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Shortcut Trick

Marked Price per pair = ₹2800.

Discounted Price (DP) per pair = 82% of 2800 = ₹2296.

Special offer allows 5 pairs at full DP and 2 pairs at 50% DP (which is 1 pair at full DP).

Total effective cost for 7 pairs = (5 + 1) × 2296 = 6 × 2296.

Total Cost = ₹13,776.

∴ The correct answer is ₹13,776.

Alternate Method

Given:

Marked Price (MP) of a pair of shoes = ₹2800

Trade Discount = 18%

Quantity 1 = 5 pairs at discounted price

Quantity 2 = 2 pairs at 50% of discounted price

Formula Used:

Discounted Price = MP × (100 − Discount%) / 100

Total Cost = (Quantity₁ × Price₁) + (Quantity₂ × Price₂)

Calculations:

⇒ Trade Discount amount = 18% of 2800 = 0.18 × 2800 = ₹504

⇒ Discounted Price (DP) per pair = 2800 − 504 = ₹2296

⇒ Cost of first 5 pairs = 5 × 2296 = ₹11,480

⇒ Cost of next 2 pairs = 2 × (50% of 2296) = 2 × 1148 = ₹2,296

⇒ Total Cost for 7 pairs = 11480 + 2296 = ₹13,776

∴ The correct answer is ₹13,776.

Additional Information

Marked Price and Selling Price

Selling Price (SP) is calculated as MP − Discount. If multiple items have different discounts, the total SP is the sum of individual SPs.

Successive Discounts

When two discounts d₁% and d₂% are applied successively, the effective discount is (d₁ + d₂ − (d₁×d₂)/100)%.

Trade Discount vs Cash Discount

Trade discount is given on the list price at the time of sale, while cash discount is offered for immediate or early payment.

Shortcut Trick

Direct calculation: Age of A = 2 × (Age of B) + 5

Given Person B is 15 years old.

Age of Person A = (2 × 15) + 5 = 30 + 5

Age of Person A = 35 years

∴ The correct answer is 35 years.

Alternate Method

Given:

Present age of Person B = 15 years

Present age of Person A = 5 years more than twice the age of Person B

Formula Used:

Age_A = (2 × Age_B) + 5

Calculations:

⇒ Age of Person B = 15 years

⇒ Twice the age of Person B = 2 × 15 = 30 years

⇒ Age of Person A = 30 + 5

⇒ Age of Person A = 35 years

∴ The correct answer is 35 years.

Additional Information

Linear Equations in Age Problems

Age relations such as 'twice as old' or 'n years older' are formulated as linear equations (y = mx + c) where m is the multiplier and c is the constant difference.

Constant Difference Rule

The difference between the ages of two individuals always remains constant, regardless of the number of years that pass.

Ratio Method

For complex age problems, ratios are used by finding a common multiplier (k) to satisfy the given conditions for different time periods.

Shortcut Trick

Break the expression into three parts using the VBODMAS rule.

Part 1: {(16/36) ÷ (16/4)} = (4/9) ÷ 4 = 1/9.

Part 2: ((2/7) × (14/6) + 5/12) = (2/3 + 5/12) = 13/12.

Part 3: (4/7) ÷ (39/12 of 12/7) = (4/7) ÷ (39/7) = 4/39.

Final: (1/9 ÷ 13/12) + 4/39 = (1/9 × 12/13) + 4/39 = 4/39 + 4/39 = 8/39.

∴ The correct answer is 8/39.

Alternate Method

Given: {(16/36) ÷ (16/4)} ÷ ((2/7) × (14/6) + (5/12)) + (4/7) ÷ (39/12) of (12/7)

Formula Used: Order of operations follows VBODMAS (Vinculum, Brackets, Of, Division, Multiplication, Addition, Subtraction).

⇒ Simplify first bracket: (16/36) ÷ (16/4) = 4/9 ÷ 4 = 4/9 × 1/4 = 1/9

⇒ Simplify inner part of second group: (2/7 × 14/6) = 2/7 × 7/3 = 2/3

⇒ Complete second bracket: 2/3 + 5/12 = 8/12 + 5/12 = 13/12

⇒ Simplify Of operation: (39/12) of (12/7) = 39/12 × 12/7 = 39/7

⇒ Remaining expression: (1/9 ÷ 13/12) + (4/7 ÷ 39/7)

⇒ Perform divisions: (1/9 × 12/13) + (4/7 × 7/39) = 4/39 + 4/39

⇒ Final Addition: 8/39

∴ The correct answer is 8/39.

Additional Information

The VBODMAS Rule

It ensures mathematical consistency by prioritizing Brackets then Of (which is a form of multiplication) over Division and Multiplication.

Fraction Division

Dividing by a fraction is equivalent to multiplying by its reciprocal (inverted form): a/b ÷ c/d = a/b × d/c.

Common Denominators

When adding fractions like 2/3 and 5/12, find the LCM of denominators (12) to create a common base before adding numerators.

Shortcut Trick

Identify common base 5 and use the Net Power rule.

Net Power = (Power of term 1 + Power of term 2) − Power of denominator.

Calculation: [3x + 2(x − 1)] − (4x − 1).

⇒ (3x + 2x − 2) − 4x + 1 = 5x − 2 − 4x + 1 = x − 1.

The resulting expression is 5^x-1.

∴ The correct answer is 5^x-1.

Alternate Method

Given: [5^3x × (5²)^(x-1)] ÷ 5^4x-1

Formula Used: (a^m)ⁿ = a^m×n | a^m × aⁿ = a^m+n | a^m ÷ aⁿ = a^m−n

Step-by-step Simplification:

⇒ [5^3x × 5^2(x-1)] ÷ 5^4x-1

⇒ [5^3x × 5^2x-2] ÷ 5^4x-1

⇒ 5^{(3x + 2x − 2)} ÷ 5^4x-1

⇒ 5^{(5x − 2)} ÷ 5^4x-1

⇒ 5^{(5x − 2) − (4x − 1)}

⇒ 5^{5x − 2 − 4x + 1}

⇒ 5^{x − 1}

∴ The correct answer is 5^x-1.

Additional Information

Product and Quotient Laws

When multiplying terms with the same base, add their exponents (a^m × aⁿ = a^m+n). When dividing, subtract them (a^m ÷ aⁿ = a^m−n).

Power of a Power Rule

Raising a power to another power requires multiplying the exponents: (a^m)ⁿ = a^m×n.

Negative Exponents

Any non-zero base raised to a negative power is the reciprocal: a⁻ⁿ = 1/aⁿ.

Zero Exponent Rule

Any non-zero base raised to the power of zero is always 1: a⁰ = 1.

Shortcut Trick

Identify the increase in average: 25,500 − 24,000 = ₹1,500.

Income of new employee = New average + (Initial number of employees × Increase in average).

Income = 25,500 + (6 × 1,500) = 25,500 + 9,000 = ₹34,500.

∴ The correct answer is ₹34,500.

Alternate Method

Given:

Initial number of employees = 6 | Initial average income = ₹24,000

Final number of employees = 7 | Final average income = ₹25,500

Formula Used:

Total Income = Average Income × Number of Employees

Calculations:

⇒ Total monthly income of 6 employees = 6 × 24,000 = ₹1,44,000

⇒ Total monthly income of 7 employees = 7 × 25,500 = ₹1,78,500

⇒ Monthly income of the new employee = Total (7 employees) − Total (6 employees)

⇒ Monthly income = 1,78,500 − 1,44,000 = ₹34,500

∴ The correct answer is ₹34,500.

Additional Information

Net Deviation Concept

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

Weighted Average

When two groups are combined, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Inclusion and Exclusion

If adding a new member increases the average, the new member's value is higher than the original average.

Shortcut Trick

Divisor = 5 × Remainder = 5 × 56 = 280

Quotient = Divisor ÷ 14 = 280 ÷ 14 = 20

Dividend = (Divisor × Quotient) + Remainder

Dividend = (280 × 20) + 56 = 5600 + 56 = 5656

∴ The correct answer is 5656.

Alternate Method

Given:

Remainder (r) = 56

Divisor (d) = 5 × Remainder

Divisor (d) = 14 × Quotient (q)

Formula Used:

Dividend = (Divisor × Quotient) + Remainder

Calculations:

⇒ Divisor = 5 × 56 = 280

⇒ 14 × Quotient = 280

⇒ Quotient = 280 ÷ 14 = 20

⇒ Dividend = (280 × 20) + 56

⇒ Dividend = 5600 + 56

⇒ Dividend = 5656

∴ The correct answer is 5656.

Additional Information

Division Algorithm

For any two whole numbers, the dividend is always equal to the product of the divisor and quotient plus the remainder.

Remainder Property

The remainder must always be greater than or equal to 0 and strictly less than the divisor (0 ≤ r < d).

Euclid's Division Lemma

It states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r.

Shortcut Trick

The ratio of surface areas of two spheres is equal to the square of the ratio of their radii.

Radius of Sphere A : Radius of Sphere B = 6 : 3 = 2 : 1

Surface Area Ratio = (2)² : (1)² = 4 : 1

∴ The correct answer is 4:1.

Alternate Method

Given:

Radius of Sphere A (r₁) = 6 cm

Radius of Sphere B (r₂) = 3 cm

Formula Used:

Surface Area (S) of a sphere = 4πr²

Calculations:

⇒ Surface Area of Sphere A (S₁) = 4 × π × (6)²

⇒ S₁ = 4 × π × 36 = 144π cm²

⇒ Surface Area of Sphere B (S₂) = 4 × π × (3)²

⇒ S₂ = 4 × π × 9 = 36π cm²

⇒ Ratio = S₁ : S₂ = 144π : 36π

⇒ Ratio = 144 / 36 = 4 / 1

∴ The correct answer is 4:1.

Additional Information

Volume of a Sphere

The volume (V) of a sphere is given by the formula V = (4/3)πr³.

Ratio of Volumes

The ratio of the volumes of two spheres is equal to the cube of the ratio of their radii: V₁/V₂ = (r₁/r₂)³.

Hemisphere Properties

For a hemisphere, the Curved Surface Area = 2πr² and the Total Surface Area = 3πr².

Shortcut Trick

Summing both conditions: (8 + 5) Mangoes + (5 + 8) Oranges = 164 + 148

⇒ 13 (Mangoes + Oranges) = ₹312

⇒ 1 (Mango + Orange) = 312 ÷ 13 = ₹24

⇒ 3 (Mangoes + Oranges) = 3 × 24 = ₹72

∴ The correct answer is ₹72.

Alternate Method

Given:

Cost of 8 mangoes and 5 oranges = ₹164

Cost of 5 mangoes and 8 oranges = ₹148

Formula Used:

Symmetrical Linear Equations: Add equations when coefficients are interchanged.

Calculations:

⇒ Let cost of 1 mango be M and 1 orange be O.

⇒ 8M + 5O = 164 --- (Equation 1)

⇒ 5M + 8O = 148 --- (Equation 2)

⇒ Adding Eq(1) and Eq(2): (8M + 5M) + (5O + 8O) = 164 + 148

⇒ 13M + 13O = 312

⇒ 13(M + O) = 312

⇒ M + O = 312 ÷ 13 = 24

⇒ Cost of 3 mangoes and 3 oranges = 3 × (M + O)

⇒ 3 × 24 = 72

∴ The correct answer is ₹72.

Additional Information

Linear Equations in Two Variables

Equations expressed in the form ax + by = c, where a and b are coefficients and c is a constant.

Symmetrical Coefficients Property

When coefficients are interchanged (ax + by = m and bx + ay = n), adding them directly gives a simplified value for (x + y).

Unitary Method

A problem-solving technique where you first find the value of a single unit and then multiply it by the required quantity.

Shortcut Trick

In a distribution with two values, the mean acts as the balance point.

Difference between first value (10) and mean (12.5) = 12.5 − 10 = 2.5

Difference between second value (15) and mean (12.5) = 15 − 12.5 = 2.5

Since the deviations are equal in magnitude (2.5), the frequencies must also be equal.

Given first frequency = 5, therefore p = 5.

∴ The correct answer is 5.

Alternate Method

Given:

Mean (χ) = 12.5

Values (x_i): 10, 15

Frequencies (f_i): 5, p

Formula Used:

Arithmetic Mean (χ) = ∑f_ix_i ÷ ∑f_i

Calculations:

⇒ 12.5 = (10 × 5 + 15 × p) ÷ (5 + p)

⇒ 12.5 × (5 + p) = 50 + 15p

⇒ 62.5 + 12.5p = 50 + 15p

⇒ 62.5 − 50 = 15p − 12.5p

⇒ 12.5 = 2.5p

⇒ p = 12.5 ÷ 2.5

⇒ p = 5

∴ The correct answer is 5.

Additional Information

Properties of Arithmetic Mean

The algebraic sum of deviations of a set of observations from their arithmetic mean is always zero. ∑f_i(x_i − χ) = 0.

Weighted Average Formula

When different values have different importance or frequency, the weighted mean is calculated as χ_w = (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n).

Change of Origin and Scale

If every observation is increased by 'a', the mean increases by 'a'. If every observation is multiplied by 'k', the mean is multiplied by 'k'.

Shortcut Trick

The Third Proportional of a and b is calculated as b² ÷ a.

First, find m: m = (20 × 20) ÷ 4 = 100.

Then, find n: n = (50 × 50) ÷ 100 = 25.

∴ The correct answer is 25.

Alternate Method

Given:

First pair: 4, 20 and m are in continuous proportion.

Second pair: m, 50 and n are in continuous proportion.

Formula Used:

If a, b, c are in continuous proportion, then a : b = b : c ⇒ c = b² ÷ a.

Calculations:

⇒ For the first set: 4 : 20 = 20 : m

⇒ m = (20 × 20) ÷ 4

⇒ m = 400 ÷ 4 = 100

⇒ For the second set: 100 : 50 = 50 : n

⇒ n = (50 × 50) ÷ 100

⇒ n = 2500 ÷ 100 = 25

∴ The correct answer is 25.

Additional Information

Mean Proportional

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

Fourth Proportional

If four numbers a, b, c, and d are in proportion (a : b :: c : d), then the fourth proportional d = (b × c) ÷ a.

Continued Proportion

Three numbers are in continued proportion if the ratio of the first to the second is equal to the ratio of the second to the third (a/b = b/c).