SSC CGL Tier 2 Quant Previous Year Paper (Held On: 18 Feb 2018)

Shortcut Trick

Total odd digits available are {1, 3, 5, 7, 9}, which equals 5 options.

Each of the 3 positions in a 3-digit number has 5 choices as all digits must be odd.

Total numbers = 5³ = 5 × 5 × 5 = 125.

∴ The correct answer is 125 numbers.

Alternate Method

Given: A three-digit number where all digits are odd.

Formula Used: Fundamental Principle of Counting (Multiplication Rule).

⇒ The set of odd digits available is {1, 3, 5, 7, 9}. The number of elements in this set is 5.

⇒ Number of choices for the hundreds place (H) = 5 (can be 1, 3, 5, 7, or 9).

⇒ Number of choices for the tens place (T) = 5 (can be 1, 3, 5, 7, or 9).

⇒ Number of choices for the units place (U) = 5 (can be 1, 3, 5, 7, or 9).

⇒ Total three-digit numbers = 5 × 5 × 5 = 125.

∴ The correct answer is 125 numbers.

Additional Information

Fundamental Principle of Counting

If an event can occur in m ways and another independent event can occur in n ways, then the total number of ways both can occur is m × n.

Number of Total 3-Digit Numbers

The total number of 3-digit numbers is 9 × 10 × 10 = 900 (Hundreds place cannot be 0).

Even Digit Constraint

If all digits must be even {0, 2, 4, 6, 8}, the hundreds place has 4 choices (cannot be 0), while tens and units have 5, totaling 4 × 5 × 5 = 100.

Shortcut Trick

To maximize one number in a fixed sum of distinct positive integers, we must minimize all other numbers.

Smallest 9 distinct positive integers = 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Sum of these 9 integers = 45.

Greatest possible number = Total Sum − Sum of the smallest 9 numbers = 100 − 45 = 55.

∴ The correct answer is 55.

Alternate Method

Given:

Total numbers (n) = 10

Sum of 10 different positive integers = 100

Integers must be distinct and positive (starting from 1).

Formula Used:

Sum of first 'n' natural numbers = n × (n + 1) ÷ 2

Calculations:

⇒ Let the 10 different positive integers be x₁, x₂, x₃, ..., x₁₀.

⇒ To maximize x₁₀, we take x₁ to x₉ as the smallest possible distinct positive integers.

⇒ Smallest integers are 1, 2, 3, 4, 5, 6, 7, 8, 9.

⇒ Sum of these 9 integers = 9 × (9 + 1) ÷ 2

⇒ Sum = (9 × 10) ÷ 2 = 45

⇒ Let the greatest number be X.

⇒ Sum of all 10 numbers = 45 + X = 100

⇒ X = 100 − 45

⇒ X = 55

∴ The correct answer is 55.

Additional Information

Sum of First n Natural Numbers

The sum of the first n positive integers is calculated using the arithmetic progression formula: S = n(n + 1) ÷ 2.

Maximization Principle

In any set of numbers with a fixed sum, to maximize one element, all other elements must be assigned their minimum possible values based on given constraints (like being positive or distinct).

Distinct vs Non-distinct Integers

If the question did not specify "different" integers, the smallest positive integers could all be 1. Then the greatest number would be 100 − (9 × 1) = 91.

Given:

N = 0.369369...

M = 0.531531...

Formula used:

0.abcabc... = $\frac{abc}{999}$

Calculations:

⇒ N = $\frac{369}{999}$

⇒ N = $\frac{41}{111}$

⇒ M = $\frac{531}{999}$

⇒ M = $\frac{59}{111}$

⇒ $\frac{1}{N} + \frac{1}{M}$ = $\frac{111}{41} + \frac{111}{59}$

⇒ $\frac{1}{N} + \frac{1}{M}$ = 111 × $(\frac{1}{41} + \frac{1}{59})$

⇒ $\frac{1}{N} + \frac{1}{M}$ = 111 × $(\frac{59 + 41}{41 \times 59})$

⇒ $\frac{1}{N} + \frac{1}{M}$ = 111 × $\frac{100}{2419}$

⇒ $\frac{1}{N} + \frac{1}{M}$ = $\frac{11100}{2419}$

∴ The correct answer is option (1).

Given:

A = $\frac{0.216 + 0.008}{0.36 + 0.04 - 0.12}$

B = $\frac{0.729 - 0.027}{0.81 + 0.09 + 0.27}$

(A² + B²)² = ?

Formula used:

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

Calculation:

A = $\frac{0.6^3 + 0.2^3}{0.6^2 + 0.2^2 - (0.6 \times 0.2)}$

⇒ A = 0.6 + 0.2 = 0.8

B = $\frac{0.9^3 - 0.3^3}{0.9^2 + 0.3^2 + (0.9 \times 0.3)}$

⇒ B = 0.9 - 0.3 = 0.6

⇒ (A² + B²)² = (0.8² + 0.6²)²

⇒ (A² + B²)² = (0.64 + 0.36)²

⇒ (A² + B²)² = (1)² = 1

∴ The correct answer is option (2).

Given:

A = $\frac{1}{1 \times 2} + \frac{1}{1 \times 4} + \frac{1}{2 \times 3} + \frac{1}{4 \times 7} + \frac{1}{3 \times 4} + \frac{1}{7 \times 10} + \dots$ up to 20 terms

Formula used:

$\frac{1}{n(n+k)} = \frac{1}{k} \times (\frac{1}{n} - \frac{1}{n+k})$

Calculations:

A = ($\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots$ 10 terms) + ($\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \dots$ 10 terms)

⇒ A = ($\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{10 \times 11}$) + ($\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \dots + \frac{1}{28 \times 31}$)

⇒ A = $(1 - \frac{1}{11}) + \frac{1}{3} \times (1 - \frac{1}{31})$

⇒ A = $\frac{10}{11} + \frac{1}{3} \times \frac{30}{31}$

⇒ A = $\frac{10}{11} + \frac{10}{31}$

⇒ A = $\frac{310 + 110}{341}$

⇒ A = $\frac{420}{341}$

∴ The correct answer is option 4.

The correct answer is 8

Key Points

• To solve this problem, we need to perform the prime factorization of each individual number on the left side of the equation to determine the exponents p, q, r, and s.
• The prime factorization of each number is as follows:
  

56 = 8 × 7 = 2³ × 7¹
75 = 25 × 3 = 5² × 3¹
60 = 4 × 3 × 5 = 2² × 3¹ × 5¹
84 = 4 × 3 × 7 = 2² × 3¹ × 7¹
210 = 2 × 3 × 5 × 7 = 2¹ × 3¹ × 5¹ × 7¹

• According to the Product Rule of Exponents (a^m × aⁿ = a^m+n), we sum the exponents for each prime base across all factors:
  

For base 2: p = 3 + 2 + 2 + 1 = 8
For base 3: q = 1 + 1 + 1 + 1 = 4
For base 5: r = 2 + 1 + 1 = 4
For base 7: s = 1 + 1 + 1 = 3

• The equation is 2⁸ × 3⁴ × 5⁴ × 7³ = 2^p × 3^q × 5^r × 7^s. Comparing both sides, we get the values: p = 8, q = 4, r = 4, and s = 3.
• The final step is to calculate the value of the given expression [(p + q) / s] + r:
  

Value = [(8 + 4) ÷ 3] + 4
Value = [12 ÷ 3] + 4
Value = 4 + 4 = 8.

• The result is 8, which makes Option 2 the correct choice.

Additional Information

• Prime Factorization is a method used to find which prime numbers multiply together to make the original number. It is an essential tool in Number Theory and simplifying complex radical expressions.
• The Fundamental Theorem of Arithmetic ensures that every composite number has a unique set of prime factors, which allows us to compare exponents when equations are expressed in prime base form.
• When performing these calculations, ensure that every factor is reduced to its prime base (e.g., 4 must be written as 2², and 8 as 2³) before summing exponents.

Given:

A = $3\frac{1}{4} \times 4\frac{1}{4} \div 34 - \frac{47}{32} + \frac{47}{16}$

B = $2\frac{1}{2} + 5\frac{1}{2} \div 55 - \frac{11}{10}$

Formula used:

BODMAS rule

Calculation:

A = $\frac{13}{4} \times \frac{17}{4} \times \frac{1}{34} - \frac{47}{32} + \frac{94}{32}$

⇒ A = $\frac{13}{32} + \frac{47}{32}$

⇒ A = $\frac{60}{32} = \frac{15}{8}$

B = $\frac{5}{2} + \frac{11}{2 \times 55} - \frac{11}{10}$

⇒ B = $\frac{25}{10} + \frac{1}{10} - \frac{11}{10}$

⇒ B = $\frac{15}{10} = \frac{3}{2}$

A - B = $\frac{15}{8} - \frac{3}{2}$

⇒ A - B = $\frac{15 - 12}{8} = \frac{3}{8}$

∴ The correct answer is option (4).

Shortcut Trick

Identify the range of multiples of 13: 13 × 8 = 104 and 13 × 30 = 390.

Number of terms (n) = (Last Multiple − First Multiple) / 13 + 1 = (390 − 104) / 13 + 1 = 23.

Sum = Average × n = [(104 + 390) / 2] × 23 = 247 × 23.

Using 247 × 23 = 5681.

∴ The correct answer is 5681.

Alternate Method

Given: Numbers between 100 and 400 divisible by 13.

Formula Used: Sum of Arithmetic Progression S_n = (n/2) × [a + l]

⇒ First term (a) = 104 (Since 100/13 = 7.69, next multiple is 13 × 8)

⇒ Last term (l) = 390 (Since 400/13 = 30.76, previous multiple is 13 × 30)

⇒ Common difference (d) = 13

⇒ Number of terms (n): l = a + (n − 1)d

⇒ 390 = 104 + (n − 1) × 13

⇒ 286 = (n − 1) × 13

⇒ n − 1 = 286 ÷ 13 = 22

⇒ n = 23

⇒ Sum S₂₃ = (23 / 2) × (104 + 390)

⇒ S₂₃ = (23 / 2) × 494

⇒ S₂₃ = 23 × 247 = 5681

∴ The correct answer is 5681.

Additional Information

General Term of an AP

The n^th term of an arithmetic progression is given by a_n = a + (n − 1)d, where a is the first term and d is the common difference.

Sum of First n Natural Numbers

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

Divisibility Rule for 13

Multiply the last digit by 4 and add it to the rest of the number. If the result is divisible by 13, the original number is also divisible by 13.

Shortcut Trick

Factorize 1728 = 2⁶ × 3³ and LCM 5184 = 2⁶ × 3⁴.

For LCM, we take the maximum power of each prime factor.

Since the power of 3 in LCM (4) is greater than in 1728 (3), K must have 3⁴.

Since the power of 2 in LCM (6) is equal to 1728 (6), K can have 2ⁿ where 0 ≤ n ≤ 6.

Possible powers of 2 for K are {0, 1, 2, 3, 4, 5, 6}, which gives 7 possibilities.

∴ The correct answer is 7 values.

Alternate Method

Given:

First number = 1728

LCM(1728, K) = 5184

Formula:

If A = p^a × q^b and B = p^x × q^y, then LCM(A, B) = p^max(a,x) × q^max(b,y)

Calculations:

⇒ Prime factorization of 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2⁶ × 3³

⇒ Prime factorization of 5184 = 1728 × 3 = 2⁶ × 3⁴

⇒ Let K = 2^a × 3^b

⇒ LCM(2⁶ × 3³, 2^a × 3^b) = 2^{max(6, a)} × 3^{max(3, b)}

⇒ Equating with the given LCM: 2^{max(6, a)} × 3^{max(3, b)} = 2⁶ × 3⁴

⇒ For 3^{max(3, b)} = 3⁴, b must be exactly 4.

⇒ For 2^{max(6, a)} = 2⁶, a can be any integer such that 0 ≤ a ≤ 6.

⇒ Possible values for a are {0, 1, 2, 3, 4, 5, 6}.

⇒ Number of possible values for K = 7 × 1 = 7.

∴ The correct answer is 7 values.

Additional Information

LCM Prime Factorization Method

LCM of two or more numbers is the product of the highest powers of all prime factors present in any of the numbers.

HCF vs LCM Relationship

For any two numbers A and B, A × B = HCF(A, B) × LCM(A, B). This is valid only for two numbers.

HCF Calculation

HCF is the product of the lowest powers of common prime factors. If HCF(1728, K) was required, we would take 2^min(6,a) × 3^min(3,b).

Shortcut Trick

Identify the terms as multiples: (3 × 3³³) × (2 × 2³³) = 6^x

Combine powers: 3³⁴ × 2³⁴ = 6^x

Apply power rule: (3 × 2)³⁴ = 6^x

Equate bases: 6³⁴ = 6^x ⇒ x = 34

∴ The correct answer is 34.

Alternate Method

Given equation: (3³³ + 3³³ + 3³³) (2³³ + 2³³) = 6^x

Formula used: a^m × aⁿ = a^m+n and aⁿ × bⁿ = (ab)ⁿ

⇒ [3³³ × (1 + 1 + 1)] × [2³³ × (1 + 1)] = 6^x

⇒ (3³³ × 3¹) × (2³³ × 2¹) = 6^x

⇒ 3^{(33 + 1)} × 2^{(33 + 1)} = 6^x

⇒ 3³⁴ × 2³⁴ = 6^x

⇒ (3 × 2)³⁴ = 6^x

⇒ 6³⁴ = 6^x

⇒ x = 34

∴ The correct answer is 34.

Additional Information

Laws of Exponents: Product Rule

When multiplying terms with the same base, the exponents are added: a^m × aⁿ = a^m+n.

Power of a Product Rule

When different bases are raised to the same power, they can be multiplied under the same power: aⁿ × bⁿ = (ab)ⁿ.

Comparison of Indices

In an equation of the form a^x = a^y, if a ≠ 0, 1, or −1, then x = y must hold true.

Shortcut Trick LCM of denominators of powers (6, 4, 3) is 12. Raise each term to the power of 12 to eliminate fractions: (651/6)12 = 652 = 4225 (171/4)12 = 173 = 4913 (121/3)12 = 124 = 20736 Since 20736 > 4913 > 4225, Statement III is correct. ∴ The correct answer is Only III.

Alternate Method Given expressions: (65)1/6, (17)1/4, (12)1/3 Method: To compare surds of different orders, make the order of the surds same by taking the LCM of the denominators of their fractional powers. ⇒ Denominators of powers are 6, 4, and 3. ⇒ LCM(6, 4, 3) = 12 ⇒ (65)1/6 = (65)(1/6)×(2/2) = (65)2/12 = (652)1/12 = (4225)1/12 ⇒ (17)1/4 = (17)(1/4)×(3/3) = (17)3/12 = (173)1/12 = (4913)1/12 ⇒ (12)1/3 = (12)(1/3)×(4/4) = (12)4/12 = (124)1/12 = (20736)1/12 ⇒ Comparing the values inside the 12th root: 20736 > 4913 > 4225 ⇒ (20736)1/12 > (4913)1/12 > (4225)1/12 ⇒ (12)1/3 > (17)1/4 > (65)1/6 ∴ The correct answer is Only III.

Additional Information Comparing Surds of Different Orders To compare surds like n√a and m√b, convert them to the same order by using the LCM of n and m as the new power. Exponent Law for Powers The property (am)n = am×n is used to simplify fractional exponents when raising them to a common multiple. Monotonic Property For any positive numbers x and y, if x > y, then xn > yn for any positive power n. This allows us to compare bases directly after equalizing exponents.

Shortcut Trick

Since P = 7 + 4√3 and 7² − (4√3)² = 49 − 48 = 1, its reciprocal is simply its conjugate.

Given PQ = 1 ⇒ Q = 1/P = 7 − 4√3.

The expression 1/P² + 1/Q² is equivalent to Q² + P².

Sum (P + Q) = (7 + 4√3) + (7 − 4√3) = 14.

Using the identity P² + Q² = (P + Q)² − 2PQ = 14² − 2(1) = 196 − 2 = 194.

∴ The correct answer is 194.

Alternate Method

Given:

P = 7 + 4√3

PQ = 1

Formula Used:

(a + b)² = a² + b² + 2ab

Rationalization: 1/(a + b) = (a − b)/(a² − b²)

Calculations:

⇒ Q = 1/P = 1/(7 + 4√3)

⇒ Rationalizing Q: Q = (7 − 4√3) / (7² − (4√3)²)

⇒ Q = (7 − 4√3) / (49 − 48) = 7 − 4√3

⇒ Required value = 1/P² + 1/Q²

⇒ Since Q = 1/P and P = 1/Q, the expression becomes Q² + P²

⇒ (P + Q) = (7 + 4√3) + (7 − 4√3) = 14

⇒ P² + Q² = (P + Q)² − 2PQ

⇒ P² + Q² = 14² − 2(1)

⇒ P² + Q² = 196 − 2 = 194

∴ The correct answer is 194.

Additional Information

Conjugate Surds

If the product of two surds is a rational number, they are called conjugate surds. For P = a + √b, the conjugate is a − √b.

Reciprocal Property

If a² − b = 1, then the reciprocal of (a + √b) is exactly its conjugate (a − √b).

Algebraic Identity

For any two numbers x and y, the sum of their squares is given by x² + y² = (x + y)² − 2xy.

Shortcut Trick

Let k = (x/y) + (y/x). The expression is (k² − 2) + 4k + 6 = k² + 4k + 4 = (k + 2)².

Given x = √5 + 1 and y = √5 − 1, the sum x + y = 2√5 and product xy = 4.

Calculate k = (x² + y²)/xy = [(x + y)² − 2xy] / xy = (20 − 8) / 4 = 3.

Substitute k = 3 into (k + 2)² ⇒ (3 + 2)² = 5² = 25.

∴ The correct answer is 25.

Alternate Method

Given: x = √5 + 1 and y = √5 − 1

Method: Algebraic simplification using identities.

⇒ x + y = (√5 + 1) + (√5 − 1) = 2√5

⇒ xy = (√5 + 1)(√5 − 1) = (√5)² − 1² = 5 − 1 = 4

⇒ x² + y² = (x + y)² − 2xy = (2√5)² − 2(4) = 20 − 8 = 12

⇒ x/y + y/x = (x² + y²) / xy = 12 / 4 = 3

⇒ x²/y² + y²/x² = (x/y + y/x)² − 2 = 3² − 2 = 7

⇒ Expression = (x²/y² + y²/x²) + 4(x/y + y/x) + 6

⇒ Expression = 7 + 4(3) + 6

⇒ Expression = 7 + 12 + 6 = 25

∴ The correct answer is 25.

Additional Information

Algebraic Identity

The expression a² + 2ab + b² can be written as (a + b)². In this problem, we converted a quadratic form in terms of (x/y + y/x) into a perfect square.

Conjugate Surds

When x = √a + b and y = √a − b, their product xy = a − b² is always a rational number, which simplifies fractional expressions.

Sum of Squares

The relation x² + y² = (x + y)² − 2xy is a fundamental tool for solving problems involving roots and symmetric expressions.

Shortcut Trick

Combine the terms over a common denominator: (x² + y² + z² + 2xy + 2yz + 2zx) ÷ xyz.

This is the algebraic expansion of (x + y + z)² ÷ xyz.

Sum: x + y + z = (2 + √3) + (2 − √3) + 1 = 5.

Product: xyz = (2 + √3)(2 − √3) × 1 = (4 − 3) × 1 = 1.

Result = 5² ÷ 1 = 25.

∴ The correct answer is 25.

Alternate Method

Given: x = 2 + √3, y = 2 − √3, z = 1

Formula Used: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ Expression = (x/yz) + (y/xz) + (z/xy) + 2(1/x + 1/y + 1/z)

⇒ Expression = (x² + y² + z²)/xyz + 2(yz + xz + xy)/xyz

⇒ Expression = (x² + y² + z² + 2xy + 2yz + 2zx) / xyz

⇒ Expression = (x + y + z)² / xyz

⇒ x + y + z = (2 + √3) + (2 − √3) + 1 = 5

⇒ xyz = (2 + √3)(2 − √3)(1) = (4 − 3)(1) = 1

⇒ Value = (5)² / 1 = 25

∴ The correct answer is 25.

Additional Information

Conjugate Surds

If two surds differ only in the sign between terms, like (a + √b) and (a − √b), their product is always a rational number: a² − b.

Algebraic Identities

The square of a trinomial is (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. This is often used to simplify cyclic symmetric expressions.

Rationalization

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

Shortcut Trick

If coefficients are rational, irrational roots occur in conjugate pairs: Roots are 5 + 3√3 and 5 − 3√3.

Sum of roots = 10 and Product of roots = 25 − 27 = −2.

Quadratic equation: x² − 10x − 2 = 0 ⇒ a = 1, b = −10, c = −2.

Value = (1² + (−10)² + (−2)²) ÷ (1 − 10 − 2) = 105 ÷ (−11) = −105/11.

∴ The correct answer is -105/11.

Alternate Method

Given:

Quadratic equation: ax² + bx + c = 0

One root (α) = 5 + 3√3

Since a, b, c are rational, the other root (β) must be the conjugate: 5 − 3√3

Formula Used:

Sum of roots (α + β) = −b/a

Product of roots (α × β) = c/a

Calculations:

⇒ Sum of roots = (5 + 3√3) + (5 − 3√3) = 10

⇒ Product of roots = (5 + 3√3)(5 − 3√3) = 5² − (3√3)² = 25 − 27 = −2

⇒ The equation is x² − (Sum)x + (Product) = 0

⇒ x² − 10x − 2 = 0

⇒ Comparing with ax² + bx + c = 0, we get a = 1, b = −10, c = −2

⇒ Value = (a² + b² + c²) ÷ (a + b + c)

⇒ Value = (1² + (−10)² + (−2)²) ÷ (1 + (−10) + (−2))

⇒ Value = (1 + 100 + 4) ÷ (1 − 10 − 2)

⇒ Value = 105 ÷ (−11) = −105/11

∴ The correct answer is -105/11.

Additional Information

Conjugate Roots Theorem

If the coefficients of a quadratic equation are rational, irrational roots of the form p + √q always occur in pairs as p ± √q.

Sum and Product of Roots

For ax² + bx + c = 0, Sum = −b/a and Product = c/a. The equation can be written as x² − (Sum)x + Product = 0 when a = 1.

Nature of Roots

The discriminant D = b² − 4ac determines the nature. If D > 0 and not a perfect square, roots are irrational and distinct.

Shortcut Trick

Use the value substitution method by assuming a = 1, b = 1, and c = 1.

x = 1/1 + 1/1 = 2, y = 1/1 + 1/1 = 2, and z = 1/1 + 1/1 = 2.

Substitute these values into the expression: xyz − x² − y² − z².

(2 × 2 × 2) − 2² − 2² − 2² ⇒ 8 − 4 − 4 − 4 = −4.

∴ The correct answer is −4.

Alternate Method

Given: x = (a/b) + (b/a), y = (b/c) + (c/b), z = (c/a) + (a/c)

Formula Used: Value Substitution and Algebraic Expansion

Let a = 1, b = 1, c = 1

⇒ x = (1/1) + (1/1) = 2

⇒ y = (1/1) + (1/1) = 2

⇒ z = (1/1) + (1/1) = 2

Now, calculating the required expression:

⇒ xyz − x² − y² − z²

⇒ (2 × 2 × 2) − 2² − 2² − 2²

⇒ 8 − 4 − 4 − 4

⇒ 8 − 12 = −4

∴ The correct answer is −4.

Additional Information

Algebraic Substitution Method

Substitution is a powerful technique where variables are replaced with suitable constants (often 1 or 0) to verify identities or find values in symmetric expressions quickly.

Symmetric Expressions

An expression is symmetric if it remains unchanged when the variables are swapped. In this problem, x, y, and z are cyclically symmetric in terms of a, b, and c.

Reciprocal Property

For any positive real number k, the expression k + 1/k ≥ 2. This property helps in estimating bounds for expressions like x, y, and z given in the question.

Shortcut Trick Apply the identity: (a + 1/a)2 = (a − 1/a)2 + 4. The equation becomes: (a − 1/a)2 + 4 − 2(a − 1/a) = 12. Let x = (a − 1/a), then x2 − 2x − 8 = 0 ⇒ (x − 4)(x + 2) = 0. If x = 4, then a = 2 + √5 or 2 − √5. If x = −2, then a = −1 + √2 or −1 − √2. None of these results match the options (1), (2), or (3). ∴ The correct answer is None of these.

Alternate Method Given: [a + (1/a)]2 − 2[a − (1/a)] = 12 Formula Used: (x + y)2 = (x − y)2 + 4xy ⇒ (a + 1/a)2 = (a − 1/a)2 + 4 × a × (1/a) = (a − 1/a)2 + 4 Substitute this into the original equation: ⇒ [(a − 1/a)2 + 4] − 2[a − (1/a)] = 12 ⇒ (a − 1/a)2 − 2(a − 1/a) + 4 − 12 = 0 ⇒ (a − 1/a)2 − 2(a − 1/a) − 8 = 0 Let k = (a − 1/a): ⇒ k2 − 2k − 8 = 0 ⇒ k2 − 4k + 2k − 8 = 0 ⇒ k(k − 4) + 2(k − 4) = 0 ⇒ (k − 4)(k + 2) = 0 ⇒ k = 4 or k = −2 Case 1: a − 1/a = 4 ⇒ a2 − 1 = 4a ⇒ a2 − 4a − 1 = 0 ⇒ a = [−(−4) ± √((−4)2 − 4 × 1 × (−1))] / (2 × 1) ⇒ a = [4 ± √(16 + 4)] / 2 = [4 ± √20] / 2 = 2 ± √5 Case 2: a − 1/a = −2 ⇒ a2 − 1 = −2a ⇒ a2 + 2a − 1 = 0 ⇒ a = [−2 ± √(22 − 4 × 1 × (−1))] / (2 × 1) ⇒ a = [−2 ± √(4 + 4)] / 2 = [−2 ± √8] / 2 = −1 ± √2 ∴ The correct answer is None of these.

Additional Information Reciprocal Square Identities (a + 1/a)2 = a2 + 1/a2 + 2 and (a − 1/a)2 = a2 + 1/a2 − 2. Relationship between Sum and Difference The difference between the square of the sum and the square of the difference is always (a + 1/a)2 − (a − 1/a)2 = 4. Quadratic Formula For any quadratic equation ax2 + bx + c = 0, the roots are calculated using x = [−b ± √(b2 − 4ac)] / 2a. Nature of Roots Roots are real and distinct if the discriminant D = b2 − 4ac > 0, and they are irrational if D is not a perfect square.

Shortcut Trick

Given x² − 4x + 1 = 0 ⇒ x + 1/x = 4.

Squaring once: x² + 1/x² = 4² − 2 = 14.

Squaring again: x⁴ + 1/x⁴ = 14² − 2 = 194.

Expression = x⁵(x⁴ − 194) + x³(x⁴ − 194).

Substitute 194 = x⁴ + 1/x⁴ ⇒ x⁴ − 194 = −1/x⁴.

Value = x⁵(−1/x⁴) + x³(−1/x⁴) = −x − 1/x = −(4).

∴ The correct answer is −4.

Alternate Method

Given: x² − 4x + 1 = 0

Formula Used: If x + 1/x = k, then x² + 1/x² = k² − 2 and x⁴ + 1/x⁴ = (k² − 2)² − 2.

Calculations:

⇒ x² + 1 = 4x

⇒ x + 1/x = 4

⇒ (x + 1/x)² = 16 ⇒ x² + 1/x² + 2 = 16

⇒ x² + 1/x² = 14

⇒ (x² + 1/x²)² = 196 ⇒ x⁴ + 1/x⁴ + 2 = 196

⇒ x⁴ + 1/x⁴ = 194

⇒ The given expression is x⁹ + x⁷ − 194x⁵ − 194x³

⇒ Factor out common terms: x⁵(x⁴ − 194) + x³(x⁴ − 194)

⇒ From x⁴ + 1/x⁴ = 194, we get x⁴ − 194 = −1/x⁴

⇒ Value = x⁵(−1/x⁴) + x³(−1/x⁴)

⇒ Value = −x − 1/x = −(x + 1/x)

⇒ Value = −4

∴ The correct answer is −4.

Additional Information

Standard Algebraic Identities

If x + 1/x = k, then x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k.

Higher Order Powers

To find x⁴ + 1/x⁴, square the result of x² + 1/x² and subtract 2.

Substitution Technique

In complex polynomial expressions, identify constants like 194 as specific functions of x (like x⁴ + 1/x⁴) to simplify terms by cancellation.

Shortcut Trick

For algebraic identities, we can substitute convenient values for variables.

Given x + y = 3, let us assume y = 0 and x = 3.

The expression to evaluate is x³ + y³ + 9xy.

Substitute the values: (3)³ + (0)³ + 9 × 3 × 0.

⇒ 27 + 0 + 0 = 27.

∴ The correct answer is 27.

Alternate Method

Given:

x + y = 3

Formula Used:

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculations:

Take the cube on both sides of the given equation x + y = 3:

⇒ (x + y)³ = 3³

⇒ x³ + y³ + 3xy(x + y) = 27

Now, substitute the value of x + y = 3 back into the equation:

⇒ x³ + y³ + 3xy × 3 = 27

⇒ x³ + y³ + 9xy = 27

∴ The correct answer is 27.

Additional Information

Standard Cubic Identity

The formula for the cube of a sum is (a + b)³ = a³ + 3a²b + 3ab² + b³, often written as a³ + b³ + 3ab(a + b).

Sum of Two Cubes

The identity for factoring the sum of two cubes is a³ + b³ = (a + b)(a² − ab + b²).

Special Condition Identity

If a + b + c = 0, then a³ + b³ + c³ = 3abc. This is useful for simplifying complex algebraic fractions and expressions.

Shortcut Trick

Simplified A = x⁴ − 1 and B = y² − 1 using the identity a² − b² = (a − b)(a + b).

For x = 2, A = 2⁴ − 1 = 15; For y = 9, B = 9² − 1 = 80.

Expression = A² + 2AB + AB² = A(A + 2B + B²).

Substitute: 15(15 + 160 + 6400) = 15 × 6575.

Result = 98625.

∴ The correct answer is 98625.

Alternate Method

Given: x = 2 and y = 9

Method: Algebraic simplification using the difference of squares identity.

⇒ A = (x⁸ − 1) / (x⁴ + 1)

⇒ A = ((x⁴)² − 1²) / (x⁴ + 1) = (x⁴ − 1)(x⁴ + 1) / (x⁴ + 1) = x⁴ − 1

⇒ A = 2⁴ − 1 = 16 − 1 = 15

⇒ B = (y⁴ − 1) / (y² + 1)

⇒ B = ((y²)² − 1²) / (y² + 1) = (y² − 1)(y² + 1) / (y² + 1) = y² − 1

⇒ B = 9² − 1 = 81 − 1 = 80

Now, calculate the value of A² + 2AB + AB²:

⇒ 15² + 2 × 15 × 80 + 15 × 80²

⇒ 225 + 2400 + 15 × 6400

⇒ 2625 + 96000 = 98625

∴ The correct answer is 98625.

Additional Information

Algebraic Identities

The identity a² − b² = (a − b)(a + b) can be extended to higher powers like a²ⁿ − b²ⁿ = (aⁿ − bⁿ)(aⁿ + bⁿ).

Factoring Groups

Polynomial expressions can often be simplified by factoring out common variables, such as A² + 2AB + AB² = A(A + 2B + B²).

Power Law

When simplifying expressions with indices, remember that (x^m)ⁿ = x^m×n and x^m ÷ xⁿ = x^m−n.

Shortcut Trick

If a large cube is cut into n³ identical small cubes, the Total Surface Area becomes n times the original area.

Given: Total small cubes = 27 = 3³, so n = 3.

New Total Area = 3 × Original Area.

Percentage Increase = (n − 1) × 100 = (3 − 1) × 100 = 200%.

∴ The correct answer is 200%.

Alternate Method

Given: A solid cube is cut into 27 identical small cubes.

Formula: Total Surface Area (TSA) of a cube = 6 × Side²

⇒ Let the side of each small cube be 1 unit.

⇒ Volume of 27 small cubes = 27 × 1³ = 27 cubic units.

⇒ Volume of the original large cube = 27 cubic units.

⇒ Side of the large cube = 27^1/3 = 3 units.

⇒ Total Surface Area of the large cube = 6 × 3² = 6 × 9 = 54 sq. units.

⇒ Total Surface Area of 27 small cubes = 27 × (6 × 1²) = 27 × 6 = 162 sq. units.

⇒ Increase in Total Surface Area = 162 − 54 = 108 sq. units.

⇒ Percentage Increase = (Increase ÷ Original TSA) × 100

⇒ Percentage Increase = (108 ÷ 54) × 100 = 2 × 100 = 200%.

∴ The correct answer is 200%.

Additional Information

Volume and Area Ratio

When a solid is divided, the total volume remains constant, but the total surface area increases because new surfaces are exposed.

Number of Cuts

To divide a cube into n³ identical cubes, a minimum of 3 × (n − 1) cuts are required (n − 1 cuts along each of the three axes).

TSA of Small vs Large Cube

Ratio of TSA of n³ small cubes to original cube is n : 1. For n = 3, Ratio = 3 : 1, representing a 200% increase.

Shortcut Trick

Volume of Square Pyramid = (1/3) × 20² × 45 = 6000 cm³

Volume of Triangular Pyramid = (1/3) × [(√3/4) × 10²] × 10√3

Volume of Triangular Pyramid = (1/3) × 25√3 × 10√3 = 250 cm³

Number of Triangular Pyramids = 6000 ÷ 250 = 24

∴ The correct answer is 24.

Alternate Method

Given: Side of square base (a₁) = 20 cm, Height (h₁) = 45 cm

Given: Side of equilateral base (a₂) = 10 cm, Height (h₂) = 10√3 cm

Formula: Volume of Pyramid = (1/3) × Base Area × Height

⇒ Volume of Square Pyramid (V₁) = (1/3) × (Side)² × h₁

⇒ V₁ = (1/3) × 20 × 20 × 45 = 400 × 15 = 6000 cm³

⇒ Volume of Triangular Pyramid (V₂) = (1/3) × (√3/4 × (Side)²) × h₂

⇒ V₂ = (1/3) × (√3/4 × 100) × 10√3

⇒ V₂ = (1/3) × 25√3 × 10√3 = (1/3) × 250 × 3 = 250 cm³

⇒ Total Number = V₁ ÷ V₂ = 6000 ÷ 250

⇒ Number = 24

∴ The correct answer is 24.

Additional Information

Volume of a Pyramid

The volume of any pyramid is given by V = (1/3) × Area of Base × Vertical Height.

Equilateral Triangle Properties

Area of an equilateral triangle = (√3/4) × side² and Height = (√3/2) × side.

Surface Area of Regular Pyramid

Total Surface Area = Base Area + (1/2) × Perimeter of Base × Slant Height.

Shortcut Trick

Term 1: (&sin;7x − &sin;5x) ÷ (cos7x + cos5x) = tan((7x − 5x) ÷ 2) = tan x

Term 2: (cos6x − cos4x) ÷ (&sin;6x + &sin;4x) = −tan((6x − 4x) ÷ 2) = −tan x

Substituting values: tan x − (−tan x) = 2 tan x

∴ The correct answer is 2tanx.

Alternate Method

Given: [(&sin;7x − &sin;5x) ÷ (cos7x + cos5x)] − [(cos6x − cos4x) ÷ (&sin;6x + &sin;4x)]

Formulas Used:

&sin;C − &sin;D = 2 cos((C+D)/2) &sin((C−D)/2)

cosC + cosD = 2 cos((C+D)/2) cos((C−D)/2)

cosC − cosD = −2 &sin((C+D)/2) &sin((C−D)/2)

&sin;C + &sin;D = 2 &sin((C+D)/2) cos((C−D)/2)

Calculations:

⇒ First part = [2 cos((7x+5x)/2) &sin((7x−5x)/2)] ÷ [2 cos((7x+5x)/2) cos((7x−5x)/2)]

⇒ First part = [2 cos 6x &sin; x] ÷ [2 cos 6x cos x] = &sin; x ÷ cos x = tan x

⇒ Second part = [−2 &sin((6x+4x)/2) &sin((6x−4x)/2)] ÷ [2 &sin((6x+4x)/2) cos((6x−4x)/2)]

⇒ Second part = [−2 &sin; 5x &sin; x] ÷ [2 &sin; 5x cos x] = −&sin; x ÷ cos x = −tan x

⇒ Final Expression = tan x − (−tan x)

⇒ Final Expression = tan x + tan x = 2 tan x

∴ The correct answer is 2tanx.

Additional Information

Transformation Formulas (Sum-to-Product)

These formulas convert the sum or difference of two sine or cosine functions into a product of functions, defined as sinC ± sinD and cosC ± cosD.

Quotient Identity

The tangent of an angle is the ratio of the sine of that angle to its cosine: tan θ = sin θ ÷ cos θ.

Negative Angle Identity

The formula for cosC − cosD can also be written as 2 &sin((C+D)/2) &sin((D−C)/2) to handle the negative sign naturally.

Shortcut Trick

Put θ = 0° to simplify the trigonometric expression.

When θ = 0°, then cos 2θ = cos 0° = 1 and sin θ = sin 0° = 0.

Expression = [(cos³ 0° + 3cos 0°) ÷ (cos⁶ 0° − sin⁶ 0°)]

Expression = [(1³ + 3 × 1) ÷ (1⁶ − 0⁶)] = (1 + 3) ÷ 1 = 4.

∴ The correct answer is 4.

Alternate Method

Given: [(cos³ 2θ + 3cos 2θ) ÷ (cos⁶ θ − sin⁶ θ)]

Formula Used:

1. a³ − b³ = (a − b)(a² + ab + b²)
2. cos 2θ = cos² θ − sin² θ
3. sin² 2θ = 4 sin² θ cos² θ

Calculation:

⇒ Numerator = cos³ 2θ + 3cos 2θ = cos 2θ(cos² 2θ + 3)

⇒ Denominator = cos⁶ θ − sin⁶ θ = (cos² θ)³ − (sin² θ)³

⇒ Denominator = (cos² θ − sin² θ)(cos⁴ θ + cos² θ sin² θ + sin⁴ θ)

⇒ Denominator = (cos 2θ)[(cos² θ + sin² θ)² − cos² θ sin² θ]

⇒ Denominator = (cos 2θ)[1 − (1/4) sin² 2θ]

⇒ Denominator = (cos 2θ)[1 − (1/4)(1 − cos² 2θ)]

⇒ Denominator = (cos 2θ)[1 − 1/4 + (1/4) cos² 2θ] = (cos 2θ)[3/4 + (1/4) cos² 2θ]

⇒ Denominator = [cos 2θ(3 + cos² 2θ)] / 4

⇒ Expression = [cos 2θ(cos² 2θ + 3)] ÷ [[cos 2θ(3 + cos² 2θ)] / 4]

⇒ Expression = 1 ÷ (1/4) = 4

∴ The correct answer is 4.

Additional Information

Double Angle Identities

cos 2θ = cos² θ − sin² θ = 2cos² θ − 1 = 1 − 2sin² θ.

Power Reduction Formulas

cos⁴ θ + sin⁴ θ = 1 − 2sin² θ cos² θ and cos⁶ θ + sin⁶ θ = 1 − 3sin² θ cos² θ.

Value Substitution Method

For trigonometric identities, substituting standard angles like 0°, 30°, 45° or 60° (where defined) helps solve complex expressions instantly.

Shortcut Trick

Let θ = π/4 + A.

Then, 3π/4 + A = π/2 + (π/4 + A) = π/2 + θ.

The expression becomes tan θ × tan(π/2 + θ).

Using the trigonometric identity tan(π/2 + θ) = −cot θ.

Product = tan θ × (−cot θ) = −(tan θ × cot θ) = −1.

∴ The correct answer is -1.

Alternate Method

Given: tan(π/4 + A) × tan(3π/4 + A)

Formula Used: tan(X + Y) = (tan X + tan Y) ÷ (1 − tan X × tan Y)

⇒ tan(π/4 + A) = (tan π/4 + tan A) ÷ (1 − tan π/4 × tan A)

⇒ tan(π/4 + A) = (1 + tan A) ÷ (1 − tan A)

⇒ tan(3π/4 + A) = (tan 3π/4 + tan A) ÷ (1 − tan 3π/4 × tan A)

⇒ tan(3π/4 + A) = (−1 + tan A) ÷ (1 − (−1) × tan A) = (tan A − 1) ÷ (1 + tan A)

⇒ Product = [(1 + tan A) ÷ (1 − tan A)] × [(tan A − 1) ÷ (1 + tan A)]

⇒ Product = [(1 + tan A) ÷ (1 − tan A)] × [−(1 − tan A) ÷ (1 + tan A)]

⇒ Product = −1

∴ The correct answer is -1.

Additional Information

Trigonometric Angle Identities

tan(π/2 + θ) = −cot θ and tan(π/2 − θ) = cot θ. Similarly, tan(π + θ) = tan θ.

Sum and Difference Formulas

tan(A ± B) = (tan A ± tan B) ÷ (1 ∓ tan A × tan B). For A = 45°, tan(45° + θ) = (1 + tan θ) ÷ (1 − tan θ).

Reciprocal Relation

tan θ × cot θ = 1 for any angle θ where the functions are defined.

Shortcut Trick

Put θ = 30° to evaluate the expression quickly.

sec 2θ = sec 60° = 2 and √(sec²θ − 1) = tan 30° = 1/√3.

cot 30° − tan 30° = √3 − 1/√3 = 2/√3.

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

Expression = [3/√3] × [1/√3] = 3/3 = 1.

∴ The correct answer is 1.

Alternate Method

Given: [(sec 2θ + 1)√(sec²θ − 1)] × 1/2 × (cotθ − tanθ)

Formula used: 1 + cos 2θ = 2 cos²θ | sin 2θ = 2 sinθ cosθ | sec²θ − 1 = tan²θ

⇒ (sec 2θ + 1) = (1/cos 2θ) + 1 = (1 + cos 2θ) / cos 2θ = (2 cos²θ) / cos 2θ

⇒ √(sec²θ − 1) = √tan²θ = tanθ = sinθ / cosθ

⇒ 1/2 × (cotθ − tanθ) = 1/2 × (cosθ/sinθ − sinθ/cosθ) = 1/2 × (cos²θ − sin²θ) / (sinθ cosθ)

⇒ 1/2 × (cos 2θ) / (1/2 sin 2θ) = cos 2θ / sin 2θ = cot 2θ

⇒ Expression = [(2 cos²θ / cos 2θ) × (sinθ / cosθ)] × (cos 2θ / sin 2θ)

⇒ Expression = (2 sinθ cosθ / cos 2θ) × (cos 2θ / sin 2θ) = (sin 2θ / cos 2θ) × (cos 2θ / sin 2θ)

⇒ Expression = tan 2θ × cot 2θ = 1

∴ The correct answer is 1.

Additional Information

Trigonometric Double Angle Identities

cos 2θ = cos²θ − sin²θ = 2 cos²θ − 1 = 1 − 2 sin²θ and sin 2θ = 2 sinθ cosθ.

Pythagorean Identities

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

Difference of Cot and Tan

The expression cot θ − tan θ always simplifies to 2 cot 2θ, which is a very useful shortcut for competitive exams.

Shortcut Trick

Reduce the angle 630° by subtracting the nearest multiple of 360°.

sin(630° + A) = sin(360° + 270° + A) = sin(270° + A).

Using the Allied Angle rule, sin(270° + A) = −cos A.

Expression: −cos A + cos A = 0.

∴ The correct answer is 0.

Alternate Method

Given: sin(630° + A) + cos A

Formula Used: sin(n × 360° + θ) = sin θ and sin(270° + θ) = −cos θ

⇒ sin(630° + A) = sin(360° + 270° + A)

⇒ sin(270° + A)

⇒ sin(3 × 90° + A)

⇒ −cos A [Since (270° + A) lies in the 4th quadrant where sine is negative]

⇒ Expression = −cos A + cos A

⇒ 0

∴ The correct answer is 0.

Additional Information

Allied Angles Rule (ASTC)

In the 4th quadrant (270° to 360°), only Cosine and Secant are positive. Sine, Tangent, Cotangent, and Cosecant are negative.

Trigonometric Function Change

Functions change at (90° ± θ) and (270° ± θ): sin → cos, tan → cot, sec → cosec. They remain the same at (180° ± θ) and (360° ± θ).

Periodicity of Sine and Cosine

The values of sin θ and cos θ repeat every 360° (or 2π radians). Thus, sin(n × 360° + θ) = sin θ for any integer n.

Shortcut Trick

Apply compound angle identities: sin(A+B) and cos(A+B).

Numerator = sin 59° cos 31° + cos 59° sin 31° = sin(59° + 31°) = sin 90° = 1.

Denominator = cos 20° cos 25° − sin 20° sin 25° = cos(20° + 25°) = cos 45° = 1/√2.

Value = 1 ÷ (1/√2) = √2.

∴ The correct answer is √2.

Alternate Method

Given: [(sin 59° cos 31° + cos 59° sin 31°) ÷ (cos 20° cos 25° − sin 20° sin 25°)]

Formulas Used:

1. sin A cos B + cos A sin B = sin(A + B)
2. cos A cos B − sin A sin B = cos(A + B)

⇒ Numerator = sin(59° + 31°) = sin 90°

⇒ Denominator = cos(20° + 25°) = cos 45°

⇒ Value = sin 90° ÷ cos 45°

⇒ Value = 1 ÷ (1/√2)

⇒ Value = 1 × √2 = √2

∴ The correct answer is √2.

Additional Information

Sine Compound Angle Formulas

Addition: sin(A + B) = sin A cos B + cos A sin B | Subtraction: sin(A − B) = sin A cos B − cos A sin B.

Cosine Compound Angle Formulas

Addition: cos(A + B) = cos A cos B − sin A sin B | Subtraction: cos(A − B) = cos A cos B + sin A sin B.

Standard Trigonometric Values

sin 90° = 1 | cos 90° = 0 | sin 45° = 1/√2 | cos 45° = 1/√2.

Tangent Compound Angle Formula

tan(A + B) = (tan A + tan B) ÷ (1 − tan A tan B).

Shortcut Trick

Use the substitution method to solve identity-based questions quickly.

Let A = 0°, B = 0°, and C = 0°.

Expression = cos(90 − 0)sin(0 − 0) + sin(90 + 0)cos(0 + 0) − sin(90 − 0)cos(0 + 0)

⇒ cos(90°)sin(0°) + sin(90°)cos(0°) − sin(90°)cos(0°)

⇒ (0 × 0) + (1 × 1) − (1 × 1) = 1 − 1 = 0

∴ The correct answer is 0.

Alternate Method

Given: cos(90 − B)sin(C − A) + sin(90 + A)cos(B + C) − sin(90 − C)cos(A + B)

Formula Used:

1. cos(90 − θ) = sin θ | 2. sin(90 + θ) = cos θ | 3. sin(90 − θ) = cos θ
2. sin(X − Y) = sin X cos Y − cos X sin Y
3. cos(X + Y) = cos X cos Y − sin X sin Y

Calculations:

⇒ sin B sin(C − A) + cos A cos(B + C) − cos C cos(A + B)

⇒ sin B(sin C cos A − cos C sin A) + cos A(cos B cos C − sin B sin C) − cos C(cos A cos B − sin A sin B)

⇒ (sin B sin C cos A − sin B cos C sin A) + (cos A cos B cos C − cos A sin B sin C) − (cos C cos A cos B − cos C sin A sin B)

⇒ sin B sin C cos A − sin B cos C sin A + cos A cos B cos C − cos A sin B sin C − cos A cos B cos C + sin A sin B cos C

⇒ (sin B sin C cos A − cos A sin B sin C) + (sin A sin B cos C − sin B cos C sin A) + (cos A cos B cos C − cos A cos B cos C)

⇒ 0 + 0 + 0 = 0

∴ The correct answer is 0.

Additional Information

Allied Angle Rules

For (90 ± θ), trigonometric ratios change (sin ↔ cos, tan ↔ cot, sec ↔ cosec). For (180 ± θ), they remain the same.

Compound Angle Identities

sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B.

Quadrant Signs (ASTC Rule)

In 1st Quad (90−θ), All are positive. In 2nd Quad (90+θ or 180−θ), only Sin and Cosec are positive.

Shortcut Trick

For a 45° angle, Height = Base. Given Height₁ = 200 m, so Base₁ = 200 m.

Total distance between trees = 400 m → Base₂ = 400 − 200 = 200 m.

For a 60° angle, Height = Base × √3.

Height₂ = 200 × √3 = 200√3 m.

∴ The correct answer is 200√3 metres.

Alternate Method

Given:

Height of first tree (h₁) = 200 m

Total distance between trees (BD) = 400 m

Angles of elevation at point P: θ₁ = 45° and θ₂ = 60°

Formula Used:

tan θ = Perpendicular / Base

Calculations:

In ΔABP:

⇒ tan 45° = AB / BP

⇒ 1 = 200 / BP

⇒ BP = 200 m

Distance between Point P and second tree:

⇒ PD = BD − BP

⇒ PD = 400 − 200 = 200 m

In ΔCDP:

⇒ tan 60° = CD / PD

⇒ √3 = h / 200

⇒ h = 200√3 m

∴ The correct answer is 200√3 metres.

Additional Information

Trigonometric Ratios for Standard Angles

tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3. These are essential for solving height and distance problems.

Angle of Depression vs. Elevation

The angle of depression from the top of the tree to the point on the road is equal to the angle of elevation from the point to the top of the tree due to alternate interior angles.

Right Triangle Properties

In a 45°-45°-90° triangle, the legs are equal (ratio 1:1:√2). In a 30°-60°-90° triangle, the ratio of sides is 1:√3:2.

Shortcut Trick

Base distance from building = Building height ÷ tan 45° = 40 ÷ 1 = 40 m.

Total height (Building + Tower) = Base distance × tan 60° = 40 × √3 = 40√3 m.

Tower height = Total height − Building height = 40√3 − 40 = 40(√3 − 1) m.

∴ The correct answer is 40(√3 − 1) m.

Alternate Method

Given:

Height of building (AB) = 40 m

Angle of depression from top of tower (C) to point (D) = 60° → Angle of elevation at D = 60°

Angle of depression from bottom of tower (B) to point (D) = 45° → Angle of elevation at D = 45°

In ΔABD (where B is top of building):

⇒ tan 45° = AB ÷ AD

⇒ 1 = 40 ÷ AD

⇒ AD = 40 m

In ΔACD (where C is top of tower):

⇒ tan 60° = AC ÷ AD

⇒ √3 = (AB + BC) ÷ 40

⇒ 40√3 = 40 + BC

⇒ BC = 40√3 − 40

⇒ BC = 40(√3 − 1)

∴ The correct answer is 40(√3 − 1) m.

Additional Information

Trigonometric Ratios

tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3. These are frequently used in height and distance problems.

Angle of Depression vs Elevation

The angle of depression from an observer to an object is equal to the angle of elevation from the object to the observer because they are alternate interior angles.

Height and Distance General Rule

If the angle of elevation changes from θ₁ to θ₂ as one moves a distance 'd' towards an object of height 'h', then h = d ÷ (cot θ₁ − cot θ₂).

Shortcut Trick

Distance moved towards tower (D) = h × (cot α − cot β)

Given: tan α = 3/4 ⇒ cot α = 4/3 and tan β = 4/3 ⇒ cot β = 3/4

560 = h × (4/3 − 3/4)

560 = h × (7/12)

h = (560 × 12) ÷ 7 = 80 × 12 = 960 m

∴ The correct answer is 960 m.

Alternate Method

Given:

Initial angle α, tan α = 3/4

Second angle β, tan β = 4/3

Distance moved (PQ) = 560 m

Formula Used:

In a right-angled triangle, tan θ = Perpendicular ÷ Base

Calculations:

⇒ Let height of tower AB = h metres.

⇒ In ΔABQ, tan β = AB / QB ⇒ 4/3 = h / QB ⇒ QB = 3h/4.

⇒ In ΔABP, tan α = AB / PB ⇒ 3/4 = h / (PQ + QB).

⇒ 3/4 = h / (560 + 3h/4).

⇒ 3 × (560 + 3h/4) = 4h.

⇒ 1680 + 9h/4 = 4h.

⇒ 4h − 9h/4 = 1680.

⇒ (16h − 9h) / 4 = 1680.

⇒ 7h / 4 = 1680.

⇒ h = (1680 × 4) ÷ 7 = 240 × 4 = 960.

∴ The correct answer is 960 m.

Additional Information

Distance-Height Relationship

When moving towards or away from a tower, the height h and distance d between two observation points are related by h = d / (cot α_smaller − cot α_larger).

Angle of Elevation

The angle of elevation is the angle between the horizontal line of sight and the object when looking upwards. As you move towards the object, the angle of elevation increases.

Standard Trigonometric Ratios

Common tan θ values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. Cot θ is the reciprocal of tan θ (1/tan θ).

⇒ Sale of milk by salesman A in state P = 200000 × 0.24 × 0.65 = 31200

⇒ Sale of milk by salesman A in state Q = 200000 × 0.10 × 0.80 = 16000

⇒ Sale of milk by salesman A in state R = 200000 × 0.17 × 0.50 = 17000

⇒ Sale of milk by salesman A in state S = 200000 × 0.13 × 0.70 = 18200

⇒ Sale of milk by salesman A in state T = 200000 × 0.22 × 0.60 = 26400

⇒ Sale of milk by salesman A in state U = 200000 × 0.14 × 0.80 = 22400

⇒ Total sale by salesman A in all the states = (31200 + 16000 + 17000 + 18200 + 26400 + 22400) = 131200 litres

∴ Average sale by salesman A in all the states = 131200/6 = 21866.67 litres

only two milkman A and B sell the milk and if sales of A is given then the sales of B can be calculated by subtracting sales of A from sales of milk

⇒ Sale of milk by salesman B in state P = 200000 × 0.24 × 0.35 = 16800

⇒ Sale of milk by salesman B in state Q = 200000 × 0.10 × 0.20 = 4000

⇒ Sale of milk by salesman A in state R = 200000 × 0.17 × 0.50 = 17000

⇒ Sale of milk by salesman A in state T = 200000 × 0.22 × 0.60 = 26400

⇒ Total sale by B in states P and Q = 16800 + 4000 = 20800 litres

⇒ Total sale by A in states R and T = 17000 + 26400 = 43400 litres

∴ Required Ratio = 20800 ∶ 43400 = 104 ∶ 217

only two milkman A and B sell the milk and if sales of A is given then the sales of B can be calculated by subtracting sales of A from sales of milk

⇒ Total sale in states Q, T and S together = 200000 × (0.10 + 0.13 + 0.22) = 90000 litres

⇒ Average sale of milk in states Q, T and S together = 90000/3 = 30000 litres

∴ Central angle = (30000/200000) × 360 = 54°

only two milkman A and B sell the milk and if sales of A is given then the sales of B can be calculated by subtracting sales of A from sales of milk

⇒ Sale of milk in state T by salesmen B = 200000 × 0.22 × 0.40 = 17600 litres

⇒ Total sale of milk in state R and S together = 200000 × (0.17 + 0.13) = 60000 litres

∴ Difference in the sale of milk in state T by salesmen B and the total sale of milk in state R and S together = 60000 - 17600 = 42400 litres

only two milkman A and B sell the milk and if sales of A is given then the sales of B can be calculated by subtracting sales of A from sales of milk

⇒ Sale of milk in state R by salesmen A = 200000 × 0.17 × 0.50 = 17000

⇒ Sale of milk in state R by salesmen B = 200000 × 0.17 × 0.50 = 17000

∴ Difference in sale = 17000 - 17000 = 0

only two milkman A and B sell the milk and if sales of A is given then the sales of B can be calculated by subtracting sales of A from sales of milk

Shortcut Trick

Total volume after mixing = 300 + 50 = 350 ml.

Final ratio of Acid : Water is 2 : 5 (Total 7 units).

Final Acid quantity = (2 ÷ 7) × 350 = 100 ml.

Since only water was added, the Acid quantity in the original 300 ml mixture remains 100 ml.

Original Water quantity = 300 − 100 = 200 ml.

Original Ratio Acid : Water = 100 : 200 = 1 : 2. Comparing with 1 : x, we get x = 2.

∴ The correct answer is 2.

Alternate Method

Given: Initial mixture = 300 ml, Water added = 50 ml, Final ratio = 2 : 5.

Formula: Quantity of component = (Ratio share ÷ Sum of ratios) × Total Quantity.

⇒ Total Final volume = 300 + 50 = 350 ml

⇒ In the final mixture, Ratio sum = 2 + 5 = 7

⇒ Amount of Acid in final mixture = (2/7) × 350 = 100 ml

⇒ Amount of Water in final mixture = (5/7) × 350 = 250 ml

⇒ Original Water = Final Water − Added Water = 250 − 50 = 200 ml

⇒ Ratio of Acid : Water in original mixture = 100 : 200 = 1 : 2

⇒ Comparing 1 : 2 with 1 : x

⇒ x = 2

∴ The correct answer is 2.

Additional Information

Mixture and Alligation Principle

When a new substance is added to a mixture, the quantity of the other non-added substance remains constant in absolute terms (mass/volume), even though its percentage or ratio share changes.

Ratio Balancing

To equate two ratios where one component is constant, multiply the entire ratio by a factor such that the parts of the constant component become identical.

Replacement Method

If a part of the mixture is replaced by another substance, the final quantity of the original substance is given by Q_final = Q_initial × [1 − (Replacement volume ÷ Total volume)]ⁿ.

Shortcut Trick

Initial Milk : Water = 11 : 2

New Milk : Water = 2 : 1

Equating Milk: Initial = 11 : 2, New = 11 : 5.5 (multiplied by 5.5)

Water added = 5.5 − 2 = 3.5 units → 35 liters

1 unit = 10 liters ⇒ Milk = 11 units = 110 liters

∴ The correct answer is 110 liters.

Alternate Method

Given:

Ratio of Milk to Water in original mixture = 11 : 2

Water added = 35 liters

New ratio of Milk to Water = 2 : 1

Calculations:

Let the original amount of Milk be 11x liters and Water be 2x liters.

According to the problem, when 35 liters of water are added:

⇒ 11x ÷ (2x + 35) = 2 ÷ 1

⇒ 11x = 2 × (2x + 35)

⇒ 11x = 4x + 70

⇒ 11x − 4x = 70

⇒ 7x = 70

⇒ x = 10

Original quantity of milk = 11x

⇒ Milk = 11 × 10 = 110 liters

∴ The correct answer is 110 liters.

Additional Information

Ratio Balancing Method

When one component in a mixture is added or removed, make the other component constant across both ratios to find the value of one unit.

Mixture Replacement Formula

Final Quantity = Initial Quantity × (1 − x/V)ⁿ, where x is quantity replaced, V is total volume, and n is number of operations.

Alligation Rule

Ratio of ingredients = (Cheaper price − Mean price) : (Mean price − Dearer price), used to find mixing proportions.

Shortcut Trick

Profit Ratio = (Capital A × Time A) : (Capital B × Time B) : (Capital C × Time C)

Ratio = (25 × 12) : (40 × 9) : (50 × 5) = 300 : 360 : 250 = 30 : 36 : 25

Sum of ratios = 30 + 36 + 25 = 91 units

91 units = ₹2,73,000 → 1 unit = ₹3,000

C profit = 25 units × 3,000 = ₹75,000

∴ The correct answer is ₹75,000.

Alternate Method

Given:

Investment of A = ₹25 lakhs, B = ₹40 lakhs, C = ₹50 lakhs

Time period: A = 12 months, B = 12 − 3 = 9 months, C = 12 − (3 + 4) = 5 months

Total Profit = ₹2,73,000

Formula Used:

Profit Ratio = (Investment₁ × Time₁) : (Investment₂ × Time₂) : (Investment₃ × Time₃)

Calculations:

⇒ Profit Ratio A : B : C = (25 × 12) : (40 × 9) : (50 × 5)

⇒ Profit Ratio A : B : C = 300 : 360 : 250

⇒ Simplified Ratio = 30 : 36 : 25

⇒ Sum of ratio parts = 30 + 36 + 25 = 91

⇒ C share of profit = (Ratio of C ÷ Sum of ratios) × Total Profit

⇒ C share = (25 ÷ 91) × 2,73,000

⇒ C share = 25 × 3,000

⇒ C share = ₹75,000

∴ The correct answer is ₹75,000.

Additional Information

Ratio of Profit

In any partnership, the profit is distributed in the ratio of the product of capital and the time period for which it is invested: P₁ : P₂ = C₁T₁ : C₂T₂.

Ratio of Capital and Time

If the profit ratio and time ratio are known, the capital ratio is C₁ : C₂ = (P₁/T₁) : (P₂/T₂). Similarly, for time, T₁ : T₂ = (P₁/C₁) : (P₂/C₂).

Working vs Sleeping Partner

A working partner manages the business and may receive a salary or additional commission from the profit before the remaining profit is shared among all partners based on investment.

Shortcut Trick

Distributed Profit = 75% of 36,000 = ₹27,000.

Working Profit Ratio (A : B : C) = 4 : 5 : 7.

Actual Profit Ratio (C is sleeping) = 4 : 5 : (7 ÷ 2) = 8 : 10 : 7.

B's share = [10 ÷ (8 + 10 + 7)] × 27,000 = (10 ÷ 25) × 27,000 = ₹10,800.

∴ The correct answer is ₹10800.

Alternate Method

Given:

Investment Ratio (A : B : C) = 4 : 5 : 7

Total Profit = ₹36,000

Reinvestment = 25%

Condition: C's share is halved (sleeping partner).

Calculations:

⇒ Amount to be distributed = Total Profit − 25% of Total Profit

⇒ Distributed Profit = 36,000 × (75/100) = ₹27,000

⇒ Ratio of profits (A : B : C) = 4 : 5 : (7 ÷ 2)

⇒ Multiplying by 2 to remove fractions = 8 : 10 : 7

⇒ Sum of ratio parts = 8 + 10 + 7 = 25 units

⇒ 25 units = ₹27,000

⇒ 1 unit = 27,000 ÷ 25 = ₹1,080

⇒ B's share = 10 units = 10 × 1,080 = ₹10,800

∴ The correct answer is ₹10800.

Additional Information

Profit Sharing Rule

In any partnership, Profit Ratio = (Capital₁ × Time₁) : (Capital₂ × Time₂). If time is same, profit ratio is simply the capital ratio.

Sleeping Partner

A partner who invests capital but does not participate in the management of the business. Often, working partners receive an extra salary or commission before the profit is split.

Net Distributed Profit

Profit distributed among partners = Gross Profit − (Salaries + Reserves/Reinvestments + Taxes).

Shortcut Trick

Total work = LCM(36, 12) = 36 units.

Efficiency of A = 36 ÷ 36 = 1 unit/day; Efficiency of B = 36 ÷ 12 = 3 units/day.

Combined efficiency = 1 + 3 = 4 units/day.

Work done in 3 days = 4 × 3 = 12 units.

Work remaining = 36 − 12 = 24 units.

Fraction of work left = 24 / 36 = 2/3.

∴ The correct answer is 2/3.

Alternate Method

Given: Time taken by A = 36 days, Time taken by B = 12 days.

Formula: Total Work = LCM of time taken; Efficiency = Total Work ÷ Time.

⇒ Total efficiency of (A + B) = 1 + 3 = 4 units/day

⇒ Work completed by A and B together in 3 days = 4 × 3 = 12 units

⇒ Remaining work = Total work − Work completed

⇒ Remaining work = 36 − 12 = 24 units

⇒ Fraction of work left = Remaining work / Total work

⇒ Fraction of work left = 24 / 36 = 2/3

∴ The correct answer is 2/3.

Additional Information

Combined Work Formula

If person A takes 'x' days and B takes 'y' days to complete a work, they will finish the work together in (x × y) / (x + y) days.

Work and Efficiency

Efficiency is inversely proportional to the time taken. Efficiency × Time = Total Work (Constant).

Fraction of Work Done

If a person can complete a work in 'D' days, then the part of work done in 1 day is 1/D and in 'n' days is n/D.

Shortcut Trick

Total Work = LCM(45, 15, 5) = 45 units

Efficiency of A = 1, B = 3, (A+B+C) = 9

Efficiency of C = 9 − (1 + 3) = 5 units/day

Time taken by C alone = 45 ÷ 5 = 9 days

∴ The correct answer is 9 days.

Alternate Method

Given: A takes 45 days, B takes 15 days, and A, B, and C together take 5 days.

Formula: Efficiency = Total Work ÷ Time

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

⇒ 9 = 1 + 3 + Efficiency of C

⇒ Efficiency of C = 9 − 4 = 5

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

⇒ Time = 45 ÷ 5 = 9 days

∴ The correct answer is 9 days.

Additional Information

Work and Wages

Wages are distributed among workers in the ratio of their efficiencies or the amount of work done by each.

Pipes and Cisterns

This concept follows the same logic as Time & Work, where an inlet pipe does positive work and an outlet pipe does negative work.

Combined Efficiency

Total efficiency of a group is the sum of individual efficiencies. If one person leaves, their efficiency is subtracted from the total.

∴ The correct answer is 9 days.

Shortcut Trick

Combined efficiency of (A + B + C) = 1/7.5 = 2/15 per day.

Efficiency of B = 1/15 per day.

Efficiency of (A + C) = Combined efficiency − Efficiency of B.

Efficiency of (A + C) = 2/15 − 1/15 = 1/15 per day.

Time for (A + C) to finish = 1 ÷ (1/15) = 15 days.

∴ The correct answer is 15 days.

Alternate Method

Given:

Time taken by (A + B + C) = 7.5 days

Time taken by B alone = 15 days

Productivity of C = 3 × Productivity of A

Formula Used:

Total Work = Efficiency × Time

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

Calculations:

⇒ Total work = LCM (7.5, 15) = 15 units.

⇒ Efficiency of (A + B + C) = 15 ÷ 7.5 = 2 units/day.

⇒ Efficiency of B = 15 ÷ 15 = 1 unit/day.

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

⇒ Efficiency of (A + C) = 2 − 1 = 1 unit/day.

⇒ Time taken by (A + C) together = Total Work ÷ Efficiency of (A + C)

⇒ Time taken by (A + C) = 15 ÷ 1 = 15 days.

∴ The correct answer is 15 days.

Additional Information

Efficiency-Time Relationship

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

Combined Efficiency Rule

The efficiency of multiple people working together is the sum of their individual efficiencies: Eff(Total) = Eff₁ + Eff₂ + Eff₃.

Productivity Ratio

If person X is n times as productive as person Y, then Efficiency(X) = n × Efficiency(Y), and Time(X) = Time(Y) ÷ n.

Shortcut Trick

LCM of 9, 12, and 36 = 36 units (Total Work).

Efficiencies: A = 4, B = 3, C = 1 (Total = 8 units/day).

Work done in 3 days = 8 × 3 = 24 units.

Remaining work = 36 − 24 = 12 units.

Days for B and C = 12 ÷ (3 + 1) = 3 days.

∴ The correct answer is 3 days.

Alternate Method

Given:

Time taken by A = 9 days

Time taken by B = 12 days

Time taken by C = 36 days

Formula Used:

Efficiency = Total Work ÷ Time

Calculations:

⇒ Combined efficiency of A, B and C = 4 + 3 + 1 = 8 units/day

⇒ Work completed by A, B and C in 3 days = 8 × 3 = 24 units

⇒ Remaining work = Total work − Completed work = 36 − 24 = 12 units

⇒ Combined efficiency of B and C = 3 + 1 = 4 units/day

⇒ Time taken by B and C to finish rest of the job = 12 ÷ 4 = 3 days

∴ The correct answer is 3 days.

Additional Information

Pipes and Cisterns

Works on the same efficiency principle where filling pipes have positive efficiency and leakage or outlet pipes have negative efficiency.

Work and Wages

Wages are divided in the ratio of the actual work done by each individual. If they work for the same duration, wages are divided in the ratio of their efficiencies.

Alternate Day Work

In problems where people work on alternate days, find the total work done in one complete cycle (e.g., 2 days or 3 days) and then divide the total work by this cycle's value.

Shortcut Trick

Total weight of P, Q, and R = 3 × 71 = 213 kg.

Sum of (P + Q) and (Q + R) = (2 × 66) + (2 × 76.5) = 132 + 153 = 285 kg.

Since Q is included in both pairs, Weight of Q = (Sum of pairs) − (Total Sum of P, Q, R).

Weight of Q = 285 − 213 = 72 kg.

∴ The correct answer is 72 kg.

Alternate Method

Given:

Average weight (P, Q, R) = 71 kg

Average weight (P, Q) = 66 kg

Average weight (Q, R) = 76.5 kg

Formula Used:

Total = Average × Number of entities

⇒ Sum of weights of P, Q, and R = 71 × 3 = 213 kg

⇒ Sum of weights of P and Q = 66 × 2 = 132 kg

⇒ Weight of R = (Sum of P, Q, R) − (Sum of P, Q) = 213 − 132 = 81 kg

⇒ Sum of weights of Q and R = 76.5 × 2 = 153 kg

⇒ Weight of Q = (Sum of Q, R) − (Weight of R) = 153 − 81 = 72 kg

∴ The correct answer is 72 kg.

Additional Information

General Average Formula

Average = (Sum of all observations) ÷ (Total number of observations).

Sum from Average

Sum of observations = Average × Count. This is the base for solving most average-related problems.

Overlap Property

When an element is part of two subsets that cover the entire set, the value of the overlapping element = (Sum of Subset 1 + Sum of Subset 2) − Total Sum.

Deviation Method

Net deviation from the average is always zero. If the average of a group is A, and two sub-groups have deviations d1 and d2, the overlapping element value = A + (d1 + d2).

Shortcut Trick

Assume a base average of ₹2250 for all sarees.

Extra cost from the 3 new sarees = 3 × (2750 − 2250) = 3 × 500 = ₹1500.

Distributed extra cost over all 8 sarees = 1500 ÷ 8 = ₹187.5.

New Average = 2250 + 187.5 = ₹2437.5.

∴ The correct answer is ₹2437.5.

Alternate Method

Given:

Average of 5 sarees = ₹2250

Average of 3 sarees = ₹2750

Formula Used:

Average = (Sum of all observations) ÷ (Total number of observations)

Weighted Average = (n₁ × A₁ + n₂ × A₂) ÷ (n₁ + n₂)

Calculations:

⇒ Total cost of first 5 sarees = 5 × 2250 = ₹11250

⇒ Total cost of next 3 sarees = 3 × 2750 = ₹8250

⇒ Total cost of all 8 sarees = 11250 + 8250 = ₹19500

⇒ Total number of sarees = 5 + 3 = 8

⇒ Combined Average = 19500 ÷ 8

⇒ Combined Average = ₹2437.5

∴ The correct answer is ₹2437.5.

Additional Information

Weighted Average Principle

When two groups with averages A₁ and A₂ and sizes n₁ and n₂ are combined, the resulting average always lies between A₁ and A₂.

Net Deviation Method

The sum of deviations of all observations from the actual average is always zero. This is useful for finding missing values in a series.

Effect of Constant Change

If every value in a data set is increased or decreased by a constant value 'k', the average also increases or decreases by 'k'.

Shortcut Trick

Total target score set by Team A = 50 × 5.3 = 265 runs.

Runs Team B needs to score in final 5 overs = 5 × 7.2 = 36 runs.

Team B current score = Total Target − Runs Needed = 265 − 36 = 229 runs.

∴ The correct answer is 229.

Alternate Method

Given:

Run rate of Team A = 5.3 runs/over

Total overs for Team A = 50 overs

Required run rate for Team B = 7.2 runs/over

Overs remaining for Team B = 5 overs

Formula:

Total Score = Average Run Rate × Number of Overs

Calculations:

⇒ Total runs scored by Team A = 50 × 5.3

⇒ Total runs scored by Team A = 265 runs

⇒ Total runs Team B needs in the remaining 5 overs = 5 × 7.2

⇒ Total runs Team B needs = 36.0 runs

⇒ Current score of Team B = Total target of Team A − Runs needed by Team B

⇒ Current score of Team B = 265 − 36

⇒ Current score of Team B = 229 runs

∴ The correct answer is 229.

Additional Information

Run Rate Formula

Run Rate is calculated as the total runs scored divided by the total overs faced: Run Rate = Total Runs ÷ Total Overs.

Required Run Rate

Required Run Rate = (Target Score − Current Score) ÷ Overs Remaining. This indicates how fast the batting team needs to score to win/tie.

Net Run Rate (NRR)

NRR = (Average runs per over scored by a team) − (Average runs per over scored against the team) over a tournament.

Shortcut Trick

Even numbers between 104 and 148 are 106, 108, ..., 146.

For an Arithmetic Progression (AP), Average = (First term + Last term) ÷ 2.

Average = (106 + 146) ÷ 2 = 252 ÷ 2 = 126.

∴ The correct answer is 126.

Alternate Method

Given: Range of numbers is between 104 and 148.

Formula: Average of n terms in AP = (a + l) ÷ 2, where a is the first term and l is the last term.

⇒ First even number (a) = 106

⇒ Last even number (l) = 146

⇒ Common difference (d) for even numbers = 2

⇒ Number of terms (n) = [(Last − First) ÷ d] + 1

⇒ n = [(146 − 106) ÷ 2] + 1 = 20 + 1 = 21

⇒ Average = (First term + Last term) ÷ 2

⇒ Average = (106 + 146) ÷ 2 = 252 ÷ 2 = 126

∴ The correct answer is 126.

Additional Information

Arithmetic Progression (AP) Property

The average of any number of terms in an AP is always equal to the mean of the first and last terms, or the middle term if n is odd.

Average of First n Even Numbers

The average of the first n consecutive even numbers (starting from 2) is given by the formula (n + 1).

Inclusive vs Exclusive

The word 'between' excludes the boundary values. If the question used 'from 104 to 148', 104 and 148 would be included as the first and last terms.

Shortcut Trick

Use the cross-multiplication method for Profit/Loss:

Buy: 4 items → ₹3 | Sell: 3 items → ₹4

CP = 3 × 3 = 9 units

SP = 4 × 4 = 16 units

Profit % = [(16 − 9) ÷ 9] × 100 = 7/9 × 100 ≈ 77.7%

∴ The correct answer is 77.7% profit.

Alternate Method

Given: CP of 4 bananas = ₹3, SP of 3 bananas = ₹4.

Formula Used: Profit % = [(Total SP − Total CP) ÷ Total CP] × 100

⇒ To compare, we make the number of bananas equal.

⇒ LCM of 4 and 3 is 12.

⇒ CP of 12 bananas = (3 ÷ 4) × 12 = ₹9

⇒ SP of 12 bananas = (4 ÷ 3) × 12 = ₹16

⇒ Profit = SP − CP = 16 − 9 = ₹7

⇒ Profit % = (Profit ÷ CP) × 100 = (7 ÷ 9) × 100

⇒ Profit % = 77.77...% ≈ 77.7%

∴ The correct answer is 77.7% profit.

Additional Information

Profit and Loss Basics

Profit % = (Profit/CP) × 100 and Loss % = (Loss/CP) × 100. Both are always calculated on the Cost Price unless specified otherwise.

Equalizing Quantity Method

In transaction problems, finding the LCM of the quantities bought and sold allows for a direct comparison of total costs and total revenues.

Relationship between CP, MP, and SP

Marked Price (MP) − Discount = Selling Price (SP). Also, CP × (1 + Profit%) = SP.

Shortcut Trick

Use successive profit multipliers: Initial CP × Multiplier₁ × Multiplier₂ = Final SP

Effective Multiplier = 1.08 × 1.12 = 1.2096

Wholesaler Cost = 8448 ÷ 1.2096

Wholesaler Cost ≈ 6984.12

∴ The correct answer is ₹6984.

Alternate Method

Given:

Profit percentage of wholesaler = 8%

Profit percentage of retailer = 12%

Price paid by customer (Final SP) = ₹8,448

Formula Used:

Final Value = Initial Value × (1 + r₁/100) × (1 + r₂/100)

Calculations:

⇒ Let the cost price for the wholesaler be CP.

⇒ Selling price of wholesaler (which is CP for retailer) = CP × (1 + 8/100) = 1.08 × CP

⇒ Selling price of retailer (which is price paid by customer) = (1.08 × CP) × (1 + 12/100)

⇒ 8448 = CP × 1.08 × 1.12

⇒ 8448 = CP × 1.2096

⇒ CP = 8448 ÷ 1.2096

⇒ CP ≈ 6984.126

∴ The correct answer is ₹6984.

Additional Information

Effective Percentage Change

When a value is changed by x% and then by y% successively, the total percentage change is [x + y + (xy/100)]%.

Profit and Loss Relation

Selling Price (SP) = Cost Price (CP) × (100 + Profit%)/100. This is the base for chain-rule calculations in multi-stage sales.

Successive Discounts

Similar to successive profits, for discounts of d₁ and d₂, the final price = MP × (1 − d₁/100) × (1 − d₂/100).

Shortcut Trick

Apply Alligation on profit percentages: (Higher % − Average %) : (Average % − Lower %)

Ratio of quantities = (16 − 14.2) : (14.2 − 10) = 1.8 : 4.2

Simplified ratio = 18 : 42 = 3 : 7

Total parts = 3 + 7 = 10 units → 2000 kg

Rice at 10% profit = (3 ÷ 10) × 2000 = 600 kg

∴ The correct answer is 600 kg.

Alternate Method

Given:

Total quantity = 2000 kg

Profit on part 1 (P₁) = 10%

Profit on part 2 (P₂) = 16%

Average profit (P_avg) = 14.2%

Calculations:

⇒ Let the quantity sold at 10% be Q₁ and at 16% be Q₂.

⇒ Q₁ : Q₂ = (16 − 14.2) : (14.2 − 10)

⇒ Q₁ : Q₂ = 1.8 : 4.2 = 3 : 7

⇒ Total ratio units = 3 + 7 = 10 units

⇒ 10 units = 2000 kg

⇒ 1 unit = 200 kg

⇒ Quantity sold at 10% (3 units) = 3 × 200 = 600 kg

∴ The correct answer is 600 kg.

Additional Information

Alligation Rule

It is used to find the ratio in which two ingredients at given prices/rates must be mixed to produce a mixture of a desired average price/rate. Ratio = (Cheaper Quantity / Dearer Quantity) = (Dearer Rate − Mean Rate) / (Mean Rate − Cheaper Rate).

Weighted Average Formula

Average Profit % = (n₁ × p₁ + n₂ × p₂) ÷ (n₁ + n₂), where n is quantity and p is profit percentage.

Successive Percentage Change

If a value changes by a% and then by b%, the effective change is (a + b + ab/100)%. This is distinct from alligation which deals with mixing parts.

Shortcut Trick The difference between the two selling prices corresponds to the total change from loss to profit. Difference in SP = 9.2 − 7.6 = 1.6 lakhs. Given, Profit = 3 × Loss. Thus, Ratio of Loss : Profit = 1 : 3. Total ratio parts = 1 + 3 = 4 parts. Value of 4 parts = 1.6 lakhs. Value of 1 part (Loss) = 1.6 ÷ 4 = 0.4 lakhs. Cost Price = SP1 + Loss = 7.6 + 0.4 = 8 lakhs. ∴ The correct answer is 8 lakhs. Alternate Method Given: Selling Price 1 (SP1) = Rs. 7.6 lakhs (Loss) Selling Price 2 (SP2) = Rs. 9.2 lakhs (Profit) Profit = 3 × Loss

Formula Used: Loss = CP − SP1 Profit = SP2 − CP Calculations: Let the loss incurred be x. ⇒ Profit = 3x ⇒ CP − 7.6 = x ... (i) ⇒ 9.2 − CP = 3x ... (ii) Substitute x from (i) into (ii): ⇒ 9.2 − CP = 3(CP − 7.6) ⇒ 9.2 − CP = 3CP − 22.8 ⇒ 9.2 + 22.8 = 3CP + CP ⇒ 32.0 = 4CP ⇒ CP = 32.0 ÷ 4 ⇒ CP = 8 lakhs ∴ The correct answer is 8 lakhs. Additional Information Profit and Loss Basics Profit occurs when SP > CP (Profit = SP − CP), and Loss occurs when CP > SP (Loss = CP − SP). Profit and Loss Percentage Profit% = (Profit ÷ CP) × 100 and Loss% = (Loss ÷ CP) × 100. Both are calculated on the Cost Price unless specified otherwise. Marked Price and Discount Discount = Marked Price (MP) − Selling Price (SP). Discount % = (Discount ÷ MP) × 100.

Shortcut Trick

Convert 25% directly to 1/4 to speed up calculation.

⇒ 25% of 1200 = 1200 ÷ 4 = 300

⇒ 0.09% of 300 = 0.09 × 3 = 0.27

∴ The correct answer is 0.27.

Alternate Method

Given:

The expression is 0.09% of 25% of 1200.

Concept Used:

a% of b = (a ÷ 100) × b

Calculations:

⇒ (0.09 ÷ 100) × (25 ÷ 100) × 1200

⇒ (9 ÷ 10000) × (1 ÷ 4) × 1200

⇒ (9 ÷ 10000) × 300

⇒ 2700 ÷ 10000

⇒ 0.27

∴ The correct answer is 0.27.

Additional Information

Fraction Equivalents of Percentages

Using fractions makes percentage calculations much faster. For instance, 25% = 1/4, 50% = 1/2, 75% = 3/4, 20% = 1/5.

BODMAS Rule

In simplification, 'of' denotes multiplication but must be solved before standard division and multiplication according to the BODMAS sequence.

Percentage Interchangeability

x% of y = y% of x. Depending on the values, swapping the numbers can make finding the percentage significantly easier to calculate.

Shortcut Trick

Let the original number be 100%.

The number becomes 116% of itself, so the increase is 16%.

According to the question, this 16% increase is equal to 20.

16% → 20

100% → (20 ÷ 16) × 100 = 1.25 × 100 = 125.

∴ The correct answer is 125.

Alternate Method

Given:

Increase in value = 20

New value = 116% of the original number

Calculations:

⇒ Let the original number be x.

⇒ According to the problem: x + 20 = 116% of x

⇒ x + 20 = (116 ÷ 100) × x

⇒ x + 20 = 1.16x

⇒ 1.16x − x = 20

⇒ 0.16x = 20

⇒ x = 20 ÷ 0.16

⇒ x = 2000 ÷ 16

⇒ x = 125

∴ The correct answer is 125.

Additional Information

Percentage Increase Formula

Percentage Increase = (Increase in Value ÷ Original Value) × 100.

Base Value Concept

In percentage problems, the original value is always considered 100% unless specified otherwise.

Fractional Equivalents

Knowing common fractions helps speed: 16% = 16/100 = 4/25. Thus, 4 units = 20, so 25 units = (20/4) × 25 = 125.

Shortcut Trick

Let the third number be 100.

First number = 50 (100 − 50) and Second number = 25 (100 − 75).

To make 25 equal to 50, we need to add 25.

Required % = (Amount to increase ÷ Second number) × 100 = (25 ÷ 25) × 100 = 100%.

∴ The correct answer is 100%.

Alternate Method

Given:

1st Number = 3rd Number − 50% of 3rd Number

2nd Number = 3rd Number − 75% of 3rd Number

Formula:

Percentage Increase = [(Difference) ÷ (Initial Value)] × 100

Calculations:

⇒ Let the third number be x = 100.

⇒ First number (A) = 100 − 50 = 50.

⇒ Second number (B) = 100 − 75 = 25.

⇒ Target is to make B = A, so we need to increase B by (A − B).

⇒ Required Increase = 50 − 25 = 25.

⇒ Percentage enhancement = (25 ÷ 25) × 100.

⇒ Percentage enhancement = 1 × 100 = 100%.

∴ The correct answer is 100%.

Additional Information

Base of Comparison

The denominator in a percentage calculation always represents the value we are comparing from. Here, we compare the increase with the second number, so 25 is the base.

Fractional Conversions

Quick calculations can be done using fractions: 50% = 1/2 and 75% = 3/4. If the third number is 4 units, the first is 2 units and the second is 1 unit.

Percentage Change Formula

% Change = [(New Value − Old Value) ÷ Old Value] × 100. A 100% increase means the value has doubled.

Shortcut Trick

Price ratio = 60 : 75 → 4 : 5.

Required expenditure = 110% of 4 = 4.4 units.

Initial consumption = 1 unit.

New consumption = 4.4 ÷ 5 = 0.88 units.

Reduction % = (1 − 0.88) × 100 = 12%.

∴ The correct answer is 12%.

Alternate Method

Given:

Initial Price (P₁) = ₹60/liter

New Price (P₂) = ₹75/liter

Expenditure Increase = 10%

Formula used:

Expenditure = Price × Consumption

⇒ Let the initial consumption be C₁ = 100 units.

⇒ Initial Expenditure (E₁) = 60 × 100 = 6000.

⇒ Target New Expenditure (E₂) = 6000 + (10% of 6000) = 6600.

⇒ New Consumption (C₂) = E₂ ÷ P₂ = 6600 ÷ 75.

⇒ C₂ = 88 units.

⇒ Reduction in consumption = Initial − New = 100 − 88 = 12 units.

⇒ Percentage reduction = (12 ÷ 100) × 100 = 12%.

∴ The correct answer is 12%.

Additional Information

Constant Expenditure Rule

If the expenditure remains constant, the price and consumption are inversely proportional (C ∝ 1/P).

Net Percentage Change Formula

% Change in Expenditure = % Change in Price + % Change in Consumption + (%P × %C)/100.

Base Change for Reduction

If Price increases by x% and Expenditure is to stay constant, Consumption must reduce by [x/(100+x)] × 100%.

Shortcut Trick

Average speed in kmph = 900 ÷ 25 = 36 kmph.

Direct conversion (kmph to m/s): Speed × 5/18.

Speed in m/s = 36 × (5/18) = 2 × 5 = 10 m/s.

∴ The correct answer is 10 m/s.

Alternate Method

Given:

Total Distance = 900 km

Total Time = 25 hours

Formula Used:

Average Speed = Total Distance ÷ Total Time

Conversion: 1 kmph = 5/18 m/s

Calculations:

⇒ Average speed = 900 ÷ 25

⇒ Average speed = 36 kmph

⇒ Speed in m/s = 36 × (5/18)

⇒ Speed = 2 × 5

⇒ Speed = 10 m/s

∴ The correct answer is 10 m/s.

Additional Information

Unit Conversion Rules

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

Average Speed

Average Speed = Total Distance ÷ Total Time. If a journey is split into two equal distances with speeds x and y, the average speed is 2xy ÷ (x + y).

Relative Speed

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

Shortcut Trick

When the distance covered is the same for two different speeds, use the Harmonic Mean formula.

Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂)

Average Speed = (2 × 18 × 30) ÷ (18 + 30)

Average Speed = 1080 ÷ 48 = 22.5 km/hr

∴ The correct answer is 22.5 km/hr.

Alternate Method

Given:

Upstream Speed (U) = 18 km/hr

Downstream Speed (D) = 30 km/hr

Formula: Average Speed = Total Distance ÷ Total Time

⇒ Let the distance for one side be the LCM of speeds (18, 30) = 90 km.

⇒ Time taken for Upstream (T₁) = Distance ÷ Upstream Speed = 90 ÷ 18 = 5 hours

⇒ Time taken for Downstream (T₂) = Distance ÷ Downstream Speed = 90 ÷ 30 = 3 hours

⇒ Total Distance = 90 + 90 = 180 km

⇒ Total Time = 5 + 3 = 8 hours

⇒ Average Speed = 180 ÷ 8

⇒ Average Speed = 22.5 km/hr

∴ The correct answer is 22.5 km/hr.

Additional Information

Speed of Boat in Still Water

It is calculated as the average of the downstream and upstream speeds: B = (D + U) ÷ 2.

Speed of the Stream

It is calculated as half the difference between downstream and upstream speeds: S = (D − U) ÷ 2.

Downstream and Upstream Relation

Downstream Speed = Speed of Boat + Speed of Current | Upstream Speed = Speed of Boat − Speed of Current.

Average Speed (General Case)

If distances d₁, d₂ are different, Average Speed = (d₁ + d₂) ÷ (t₁ + t₂).

Shortcut Trick

Relative speed of cyclists moving towards each other = Sum of their speeds.

Relative Speed = 21 + 24 = 45 km/hr.

Time = 1 hour 12 minutes = 1.2 hours.

Initial Distance = Relative Speed × Time = 45 × 1.2 = 54 km.

∴ The correct answer is 54 km.

Alternate Method

Given:

Speed of cyclist A = 21 km/hr

Speed of cyclist B = 24 km/hr

Time of meeting = 1 hour 12 minutes

Formula Used:

Distance = Relative Speed × Time

Relative Speed (Opposite Direction) = S₁ + S₂

Calculations:

⇒ Convert time into hours: 1 hr 12 min = 1 + 12/60 = 1 + 1/5 = 6/5 hours

⇒ Time = 1.2 hours

⇒ Relative Speed = 21 + 24 = 45 km/hr

⇒ Total Distance = Relative Speed × Time

⇒ Distance = 45 × 1.2

⇒ Distance = 54 km

∴ The correct answer is 54 km.

Additional Information

Relative Speed in Same Direction

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

Conversion of Units

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

Time of Meeting

Time taken to meet = Initial distance between objects ÷ Relative Speed.

Shortcut Trick

Stoppage time per hour = (Difference in speeds ÷ Speed excluding stoppages) × 60 minutes

Difference in speeds = 60 km/h − 45 km/h = 15 km/h

Stoppage time = (15 ÷ 60) × 60 minutes

Stoppage time = 15 minutes

∴ The correct answer is 15 minutes.

Alternate Method

Given:

Speed of bus (Excluding stoppages) = 60 km/h

Speed of bus (Including stoppages) = 45 km/h

Formula Used:

Stoppage time per hour = (Distance lost due to stoppage ÷ Original speed) × 60

Calculations:

⇒ Distance covered in 1 hour without stoppage = 60 km

⇒ Distance covered in 1 hour with stoppage = 45 km

⇒ Distance not covered due to stoppage = 60 km − 45 km = 15 km

⇒ Time taken to cover this 15 km at 60 km/h = Distance ÷ Speed

⇒ Time = 15 ÷ 60 hours

⇒ Time in minutes = (15 ÷ 60) × 60

⇒ Time = 15 minutes

∴ The correct answer is 15 minutes.

Additional Information

General Stoppage Rule

Rest time per hour = (Faster Speed − Slower Speed) ÷ Faster Speed. Multiply by 60 to get the value in minutes.

Average Speed

Average Speed = Total Distance ÷ Total Time. Including stoppages reduces the average speed because the total time increases while the distance remains the same.

Unit Conversion

To convert km/h to m/s, multiply by 5/18. To convert hours to minutes, multiply by 60.

Shortcut Trick

Annual Simple Interest Rate = 15% ÷ 3 years = 5% p.a.

Effective Compound Interest rate for 3 years at 5% = 15.7625%.

CI = 15 lakhs × 15.7625% = 2.364375 lakhs.

Approximate value = ₹2.36 lakhs.

∴ The correct answer is ₹2.36 lakhs.

Alternate Method

Given: Principal (P) = ₹15 lakhs, Time (T) = 3 years, SI increase = 15%.

Formula used: CI = P[(1 + R/100)^T − 1]

⇒ Simple Interest Rate (R) = 15% ÷ 3 = 5% p.a.

⇒ Amount (A) = 15 × (1 + 5/100)³

⇒ A = 15 × (1.05)³

⇒ A = 15 × 1.157625 = 17.364375 lakhs.

⇒ CI = Amount − Principal = 17.364375 − 15

⇒ CI = 2.364375 lakhs ≈ ₹2.36 lakhs.

∴ The correct answer is ₹2.36 lakhs.

Additional Information

Successive Rate Formula

For two years, effective CI rate is [x + y + (xy/100)]%. For 5%, it is 5 + 5 + 0.25 = 10.25%.

CI-SI Difference for 3 Years

The difference between CI and SI for 3 years is given by P(R/100)² × (3 + R/100).

3-Year Ratio Method

Compound Interest for 3 years can be calculated using the ratio 3 : 3 : 1 applied to interests on interests.

Shortcut Trick

Interest earned in the 3rd year = ₹10,272 − ₹9,600 = ₹672.

In compound interest, the amount at the end of the 2nd year acts as the principal for the 3rd year.

Rate % = (Interest ÷ Principal) × 100

Rate % = (672 ÷ 9,600) × 100 = 7%

∴ The correct answer is 7%.

Alternate Method

Given:

Amount after 2 years (A₂) = ₹9,600

Amount after 3 years (A₃) = ₹10,272

Formula Used:

A = P(1 + R/100)ⁿ

For consecutive years: A_n+1 = A_n(1 + R/100)

Calculations:

⇒ A₃ = A₂ × (1 + R/100)

⇒ 10,272 = 9,600 × (1 + R/100)

⇒ 1 + R/100 = 10,272 ÷ 9,600

⇒ 1 + R/100 = 1.07

⇒ R/100 = 1.07 − 1

⇒ R/100 = 0.07

⇒ R = 7

∴ The correct answer is 7%.

Additional Information

Successive Year Interest

In compound interest, the difference between the amounts of two consecutive years is the interest on the amount of the preceding year. R = [(A_n+1 − A_n) ÷ A_n] × 100.

Compound Interest Formula

The general formula for amount is A = P(1 + R/100)ⁿ, where P is principal, R is rate per annum, and n is time in years.

SI vs CI for 2 Years

The difference between Compound Interest and Simple Interest for 2 years is given by D = P × (R/100)².

Shortcut Trick

Net effective interest rate for 2 years = 5 + 6 + (5 × 6) ÷ 100 = 11.3%

Total amount percentage = 100% (Principal) + 11.3% (Interest) = 111.3%

Given: 111.3% of x = Rs 24,486

Value of x = 24,486 ÷ 1.113

Value of x = 22,000

∴ The correct answer is Rs 22000.

Alternate Method

Given:

Principal = Rs x

Rate for year 1 (R₁) = 5%

Rate for year 2 (R₂) = 6%

Amount (A) after 2 years = Rs 24,486

Formula Used:

Amount (A) = P × (1 + R₁/100) × (1 + R₂/100)

Calculations:

⇒ 24,486 = x × (1 + 5/100) × (1 + 6/100)

⇒ 24,486 = x × (105/100) × (106/100)

⇒ 24,486 = x × (21/20) × (53/50)

⇒ 24,486 = x × (1113/1000)

⇒ x = (24,486 × 1000) ÷ 1113

⇒ x = 22 × 1000

⇒ x = 22,000

∴ The correct answer is Rs 22000.

Additional Information

Net Effective Rate for Successive Years

For two successive years with interest rates r₁% and r₂%, the combined effective interest rate is (r₁ + r₂ + r₁r₂/100)%.

General Amount Formula for CI

When interest rates are different for each year, Amount = P × (1 + r₁/100) × (1 + r₂/100) × ... × (1 + r_n/100).

Compound Interest (CI)

Compound Interest is calculated on the principal plus any accumulated interest from previous periods, given by CI = Amount − Principal.

Shortcut Trick

Effective rate for annual compounding = 40%

For quarterly compounding: Rate = 40% ÷ 4 = 10% and Time = 4 quarters

Effective rate for 4 periods of 10% = 46.41% (Using successive percentage formula)

Difference in interest = (46.41% − 40%) of ₹10,000 = 6.41% of 10,000

Difference = 0.0641 × 10,000 = ₹641

∴ The correct answer is ₹641.

Alternate Method

Given: Principal (P) = ₹10,000, Rate (R) = 40% p.a., Time (T) = 1 year.

Formula Used: Amount (A) = P(1 + r/100)ⁿ; CI = A − P

Case 1: Compounded Annually

⇒ CI₁ = P × R × T / 100 = 10,000 × 40 × 1 / 100

⇒ CI₁ = ₹4,000

Case 2: Compounded Quarterly

⇒ Quarterly Rate (r) = 40 ÷ 4 = 10% per quarter

⇒ Number of quarters (n) = 1 × 4 = 4

⇒ Amount (A) = 10,000 × (1 + 10/100)⁴

⇒ A = 10,000 × (1.1)⁴ = 10,000 × 1.4641

⇒ A = ₹14,641

⇒ CI₂ = 14,641 − 10,000 = ₹4,641

Difference in CI:

⇒ Difference = CI₂ − CI₁ = 4,641 − 4,000

⇒ Difference = ₹641

∴ The correct answer is ₹641.

Additional Information

Effective Interest Rate

The effective annual rate increases with the frequency of compounding. For n periods at rate r, the effective rate is [(1 + r/100)ⁿ − 1] × 100%.

Compounding Frequency Rules

Half-yearly: Rate = R/2, Time = 2T. Quarterly: Rate = R/4, Time = 4T. Monthly: Rate = R/12, Time = 12T.

Successive Percentage Formula

For two successive changes of a% and b%, the net change is [a + b + (ab/100)]%. This can be used iteratively for compound interest.

Shortcut Trick

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

Here, x = 40% and y = 40%.

Equivalent Discount = (40 + 40 − (40 × 40)/100)%

Equivalent Discount = (80 − 1600/100)% = (80 − 16)% = 64%

∴ The correct answer is 64%.

Alternate Method

Given:

First Discount (D₁) = 40%

Second Discount (D₂) = 40%

Formula Used:

Final Price = Marked Price × (1 − D₁/100) × (1 − D₂/100)

Single Discount % = [(Marked Price − Final Price) / Marked Price] × 100

Calculations:

⇒ Let the Marked Price (MP) = ₹100

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

⇒ Price after 2nd discount = 60 − (40% of 60) = 60 − 24 = ₹36

⇒ Total Discount = MP − Final Selling Price = 100 − 36 = 64

⇒ Equivalent Discount Percentage = (64 / 100) × 100 = 64%

∴ The correct answer is 64%.

Additional Information

Successive Discount Formula

For two successive discounts of a% and b%, the effective discount is (a + b − ab/100)%. This is always less than the sum of individual discounts.

Three Successive Discounts

For three discounts a%, b%, and c%, the net discount is [1 − (1 − a/100)(1 − b/100)(1 − c/100)] × 100%. Alternatively, apply the two-discount formula twice.

Fraction Method for Discounts

Convert percentage to fractions (e.g., 40% = 2/5). If the initial value is 5, after a 2/5 discount, the value becomes 3. Successive discounts are calculated by multiplying the ratios: (3/5) × (3/5) = 9/25. Discount = (25 − 9)/25 = 16/25 = 64%.

Shortcut Trick

Ratio of MP : SP = 25,000 : 18,000 = 25 : 18

Discount units = 25 − 18 = 7 units

Discount % = (7 ÷ 25) × 100

Discount % = 7 × 4 = 28%

∴ The correct answer is 28%.

Alternate Method

Given:

Marked Price (MP) = ₹25,000

Selling Price (SP) = ₹18,000

Formula Used:

Discount = MP − SP

Discount % = (Discount ÷ MP) × 100

Calculations:

⇒ Discount = 25,000 − 18,000

⇒ Discount = ₹7,000

⇒ Discount % = (7,000 ÷ 25,000) × 100

⇒ Discount % = (7 ÷ 25) × 100

⇒ Discount % = 28%

∴ The correct answer is 28%.

Additional Information

Marked Price (MP)

It is also known as the List Price or Tag Price. It is the price printed on the label of an item.

Selling Price (SP)

The actual price at which a product is sold after deducting the discount from the Marked Price. SP = MP − Discount.

Relation between CP and MP

MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%). This is useful for finding the markup percentage.

Shortcut Trick

Marked Price (MP) = ₹10,000

Price after first 12% discount = 88% of 10,000 = ₹8,800

Final Selling Price (SP) = ₹8,360

Additional discount amount = 8,800 − 8,360 = ₹440

Additional discount % = (440 ÷ 8,800) × 100 = 5%

∴ The correct answer is 5%.

Alternate Method

Given:

Marked Price (MP) = ₹10,000

First discount (d₁) = 12%

Final Selling Price (SP) = ₹8,360

Formula Used:

Selling Price = MP × (100 − d₁)/100 × (100 − d₂)/100

Calculations:

⇒ Price after 1st discount = 10,000 × (100 − 12) ÷ 100

⇒ Price after 1st discount = 10,000 × 0.88 = ₹8,800

⇒ Let additional discount be x%.

⇒ 8,360 = 8,800 × (100 − x) ÷ 100

⇒ 8,360 ÷ 88 = 100 − x

⇒ 95 = 100 − x

⇒ x = 100 − 95

⇒ x = 5%

∴ The correct answer is 5%.

Additional Information

Successive Discount Formula

If two successive discounts are a% and b%, the equivalent single discount is given by (a + b − ab/100)%.

Marked Price vs Selling Price

Marked Price (MP) is the tag price, and Selling Price (SP) = MP − Discount. Discount is always calculated on the MP unless stated otherwise.

Effective Discount Rate

The effective discount rate can be calculated as [(Initial MP − Final SP) ÷ Initial MP] × 100.