Quantitative Aptitude PYQs

Shortcut Trick

Use complementary angle relations: cot 72° = tan 18° and tan 72° = cot 18°.

Group terms: (sec² 18° − tan² 18°) + (cot² 18° − cosec² 18°) + sin² 30° + cosec² 60°.

Apply identities:

(1) + (−1) + sin² 30° + cosec² 60°.

Value = 0 + (1/2)² + (2/√3)² = 1/4 + 4/3.

Total = (3 + 16) / 12 = 19/12.

∴ The correct answer is 19/12.

Alternate Method

Given: sec² 18° − cot² 72° + sin² 30° + cosec² 60° − cosec² 18° + tan² 72°

Formula Used: cot(90° − θ) = tan θ | tan(90° − θ) = cot θ | sec² θ − tan² θ = 1 | cosec² θ − cot² θ = 1

⇒ sec² 18° − tan²(90° − 72°) + sin² 30° + cosec² 60° − cosec² 18° + cot²(90° − 72°)

⇒ sec² 18° − tan² 18° + sin² 30° + cosec² 60° − cosec² 18° + cot² 18°

⇒ (sec² 18° − tan² 18°) + sin² 30° + cosec² 60° − (cosec² 18° − cot² 18°)

⇒ 1 + (1/2)² + (2/√3)² − 1

⇒ 1/4 + 4/3

⇒ (3 + 16) / 12

⇒ 19/12

∴ The correct answer is 19/12.

Additional Information

Complementary Angle Identities

For angles adding to 90°, ratios convert: sin ↔ cos, tan ↔ cot, sec ↔ cosec.

Pythagorean Identities

The three basic identities are sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, and cosec²θ − cot²θ = 1.

Reciprocal Trigonometric Values

Standard values for common angles: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3, and their reciprocals.

Shortcut Trick

Mode is the observation that occurs most frequently in the data set.

Count occurrences: 68 appears 4 times.

Count occurrences: 85 appears 3 times.

Count occurrences: 64 appears 3 times.

Other values (72, 76, etc.) occur only 1 or 2 times.

∴ The correct answer is 68.

Alternate Method

Given data: 85, 73, 89, 82, 67, 61, 78, 69, 86, 68, 64, 76, 64, 85, 68, 85, 68, 81, 68, 72, 64, 72, 76, 84, 88

Formula Used: Mode = Value with the maximum frequency.

⇒ Frequency of 61 = 1

⇒ Frequency of 64 = 3

⇒ Frequency of 68 = 4

⇒ Frequency of 72 = 2

⇒ Frequency of 76 = 2

⇒ Frequency of 85 = 3

⇒ All other values (67, 69, 73, 78, 81, 82, 84, 86, 88, 89) occur 1 time each.

⇒ The value 68 occurs the most number of times (4).

∴ The correct answer is 68.

Additional Information

Empirical Formula

For a moderately skewed distribution, the relationship between Mean, Median, and Mode is: Mode = (3 × Median) − (2 × Mean).

Bimodal and Multimodal Data

If two or more values have the same maximum frequency, the data set is said to be bimodal (two modes) or multimodal (multiple modes).

Mean and Median

Mean is the average calculated by Σx ÷ n. Median is the middle value of the data set when arranged in ascending or descending order.

Shortcut Trick

The average of the first 'n' odd natural numbers is always equal to 'n'.

Here, the total number of terms (n) = 16.

Average = 16.

∴ The correct answer is 16.

Alternate Method

Given: n = 16 odd natural numbers.

Formula: Average of an Arithmetic Progression (A.P.) = (First term + Last term) ÷ 2

⇒ First odd natural number (a) = 1

⇒ 16^th odd natural number (l) = a + (n − 1) × d = 1 + (16 − 1) × 2 = 31

⇒ Total Sum = (n ÷ 2) × (a + l) = (16 ÷ 2) × (1 + 31) = 8 × 32 = 256

⇒ Average = Total Sum ÷ n = 256 ÷ 16 = 16

∴ The correct answer is 16.

Additional Information

Average of Even Numbers

The average of the first n even natural numbers is always (n + 1).

Average of Consecutive Natural Numbers

The average of the first n natural numbers is given by (n + 1) ÷ 2.

Sum of First n Odd Numbers

The sum of the first n odd natural numbers is n².

Average of Squares

The average of squares of first n natural numbers is [(n + 1)(2n + 1)] ÷ 6.

Shortcut Trick

Let the total number of registered voters be X.

Apply the successive percentage factors: X × (Cast%) × (Valid%) × (Candidate Share%) = 9261

⇒ X × 75/100 × 98/100 × 75/100 = 9261

⇒ X × 3/4 × 49/50 × 3/4 = 9261

⇒ X × 441/800 = 9261 ⇒ X = 21 × 800 = 16800

∴ The correct answer is 16,800.

Alternate Method

Given:

Votes cast = 75% of Registered Voters

Invalid votes = 2% of cast votes ⇒ Valid votes = 98% of cast votes

Candidate share = 75% of valid votes = 9261

Formula: Registered Voters = (Winner Votes × 100 × 100 × 100) ÷ (Cast% × Valid% × Share%)

⇒ Total Voters = (9261 × 100 × 100 × 100) ÷ (75 × 98 × 75)

⇒ Total Voters = (9261 × 4 × 50 × 4) ÷ (3 × 49 × 3)

⇒ Total Voters = (9261 × 800) ÷ (9 × 49)

⇒ Total Voters = (9261 × 800) ÷ 441

⇒ Total Voters = 21 × 800

⇒ Total Voters = 16800

∴ The correct answer is 16,800.

Additional Information

Election Vote Distribution

Total Registered = Cast + Not Cast; Cast = Valid + Invalid. Calculations are always performed on Valid votes unless specified otherwise.

Successive Percentage Method

When multiple changes occur on a base value, the final value = Initial × (a/100) × (b/100) × (c/100). This is the fastest way to solve election and population problems.

Valid vs Invalid Ratio

If p% are invalid, then valid votes = Total Cast × (100 − p)/100.

Shortcut Trick

Effective successive discount of 30% and 20% = 30 + 20 − (30 × 20)/100 = 44%

Difference in discount percentages = 46% − 44% = 2%

Given that 2% of the Marked Price = ₹256

Marked Price (100%) = (256 ÷ 2) × 100 = ₹12,800

∴ The correct answer is ₹12,800.

Alternate Method

Given:

Single discount = 46%

Successive discounts = 30% and 20%

Difference in discount amount = ₹256

Formula Used:

Equivalent discount for two successive discounts a% and b% = (a + b − ab/100)%

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

⇒ Effective discount of 30% and 20% = (30 + 20 − (30 × 20)/100)%

⇒ Effective discount = 50 − 6 = 44%

⇒ Difference between first and second discount scenarios = 46% − 44% = 2%

⇒ According to the problem: 2% of Marked Price = 256

⇒ Marked Price = (256 × 100) ÷ 2

⇒ Marked Price = 128 × 100 = 12,800

∴ The correct answer is ₹12,800.

Additional Information

Single Equivalent Discount Formula

For two successive discounts of x% and y%, the single equivalent discount is (x + y − xy/100)%.

Relationship between MP and SP

The Selling Price after a discount d% on Marked Price is calculated as: SP = MP × (100 − d)/100.

Successive Discounts (Three or more)

For three successive discounts d₁, d₂, and d₃, the final SP is MP × (1 − d₁/100) × (1 − d₂/100) × (1 − d₃/100).

Shortcut Trick

When the distance is the same for both journeys, use the Harmonic Mean formula:

Average Speed = (2 × s₁ × s₂) / (s₁ + s₂)

⇒ Average Speed = (2 × 10 × 40) / (10 + 40)

⇒ Average Speed = 800 / 50 = 16 km/hr.

∴ The correct answer is 16 km/hr.

Alternate Method

Given:

Speed while going (s₁) = 10 km/hr

Speed while returning (s₂) = 40 km/hr

Formula Used:

Average Speed = Total Distance ÷ Total Time

Let the distance between home and school be the LCM of 10 and 40, which is 40 km.

⇒ Distance (D) = 40 km

⇒ Time taken while going (t₁) = Distance / s₁ = 40 / 10 = 4 hours

⇒ Time taken while returning (t₂) = Distance / s₂ = 40 / 40 = 1 hour

⇒ Total Distance = 40 + 40 = 80 km

⇒ Total Time = 4 + 1 = 5 hours

⇒ Average Speed = 80 / 5 = 16 km/hr

∴ The correct answer is 16 km/hr.

Additional Information

Average Speed for different distances

When distances are not equal, use the general formula: Average Speed = (d₁ + d₂ + ... + d_n) / (t₁ + t₂ + ... + t_n).

Three equal distances

If a journey is divided into three equal parts covered at speeds x, y, and z, the average speed is (3xyz) / (xy + yz + zx).

Speed Conversion

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

Shortcut Trick

Total deduction = 29 + 70 + 28 = ₹127

Remaining amount = 13,525 − 127 = ₹13,398

Sum of ratio parts = 19 + 6 + 8 = 33 units

1 unit = 13398 ÷ 33 = 406

Difference (B & C) = |(8 × 406 + 28) − (6 × 406 + 70)| = 812 − 42 = 770

∴ The correct answer is ₹770.

Alternate Method

Given:

Total amount = ₹13,525

Deductions from A, B, C = ₹29, ₹70, ₹28

Ratio after deduction = 19 : 6 : 8

Sum of remaining amount = 13,525 − (29 + 70 + 28)

⇒ 13,525 − 127 = 13,398

⇒ Let the shares after deduction be 19x, 6x, and 8x.

⇒ 19x + 6x + 8x = 13,398

⇒ 33x = 13,398

⇒ x = 13,398 ÷ 33 = 406

⇒ B's original share = 6x + 70 = 6 × 406 + 70 = 2,436 + 70 = 2,506

⇒ C's original share = 8x + 28 = 8 × 406 + 28 = 3,248 + 28 = 3,276

⇒ Difference = 3,276 − 2,506 = 770

∴ The correct answer is ₹770.

Additional Information

Ratio and Proportion Basics

A ratio a : b indicates that for every 'a' units of the first quantity, there are 'b' units of the second. To find individual values, divide the total sum by the sum of ratio parts (a + b) to find the value of one unit.

Handling Deductions in Ratios

When fixed amounts are deducted from shares to reach a specific ratio, subtract the total deduction from the grand total before applying the ratio distribution formula.

Unitary Method in Ratios

Value of 'n' parts = (Total Value ÷ Total Ratio Parts) × n. This is the fastest way to calculate individual components in banking exams.

Shortcut Trick

Ratio of quantities = 10 : 15 = 2 : 3

Weighted Average CP = [(2 × 20) + (3 × 30)] ÷ (2 + 3)

⇒ 130 ÷ 5 = ₹ 26 per kg

Selling price for 10% profit = 26 × 110% = ₹ 28.60

∴ The correct answer is ₹ 28.60.

Alternate Method

Given:

Quantity₁ = 10 kg, Price₁ = ₹ 20/kg

Quantity₂ = 15 kg, Price₂ = ₹ 30/kg

Desired Profit = 10%

Formula Used:

Total CP = (Q₁ × P₁) + (Q₂ × P₂)

Selling Price (SP) = CP_avg × (100 + Profit%)/100

⇒ Total cost of 10 kg wheat = 10 × 20 = ₹ 200

⇒ Total cost of 15 kg wheat = 15 × 30 = ₹ 450

⇒ Total CP of 25 kg mixture = 200 + 450 = ₹ 650

⇒ Cost price per kg of mixture = 650 ÷ 25 = ₹ 26

⇒ To earn 10% profit, SP = 26 + (10% of 26)

⇒ SP = 26 + 2.60 = ₹ 28.60

∴ The correct answer is ₹ 28.60.

Additional Information

Weighted Average

The average price of a mixture is calculated as (n₁c₁ + n₂c₂) ÷ (n₁ + n₂), where n is quantity and c is cost.

Profit Percentage

Profit is always calculated on the Cost Price (CP). Selling Price = CP × (100 + Profit%) ÷ 100.

Alligation Rule

It is a shorter way to find the ratio in which two ingredients are mixed: (Higher Price − Mean Price) : (Mean Price − Lower Price).

Shortcut Trick

Total capacity = LCM of (15, 35, 60) = 420 units.

Efficiencies: A = 420/15 = 28 | B = 420/35 = 12 | C = −420/60 = −7 (Outlet).

Combined Efficiency (A + B + C) = 28 + 12 − 7 = 33 units/min.

Required time = 420 ÷ 33 = 140/11 minutes.

Time = 12⁸/₁₁ minutes.

∴ The correct answer is 12⁸/₁₁ minutes.

Alternate Method

Given:

Time for Pipe A = 15 mins

Time for Pipe B = 35 mins

Time for Pipe C (Emptying) = 1 hour = 60 mins

Formula used: Net Efficiency = Efficiency of A + Efficiency of B − Efficiency of C

Combined efficiency of A + B + C = 28 + 12 − 7 = 33 units/min

⇒ Total capacity = LCM(15, 35, 60) = 420 units

⇒ Time to fill cistern = Capacity ÷ Combined Efficiency

⇒ Time = 420 ÷ 33

⇒ Time = 140 ÷ 11

⇒ Time = 12⁸/₁₁ minutes

∴ The correct answer is 12⁸/₁₁ minutes.

Additional Information

Inlet and Outlet Pipes

Work done by an inlet pipe is always taken as positive, whereas work done by an outlet pipe is taken as negative.

Net Work Formula

If pipe A fills in x hours and pipe B empties in y hours (y > x), then the part filled in 1 hour is (1/x − 1/y).

Efficiency Relation

Efficiency is inversely proportional to the time taken to complete the work (Efficiency ∝ 1/Time).

Shortcut Trick

Convert both values to a common form (percentage).

66% of 1^st Number = (3/5 × 100)% of 2^nd Number

66% of 1^st Number = 60% of 2^nd Number

Ratio (1^st : 2^nd) = Inverse of their coefficients = 60 : 66

Simplifying by dividing by 6 ⇒ 10 : 11

∴ The correct answer is 10 : 11.

Alternate Method

Given:

Percentage of first number = 66%

Fraction of second number = 3/5

Formula:

Value = (Percentage/100) × Number | Ratio A : B = A/B

Let the first number be x and the second number be y.

⇒ 66% of x = (3/5) of y

⇒ (66/100) × x = (3/5) × y

⇒ (33/50) × x = (3/5) × y

⇒ x/y = (3/5) ÷ (33/50)

⇒ x/y = (3/5) × (50/33)

⇒ x/y = (1 × 10) / (1 × 11)

⇒ x/y = 10/11

∴ The correct answer is 10 : 11.

Additional Information

Percentage to Fraction Conversion

To convert x% to a fraction, divide by 100: x/100. Conversely, multiply a fraction by 100 to get the percentage.

Ratio Balancing Rule

If m × A = n × B, then the ratio A : B is equal to n : m. This is known as the inverse coefficient rule.

Simplifying Ratios

A ratio a : b can be simplified by dividing both terms by their Highest Common Factor (HCF) to reach the simplest form.

Shortcut Trick

Income ratio = 8 : 5 and Expenditure ratio = 2 : 1

Savings are same (₹160), use cross-product method:

| (8 × 1) − (5 × 2) | units = | 160 × (1 − 2) |

⇒ | 8 − 10 | units = | −160 | ⇒ 2 units = 160

⇒ 1 unit = 80

Difference in income = (8 − 5) = 3 units = 3 × 80 = ₹240

∴ The correct answer is ₹240.

Alternate Method

Given: Ratio of Incomes = 8 : 5, Ratio of Expenditures = 2 : 1, Individual Savings = ₹160.

Formula: Income − Expenditure = Savings

Let the incomes be 8x and 5x.

Let the expenditures be 2y and y.

⇒ 8x − 2y = 160 ...

(i) ⇒ 5x − y = 160 ⇒ y = 5x − 160 ...

(ii) Substituting the value of y from equation (ii) into equation (i):

⇒ 8x − 2(5x − 160) = 160

⇒ 8x − 10x + 320 = 160

⇒ −2x = 160 − 320

⇒ −2x = −160 ⇒ x = 80

⇒ Monthly income of first person = 8 × 80 = ₹640

⇒ Monthly income of second person = 5 × 80 = ₹400

⇒ Difference = 640 − 400 = ₹240

∴ The correct answer is ₹240.

Additional Information

Income, Expenditure and Savings

The basic relationship is Income = Expenditure + Savings. If savings are equal for both parties, the difference in income units is proportional to the difference in expenditure units after scaling.

Ratio Balancing

To solve without algebra, multiply the expenditure ratio by a factor so that the difference (Income Unit − Expenditure Unit) is equal for both persons.

Proportion Rule

If A:B is the ratio of two quantities, the difference is expressed as |A − B| units. To find the real value, multiply this unit difference by the value of 1 unit.

Shortcut Trick

Total Surface Area (TSA) ratio units = 2 × (4×3 + 3×2 + 2×4) = 52 units²

Scaling factor squared (x²) = Total Area ÷ Ratio units = 832 ÷ 52 = 16

Linear scaling factor (x) = √16 = 4 cm

Volume ratio units = 4 × 3 × 2 = 24 units³

Actual Volume = 24 × (x)³ = 24 × 4³ = 24 × 64 = 1536 cm³

∴ The correct answer is 1536 cm³.

Alternate Method

Given:

Ratio of length (L) : breadth (B) : height (H) = 4 : 3 : 2

Total Surface Area (TSA) = 832 cm²

Formula Used:

Total Surface Area of cuboid = 2(LB + BH + HL)

Volume of cuboid = L × B × H

Calculations:

⇒ Let Length (L) = 4x, Breadth (B) = 3x, and Height (H) = 2x

⇒ 2[(4x × 3x) + (3x × 2x) + (2x × 4x)] = 832

⇒ 2[12x² + 6x² + 8x²] = 832

⇒ 2[26x²] = 832

⇒ 52x² = 832

⇒ x² = 832 ÷ 52 = 16

⇒ x = √16 = 4

⇒ L = 4 × 4 = 16 cm

⇒ B = 3 × 4 = 12 cm

⇒ H = 2 × 4 = 8 cm

⇒ Volume = 16 × 12 × 8

⇒ Volume = 1536 cm³

∴ The correct answer is 1536 cm³.

Additional Information

Lateral Surface Area (LSA)

The surface area of the four vertical walls of the cuboid is 2h(l + b).

Diagonal of a Cuboid

The length of the longest rod that can fit in a cuboid is √(l² + b² + h²).

Properties of a Cube

A cube is a special cuboid where l = b = h = a. TSA = 6a² and Volume = a³.

Shortcut Trick

Ratio (A : B) 9 yrs ago = 3 : 1 (Difference = 2)

Ratio (A : B) 7 yrs hence = 2 : 1 × 2 = 4 : 2 (Difference = 2)

Total time gap = 9 + 7 = 16 years

Unit difference in B’s ratio = 2 − 1 = 1 unit

⇒ 1 unit = 16 years

Bhawna’s age 9 yrs ago = 1 unit = 16 years

Bhawna’s present age = 16 + 9 = 25 years

∴ The correct answer is 25 years.

Alternate Method

Given: 9 years ago Asha = 3 × Bhawna. 7 years hence Asha = 2 × Bhawna.

Let the present age of Bhawna be B and Asha be A.

⇒ (A − 9) = 3(B − 9)

⇒ A − 9 = 3B − 27

⇒ A = 3B − 18 — (i)

⇒ (A + 7) = 2(B + 7)

⇒ A + 7 = 2B + 14

⇒ A = 2B + 7 — (ii)

Equating (i) and (ii):

⇒ 3B − 18 = 2B + 7

⇒ 3B − 2B = 7 + 18

⇒ B = 25 years

∴ The correct answer is 25 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals always remains the same, regardless of how many years pass.

Ratio Balancing

In age problems involving ratios, multiply the ratios by the difference of the other ratio to equalize the unit gaps between years.

Average Age Property

If the average age of a group of n people is X, then after k years, the new average age will be X + k.

Shortcut Trick

To minimize sections, maximize students per section using HCF.

Students per section = HCF(156, 143)

156 = 13 × 12 and 143 = 13 × 11 ⇒ HCF = 13

Number of sections = (156 ÷ 13) + (143 ÷ 13)

Number of sections = 12 + 11 = 23

∴ The correct answer is 23.

Alternate Method

Given:

Number of boy students = 156

Number of girl students = 143

Condition: Each section must have same number of students and contain only boys or girls.

Formula Used:

To have the fewest sections, the size of each section must be the Highest Common Factor (HCF) of the number of boys and girls.

Total Sections = (Number of Boys ÷ HCF) + (Number of Girls ÷ HCF)

Calculations:

⇒ Prime factors of 156 = 2 × 2 × 3 × 13

⇒ Prime factors of 143 = 11 × 13

⇒ Common factor (HCF) = 13

⇒ Number of boy sections = 156 ÷ 13 = 12

⇒ Number of girl sections = 143 ÷ 13 = 11

⇒ Total number of sections = 12 + 11

⇒ Total sections = 23

∴ The correct answer is 23.

Additional Information

Highest Common Factor (HCF)

It is the largest positive integer that divides each of the integers without a remainder. It is used to find the maximum size for equal grouping.

Least Common Multiple (LCM)

LCM is the smallest positive integer that is divisible by both a and b. It is used for problems involving cycles or meeting times.

HCF × LCM Property

For two numbers a and b, Product of numbers = HCF(a, b) × LCM(a, b).

Shortcut Trick

Unit digit calculation: Unit digit of 13³ is 7 (3³ = 27).

Unit digit of 8³ is 2 (8³ = 512).

Difference in unit digits = 7 − 2 = 5.

Comparing options, only 1685 and 1725 end in 5.

Since 13³ − 8³ = 2197 − 512 = 1685.

∴ The correct answer is 1685.

Alternate Method

Given: 13³ − 8³

Formula used: a³ − b³ = (a − b)(a² + ab + b²)

Here, a = 13 and b = 8

⇒ (13 − 8) × (13² + 13 × 8 + 8²)

⇒ 5 × (169 + 104 + 64)

⇒ 5 × 337

⇒ 1685

∴ The correct answer is 1685.

Additional Information

Sum of Cubes Identity

The sum of two cubes is given by a³ + b³ = (a + b)(a² − ab + b²).

Difference of Squares

The difference of two squares is given by a² − b² = (a − b)(a + b).

Common Cubes

11³ = 1331, 12³ = 1728, 13³ = 2197, 14³ = 2744, 15³ = 3375.

Shortcut Trick

Surface area of a sphere = 4 × Area of its great circle

Area of great circle = π × (radius)² = π × 3² = 9π

Surface Area = 4 × 9π = 36π cm²

∴ The correct answer is 36 π cm².

Alternate Method

Given: Radius of the sphere (r) = 3 cm

Formula Used: Total Surface Area of a Sphere = 4πr²

⇒ Surface Area = 4 × π × (3)²

⇒ Surface Area = 4 × π × 9

⇒ Surface Area = 36π cm²

∴ The correct answer is 36 π cm².

Additional Information

Volume of a Sphere

The amount of space occupied by a sphere is given by V = (4/3)πr³.

Surface Area of a Hemisphere

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr².

Ratio of Surface Area to Volume

For a sphere, the ratio of its surface area to its volume is 3/r.

Shortcut Trick

Single equivalent discount formula for two successive discounts a% and b%:

Equivalent Discount = [a + b − (a × b / 100)]%

⇒ Equivalent Discount = [28 + 16 − (28 × 16 / 100)]%

⇒ Equivalent Discount = [44 − 4.48]% = 39.52%

∴ The correct answer is 39.52%.

Alternate Method

Given: First discount (d₁) = 28%, Second discount (d₂) = 16%

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

⇒ Let the Marked Price (MP) be 100 units.

⇒ Price after 1^st discount = 100 − (28% of 100) = 100 − 28 = 72 units.

⇒ Price after 2^nd discount = 72 − (16% of 72)

⇒ Second discount amount = (16 × 72) / 100 = 1152 / 100 = 11.52 units.

⇒ Final Selling Price (SP) = 72 − 11.52 = 60.48 units.

⇒ Total Discount = MP − SP = 100 − 60.48 = 39.52 units.

⇒ Single Equivalent Discount Percentage = (39.52 / 100) × 100 = 39.52%

∴ The correct answer is 39.52%.

Additional Information

Three Successive Discounts

If three discounts a%, b%, and c% are given, the equivalent discount is 1 − [(1 − a/100)(1 − b/100)(1 − c/100)] × 100.

Successive Percentage Increase

For two successive increases of x% and y%, the effective increase is [x + y + (xy/100)]%.

Relation between SP, MP and Discount

Selling Price (SP) = Marked Price (MP) × (100 − Discount%)/100.

Shortcut Trick

Effective Rate of interest for 2 years = R + R + (R × R)/100

Effective Rate = 15 + 15 + (15 × 15)/100 = 32.25%

Total Interest = 15000 × 32.25%

Total Interest = 150 × 32.25 = 4837.50

∴ The correct answer is ₹4,837.50.

Alternate Method

Given:

Principal (P) = ₹15,000

Rate (R) = 15% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Compound Interest (CI) = A − P

⇒ Amount (A) = 15000 × (1 + 15/100)²

⇒ A = 15000 × (1.15)²

⇒ A = 15000 × 1.3225

⇒ A = 19837.50

⇒ Interest = 19837.50 − 15000

⇒ Interest = 4837.50

∴ The correct answer is ₹4,837.50.

Additional Information

Compound Interest Formula

CI = P[(1 + R/100)ⁿ − 1], where P is principal, R is rate, and n is time in years.

Difference between CI and SI

The difference for 2 years is given by CI − SI = P(R/100)².

Half-Yearly Compounding

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

Shortcut Trick

Solve following BODMAS order from inner to outer brackets:

Parentheses: (21 − 11 + 9) = 19

Curly Braces: {31 − 19} = 12

Square Brackets: [25 − 12] = 13

Final calculation: 35 − 13 = 22

∴ The correct answer is 22.

Alternate Method

Given expression: 35 − [25 − {31 − (21 − 11 + 9)}]

Formula used: BODMAS Rule (Brackets, Of, Division, Multiplication, Addition, Subtraction)

⇒ 35 − [25 − {31 − (10 + 9)}]

⇒ 35 − [25 − {31 − 19}]

⇒ 35 − [25 − 12]

⇒ 35 − 13

⇒ 22

∴ The correct answer is 22.

Additional Information

BODMAS / VBODMAS Rule

It defines the order of operations: V (Vinculum/Bar), B (Brackets), O (Of), D (Division), M (Multiplication), A (Addition), S (Subtraction).

Types of Brackets

Brackets must be opened in this specific order: Vinculum (line), then Circular brackets (), then Curly braces {}, and finally Square brackets [].

Addition and Subtraction Priority

When only addition and subtraction remain, operations are performed strictly from left to right to ensure accuracy.

Shortcut Trick

Net Discount Formula for two successive discounts a% and b%:

Net Discount = [a + b − (a × b / 100)]%

Here, a = 40 and b = 15

Net Discount = [40 + 15 − (40 × 15 / 100)]% = [55 − 6]% = 49%

∴ The correct answer is 49%.

Alternate Method

Given:

First discount (d₁) = 40%

Second discount (d₂) = 15%

Formula: Final Selling Price = Marked Price × (1 − d₁/100) × (1 − d₂/100)

Let the Marked Price (MP) be 100 units.

⇒ Price after 1^st discount = 100 − (40% of 100)

⇒ Price after 1^st discount = 100 − 40 = 60 units

⇒ Price after 2^nd discount = 60 − (15% of 60)

⇒ Price after 2^nd discount = 60 − (15/100 × 60)

⇒ Price after 2^nd discount = 60 − 9 = 51 units

⇒ Single Equivalent Discount = (MP − Final Price) / MP × 100

⇒ Single Equivalent Discount = (100 − 51) / 100 × 100

⇒ Single Equivalent Discount = 49%

∴ The correct answer is 49%.

Additional Information

Three Successive Discounts

For three successive discounts a%, b%, and c%, calculate the net of a and b first, then use that result with c using the formula x + y − xy/100.

Marked Price and Selling Price Relation

Selling Price = MP × (100 − d₁)/100 × (100 − d₂)/100. This is useful for large number calculations.

Successive Percentage Increase

If values increase successively, use the formula: a + b + (ab / 100)% for the overall percentage change.