RRB Group D Mathematics Previous Year Question Paper - All 17 Questions Solved out of 25 questions
Welcome aspirants! This comprehensive study post contains all 17 solved questions discussed in our video class for the RRB Group D Mathematics shift paper. Every question features standard mathematical solutions along with Wisdom Helps mental math shortcuts designed for high-speed calculation during competitive exams.
RRB Group D Previous Year Mathematics Solved Paper on YouTube
📌 Key Formula & Concept Reference Sheet
| Mathematics Topic | Core Formula / Wisdom Helps Speed Shortcut |
|---|---|
| Average | Total Sum = n × Average |
| Average Speed | Avg Speed = 2xy / (x + y) (When distance is equal) |
| LCM & HCF | LCM × HCF = Number1 × Number2 |
| Compound Interest (2 Yrs) | Effective Interest Rate = x + y + (x × y) / 100 % |
| Promotional Discount % | Discount % = [Free Items / Total Items] × 100 |
| Time & Work | Total Work = Efficiency × Time |
📝 All 17 Solved Questions from the Exam Paper
Question: 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. Find the age of the teacher (in years).
Sum of ages of 26 students = 26 × 26 = 676 years
Sum of ages including teacher (27 people) = 27 × 27 = 729 years
Teacher's Age = 729 − 676 = 53 years
New Average = 27
Increase for existing 26 students = 26 × 1 = 26
Teacher's Age = 27 + 26 = 53 Years
Question: If 20% of a number is added to 90, then the result is the same number. Find 80% of that same number.
Let the number be x.
20% of x = (1/5)x
Given: (1/5)x + 90 = x ⇒ x − (1/5)x = 90 ⇒ (4/5)x = 90
x = (90 × 5) / 4 = 112.5
Now, 80% of the number = 112.5 × (80 / 100) = 90
Question: Smriti goes to a shopping mall at a speed of 21 km/h and returns at a speed of 69 km/h along the same route. Find her average speed for the entire journey.
Distance is constant, so apply:
Average Speed = 2xy / (x + y)
= (2 × 21 × 69) / (21 + 69) = 2898 / 90 = 32.2 km/h
Question: An article was bought for ₹8,900. It was marked up 40% above cost price and sold at a discount of 5% on the marked price. What was the overall profit percentage?
Ignore ₹8,900 as we need percentage profit.
Let Cost Price (CP) = 100%
Marked Price (MP) = 140%
Discount = 5% of 140 = 7%
Selling Price (SP) = 140 − 7 = 133%
Profit Percentage = 133% − 100% = 33%
Question: Given p = 6 and q = 3, evaluate the value of the algebraic expression:
p3 − 3p2 + 3p − 9 + 3q2 − q3.
Substitute p = 6 and q = 3 directly:
p3 = 216
− 3p2 = −3(36) = −108
+ 3p = 3(6) = 18
− 9
+ 3q2 = 3(9) = 27
− q3 = −27
Notice +27 and −27 cancel out.
Remaining = 216 − 108 + 18 − 9 = 234 − 117 = 117
Question: Which of the given ratios is equivalent to 3 : 2?
Check ratio option 54 : 36.
Dividing both numbers by 18 gives 54 / 18 = 3 and 36 / 18 = 2.
Result = 3 : 2
Question: 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 the cost of the TV decreases by 76%, find the percentage change in the total cost of purchasing 5 washing machines and 2 TVs.
Let original TV Cost = 100 ⇒ Washing Machine Cost = 60
Original Cost for 5 WMs + 2 TVs = (5 × 60) + (2 × 100) = 300 + 200 = 500
New Washing Machine Price = 60 × 1.52 = 91.2
New TV Price = 100 × (1 − 0.76) = 24
New Cost for 5 WMs + 2 TVs = (5 × 91.2) + (2 × 24) = 456 + 48 = 504
Increase = 504 − 500 = 4
Percentage Increase = (4 / 500) × 100 = 0.8% Increase
Question: If tan θ = 12 / 5, what is the value of sec θ?
tan θ = Opposite / Adjacent = 12 / 5
Using Pythagorean Triplet (5, 12, 13), Hypotenuse = 13.
sec θ = Hypotenuse / Adjacent = 13 / 5
Question: The Total Surface Area (TSA) of a solid right circular cylinder is 542. If its Curved Surface Area (CSA) is 2/5 of its TSA, find its CSA.
CSA = (2 / 5) × 542
Multiply numerator and denominator by 2 to get a denominator of 10:
CSA = (4 × 542) / 10 = 2168 / 10 = 216.8
Question: Two numbers have an HCF of 18 and an LCM of 1512. If one number is 126, find the positive difference between the two numbers.
Formula: HCF × LCM = Number1 × Number2
18 × 1512 = 126 × Number2
Number2 = (18 × 1512) / 126 = 216
Difference = 216 − 126 = 90
Question: Express 1/20 as a percentage of 10/40.
Required Percentage = [(1/20) / (10/40)] × 100
= (1 / 20) × (40 / 10) × 100
= (40 / 200) × 100 = 20%
Question: Out of 100 marks, a group of 21 students scored the following marks in a test: 90, 85, 95, 85, 94, 85, 90, 85... Find the Mode of this dataset.
Mode is defined as the most frequently occurring value in a given dataset.
Counting the frequency: The score 85 appears 4 times (maximum frequency).
Mode = 85
Question: A fruit vendor bought 150 mangoes. He sold them such that the selling price of 120 mangoes equals the cost price of 150 mangoes. Find his profit percentage.
120 × SP = 150 × CP
SP / CP = 150 / 120 = 5 / 4
Cost Price = 4, Selling Price = 5 ⇒ Profit = 1
Profit % = (1 / 4) × 100 = 25%
Question: Three pipes A, B, and C can individually fill an empty tank in 40 minutes, 20 minutes, and 30 minutes respectively. Pipes A, B, and C discharge chemicals P, Q, and R. If all three pipes are opened together when the tank is empty, what is the proportion of chemical Q in the liquid after 9 minutes?
Total Tank Capacity = LCM(40, 20, 30) = 120 Units
Pipe A (P) Efficiency = 120 / 40 = 3 units/min
Pipe B (Q) Efficiency = 120 / 20 = 6 units/min
Pipe C (R) Efficiency = 120 / 30 = 4 units/min
Total filling rate = 3 + 6 + 4 = 13 units/min
In 9 minutes:
Amount of Q filled = 9 × 6
Total liquid filled = 9 × 13
Proportion of Q = (9 × 6) / (9 × 13) = 6 / 13
Question: A and B working together can finish a piece of work in 6 days. B alone can complete the same work in 12 days. How many days will A alone take to finish double the original work?
Let Total Work = LCM(6, 12) = 12 Units
Efficiency of (A + B) = 12 / 6 = 2 units/day
Efficiency of B = 12 / 12 = 1 unit/day
Efficiency of A = 2 − 1 = 1 unit/day
Double the original work = 12 × 2 = 24 units
Time required for A = 24 / 1 = 24 Days
Question: A store offers a scheme: "Buy 6, Get 4 Free". Find the equivalent discount percentage offered on the transaction.
Discount % = [Free Quantity / Total Quantity] × 100
Total Quantity = 6 + 4 = 10
Discount Percentage = (4 / 10) × 100 = 40%
Question: Find the total amount accumulated on a principal sum of ₹6,800 at 15% per annum interest compounded annually for 2 years.
Effective CI Rate for 2 years = 15 + 15 + (15 × 15)/100 = 32.25%
Total Amount Rate = 100% + 32.25% = 132.25%
Total Amount = 6800 × (132.25 / 100) = 68 × 132.25 = 8993
Digit Sum Verification: Digit Sum of 68 = 6+8 = 14 ⇒ 5. Digit Sum of 132.25 = 1+3+2+2+5 = 13 ⇒ 4. Product Digit Sum = 5 × 4 = 20 ⇒ 2. Digit Sum of option 8993 = 8+9+9+3 = 29 ⇒ 2 (Matches!).
💡 Wisdom Helps Exam Strategy for Railway Mathematics
- Master Percentage Fractions: Convert fractional values like 1/5 = 20% or 1/4 = 25% mentally to bypass scratchpad work.
- Use Option Elimination & Digit Sum: For heavy multiplication in Compound Interest or Mensuration, verify the Digit Sum of choices.
- Daily Practice Routine: Work through at least one full previous year shift paper daily to build question recognition speed.
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