Quantitative Aptitude PYQs

Shortcut Trick

Successive percentage change formula: Net Change = a + b + (a × b) ÷ 100.

Here, markup a = 76% and discount b = −25%.

Profit% = 76 − 25 − (76 × 25) ÷ 100.

Profit% = 51 − (1900 ÷ 100) = 51 − 19 = 32%.

∴ The correct answer is 32%.

Alternate Method

Given: Marked Price (MP) is 76% more than Cost Price (CP). Discount is 25%.

Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = MP × (100 − Discount%) ÷ 100 | Profit% = (SP − CP) ÷ CP × 100.

⇒ Let Cost Price (CP) = 100.

⇒ Marked Price (MP) = 100 + 76% of 100 = 176.

⇒ Selling Price (SP) = MP − 25% of MP = 176 − (25 ÷ 100 × 176).

⇒ SP = 176 − 44 = 132.

⇒ Profit = SP − CP = 132 − 100 = 32.

⇒ Profit% = (32 ÷ 100) × 100 = 32%.

∴ The correct answer is 32%.

Additional Information

Marked Price and Discount

The price listed on the product is the Marked Price. Discount is always calculated on the MP to arrive at the Selling Price.

Profit and Markup

Markup is the increase over CP to set MP. Profit is the actual gain over CP after selling. Profit% = Markup% − Discount% − (Markup% × Discount% ÷ 100).

Shortcut Trick

For a number to be divisible by 3, the sum of its digits must be divisible by 3.

Sum = 4 + a + 2 + b = 6 + a + b.

Since 6 is already divisible by 3, (a + b) must be 0, 3, 6, or 9.

Checking options: (0+0)=0 (Yes), (3+1)=4 (No), (2+0)=2 (No), (0+1)=1 (No).

∴ The correct answer is a=0,b=0.

Alternate Method

Given: The number 4a2b is divisible by 3.

Property: A number is divisible by 3 if ∑ digits ÷ 3 leaves remainder 0.

⇒ Sum = 4 + a + 2 + b = 6 + a + b

⇒ For divisibility, (6 + a + b) mod 3 = 0

⇒ Let's test Option 1: a = 0, b = 0 ⇒ Sum = 6 + 0 + 0 = 6 (Divisible by 3)

⇒ Let's test Option 2: a = 3, b = 1 ⇒ Sum = 6 + 3 + 1 = 10 (Not divisible)

⇒ Let's test Option 3: a = 2, b = 0 ⇒ Sum = 6 + 2 + 0 = 8 (Not divisible)

⇒ Let's test Option 4: a = 0, b = 1 ⇒ Sum = 6 + 0 + 1 = 7 (Not divisible)

∴ The correct answer is a=0,b=0.

Additional Information

Divisibility Rule of 9

Similar to 3, a number is divisible by 9 if the sum of its digits is divisible by 9.

Divisibility Rule of 3

If the sum of digits is a multiple of 3, the whole number is divisible by 3.

Divisibility Rule of 6

A number is divisible by 6 if it is divisible by both 2 (even number) and 3 (sum of digits rule).

Shortcut Trick

Initial population = 100%

After 38% left, Remaining = 62%

After 50% of remaining left, Final = 62% × (1/2) = 31%

Ratio = Initial : Present = 100 : 31

∴ The correct answer is 100:31.

Alternate Method

Given:

Initial Population = P

Population left due to water crisis = 38%

Remaining Population = 100% − 38% = 62%

Population left for jobs = 50% of 62%

Present Population = 1550

⇒ Present Population = 62% × (100% − 50%) of Initial

⇒ Present Population = 62% × 50% = 31% of Initial

⇒ Ratio = Initial : Present = 100% : 31%

⇒ Ratio = 100 : 31

∴ The correct answer is 100:31.

Additional Information

Successive Percentage Change

When a value is changed by x% and then by y% successively, the final value is Initial × (1 ± x/100) × (1 ± y/100).

Remaining Value Concept

If a% is removed from a total, the remainder is (100 − a)% of the total. Always apply the second percentage on this remainder.

Ratio of Initial to Final

Ratio = 1 : [(1 − r₁) × (1 − r₂)] where r is the rate of decrease in decimal form.

Shortcut Trick

Volume of Cube = 10³ = 1000 cm³.

Product of Cuboid ratio parts (1 : 2 : 4) = 1 × 2 × 4 = 8.

Volume per unit cube = 1000 ÷ 8 = 125.

Scale factor (x) = √³(125) = 5.

Longest side = 4 units = 4 × 5 = 20 cm.

∴ The correct answer is 20 cm.

Alternate Method

Given:

Side of the cube (a) = 10 cm

Ratio of cuboid dimensions (l : b : h) = 4 : 2 : 1 (Let sides be 4x, 2x, and x)

Formula Used:

Volume of Cube = a³

Volume of Cuboid = length × breadth × height

Calculations:

⇒ Volume of cube = 10 × 10 × 10 = 1000 cm³

⇒ Volume of cuboid = (4x) × (2x) × (x) = 8x³

⇒ According to the question: 8x³ = 1000

⇒ x³ = 1000 ÷ 8

⇒ x³ = 125

⇒ x = √³(125) = 5 cm

⇒ Dimensions of the cuboid are:

⇒ Length = 4 × 5 = 20 cm

⇒ Breadth = 2 × 5 = 10 cm

⇒ Height = 1 × 5 = 5 cm

⇒ The longest side is 20 cm.

∴ The correct answer is 20 cm.

Additional Information

Total Surface Area of a Cuboid

The total surface area is calculated as 2(lb + bh + hl), where l, b, and h are length, breadth, and height respectively.

Diagonal of a Cuboid

The longest rod that can be placed inside a cuboid is its diagonal, given by √(l² + b² + h²).

Volume and Area of a Cube

For a cube with side 'a', Volume = a³ and Total Surface Area = 6a².

Shortcut Trick

Total distance to cross = Sum of lengths of both trains = 350 + 350 = 700 m.

Relative speed (opposite direction) = Sum of individual speeds = 700 ÷ 14 = 50 m/s.

Since speeds are equal, speed of each train = 50 ÷ 2 = 25 m/s.

∴ The correct answer is 25 m/s.

Alternate Method

Given: Length of each train = 350 m, Time taken = 14 seconds, Opposite directions, Speeds are equal.

Formula Used: Relative Speed = (L₁ + L₂) ÷ Time

Let the speed of each train be v m/s.

⇒ Relative speed in opposite direction = v + v = 2v

⇒ Total Distance = Length of Train 1 + Length of Train 2 = 350 + 350 = 700 m

⇒ Time = 14 seconds

⇒ 2v = 700 ÷ 14

⇒ 2v = 50

⇒ v = 25 m/s

∴ The correct answer is 25 m/s.

Additional Information

Relative Speed Concept

When two objects move in opposite directions, their relative speed is the sum of their individual speeds (S₁ + S₂). If moving in the same direction, the relative speed is the difference (|S₁ − S₂|).

Speed Conversion

To convert speed from km/hr to m/s, multiply by 5/18. To convert from m/s to km/hr, multiply by 18/5.

Train Crossing Objects

To cross a stationary pole or man, distance = length of train. To cross a platform or another train, distance = length of train + length of object/train.

Shortcut Trick

Total amount to be divided after deductions = 1404 − (47 + 26 + 29) = 1302.

New ratio = 10 : 14 : 18 → Total units = 10 + 14 + 18 = 42 units.

1 unit = 1302 ÷ 42 = 31.

Vidya's share (after deduction) = 14 × 31 = 434.

Actual share of Vidya = 434 + 26 = 460.

∴ The correct answer is ₹460.

Alternate Method

Given: Total amount = ₹1,404. Deductions = ₹47, ₹26, and ₹29. New ratio = 10 : 14 : 18.

Formula Used: Adjusted Total = Total − Sum of Deductions; Individual Share = (Ratio part ÷ Total parts) × Adjusted Total + Deduction.

⇒ Total Deductions = 47 + 26 + 29 = 102

⇒ Remaining amount = 1404 − 102 = 1302

⇒ Sum of ratio parts = 10 + 14 + 18 = 42

⇒ Vidya's share after deduction = (14 ÷ 42) × 1302

⇒ Vidya's share after deduction = (1 ÷ 3) × 1302 = 434

⇒ Actual share of Vidya = 434 + 26 = 460

∴ The correct answer is ₹460.

Additional Information

Ratio and Proportion

If a quantity A is divided in ratio x : y, then the parts are (x / (x+y)) × A and (y / (x+y)) × A.

Deduction problems

Always subtract the fixed deductions from the total before applying the ratio, then add the respective deduction back to find the original amount.

Shortcut Trick

Total Votes = 3,75,000

Valid Votes = 75% of 3,75,000

Other candidate % of valid votes = 100% − 40% = 60%

Required Votes = 3,75,000 × 0.75 × 0.60 = 3,75,000 × 0.45

∴ The correct answer is 1,68,750.

Alternate Method

Given:

Total Votes = 3,75,000

Invalid Votes = 25%

Candidate 1 = 40% of valid votes

⇒ Valid Votes = 3,75,000 − (25% of 3,75,000)

⇒ Valid Votes = 3,75,000 × (75/100) = 2,81,250

⇒ Candidate 2 % of valid votes = 100% − 40% = 60%

⇒ Candidate 2 Votes = 2,81,250 × (60/100)

⇒ Candidate 2 Votes = 2,81,250 × 0.6 = 1,68,750

∴ The correct answer is 1,68,750.

Additional Information

Percentage Change

To find the percentage of a remaining value, multiply the total by the product of remaining percentage fractions: Final = Total × (1 − r₁) × (1 − r₂).

Election Problems

Always identify the base for percentages: whether it is % of Total Votes, Enrolled Voters, or Valid Votes.

Valid vs Invalid

Valid Votes = Total Votes − (Invalid Votes + Votes not cast).

Shortcut Trick

We know that 1 + tan² A = sec² A.

The expression 4 + 4tan² A can be written as 4(1 + tan² A) = 4sec² A.

Since sec A = 1/cos A, then sec A = 5/4.

⇒ 4 × (5/4)² = 4 × 25/16 = 25/4.

∴ The correct answer is 25/4.

Alternate Method

Given: cos A = 4/5 (Base/Hypotenuse)

Formula Used: sin² A + cos² A = 1 OR tan A = Perpendicular/Base

⇒ In a right-angled triangle, if Base = 4 and Hypotenuse = 5, then Perpendicular = √(5² − 4²) = 3.

⇒ tan A = Perpendicular / Base = 3/4.

⇒ Value to find = 4 + 4(3/4)²

⇒ 4 + 4(9/16) = 4 + 9/4

⇒ (16 + 9) / 4 = 25/4.

∴ The correct answer is 25/4.

Additional Information

Trigonometric Identities

Standard Pythagorean identities: sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.

Reciprocal Relations

sec θ = 1/cos θ, cosec θ = 1/sin θ, and cot θ = 1/tan θ.

Pythagorean Triplets

Common sets of integers satisfying a² + b² = c² like (3, 4, 5), (5, 12, 13), and (8, 15, 17) help in rapid calculation.

Shortcut Trick

The age difference between two people always remains constant (12 years).

6 years ago, the ratio of Sneha to Aryan was 3 :

1. Difference in ratio parts = 3 − 1 = 2 units → 12 years.

1 unit = 6 years ∴ Sneha's age 6 years ago = 3 × 6 = 18 years.

Present age of Sneha = 18 + 6 = 24 years.

∴ The correct answer is 24 years.

Alternate Method

Given: Difference in ages = 12 years; 6 years ago, Sneha = 3 × Aryan.

Let Aryan's age 6 years ago be x.

Then Sneha's age 6 years ago = 3x.

⇒ 3x − x = 12

⇒ 2x = 12

⇒ x = 6

⇒ Sneha's age 6 years ago = 3 × 6 = 18 years.

⇒ Sneha's present age = 18 + 6 = 24 years.

∴ The correct answer is 24 years.

Additional Information

Constant Age Difference

The numerical difference between the ages of two persons remains the same at any point in time (past, present, or future).

Ratio and Proportionality

If a ratio changes over time, equate the difference in ratio parts to the actual constant age difference to find the value of one unit.

Shortcut Trick

Take 2^x−1 common

2^x−1(1 + 2²) = 160

⇒ 2^x−1 × 5 = 160 ⇒ 2^x−1 = 32

32 = 2⁵ ⇒ x − 1 = 5 ⇒ x = 6

∴ Answer = 2x + 1 = 13

Alternate Method

Given:

2^x−1 + 2^x+1 = 160

Formula used:

a^m × aⁿ = a^m+n

Calculation:

⇒ 2^x+1 = 2^x−1 × 2²

⇒ So equation becomes:

⇒ 2^x−1 + 4 × 2^x−1 = 160

⇒ 5 × 2^x−1 = 160

⇒ 2^x−1 = 32

⇒ 32 = 2⁵

⇒ x − 1 = 5 ⇒ x = 6

⇒ Required value = 2x + 1 = 2×6 + 1 = 13

∴ The correct answer is 13.

Additional Information

Laws of Exponents

a^m × aⁿ = a^m+n

Common Factor Method

Take common terms to simplify exponential equations

Power Equality

If bases are same, equate powers

Standard Powers

2⁵ = 32, useful for quick solving

Shortcut Trick

Since the distance segments are equal (300 km each), use the harmonic mean formula for average speed.

Average Speed = [2 × S₁ × S₂] ÷ [S₁ + S₂]

60 = [2 × 42 × S₂] ÷ [42 + S₂]

60 × (42 + S₂) = 84 × S₂ ⇒ 2520 + 60S₂ = 84S₂ ⇒ 24S₂ = 2520

S₂ = 2520 ÷ 24 = 105 km/hr.

∴ The correct answer is 105 km/hr.

Alternate Method

Given: Total Distance = 600 km, Average Speed = 60 km/hr, First 300 km Speed = 42 km/hr.

Formula Used: Total Time = Total Distance ÷ Average Speed

⇒ Total Time required = 600 ÷ 60 = 10 hours

⇒ Time spent for first 300 km = 300 ÷ 42 = 50 ÷ 7 hours

⇒ Remaining Time = 10 − (50 ÷ 7) = (70 − 50) ÷ 7 = 20 ÷ 7 hours

⇒ Required speed for remaining 300 km = 300 ÷ (20 ÷ 7)

⇒ Speed = (300 × 7) ÷ 20 = 15 × 7 = 105 km/hr

∴ The correct answer is 105 km/hr.

Additional Information

Average Speed (Equal Distances)

When a journey is divided into 2 equal distances with speeds x and y, the average speed is 2xy ÷ (x + y).

Average Speed (Equal Time)

When a journey is divided into 2 equal time intervals with speeds x and y, the average speed is (x + y) ÷ 2.

General Average Speed Formula

Average Speed = Total Distance Covered ÷ Total Time Taken.

Shortcut Trick

Extra runs scored above the previous average = 209 − 51 = 158.

Increase in average = Extra runs ÷ Total innings = 158 ÷ 15.

Increase = 10.53.

Rounding 10.53 to the nearest whole number gives 11.

∴ The correct answer is 11.

Alternate Method

Given: Average of 14 innings = 51 | 15^th inning score = 209.

Formula: New Average = (Old Sum + New Score) ÷ New Number of Innings.

⇒ Total runs in 14 innings = 14 × 51 = 714.

⇒ Total runs after 15 innings = 714 + 209 = 923.

⇒ New average = 923 ÷ 15 = 61.533...

⇒ Increase in average = 61.533 − 51 = 10.533...

⇒ Nearest whole number = 11.

∴ The correct answer is 11.

Additional Information

Average Formula

Average = Total Sum of Observations ÷ Total Number of Observations.

Net Deviation Rule

If a new value is added, the change in average = (New Value − Old Average) ÷ New Total Count.

Rounding Rules

For nearest whole number, if the decimal part is 0.5 or more, round up; if less than 0.5, round down.

Shortcut Trick

Identify like terms in the expression, which are terms with the same variable and power.

The like terms are x² and x².

Combine them by adding their coefficients: 1x² + 1x² = 2x².

The remaining term 2x stays as it is since it has no other matching term.

∴ The correct answer is 2x² + 2x.

Alternate Method

Given: x² + 2x + x²

To simplify an algebraic expression, we group terms with the same degree (power).

⇒ (x² + x²) + 2x

⇒ 2x² + 2x

The expression is now in its simplest form because x² and x are not like terms.

∴ The correct answer is 2x² + 2x.

Additional Information

Like Terms

Terms that have the same variables raised to the same exponents. Only these terms can be added or subtracted together.

Coefficients and Variables

In a term like 2x², 2 is the coefficient and x is the variable.

Degree of Polynomial

The degree is the highest power of the variable in the polynomial. For 2x² + 2x, the degree is 2 (quadratic).

Shortcut Trick
Sum of digits = 15
Difference of digits = 3
Greater digit = (Sum + Difference) ÷ 2 = (15 + 3) ÷ 2 = 9
Smaller digit = (Sum − Difference) ÷ 2 = (15 − 3) ÷ 2 = 6
Product = 9 × 6 = 54
∴ The correct answer is 54.

Alternate Method
Given: Sum of two digits = 15, Difference = 3.
Formula Used: Product = Digit1 × Digit2

Let the digits be x and y, where x > y.
⇒ x + y = 15
⇒ x − y = 3
⇒ Adding both equations: 2x = 18 ⇒ x = 9
⇒ Substituting x in first equation: 9 + y = 15 ⇒ y = 6
⇒ Product of digits = 9 × 6 = 54
∴ The correct answer is 54.

Additional Information
Sum and Difference of Numbers
If the sum (S) and difference (D) of two numbers are known, the larger number is (S+D)/2 and the smaller number is (S−D)/2.
Number Representation
A two-digit number with tens digit 't' and units digit 'u' is expressed as 10t + u.
Sum of Digits Rule
The sum of digits is used in divisibility tests for 3 and 9; a number is divisible by 9 if the sum of its digits is divisible by 9.

Shortcut Trick

The mode is defined as the value that appears most frequently in a given data set.

In the given observations, identify the frequency of each number:

Number 3 appears 5 times (at positions 1, 5, 11, 14, 17).

Number 4 appears 3 times (at positions 2, 3, 16).

All other numbers (1, 2, 5, 6, 7, 8, 9) appear fewer than 3 times.

∴ The correct answer is 3.

Alternate Method

Given: Observations = 3, 4, 4, 7, 3, 6, 2, 2, 6, 1, 3, 5, 9, 3, 8, 4, 3.

To find the mode, we construct a frequency distribution table:

⇒ Observation 1: Frequency = 1

⇒ Observation 2: Frequency = 2

⇒ Observation 3: Frequency = 5

⇒ Observation 4: Frequency = 3

⇒ Observation 5: Frequency = 1

⇒ Observation 6: Frequency = 2

⇒ Observation 7: Frequency = 1

⇒ Observation 8: Frequency = 1

⇒ Observation 9: Frequency = 1

⇒ The observation with the maximum frequency is 3.

∴ The correct answer is 3.

Additional Information

Measures of Central Tendency

Mean is the average of all observations. Median is the middle value when data is arranged in ascending/descending order. Mode is the most frequent value.

Empirical Relationship

For a moderately skewed distribution: 3 × Median = Mode + 2 × Mean.

Types of Mode

A data set can have one mode (unimodal), two modes (bimodal), or multiple modes (multimodal). If no value repeats, there is no mode.

Shortcut Trick

Let A take x hours. Then B takes x + 12 hours.

Combined time formula: (x × (x + 12)) ÷ (x + (x + 12)) = 8.

Check options: If x = 12, then (12 × 24) ÷ (12 + 24) = 288 ÷ 36 = 8 hours.

The condition is satisfied perfectly with 12.

∴ The correct answer is 12 hours.

Alternate Method

Given:

Time taken by (A + B) = 8 hours

Let time taken by Pipe A = t hours

Time taken by Pipe B = (t + 12) hours

Formula Used:

1/A + 1/B = 1/Total Time

Calculations:

⇒ 1/t + 1/(t + 12) = 1/8

⇒ (t + 12 + t) / (t(t + 12)) = 1/8

⇒ (2t + 12) / (t² + 12t) = 1/8

⇒ 8(2t + 12) = t² + 12t

⇒ 16t + 96 = t² + 12t

⇒ t² − 4t − 96 = 0

⇒ t² − 12t + 8t − 96 = 0

⇒ t(t − 12) + 8(t − 12) = 0

⇒ (t − 12)(t + 8) = 0

⇒ t = 12 (as time cannot be negative)

∴ The correct answer is 12 hours.

Additional Information

Pipes and Cisterns Efficiency

The rate of work done by a pipe is inversely proportional to the time taken. If a pipe fills a tank in x hours, its efficiency is 1/x per hour.

Inlet and Outlet Pipes

Inlet pipes do positive work (addition), while outlet/leakage pipes do negative work (subtraction) in the net efficiency calculation.

Time and Work Relation

The basic concept of Pipes and Cisterns is identical to Time and Work, where pipes are treated as workers and the tank capacity as total work.

Shortcut Trick

Savings % = 100% − Spending % = 100% − 75% = 25%.

Monthly Savings = 25% of ₹40,000.

Monthly Savings = (1/4) × 40,000 = ₹10,000.

∴ The correct answer is ₹10,000.

Alternate Method

Given:

Monthly Income = ₹40,000

Expenditure Percentage = 75%

Formula Used:

Savings = Income − Expenditure

Calculations:

⇒ Expenditure = 75% of 40,000

⇒ Expenditure = (75/100) × 40,000

⇒ Expenditure = 0.75 × 40,000 = ₹30,000

⇒ Savings = 40,000 − 30,000

⇒ Savings = ₹10,000

∴ The correct answer is ₹10,000.

Additional Information

Percentage Change

Income, Expenditure, and Savings are related as: Income = Expenditure + Savings. If any two are changed by a certain %, the third can be found using the same linear relation.

Base Value Concept

In income-based problems, the Income is always considered as the Base (100%) unless specified otherwise.

Fractional Equivalents

Using fractions speeds up calculations: 75% = 3/4, 25% = 1/4, 50% = 1/2, 20% = 1/5.

Shortcut Trick
Loss of 20% means the Selling Price (SP) is 80% of the Cost Price (CP).
Using the unitary method: 80% ⇒ ₹240.
1% ⇒ 240 ÷ 80 = ₹3.
100% (CP) ⇒ 3 × 100 = ₹300.
∴ The correct answer is ₹300.

Alternate Method
Given: Selling Price (SP) = ₹240, Loss% = 20%.
Formula: CP = [SP × 100] ÷ [100 − Loss%]

⇒ CP = [240 × 100] ÷ [100 − 20]
⇒ CP = [24000] ÷ 80
⇒ CP = 300
∴ The correct answer is ₹300.

Additional Information
Cost Price & Selling Price
The price at which an article is purchased is Cost Price (CP), and the price at which it is sold is Selling Price (SP).
Loss Calculation
When SP < CP, there is a loss. Loss = CP − SP and Loss% = (Loss ÷ CP) × 100.

Shortcut Trick

Net efficiency = Filling efficiency − Emptying efficiency

Efficiency of Pipe 1 = 12 ÷ 3 = 4 units/min

Efficiency of Pipe 2 = 12 ÷ 12 = 1 unit/min

Combined efficiency = 4 − 1 = 3 units/min

Time to fill half tank (6 units) = 6 ÷ 3 = 2 minutes

∴ The correct answer is 2 min.

Alternate Method

Given:

Filling time of Pipe A = 3 minutes

Emptying time of Pipe B = 12 minutes

Formula Used:

Total Work = LCM of individual times; Efficiency = Total Work ÷ Time

Calculations:

⇒ Total capacity of the tank = 12 units

⇒ Half of the tank capacity = 12 ÷ 2 = 6 units

⇒ Combined efficiency of (A + B) = 4 − 1 = 3 units/min

⇒ Time taken to fill half tank = 6 ÷ 3

⇒ Time = 2 minutes

∴ The correct answer is 2 min.

Additional Information

Net Work Done

When one pipe fills and another empties, the net work per unit time is the difference of their efficiencies: (1/x − 1/y).

Pipes and Cisterns Logic

Filling pipes have positive efficiency while emptying pipes (leakage) have negative efficiency.

Fractional Work

To find time for a fraction of work, multiply the Total Work by that fraction before dividing by the combined efficiency.

Shortcut Trick

Effective discount = a + b − (a × b ÷ 100) = 20 + 25 − (20 × 25 ÷ 100) = 45 − 5 = 40%.

If Marked Price (MP) is 100%, then Selling Price (SP) = 100% − 40% = 60%.

Given 60% = ₹48,000.

Therefore, 100% = (48,000 ÷ 60) × 100 = 800 × 100 = ₹80,000.

∴ The correct answer is ₹80,000.

Alternate Method

Given → Selling Price (SP) = ₹48,000 | Discount 1 (d₁) = 20% | Discount 2 (d₂) = 25%.

Formula: SP = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ 48,000 = MP × (80/100) × (75/100)

⇒ 48,000 = MP × (4/5) × (3/4)

⇒ 48,000 = MP × (3/5)

⇒ MP = (48,000 × 5) ÷ 3

⇒ MP = 16,000 × 5 = 80,000

∴ The correct answer is ₹80,000.

Additional Information

Marked Price & Discount

The Marked Price (MP) is the price printed on the label. Discount is always calculated on the MP. SP = MP − Discount.

Successive Discount Formula

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