Cracking the Mathematics section of the Railway Recruitment Board (RRB) exams, be it NTPC, Group D, or ALP, is a crucial step towards securing a coveted position in Indian Railways. This section tests your speed, accuracy, and fundamental understanding of mathematical concepts. With lakhs of candidates competing, mastering the most frequently asked questions can give you a significant edge. This comprehensive 2000-word guide provides you with the 20 most important types of questions, complete with detailed solutions, conceptual explanations, and time-saving tricks to boost your performance.
Decoding the Railway Mathematics Syllabus
Before diving into the questions, it’s essential to understand the battlefield. The mathematics syllabus for most railway exams is based on the 10th-grade level, ensuring a level playing field for all aspirants. The core topics that consistently appear across RRB NTPC, Group D, and ALP exams are:
- Number System: Includes HCF, LCM, Decimals, Fractions, and Simplification using the BODMAS rule.
- Arithmetic: This is the heart of the syllabus, covering Percentage, Ratio and Proportion, Profit and Loss, Time and Work, Time, Speed & Distance, and Simple & Compound Interest.
- Mensuration: Questions on area, perimeter, and volume of basic geometric shapes (2D and 3D).
- Advanced Maths: Elementary Algebra, Geometry, and Trigonometry.
- Statistics: Basic questions on mean, median, mode, and data interpretation.
Now, let’s tackle the 20 most important questions, categorized by topic for a structured learning experience.
Category 1: Number System & Simplification
This is the most fundamental area and often the starting point for the Maths paper.
Question 1: The sum of two numbers is 40 and their product is 375. What is the sum of their reciprocals?
Solution:
Let the two numbers be x and y.
Given: x + y = 40
Given: x * y = 375
We need to find the sum of their reciprocals, which is (1/x) + (1/y).
To add these fractions, we find a common denominator: (1/x) + (1/y) = (y + x) / (x * y).
Now, substitute the given values: (40) / (375).
Simplify the fraction: 40/375 = 8/75.
Answer: 8/75
Concept Corner: This question tests your ability to manipulate algebraic expressions. The key is to express the target value (sum of reciprocals) in terms of the given values (sum and product of the numbers).
Pro Tip: Remember this direct formula: The sum of reciprocals of two numbers is equal to their sum divided by their product. This saves you from the initial algebraic steps.
Question 2: Simplify the expression: 108 ÷ 36 of 1/4 + 2/5 × 3 ¼
Solution:
We use the ‘BODMAS’ rule (Brackets, Of, Division, Multiplication, Addition, Subtraction).
Step 1 (Of): 36 of 1/4 = 36 × 1/4 = 9.
The expression becomes: 108 ÷ 9 + 2/5 × 3 ¼.
Step 2 (Convert Mixed Fraction): 3 ¼ = 13/4.
The expression is now: 108 ÷ 9 + 2/5 × 13/4.
Step 3 (Division): 108 ÷ 9 = 12.
The expression is now: 12 + 2/5 × 13/4.
Step 4 (Multiplication): 2/5 × 13/4 = 26/20 = 13/10.
The expression is now: 12 + 13/10.
Step 5 (Addition): 12 + 1.3 = 13.3.
Answer: 13.3
Concept Corner: The BODMAS rule dictates the order of operations in mathematics. The word ‘of’ is treated as a priority multiplication before any division or standard multiplication.
Pro Tip: Always solve the ‘of’ part first. A common mistake is to perform the division before handling the ‘of’, leading to an incorrect answer.
Question 3: The HCF and LCM of two numbers are 12 and 336 respectively. If one of the numbers is 84, find the other number.
Solution:
We use the fundamental property connecting HCF, LCM, and the two numbers.
Formula: Product of two numbers = Product of their HCF and LCM.
Let the other number be ‘x’.
84 * x = 12 * 336
x = (12 * 336) / 84
x = (1 * 336) / 7
x = 48
Answer: The other number is 48.
Concept Corner: This principle is a cornerstone of number theory and is extremely useful for solving such problems directly. It establishes a relationship between two numbers and their highest common factor and least common multiple.
Pro Tip: In the exam, don’t multiply 12 and 336 first. Look for opportunities to simplify the division, as shown in the solution. It makes calculations much faster.
Category 2: Percentages, Profit & Loss
This category is crucial for business-related and comparison-based problems.
Question 4: If the price of sugar is increased by 25%, by what percentage must a household reduce its consumption to keep the expenditure unchanged?
Solution:
Let the initial price be P and initial consumption be C. Initial Expenditure = P * C.
New Price = P * (1 + 25/100) = 1.25P.
Let the new consumption be C’. The expenditure must remain the same.
So, 1.25P * C’ = P * C
C’ = (P * C) / 1.25P = C / 1.25 = 0.8C.
The reduction in consumption is C – C’ = C – 0.8C = 0.2C.
Percentage reduction = (Reduction / Initial Consumption) * 100
= (0.2C / C) * 100 = 20%.
Answer: 20%
Concept Corner: This is a classic problem of inverse proportionality. When two quantities (price and consumption) have a constant product (expenditure), an increase in one leads to a proportional decrease in the other.
Pro Tip: Use the direct formula for such cases: Percentage Reduction = [r / (100 + r)] * 100, where ‘r’ is the percentage increase. Here, r = 25. So, [25 / (100 + 25)] * 100 = (25/125) * 100 = 20%.
Question 5: A shopkeeper sells an article at a profit of 20%. If he had bought it for 10% less and sold it for ₹18 less, he would have gained 30%. Find the cost price.
Solution:
Let the original Cost Price (CP) be ₹x.
Original Selling Price (SP) = x + 20% of x = 1.2x.
New CP = x – 10% of x = 0.9x.
New SP = Original SP – 18 = 1.2x – 18.
New Profit = 30% on the New CP.
So, New SP = New CP * (1 + 30/100) = 0.9x * 1.3 = 1.17x.
Now, equate the two expressions for the New SP:
1.2x – 18 = 1.17x
0.03x = 18
x = 18 / 0.03 = 1800 / 3 = 600.
Answer: The cost price is ₹600.
Concept Corner: This problem involves creating and solving linear equations based on changing conditions of cost price, selling price, and profit percentage.
Pro Tip: Be careful to calculate the new profit percentage (30%) on the new cost price (0.9x), not the original one. This is a common trap.
Question 6: A dishonest dealer professes to sell his goods at cost price but uses a weight of 900 grams for a 1 kg weight. Find his gain percent.
Solution:
The dealer sells 900 grams but charges for 1000 grams (1 kg).
The cost price (CP) is for what he gives (900g).
The selling price (SP) is for what he claims to give (1000g).
Let the cost of 1 gram be ₹1. CP = ₹900, SP = ₹1000.
Gain = SP – CP = 1000 – 900 = 100.
Gain Percent = (Gain / CP) * 100
= (100 / 900) * 100 = (1/9) * 100 = 11.11%.
Answer: 11.11%
Concept Corner: In problems involving false weights, the profit is calculated on the actual quantity of goods sold (the true weight), which acts as the cost price.
Pro Tip: Use the formula: Gain % = [(True Weight – False Weight) / False Weight] * 100. This gives you the answer in one step: [(1000 – 900) / 900] * 100.
Category 3: Time, Work, and Distance
These are application-based topics that frequently appear in various forms.
Question 7: A can do a piece of work in 15 days and B can do it in 20 days. They work together for 4 days and then A leaves. In how many days will B complete the remaining work?
Solution:
Find the total work using LCM of 15 and 20. LCM(15, 20) = 60 units.
A’s one-day work = 60 / 15 = 4 units/day.
B’s one-day work = 60 / 20 = 3 units/day.
(A+B)’s one-day work = 4 + 3 = 7 units/day.
Work done in 4 days = 7 units/day * 4 days = 28 units.
Remaining work = 60 – 28 = 32 units.
Time taken by B to complete the remaining work = Remaining Work / B’s efficiency.
= 32 / 3 = 10 ⅔ days.
Answer: 10 ⅔ days.
Concept Corner: The LCM method is the most efficient way to solve Time and Work problems. The LCM represents the total quantum of work, and each person’s efficiency is the amount of work they do per day.
Pro Tip: Always convert days/hours into efficiency (work per unit time). This makes combining efforts (addition) or finding remaining work (subtraction) straightforward.
Question 8: A train 150m long is running at a speed of 90 km/hr. How much time will it take to cross a platform 200m long?
Solution:
First, convert the speed to m/s: Speed = 90 km/hr * (5/18) = 5 * 5 = 25 m/s.
Total distance to be covered = Length of train + Length of platform.
Total Distance = 150m + 200m = 350m.
Time = Total Distance / Speed.
Time = 350 / 25 = 14 seconds.
Answer: 14 seconds.
Concept Corner: When a train crosses a platform (or a bridge or another train), the total distance covered is the sum of the length of the train and the length of the object it is crossing.
Pro Tip: Remember the conversion factor: to convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5. This is a must-know shortcut.
Question 9: Two pipes A and B can fill a tank in 12 minutes and 15 minutes respectively. If both are opened together and after 3 minutes, pipe A is closed, how much more time will it take for pipe B to fill the tank?
Solution:
Total capacity of the tank = LCM(12, 15) = 60 units.
Efficiency of A = 60 / 12 = 5 units/min.
Efficiency of B = 60 / 15 = 4 units/min.
Combined efficiency (A+B) = 5 + 4 = 9 units/min.
Water filled in first 3 minutes = 9 units/min * 3 min = 27 units.
Remaining capacity = 60 – 27 = 33 units.
Time taken by B to fill the remaining part = Remaining Capacity / B’s efficiency.
= 33 / 4 = 8.25 minutes or 8 minutes 15 seconds.
Answer: 8.25 minutes.
Concept Corner: ‘Pipes and Cistern’ problems are a sub-type of ‘Time and Work’. The same LCM method applies. If a pipe is emptying the tank, its efficiency is treated as negative.
Pro Tip: Visualize the problem. The total work is filling the 60-unit tank. Calculate how much is done, then calculate how much is left and who is doing it.
Category 4: Interest & Ratio
Question 10: Find the difference between the simple interest and compound interest on ₹5000 for 2 years at 4% per annum.
Solution:
Simple Interest (SI) = (P * R * T) / 100 = (5000 * 4 * 2) / 100 = ₹400.
Compound Interest (CI) = P * [(1 + R/100)^T – 1]
= 5000 * [(1 + 4/100)^2 – 1]
= 5000 * [(1.04)^2 – 1] = 5000 * [1.0816 – 1] = 5000 * 0.0816 = ₹408.
Difference = CI – SI = 408 – 400 = ₹8.
Answer: ₹8.
Concept Corner: The difference between CI and SI for the second year arises because CI is calculated on the principal plus the interest accrued in the first year, whereas SI is calculated only on the principal.
Pro Tip: For a 2-year period, use the direct formula: Difference = P * (R/100)^2. Difference = 5000 * (4/100)^2 = 5000 * (1/25)^2 = 5000 / 625 = ₹8. This is a massive time-saver.
Question 11: The ratio of two numbers is 3:4 and their HCF is 4. What is their LCM?
Solution:
Let the numbers be 3x and 4x. Since the highest common factor (HCF) is 4, x must be 4. So, the numbers are 3 * 4 = 12 and 4 * 4 = 16. Now, find the LCM of 12 and 16. 12 = 2 * 2 * 3 16 = 2 * 2 * 2 * 2 LCM(12, 16) = 2 * 2 * 2 * 2 * 3 = 48. Answer: 48.
Concept Corner: When numbers are given in a ratio (like a:b) and their HCF is ‘h’, the numbers are ‘ah’ and ‘bh’. Their LCM will be ‘abh’.
Pro Tip: Directly use the formula LCM = HCF * a * b, where ‘a’ and ‘b’ are the ratio parts. LCM = 4 * 3 * 4 = 48.
Question 12: A bag contains 50p, 25p, and 10p coins in the ratio 5:9:4, amounting to ₹206. Find the number of coins of each type.
Solution:
Let the number of 50p, 25p, and 10p coins be 5x, 9x, and 4x respectively. Convert the value of each set of coins to Rupees. Value of 50p coins = 5x * 0.50 = ₹2.5x Value of 25p coins = 9x * 0.25 = ₹2.25x Value of 10p coins = 4x * 0.10 = ₹0.40x Total Value = 2.5x + 2.25x + 0.40x = 5.15x Given, total value = ₹206. 5.15x = 206 => x = 206 / 5.15 = 40. Number of 50p coins = 5x = 5 * 40 = 200. Number of 25p coins = 9x = 9 * 40 = 360. Number of 10p coins = 4x = 4 * 40 = 160. Answer: 200 (50p), 360 (25p), 160 (10p).
Concept Corner: This problem combines the concept of ratios with basic linear equations. The key is to convert the ratio of the number of coins into a ratio of their values.
Pro Tip: To avoid decimals, you can convert everything to Paise. Total value = 20600 Paise. Value of coins = (5x * 50) + (9x * 25) + (4x * 10) = 250x + 225x + 40x = 515x. Then 515x = 20600, which gives x = 40.
Category 5: Mensuration, Algebra & Geometry
Question 13: The length of a rectangular park is 12m more than its breadth. If the perimeter of the park is 200m, what is its area?
Solution:
Let the breadth (B) be x meters. Then the length (L) is (x + 12) meters. Perimeter = 2 * (L + B) = 200m. 2 * ( (x+12) + x ) = 200 2 * (2x + 12) = 200 4x + 24 = 200 4x = 176 => x = 44. So, Breadth = 44m. Length = 44 + 12 = 56m. Area = L * B = 56 * 44 = 2464 sq. m. Answer: 2464 sq. m.
Concept Corner: This question requires knowledge of the basic formulas for the perimeter and area of a rectangle and the ability to set up and solve a linear equation.
Pro Tip: From 2*(L+B) = 200, you know L+B = 100. You also know L-B = 12. You can solve this pair of linear equations quickly to find L=56 and B=44.
Question 14: If x + 1/x = 5, what is the value of x² + 1/x²?
Solution:
We are given x + 1/x = 5. We know the algebraic identity: (a + b)² = a² + b² + 2ab. Let a = x and b = 1/x. (x + 1/x)² = x² + (1/x)² + 2 * x * (1/x) (x + 1/x)² = x² + 1/x² + 2 Substitute the given value: (5)² = x² + 1/x² + 2 25 = x² + 1/x² + 2 x² + 1/x² = 25 – 2 = 23. Answer: 23.
Concept Corner: This is a classic question from elementary algebra that tests your knowledge of squaring binomials.
Pro Tip: Remember the formula: If x + 1/x = k, then x² + 1/x² = k² – 2. This allows you to solve the problem in seconds.
Question 15: The angles of a triangle are in the ratio 2:3:4. Find the largest angle.
Solution:
Let the angles be 2x, 3x, and 4x. The sum of angles in a triangle is always 180°. 2x + 3x + 4x = 180° 9x = 180° => x = 20°. The angles are: 2x = 2 * 20 = 40° 3x = 3 * 20 = 60° 4x = 4 * 20 = 80° The largest angle is 80°. Answer: 80°.
Concept Corner: This question tests the fundamental property of triangles – the angle sum property – combined with the concept of ratios.
Pro Tip: To find the largest angle directly, take its ratio part (4) and divide by the sum of ratio parts (2+3+4=9), then multiply by 180. Largest Angle = (4/9) * 180 = 80°.
More High-Frequency Questions
Question 16 (Average Speed): A man travels from A to B at 20 km/hr and returns from B to A at 30 km/hr. Find his average speed for the entire journey.
- Solution: Use the formula 2xy/(x+y) for two equal distances. Average Speed = (2 * 20 * 30) / (20 + 30) = 1200 / 50 = 24 km/hr.
- Pro Tip: Do not take the simple average of the speeds. This is a common error. The correct formula accounts for the time taken for each part of the journey.
Question 17 (Age Problems): The ratio of the present ages of A and B is 5:7. Eight years ago, their ages were in the ratio 7:13. Find their present ages.
- Solution: Let present ages be 5x and 7x. Eight years ago, they were 5x-8 and 7x-8. (5x-8)/(7x-8) = 7/13. Cross-multiply: 13(5x-8) = 7(7x-8) => 65x – 104 = 49x – 56 => 16x = 48 => x=3. Present ages are 5*3=15 years and 7*3=21 years.
- Concept: This is another problem solvable with linear equations based on ratios.
Question 18 (Trigonometry): If sin(θ) = 5/13, find the value of tan(θ).
- Solution: sin(θ) = Perpendicular/Hypotenuse. So, P=5, H=13. Using Pythagoras theorem, B = √(H² – P²) = √(13² – 5²) = √(169-25) = √144 = 12. tan(θ) = Perpendicular/Base = 5/12.
- Concept: This tests basic trigonometric ratios and the Pythagorean theorem.
Question 19 (Elementary Statistics): Find the mean of the first 10 prime numbers.
- Solution: The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Sum = 129. Number of terms = 10. Mean = Sum / Number of terms = 129 / 10 = 12.9.
- Concept: Requires knowledge of what prime numbers are and the formula for the arithmetic mean.
Question 20 (Mixtures): In what ratio must a grocer mix two varieties of tea worth ₹60/kg and ₹65/kg so that by selling the mixture at ₹68.20/kg, he may gain 10%?
- Solution: SP of mixture = ₹68.20. Gain = 10%. CP of mixture = SP / (1 + Gain%) = 68.20 / 1.1 = ₹62. Now use alligation.
(60) (65)
\ /
(62)
/ \
(65-62=3) (62-60=2)
The required ratio is 3:2. - Concept: This is a classic alligation problem. The key is to find the cost price of the mixture first before applying the alligation rule.