Quantitative Aptitude PYQs

Shortcut Trick

Effective rate for 2 years at 20% p.a. = 20 + 20 + (20 × 20) ÷ 100 = 44%.

Compound Interest = 44% of ₹3,000.

CI = 3,000 × 0.44 = ₹1,320.

∴ The correct answer is ₹1,320.

Alternate Method

Given: Principal (P) = ₹3,000, Rate (R) = 20%, Time (n) = 2 years.

Formula Used: Amount (A) = P(1 + R ÷ 100)ⁿ and CI = A − P

⇒ A = 3,000 × (1 + 20 ÷ 100)²

⇒ A = 3,000 × (120 ÷ 100) × (120 ÷ 100)

⇒ A = 3,000 × 1.2 × 1.2

⇒ A = 3,000 × 1.44

⇒ A = ₹4,320

⇒ CI = 4,320 − 3,000

⇒ CI = ₹1,320

∴ The correct answer is ₹1,320.

Additional Information

CI and SI Difference (for 2 years)

The difference between Compound Interest and Simple Interest for 2 years is given by: D = P × (R ÷ 100)².

Effective Percentage Rate

For successive growth of x% and y%, the total growth is x + y + (xy ÷ 100)%. For 2 years at rate R, it is 2R + (R² ÷ 100)%.

Compounding Frequencies

If interest is compounded half-yearly, the rate becomes R ÷ 2 and the time period becomes 2n.

Shortcut Trick

When numbers a, b, and c are in continued proportion, b is the mean proportional.

Logic: b = √(a × c)

Applying the trick: x = √(2 × 18) = √36 = 6

∴ The correct answer is 6.

Alternate Method

Given:

The numbers 2, x, and 18 are in continued proportion.

Formula Used:

Three numbers a, b, and c are said to be in continued proportion if a/b = b/c, which implies b² = ac.

Calculations:

⇒ 2 ÷ x = x ÷ 18

⇒ x² = 2 × 18

⇒ x² = 36

⇒ x = √36

⇒ x = 6 (as we need the positive value)

∴ The correct answer is 6.

Additional Information

Mean Proportional

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

Third Proportional

If a, b, c are in continued proportion, then c is called the third proportional to a and b, where c = b²/a.

Fourth Proportional

If a:b = c:d, then d is called the fourth proportional to a, b, and c, calculated as d = (b × c) / a.

Shortcut Trick

Total volume of spherical tank = (4/3) × π × r³.

Given r = 2 m, Total Volume = (4/3) × 3.14159 × 8 ≈ 33.51 m³.

Water volume = 80% of 33.51 = 0.8 × 33.51 ≈ 26.81 m³.

Rounding to the nearest cubic meter, we get 27 m³.

∴ The correct answer is 27 m³.

Alternate Method

Given:

Radius of the sphere (r) = 2 m

Filled percentage = 80%

Formula Used:

Volume of a sphere = (4/3) × π × r³

Calculations:

⇒ Total volume of the tank = (4/3) × (22/7) × (2)³

⇒ Total volume = (4/3) × (22/7) × 8 = 704 / 21 m³

⇒ Total volume ≈ 33.5238 m³

⇒ Volume of water = 80% of 33.5238

⇒ Volume of water = (80 / 100) × 33.5238

⇒ Volume of water = 0.8 × 33.5238 = 26.819 m³

⇒ Rounded to nearest cubic meter = 27 m³

∴ The correct answer is 27 m³.

Additional Information

Surface Area of a Sphere

The total surface area is given by the formula A = 4 × π × r².

Volume of a Hemisphere

A hemisphere is half of a sphere; its volume is V = (2/3) × π × r³.

Volume to Liters Conversion

In practical applications, 1 cubic meter (m³) of volume is equivalent to 1,000 liters of water.

Shortcut Trick

Ratio of goods: 50% (Part 1) : 20% (40% of 50%, Part 2) : 30% (Rest, Part 3).

Simplified ratio = 5 : 2 : 3.

Apply weighted average for profit/loss: x = [(5 × 12) + (2 × 16) + (3 × −19)] ÷ (5 + 2 + 3).

x = [60 + 32 − 57] ÷ 10 = 35 ÷ 10 = 3.5.

∴ The correct answer is 3.5.

Alternate Method

Given:

Total goods = 100 units (Assumed CP = ₹100)

Part 1: 50% of 100 = 50 units at 12% profit

Remaining: 100 − 50 = 50 units

Part 2: 40% of remaining = 0.40 × 50 = 20 units at 16% profit

Part 3: Still remaining = 50 − 20 = 30 units at 19% loss

Calculations:

⇒ Profit from Part 1 = 50 × 12 ÷ 100 = 6

⇒ Profit from Part 2 = 20 × 16 ÷ 100 = 3.2

⇒ Loss from Part 3 = 30 × 19 ÷ 100 = 5.7

⇒ Total Net Profit = 6 + 3.2 − 5.7 = 3.5

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

⇒ x = (3.5 ÷ 100) × 100 = 3.5

∴ The correct answer is 3.5.

Additional Information

Weighted Average Profit/Loss

When goods are divided into parts, the net profit/loss is the sum of (weight × percentage) for each part divided by the total weight. Loss is always substituted as a negative value.

Successive Percentage Change

If a value is changed by a% and then by b%, the effective change is (a + b + ab ÷ 100)%. This is different from weighted averages used for parts of a whole.

Base Value Selection

Assuming a base value of 100 for calculations involving percentages simplifies arithmetic and helps avoid complex decimal fractions during competitive exams.

Shortcut Trick

46% of Maximum Marks = Girish score + 81 = 448 + 81 = 529

1% of Maximum Marks = 529 ÷ 46 = 11.5

Friend's Percentage = Friend's Score ÷ Value of 1%

Friend's Percentage = 805 ÷ 11.5 = 70%

∴ The correct answer is 70%.

Alternate Method

Given:

Girish's marks = 448

Difference to 46% = 81 marks

Friend's marks = 805

Formula Used:

Percentage of marks = (Obtained Marks ÷ Maximum Marks) × 100

Calculations:

⇒ Let Maximum Marks be 'M'.

⇒ 46% of M − 81 = 448

⇒ 0.46 × M = 448 + 81

⇒ 0.46 × M = 529

⇒ M = 529 ÷ 0.46 = 1150

⇒ Friend's Percentage = (805 ÷ 1150) × 100

⇒ Friend's Percentage = 80500 ÷ 1150 = 8050 ÷ 115

⇒ Friend's Percentage = 70%

∴ The correct answer is 70%.

Additional Information

Percentage Change

Percentage Increase/Decrease = (Difference ÷ Original Value) × 100.

Fraction to Percentage

To convert a fraction to a percentage, multiply by 100. Common values: 1/4 = 25%, 1/5 = 20%, 3/4 = 75%.

Base Value Concept

In problems involving percentages, the Maximum Marks or Total Value always acts as the base (100%).

Shortcut Trick

Total amount to be divided after deductions = 2288 − (39 + 47 + 40) = 2162.

Sum of ratio parts = 17 + 15 + 14 = 46 units.

Value of 46 units = 2162 ⇒ 1 unit = 2162 ÷ 46 = 47.

Meera’s share after deduction = 15 units = 15 × 47 = 705.

∴ The correct answer is ₹705.

Alternate Method

Given: Total sum = ₹2,288. Deductions for Pankaj, Meera, and Ashok are ₹39, ₹47, and ₹40 respectively. The ratio of their shares after deduction is 17 : 15 : 14.

Formula: Individual Share = (Individual Ratio ÷ Sum of Ratios) × Remaining Total

⇒ Total deduction = 39 + 47 + 40 = ₹126

⇒ Remaining sum = 2288 − 126 = ₹2,162

⇒ Ratio sum = 17 + 15 + 14 = 46

⇒ Let the shares after deduction be 17x, 15x, and 14x

⇒ 46x = 2162

⇒ x = 2162 ÷ 46 = 47

⇒ Meera’s share after deduction = 15x

⇒ Meera’s share after deduction = 15 × 47 = 705

∴ The correct answer is ₹705.

Additional Information

Ratio Distribution Rule

To divide a quantity A in ratio a:b:c, the shares are [a/(a+b+c)]×A, [b/(a+b+c)]×A, and [c/(a+b+c)]×A.

Original Share Calculation

The original share of a person is found by adding the deducted amount back to their share after deduction (Original = Share after deduction + Deducted amount).

Effective Ratio

When fixed amounts are added or subtracted from shares, the final ratio changes; calculations must always be performed on the adjusted total amount.

Shortcut Trick

Using the net deviation method relative to the average of 15:

Total deviation must be zero for the 4 numbers.

Deviation of known numbers: (6.5 − 15) + (10.8 − 15) + (20 − 15) = −8.5 − 4.2 + 5 = −7.7

The fourth number a must compensate this: a = 15 + 7.7 = 22.7

∴ The correct answer is 22.7.

Alternate Method

Given: Observations = 6.5, 10.8, 20, a | Total observations = 4 | Average = 15

Formula Used: Average = (Sum of all observations) ÷ (Total number of observations)

⇒ Sum of observations = Average × Total number of observations

⇒ 6.5 + 10.8 + 20 + a = 15 × 4

⇒ 37.3 + a = 60

⇒ a = 60 − 37.3

⇒ a = 22.7

∴ The correct answer is 22.7.

Additional Information

Arithmetic Mean

The average or Arithmetic Mean is the central value of a set of numbers, calculated by dividing the total sum by the count of items.

Net Deviation Principle

The sum of the differences between each number in a set and the mean of the set is always zero.

Sum Formula

To find an unknown value in a sequence when the average is known: Sum = Average × Total Number of items.

Given:

Expression: ((12/24) / (12/6)) ÷ (2/9 × 18/6 + 4/8) + 6/9 ÷ 28/11 of 11/9

Formula used:

BODMAS Rule: Brackets, Of, Division, Multiplication, Addition, Subtraction

Calculations:

Applying the BODMAS rule to find the simplified value:

⇒ ((1/2) / 2) ÷ (2/9 × 3 + 1/2) + 6/9 ÷ (28/11 × 11/9)

⇒ 1/4 ÷ (2/3 + 1/2) + 2/3 ÷ 28/9

⇒ 1/4 ÷ 7/6 + (2/3 × 9/28)

⇒ (1/4 × 6/7) + (1/1 × 3/14)

⇒ 6/28 + 3/14

⇒ 3/14 + 3/14

⇒ 6/14

⇒ 3/7

∴ The correct answer is option (4).

Shortcut Trick

Present ratio of Father and Son = 3 : 1 (Difference = 2 units)

Ratio after 15 years = 2 : 1 (Difference = 1 unit)

Make the difference equal by multiplying future ratio by 2 ⇒ 4 : 2

Increase in units (4 − 3 or 2 − 1) = 1 unit → 15 years

Present age of Father = 3 units × 15 = 45 years

∴ The correct answer is 45 years.

Alternate Method

Given:

Let the present age of the Son be x years.

Present age of the Father = 3x years.

After 15 years, their ages will be (x + 15) and (3x + 15) respectively.

Calculations:

⇒ 3x + 15 = 2(x + 15)

⇒ 3x + 15 = 2x + 30

⇒ 3x − 2x = 30 − 15

⇒ x = 15

⇒ Father present age = 3x = 3 × 15 = 45 years

∴ The correct answer is 45 years.

Additional Information

Constant Age Gap

The mathematical difference between the ages of two individuals is a constant value and does not change over time.

Ratio Normalization

When solving age problems using ratios, if the time interval added is the same for both people, the difference between the ratio parts must be made equal.

Linear Equation Setup

For a person currently A years old, the age n years ago was (A − n) and the age n years later will be (A + n).

Shortcut Trick

Rate of interest = [(Difference in Amounts) ÷ (Amount after 2 years)] × 100

Rate (R) = [(8,651.50 − 7,865) ÷ 7,865] × 100 = (786.5 ÷ 7,865) × 100 = 10%

For R = 10%, Multiplying Factor for 2 years = (1.1)² = 1.21

Sum (P) = 7,865 ÷ 1.21 = ₹6,500

∴ The correct answer is ₹6,500.

Alternate Method

Given:

Amount after 2 years (A₂) = ₹7,865

Amount after 3 years (A₃) = ₹8,651.50

Time gap = 1 year

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

The interest for the 3^rd year is the compound interest earned on the amount at the end of the 2^nd year.

⇒ Interest = A₃ − A₂

⇒ Interest = 8,651.50 − 7,865 = ₹786.50

⇒ Rate (R) = (786.50 ÷ 7,865) × 100 = 10%

Now, using the formula for 2 years:

⇒ 7,865 = P(1 + 10/100)²

⇒ 7,865 = P(1.1)²

⇒ 7,865 = P × 1.21

⇒ P = 7,865 ÷ 1.21

⇒ P = 6,500

∴ The correct answer is ₹6,500.

Additional Information

Successive Amount Property

In compound interest, the amount after n years becomes the Principal for the (n+1)^th year.

Constant Ratio in CI

The ratio of amounts for equal time intervals is constant: A₂/A₁ = A₃/A₂ = (1 + R/100).

SI vs CI for 2 Years

The difference between Simple Interest and Compound Interest for 2 years is given by D = P(R/100)².

Shortcut Trick

In an isosceles triangle, the angles opposite to equal sides are equal. Since 110° is an obtuse angle, it cannot be repeated (as 110° + 110° > 180°).

Remaining sum for two equal angles = 180° − 110° = 70°

Individual base angles = 70° ÷ 2 = 35°

∴ The correct answer is 35degree and 35degree.

Alternate Method

Given:

One angle of an isosceles triangle = 110°

Formula Used:

Sum of all internal angles of a triangle = 180°

Calculations:

Let the equal angles of the isosceles triangle be x.

In any triangle, sum of angles = 180°

⇒ 110° + x + x = 180°

⇒ 110° + 2x = 180°

⇒ 2x = 180° − 110°

⇒ 2x = 70°

⇒ x = 70° ÷ 2

⇒ x = 35°

∴ The correct answer is 35degree and 35degree.

Additional Information

Angle Sum Property

The sum of the three interior angles of any triangle is always 180°.

Isosceles Triangle Property

In an isosceles triangle, the angles opposite to the equal sides are equal in measure.

Obtuse Isosceles Triangle

A triangle can have only one obtuse angle (greater than 90°). In an isosceles triangle, the obtuse angle must be the vertex angle.

Shortcut Trick

Net percentage change = a + b + (a × b)/100 %

Here, a = −10% (decrease) and b = +20% (increase)

⇒ Net change = −10 + 20 + (−10 × 20)/100 = 8% net increase

⇒ Final population = 5000 × 108% = 5400

∴ The correct answer is 5400.

Alternate Method

Given:

⇒ Initial population = 5,000

⇒ First year decrease = 10%

⇒ Second year increase = 20%

Formula Used:

⇒ Final Population = Initial Population × (1 − Decrease%/100) × (1 + Increase%/100)

    -10% +20%

Shortcut Trick

The difference between a two-digit number and its reversed version is always 9 × (difference of its digits).

Difference of digits = 27 ÷ 9 = 3.

Sum of digits is given as 11.

Digits are (11 + 3) ÷ 2 = 7 and (11 − 3) ÷ 2 = 4.

Since the reversed number is larger, the original number must be the smaller one, 47.

∴ The correct answer is 47.

Alternate Method

Given: Sum of digits = 11 | (Reversed Number − Original Number) = 27.

Let the tens digit be x and the units digit be y.

⇒ Original Number = 10x + y

⇒ x + y = 11 --- (Eq. 1)

⇒ Reversed Number = 10y + x

⇒ (10y + x) − (10x + y) = 27

⇒ 9y − 9x = 27 ⇒ y − x = 3 --- (Eq. 2)

⇒ Adding (Eq. 1) and (Eq. 2): 2y = 14 ⇒ y = 7

⇒ Substituting y in (Eq. 1): x + 7 = 11 ⇒ x = 4

⇒ Original Number = 10(4) + 7 = 47

∴ The correct answer is 47.

Additional Information

Difference between a number and its reverse

For a two-digit number, (10x + y) − (10y + x) = 9(x − y). This means the difference is always a multiple of 9.

Sum of a number and its reverse

For a two-digit number, (10x + y) + (10y + x) = 11(x + y). This means the sum is always a multiple of 11.

General Form

Any two-digit number with digits a and b can be expressed in expanded form as 10a + b.

Shortcut Trick

Assume total distance AB = 108 km (LCM of 3, 2, 9, 12, 18).

Onward time (A to B) = (36 km ÷ 12 km/hr) + (72 km ÷ 18 km/hr) = 3 + 4 = 7 hours.

Return time (B to A) = (54 km ÷ 9 km/hr) + (54 km ÷ 18 km/hr) = 6 + 3 = 9 hours.

Total distance = 108 + 108 = 216 km | Total time = 7 + 9 = 16 hours.

Average Speed = 216 ÷ 16 = 13.5 km/hr.

∴ The correct answer is 13.5 km/hr.

Alternate Method

Given:

Let the one-way distance between A and B be D.

Onward: 1/3 D at 12 km/hr and 2/3 D at 18 km/hr.

Return: 1/2 D at 9 km/hr and 1/2 D at 18 km/hr.

Formula: Average Speed = Total Distance ÷ Total Time

⇒ Total Distance = 2D

⇒ Time for onward journey (T₁) = (D/3 ÷ 12) + (2D/3 ÷ 18) = D/36 + D/27 = (3D + 4D)/108 = 7D/108

⇒ Time for return journey (T₂) = (D/2 ÷ 9) + (D/2 ÷ 18) = D/18 + D/36 = (2D + D)/36 = 3D/36 = D/12

⇒ Total Time = T₁ + T₂ = 7D/108 + D/12 = (7D + 9D)/108 = 16D/108 = 4D/27

⇒ Average Speed = 2D ÷ (4D/27) = (2 × 27) ÷ 4 = 54 ÷ 4 = 13.5 km/hr

∴ The correct answer is 13.5 km/hr.

Additional Information

Average Speed for Equal Distances

If a certain distance is covered at speed 'x' and the same distance is covered at speed 'y', then Average Speed = 2xy ÷ (x + y).

Average Speed for Equal Time

If a body travels at speed 'x' for time 't' and speed 'y' for the same time 't', then Average Speed = (x + y) ÷ 2.

Unit Conversion

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

Shortcut Trick

To find the median, arrange the data in ascending order.

Ordered list: 38, 39, 41, 42, 44, 45, 46, 47, 48, 49, 50, 52, 53, 55, 60.

For 15 observations (odd), the median is the middle term at position (15 + 1) ÷ 2 = 8^th.

The 8^th value in the sorted list is 47.

∴ The correct answer is 47 min.

Alternate Method

Given:

Number of students (n) = 15

Test completion times: 42, 38, 55, 47, 50, 44, 60, 39, 48, 52, 45, 41, 49, 46, 53

Formula Used:

Median (for odd n) = ((n + 1) ÷ 2)^th observation

Calculations:

⇒ Arrange the data in ascending order:

⇒ 38, 39, 41, 42, 44, 45, 46, 47, 48, 49, 50, 52, 53, 55, 60

⇒ Median position = (15 + 1) ÷ 2

⇒ Median position = 8^th term

⇒ Counting to the 8^th term in the ordered list:

⇒ 8^th term = 47

∴ The correct answer is 47 min.

Additional Information

Median for Even Observations

If n is even, the median is the arithmetic mean of the (n ÷ 2)^th and ((n ÷ 2) + 1)^th terms.

Empirical Relationship

For a moderately skewed distribution, the relationship between central tendencies is: Mode = 3 Median − 2 Mean.

Mean Calculation

The Arithmetic Mean is calculated as the sum of all observations divided by the total number of observations (Σx ÷ n).

Shortcut Trick

Compare the effective discount percentages for each scheme:

Scheme A: 5 + 21 − (5 × 21 ÷ 100) = 24.95%

Scheme B: Clearly much higher than 38% (first discount alone is 38%).

Scheme C: Average discount = [(6 × 8) + (1 × 27)] ÷ 7 = 75 ÷ 7 ≈ 10.71%

Scheme D: [Free items ÷ Total items] × 100 = 3 ÷ 10 × 100 = 30%

Comparing these, Scheme C offers the minimum discount.

∴ The correct answer is Scheme C.

Alternate Method

Assume the Marked Price (MP) of each toy is ₹100. For 7 toys, total MP = ₹700.

Formula Used:

⇒ Net Discount = Total MP − Total SP

⇒ Discount % = (Discount ÷ MP) × 100

Calculations:

⇒ Scheme A: Net Discount % = 5 + 21 − (5 × 21)/100 = 24.95%.

⇒ Scheme B: This involves three successive discounts (38%, 33%, 21%). Even the first discount (38%) is greater than the others, making it highly beneficial.

⇒ Scheme C: Discount on first 6 toys = 6 × 8% of 100 = ₹48. Discount on 7^th toy = 1 × 27% of 100 = ₹27. Total Discount = 48 + 27 = ₹75. Discount % = (75 ÷ 700) × 100 = 10.71%.

⇒ Scheme D: Buy 7 get 3 free implies 3 free out of 10 total toys. Discount % = (3 ÷ 10) × 100 = 30%.

⇒ Comparing all, 10.71% is the lowest discount value, making it the least beneficial to the customer.

∴ The correct answer is Scheme C.

Additional Information

Successive Discount Formula

For two successive discounts of a% and b%, the equivalent single discount is given by [a + b − (ab ÷ 100)]%.

Buy X Get Y Free Discount

The effective discount percentage in a 'Buy X Get Y Free' offer is calculated as [Y ÷ (X + Y)] × 100.

Marked Price and Selling Price

Selling Price (SP) = Marked Price (MP) × [ (100 − Discount %) ÷ 100 ]. The higher the discount, the lower the SP and the more beneficial it is for the buyer.

Shortcut Trick

When the Cost Price (CP) of two items is the same, the overall profit or loss percentage is the simple arithmetic average of the individual percentages.

Overall Percentage = (Profit% + Loss%) ÷ 2

Net Percentage = (+20% − 10%) ÷ 2 = 10% ÷ 2 = 5%

Positive result indicates a gain.

∴ The correct answer is 5% gain.

Alternate Method

Given:

CP of first item = ₹100, Profit = 20%

CP of second item = ₹100, Loss = 10%

Formula Used:

Selling Price (SP) = CP × (100 ± P/L)%

Overall % = (Total SP − Total CP) ÷ Total CP × 100

⇒ Total CP = 100 + 100 = ₹200

⇒ SP of first item = 100 + 20% of 100 = 100 + 20 = ₹120

⇒ SP of second item = 100 − 10% of 100 = 100 − 10 = ₹90

⇒ Total SP = 120 + 90 = ₹210

⇒ Overall Gain = Total SP − Total CP = 210 − 200 = ₹10

⇒ Overall Gain % = (10 ÷ 200) × 100 = 5%

∴ The correct answer is 5% gain.

Additional Information

Same Selling Price Rule

If two articles are sold at the same Selling Price (SP), one at x% profit and another at x% loss, there is always an overall loss of (x/10)²%.

Profit and Loss Basics

Profit % = (Profit/CP) × 100 and Loss % = (Loss/CP) × 100. Unless specified, Cost Price (CP) is always the base for calculations.

Successive Percentage Change

For two successive changes of a% and b%, the effective change is given by the formula [a + b + (ab/100)]%.

Shortcut Trick

Find the LCM of the given divisors: 14, 15, and 8.

LCM(14, 15, 8) = 840.

The largest 4-digit multiple of 840 is 840 × 11 = 9240.

Add the required remainder (4) to this multiple: 9240 + 4 = 9244.

∴ The correct answer is 9244.

Alternate Method

Given:

Divisors = 14, 15, 8

Remainder = 4

Formula Used:

Required Number = (Largest 4-digit multiple of LCM) + Remainder

⇒ Prime factorization: 14 = 2 × 7, 15 = 3 × 5, 8 = 2³.

⇒ LCM = 2³ × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840.

⇒ Largest 4-digit number = 9999.

⇒ Dividing 9999 by 840: 9999 ÷ 840 = 11 (Quotient) and 759 (Remainder).

⇒ Largest 4-digit multiple of 840 = 9999 − 759 = 9240.

⇒ Required number = 9240 + 4 = 9244.

∴ The correct answer is 9244.

Additional Information

LCM and Remainder

When a number N is divided by a, b, and c leaving the same remainder r, the general form is N = k × LCM(a, b, c) + r, where k is an integer.

Different Remainders

If N divided by a, b, c leaves remainders r₁, r₂, r₃ such that (a−r₁) = (b−r₂) = (c−r₃) = d, then N = k × LCM(a, b, c) − d.

LCM Calculation

LCM is the product of the highest powers of all prime factors involved in the given numbers.

Shortcut Trick

Equivalent discount for successive discounts of 20% and 12% = 20 + 12 − (20 × 12 / 100) = 32 − 2.4 = 29.6%

Difference in discount percentage = 30% (Flat) − 29.6% (Successive) = 0.4%

Difference in Selling Price = 0.4% of Marked Price

Difference = 0.4% × 80,000 = 0.004 × 80,000 = 320

∴ The correct answer is ₹320.

Alternate Method

Given:

Marked Price (MP) = ₹80,000

Option 1: Flat discount of 30%

Option 2: Successive discounts of 20% and 12%

Formula Used:

Selling Price (SP) = MP × (100 − D%) / 100

For successive discounts d₁ and d₂, Net SP = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ Selling Price 1 (Option 1) = 80,000 × (70 / 100) = ₹56,000

⇒ Price after first discount in Option 2 = 80,000 × (80 / 100) = ₹64,000

⇒ Selling Price 2 (Option 2) = 64,000 × (88 / 100) = ₹56,320

⇒ Difference in Selling Price = ₹56,320 − ₹56,000 = ₹320

Thus, the difference between the two options is ₹320, which matches Option (4).

∴ The correct answer is ₹320.

Additional Information

Successive Discount Formula

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

Marked Price and Selling Price

Selling Price is the price at which an item is sold after applying discounts on the Marked Price (MP): SP = MP − Discount.

Comparison of Discounts

A single flat discount is usually greater than or equal to successive discounts of the same total numerical value (e.g., 30% is better for the customer than 20% + 10%).

Shortcut Trick

The time interval for both to sound together is the LCM of their individual intervals.

LCM(12, 18) = 36 minutes.

Total duration from 9:00 AM to 11:00 AM = 2 hours = 120 minutes.

Number of times they sound again = 120 ÷ 36 = 3 (integer quotient).

∴ The correct answer is 3.

Alternate Method

Given:

Frequency of school drum = every 12 min

Frequency of school bell = every 18 min

First simultaneous sound = 9:00 AM

End time = 11:00 AM

Calculations:

⇒ Prime factors of 12 = 2 × 2 × 3

⇒ Prime factors of 18 = 2 × 3 × 3

⇒ LCM(12, 18) = 2 × 2 × 3 × 3 = 36 minutes

⇒ The instruments sound together at intervals of 36 minutes.

⇒ Time instances: 9:00 AM + 36 min = 9:36 AM

⇒ 9:36 AM + 36 min = 10:12 AM

⇒ 10:12 AM + 36 min = 10:48 AM

⇒ 10:48 AM + 36 min = 11:24 AM (which is after 11:00 AM)

⇒ The soundings occurring again before 11:00 AM are at 9:36, 10:12, and 10:48.

∴ The correct answer is 3.

Additional Information

LCM for Simultaneous Events

To find when multiple cyclic events occur together, find the Least Common Multiple (LCM) of their time periods.

Total Occurrences Formula

Total times = (Total Duration ÷ LCM) + 1, if the starting point is included. If only again is asked, exclude the start.

Time Conversion

Always convert time intervals to a common unit (usually minutes or seconds) before calculating the LCM.