SSC CGL Tier 2 Quant Previous Year Paper (Held On: 09 March 2018)

Shortcut Trick

Unit digit of product = Unit digit of (3 × 6 × N) = Unit digit of (18 × N) = Unit digit of (8 × N).

Given that the unit digit is (N + 2), we check the options for the condition: UD(8 × N) = N + 2.

Testing N = 6: Unit digit of (8 × 6) = Unit digit of 48 = 8.

Value of (N + 2) = 6 + 2 = 8.

Since 8 = 8, the condition is satisfied when N = 6.

∴ The correct answer is 6.

Alternate Method

Given:

Numbers: 433, 456, 43N

Product: 433 × 456 × 43N

Unit digit of product = N + 2

Formula Used:

The unit digit of a product depends only on the unit digits of the numbers being multiplied.

Calculations:

⇒ Unit digit of 433 = 3

⇒ Unit digit of 456 = 6

⇒ Unit digit of 43N = N

⇒ Unit digit of (433 × 456 × 43N) = Unit digit of (3 × 6 × N)

⇒ Unit digit of (18 × N) = Unit digit of (8 × N)

⇒ According to the question: Unit digit of (8 × N) = N + 2

⇒ If N = 1: UD(8 × 1) = 8; N + 2 = 3 (8 ≠ 3)

⇒ If N = 3: UD(8 × 3) = 4; N + 2 = 5 (4 ≠ 5)

⇒ If N = 8: UD(8 × 8) = 4; N + 2 = 10 → UD = 0 (4 ≠ 0)

⇒ If N = 6: UD(8 × 6) = 8; N + 2 = 8 (8 = 8)

⇒ Since the condition is satisfied for N = 6, the value of N is 6.

∴ The correct answer is 6.

Additional Information

Unit Digit of Powers (Cyclicity)

The unit digit of any number raised to a power repeats in a cycle. For digits 2, 3, 7, and 8, the cyclicity is 4; for 4 and 9, it is 2; and for 0, 1, 5, and 6, it is 1.

Product Unit Digit Rule

To find the unit digit of a large product, multiply only the unit digits of the individual factors. The unit digit of the final product will be the unit digit of this multiplication.

Special Case for 5

If any factor in a product has a unit digit of 5 and there is at least one even number in the product, the unit digit of the entire product will always be 0.

Shortcut Trick

Let x = 12345.

Then, N = x² + x + (x + 1).

N = x² + 2x + 1 = (x + 1)².

√N = x + 1 = 12345 + 1 = 12346.

∴ The correct answer is 12346.

Alternate Method

Given: N = (12345)² + 12345 + 12346

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

⇒ N = (12345)² + 12345 + (12345 + 1)

⇒ N = (12345)² + 2 × 12345 + 1

⇒ N = (12345)² + 2 × (12345) × (1) + (1)²

⇒ N = (12345 + 1)²

⇒ N = (12346)²

⇒ √N = √(12346)²

⇒ √N = 12346

∴ The correct answer is 12346.

Additional Information

Algebraic Identity: Square of a Sum

The expression a² + 2ab + b² is always equal to (a + b)². In this problem, a = 12345 and b = 1.

Related Identity: Square of a Difference

Similarly, a² − 2ab + b² = (a − b)², which is useful for simplifying expressions with subtraction.

Difference of Squares

The identity a² − b² = (a − b)(a + b) is another fundamental tool for simplification and factoring.

Shortcut Trick

For Statement I: Observe that each fraction term simplifies to the same value: 3/20 = 0.15.

Sum = 0.15 × 4 = 0.6. Thus, Statement I is TRUE.

For Statement II: Observe the position of significant digits. 0.01 (2nd place) + 0.0001 (4th place) + 0.000001 (6th place).

Sum = 0.010101. Thus, Statement II is TRUE.

∴ The correct answer is Both I and II.

Alternate Method

Given expressions:

I. (0.03/0.2) + (0.003/0.02) + (0.0003/0.002) + (0.00003/0.0002)

II. (0.01) + (0.01)² + (0.001)²

Step-by-step calculation for Statement I:

⇒ 0.03/0.2 = 3/20 = 0.15

⇒ 0.003/0.02 = 3/20 = 0.15

⇒ 0.0003/0.002 = 3/20 = 0.15

⇒ 0.00003/0.0002 = 3/20 = 0.15

⇒ Total Sum = 0.15 + 0.15 + 0.15 + 0.15 = 0.60

Statement I is verified as TRUE.

Step-by-step calculation for Statement II:

⇒ 0.01 = 0.01

⇒ (0.01)² = 0.0001

⇒ (0.001)² = 0.000001

⇒ Total Sum = 0.01 + 0.0001 + 0.000001 = 0.010101

Statement II is verified as TRUE.

∴ The correct answer is Both I and II.

Additional Information

Decimal Power Rule

When a decimal number with n decimal places is raised to the power p, the resulting number has n × p decimal places.

Division of Decimals

To divide decimals, shift the decimal point in both the numerator and denominator to the right by the same number of places until the denominator is an integer.

Decimal Addition

Always align the decimal points vertically before adding or subtracting to ensure correct place value summation.

Shortcut Trick

Evaluate each term by shifting the decimal: 1 / (0.1)² = 1/0.01 = 100

For the second term: 1 / (0.01)² = 1/0.0001 = 10000

For the third term: 1 / (0.5)² = 1/0.25 = 4

For the fourth term: 1 / (0.05)² = 1/0.0025 = 400

Total Sum = 100 + 10000 + 4 + 400 = 10504

∴ The correct answer is 10504.

Alternate Method

Given Expression: 1 / (0.1)² + 1 / (0.01)² + 1 / (0.5)² + 1 / (0.05)²

Step-by-step simplification:

⇒ 1 / 0.01 + 1 / 0.0001 + 1 / 0.25 + 1 / 0.0025

⇒ (1 × 100) / 1 + (1 × 10000) / 1 + (1 × 100) / 25 + (1 × 10000) / 25

⇒ 100 + 10000 + 4 + 400

⇒ 10100 + 404

⇒ 10504

∴ The correct answer is 10504.

Additional Information

Squaring Decimals

When squaring a decimal, the result has twice the number of decimal places as the original. For example, (0.05)² has two decimal places, so the result 0.0025 has four decimal places.

Reciprocals of Decimals

1 / 0.x is equivalent to 10/x. Similarly, 1 / 0.00x is equivalent to 1000/x. This conversion helps in rapid simplification of fractions involving decimals.

BODMAS Rule

In simplification, always follow the order: Brackets, Orders (Powers/Roots), Division, Multiplication, Addition, and Subtraction.

Shortcut Trick

For Statement I: The product (1 + 1/2)(1 + 1/3)...(1 + 1/n) simplifies to (n + 1) ÷ 2.

Value = (998 + 1) ÷ 2 = 499.5. Since 499.5 > 497, Statement I is True.

For Statement II: Comparing the whole numbers, LHS = 14 + 5 − 2 = 17 and RHS = 11 + 12 − 7 = 16.

Since 17 > 16 and fractional values (0.5 vs 0.25) support this, Statement II is also True.

∴ The correct answer is Both I and II.

Alternate Method

Given:

Statement I: (1 + 1/2)(1 + 1/3)(1 + 1/4) ... (1 + 1/998) > 497

Statement II: 14 3/4 + 5 1/4 − 2 1/2 > 11 1/8 + 12 3/8 − 7 1/4

Checking Statement I:

⇒ (1 + 1/2) × (1 + 1/3) × (1 + 1/4) × ... × (1 + 1/998)

⇒ (3/2) × (4/3) × (5/4) × ... × (999/998)

⇒ By telescoping cancellation, intermediate numerators and denominators cancel out.

⇒ Result = 999 ÷ 2 = 499.5

⇒ 499.5 > 497 is True.

Checking Statement II:

⇒ LHS = (14 + 5 − 2) + (3/4 + 1/4 − 1/2)

⇒ LHS = 17 + (1 − 0.5) = 17.5

⇒ RHS = (11 + 12 − 7) + (1/8 + 3/8 − 1/4)

⇒ RHS = 16 + (4/8 − 2/8) = 16 + 2/8 = 16.25

⇒ 17.5 > 16.25 is True.

∴ The correct answer is Both I and II.

Additional Information

Telescoping Products

A product where each term cancels out a part of the subsequent term, leaving only the first denominator and the last numerator, such as Π(1 + 1/n) = (n+1)/initial_denominator.

Mixed Fraction Addition

To solve quickly, separate the whole number parts and the fractional parts: (A + a/b) + (C + c/d) = (A + C) + (a/b + c/d).

Inequality Rules

When comparing two expressions, if the whole number part of one is significantly larger than the other while fractional differences are small, the larger whole number determines the greater value.

Shortcut Trick Statement I: Equalize numerators for 3/110 and 9/308 → 9/330 < 9/308 (True). Cross-multiply 9/308 and 7/225 → 9 × 225 (2025) < 7 × 308 (2156) (True). Statement II: Sum is approximately 100 × 6 = 600. Given value 279 is too low. Exact calculation for II: (99 × 6) + (1+2+3+4+5+6)/7 = 594 + 21/7 = 594 + 3 = 597. Only Statement I is correct. ∴ The correct answer is Only I. Alternate Method Checking Statement I: 3/110 < 9/308 < 7/225 ⇒ First comparison: 3/110 = (3 × 3) / (110 × 3) = 9/330 ⇒ Since 330 > 308, 9/330 < 9/308 is True. ⇒ Second comparison: 9/308 and 7/225 ⇒ Cross-multiplying: 9 × 225 = 2025 and 7 × 308 = 2156 ⇒ Since 2025 < 2156, 9/308 < 7/225 is True. Statement I is TRUE. Checking Statement II: 99(1/7) + 99(2/7) + 99(3/7) + 99(4/7) + 99(5/7) + 99(6/7) = 279 ⇒ Total terms = 6 ⇒ Sum = (99 + 99 + 99 + 99 + 99 + 99) + (1/7 + 2/7 + 3/7 + 4/7 + 5/7 + 6/7) ⇒ Sum = (99 × 6) + (1+2+3+4+5+6)/7 ⇒ Sum = 594 + 21/7 = 594 + 3 = 597 ⇒ 597 ≠ 279. Statement II is FALSE. ∴ The correct answer is Only I. Additional Information Comparison of Fractions For two fractions a/b and c/d, if a × d < b × c, then a/b < c/d. Also, for same numerators, the fraction with the smaller denominator is larger. Mixed Fraction Property A mixed fraction x(y/z) can be written as the sum of its integer part and fractional part: x + y/z. Sum of First n Natural Numbers The sum of the first n natural numbers is given by the formula: Sn = n(n + 1) / 2.

Shortcut Trick

The expression f(x) = 1/x − 1/(x + 1) is a telescoping series term.

Summing from x = 1 to x = 10: (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + ... + (1/10 − 1/11).

In a telescoping sum, all intermediate terms cancel out, leaving only the first and last parts.

Sum = First part of f(1) − Last part of f(10) = 1/1 − 1/11.

Sum = 10/11.

∴ The correct answer is 10/11.

Alternate Method

Given:

f(x) = (1/x) − 1/(x + 1)

Required sum: f(1) + f(2) + f(3) + ... + f(10)

Calculations:

⇒ f(1) = 1/1 − 1/2

⇒ f(2) = 1/2 − 1/3

⇒ f(3) = 1/3 − 1/4

... ... ...

⇒ f(10) = 1/10 − 1/11

⇒ Sum = (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + ... + (1/10 − 1/11)

⇒ Sum = 1/1 + (− 1/2 + 1/2) + (− 1/3 + 1/3) + ... + (− 1/10 + 1/10) − 1/11

⇒ Sum = 1 − 1/11

⇒ Sum = (11 − 1) / 11

⇒ Sum = 10/11

∴ The correct answer is 10/11.

Additional Information

Telescoping Series

A telescoping series is a series where the partial sums have only a fixed number of terms after cancellation. The general form is ∑ [g(n) − g(n+1)] = g(1) − g(N+1).

Partial Fraction Decomposition

This method is often used to break down expressions like 1 / [n × (n+1)] into simpler parts: 1/n − 1/(n+1) to facilitate summation.

General Summation Formula

For a series with a difference of k in denominators, f(n) = 1 / [n × (n+k)], the sum is (1/k) × [ (1/1 + 1/2 + ... + 1/k) − (1/(N+1) + ... + 1/(N+k)) ].

Shortcut Trick

Factor out the smallest power: N = 4¹¹(1 + 4¹ + 4² + 4³)

Simplify the bracket: N = 4¹¹ × (1 + 4 + 16 + 64) = 4¹¹ × 85

Convert to prime bases: (2²)¹¹ × (5¹ × 17¹) = 2²² × 5¹ × 17¹

Number of factors = (22 + 1) × (1 + 1) × (1 + 1) = 23 × 2 × 2 = 92

∴ The correct answer is 92.

Alternate Method

Given: N = 4¹¹ + 4¹² + 4¹³ + 4¹⁴

Formula Used: For a number N = a^p × b^q × c^r (where a, b, c are prime), total factors = (p + 1)(q + 1)(r + 1).

⇒ N = 4¹¹(1 + 4¹ + 4² + 4³)

⇒ N = 4¹¹ × (1 + 4 + 16 + 64)

⇒ N = 4¹¹ × 85

⇒ N = (2²)¹¹ × 5¹ × 17¹

⇒ N = 2²² × 5¹ × 17¹

⇒ Total Factors = (22 + 1) × (1 + 1) × (1 + 1)

⇒ Total Factors = 23 × 2 × 2 = 92

∴ The correct answer is 92.

Additional Information

Total Number of Factors

If the prime factorization is p₁^a × p₂^b × p₃^c, the total number of factors is (a + 1)(b + 1)(c + 1).

Sum of Factors

The sum of factors is given by [(p₁^a+1 − 1)/(p₁ − 1)] × [(p₂^b+1 − 1)/(p₂ − 1)] × ...

Number of Prime Factors

The total number of prime factors is the sum of the exponents of the prime bases: a + b + c.

Shortcut Trick

Convert N to prime base: N = 9⁹ = (3²)⁹ = 3¹⁸.

For a divisor to be a perfect cube, the power of its prime factor must be a multiple of 3.

Available power of 3 is from 0 to 18.

Multiple of 3 in the range [0, 18] are 0, 3, 6, 9, 12, 15, 18.

Shortcut formula: [Total Exponent ÷ 3] + 1 = [18 ÷ 3] + 1 = 6 + 1 = 7.

∴ The correct answer is 7.

Alternate Method

Given: N = 9⁹

Formula Used: If a number is expressed as N = p^a × q^b..., the number of perfect n^th power divisors is ([a/n] + 1) × ([b/n] + 1)... where [ ] denotes the greatest integer function.

⇒ N = 9⁹ = (3²)⁹ = 3¹⁸

⇒ For a divisor to be a perfect cube, it must be in the form 3^k where k is a multiple of 3.

⇒ The possible values for k such that 0 ≤ k ≤ 18 are 0, 3, 6, 9, 12, 15, 18.

⇒ Counting the values: {0, 3, 6, 9, 12, 15, 18} contains 7 elements.

∴ The correct answer is 7.

Additional Information

Number of Divisors

For a number N = p^a × q^b, the total number of divisors is (a + 1)(b + 1).

Perfect Square Divisors

To find perfect square divisors of p^a, we count multiples of 2 in the exponent range: [a/2] + 1.

Prime Factorization Rule

Always convert the base to a prime number (like 9 to 3²) before applying divisor formulas.

Shortcut Trick

Factorize the expression: N = 3¹³(3 + 1) − 12 = 4 × 3¹³ − 12 = 12(3¹² − 1)

Apply identity a² − b² = (a − b)(a + b): N = 12(3⁶ − 1)(3⁶ + 1)

Calculate powers: 3⁶ = 729, so N = 12 × (729 − 1) × (729 + 1)

Simplify: N = 12 × 728 × 730 = 12 × 728 × 10 × 73

Check primes: 73 is a prime number and is the largest among the factors 2, 3, 5, 7, 13, and 73.

∴ The correct answer is 73.

Alternate Method

Given:

N = 3¹⁴ + 3¹³ − 12

Formula Used:

1. a^m × aⁿ = a^m+n
2. a² − b² = (a − b)(a + b)

Calculations:

⇒ N = 3¹³ × 3¹ + 3¹³ − 12

⇒ N = 3¹³(3 + 1) − 12

⇒ N = 4 × 3¹³ − 12

⇒ N = 12 × 3¹² − 12 = 12(3¹² − 1)

⇒ N = 12((3⁶)² − 1²)

⇒ N = 12(3⁶ − 1)(3⁶ + 1)

⇒ N = 12(729 − 1)(729 + 1)

⇒ N = 12 × 728 × 730

⇒ Factorizing 728: 728 = 8 × 91 = 8 × 7 × 13

⇒ Factorizing 730: 730 = 10 × 73

⇒ Prime factors of N: 2, 3, 5, 7, 13, 73

⇒ The largest prime factor is 73.

∴ The correct answer is 73.

Additional Information

Factorization of aⁿ − bⁿ

For any positive integer n, aⁿ − bⁿ is always divisible by (a − b). If n is even, it is also divisible by (a + b).

Common Prime Factors

Prime numbers are numbers greater than 1 with no divisors other than 1 and themselves. Common small primes include 2, 3, 5, 7, 11, 13, 17, 19, 71, 73, 79, etc.

Exponential Laws

x^a − x^b = x^b(x^a−b − 1), which is useful for factoring out common terms in exponential expressions.

Shortcut Trick For Statement I: Recognize the pattern √12...n...21 = n times 1 (e.g., √121 = 11, √12321 = 111). Sum = 11 + 111 + 1111 = 1233. (True) For Statement II: Decimal square root rule — √0.64 = 0.8 and √0.36 = 0.6. Sum = 0.8 + 8 + 6 + 0.6 = 15.4. Since 15.4 > 15, it is True. ∴ The correct answer is Both I and II.

Alternate Method Given: I. √121 + √12321 + √1234321 = 1233 II. √0.64 + √64 + √36 + √0.36 > 15 Calculation for Statement I: ⇒ √121 = 11 ⇒ √12321 = 111 ⇒ √1234321 = 1111 ⇒ L.H.S = 11 + 111 + 1111 ⇒ L.H.S = 1233 ⇒ 1233 = 1233 (Statement I is TRUE) Calculation for Statement II: ⇒ √0.64 = 0.8 ⇒ √64 = 8 ⇒ √36 = 6 ⇒ √0.36 = 0.6 ⇒ L.H.S = 0.8 + 8 + 6 + 0.6 ⇒ L.H.S = 15.4 ⇒ 15.4 > 15 (Statement II is TRUE) ∴ The correct answer is Both I and II.

Additional Information Squares of numbers consisting only of 1s When a number like 111... (n times) is squared, the result is a palindromic number 123...(n)...321. For example, 11112 = 1234321. Square Roots of Decimals The square root of a decimal number has half the number of decimal places as the original number. For example, √0.0064 (4 places) = 0.08 (2 places). BODMAS Rule Always solve mathematical expressions in the order: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Shortcut Trick

Group the main terms: (2 + √2) + (2 − √2) = 4.

Combine the reciprocal terms: 1/(2 + √2) + 1/(2 − √2) = (2 − √2 + 2 + √2) / (4 − 2).

The fraction part simplifies to 4 ÷ 2 = 2.

The total value is the sum of both parts: 4 + 2 = 6.

∴ The correct answer is 6.

Alternate Method

Given expression: (2 + √2) + (1 ÷ (2 + √2)) + (1 ÷ (2 − √2)) + (2 − √2)

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

⇒ Rearrange the expression to group similar terms: [(2 + √2) + (2 − √2)] + [1 ÷ (2 + √2) + 1 ÷ (2 − √2)]

⇒ Simplify the first group: 2 + √2 + 2 − √2 = 4

⇒ Simplify the second group using a common denominator: [(2 − √2) + (2 + √2)] ÷ [(2 + √2) × (2 − √2)]

⇒ The numerator of the fraction: 2 − √2 + 2 + √2 = 4

⇒ The denominator of the fraction: (2)² − (√2)² = 4 − 2 = 2

⇒ The value of the fractional part: 4 ÷ 2 = 2

⇒ Total value of the expression: 4 + 2 = 6

∴ The correct answer is 6.

Additional Information

Rationalization of Denominator

To rationalize a denominator containing a surd, multiply both the numerator and denominator by the conjugate of the denominator.

Conjugate Surds

For a binomial surd (a + √b), its conjugate is (a − √b). Their product (a + √b)(a − √b) = a² − b is always a rational number.

Algebraic Identities

The difference of squares formula (a + b)(a − b) = a² − b² is essential for simplifying complex surd expressions and fractions.

Shortcut Trick

Difference of numbers (x − y) = (Difference of squares) ÷ (Sum of numbers)

x − y = 56 ÷ 14 = 4

Sum of squares = [(Sum)² + (Difference)²] ÷ 2

x² + y² = (14² + 4²) ÷ 2 = (196 + 16) ÷ 2 = 106

∴ The correct answer is 106.

Alternate Method

Given: Sum (x + y) = 14 and Difference of squares (x² − y²) = 56.

Formula Used: x² − y² = (x + y)(x − y) and x² + y² = [(x + y)² + (x − y)²] ÷ 2

⇒ (x + y)(x − y) = 56

⇒ 14 × (x − y) = 56

⇒ x − y = 4

⇒ 2x = (x + y) + (x − y) = 14 + 4 = 18 → x = 9

⇒ 2y = (x + y) − (x − y) = 14 − 4 = 10 → y = 5

⇒ Sum of squares = 9² + 5² = 81 + 25 = 106

∴ The correct answer is 106.

Additional Information

Algebraic Identity for Squares

The difference of squares of two numbers is equal to the product of their sum and their difference: a² − b² = (a + b)(a − b).

Relationship between Sum, Difference, and Squares

The sum of squares of two numbers can be expressed as: x² + y² = [(x + y)² + (x − y)²] ÷ 2.

Finding Numbers from Sum and Difference

If S is the sum and D is the difference, then the larger number is (S + D) ÷ 2 and the smaller number is (S − D) ÷ 2.

Shortcut Trick

Let x = 1006.

The expression is x² − (x + 1)(x − 1) + (x + 2)(x − 2) − (x + 3)(x − 3).

Applying the identity (a + b)(a − b) = a² − b²:

Net Value = x² − (x² − 1) + (x² − 4) − (x² − 9).

Net Value = 1 − 4 + 9 = 6.

∴ The correct answer is 6.

Alternate Method

Given:

1006² − 1007 × 1005 + 1008 × 1004 − 1009 × 1003

Formula Used:

(a + b)(a − b) = a² − b²

Calculations:

⇒ 1006² − (1006 + 1)(1006 − 1) + (1006 + 2)(1006 − 2) − (1006 + 3)(1006 − 3)

⇒ 1006² − (1006² − 1²) + (1006² − 2²) − (1006² − 3²)

⇒ 1006² − (1006² − 1) + (1006² − 4) − (1006² − 9)

⇒ 1006² − 1006² + 1 + 1006² − 4 − 1006² + 9

⇒ (1006² − 1006² + 1006² − 1006²) + (1 − 4 + 9)

⇒ 0 + 6

⇒ 6

∴ The correct answer is 6.

Additional Information

Algebraic Identities

Difference of squares: (a + b)(a − b) = a² − b² is a key identity for simplifying products of numbers centered around a common base.

BODMAS Rule

Always solve mathematical expressions following the order: Brackets, Orders (Powers/Roots), Division, Multiplication, Addition, and Subtraction.

Sum of Squares Identity

Expansion of (a ± b)² = a² ± 2ab + b² is often used in conjunction with the difference of squares in competitive exam simplifications.

Shortcut Trick

Rearrange the given equation: a² + b² − 6a − 4b + 13 = 0

For an equation of the form a² + b² + Aa + Bb + C = 0, the values are a = −A ÷ 2 and b = −B ÷ 2.

Value of a = −(−6) ÷ 2 = 3

Value of b = −(−4) ÷ 2 = 2

Required sum: a + b = 3 + 2 = 5

∴ The correct answer is 5.

Alternate Method

Given: a² + b² = 4b + 6a − 13

Formula used: (x − y)² = x² − 2xy + y²

⇒ a² + b² − 6a − 4b + 13 = 0

⇒ a² − 6a + 9 + b² − 4b + 4 = 0

⇒ (a² − 2 × a × 3 + 3²) + (b² − 2 × b × 2 + 2²) = 0

⇒ (a − 3)² + (b − 2)² = 0

⇒ For real numbers, if the sum of squares is zero, then each term must be zero.

⇒ a − 3 = 0 ⇒ a = 3

⇒ b − 2 = 0 ⇒ b = 2

⇒ a + b = 3 + 2 = 5

∴ The correct answer is 5.

Additional Information

Sum of Squares Property

If x² + y² + z² + ... = 0 for real numbers x, y, z, then x = 0, y = 0, z = 0, and so on.

Algebraic Identities

Standard identities include (a+b)² = a²+2ab+b² and (a−b)² = a²−2ab+b².

Completing the Square

To convert x² + kx into a perfect square, add and subtract (k/2)² from the expression.

Shortcut Trick

Since xy = 12 and x, y are positive integers, check factors of 12: (1, 12), (2, 6), (3, 4).

Testing x = 3 and y = 4 in the equation x⁴ + y⁴ + x²y² = 481.

3⁴ + 4⁴ + (3 × 4)² = 81 + 256 + 144 = 481. This satisfies the equation.

Required value = x² − xy + y² = 3² − 12 + 4² = 9 − 12 + 16 = 13.

∴ The correct answer is 13.

Alternate Method

Given: x⁴ + y⁴ + x²y² = 481 and xy = 12

Formula Used: x⁴ + x²y² + y⁴ = (x² + y²)² − x²y²

⇒ x⁴ + x²y² + y⁴ = 481

⇒ (x² + y²)² − (xy)² = 481

⇒ (x² + y²)² − (12)² = 481

⇒ (x² + y²)² − 144 = 481

⇒ (x² + y²)² = 481 + 144 = 625

⇒ x² + y² = √625 = 25 (As x, y are positive integers)

⇒ Required value = x² − xy + y²

⇒ Value = (x² + y²) − xy = 25 − 12

⇒ Value = 13

∴ The correct answer is 13.

Additional Information

Polynomial Identity

The expression x⁴ + x²y² + y⁴ can be factored as (x² + xy + y²)(x² − xy + y²).

Sum and Difference of Squares

(x² + y²)² = x⁴ + y⁴ + 2x²y², which helps in rearranging terms for biquadratic equations.

Algebraic Relationships

The value of x² + y² can also be expressed as (x + y)² − 2xy or (x − y)² + 2xy.

Given:

A = 1 + 2^P

B = 1 + 2^-P

Formula used:

x^-n = $\frac{1}{x^n}$

Calculations:

A = 1 + 2^P

⇒ 2^P = A - 1

B = 1 + 2^-P

⇒ B = 1 + $\frac{1}{2^P}$

⇒ B = 1 + $\frac{1}{A - 1}$

⇒ B = $\frac{A - 1 + 1}{A - 1}$

⇒ B = $\frac{A}{A - 1}$

∴ The correct answer is option (3).

Shortcut Trick From ax2 + bx + c = 0, we have A + B = −b/a and AB = c/a. New roots: R1 = AB + (A + B) = (c − b)/a and R2 = AB − (A + B) = (c + b)/a. Sum of new roots = R1 + R2 = 2c/a. Product of new roots = R1R2 = (c2 − b2)/a2. Equation format: a2x2 − (a2 × Sum)x + (a2 × Product) = 0. ∴ The correct answer is a2x2 − 2acx + c2 − b2 = 0. Alternate Method Given equation: ax2 + bx + c = 0 Roots: A and B Formula used: Sum of roots = −coeff(x)/coeff(x2) and Product = constant/coeff(x2) ⇒ A + B = −b/a ⇒ AB = c/a ⇒ Let the new roots be α and β. ⇒ α = AB + A + B = AB + (A + B) = c/a + (−b/a) = (c − b)/a ⇒ β = AB − A − B = AB − (A + B) = c/a − (−b/a) = (c + b)/a ⇒ Sum of new roots (α + β) = (c − b)/a + (c + b)/a = 2c/a ⇒ Product of new roots (αβ) = [(c − b)/a] × [(c + b)/a] = (c2 − b2)/a2 ⇒ The required quadratic equation is x2 − (α + β)x + αβ = 0 ⇒ x2 − (2c/a)x + (c2 − b2)/a2 = 0 ⇒ Multiplying the entire equation by a2: ⇒ a2x2 − 2acx + c2 − b2 = 0 ∴ The correct answer is a2x2 − 2acx + c2 − b2 = 0. Additional Information Sum and Product of Roots For a quadratic equation ax2 + bx + c = 0, the sum of roots is −b/a and the product of roots is c/a. Formation of Equation If S is the sum of roots and P is the product of roots, the quadratic equation is given by x2 − Sx + P = 0. Nature of Roots The nature of roots depends on the discriminant D = b2 − 4ac. If D > 0, roots are real and distinct; if D = 0, roots are real and equal; if D < 0, roots are imaginary.

Shortcut Trick

Substitute q = 0 in the given equation: √(1−p²) = √3/2.

Squaring both sides gives 1−p² = 3/4 ⇒ p² = 1/4.

The target expression simplifies to √(4p² + 4q²).

Substituting q = 0 and p² = 1/4: √(4 × 1/4 + 0) = √1 = 1.

∴ The correct answer is 1.

Alternate Method

Given:

√((1−p²)(1−q²)) = √3/2 and q = 0

Formula Used:

Algebraic substitution and simplification of radical expressions.

Calculations:

⇒ √((1−p²)(1−0²)) = √3/2

⇒ √(1−p²) = √3/2

⇒ 1 − p² = 3/4 (by squaring both sides)

⇒ p² = 1 − 3/4 = 1/4

⇒ Required value = √((2p² + 2q² + 2pq) + (2p² + 2q² − 2pq))

⇒ Value = √(2p² + 2q² + 2pq + 2p² + 2q² − 2pq)

⇒ Value = √(4p² + 4q²)

⇒ Value = √(4 × (1/4) + 4 × 0²)

⇒ Value = √(1 + 0) = 1

∴ The correct answer is 1.

Additional Information

BODMAS Rule

Always solve expressions in the order of Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Square Root Properties

The property √(a × b) = √a × √b applies for all non-negative real numbers a and b, helping simplify complex radicals.

Algebraic Identities

Basic identities like (a+b)² = a² + b² + 2ab and (a−b)² = a² + b² − 2ab are fundamental for simplifying polynomial roots.

Shortcut Trick Let (a + b) = x. The equation becomes x2 − 2x − 80 = 0. Solving the quadratic gives x = 10 or x = −8. Since ab = 16, factors of 16 that add to 10 are 8 and 2. For a = 8 and b = 2: 3a − 19b = 3 × 8 − 19 × 2 = 24 − 38 = −14. This matches Option 2. ∴ The correct answer is −14. Alternate Method Given: (a + b)2 − 2(a + b) = 80 and ab = 16 Formula: If sum S and product P of two numbers are known, the numbers are roots of t2 − St + P = 0. ⇒ (a + b)2 − 2(a + b) − 80 = 0 ⇒ Let k = a + b, then k2 − 2k − 80 = 0 ⇒ (k − 10)(k + 8) = 0 ⇒ k = 10 or k = −8 ⇒ Case 1: If a + b = 10 and ab = 16, roots of t2 − 10t + 16 = 0 are 8 and 2. ⇒ If a = 8 and b = 2: 3a − 19b = 3 × 8 − 19 × 2 = 24 − 38 = −14 ⇒ Case 2: If a + b = −8 and ab = 16, roots of t2 + 8t + 16 = 0 are −4 and −4. ⇒ If a = −4 and b = −4: 3a − 19b = 3 × (−4) − 19 × (−4) = −12 + 76 = 64 ∴ The correct answer is −14. Additional Information Quadratic Formula For ax2 + bx + c = 0, roots are found by x = [−b ± √(b2 − 4ac)] / 2a. Algebraic Identities (a + b)2 = a2 + b2 + 2ab and (a − b)2 = a2 + b2 − 2ab are fundamental for solving system of equations. Nature of Roots Discriminant D = b2 − 4ac determines if roots are real and distinct (D > 0), real and equal (D = 0), or imaginary (D < 0).

Shortcut Trick

Since x, y, z are natural numbers, x^{y + z} = 1 implies x = 1.

Substitute x = 1: y^{1 + z} = 1024 and z^{1 + y} = 729.

Known powers: 2¹⁰ = 1024 and 9³ = 729.

Comparing these, we get y = 2 and z = 9 (Check: 9^{1 + 2} = 729).

Required value: (9 + 1)^{2 + 1 + 1} = 10⁴ = 10000.

∴ The correct answer is 10000.

Alternate Method

Given values:

x^{y + z} = 1

y^{x + z} = 1024

z^{x + y} = 729

Calculations:

⇒ For x^{y + z} = 1, since y + z ≥ 2 (natural numbers), x must be 1.

⇒ Substitute x = 1 in y^{x + z} = 1024: y^{1 + z} = 1024.

⇒ Prime factorizing 1024: 2¹⁰ = 4⁵ = 32².

⇒ If y = 2, then 1 + z = 10 → z = 9.

⇒ If y = 4, then 1 + z = 5 → z = 4.

⇒ If y = 32, then 1 + z = 2 → z = 1.

⇒ Now check these pairs in the third equation z^{1 + y} = 729.

⇒ For (y=2, z=9): 9^{1 + 2} = 9³ = 729. (Satisfied)

⇒ Thus, x = 1, y = 2, and z = 9.

⇒ Target expression: (z + 1)^{y + x + 1}

⇒ (9 + 1)^{2 + 1 + 1} = 10⁴

⇒ 10 × 10 × 10 × 10 = 10000.

∴ The correct answer is 10000.

Additional Information

Laws of Indices

Key rules include a^m × aⁿ = a^m+n and (a^m)ⁿ = a^mn. If a^x = a^y and a ≠ 1, 0, −1, then x = y.

Special case for base 1

1 raised to any real power is always 1 (1ⁿ = 1). Conversely, if xⁿ = 1 and n ≠ 0, then x must be 1 for natural numbers.

Powers of 2 and 3

Commonly used values: 2¹⁰ = 1024, 3⁴ = 81, 3⁵ = 243, 3⁶ = 729, 9³ = 729.

Shortcut Trick

Product xyz = [(x+y+z)³ − 3(x+y+z)(x²+y²+z²) + 2(x³+y³+z³)] ÷ 6

xyz = [1³ − 3(1)(2) + 2(3)] ÷ 6

xyz = [1 − 6 + 6] ÷ 6

xyz = 1 ÷ 6 = 1/6

∴ The correct answer is 1/6.

Alternate Method

Given:

x + y + z = 1

x² + y² + z² = 2

x³ + y³ + z³ = 3

Formula Used:

1. (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

2. x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − (xy + yz + zx))

Calculations:

⇒ (1)² = 2 + 2(xy + yz + zx)

⇒ 1 − 2 = 2(xy + yz + zx)

⇒ xy + yz + zx = −1/2

⇒ 3 − 3xyz = 1 × [2 − (−1/2)]

⇒ 3 − 3xyz = 2 + 1/2

⇒ 3 − 3xyz = 5/2

⇒ 3xyz = 3 − 5/2

⇒ 3xyz = 1/2

⇒ xyz = 1/6

∴ The correct answer is 1/6.

Additional Information

Sum of Cubes Identity

The expression a³ + b³ + c³ − 3abc is always equal to (a + b + c)(a² + b² + c² − ab − bc − ca).

Special Case for xyz

If a + b + c = 0, then a³ + b³ + c³ = 3abc. This is a frequently used property in competitive exams.

Square of Trinomial

The identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) helps find the sum of products taken two at a time.

Shortcut Trick

The angle formed by the internal bisectors of two angles of a triangle at the incenter O is given by the formula:

∠QOR = 90° + (∠P ÷ 2)

Given ∠P = 70°

∠QOR = 90° + (70° ÷ 2) = 90° + 35° = 125°

∴ The correct answer is 125°.

Alternate Method

Given: ∠QPR = 70°, O is the intersection of internal bisectors of ∠Q and ∠R.

Formula: Angle at the incenter ∠QOR = 90° + (∠P / 2)

⇒ In ΔPQR, ∠P + ∠Q + ∠R = 180°

⇒ ∠Q + ∠R = 180° − 70° = 110°

⇒ Divide by 2: (∠Q/2) + (∠R/2) = 55°

⇒ In ΔOQR, ∠QOR + ∠OQR + ∠ORQ = 180°

⇒ ∠QOR + (∠Q/2) + (∠R/2) = 180°

⇒ ∠QOR + 55° = 180°

⇒ ∠QOR = 180° − 55° = 125°

∴ The correct answer is 125°.

Additional Information

Incenter Properties

The incenter is the point of intersection of internal angle bisectors and is equidistant from all sides of the triangle.

External Bisector Angle

The angle formed by external bisectors of ∠Q and ∠R is given by: 90° − (∠P / 2).

Circumcenter Angle

The angle formed by the sides at the circumcenter is twice the angle at the opposite vertex: ∠QCR = 2 × ∠P.

Shortcut Trick

For any triangle where medians to sides b and c are perpendicular, the relation is b² + c² = 5a².

In the isosceles ΔPQR, PQ = PR. Let PQ = PR = b = c and QR = a.

Applying the theorem: PQ² + PR² = 5 × QR².

Since PQ = PR ⇒ 2 × PQ² = 5 × QR².

Rearranging the terms: PQ² / QR² = 5/2.

∴ The correct answer is 5/2.

Alternate Method

Given: In ΔPQR, PQ = PR. RS and QT are medians to sides PQ and PR. RS ⊥ QT.

Formula Used: Centroid G divides medians in ratio 2:1. In an isosceles triangle, medians to equal sides are equal (RS = QT).

⇒ Let medians RS = QT = 3x.

⇒ In ΔGQR, G is the centroid, so GQ = GR = (2/3) × 3x = 2x.

⇒ Since ∠G = 90°, by Pythagoras: GQ² + GR² = QR².

⇒ (2x)² + (2x)² = QR² ⇒ 8x² = QR².

⇒ In ΔGQS, GS = (1/3) × 3x = x. By Pythagoras: GQ² + GS² = QS².

⇒ (2x)² + x² = (PQ/2)² ⇒ 5x² = PQ²/4 ⇒ 20x² = PQ².

⇒ (PQ/QR)² = PQ² / QR² = 20x² / 8x² = 5/2.

∴ The correct answer is 5/2.

Additional Information

Perpendicular Medians Property

In any triangle, if the medians to sides b and c are perpendicular, then the relation between the sides is given by b² + c² = 5a².

Apollonius Theorem

For a triangle with side lengths a, b, c and median m_a to side a: b² + c² = 2(m_a² + (a/2)²).

Centroid Theorem

The centroid G of a triangle divides each median in the ratio 2:1 from the vertex to the midpoint of the opposite side.

Shortcut Trick

According to the Midpoint Theorem, a line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of its length.

Since S and T are midpoints of PQ and PR, then ST ∥ QR (Statement III) and ST = 1/2(QR) (Statement II).

In ΔPST and ΔPQR, ∠P is common and PS/PQ = PT/PR = 1/2, so ΔPST ∼ ΔPQR by SAS similarity (Statement I).

Thus, all three statements are mathematically derived from the Midpoint Theorem properties.

∴ The correct answer is All I, II and III.

Alternate Method

Given: S is the midpoint of PQ and T is the midpoint of PR in ΔPQR.

Formula: Basic Proportionality Theorem and SAS Similarity Criterion.

Calculations:

⇒ In ΔPQR, S is midpoint of PQ ⇒ PS = SQ = 1/2(PQ)

⇒ T is midpoint of PR ⇒ PT = TR = 1/2(PR)

⇒ In ΔPST and ΔPQR:

⇒ PS/PQ = 1/2 and PT/PR = 1/2

⇒ ∠SPT = ∠QPR (Common angle)

⇒ ∴ ΔPST ∼ ΔPQR (By SAS Similarity) → Statement I is TRUE.

⇒ Since ΔPST ∼ ΔPQR, the corresponding angles are equal: ∠PST = ∠PQR.

⇒ Since corresponding angles are equal, ST ∥ QR → Statement III is TRUE.

⇒ Also, the ratio of corresponding sides is equal: ST/QR = PS/PQ = 1/2

⇒ ST = 1/2(QR) → Statement II is TRUE.

∴ The correct answer is All I, II and III.

Additional Information

Converse of Midpoint Theorem

If a line is drawn through the midpoint of one side of a triangle and is parallel to another side, it bisects the third side.

Area Ratio of Similar Triangles

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Area(ΔPST) : Area(ΔPQR) = (1/2)² = 1 : 4.

Perimeter Relation

For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides. Perimeter(ΔPST) = 1/2 × Perimeter(ΔPQR).

Shortcut Trick

To find the ratio A ∶ B ∶ C when 6A = 4B = 9C, take the LCM of the coefficients.

LCM of 6, 4, and 9 = 36.

Divide the LCM by each coefficient to get the ratio parts:

A = 36 ÷ 6 = 6 | B = 36 ÷ 4 = 9 | C = 36 ÷ 9 = 4.

∴ The correct answer is 6 ∶ 9 ∶ 4.

Alternate Method

Given: 6A = 4B = 9C

Formula Used: If kA = mB = nC, then A ∶ B ∶ C = 1/k ∶ 1/m ∶ 1/n.

Let 6A = 4B = 9C = k

⇒ A = k/6

⇒ B = k/4

⇒ C = k/9

⇒ A ∶ B ∶ C = k/6 ∶ k/4 ∶ k/9

⇒ A ∶ B ∶ C = 1/6 ∶ 1/4 ∶ 1/9

To eliminate fractions, multiply by the LCM of 6, 4, 9 which is 36:

⇒ A ∶ B ∶ C = (1/6 × 36) ∶ (1/4 × 36) ∶ (1/9 × 36)

⇒ A ∶ B ∶ C = 6 ∶ 9 ∶ 4

∴ The correct answer is 6 ∶ 9 ∶ 4.

Additional Information

Inverse Ratio

The inverse ratio of a ∶ b is 1/a ∶ 1/b, which simplifies to b ∶ a.

Compound Ratio

For two ratios a ∶ b and c ∶ d, the compound ratio is calculated as (a × c) ∶ (b × d).

Third Proportional

In a proportion a ∶ b ∷ b ∶ c, the value c is the third proportional, where c = b²/a.

Shortcut Trick

If three numbers a, b, and c are in continued proportion, then a : b :: b : c.

The third proportional formula is c = b² ÷ a.

Given: a = 10, b = 25

Third Proportional = (25 × 25) ÷ 10 = 625 ÷ 10 = 62.5

∴ The correct answer is 62.5.

Alternate Method

Given:

First number (a) = 10

Second number (b) = 25

Let the third proportional be x.

Formula Used:

a : b :: b : x ⇒ a/b = b/x

According to the continued proportion:

⇒ 10 ÷ 25 = 25 ÷ x

⇒ 10 × x = 25 × 25

⇒ 10x = 625

⇒ x = 625 ÷ 10

⇒ x = 62.5

∴ The correct answer is 62.5.

Additional Information

Mean Proportional

If b is the mean proportional between a and c, then b = √(a × c).

Fourth Proportional

In the proportion a : b :: c : d, d is the fourth proportional and is calculated as d = (b × c) ÷ a.

Continued Proportion

Three numbers a, b, c 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).

Shortcut Trick

Ratio of coins → 1₹ : 50p : 10p = 6 : 9 : 10

Total relative value = (6 × 1) + (9 × 0.5) + (10 × 0.1) = 6 + 4.5 + 1 = 11.5 units

Given total value = ₹34.5 ∴ Multiplier = 34.5 ÷ 11.5 = 3

Number of 10-paise coins = 10 × 3 = 30

∴ The correct answer is 30.

Alternate Method

Given:

Total amount in purse = ₹34.5

Ratio of number of coins = 6 : 9 : 10

Denominations = ₹1, 50-paise (₹0.5), 10-paise (₹0.1)

Let the number of 1-rupee, 50-paise, and 10-paise coins be 6x, 9x, and 10x respectively.

⇒ Total value = (6x × 1) + (9x × 0.5) + (10x × 0.1) = 34.5

⇒ 6x + 4.5x + 1x = 34.5

⇒ 11.5x = 34.5

⇒ x = 34.5 ÷ 11.5

⇒ x = 3

⇒ Number of 10-paise coins = 10x

⇒ Number of 10-paise coins = 10 × 3 = 30

∴ The correct answer is 30.

Additional Information

Value of Coins Formula

Total Value = Σ(Number of coins × Face value of each coin). Ensure all denominations are in the same unit (either Rupees or Paise).

Unit Conversion

To convert paise to rupees, divide by 100 (e.g., 50p = ₹0.50). To convert rupees to paise, multiply by 100.

Ratio of Number vs Ratio of Value

If ratio of values is given as V₁:V₂ and face values are F₁, F₂, then ratio of number of coins = (V₁/F₁) : (V₂/F₂).

Shortcut Trick

For four numbers a, b, c, and d, the number to be added (x) = |(ad − bc) ÷ ((a + d) − (b + c))|

Here a = 103, b = 135, c = 110, d = 144

Numerator = |(103 × 144) − (135 × 110)| = |14832 − 14850| = 18

Denominator = (103 + 144) − (135 + 110) = 247 − 245 = 2

x = 18 ÷ 2 = 9

∴ The correct answer is 9.

Alternate Method

Given: Numbers are 103, 135, 110, and 144.

Formula: If a, b, c, and d are in proportion, then (a + x)/(b + x) = (c + x)/(d + x).

⇒ (103 + x) × (144 + x) = (110 + x) × (135 + x)

⇒ 14832 + 103x + 144x + x² = 14850 + 110x + 135x + x²

⇒ 14832 + 247x = 14850 + 245x

⇒ 247x − 245x = 14850 − 14832

⇒ 2x = 18

⇒ x = 9

∴ The correct answer is 9.

Additional Information

Mean Proportional

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

Third Proportional

If a, b, and c are in continued proportion (a:b = b:c), then c is the third proportional and c = b² ÷ a.

Invertendo and Alternendo

If a/b = c/d, then b/a = d/c (Invertendo) and a/c = b/d (Alternendo).

Componendo and Dividendo

If a/b = c/d, then (a + b)/(a − b) = (c + d)/(c − d).

Shortcut Trick

Revenue Ratio = (Price Ratio) × (Visitor Ratio)

Revenue Ratio = (11 × 8) : (12 × 7) = 88 : 84

Given: 88 units = ₹1,76,000 ⇒ 1 unit = ₹2,000

New Revenue = 84 units = 84 × 2,000 = ₹1,68,000

∴ The correct answer is ₹168000.

Alternate Method

Given:

Initial Price : Final Price = 11 : 12

Initial Visitors : Final Visitors = 8 : 7

Initial Revenue = ₹1,76,000

Formula:

Total Revenue = Ticket Price × Number of Visitors

⇒ Ratio of Daily Revenues = (11 × 8) : (12 × 7)

⇒ Ratio of Daily Revenues = 88 : 84

⇒ Let the initial revenue be 88x and final revenue be 84x.

⇒ 88x = 1,76,000

⇒ x = 1,76,000 ÷ 88

⇒ x = 2,000

⇒ New Daily Revenue = 84x = 84 × 2,000

⇒ New Daily Revenue = ₹1,68,000

∴ The correct answer is ₹168000.

Additional Information

Revenue and Ratio

Total Revenue is directly proportional to both the price per unit and the quantity sold (Revenue = Price × Consumption).

Successive Change in Ratio

When multiple factors affect a result, the net ratio is the product of individual ratios: (a₁/a₂) × (b₁/b₂) = Result₁/Result₂.

Percentage Conversion

If price increases by x% and visitors decrease by y%, net change % = [x − y − (xy/100)]%.

Shortcut Trick

Total Average Weight = 74 kg

Deviation for (X + Y) = 2 × (68 − 74) = −12 kg

Deviation for (Y + Z) = 2 × (78 − 74) = +8 kg

Weight of Y = Base Average + Total Deviation = 74 + (−12 + 8) = 74 − 4 = 70 kg

∴ The correct answer is 70 kg.

Alternate Method

Given: Average weight of (X, Y, Z) = 74 kg, (X, Y) = 68 kg, and (Y, Z) = 78 kg.

Formula Used: Total Weight = Average Weight × Number of persons

⇒ Total weight of X, Y and Z = 74 × 3 = 222 kg

⇒ Total weight of X and Y = 68 × 2 = 136 kg

⇒ Total weight of Y and Z = 78 × 2 = 156 kg

⇒ Sum of (X + Y) and (Y + Z) = (X + Y) + (Y + Z) = X + 2Y + Z

⇒ X + 2Y + Z = 136 + 156 = 292 kg

⇒ Weight of Y = (X + 2Y + Z) − (X + Y + Z) = 292 − 222

⇒ Weight of Y = 70 kg

∴ The correct answer is 70 kg.

Additional Information

Average Concept

Average is the sum of observations divided by the number of observations. Average = Σx ÷ n.

Overlap Rule in Averages

If an item is included in two different overlapping subsets, its value = (Sum of Subset 1 + Sum of Subset 2) − Total Sum of all items.

Weighted Average

When combining groups, the new average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), where A represents the respective averages.

Shortcut Trick

Sum of 3 numbers = 26 × 3 = 78.

Ratio of First number : (Sum of other two) = 2 : 11.

Total units = 2 + 11 = 13 units.

13 units = 78 ⇒ 1 unit = 6.

First number = 2 units = 2 × 6 = 12.

∴ The correct answer is 12.

Alternate Method

Given:

Average of 3 numbers = 26

Relationship: First number = (2/11) × Sum of other two numbers

Formula Used:

Total Sum = Average × Number of terms

Ratio distribution: Value = (Specific Ratio / Total Ratio) × Total Sum

Calculations:

⇒ Total sum of 3 numbers = 26 × 3 = 78

⇒ Let the first number be A and the sum of the other two be (B + C).

⇒ A = (2/11) × (B + C)

⇒ A / (B + C) = 2 / 11

⇒ Ratio A : (B + C) = 2 : 11

⇒ Total Ratio = 2 + 11 = 13

⇒ First number (A) = (2 / 13) × 78

⇒ A = 2 × 6 = 12

∴ The correct answer is 12.

Additional Information

Average Fundamentals

The average of n numbers is their sum divided by n. Sum = Average × n.

Ratio and Proportion

If a total quantity T is divided in the ratio x : y, the parts are [x / (x+y)] × T and [y / (x+y)] × T.

Combined Average

If two groups have averages A₁ and A₂ with n₁ and n₂ terms, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Shortcut Trick

Ratio of Boys to Girls = 20 : 30 = 2 : 3.

Deviation of Boys' average from class average = 54 − 48.6 = +5.4.

Total deviation for 2 units of boys = 2 × 5.4 = +10.8.

This positive deviation must be balanced by 3 units of girls: −10.8 ÷ 3 = −3.6.

Average weight of girls = Class Average + Girl's deviation = 48.6 − 3.6 = 45 kg.

∴ The correct answer is 45 kg.

Alternate Method

Given:

Total students = 50

Average weight of class = 48.6 kg

Number of boys = 20, Average weight of boys = 54 kg

Formula Used:

Total Weight = Average × Number of students

Average of a group = Total weight ÷ Total number of students

⇒ Total weight of the class = 50 × 48.6 = 2430 kg

⇒ Total weight of 20 boys = 20 × 54 = 1080 kg

⇒ Number of girls in the class = 50 − 20 = 30

⇒ Total weight of 30 girls = Total weight of class − Total weight of boys

⇒ Total weight of 30 girls = 2430 − 1080 = 1350 kg

⇒ Average weight of girls = 1350 ÷ 30

⇒ Average weight of girls = 45 kg

∴ The correct answer is 45 kg.

Additional Information

Weighted Average Formula

The combined average of two groups is given by A_avg = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂), where n is the number and a is the average.

Alligation Method

The ratio of the number of items in two groups is inversely proportional to the difference between their individual averages and the mean average.

Deviation Concept

In any distribution, the sum of deviations from the arithmetic mean is always zero. This is used to find missing values quickly.

Shortcut Trick

The sequence of odd numbers 113, 115, 117, ..., 159 forms an Arithmetic Progression (AP).

For any set of numbers in AP, the Average = (First term + Last term) ÷ 2.

Average = (113 + 159) ÷ 2 = 272 ÷ 2 = 136.

∴ The correct answer is 136.

Alternate Method

Given: First odd number (a) = 113, Last odd number (l) = 159, Common difference (d) = 2.

Formula Used: Average = Sum ÷ Number of terms (n); n = [(l − a) ÷ d] + 1; Sum = (n ÷ 2) × (a + l)

⇒ n = [(159 − 113) ÷ 2] + 1

⇒ n = [46 ÷ 2] + 1 = 23 + 1 = 24 terms

⇒ Sum = (24 ÷ 2) × (113 + 159)

⇒ Sum = 12 × 272 = 3264

⇒ Average = 3264 ÷ 24 = 136

∴ The correct answer is 136.

Additional Information

Average of First n Odd Numbers

The average of the first n odd natural numbers starting from 1 is always equal to n.

Average of First n Even Numbers

The average of the first n even natural numbers starting from 2 is equal to (n + 1).

Sum of Squares of First n Natural Numbers

Sum = [n(n + 1)(2n + 1)] ÷ 6; Average = [(n + 1)(2n + 1)] ÷ 6.

Shortcut Trick

Required SP per kg = [CP per kg × (100 + Profit %)] ÷ (100 − Waste %)

Required SP per kg = [30 × (100 + 40)] ÷ (100 − 20)

Required SP per kg = [30 × 140] ÷ 80 = 4200 ÷ 80

Required SP per kg = ₹52.5

∴ The correct answer is ₹52.5.

Alternate Method

Given:

Cost Price (CP) of Jowar = ₹30 per kg

Wastage = 20%

Overall Profit required = 40%

Formula Used:

Total Selling Price = Total Cost Price × (1 + Profit/100)

Selling Price per kg = Total Selling Price ÷ Usable Quantity

⇒ Let us assume the total quantity purchased = 100 kg

⇒ Total Cost Price = 100 kg × ₹30/kg = ₹3000

⇒ Quantity remaining after 20% wastage = 100 − (20% of 100) = 80 kg

⇒ Total Selling Price needed for 40% profit = 3000 + (40/100 × 3000)

⇒ Total Selling Price = 3000 + 1200 = ₹4200

⇒ Selling price per kg of usable grain = 4200 ÷ 80

⇒ Selling price per kg = 420 ÷ 8 = ₹52.5

∴ The correct answer is ₹52.5.

Additional Information

Effect of Wastage on Cost

Wastage increases the effective cost price of the remaining items. Effective CP = Total Cost ÷ Usable Quantity.

Profit Percentage Formula

Profit % = [(Total SP − Total CP) ÷ Total CP] × 100. Always calculate profit on the original investment (CP).

Mark-up Percentage

Mark-up is the amount by which the cost of an item is increased to arrive at the selling price. SP = CP × (1 + Mark-up %).

Shortcut Trick

8% Loss → Selling Price is 92% of CP.

16% Profit → Required Selling Price is 116% of CP.

92% → ₹69

116% → (69 ÷ 92) × 116 = 0.75 × 116 = ₹87.

∴ The correct answer is ₹87.

Alternate Method

Given:

Initial Selling Price (SP₁) = ₹69

Loss percentage = 8%

Desired Profit percentage = 16%

Formula Used:

Cost Price (CP) = [SP × 100] ÷ [100 − Loss%]

New Selling Price (SP₂) = [CP × (100 + Profit%)] ÷ 100

Calculations:

⇒ CP = (69 × 100) ÷ (100 − 8)

⇒ CP = (69 × 100) ÷ 92

⇒ CP = 6900 ÷ 92 = ₹75

⇒ SP₂ = [75 × (100 + 16)] ÷ 100

⇒ SP₂ = (75 × 116) ÷ 100

⇒ SP₂ = 3 × 116 ÷ 4

⇒ SP₂ = 3 × 29 = ₹87

∴ The correct answer is ₹87.

Additional Information

Relation between CP and MP

The ratio of Cost Price to Marked Price is given by CP/MP = (100 − Discount%) / (100 + Profit%).

Successive Discounts

Two successive discounts of d₁% and d₂% are equivalent to a single discount of [d₁ + d₂ − (d₁ × d₂)/100]%.

Profit or Loss Percentage

Profit or Loss is always calculated on the Cost Price (CP) unless stated otherwise in the question.

Shortcut Trick

If CP of 'A' items = SP of 'B' items, then CP : SP = B : A.

Here, CP of 25 items = SP of x items ⇒ CP : SP = x : 25.

Profit = 25% ⇒ Ratio of CP : SP = 100 : 125 = 4 : 5.

Comparing ratios: 5 units = 25 items ⇒ 1 unit = 5 items.

Value of x = 4 units = 4 × 5 = 20 items.

∴ The correct answer is 20 items.

Alternate Method

Given: Cost Price (CP) of 25 items = Selling Price (SP) of x items. Profit percentage = 25%.

Formula Used: Profit% = [(Items bought − Items sold) ÷ Items sold] × 100

⇒ Let CP of 1 item = ₹1.

⇒ Total CP = CP of 25 items = ₹25.

⇒ Revenue earned by selling x items = CP of 25 items = ₹25.

⇒ SP of x items = ₹25 ⇒ SP of 1 item = 25/x.

⇒ Profit% = [(SP − CP) ÷ CP] × 100

⇒ 25 = [(25/x − 1) ÷ 1] × 100

⇒ 25/100 = 25/x − 1

⇒ 0.25 + 1 = 25/x

⇒ 1.25 = 25/x

⇒ x = 25 ÷ 1.25 = 20 items.

∴ The correct answer is 20 items.

Additional Information

Quantity Based Profit/Loss

When the cost price of M articles is equal to the selling price of N articles, the profit or loss percentage is given by [(M − N) ÷ N] × 100. Positive value indicates profit, and negative indicates loss.

Inverse Relationship

For a fixed total amount, the price per item is inversely proportional to the number of items. If the number of items sold is less than items bought for the same price, there is always a profit.

Successive Discount

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

Shortcut Trick

Profit = 15% = 3/20 in fraction form.

Ratio of CP : SP = 20 : (20 + 3) = 20 : 23.

Given SP = 23 units = ₹7130.

1 unit = 7130 ÷ 23 = 310.

CP = 20 units = 20 × 310 = ₹6200.

∴ The correct answer is ₹6200.

Alternate Method

Given: Selling Price (SP) = ₹7130, Profit Percentage = 15%

Formula Used: Cost Price (CP) = [Selling Price × 100] ÷ [100 + Profit %]

⇒ CP = [7130 × 100] ÷ [100 + 15]

⇒ CP = 713000 ÷ 115

⇒ CP = 6200

∴ The correct answer is ₹6200.

Additional Information

Cost Price in case of Loss

When an item is sold at a loss, the Cost Price is calculated as: CP = [SP × 100] ÷ [100 − Loss %].

Marked Price and Discount

Selling Price (SP) = Marked Price (MP) − Discount. Relationship between CP and MP is: CP ÷ MP = [100 − Discount %] ÷ [100 + Profit %].

Effective Profit/Loss

For successive changes of a% and b%, the effective change is [a + b + (a × b) ÷ 100] %.

Given:

0.02% of 150% of 600

Formula used:

x% of y = $\frac{x}{100} \times y$

Calculations:

? = $\frac{0.02}{100} \times \frac{150}{100} \times 600$

⇒ ? = 0.0002 × 1.5 × 600

⇒ ? = 0.0002 × 900

⇒ ? = 0.18

∴ The correct answer is option (1).

Shortcut Trick

Original number = 100%.

New number = 125%.

Increase in percentage = 125% − 100% = 25%.

According to the question, 25% = 40.

Original number (100%) = 40 × 4 = 160.

∴ The correct answer is 160.

Alternate Method

Given:

Increase in value = 40

New value = 125% of original number

Formula Used:

New Number = Original Number + Increase

Let the number be x.

⇒ x + 40 = 125% of x

⇒ x + 40 = (125 ÷ 100) × x

⇒ x + 40 = 1.25x

⇒ 1.25x − x = 40

⇒ 0.25x = 40

⇒ x = 40 ÷ 0.25

⇒ x = 40 × (100 ÷ 25)

⇒ x = 40 × 4

⇒ x = 160

∴ The correct answer is 160.

Additional Information

Percentage to Fraction Conversion

125% can be written as 125/100 = 5/4. This means the original number is 4 units and the new number is 5 units.

Percentage Increase Formula

Percentage Increase = (Increase / Original Value) × 100. In this case, 25 = (40 / Original) × 100.

Base Value Concept

Whenever a value is compared to "itself" in percentages, the "itself" part is always considered the base, which is 100%.

Shortcut Trick

Total Marks = 300

Passing Marks = Student score + Marks needed to reach passing grade

Passing Marks = 75 + 6 = 81

Passing Percentage = 81 ÷ (300 ÷ 100) = 81 ÷ 3 = 27%

∴ The correct answer is 27%.

Alternate Method

Given:

Total Maximum Marks = 300

Marks Obtained by Student = 75

Shortfall for Passing = 6 marks

Formula Used:

Percentage = (Part ÷ Whole) × 100

Calculations:

⇒ Passing Marks = 75 + 6

⇒ Passing Marks = 81

⇒ Passing Percentage = (Passing Marks ÷ Total Marks) × 100

⇒ Passing Percentage = (81 ÷ 300) × 100

⇒ Passing Percentage = 81 ÷ 3

⇒ Passing Percentage = 27%

∴ The correct answer is 27%.

Additional Information

Percentage Calculation

To find what percentage x is of y, use the formula (x ÷ y) × 100. Always ensure the units of x and y are the same.

Fraction to Percentage Conversion

Common fractions like 1/4 = 25%, 1/5 = 20%, and 1/10 = 10% can be used to speed up calculations in competitive exams.

Passing Marks Concept

Passing Marks = Marks Obtained + Marks failed by = Total Marks × Passing Percentage ÷ 100.

Shortcut Trick

Tax Amount = Income × Tax Rate. If Tax remains the same, Income is inversely proportional to the Tax Rate.

Ratio of Tax Rates (Old : New) = 20 : 16 ⇒ 5 : 4

Ratio of Incomes (Old : New) = 4 : 5 (Inverse ratio)

Increase in Income = 5 − 4 = 1 unit

Given 1 unit = ₹2 lakhs, so New Income (5 units) = 5 × 2 = ₹10 lakhs.

∴ The correct answer is ₹10 lakhs.

Alternate Method

Given:

Increase in income = ₹2 lakhs

Old tax rate = 20% | New tax rate = 16%

Tax amount remains unchanged.

Calculations:

⇒ Let the original annual income be x lakhs.

⇒ Increased annual income = (x + 2) lakhs.

⇒ According to the question: 20% of x = 16% of (x + 2)

⇒ (20 ÷ 100) × x = (16 ÷ 100) × (x + 2)

⇒ 20x = 16(x + 2)

⇒ 20x = 16x + 32

⇒ 20x − 16x = 32

⇒ 4x = 32

⇒ x = 32 ÷ 4 = 8 lakhs

⇒ Increased income = x + 2 = 8 + 2 = 10 lakhs.

∴ The correct answer is ₹10 lakhs.

Additional Information

Inverse Proportion Rule

When the product of two variables is constant (Income × Rate = Tax), then the variables are inversely proportional to each other. If one increases, the other must decrease in the same ratio.

Percentage Base Change

If a value is reduced by r%, to restore it to the original value, the new value must be increased by [r ÷ (100 − r)] × 100%.

Taxable Income

Total Tax = (Gross Income − Exemptions) × Tax Rate. In basic quantitative aptitude problems, exemptions are usually assumed to be zero unless specified.

Shortcut Trick

Speed in kmph = Distance ÷ Time = 630 ÷ 20 = 31.5 kmph.

Conversion factor from kmph to m/s = 5/18.

Speed in m/s = 31.5 × 5/18 = 1.75 × 5 = 8.75 m/s.

∴ The correct answer is 8.75 m/s.

Alternate Method

Given:

Total Distance = 630 km

Total Time = 20 hours

Formula Used:

Average Speed = Total Distance ÷ Total Time

Conversion: Speed (m/s) = Speed (kmph) × 5/18

Calculations:

⇒ Average Speed (kmph) = 630 / 20

⇒ Average Speed (kmph) = 31.5 kmph

⇒ Average Speed (m/s) = 31.5 × 5/18

⇒ Average Speed (m/s) = (31.5 ÷ 9) × (5 ÷ 2)

⇒ Average Speed (m/s) = 3.5 × 2.5

⇒ Average Speed (m/s) = 8.75 m/s

∴ The correct answer is 8.75 m/s.

Additional Information

Unit Conversion Rules

To convert km/h to m/s, multiply the speed by 5/18. To convert m/s to km/h, multiply by 18/5.

Average Speed for Constant Distances

If a journey is divided into two equal distances traveled at speeds x and y, the average speed is given by (2 × x × y) ÷ (x + y).

Relative Speed

When two objects move in the same direction, relative speed is |s₁ − s₂|. In opposite directions, it is s₁ + s₂.

Shortcut Trick

When the same distance is covered at two different speeds, x and y, the Average Speed = (2 × x × y) ÷ (x + y).

Here, Upstream speed (x) = 48 km/hr and Downstream speed (y) = 80 km/hr.

Average Speed = (2 × 48 × 80) ÷ (48 + 80)

Average Speed = 7680 ÷ 128 = 60 km/hr.

∴ The correct answer is 60 km/hr.

Alternate Method

Given: Upstream speed = 48 km/hr, Downstream speed = 80 km/hr.

Formula Used: Average Speed = Total Distance ÷ Total Time

⇒ Let the distance for one side of the journey be D km.

⇒ Total distance covered in the round trip = D + D = 2D km.

⇒ Time taken to travel upstream (t₁) = Distance ÷ Speed = D ÷ 48 hours.

⇒ Time taken to travel downstream (t₂) = Distance ÷ Speed = D ÷ 80 hours.

⇒ Total Time = D ÷ 48 + D ÷ 80

⇒ Taking LCM of 48 and 80, which is 240: Total Time = (5D + 3D) ÷ 240 = 8D ÷ 240 = D ÷ 30 hours.

⇒ Average Speed = Total Distance ÷ Total Time = 2D ÷ (D ÷ 30)

⇒ Average Speed = 2D × 30 ÷ D = 60 km/hr.

∴ The correct answer is 60 km/hr.

Additional Information

Average Speed for Equal Distances

When a person travels equal distances at speeds S₁, S₂, ..., S_n, the average speed is the harmonic mean of the speeds.

Boats and Streams Speed Formulas

Speed of Boat in Still Water = (Downstream Speed + Upstream Speed) ÷ 2 | Speed of Stream = (Downstream Speed − Upstream Speed) ÷ 2.

Weighted Average Speed

If distances d₁ and d₂ are covered at speeds v₁ and v₂, then Average Speed = (d₁ + d₂) ÷ (d₁/v₁ + d₂/v₂).

Shortcut Trick

Convert Speed to m/s: Speed = 2520 × 5/18 = 700 m/s.

Distance = Speed × Time = 700 × 0.2.

Distance = 140 m.

∴ The correct answer is 140 m.

Alternate Method

Given:

Speed of the bullet = 2520 km/hr

Time taken to hit the target = 0.2 seconds

Formula Used:

Distance = Speed × Time

1 km/hr = 5/18 m/s

Calculations:

⇒ Speed in m/s = 2520 × (5/18)

⇒ Speed = 140 × 5 = 700 m/s

⇒ Distance = Speed × Time

⇒ Distance = 700 m/s × 0.2 s

⇒ Distance = 140 m

∴ The correct answer is 140 m.

Additional Information

Unit Conversion Trick

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

Average Speed

Average Speed = Total Distance ÷ Total Time. In cases of uniform speed, it is equal to the constant speed.

Proportionality Rules

If Speed is constant, Distance ∝ Time. If Time is constant, Distance ∝ Speed. If Distance is constant, Speed ∝ 1/Time.

Shortcut Trick

Relative speed (opposite direction) = 55 + 80 = 135 km/hr

Time = 1 hour 36 minutes = 1 + 36/60 = 1.6 hours

Distance between X and Y = Relative Speed × Time

Distance = 135 × 1.6 = 216 km

∴ The correct answer is 216 km.

Alternate Method

Given:

Speed of Train A = 55 km/hr

Speed of Train B = 80 km/hr

Time taken to meet = 1 hour 36 minutes

Formula Used:

Distance = Relative Speed × Time

Relative Speed (Opposite Direction) = S₁ + S₂

Calculations:

⇒ Convert time to hours: 1 hr 36 min = 1 + 36/60 hours

⇒ Time = 1 + 0.6 = 1.6 hours

⇒ Relative Speed = 55 + 80 = 135 km/hr

⇒ Total Distance = 135 × 1.6

⇒ Total Distance = 135 × (16/10)

⇒ Total Distance = 27 × 8 = 216 km

∴ The correct answer is 216 km.

Additional Information

Relative Speed in Same Direction

When two objects move in the same direction, relative speed is the difference of their individual speeds: S_rel = |S₁ − S₂|.

Relative Speed in Opposite Direction

When two objects move towards each other or away from each other in opposite directions, relative speed is the sum of their individual speeds: S_rel = S₁ + S₂.

Time, Speed, and Distance Relation

The fundamental relation is Distance = Speed × Time. Ensure all units are consistent (e.g., all in km and hours, or all in meters and seconds).

Unit Conversion Shortcut

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

Shortcut Trick

Rate of interest = 16% increase ÷ 2 years = 8% per annum.

Effective compound interest rate for 2 years = 8% + 8% + (8 × 8) ÷ 100 = 16.64%.

Compound Interest = 16.64% of ₹25,000

CI = 16.64 × 250 = ₹4160.

∴ The correct answer is ₹4160.

Alternate Method

Given:

Principal (P) = ₹25,000

Simple interest for 2 years increases the principal by 16%.

Formula Used:

Rate (R) = Total SI % ÷ Time

Tree Method for 2 years CI = (2 × SI for 1 year) + (Interest on 1st year's SI)

8% 8%

P = ₹25,000

Yr 1 SI = ₹2,000

Yr 2 SI = ₹2,000

8% on ₹2000

• ₹160

Total CI = ₹4,160

Calculations:

⇒ Total Simple Interest = 16%

⇒ Rate of Interest (R) = 16% ÷ 2 = 8% per annum.

⇒ First year SI = 8% of 25,000 = ₹2000

⇒ Second year SI = ₹2000

⇒ Interest on first year's SI = 8% of 2000 = ₹160

⇒ Total Compound Interest = 2000 + 2000 + 160 = ₹4160.

∴ The correct answer is ₹4160.

Additional Information

Effective CI Rate

For 2 years, the effective compound interest percentage can be calculated directly using the successive percentage formula: 2R + (R² ÷ 100).

Difference Between CI and SI

For 2 years, the difference between Compound Interest and Simple Interest on a principal P is purely the interest earned on the first year's interest, given by: Difference = P × (R ÷ 100)².

Rule of 72

A quick mental math shortcut to find the approximate time needed to double an investment at a fixed annual compound interest rate is: Time ≈ 72 ÷ R.

Shortcut Trick

Compound interest for any year is the interest on the amount at the end of the previous year.

For consecutive years, CI_n+1 = CI_n + (Rate% of CI_n).

CI for 3^rd year = CI for 2^nd year + 12% of CI for 2^nd year.

⇒ CI for 3^rd year = 250 + (12% of 250) = 250 + 30 = 280.

∴ The correct answer is ₹280.

Alternate Method

Given:

Compound Interest for 2^nd year = ₹250

Rate of interest (R) = 12% per annum

Formula Used:

CI for (n+1)^th year = CI for n^th year × (1 + R/100)

Calculations:

⇒ Interest for 3^rd year = Interest on 2^nd year Principal + Interest on 2^nd year Interest

⇒ Since CI for 2^nd year is ₹250, the interest for 3^rd year will include this ₹250 plus 12% interest on it.

⇒ CI for 3^rd year = 250 × (1 + 12/100)

⇒ CI for 3^rd year = 250 × (112/100)

⇒ CI for 3^rd year = 250 × 1.12

⇒ CI for 3^rd year = 280

∴ The correct answer is ₹280.

Additional Information

Consecutive Years CI Property

In compound interest, the difference between the CI of two consecutive years (say n and n+1) is equal to the simple interest on the CI of the n^th year. Formula: CI_n+1 − CI_n = (CI_n × R × 1) / 100.

Effective Rate of Interest

For two years at rate R%, the effective CI rate is R + R + (R × R)/100. This helps in comparing SI and CI directly.

CI vs SI for 1st Year

For the first year, if the interest is compounded annually, the Simple Interest and Compound Interest are always equal on the same principal and rate.

Shortcut Trick

Yearly rate = 8% for 1 year.

Half-yearly rate = 4% per half-year. For 1 year (2 periods), effective rate = 4 + 4 + (4 × 4)/100 = 8.16%.

Difference in interest rates = 8.16% − 8% = 0.16%.

Required Difference = 0.16% of ₹12,500 = (0.16 ÷ 100) × 12,500 = 125 × 0.16 = ₹20.

∴ The correct answer is ₹20.

Alternate Method

Given: Principal (P) = ₹12,500, Rate (R) = 8% p.a., Time (T) = 1 year.

Formula Used: CI = P[(1 + r/100)ⁿ − 1]

Case 1: Compounded Yearly

⇒ Time (n) = 1 year, Rate (r) = 8%

⇒ CI_yearly = 12,500 × (8 ÷ 100) = ₹1,000

Case 2: Compounded Half-yearly

⇒ Time (n) = 2 half-years, Rate (r) = 8 ÷ 2 = 4%

⇒ Amount = 12,500 × (1 + 4 ÷ 100)²

⇒ Amount = 12,500 × (1.04)² = 12,500 × 1.0816 = ₹13,520

⇒ CI_half-yearly = 13,520 − 12,500 = ₹1,020

Calculating Difference:

⇒ Difference = 1,020 − 1,000 = ₹20

∴ The correct answer is ₹20.

Additional Information

Effective Rate of Interest

When interest is compounded k times a year, the effective annual rate is (1 + r/k)^k − 1.

Difference between CI and SI for 2 periods

The difference for two compounding periods is given by P × (r/100)². In this problem, half-yearly compounding for 1 year acts as 2 periods.

Compounding Frequency Rule

For half-yearly compounding, divide the annual rate by 2 and multiply the number of years by 2. For quarterly, divide rate by 4 and multiply years by 4.

Shortcut Trick

Rate 8% = 2/25. The ratio of Principal to Amount for 1 year is 25 : 27.

For 2 years, the ratio becomes (25)² : (27)² = 625 : 729.

729 units = ₹72,900 → 1 unit = ₹100.

Principal (625 units) = 625 × 100 = ₹62500.

∴ The correct answer is ₹62500.

Alternate Method

Given:

Amount (A) = ₹72,900

Rate (R) = 8% p.a.

Time (n) = 2 years

Formula Used:

A = P(1 + R/100)ⁿ

Calculations:

⇒ 72900 = P(1 + 8/100)²

⇒ 72900 = P(1 + 2/25)²

⇒ 72900 = P(27/25)²

⇒ 72900 = P × (729 / 625)

⇒ P = (72900 × 625) / 729

⇒ P = 100 × 625

⇒ P = 62500

∴ The correct answer is ₹62500.

Additional Information

Effective Rate for 2 Years

The effective compound interest rate for 2 years at R% is given by (2R + R²/100)%. For 8%, it is 16 + 0.64 = 16.64%.

Difference between CI and SI

For 2 years, the difference (D) between CI and SI is P(R/100)².

Compounding Periods

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

Shortcut Trick

Given Q = −R ⇒ Q + R = 0°. From P + Q + R = 60°, we get P = 60°.

Factor out (&cos;P − &sin;P) to get: (&cos;P − &sin;P) × (&cos;Q &cos;R − &sin;Q &sin;R).

This is the compound angle form: (&cos;P − &sin;P) × &cos;(Q + R).

Substitute values: (&cos;60° − &sin;60°) × &cos;0° = (1/2 − √3/2) × 1.

∴ The correct answer is (1 − √3) / 2.

Alternate Method

Given: P + Q + R = 60° and Q = −R.

Formula Used: &cos;(A + B) = &cos;A &cos;B − &sin;A &sin;B

⇒ Q + R = 0°

⇒ P + (Q + R) = 60° ⇒ P = 60°

⇒ Expression = &cos;Q &cos;R (&cos;P − &sin;P) + &sin;Q &sin;R (&sin;P − &cos;P)

⇒ Expression = &cos;Q &cos;R (&cos;P − &sin;P) − &sin;Q &sin;R (&cos;P − &sin;P)

⇒ Expression = (&cos;P − &sin;P) (&cos;Q &cos;R − &sin;Q &sin;R)

⇒ Expression = (&cos;P − &sin;P) &cos;(Q + R)

⇒ Expression = (&cos;60° − &sin;60°) &cos;(0°)

⇒ Expression = (1/2 − √3/2) × 1

⇒ Expression = (1 − √3) / 2

∴ The correct answer is (1 − √3) / 2.

Additional Information

Compound Angle Identities

The cosine of a sum is given by &cos;(A + B) = &cos;A &cos;B − &sin;A &sin;B, and the cosine of a difference is &cos;(A − B) = &cos;A &cos;B + &sin;A &sin;B.

Special Values

Standard trigonometric values: &sin;60° = √3/2, &cos;60° = 1/2, &sin;0° = 0, and &cos;0° = 1.

Negative Angle Identities

Cosine is an even function: &cos;(−θ) = &cos;θ. Sine is an odd function: &sin;(−θ) = −&sin;θ.

Shortcut Trick

Substitute θ = 45° directly into the given expression.

The term (90 − θ) becomes (90 − 45) = 45°.

Numerator = 1 − tan²(45°) = 1 − (1)² = 0.

Since the numerator is 0 and the denominator cos²(45°) is non-zero, the fraction value is 0.

Expression = 0 − 1 = −1.

∴ The correct answer is −1.

Alternate Method

Given: θ = 45°

Formula Used: tan(90 − θ) = cot θ and cos(90 − θ) = sin θ

⇒ The expression is: [1 − tan²(90 − 45°)] ÷ cos²(90 − 45°) − 1

⇒ [1 − tan²(45°)] ÷ cos²(45°) − 1

⇒ We know that tan 45° = 1 and cos 45° = 1/√2

⇒ [1 − (1)²] ÷ (1/√2)² − 1

⇒ [1 − 1] ÷ (1/2) − 1

⇒ 0 ÷ 0.5 − 1

⇒ 0 − 1 = −1

∴ The correct answer is −1.

Additional Information

Complementary Angle Identities

For any angle θ, sin(90 − θ) = cos θ, cos(90 − θ) = sin θ, and tan(90 − θ) = cot θ.

Trigonometric Values of 45°

At 45°, sin and cos are equal (1/√2), and tan and cot are equal (1).

Pythagorean Identities

Standard identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Shortcut Trick

Substitute x = 45°, so (90 − x) = 45°.

Values at 45°: cot 45° = 1 and cosec 45° = √2.

Expression = [1 + 2(1)² − 2(√2)(1)] / [√2 − 1] = [3 − 2√2] / [√2 − 1].

Simplifying: [(√2 − 1)²] / [√2 − 1] = √2 − 1.

Option (4): sec 45° − tan 45° = √2 − 1, which matches the result.

∴ The correct answer is secx − tanx.

Alternate Method

Given: [1 + 2cot²(90 − x) − 2cosec(90 − x) × cot(90 − x)] / [cosec(90 − x) − cot(90 − x)]

Formula Used: cot(90 − x) = tanx | cosec(90 − x) = secx | 1 + tan²x = sec²x

⇒ Replacing (90 − x) ratios with x ratios:

⇒ Numerator = 1 + 2tan²x − 2secx × tanx

⇒ Numerator = (1 + tan²x) + tan²x − 2secx × tanx

⇒ Numerator = sec²x + tan²x − 2secx × tanx

⇒ Numerator = (secx − tanx)²

⇒ Expression = (secx − tanx)² / (secx − tanx)

⇒ Expression = secx − tanx

∴ The correct answer is secx − tanx.

Additional Information

Complementary Angle Identities

sin(90 − x) = cosx, cos(90 − x) = sinx, tan(90 − x) = cotx, and sec(90 − x) = cosecx.

Pythagorean Identities

sin²x + cos²x = 1, 1 + tan²x = sec²x, and 1 + cot²x = cosec²x.

Reciprocal Relations

secx = 1/cosx, cosecx = 1/sinx, and tanx = sinx/cosx = 1/cotx.

Shortcut Trick

Use value substitution method: Let θ = 45°.

Substitute in the expression: sin(180° − 45°) × sin(90° − 45°) − [cot(90° − 45°) / (1 + tan²45°)]

⇒ sin 135° × sin 45° − [cot 45° / (1 + 1²)]

⇒ (1/√2) × (1/√2) − [1 / 2] = 1/2 − 1/2 = 0.

∴ The correct answer is 0.

Alternate Method

Given: sin(180 − θ) × sin (90 − θ) − [cot(90 − θ) / (1 + tan²θ)]

Using the identities shown in the diagram:

⇒ sin(180 − θ) = sin θ

⇒ sin(90 − θ) = cos θ

⇒ cot(90 − θ) = tan θ

⇒ 1 + tan²θ = sec²θ

Substitute these into the main expression:

⇒ (sin θ × cos θ) − [tan θ / sec²θ]

⇒ sin θ cos θ − [(sin θ / cos θ) / (1 / cos²θ)]

⇒ sin θ cos θ − [(sin θ / cos θ) × cos²θ]

⇒ sin θ cos θ − (sin θ × cos θ)

⇒ 0

∴ The correct answer is 0.

Additional Information

Complementary Angle Rules

For any angle θ, sin(90 − θ) = cosθ, tan(90 − θ) = cotθ, and sec(90 − θ) = cosecθ.

Supplementary Angle Rules

For angles in the second quadrant, sin(180 − θ) = sinθ, cos(180 − θ) = −cosθ, and tan(180 − θ) = −tanθ.

Pythagorean Identities

Standard identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Shortcut Trick

For an elevation of 45°, the horizontal distance (Base) is equal to the height of the object (House).

⇒ Distance of point P from the house = 25 m.

For an elevation of 60°, the Total height (House + Pole) = Base × √3.

⇒ Total height = 25 × √3 = 25√3 m.

Height of pole = Total height − House height = 25√3 − 25 = 25(√3 − 1) m.

∴ The correct answer is 25(√3 − 1) m.

Alternate Method

Given:

Height of the house (BC) = 25 m.

Angle of elevation of the top of house (B) from P = 45°.

Angle of elevation of the top of pole (A) from P = 60°.

In right-angled triangle PBC:

⇒ tan 45° = BC / PC

⇒ 1 = 25 / PC

⇒ PC = 25 m

In right-angled triangle PAC:

⇒ tan 60° = AC / PC

⇒ √3 = (AB + BC) / 25

⇒ 25√3 = h + 25

⇒ h = 25√3 − 25

⇒ h = 25(√3 − 1) m

∴ The correct answer is 25(√3 − 1) m.

Additional Information

Trigonometric Ratios

In a right triangle, tan θ = Perpendicular / Base. This ratio is primarily used to relate the height of an object to its distance from a point.

Standard Values

tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3. Memorizing these is essential for quick calculations in heights and distances.

Height Difference Formula

If two objects are on top of each other and observed from distance 'd', the height of the top object 'h' is given by h = d × (tan θ₂ − tan θ₁), where θ₂ is the larger angle.

Shortcut Trick

Length of the ladder = Base × sec(15°)

Since sec(15°) = 1 / cos(15°) and cos(15°) = (√6 + √2) / 4

sec(15°) = 4 / (√6 + √2) = √6 − √2 (by rationalizing)

Length = 5 × (√6 − √2) = 5√6 − 5√2 m

∴ The correct answer is 5√6 − 5√2 m.

Alternate Method

Given:

Distance of the foot of the ladder from the wall (Base) = 5 m

Angle of elevation (θ) = 15°

Formula Used:

cos(θ) = Base / Hypotenuse (Length of ladder)

Calculations:

⇒ cos(15°) = 5 / L

⇒ L = 5 / cos(15°)

⇒ cos(15°) = cos(45° − 30°) = cos(45°)cos(30°) + sin(45°)sin(30°)

⇒ cos(15°) = (1/√2 × √3/2) + (1/√2 × 1/2) = (√3 + 1) / 2√2

⇒ L = 5 / [(√3 + 1) / 2√2] = 10√2 / (√3 + 1)

⇒ Rationalizing: L = [10√2 × (√3 − 1)] / [(√3 + 1) × (√3 − 1)]

⇒ L = (10√6 − 10√2) / (3 − 1) = (10√6 − 10√2) / 2

⇒ L = 5√6 − 5√2 m

∴ The correct answer is 5√6 − 5√2 m.

Additional Information

Trigonometric Ratios for 15° and 75°

sin 15° = cos 75° = (√6 − √2)/4 and cos 15° = sin 75° = (√6 + √2)/4.

Tangent Ratio for 15°

tan 15° = 2 − √3 and cot 15° = 2 + √3.

Angle of Elevation and Depression

The angle of elevation from an observer's eye to an object is always equal to the angle of depression from the object to the observer.

Shortcut Trick

Horizontal distance (d) = height × (cot θ_smaller − cot θ_larger)

d = 1.8 × (cot 30° − cot 60°) = 1.8 × (√3 − 1/√3) = 1.8 × (2/√3) = 3.6/√3 = 1.2√3 km

Time = 20 seconds = 20 ÷ 3600 hours = 1/180 hour

Speed = Distance ÷ Time = 1.2√3 × 180 = 216√3 km/hr

∴ The correct answer is 216√3 km/hr.

Alternate Method

Given:

Height of aeroplane (h) = 1.8 km

Initial angle of elevation (θ₁) = 60°

Final angle of elevation (θ₂) = 30°

Time taken (t) = 20 seconds

Calculations:

Let X be the point on the ground. Let P1 and P2 be the positions of the plane.

In ΔXP1M (where M is projection of P1 on ground):

⇒ tan 60° = P1M / XM

⇒ √3 = 1.8 / XM

⇒ XM = 1.8 / √3 km

In ΔXP2N (where N is projection of P2 on ground):

⇒ tan 30° = P2N / XN

⇒ 1 / √3 = 1.8 / XN

⇒ XN = 1.8√3 km

Distance covered by plane (d) = XN − XM

⇒ d = 1.8√3 − 1.8/√3

⇒ d = (1.8 × 3 − 1.8) / √3 = 3.6 / √3 = 1.2√3 km

Speed = Distance / Time

⇒ Speed = (1.2√3) / (20/3600) km/hr

⇒ Speed = 1.2√3 × 180

⇒ Speed = 216√3 km/hr

∴ The correct answer is 216√3 km/hr.

Additional Information

Trigonometric Identities

tan θ = Perpendicular / Base and cot θ = Base / Perpendicular. Common values: tan 30° = 1/√3, tan 60° = √3.

Speed Conversion

To convert speed from m/s to km/hr, multiply by 18/5. To convert seconds to hours, divide by 3600.

Angle of Elevation

It is the angle between the horizontal line from the observer and the line of sight to an object above the horizontal line.

Rice produced by state X = 48% and that of by state H = 68%

⇒ Quantity of rice produced by state X = 0.48 × 625000 = 300000

⇒ R1 type of rice produced by state X = 300000 × 0.6 = 180000

Now

⇒ Quantity of rice produced by state H = 0.68 × 625000 = 425000

⇒ R1 type of rice produced by state H = 425000 × 0.6 = 255000

⇒ Remaining quantity of rice = 425000 - 255000 = 170000

⇒ R2 type of rice produced by state H = 6/17 × 170000 = 60000

∴ Required difference = 180000 - 60000 = 120000

Total production of maize by state X and T = 625000 × (0.52 + 0.74) = 787500

Now

⇒ Quantity of rice produced by state S = 625000 × 0.48 = 300000

⇒ (R2 and R3) type of rice produced by state S = 300000 × 0.45 = 135000

⇒ R2 type of rice produced in state S = 4/9 × 135000 = 60000

And

⇒ Quantity of rice produced by state R = 625000 × 0.38 = 237500

⇒ (R2 and R3) type of rice produced by state R = 237500 × 0.4 = 95000

⇒ R2 type of rice produced in state R = 9/19 × 95000 = 45000

⇒ Total production of R2 type of Rice by state S and R = 60000 + 45000 = 105000

∴ Required sum = 787500 + 105000 = 892500

Quantity of rice produced by state X = 0.48 × 625000 = 300000

⇒ (R2 and R3) type of rice produced by state X = 300000 × 0.40 = 120000

⇒ R3 type rice produced in state X = 5/8 × 120000 = 75000

And

⇒ Quantity of rice produced by state S = 625000 × 0.48 = 300000

⇒ R1 type rice produced in state S = 300000 × 0.55 = 165000

∴ Required percentage = (75000/165000) × 100 = 45.45%

A = Average of the R3 type of rice produced by state H, R, S and X together.

B = Difference between the R2 type of rice produced by state T and R1 type of rice produced by state R.

⇒ R3 type of rice produced by state H = 11/17 × 0.4 × 0.68 × 625000 = 110000

⇒ R3 type of rice produced by state R = 10/19 × 0.4 × 0.38 × 625000 = 50000

⇒ R3 type of rice produced by state S = 5/9 × 0.45 × 0.48 × 625000 = 75000

⇒ R3 type of rice produced by state X = 5/8 × 0.4 × 0.48 × 625000 = 75000

⇒ A = Average of the R3 type of rice produced by state H, R, S and X together

⇒ A = (110000 + 50000 + 75000 + 75000) /4 = 77500

⇒ R2 type of rice produced by state T = 3/13 × 0.2 × 0.26 × 625000 = 7500

⇒ R1 type of rice produced by state R = 0.6 × 0.38 × 625000 = 142500

⇒ B = Difference between the R2 type of rice produced by state T and R1 type of rice produced by state R.

⇒ B = 142500 - 7500 = 135000

∴ B - A = 135000 - 77500 = 57500

F = Total production of R2 type of rice by all the states.

⇒ F = (6/17 × 0.4 × 0.68 + 9/19 × 0.4 × 0.38 + 3/8 × 0.4 × 0.48 + 4/9 × 0.45 × 0.48 + 3/13 × 0.2 × 0.26) × 625000

⇒ (0.096 + 0.072 + 0.072 + 0.096 + 0.012) × 625000

⇒ 0.348 × 625000

⇒ 217500

K = Average of the total production of R1 type of rice by all the states.

⇒ Total production of R1 type of rice by all the states = (0.6 × 0.68 + 0.6 × 0.38 + 0.6 × 0.48 + 0.55 × 0.48 + 0.8 × 0.26) × 625000

⇒ (0.408 + 0.228 + 0.288 + 0.264 + 0.208) × 625000

⇒ (1.396) × 625000

⇒ 872500

⇒ K = 872500/5 = 174500

∴ K/F = 174500/217500 = 0.802

Shortcut Trick

Acid concentrations: S₁ = 1/5, S₂ = 3/20, Resultant = 19/110.

By Alligation, Ratio of Volumes (V₁ : V₂) = (19/110 − 3/20) : (1/5 − 19/110).

V₁ : V₂ = (38 − 33)/220 : (44 − 38)/220 = 5 : 6.

Given V₁ = 100x and V₂ = 200y ⇒ 100x / 200y = 5 / 6.

x / 2y = 5 / 6 ⇒ x / y = 10 / 6 = 5 / 3.

∴ The correct answer is 5 ∶ 3.

Alternate Method

Given:

Beaker 1: Vol = 100 ml, Acid ∶ Water = 1 ∶ 4

Beaker 2: Vol = 200 ml, Acid ∶ Water = 3 ∶ 17

Final mixture: Acid ∶ Water = 19 ∶ 91

Formula Used:

Total Acid / Total Water = Final Acid Ratio / Final Water Ratio

Calculations:

⇒ Total Acid in x beakers = x × 100 × (1/5) = 20x ml

⇒ Total Water in x beakers = x × 100 × (4/5) = 80x ml

⇒ Total Acid in y beakers = y × 200 × (3/20) = 30y ml

⇒ Total Water in y beakers = y × 200 × (17/20) = 170y ml

⇒ Resulting Ratio: (20x + 30y) / (80x + 170y) = 19 / 91

⇒ 91 × (20x + 30y) = 19 × (80x + 170y)

⇒ 1820x + 2730y = 1520x + 3230y

⇒ 1820x − 1520x = 3230y − 2730y

⇒ 300x = 500y

⇒ x / y = 500 / 300 = 5 / 3

∴ The correct answer is 5 ∶ 3.

Additional Information

Rule of Alligation

It states that (Quantity of Cheaper / Quantity of Dearer) = (Price of Dearer − Mean Price) / (Mean Price − Price of Cheaper).

Weighted Average Formula

Resultant Concentration C = (n₁c₁ + n₂c₂) / (n₁ + n₂), where n is the quantity and c is the concentration.

Mixture Ratios

When two mixtures of components A and B are mixed, the final ratio is found by summing the individual amounts of A and B separately.

Shortcut Trick

Apply the Alligation rule to find the ratio directly.

Left side (20% solution) difference = 40 − 28 = 12.

Right side (40% solution) difference = 28 − 20 = 8.

Ratio of 20% to 40% solutions = 12 ∶ 8 = 3 ∶ 2.

∴ The correct answer is 3 ∶ 2.

Alternate Method

Given:

Concentration of first solution (c) = 20%

Concentration of second solution (d) = 40%

Mean concentration required (m) = 28%

Formula Used:

Quantity of Cheaper / Quantity of Dearer = (d − m) / (m − c)

Calculations:

⇒ Ratio = (40 − 28) ∶ (28 − 20)

⇒ Ratio = 12 ∶ 8

⇒ Dividing both sides by their HCF (4):

⇒ Ratio = 3 ∶ 2

∴ The correct answer is 3 ∶ 2.

Additional Information

Rule of Alligation

It is a modified form of the weighted average used to find the ratio in which two ingredients at different prices/concentrations are mixed to produce a mixture of a desired price/concentration.

Weighted Average Formula

The mean concentration m is given by m = (n₁c₁ + n₂c₂) ÷ (n₁ + n₂), where n represents the quantities and c represents the concentrations.

Application in Profit and Loss

The same principle can be applied to find the ratio of quantities when two parts are sold at different profit or loss percentages to gain a specific overall profit/loss.

Shortcut Trick

Let investment of A and B be x lakhs. Time for A = 12 months, B = 8 months, C = 8 months.

Profit Ratio A : (A+B+C) → 24,000 : 62,400 = 5 : 13.

Investment Ratio A : (A+B+C) → 12x : (12x + 8x + 3.5 × 8) = 12x : (20x + 28).

By equating: 12x / (20x + 28) = 5 / 13 ⇒ 156x = 100x + 140 ⇒ 56x = 140 ⇒ x = 2.5 lakhs.

C's share = (28 / 78) × 62400 = 28 × 800 = 22400.

∴ The correct answer is ₹22400.

Alternate Method

Given:

Investment of A = Investment of B = x

Investment of C = ₹3.5 lakhs

Time period: A = 12 months, B = 8 months (leaves 4 months after C joins at month 4), C = 8 months (12 − 4)

Total Profit = ₹62,400 | A's Profit = ₹24,000

Formula Used: Profit Share Ratio = (Investment₁ × Time₁) : (Investment₂ × Time₂)

Calculations:

⇒ Ratio of Profit = (x × 12) : (x × 8) : (3.5 × 8)

⇒ Ratio A : B : C = 12x : 8x : 28

⇒ A's Share / Total Profit = 12x / (12x + 8x + 28)

⇒ 24000 / 62400 = 12x / (20x + 28)

⇒ 5 / 13 = 12x / (20x + 28)

⇒ 5 × (20x + 28) = 13 × 12x

⇒ 100x + 140 = 156x

⇒ 56x = 140 ⇒ x = 2.5 lakhs

⇒ C's Share = [28 / (12×2.5 + 8×2.5 + 28)] × 62400

⇒ C's Share = [28 / (30 + 20 + 28)] × 62400

⇒ C's Share = [28 / 78] × 62400

⇒ C's Share = 28 × 800

⇒ C's Share = 22400

∴ The correct answer is ₹22400.

Additional Information

Partnership Profit Ratio

Profit is always distributed in the ratio of the product of Capital and Time (P = C × T). If time is constant, Profit ∝ Capital.

Working vs. Sleeping Partner

A working partner manages the business and may receive a salary or extra commission before the profit is distributed according to the investment ratio.

Change in Investment

If investments change mid-way, calculate the effective investment as (C₁ × T₁) + (C₂ × T₂) for each partner.

Shortcut Trick

Total profit = ₹60,000.

Remaining profit for distribution = 60% of 60,000 = ₹36,000.

Ratio of A : B = 3 : 7 → Total parts = 10.

A's share = (3/10) × 36,000 = 3 × 3,600 = 10,800.

∴ The correct answer is ₹10800.

Alternate Method

Given:

Ratio of investment (A : B) = 3 : 7

Total annual profit = ₹60,000

Reinvestment percentage = 40%

Formula:

Share of a partner = (Individual Ratio ÷ Sum of Ratios) × Total Distributable Profit

Calculations:

⇒ Total Profit = ₹60,000

⇒ Reinvested Amount = 40% of 60,000 = (40/100) × 60,000 = ₹24,000

⇒ Distributable Profit = Total Profit − Reinvested Amount

⇒ Distributable Profit = 60,000 − 24,000 = ₹36,000

⇒ Total Ratio Parts = 3 + 7 = 10 units

⇒ Value of 10 units = ₹36,000

⇒ Value of 1 unit = 36,000 ÷ 10 = ₹3,600

⇒ A's Share = 3 units = 3 × 3,600 = ₹10,800

∴ The correct answer is ₹10800.

Additional Information

Partnership Profit Rule

Profit of a partner is directly proportional to the product of their capital and the time for which it is invested: Profit Ratio = (C₁ × T₁) : (C₂ × T₂).

Effective Distributable Profit

If a certain percentage (x%) is reinvested or donated, the distributable profit is (100 − x)% of the total profit.

Working vs Sleeping Partner

A working partner may receive an additional salary or percentage of profit for managing the business before the remaining profit is shared in the investment ratio.

Shortcut Trick

Total work = LCM(72, 90) = 360 units.

Efficiency: A = 360/72 = 5, B = 360/90 = 4; Total efficiency = 9 units/day.

Work done in 10 days = 9 × 10 = 90 units.

Remaining work = 360 − 90 = 270 units.

Fraction of work left = 270 ÷ 360 = 3/4.

∴ The correct answer is 3/4.

Alternate Method

Given: A completes work in 72 days and B in 90 days.

Formula: Efficiency = Total Work ÷ Time

⇒ Combined Efficiency of A and B = 5 + 4 = 9 units/day.

⇒ Work completed by A and B together in 10 days = 10 × 9 = 90 units.

⇒ Remaining work = Total Work − Work Completed

⇒ Remaining work = 360 − 90 = 270 units.

⇒ Fraction of work left = Remaining Work ÷ Total Work

⇒ Fraction = 270 ÷ 360 = 3/4.

∴ The correct answer is 3/4.

Additional Information

One Day's Work Principle

If a person completes a task in 'n' days, then the work done in 1 day is 1/n fraction of the total work.

Work and Wages Relationship

Wages are always distributed in the ratio of the work done by each person, which is proportional to their efficiency if they work for the same duration.

Negative Work in Pipes and Cisterns

In similar problems involving inlet and outlet pipes, the outlet pipe's efficiency is taken as negative work since it empties the tank.

Shortcut Trick

Efficiency ratio of A : B = 3 : 1.

Efficiency of (A + B) = Efficiency (A + B + C) − Efficiency C = (1/16) − (1/48) = 1/24.

Since A is thrice as efficient as B, B's share of efficiency in (A + B) is 1/(3 + 1) = 1/4.

B's individual efficiency = (1/24) × (1/4) = 1/96.

Time taken by B alone = 96 days.

∴ The correct answer is 96 days.

Alternate Method

Given: C takes 48 days, A + B + C take 16 days, and A is thrice as efficient as B.

Let Total Work be the LCM of 48 and 16.

⇒ Efficiency of (A + B) = Efficiency(A + B + C) − Efficiency(C) = 3 − 1 = 2 units/day.

⇒ Given A's efficiency = 3 × B's efficiency.

⇒ Let B's efficiency = x, then A's efficiency = 3x.

⇒ 3x + x = 2 ⇒ 4x = 2 ⇒ x = 2/4 = 0.5 units/day.

⇒ Efficiency of B = 0.5 units/day.

⇒ Time taken by B alone = Total Work ÷ Efficiency of B = 48 ÷ 0.5 = 96 days.

∴ The correct answer is 96 days.

Additional Information

Relation between Time and Efficiency

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

Work and Wages

Wages are always distributed in the ratio of the work done by the individuals or the ratio of their efficiencies.

Negative Work

In problems involving pipes and cisterns, an outlet pipe performs negative work, effectively reducing the net efficiency.

Shortcut Trick

Total Work = LCM(60, 12) = 60 units.

Efficiency of A = 60 ÷ 60 = 1; Efficiency of (A+B+C) = 60 ÷ 12 = 5.

Efficiency of (B+C) = 5 − 1 = 4 units/day.

Since C is 5 times as productive as B, efficiency ratio B : C = 1 : 5.

Efficiency of B = 4 × (1 ÷ 6) = 2/3 units/day.

Time taken by B = 60 ÷ (2/3) = 90 days.

∴ The correct answer is 90 days.

Alternate Method

Given:

Time taken by A = 60 days

Time taken by (A + B + C) = 12 days

Efficiency relationship: Efficiency of C = 5 × Efficiency of B

Calculations:

⇒ Efficiency of (B + C) = Efficiency(A + B + C) − Efficiency(A)

⇒ Efficiency of (B + C) = 5 − 1 = 4 units/day

⇒ Let Efficiency of B be x. Then Efficiency of C = 5x.

⇒ x + 5x = 4

⇒ 6x = 4

⇒ x = 4 ÷ 6 = 2/3

⇒ Efficiency of B = 2/3 units/day

⇒ Time taken by B alone = Total Work ÷ Efficiency of B

⇒ Time taken by B alone = 60 ÷ (2/3)

⇒ Time taken by B alone = 60 × (3 ÷ 2) = 90 days

∴ The correct answer is 90 days.

Additional Information

Work and Efficiency Relation

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

Combined Efficiency

When individuals work together, their individual efficiencies are added to find the total rate of work: E_total = E₁ + E₂ + ... + E_n.

Relative Productivity

If A is 'n' times as productive as B, then Efficiency of A = n × Efficiency of B, and Time taken by A = (1/n) × Time taken by B.

Shortcut Trick

Since the time duration for both individuals is the same (9 days), we can directly sum their individual percentage contributions.

Work done by A in 9 days = 50%

Work done by B in 9 days = 25%

Total work done together in 9 days = 50% + 25% = 75%

∴ The correct answer is 75%.

Alternate Method

Given:

A completes 50% work in 9 days → 100% work in 18 days.

B completes 25% work in 9 days → 100% work in 36 days.

Formula Used:

Total Work = LCM of individual days; Efficiency = Total Work ÷ Days

Calculations:

⇒ Total Efficiency of (A + B) = 2 + 1 = 3 units/day

⇒ Work completed by (A + B) in 9 days = Efficiency × Time

⇒ Work completed = 3 × 9 = 27 units

⇒ Percentage of work completed = (Work Done ÷ Total Work) × 100

⇒ Percentage = (27 ÷ 36) × 100 = (3 ÷ 4) × 100 = 75%

∴ The correct answer is 75%.

Additional Information

Work and Efficiency Relation

Efficiency is inversely proportional to the time taken to complete a job. If A is twice as efficient as B, A will take half the time B takes.

Total Work Formula

Total Work = Efficiency × Time. When multiple people work together, their individual efficiencies are added: Total Efficiency = E₁ + E₂ + ... + E_n.

Fractional Work Method

If a person can do a piece of work in x days, then the work done in 1 day = 1/x. For 'n' days, the work done is n/x.

Shortcut Trick

Net Successive Discount = x + y − (x × y / 100)%

Given: x = 60 and y = 60

Net Discount = 60 + 60 − (60 × 60 / 100)

Net Discount = 120 − 36 = 84%

∴ The correct answer is 84%.

Alternate Method

Given:

First discount rate (d₁) = 60%

Second discount rate (d₂) = 60%

Formula Used:

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

Calculations:

⇒ Let the Marked Price (MP) = 100 units

⇒ Price after 1st discount = 100 − (60% of 100) = 40 units

⇒ Price after 2nd discount = 40 − (60% of 40) = 40 − 24 = 16 units

⇒ Total Discount amount = MP − Final Selling Price = 100 − 16 = 84 units

⇒ Equivalent Single Discount % = (84 / 100) × 100 = 84%

∴ The correct answer is 84%.

Additional Information

Successive Discounts for Three Rates

For three successive discounts of a%, b%, and c%, the net discount is 100 − [100 × (100−a)/100 × (100−b)/100 × (100−c)/100].

Relationship between CP, MP and Discount

The ratio of Cost Price to Marked Price is given by CP/MP = (100 − Discount%)/(100 + Profit%).

Successive Percentage Change

For a successive increase of x% and y%, the net change is [x + y + (x × y)/100]%. For discounts, use negative values for x and y.

Shortcut Trick

Discount = Marked Price − Selling Price = 480 − 400 = 80

Discount Fraction = Discount ÷ MP = 80 ÷ 480 = 1/6

As per the percentage-fraction table, 1/6 = 16.67%

∴ The correct answer is 16.67%.

Alternate Method

Given: Marked Price (MP) = Rs. 480, Selling Price (SP) = Rs. 400

Formula Used: Discount % = [(MP − SP) ÷ MP] × 100

⇒ Discount = 480 − 400 = 80

⇒ Effective Discount % = (80 ÷ 480) × 100

⇒ Effective Discount % = (1 ÷ 6) × 100

⇒ Effective Discount % = 16.666...

⇒ Effective Discount % ≈ 16.67%

∴ The correct answer is 16.67%.

Additional Information

Marked Price (MP) vs Selling Price (SP)

Discount is the reduction offered on the Marked Price to arrive at the Selling Price. Formula: SP = MP − Discount.

Successive Discounts

When two discounts a% and b% are given, the net discount is calculated as [a + b − (a × b)/100]%.

Markup Percentage

The percentage by which the Cost Price (CP) is increased to set the MP. Markup % = [(MP − CP) ÷ CP] × 100.

Shortcut Trick

Price after first 5% discount = 25,000 × 0.95 = ₹23,750.

Final Selling Price = ₹20,900.

Ratio of Discounted Price to SP = 23,750 : 20,900 = 2,375 : 2,090 = 1 : 0.88.