Quantitative Aptitude PYQs

Shortcut Trick

Assume a base average of 7.

Calculate deviations from 7: (6−7), (8−7), (7−7), (9−7), (5−7) ⇒ −1, +1, 0, +2, −2.

Net deviation = −1 + 1 + 0 + 2 − 2 = 0.

Required Mean = Base Average + (Net Deviation ÷ Count) = 7 + (0 ÷ 5) = 7.

∴ The correct answer is 7.

Alternate Method

Given:

Scores of five friends: 6, 8, 7, 9, and 5.

Total number of friends (n) = 5.

Formula Used:

Arithmetic Mean = (Sum of all observations) ÷ (Total number of observations)

Calculations:

⇒ Sum of scores = 6 + 8 + 7 + 9 + 5

⇒ Sum of scores = 35

⇒ Arithmetic Mean = 35 ÷ 5

⇒ Arithmetic Mean = 7

∴ The correct answer is 7.

Additional Information

Weighted Arithmetic Mean

When observations have different levels of importance, the mean is calculated as Σ(w_i × x_i) ÷ Σw_i, where w is the weight.

Properties of Mean

The sum of deviations of all observations from their arithmetic mean is always zero.

Median vs Mean

While the mean is the average, the median is the middle value of a sorted data set. For the set (5, 6, 7, 8, 9), the median is also 7.

Shortcut Trick

If the sum of three numbers a + b + c = 0, then a³ + b³ + c³ = 3abc.

Check the sum of given numbers: 22 + (−15) + (−7) = 22 − 22 = 0.

Apply the trick: 3 × 22 × (−15) × (−7).

Calculation: 66 × 105 = 6930.

∴ The correct answer is 6930.

Alternate Method

Given Expression: 22³ + (−15)³ + (−7)³

Let a = 22, b = −15, and c = −7.

Formula Used:

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Calculations:

⇒ First, find the sum (a + b + c):

⇒ a + b + c = 22 + (−15) + (−7)

⇒ a + b + c = 22 − 15 − 7 = 0

⇒ Since a + b + c = 0, the identity simplifies to: a³ + b³ + c³ = 3abc

⇒ Substituting the values: 3 × 22 × (−15) × (−7)

⇒ = 3 × 22 × 105

⇒ = 66 × 105

⇒ = 66 × (100 + 5) = 6600 + 330 = 6930

∴ The correct answer is 6930.

Additional Information

The Cubic Identity

The expression a³ + b³ + c³ − 3abc is always zero if a + b + c = 0 or if a = b = c.

Square of Trinomial

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca). This is used to find individual squares or sum of products.

Difference of Cubes

x³ − y³ = (x − y)(x² + xy + y²), useful for factoring two-term cubic expressions.

Shortcut Trick

Let x, y, and z be the number of ₹10, ₹5, and ₹2 coins respectively.

Interchange ₹10 & ₹2: Increase = (10 − 2) × (z − x) = 304 − 120 ⇒ 8(z − x) = 184 ⇒ z − x = 23

Interchange ₹5 & ₹2: Increase = (5 − 2) × (z − y) = 165 − 120 ⇒ 3(z − y) = 45 ⇒ z − y = 15

Substitute x = z − 23 and y = z − 15 in original eq: 10(z − 23) + 5(z − 15) + 2z = 120

17z − 305 = 120 ⇒ 17z = 425 ⇒ z = 25

∴ The correct answer is 25 coins.

Alternate Method

Given:

Total amount = ₹120

Denominations = ₹10, ₹5, ₹2

New amount (₹10 ↔ ₹2) = ₹304

New amount (₹5 ↔ ₹2) = ₹165

Formula Used:

Total Value = (Number of coins × Value of each coin)

Let the number of coins be x, y, and z for ₹10, ₹5, and ₹2 respectively.

⇒ 10x + 5y + 2z = 120 −−− (1)

When ₹10 and ₹2 are interchanged:

⇒ 2x + 5y + 10z = 304 −−− (2)

When ₹5 and ₹2 are interchanged:

⇒ 10x + 2y + 5z = 165 −−− (3)

Subtract (1) from (2):

⇒ (2x + 5y + 10z) − (10x + 5y + 2z) = 304 − 120

⇒ 8z − 8x = 184

⇒ z − x = 23 ⇒ x = z − 23 −−− (4)

Subtract (1) from (3):

⇒ (10x + 2y + 5z) − (10x + 5y + 2z) = 165 − 120

⇒ 3z − 3y = 45

⇒ z − y = 15 ⇒ y = z − 15 −−− (5)

Substitute x and y in (1):

⇒ 10(z − 23) + 5(z − 15) + 2z = 120

⇒ 10z − 230 + 5z − 75 + 2z = 120

⇒ 17z − 305 = 120

⇒ 17z = 425

⇒ z = 25

∴ The correct answer is 25 coins.

Additional Information

Linear Equations in Three Variables

Equations of the form ax + by + cz = d require at least three independent constraints to find unique values for all variables.

Coin Problems Logic

The total value is the sum of (Quantity × Face Value). Interchanging quantities of denominations p and q changes the total by (p − q) × (Difference in counts).

Substitution Method

In multi-variable systems, expressing all variables in terms of one (e.g., everything in terms of z) simplifies the solution process significantly.

Shortcut Trick

If the ratio of three numbers is a : b : c and their LCM is given, then HCF = LCM ÷ LCM(a, b, c).

LCM of ratio parts (4, 3, 13) = 4 × 3 × 13 = 156 (since they are co-prime).

HCF = 4524 ÷ 156 = 29.

∴ The correct answer is 29.

Alternate Method

Given:

Ratio of three numbers = 4 : 3 : 13

LCM of these numbers = 4524

Let the HCF of the three numbers be x.

The three numbers can be represented as 4x, 3x, and 13x.

The LCM of (4x, 3x, 13x) is calculated as:

⇒ LCM = x × LCM(4, 3, 13)

Since 4, 3, and 13 have no common factors other than 1 (they are pairwise co-prime):

⇒ LCM(4, 3, 13) = 4 × 3 × 13 = 156

Now, equating to the given LCM value:

⇒ 156x = 4524

⇒ x = 4524 ÷ 156

⇒ x = 29

The HCF of the three numbers is x = 29.

∴ The correct answer is 29.

Additional Information

Relationship for Two Numbers

For any two numbers a and b, the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF × LCM.

HCF of Fractions

HCF of (a/b, c/d, e/f) = HCF(a, c, e) ÷ LCM(b, d, f).

Co-prime Numbers

Two or more numbers are called co-prime if their HCF is 1. The LCM of a set of co-prime numbers is simply their product.

LCM of Fractions

LCM of (a/b, c/d, e/f) = LCM(a, c, e) ÷ HCF(b, d, f).

Shortcut Trick

Net multiplier = (1 − 0.20) × (1 + 0.05) = 0.8 × 1.05 = 0.84

Final Price = Marked Price × 0.84

Final Price = 2000 × 0.84 = 1680

∴ The correct answer is ₹1,680.

Alternate Method

Given:

Marked Price (MP) = ₹2,000

Discount = 20%

GST = 5%

Calculations:

⇒ Discount = 20% of 2000 = (20/100) × 2000 = 400

⇒ Price after Discount = 2000 − 400 = 1600

⇒ GST = 5% of 1600 = (5/100) × 1600 = 80

⇒ Final Price = 1600 + 80 = 1680

∴ The correct answer is ₹1,680.

Additional Information

Net Percentage Change

When a value is changed successively by x% and y%, the net effective change is (x + y + (xy/100))%. Note that a discount is a negative change.

Marked Price (MP)

MP is the price listed on the product. Discounts are always calculated on this value: Selling Price (SP) = MP × (100 − D%)/100.

GST Application

Taxes like GST are calculated on the actual selling price (the price after applying all discounts), not on the original marked price.

Shortcut Trick

Third Proportional c of two numbers a and b is calculated as c = b² ÷ a.

Given a = 0.25 and b = 0.15.

Third Proportional = (0.15 × 0.15) ÷ 0.25

= 0.0225 ÷ 0.25 = 0.09

∴ The correct answer is 0.09.

Alternate Method

Given:

First number (a) = 0.25

Second number (b) = 0.15

Formula Used:

If a, b, and x are in continued proportion, then a : b :: b : x.

This implies a/b = b/x, where x is the third proportional.

Calculations:

⇒ 0.25 ÷ 0.15 = 0.15 ÷ x

⇒ x = (0.15 × 0.15) ÷ 0.25

⇒ x = 0.0225 ÷ 0.25

⇒ To simplify, multiply both numerator and denominator by 10000:

⇒ x = 225 ÷ 2500

⇒ x = 9 ÷ 100

⇒ x = 0.09

∴ The correct answer is 0.09.

Additional Information

Mean Proportional

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

Fourth Proportional

The fourth proportional d of three numbers a, b, and c is calculated as d = (b × c) ÷ a where a : b = c : d.

Continued Proportion

A sequence of numbers is in continued proportion if the ratio of any two consecutive terms is constant, i.e., a/b = b/c = c/d.

Shortcut Trick

Total units of work = LCM(9, 18, 18) = 18 units.

Efficiencies: A = 2, B = 1, and C = 1.

In the first x days, all three work: Efficiency = 2 + 1 + 1 = 4.

In the next x days, only B and C work: Efficiency = 1 + 1 = 2.

Total work done = (4 × x) + (2 × x) = 6x = 18.

∴ The correct answer is 3.

Alternate Method

Given:

Time taken by A = 9 days

Time taken by B = 18 days

Time taken by C = 18 days

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Total Work = LCM of 9, 18, and 18 = 18 units.

⇒ Efficiency of (A + B + C) = 2 + 1 + 1 = 4 units/day.

⇒ Efficiency of (B + C) = 1 + 1 = 2 units/day.

⇒ Work done by all three in x days = 4 × x = 4x units.

⇒ Work done by B and C in the next x days = 2 × x = 2x units.

⇒ Total work = 4x + 2x = 6x units.

⇒ 6x = 18

⇒ x = 18 ÷ 6 = 3

∴ The correct answer is 3.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a task. Efficiency = Work ÷ Time.

Combined Efficiency

When multiple people work together, their individual efficiencies are added to find the total work done per day.

LCM Method for Total Work

Assuming the Total Work as the LCM of the given time periods allows for calculation with whole numbers instead of fractions.

Shortcut Trick

To find the least beneficial scheme, identify the one with the lowest effective discount percentage.

(A) Successive 34% and 37% → 34 + 37 − (34 × 37 / 100) = 71 − 12.58 = 58.42%

(B) Successive 36%, 15%, 14% → 100 × 0.64 × 0.85 × 0.86 = 46.78% price → 53.22% discount

(C) Average discount for 9 toys → (6 × 33% + 3 × 9%) / 9 = (198% + 27%) / 9 = 225% / 9 = 25%

(D) 3 toys free on 9 toys bought → (Free / Total) × 100 = (3 / 9) × 100 = 33.33% (or 3/12 = 25%)

Comparing all, Scheme C provides the minimum discount of 25%.

∴ The correct answer is Scheme C.

Alternate Method

Assume the Marked Price (MP) of 1 toy is ₹100. A customer wants to buy 9 toys, so the total MP = ₹900.

Formula Used: Effective Discount % = (Total Discount / Total MP) × 100

⇒ Scheme A: Net SP = 900 × (1 − 0.34) × (1 − 0.37) = 900 × 0.66 × 0.63 = ₹374.22

⇒ Scheme B: Net SP = 900 × 0.64 × 0.85 × 0.86 = ₹421.06

⇒ Scheme C: Discount = (6 × 33) + (3 × 9) = 198 + 27 = ₹225 | Net SP = 900 − 225 = ₹675

⇒ Scheme D: Buying 9 toys means getting 3 free out of the 9. Customer pays for 6 toys. Net SP = 6 × 100 = ₹600

⇒ Comparing the final Selling Prices, Scheme C results in the highest payment (₹675), making it the least beneficial scheme for the customer.

∴ The correct answer is Scheme C.

Additional Information

Successive Discount Formula

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

Buy X Get Y Free

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

Effective Discount for Bulk Purchase

Net Discount % = (Total Discount Value / Total Marked Price) × 100. This is useful for tiered discount schemes.

Shortcut Trick

Present ratio (Amit : Sumit) = 2 : 1 | Ratio 5 years ago = 5 : 2.

Equate the age difference (2−1=1 and 5−2=3) by multiplying the present ratio by 3.

New Present ratio = 6 : 3 | Ratio 5 years ago = 5 : 2.

Difference in ratio parts (6−5 or 3−2) = 1 unit.

1 unit = 5 years. Therefore, Sumit present age = 3 units = 3 × 5 = 15 years.

∴ The correct answer is 15 years.

Alternate Method

Given:

Present age of Amit is twice that of Sumit.

Five years ago, the ratio of their ages was 5 : 2.

Formula Used:

Age ratio equation: (Age₁ − t) / (Age₂ − t) = Ratio_past

Let the Present age of Sumit be x years.

Then, the Present age of Amit = 2x years.

⇒ Age of Sumit five years ago = x − 5

⇒ Age of Amit five years ago = 2x − 5

⇒ (2x − 5) / (x − 5) = 5 / 2

⇒ 2 × (2x − 5) = 5 × (x − 5)

⇒ 4x − 10 = 5x − 25

⇒ 5x − 4x = 25 − 10

⇒ x = 15

∴ The correct answer is 15 years.

Additional Information

Constant Age Difference

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

Ratio Balancing

When the time gap is the same for both people, make the difference between the ratio terms equal to solve the problem faster without variables.

Concept of 'n' years ago/hence

If present age is x, age 'n' years ago is (x − n) and age 'n' years hence is (x + n).

Shortcut Trick

Solve the innermost bracket first: (26 − 13 + 5) = 18.

Multiply the result by the value in the braces: −9 × 18 = −162.

Apply the final addition: 30 + (−162) = −132.

∴ The correct answer is -132.

Alternate Method

Given: 30 + [{−9 × (26 − 13 + 5)}]

Formula Used: BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction).

⇒ 30 + [{−9 × (13 + 5)}]

⇒ 30 + [{−9 × 18}]

⇒ 30 + [−162]

⇒ 30 − 162

⇒ −132

∴ The correct answer is -132.

Additional Information

BODMAS Rule

It provides the correct sequence for solving mathematical operations: Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Types of Brackets

Solve in the following priority order: Parentheses ( ) → Curly Braces { } → Square Brackets [ ].

Sign Rules for Multiplication

Multiplying two different signs results in a negative value: (+) × (−) = (−). Multiplying two same signs results in a positive value: (−) × (−) = (+).

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).

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

⇒ 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.

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.

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 = 540^1/3.

⇒ 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).

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 1^st discount = 1800 − (25% of 1800)

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

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

⇒ 2^nd 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.

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(component_i / total_i).

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).

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.

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 1^st 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.

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 1^st 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.

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.