. Formulate Equations:

• Case 1 (Speed increases):

• D = [S * (S + 14)] / 14 * (35/60)

• Case 2 (Speed decreases):

• D = [S * (S - 10)] / 10 * (30/60)

2. Equate Distances to find Original Speed (S): [S * (S + 14)] / 14 * (35/60) = [S * (S - 10)] / 10 * (30/60)

• Simplifying and solving for S: S = 130 km/h.

3. Calculate Total Distance: Substitute S = 130 into either equation.

• Using Case 1: D = [130 * (130 + 14)] / 14 * (35/60)

• D = [130 * 144] / 14 * (35/60)

• D = 780 km.

Why Watch Shubham Sir's Exam Review?

Candidates preparing for the upcoming RRB Group D shifts can benefit from Shubham Sir's exam review. The review provides useful information based on questions shared by candidates who appeared in the examination.

Learn Memory-Based Questions

The review includes memory-based questions collected from different shifts. These questions help candidates understand the type of questions that are being asked in the examination. Practising similar questions can improve preparation.

Understand Easy Solution Methods

Shubham Sir explains Maths questions using simple and easy methods. These shortcut techniques help candidates solve questions in less time while maintaining accuracy.

Get Topic-Wise Coverage

The review explains questions from different Maths topics such as Simplification, Percentage, Profit and Loss, Ratio and Proportion, Average, Time and Work, Mensuration, Geometry, Algebra, and Data Interpretation. This helps candidates identify the most important chapters for revision.

Prepare Better for Upcoming Shifts

Watching the exam review can help candidates plan their revision more effectively. It provides a clear idea of the latest question pattern and important topics. Candidates can use this information to focus on areas that are more likely to appear in the upcoming shifts.