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Shortcut Trick

The Effective Rate of Interest for 2 years at 10% per annum = 10% × 2 = 20%.

Given that 20% of the Principal (P) = ₹700.

To find the total Principal (100%), we multiply by 5.

⇒ P = 700 × 5 = 3500.

∴ The correct answer is ₹3500.

Alternate Method

Given:

Simple Interest (SI) = ₹700

Time (T) = 2 years

Rate of Interest (R) = 10% per annum

Formula Used:

SI = (P × R × T) ÷ 100

Calculations:

⇒ 700 = (P × 10 × 2) ÷ 100

⇒ 700 = (P × 20) ÷ 100

⇒ 700 = P ÷ 5

⇒ P = 700 × 5

⇒ P = 3500

∴ The correct answer is ₹3500.

Additional Information

Total Amount Formula

The total sum returned after adding interest is Amount (A) = P + SI = P × (1 + RT/100).

Constant Nature of SI

Unlike compound interest, Simple Interest remains the same every year because it is always calculated on the original Principal.

SI vs CI Difference

For a period of 2 years, the difference between Compound Interest and Simple Interest is given by D = P × (R/100)².

Shortcut Trick

Relative cost per ratio set = (280 × 3) + (180 × 4) + (105 × 7) = 840 + 720 + 735 = 2295

Value of 1 ratio unit = Total cost ÷ Relative cost = 6885 ÷ 2295 = 3

Number of type Q articles = 4 units × 3 = 12

∴ The correct answer is 12.

Alternate Method

Given:

Prices of type P, Q, R = ₹280, ₹180, ₹105

Quantity ratio = 3 : 4 : 7

Total amount spent = ₹6,885

Formula Used:

Total Cost = (Price_P × Quantity_P) + (Price_Q × Quantity_Q) + (Price_R × Quantity_R)

Calculations:

⇒ Let the number of articles of type P, Q, and R be 3x, 4x, and 7x respectively.

⇒ Total spending = (280 × 3x) + (180 × 4x) + (105 × 7x)

⇒ 6885 = 840x + 720x + 735x

⇒ 6885 = 2295x

⇒ x = 6885 ÷ 2295

⇒ x = 3

⇒ Number of articles of type Q = 4x = 4 × 3 = 12

∴ The correct answer is 12.

Additional Information

Ratio Scaling Factor

When a ratio A : B : C is given, the actual values can be expressed as Ax, Bx, and Cx, where 'x' is the constant of proportionality or scaling factor.

Weighted Average Cost

The total cost of different items is the sum of products of their individual prices and their respective quantities: Total = Σ(Price_i × Quantity_i).

Unitary Method in Ratios

To find the value of one part, divide the total quantity by the sum of the ratio parts weighted by their respective values.

Shortcut Trick

To simplify quickly, substitute a convenient value for the variable n. Let n = 0.

Expression = (2⁰⁺³ − 2 × 2⁰) ÷ (2 × 2⁰⁺²)

Expression = (2³ − 2 × 1) ÷ (2 × 2²) = (8 − 2) ÷ (2 × 4)

Expression = 6 ÷ 8 = 3/4

∴ The correct answer is 3/4.

Alternate Method

Given: (2ⁿ⁺³ − 2 × 2ⁿ) ÷ (2 × 2ⁿ⁺²)

Method: Factor out the common term 2ⁿ from both the numerator and the denominator using the Law of Exponents (a^m+n = a^m × aⁿ).

⇒ (2ⁿ × 2³ − 2 × 2ⁿ) ÷ (2 × 2ⁿ × 2²)

⇒ [2ⁿ × (2³ − 2)] ÷ [2ⁿ × (2 × 2²)]

⇒ (2³ − 2) ÷ (2 × 2²)

⇒ (8 − 2) ÷ (2 × 4)

⇒ 6 ÷ 8

⇒ 3/4

∴ The correct answer is 3/4.

Additional Information

Product Rule of Exponents

When multiplying terms with the same base, add their powers: a^m × aⁿ = a^m+n.

Quotient Rule of Exponents

When dividing terms with the same base, subtract their powers: a^m ÷ aⁿ = a^m−n.

Zero Exponent Property

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

Shortcut Trick

Let the total cast votes be 100%.

Candidate X got 40%, so the remaining 60% is shared by 3 candidates → 20% each.

Difference between Candidate X and others = 40% − 20% = 20%.

20% of cast votes = 480 → 100% of cast votes = 480 × 5 = 2400.

Given 60% of registered voters = 2400, then 100% = (2400 ÷ 60) × 100 = 4000.

∴ The correct answer is 4000.

Alternate Method

Given:

⇒ Total Registered Voters = 100%

⇒ Cast Votes = 60% of Registered Voters

⇒ Candidate X = 40% of Cast Votes

⇒ Others = Equal split of remaining (100% − 40%) = 20% of Cast Votes each

Calculations:

⇒ Let Total Registered Voters = V

⇒ Cast Votes = 0.60V

⇒ Votes for X = 0.40 × (0.60V) = 0.24V

⇒ Total remaining cast votes = 0.60V − 0.24V = 0.36V

⇒ Votes for each other candidate = 0.36V ÷ 3 = 0.12V

⇒ Given difference: 0.24V − 0.12V = 480

⇒ 0.12V = 480

⇒ V = 480 ÷ 0.12

⇒ V = 4000

∴ The correct answer is 4000.

Additional Information

Percentage of a Percentage

When a value is a percentage of another percentage, the effective percentage of the original base is (A × B) ÷ 100.

Registered vs Cast Votes

Registered voters (Voter List) is the 100% base. Cast votes are the actual subset used for candidate distribution.

Base Change Rule

To find the total from a part: Total = (Value ÷ Percentage) × 100.

Shortcut Trick

Selling Price (SP) of X = 2000 + 322 = 2322.

Multiplication factor for X after 28% and 90% discounts = 0.72 × 0.10 = 0.072.

Marked Price (MP) = 2322 ÷ 0.072 = 32250.

SP of Y after 88% discount = 32250 × 0.12 = 3870.

Profit % on Y = [(3870 − 2000) ÷ 2000] × 100 = (1870 ÷ 20) = 93.5%.

∴ The correct answer is 93.5%.

Alternate Method

Given:

Cost Price (CP) of X and Y = ₹2,000 each.

Profit on X = ₹322.

Discounts on X = 28% and 90% (successive).

Discount on Y = 88% (single).

Formula Used:

SP = CP + Profit

SP = MP × (100 − D₁)/100 × (100 − D₂)/100

Profit % = (Profit ÷ CP) × 100

Calculations:

⇒ Selling Price of X (SP_X) = CP + Profit = 2000 + 322 = ₹2,322

⇒ SP_X = MP × (1 − 0.28) × (1 − 0.90)

⇒ 2322 = MP × 0.72 × 0.10

⇒ 2322 = MP × 0.072

⇒ MP = 2322 ÷ 0.072 = ₹32,250

⇒ Selling Price of Y (SP_Y) = MP × (1 − 0.88)

⇒ SP_Y = 32250 × 0.12 = ₹3,870

⇒ Profit on Y = SP_Y − CP = 3870 − 2000 = ₹1,870

⇒ Profit % on Y = (1870 ÷ 2000) × 100 = 93.5%

∴ The correct answer is 93.5%.

Additional Information

Successive Discounts

If two successive discounts are d₁% and d₂%, the single equivalent discount is (d₁ + d₂ − d₁d₂/100)%.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). SP = MP − Discount.

Profit Percentage

Profit % is always calculated on the Cost Price (CP). Profit % = [(SP − CP) ÷ CP] × 100.

Shortcut Trick

Profit per kg = (Selling Price ÷ Selling Quantity) − (Cost Price ÷ Buying Quantity)

Profit per kg = (50 ÷ 5) − (15 ÷ 3) = 10 − 5 = ₹5

Required Quantity = Total Profit ÷ Profit per kg

Required Quantity = 125 ÷ 5 = 25 kg

∴ The correct answer is 25 kg.

Alternate Method

Given:

Cost Price (CP) of 3 kg Mangoes = ₹15

Selling Price (SP) of 5 kg Mangoes = ₹50

Target Profit = ₹125

Formula Used:

Profit = Selling Price (SP) − Cost Price (CP)

Quantity = Total Profit ÷ Profit per unit

Calculations:

⇒ CP of 1 kg Mangoes = 15 ÷ 3 = ₹5

⇒ SP of 1 kg Mangoes = 50 ÷ 5 = ₹10

⇒ Profit on 1 kg Mangoes = 10 − 5 = ₹5

⇒ Let the Quantity to be sold be 'x' kg

⇒ Total Profit = Profit per kg × Quantity

⇒ 125 = 5 × x

⇒ x = 125 ÷ 5

⇒ x = 25 kg

∴ The correct answer is 25 kg.

Additional Information

Profit Percentage

It is the profit expressed as a percentage of the Cost Price: Profit% = (Profit ÷ CP) × 100.

Relationship between SP and CP

When there is a profit, Selling Price (SP) = CP × (100 + Profit%) ÷ 100.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). SP = MP − Discount.

Shortcut Trick

Total Work = LCM(16, 24) = 48 units.

Efficiencies: A = 3 units/day, B = 2 units/day.

Work done together in 5 days = (3 + 2) × 5 = 25 units.

Remaining work = 48 − 25 = 23 units.

Additional days for B = 23 ÷ 2 = 11.5 days.

∴ The correct answer is 11.5 days.

Alternate Method

Given:

Time taken by A = 16 days

Time taken by B = 24 days

Working together duration = 5 days

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Total Work (LCM of 16 and 24) = 48 units

⇒ Efficiency of A = 48 ÷ 16 = 3 units/day

⇒ Efficiency of B = 48 ÷ 24 = 2 units/day

⇒ Combined Efficiency of (A + B) = 3 + 2 = 5 units/day

⇒ Work completed by A and B in 5 days = 5 × 5 = 25 units

⇒ Remaining Work = 48 − 25 = 23 units

⇒ Time taken by B to complete remaining work = Remaining Work ÷ Efficiency of B

⇒ Time = 23 ÷ 2 = 11.5 days

∴ The correct answer is 11.5 days.

Additional Information

Efficiency and Time Relation

Efficiency is inversely proportional to the time taken to complete the work (E ∝ 1/T). If ratio of time is x:y, the ratio of efficiency is y:x.

Pipes and Cisterns

This concept follows the same logic where inlet pipes have positive efficiency and outlet (leakage) pipes have negative efficiency.

Combined Work Formula

If A completes work in 'a' days and B in 'b' days, they together complete it in (a × b) ÷ (a + b) days.

Work and Wages

Wages are always distributed in the ratio of the actual work done by each individual, which is (Efficiency × Time Active).

Shortcut Trick

Use the Successive Division Method: Divide the larger number by the smaller one until the remainder is zero.

148 ÷ 64 → Remainder = 20

64 ÷ 20 → Remainder = 4

20 ÷ 4 → Remainder = 0

The last divisor which leaves a remainder of zero is the HCF.

∴ The correct answer is 4.

Alternate Method

Given: The numbers are 64 and 148.

Formula: HCF is the product of the lowest powers of common prime factors.

⇒ Prime factorization of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶

⇒ Prime factorization of 148 = 2 × 2 × 37 = 2² × 37

⇒ Common prime factor is 2.

⇒ Lowest power of common factor = 2²

⇒ HCF = 2 × 2 = 4

∴ The correct answer is 4.

Additional Information

HCF and LCM Relation

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(a, b) × LCM(a, b).

Co-prime Numbers

Two numbers are called co-prime if their Highest Common Factor (HCF) is 1. Example: 8 and 15 are co-prime numbers.

HCF of Fractions

The HCF of fractions is calculated as: HCF of Numerators ÷ LCM of Denominators.

Shortcut Trick

Use the algebraic identity: (a + b)² − (a − b)² = 4ab.

Here, let a = 2pq and b = 11q.

Applying the values into the formula: 4 × (2pq) × (11q).

⇒ 4 × 2 × 11 × p × q × q = 88pq².

∴ The correct answer is 88pq².

Alternate Method

Given: (2pq + 11q)² − (2pq − 11q)²

Formula Used: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b²

⇒ Expansion of (2pq + 11q)² = (2pq)² + 2(2pq)(11q) + (11q)² = 4p²q² + 44pq² + 121q²

⇒ Expansion of (2pq − 11q)² = (2pq)² − 2(2pq)(11q) + (11q)² = 4p²q² − 44pq² + 121q²

⇒ Subtracting the two results: (4p²q² + 44pq² + 121q²) − (4p²q² − 44pq² + 121q²)

⇒ 4p²q² − 4p²q² + 121q² − 121q² + 44pq² + 44pq²

⇒ 88pq²

∴ The correct answer is 88pq².

Additional Information

Sum of squares identity

(a + b)² + (a − b)² = 2(a² + b²). This identity simplifies expressions involving the addition of symmetric squared binomials.

Difference of two squares

(a + b)(a − b) = a² − b². This is a fundamental algebraic identity used for factoring and simplification.

Square of a binomial

(a + b)² = a² + 2ab + b². The middle term (2ab) is key when performing algebraic expansion or finding roots.

Shortcut Trick

Use the Empirical Relationship formula: Mode = 3 × Median − 2 × Mean.

Direct substitution: Mode = 3 × 61 − 2 × 78.

Mode = 183 − 156 = 27.

∴ The correct answer is 27.

Alternate Method

Given:

Mean = 78

Median = 61

Formula Used:

The relationship between the measures of central tendency for a moderately skewed distribution is:

Mode = 3 × Median − 2 × Mean

Calculations:

⇒ Mode = 3(61) − 2(78)

⇒ Mode = 183 − 156

⇒ Mode = 27

∴ The correct answer is 27.

Additional Information

Empirical Relationship

This formula (Mode = 3 Median − 2 Mean) is specifically applicable to moderately skewed distributions.

Symmetric Distribution

In a perfectly symmetrical distribution (like the Normal Distribution), the Mean, Median, and Mode are all equal: Mean = Median = Mode.

Skewness Identification

If Mean > Median > Mode, the distribution is Positively Skewed. If Mean < Median < Mode, it is Negatively Skewed.

Mean Formula

Mean = (∑ f_ix_i) ÷ (∑ f_i), which represents the arithmetic average of the data set.