RRB Group D 2026 Exam Review 17 Aug All Shift Maths By Shubham Sir

RRB Group D 2026 Exam Review 17 Aug All Shift Maths By Shubham Sir: The RRB Group D 2026 Mathematics paper held on August 17, 2026, included questions from arithmetic, algebra, geometry, trigonometry, number system, percentage, ratio-proportion, speed-time-distance, compound interest and average.

This exam review by Shubham Sir covers important questions, concepts and step-by-step solutions to help candidates understand the difficulty level and improve their preparation.

Also Read: 

• RRB Group D Admit Card 2026 Out

• RRB Group D City Intimation Slip 2026

RRB Group D 2026 Maths Exam Review: 17 August

The August 17 Mathematics questions tested candidates on both basic concepts and calculation-based applications. Important topics included rhombus, trigonometric identities, speed and distance, divisibility rules, percentage, discounts, trains, ratio and proportion, compound interest, linear equations and averages.

Below are the key questions and their solutions from the RRB Group D 2026 August 17 shift.

Rhombus Perimeter Calculation

Question: A rhombus has diagonals of 224 m and 420 m. Find its perimeter.

The diagonals of a rhombus bisect each other at right angles.

• Half of 224 = 112 m

• Half of 420 = 210 m

• The ratio is 112:210 = 8:15

• Using the Pythagorean triplet 8:15:17, the side = 17 × 14 = 238 m

• Perimeter = 4 × 238 = 952 m

Answer: 952 m

Trigonometric Identity-Based Question

Question: If 6x = cosec θ and 6/x = cot θ, find 18(x² − 1/x²).

Squaring both equations:

• 36x² = cosec²θ

• 36/x² = cot²θ

Using the identity:

cosec²θ − cot²θ = 1

Therefore,

36x² − 36/x² = 1

So,

x² − 1/x² = 1/36

Hence,

18(x² − 1/x²) = 18/36 = 1/2

Answer: 1/2

Speed, Time and Distance

Question: A person has to cover 907 km in 14 hours. He travels the first 5 hours at 32 km/h and covers the next 81 km at 27 km/h. At what speed should he travel for the remaining distance?

First 5-hour distance:

32 × 5 = 160 km

Time taken to cover 81 km:

81/27 = 3 hours

Thus:

• Distance covered = 160 + 81 = 241 km

• Time used = 5 + 3 = 8 hours

• Remaining distance = 907 − 241 = 666 km

• Remaining time = 14 − 8 = 6 hours

Required speed:

666/6 = 111 km/h

Answer: 111 km/h

Divisibility Rule for 4

Question: Which digit should replace x in 2x2 so that the number is divisible by 4?

A number is divisible by 4 if its last two digits are divisible by 4.

The last two digits are x2. Among the possible digits, 12 is divisible by 4.

Therefore:

x = 1

Answer: 1

Number Pattern and Square Root

Question: Find √(47² + 47 + 48).

Observe:

n² + n + (n + 1) = n² + 2n + 1 = (n + 1)²

For n = 47:

√(47² + 47 + 48) = √48² = 48

Answer: 48

Overall Discount Percentage

Question: A television marked at ₹50,000 is discounted by 15% on 70% of its price and by 5% on the remaining 30%. Find the overall discount percentage.

Discount on 70%:

15% × 70% = 10.5%

Discount on 30%:

5% × 30% = 1.5%

Total discount:

10.5% + 1.5% = 12%

Answer: 12%

Train Problem: Relative Speed

Train X is 200 m long and travels at 72 km/h. Train Y is 300 m long. The trains cross each other in 15 seconds while travelling in opposite directions.

Convert 72 km/h into m/s:

72 × 5/18 = 20 m/s

Total distance while crossing:

200 + 300 = 500 m

Relative speed:

500/15 = 100/3 m/s

Therefore:

100/3 = 20 + Speed of Y

Speed of Y = 40/3 m/s

Converting into km/h:

40/3 × 18/5 = 48 km/h

If both trains move in the same direction, their relative speed is:

72 − 48 = 24 km/h

24 km/h = 20/3 m/s

Time required to overtake:

500 ÷ (20/3) = 75 seconds

Answer: Train Y = 48 km/h; overtaking time = 75 seconds

Successive Percentage Changes

Question: An item is marked at ₹4,000. A discount of 25% is offered, after which its price is increased by 20%. Find the final price.

After 25% discount:

₹4,000 × 75/100 = ₹3,000

After 20% increase:

₹3,000 × 120/100 = ₹3,600

Answer: ₹3,600

Ratio of Speeds and Distance

Question: The speeds of two cars are in the ratio 5:7. The faster car covers 140 km in 2 hours. How much distance will the slower car cover in the same time?

Speed of faster car:

140/2 = 70 km/h

Since the speed ratio is 5:7:

Slower car's speed = 70 × 5/7 = 50 km/h

Distance covered in 2 hours:

50 × 2 = 100 km

Answer: 100 km

Ratio and Proportion After Donation

Question: ₹2,150 is divided between Sukhdev and Baldev in the ratio 20:23. If both donate ₹100 from their shares, find their new ratio.

Total parts:

20 + 23 = 43

Value of one part:

₹2,150/43 = ₹50

Sukhdev's share:

20 × 50 = ₹1,000

Baldev's share:

23 × 50 = ₹1,150

After donating ₹100 each:

• Sukhdev = ₹900

• Baldev = ₹1,050

New ratio:

900:1050 = 6:7

Answer: 6:7

Comparing Discount Schemes

Consider the following schemes:

Scheme 1: Two successive discounts of 24% and 27%

Effective discount:

24 + 27 − (24 × 27)/100 = 44.52%

Scheme 2: Buy 5, Get 3 Free

Total items = 8

Free items = 3

Discount = 3/8 × 100 = 37.5%

Scheme 3: Buy 6, Get 10

This means 4 items are free.

Discount = 4/10 × 100 = 40%

Therefore, the minimum discount is 37.5%, offered under Scheme 2.

Answer: Scheme 2

Speed and Time: Inverse Proportion

Question: A woman decreases her speed by 25%. By what fraction of her usual time will she be late?

A 25% decrease means the new speed becomes:

100% − 25% = 75%

Therefore, speed ratio:

Original : New = 4:3

Since speed and time are inversely proportional:

Original Time : New Time = 3:4

Increase in time = 4 − 3 = 1 unit

Fraction of usual time by which she is late:

1/3

Answer: 1/3 of her usual time

Compound Interest for the Second Year

Question: The compound interest for the second year is ₹5,214 at an annual interest rate of 10%. Find the principal.

Interest during the second year is calculated on the amount after the first year.

Therefore, second-year interest = 10% of P + 10% of 10% of P

= 10% + 1%

= 11% of P

Thus:

11% of P = ₹5,214

P = 5,214 × 100/11

P = ₹47,400

Answer: ₹47,400

Trigonometric Ratios of Complementary Angles

Question: If sin 6A = cos 12A, find tan 9A + cot 9A.

Using:

sin θ = cos(90° − θ)

Therefore:

6A + 12A = 90°

18A = 90°

A = 5°

So:

9A = 45°

Therefore:

tan 45° + cot 45°

= 1 + 1

= 2

Answer: 2

Simplification Question

One of the questions involved a sequence of calculations beginning with:

2.2 − 1.8 = 0.4

Following the subsequent operations step by step, the intermediate values obtained were 2.8, 0.5 and 6, leading to the final value:

Answer: 4.5

Linear Equation and Ratio

Question: The sum of two numbers is 72. If the larger number is increased by 8 and the smaller number is decreased by 6, their new ratio becomes 3:2. Find the smaller number.

Let the larger number be x and the smaller number be y.

x + y = 72

Also,

(x + 8)/(y − 6) = 3/2

Cross multiplication gives:

2(x + 8) = 3(y − 6)

2x + 16 = 3y − 18

2x − 3y = −34

From x + y = 72:

x = 72 − y

Substituting:

2(72 − y) − 3y = −34

144 − 5y = −34

5y = 178

y = 35.6

Answer: 35.6

Mean and Average

Question: The average of five numbers 1, 2, 3, 5 and x is 2.4. Find x.

Sum of the numbers:

1 + 2 + 3 + 5 + x = 11 + x

Using the average formula:

(11 + x)/5 = 2.4

11 + x = 12

Therefore:

x = 1

Answer: 1

RRB Group D 2026 Maths Exam Review: Key Topics

The August 17 Mathematics questions covered a wide range of topics. Candidates should focus on the following areas for upcoming shifts and preparation:

• Percentage and successive percentage

• Profit, loss and discount

• Ratio and proportion

• Speed, time and distance

• Train problems

• Compound interest

• Average and mean

• Linear equations

• Number system and divisibility

• Geometry

• Trigonometry

• Simplification

• Basic algebra

RRB Group D 2026 Exam Review: Conclusion

The RRB Group D 2026 Maths paper of August 17 featured questions based largely on fundamental concepts, formulas and calculation skills. Candidates preparing for the examination should revise percentage, ratio, speed-time-distance, trigonometry, geometry, number system and arithmetic regularly. Practising similar questions with shortcuts can help improve both accuracy and time management in the RRB Group D exam.