Search Previous Year Questions

Find real questions from previous competitive exam papers with answers and solutions.

Showing 10 question(s)
Clear

Question 271

RRB NTPC 2026• Mathematics• Mathematics

The third proportional to 75 and 45 is the same as the fourth proportional to 10, 5x and 6x. Find the value of x.

  1. AOption (a)

    3

  2. BOption (b)

    4

  3. COption (c)

    5

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (a)A.

3

Why this answer is correct

Shortcut Trick

Third proportional of 75 and 45 = (45 × 45) ÷ 75 = 27

Fourth proportional of 10, 5x, and 6x = (5x × 6x) ÷ 10 = 3x2

Equating both: 3x2 = 27 ⇒ x2 = 9 ⇒ x = 3

∴ The correct answer is 3.

Alternate Method

Given:

First set of numbers: 75, 45 (to find Third Proportional)

Second set of numbers: 10, 5x, 6x (to find Fourth Proportional)

Formula Used:

  1. Third proportional (c) of a and b is given by: a/b = b/c ⇒ c = b2/a
  2. Fourth proportional (d) of a, b, and c is given by: a/b = c/d ⇒ d = (b × c)/a

Calculations:

⇒ Let the third proportional to 75 and 45 be 'c'.

⇒ c = (45 × 45) ÷ 75

⇒ c = 2025 ÷ 75 = 27

⇒ Let the fourth proportional to 10, 5x, and 6x be 'd'.

⇒ 10 / 5x = 6x / d

⇒ d = (5x × 6x) ÷ 10

⇒ d = 30x2 ÷ 10 = 3x2

⇒ As per the question: c = d

⇒ 27 = 3x2

⇒ x2 = 9

⇒ x = √9 = 3

∴ The correct answer is 3.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is √(a × b).

Continued Proportion

Three numbers a, b, c are said to be in continued proportion if a : b = b : c.

Compounded Ratio

For two ratios a : b and c : d, the compounded ratio is (a × c) : (b × d).

Question 272

RRB NTPC 2026• Mathematics• Mathematics

2 men and 8 women can complete a piece of work in 13 days, and 3 men and 19 women can complete the same work in 8 days. The work of how many women is equivalent to that of one man?

  1. AOption (a)

    23

  2. BOption (b)

    29

  3. COption (c)

    25

  4. DOption (d)

    24

View answer & explanation

Correct answer

Answer: (d)D.

24

Why this answer is correct

Shortcut Trick

Equate the total work by multiplying the number of days with the group efficiency.

13 × (2M + 8W) = 8 × (3M + 19W)

⇒ 26M + 104W = 24M + 152W

⇒ 26M − 24M = 152W − 104W

⇒ 2M = 48W

⇒ 1M = 24W

∴ The correct answer is 24.

Alternate Method

Given: Group 1 (2 men + 8 women) takes 13 days. Group 2 (3 men + 19 women) takes 8 days.

Formula Used: Total Work = (Efficiency of Men + Efficiency of Women) × Number of Days

PersonDaysTotal WorkEfficiency
Group 1 (2M + 8W)131048
Group 2 (3M + 19W)813

⇒ 2M + 8W = 8 ⇒ M + 4W = 4 ⇒ M = 4 − 4W — (Eq. 1)

⇒ 3M + 19W = 13 — (Eq. 2)

Substitute M from (Eq. 1) into (Eq. 2):

⇒ 3 × (4 − 4W) + 19W = 13

⇒ 12 − 12W + 19W = 13

⇒ 7W = 13 − 12 ⇒ 7W = 1

⇒ W = 1/7

Substitute W in (Eq. 1):

⇒ M = 4 − 4 × (1/7) = 4 − 4/7 = 24/7

⇒ Man : Woman ratio = (24/7) : (1/7) = 24 : 1

⇒ 1 Man = 24 Women

∴ The correct answer is 24.

Additional Information

Man-Day Formula

If M1 men do W1 work in D1 days and M2 men do W2 work in D2 days, then (M1 × D1)/W1 = (M2 × D2)/W2.

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a task. Efficiency × Time = Constant (Work).

Work and Wages

Wages are distributed in the ratio of the work done by each individual, which is the product of their efficiency and the time they worked.

Question 273

RRB NTPC 2026• Mathematics• Mathematics

The price of an article increases by 25%, then decreases by 20%, and finally increases by 15%. What is the overall percentage change in the price?

  1. AOption (a)

    10%

  2. BOption (b)

    15%

  3. COption (c)

    20%

  4. DOption (d)

    25%

View answer & explanation

Correct answer

Answer: (b)B.

15%

Why this answer is correct

Shortcut Trick

Use the Multiplying Factor (MF) method for successive changes.

Net MF = 1.25 (25% increase) × 0.80 (20% decrease) × 1.15 (15% increase)

Net MF = (1.25 × 0.80) × 1.15 = 1.00 × 1.15 = 1.15

Overall change = (1.15 − 1) × 100 = 15%

∴ The correct answer is 15%.

Alternate Method

Given:

First increase = 25%, Second decrease = 20%, Third increase = 15%

Formula Used:

Final Price = Initial Price × (1 ± r1/100) × (1 ± r2/100) × (1 ± r3/100)

⇒ Let the Initial Price of the article be 100.

⇒ After 25% increase, price = 100 + (25% of 100) = 125.

⇒ After 20% decrease, price = 125 − (20% of 125) = 125 − 25 = 100.

⇒ After 15% increase, price = 100 + (15% of 100) = 115.

⇒ Net percentage change = [(Final Price − Initial Price) / Initial Price] × 100

⇒ Net percentage change = [(115 − 100) / 100] × 100 = 15%.

∴ The correct answer is 15%.

Additional Information

Net Successive Change Formula

For two successive percentage changes a% and b%, the net percentage change is given by the formula: (a + b + (ab/100))%.

Fractional Equivalence

Converting percentages to fractions (e.g., 25% = 1/4, 20% = 1/5) can often simplify complex successive change calculations.

Base Independence

The final percentage change in successive calculations is independent of the actual initial value of the quantity.

Question 274

RRB NTPC 2026• Mathematics• Mathematics

A cube is melted and recast into a cuboid whose dimensions are in the ratio 4:5:8. If the height of the cuboid is 12 cm, then the edge of the original cube is (correct to 2 decimal places)

  1. AOption (a)

    15.34 space cm

  2. BOption (b)

    9.44 space cm

  3. COption (c)

    12.12 space cm

  4. DOption (d)

    8.14 space cm

View answer & explanation

Correct answer

Answer: (d)D.

8.14 space cm

Why this answer is correct

Shortcut Trick

Ratio of cuboid dimensions = 4 : 5 : 8.

Given height (8 units) = 12 cm ⇒ 1 unit = 1.5 cm.

Volume of cuboid = (4 × 1.5) × (5 × 1.5) × (8 × 1.5) = 6 × 7.5 × 12 = 540 cm3.

Since Volume of Cube = Volume of Cuboid ⇒ a3 = 540.

Edge a = 3√540 ≈ 8.14 cm.

∴ The correct answer is 8.14 cm.

Alternate Method

Given: Ratio of dimensions = 4 : 5 : 8, Height = 12 cm.

Formula: Volume of cuboid = Length × Breadth × Height; Volume of cube = (Edge)3.

⇒ Let the dimensions of the cuboid be 4x, 5x, and 8x.

⇒ Given Height = 8x = 12 cm.

⇒ x = 12 ÷ 8 = 1.5 cm.

⇒ Length = 4 × 1.5 = 6 cm and Breadth = 5 × 1.5 = 7.5 cm.

⇒ Volume of Cuboid = 6 × 7.5 × 12 = 540 cm3.

⇒ When a cube is melted to form a cuboid, their volumes remain equal.

⇒ Volume of Cube = 540 cm3.

⇒ (Edge)3 = 540.

⇒ Edge = 5401/3.

⇒ Since 83 = 512 and 93 = 729, the value is slightly more than 8.

⇒ 8.14 × 8.14 × 8.14 ≈ 539.35 ≈ 540.

⇒ Edge ≈ 8.14 cm.

∴ The correct answer is 8.14 cm.

Additional Information

Volume and Surface Area of Cube

Volume = a3 and Total Surface Area = 6a2, where 'a' is the edge of the cube.

Volume of Cuboid

Volume = Length × Breadth × Height. Total Surface Area = 2(lb + bh + hl).

Recasting Rule

When one 3D shape is melted and recast into another, the total volume remains constant, provided there is no wastage.

Diagonal Length

Diagonal of a cube = a√3 and diagonal of a cuboid = √(l2 + b2 + h2).

Question 275

RRB NTPC 2026• Mathematics• Mathematics

A dealer buys an article marked at 1800 after successive discounts of 25% and 6236\frac{2}{3}%. Find the cost price of the article to the dealer.

  1. AOption (a)

    ₹1260

  2. BOption (b)

    ₹1275

  3. COption (c)

    ₹1250

  4. DOption (d)

    ₹1290

View answer & explanation

Correct answer

Answer: (a)A.

₹1260

Why this answer is correct

Shortcut Trick

Convert percentage discounts to fractions: 25% = 1/4 and 6(2/3)% = 1/15.

Multipliers for discounts: (1 − 1/4) = 3/4 and (1 − 1/15) = 14/15.

Effective Multiplier = 3/4 × 14/15 = 7/10.

Cost Price = 1800 × 7/10 = 180 × 7 = 1260.

∴ The correct answer is ₹1260.

Alternate Method

Given:

Marked Price (MP) = ₹1800

First Discount = 25%

Second Discount = 6(2/3)% = 20/3%

Formula Used:

Selling Price = MP × (100 − d1)% × (100 − d2)%

⇒ Price after 1st discount = 1800 − (25% of 1800)

⇒ Price after 1st discount = 1800 − 450 = ₹1350

⇒ Price after 2nd discount = 1350 − (20/3% of 1350)

⇒ 2nd discount amount = (20/300) × 1350 = 1/15 × 1350 = ₹90

⇒ Final Cost Price (CP) to the dealer = 1350 − 90 = ₹1260

∴ The correct answer is ₹1260.

Additional Information

Successive Discount Formula

For two successive discounts of a% and b%, the equivalent single discount is given by (a + b − (a × b)/100)%.

Marked Price and Selling Price

Selling Price = Marked Price × (100 − Discount%)/100. Discount is always calculated on the Marked Price.

Fractional Conversions

Using fractions like 6(2/3)% = 1/15 or 33(1/3)% = 1/3 can significantly speed up calculations in competitive exams.

Question 276

RRB NTPC 2026• Mathematics• Mathematics

Two alloys A and B contain copper and zinc in the ratio 3:4 and 5:9, respectively. If equal weights of both alloys are melted together to form a new alloy, what is the ratio of copper to zinc in the new alloy?

  1. AOption (a)

    8:13

  2. BOption (b)

    12:19

  3. COption (c)

    9:16

  4. DOption (d)

    11:17

View answer & explanation

Correct answer

Answer: (d)D.

11:17

Why this answer is correct

Shortcut Trick

Total units in Alloy A = 3 + 4 = 7; Alloy B = 5 + 9 = 14.

To make weights equal, take LCM(7, 14) = 14 units for each alloy.

In A (14 units): Copper = 3 × 2 = 6, Zinc = 4 × 2 = 8.

In B (14 units): Copper = 5 × 1 = 5, Zinc = 9 × 1 = 9.

Final Ratio (Cu:Zn) = (6 + 5) : (8 + 9) = 11 : 17.

∴ The correct answer is 11:17.

Alternate Method

Given:

Ratio of Copper to Zinc in Alloy A = 3 : 4

Ratio of Copper to Zinc in Alloy B = 5 : 9

Weights of both alloys are equal.

Let the weight of each alloy be 1 unit.

Quantity of Copper in Alloy A = 3/7

Quantity of Zinc in Alloy A = 4/7

Quantity of Copper in Alloy B = 5/14

Quantity of Zinc in Alloy B = 9/14

⇒ Total Copper in new alloy = 3/7 + 5/14

⇒ Total Copper = (6 + 5) / 14 = 11/14

⇒ Total Zinc in new alloy = 4/7 + 9/14

⇒ Total Zinc = (8 + 9) / 14 = 17/14

⇒ Required Ratio = (11/14) : (17/14)

⇒ Required Ratio = 11 : 17

∴ The correct answer is 11:17.

Additional Information

Mixture of Multiple Alloys

When mixing 'n' alloys of equal weight, the final ratio is the sum of respective fractions: Σ(componenti / totali).

Unequal Weight Mixing

If weights are in ratio m:n, multiply the component fractions of Alloy A by 'm' and Alloy B by 'n' before adding.

Rule of Alligation

Used to find the ratio in which two mixtures must be mixed to get a desired final concentration: (Higher value − Mean) / (Mean − Lower value).

Question 277

RRB NTPC 2026• Mathematics• Mathematics

In a division sum, the divisor is 2 times the quotient and 3 times the remainder. If the remainder is 10, what is the dividend?

  1. AOption (a)

    456

  2. BOption (b)

    464

  3. COption (c)

    465

  4. DOption (d)

    460

View answer & explanation

Correct answer

Answer: (d)D.

460

Why this answer is correct

Shortcut Trick

Remainder = 10

Divisor = 3 × 10 = 30

Quotient = 30 ÷ 2 = 15

Dividend = (30 × 15) + 10 = 460

∴ The correct answer is 460.

Alternate Method

Given:

Remainder = 10

Divisor = 2 × Quotient

Divisor = 3 × Remainder

Formula Used:

Dividend = (Divisor × Quotient) + Remainder

⇒ Divisor = 3 × 10 = 30

⇒ 30 = 2 × Quotient ⇒ Quotient = 30 ÷ 2 = 15

⇒ Dividend = (30 × 15) + 10

⇒ Dividend = 450 + 10 = 460

∴ The correct answer is 460.

Additional Information

Division Algorithm

The fundamental relation in arithmetic is Dividend = (Divisor × Quotient) + Remainder.

Remainder Rule

The remainder must always be a non-negative integer and strictly less than the divisor (0 ≤ Remainder < Divisor).

Perfect Divisibility

If a number is perfectly divisible by another, the remainder is always zero, meaning Dividend = Divisor × Quotient.

Question 278

RRB NTPC 2026• Mathematics• Mathematics

From a regular octagon of side 4 units, a largest rectangle is cut out. From the remaining pieces, two largest rectangles are cut out. The remaining area of the octagon is _____ square units.

  1. AOption (a)

    32

  2. BOption (b)

    24

  3. COption (c)

    8

  4. DOption (d)

    16

View answer & explanation

Correct answer

Answer: (d)D.

16

Why this answer is correct

Shortcut Trick

A regular octagon of side 4 units can be decomposed into a central square of area 42 = 16, 4 rectangles of area 4 × 2√2, and 4 corner triangles.

The largest rectangle (central strip) consists of the central square and two rectangles.

From the remaining 2 rectangles and 4 triangles, the two largest rectangles cut out are simply the two rectangles themselves.

The remaining area consists of the 4 corner right-angled triangles.

Area of 4 corner triangles = 2 × (side)2 ÷ 2 = (side)2 = 42 = 16 square units.

∴ The correct answer is 16 square units.

Alternate Method

Given:

Side of the regular octagon (a) = 4 units

Formula Used:

Area of regular octagon = 2 × (1 + √2) × a2

Corner triangle leg (x) = a / √2

Calculations:

⇒ Area of Octagon = 2 × (1 + √2) × 42 = 32(1 + √2) = 32 + 32√2

⇒ Corner triangle leg (x) = 4 / √2 = 2√2

⇒ Largest Rectangle (R1) = a × (a + 2x) = 4 × (4 + 4√2) = 16 + 16√2

⇒ Remaining Area after 1st cut = (32 + 32√2) − (16 + 16√2) = 16 + 16√2

⇒ This remaining area consists of two trapezoidal caps. Each cap area = 8 + 8√2

⇒ Largest rectangle from each cap (R2) = a × x = 4 × 2√2 = 8√2

⇒ Total area of two rectangles cut from pieces = 2 × 8√2 = 16√2

⇒ Final remaining area = (16 + 16√2) − 16√2 = 16

∴ The correct answer is 16 square units.

Additional Information

Regular Octagon Properties

Each internal angle is 135°, and the total area can be seen as the area of a bounding square (a + 2x)2 minus four corner triangles of area x2/2 each.

Largest Inscribed Rectangle

The largest rectangle inscribed in a regular octagon has an area exactly equal to half the area of the octagon, given by Area = a2(1 + √2).

Pythagoras in Octagons

The corner triangles formed when an octagon is inscribed in a square are isosceles right-angled triangles where the hypotenuse is the side of the octagon (a) and legs are x = a/√2.

Question 279

RRB NTPC 2026• Mathematics• Mathematics

A sum of money earns ₹2,400 as simple interest in 2 years at 12% per annum. Find the sum.

  1. AOption (a)

    ₹10,000

  2. BOption (b)

    ₹18,000

  3. COption (c)

    ₹12,000

  4. DOption (d)

    ₹24,000

View answer & explanation

Correct answer

Answer: (a)A.

₹10,000

Why this answer is correct

Shortcut Trick

The effective rate for Simple Interest is calculated as Rate × Time.

Effective interest percentage = 12% × 2 = 24%.

Given that 24% of the Sum is equivalent to ₹2,400.

Sum (Principal) = (2400 ÷ 24) × 100 = 100 × 100 = ₹10,000.

∴ The correct answer is ₹10,000.

Alternate Method

Given:

Simple Interest (SI) = ₹2,400

Rate (R) = 12% per annum

Time (T) = 2 years

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ 2400 = (P × 12 × 2) ÷ 100

⇒ 2400 = (P × 24) ÷ 100

⇒ 2400 × 100 = 24P

⇒ 240,000 = 24P

⇒ P = 240,000 ÷ 24

⇒ P = 10,000

∴ The correct answer is ₹10,000.

Additional Information

Amount in Simple Interest

The total sum of money after a period is called the Amount (A), which is the sum of the Principal (P) and Simple Interest (SI): A = P + SI.

Principal Formula

To find the Principal when SI, Rate, and Time are known: P = (100 × SI) ÷ (R × T).

SI vs CI for 1st Year

For the first year, if the interest is calculated annually, the Simple Interest and Compound Interest on a specific sum at the same rate are always equal.

Question 280

RRB NTPC 2026• Mathematics• Mathematics

A milkman adds two litres of water to every eight litres of milk and sells it at 50% more than his cost price of milk. Find the profit percentage of the milkman.

  1. AOption (a)

    87.5%

  2. BOption (b)

    68.5%

  3. COption (c)

    50%

  4. DOption (d)

    75%

View answer & explanation

Correct answer

Answer: (a)A.

87.5%

Why this answer is correct

Shortcut Trick

Ratio of Milk to Water = 8 : 2 = 4 : 1. Total Quantity = 5 units.

Total CP = Quantity of Milk × CP = 4 × 1 = 4 units.

Total SP = Total Quantity × (CP + 50% increase) = 5 × 1.5 = 7.5 units.

Profit = 7.5 − 4 = 3.5 units.

Profit % = (3.5 ÷ 4) × 100 = 87.5%.

∴ The correct answer is 87.5%.

Alternate Method

Given:

Milk quantity = 8 litres | Water added = 2 litres | SP = 50% more than CP.

Let the Cost Price (CP) of 1 litre of milk = ₹100.

⇒ Total CP of 8 litres of milk = 8 × 100 = ₹800.

⇒ Total quantity of the mixture = 8 + 2 = 10 litres.

⇒ Selling Price (SP) per litre = 100 + 50% of 100 = ₹150.

⇒ Total Selling Price (SP) = 10 × 150 = ₹1500.

⇒ Profit = Total SP − Total CP = 1500 − 800 = ₹700.

⇒ Profit Percentage = (Profit ÷ Total CP) × 100

⇒ Profit % = (700 ÷ 800) × 100

⇒ Profit % = 87.5%

∴ The correct answer is 87.5%.

Additional Information

Dishonest Dealer Concept

Profit % = [ (True Value − Error) ÷ Error ] × 100 is used when the weight is manipulated, but for mixture problems, use total CP and total SP.

Successive Change Formula

Total change % = a + b + (ab ÷ 100), where 'a' is the percentage increase due to quantity (water addition) and 'b' is the price markup percentage.

Effective Profit from Adulteration

If a milkman adds water and sells at CP, profit % = (Water ÷ Milk) × 100. If he also marks up the price, the net profit is calculated on the initial cost of milk only.