Home Test › RRB Group D 2025 Question Paper – 27 November Shift 1 | Set 2 (Q26–Q50) Mathematics

RRB Group D 2025 Question Paper – 27 November Shift 1 | Set 2 (Q26–Q50) Mathematics

RRB Group D 2025 Question Paper – 27 November Shift 1 | Set 2 (Q26–Q50) Mathematics

✍️ Prakash Joshi 📅 February 27, 20262 views 📥 Download PDF

This is Set 2 of the RRB Group D 2025 Exam conducted on 27 November 2025 (Shift 1). It contains Questions 26 to 50 (Mathematics section) in MCQ format based on the actual exam level. Attempt this section to evaluate your performance for CBT 1.

Score: 0 / 25

Questions

Q1. P is any point inside the rectangle ABCD. If PA = 98 cm, PB = 91 cm and PC = 21 cm, then the length of PD (in cm) is equal to:
PA² + PC² = PB² + PD² ⇒ 98² + 21² = 91² + PD² ⇒ PD = 42.
Q2. The vertices of a convex pentagon (in order) are P(0,0), Q(6,0), R(8,3), S(4,7), T(0,4). Find the area of the pentagon.
Coordinate polygon area formula ½|Σ(x1y2 − y1x2)| ⇒ Area = 42 sq units.
Q3. A metal rod is divided into three parts A, B and C. The lengths of A and B are in the ratio 9 : 20, and the lengths of C and B are in the ratio 5 : 11. If the difference between the lengths of A and C is 10 cm, find the total length (in cm) of the metal rod.
Let B = 220k ⇒ A = 99k and C = 100k ⇒ Difference = 10 ⇒ k = 1 ⇒ Total = 419.
Q4. Soniya got married 15 years ago. Her present age is 8/5 times her age at the time of her marriage. Her sister was 8 years younger to her at the time of her marriage. Find the present age of her sister.
Let marriage age = x ⇒ x + 15 = 8/5 x ⇒ x = 25 ⇒ Present age = 40 ⇒ Sister = 32.
Q5. The average age of 26 students of a class is 26 years. If the age of the teacher is also included, the average age of the whole group becomes 27 years. The age (in years) of the teacher is:
Students total = 26×26 = 676 ⇒ With teacher = 27×27 = 729 ⇒ Teacher age = 53.
Q6. If Δ × 2.2 + 0.33 × 3/4 − 1/20 × 0.55 = 0.99, then the value of Δ is:
0.33 × 3/4 = 0.2475, 1/20 × 0.55 = 0.0275, so 2.2Δ + 0.22 = 0.99, hence 2.2Δ = 0.77, Δ = 0.35.
Q7. If 20% of a number is added to 90, then the result is the same number. 80% of the same number is:
x = 90 + 0.2x ⇒ 0.8x = 90 ⇒ x = 112.5 ⇒ 80% = 90.
Q8. Smriti goes to a shopping mall at a speed of 21 km/hr and returns at a speed of 69 km/hr. Find her average speed (in km/hr) for the entire trip.
Average speed = 2xy/(x+y) ⇒ 2×21×69/90 = 32.2 km/hr.
Q9. An article was bought for ₹8,900. Its price was marked up by 40%. Thereafter, it was sold at a discount of 5% on the marked price. What was the profit percentage on the transaction?
MP = 8900×1.4 ⇒ SP = MP×0.95 ⇒ Profit ≈ 33%.
Q10. If p = 6, q = −3, then the value of p³ – 3p² + 3p + 3q + 3q² + q³ is:
Substitute p = 6 and q = −3 ⇒ Simplified value = 61.
Q11. Which of the following ratios is equivalent to 3 : 2?
3:2 = 1.5 and 54:36 = 1.5 ⇒ Equivalent ratio.
Q12. The cost of a washing machine is 40% less than the cost of a TV. If the cost of the washing machine increases by 52% and that of the TV decreases by 76%, then what is the percentage change in the total cost of 5 washing machines and 2 TVs?
Assume TV = 100 and WM = 60 ⇒ Apply % changes ⇒ Net decrease = 3%.
Q13. If tanθ = 12⁄5, what is the value of secθ?
If tanθ = 12⁄5, take opposite = 12 and adjacent = 5. Hypotenuse = √(12² + 5²) = 13. So secθ = hypotenuse⁄adjacent = 13⁄5.
Q14. The total surface area of a solid right circular cylinder is 542 cm². Its curved surface area is two-fifth of its total surface area. Find the curved surface area of the cylinder.
CSA = 2/5 × 542 = 216.8 cm².
Q15. Two numbers have an HCF of 18 and an LCM of 1512. If one of the numbers is 126, find the positive difference between these number.
Other number = (HCF×LCM)/126 = 216 ⇒ Difference = 90.
Q16. 1⁄20 is what percentage of 10⁄40?
10⁄40 = 1⁄4. So (1⁄20) ÷ (1⁄4) = (1⁄20) × 4 = 4⁄20 = 1⁄5 = 20%.
Q17. In statistics out of 100, the marks of 21 students in final exams are as 90, 95, 95, 94, 90, 85, 84, 83, 85, 81, 92, 93, 82, 78, 79, 81, 80, 82, 85, 76, 85, then the mode of data is:
85 appears maximum times ⇒ Mode = 85.
Q18. A fruit vendor bought 150 mangoes. He sold them at such a price that the selling price of 120 mangoes equals the cost price of 150 mangoes. Find his profit percentage.
SP per mango = 150/120 = 1.25 CP ⇒ Profit = 25%.
Q19. Three pipes, A, B and C, can fill a tank from empty to full in 40 minutes, 20 minutes and 30 minutes, respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R, respectively. What is the proportion of solution Q in the liquid in the tank after 9 minutes?
Rates: A = 1⁄40, B = 1⁄20, C = 1⁄30. Total rate = 13⁄120 per minute. In 9 minutes, total filled = 117⁄120 = 39⁄40. Amount by B in 9 minutes = 9 × 1⁄20 = 9⁄20. Proportion of Q = (9⁄20) ÷ (39⁄40) = 6⁄13.
Q20. Two-fifth of a container is filled with blue liquid, one-third of the remaining container is filled with black liquid, five-sixth of the still remaining container is filled with yellow liquid and the remaining 4.2 litres is of white colour. Find the total capacity of the container.
Remaining fraction = 1/15 ⇒ Total capacity = 4.2×15 = 63 L.
Q21. A 325 m long train overtakes a man moving at a speed of 5 km/hr (in the same direction) in 45 seconds. How much time (in seconds) will it take this train to completely cross another 440 m long train, moving in the opposite direction at a speed of 20 km/hr?
Relative speed = 26 km/hr ⇒ Opposite total = 46 km/hr ⇒ Time ≈ 52 sec.
Q22. Simplify: [3 1⁄4 ÷ {1 1⁄4 − 0.5(2 1⁄2 − 1⁄12)}] ÷ (5 × 1⁄15)
Convert mixed fractions: 3 1⁄4 = 13⁄4, 1 1⁄4 = 5⁄4, 2 1⁄2 = 5⁄2. Inside bracket: (5⁄2 − 1⁄12) = 29⁄12. Then 0.5 × 29⁄12 = 29⁄24. So 5⁄4 − 29⁄24 = 1⁄24. Thus 13⁄4 ÷ 1⁄24 = 78. Now divide by (5 × 1⁄15) = 1⁄3. So 78 ÷ 1⁄3 = 234.
Q23. A and B working together can do a piece of work in 6 days. B alone can do the same work in 12 days. How long (in days) will A alone take to do double the work?
A rate = 1/12 ⇒ Double work = 2 units ⇒ Time = 24 days.
Q24. A scheme like ‘Buy 6, Get 4 Free’ on the same kind of articles with the same MRP attracts a _____ % discount.
Buy 10 for price of 6 ⇒ Discount = 4/10 = 40%.
Q25. The amount on a sum of ₹6,800 at 15% per annum compound interest, compounded annually, in 2 years' time, will be:
Amount = 6800×(1.15)² ⇒ 8993 approx.

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