RRB Group D 2026 Paper 14 Aug All Shift Exam Details: Maths by Shubham Sir

The RRB Group D 2026 Maths paper conducted on August 14 featured questions from important topics such as Profit and Loss, Average, Ratio and Proportion, Simple Interest, Time and Work, Geometry, Mensuration, Fractions, and Percentage. The paper included both formula-based and calculation-oriented questions, testing candidates’ speed and accuracy.

In this article, Shubham Sir provides a detailed shift-wise analysis of the RRB Group D Maths paper, along with important concepts, formulas, and quick-solving approaches to help candidates understand the questions and improve their preparation. 

RRB Group D 2026 Maths Paper Analysis: August 14

The RRB Group D 2026 Maths paper tested candidates on a wide range of quantitative aptitude concepts. Questions were based on both direct formula application and calculation-based problems.

The major topics covered included:

• Profit and Loss

• Average

• Ratio and Proportion

• Simple Interest

• Time, Work and Earnings

• Geometry

• Mensuration

• Fractions

• Percentage

• Number-based problems

The following section explains the important concepts and approaches discussed by Shubham Sir for the August 14 all-shift paper.

Profit and Loss: Car Sale Problem

Profit percentage is calculated on the Cost Price (CP).

Formula:

Profit % = (Profit / CP) × 100

For example, if the Cost Price of a car is ₹5,50,000 and its Selling Price is ₹6,60,000:

• CP = ₹5,50,000

• SP = ₹6,60,000

• Profit = ₹6,60,000 − ₹5,50,000 = ₹1,10,000

Therefore,

Profit % = (1,10,000 / 5,50,000) × 100 = 20%

Thus, the profit is 20%.

Arithmetic Mean: Deviation Method

The assumed mean or deviation method can be used to calculate the average quickly, especially when the numbers are close to a convenient value.

Steps:

1. Choose a convenient assumed average.

2. Find the deviation of each number from the assumed average.

3. Add all the deviations.

4. Divide the total deviation by the number of observations.

5. Add the average deviation to the assumed average.

For example, if 50 is taken as the assumed mean and the total deviation is +10 for 10 observations:

Average deviation = 10/10 = 1

Therefore,

Actual Mean = 50 + 1 = 51

Volume of Cylinders: Melting and Recasting

When a solid object is melted and recast into another shape, its total volume remains unchanged.

The volume of a cylinder is:

V = πr²h

Suppose an original cylinder has radius R and height H. Its volume is:

V = πR²H

If it is recast into three cylinders, each having radius R/2 and height H, the volume of each new cylinder will be:

π(R/2)²H = 1/4 πR²H

Therefore, the total volume of three new cylinders is:

3/4 πR²H

Compared with the original volume, the new cylinders use 3/4 of the original volume. Hence, the unused volume is:

1 − 3/4 = 1/4

Therefore, 1/4 of the original volume remains unused.

Money Denominations Problem

Questions involving different denominations can be solved by forming equations or testing suitable values.

For example, suppose Soumya has ₹24,800 in ₹500 and ₹100 notes, and the number of ₹500 notes is four more than the number of ₹100 notes.

If the number of ₹100 notes is 38:

• Value of ₹100 notes = 38 × ₹100 = ₹3,800

• Number of ₹500 notes = 38 + 4 = 42

• Value of ₹500 notes = 42 × ₹500 = ₹21,000

• Total = ₹3,800 + ₹21,000 = ₹24,800

Therefore, the number of ₹100 notes is 38.

Ratio and Proportion: Comparing Red Balls

When comparing the proportion of a particular component in different ratios, the ratios can be converted to a common total.

Consider three bags:

• Bag A: Red : Blue = 2 : 3

• Bag B: Red : Blue = 7 : 3

• Bag C: Red : Blue = 8 : 7

The total parts are:

• A = 5

• B = 10

• C = 15

The LCM of 5, 10 and 15 is 30.

Converting each ratio to a total of 30:

• A = 12 : 18 → Red = 12

• B = 21 : 9 → Red = 21

• C = 16 : 14 → Red = 16

Therefore, Bag B has the highest proportion of red balls, with 21 red parts out of 30.

Profit Calculation with Changed Selling Price

Suppose an article is sold at 9/10 of its usual Selling Price and still gives a 20% profit.

Therefore:

9/10 × Original SP = 120% of CP

or

9/10 × Original SP = 6/5 CP

Hence,

Original SP = (6/5) × (10/9) × CP

Original SP = 4/3 CP

Therefore, the original Selling Price is 133⅓% of CP.

Thus, the profit percentage at the original Selling Price is:

133⅓% − 100% = 33⅓%

Distribution of Money in a Ratio

Suppose money is divided among P, Q and R in the ratio:

P : Q : R = 7 : 9 : 4

If P receives ₹1,500 more than R:

• Difference in ratio = 7 − 4 = 3 units

• 3 units = ₹1,500

• 1 unit = ₹500

The difference between Q and P is:

9 − 7 = 2 units

Therefore:

Difference = 2 × ₹500 = ₹1,000

Hence, Q receives ₹1,000 more than P.

Simple Interest: Finding the Rate

The formula for Simple Interest is:

SI = (P × R × T) / 100

If the Simple Interest earned in 12 years is 3/5 of the principal:

SI = 3P/5

Therefore:

3P/5 = (P × R × 12) / 100

Cancelling P:

3/5 = 12R/100

Solving for R:

R = 5%

Therefore, the rate of Simple Interest is 5% per annum.

Election Problem: Percentage of Votes

Suppose a candidate receives 41% of the total votes and loses the election by 97,632 votes.

The winning candidate receives:

100% − 41% = 59%

Difference in votes:

59% − 41% = 18%

Therefore:

18% = 97,632 votes

So,

1% = 97,632 ÷ 18 = 5,424 votes

The winning candidate received:

59 × 5,424 = 3,20,016 votes

Therefore, the winning candidate secured 3,20,016 votes.

Work, Time and Earnings

Work and earnings questions can be solved using the relationship between the number of persons, working hours, days and earnings.

Suppose 15 persons working 12 hours per day earn ₹18,000 per week. Find the earnings of 18 persons working 9 hours per day for the same number of days.

Using the proportional relationship: Earnings ∝ Persons × Hours

Therefore:

x = (18 × 9 × 18,000) / (15 × 12)

x = ₹16,200

Hence, the earnings are ₹16,200.

Geometry: Largest Circle Inside a Rectangle

The diameter of the largest circle that can be completely inscribed inside a rectangle is equal to the shorter side of the rectangle.

For a rectangle measuring 196 m × 271 m:

• Shorter side = 196 m

• Diameter = 196 m

• Radius = 196/2 = 98 m

Circumference:

C = 2πr

Taking π = 22/7:

C = 2 × 22/7 × 98

C = 616 m

Therefore, the circumference of the largest circle is 616 metres.

Comparing Fractions

Several methods can be used to identify the largest fraction.

1. Common Denominator

Convert all fractions to a common denominator and compare their numerators.

2. Cross-Multiplication

To compare a/b and c/d, compare:

a × d and b × c

The fraction corresponding to the larger product is greater.

3. Decimal Method

Convert each fraction into decimal form and compare the values.

4. Difference from 1

For fractions close to 1, calculate their difference from 1. The fraction with the smaller difference is larger.

For example:

46/51 = 1 − 5/51

Whereas:

48/90 = 1 − 42/90

Since 5/51 is smaller than 42/90, 46/51 is the larger fraction.

Simple Interest: Direct Formula

For a principal of ₹3,000, rate of interest of 8% per annum, and time of 6 years:

SI = (P × R × T) / 100

SI = (3000 × 8 × 6) / 100

SI = ₹1,440

Therefore, the Simple Interest is ₹1,440.

Profit and Loss: Change in Selling Price

Suppose an article is sold at a 10% loss. If its Selling Price is increased by ₹120, it results in a 5% profit.

At 10% loss:

SP₁ = 90% of CP

At 5% profit:

SP₂ = 105% of CP

Difference:

SP₂ − SP₁ = 105% − 90% = 15% of CP

Given that this difference is ₹120:

15% of CP = ₹120

Therefore:

CP = ₹120 × 100/15 = ₹800

Initial Selling Price:

SP₁ = 90% of ₹800 = ₹720

Hence:

• Cost Price = ₹800

• Initial Selling Price = ₹720

• Initial Loss = ₹80