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Side of square = √9 = 3 cm ⇒ Perimeter = 4 × 3 = 12 cm.

Side of regular hexagon = 12 ÷ 6 = 2 cm.

Area of hexagon = 6 × (Area of equilateral triangle with side 2 cm).

Area = 6 × (√3 ÷ 4 × 2²) = 6 × √3 = 6√3 cm².

∴ The correct answer is 6√3 cm².

Alternate Method

Given:

Area of square = 9 cm²

Perimeter of square = Perimeter of regular hexagon

Formula Used:

1. Area of square = a² (where a = side)
2. Perimeter of square = 4a
3. Perimeter of regular hexagon = 6s (where s = side)
4. Area of regular hexagon = (3√3 ÷ 2) × s²

Calculations:

⇒ a² = 9 cm²

⇒ a = √9 = 3 cm

⇒ Perimeter of square = 4 × 3 = 12 cm

⇒ Perimeter of hexagon = 6s = 12 cm

⇒ s = 12 ÷ 6 = 2 cm

⇒ Area of hexagon = (3√3 ÷ 2) × 2²

⇒ Area of hexagon = (3√3 ÷ 2) × 4

⇒ Area of hexagon = 3√3 × 2 = 6√3 cm²

∴ The correct answer is 6√3 cm².

Additional Information

Regular Hexagon Area

A regular hexagon is composed of 6 equilateral triangles. Area = 6 × (√3/4 × s²) = (3√3/2)s².

Internal Angles

Each internal angle of a regular hexagon is 120° and the sum of all internal angles is 720°.

Circumradius and Inradius

For a regular hexagon of side s, the circumradius (R) = s and the inradius (r) = (√3/2)s.

Shortcut Trick

Difference in percentage = 84% − 45% = 39%

Total difference in marks = 62 (pass) + 35.5 (fail) = 97.5 marks

Since 39% corresponds to 97.5, then 1% = 97.5 ÷ 39 = 2.5

Maximum Marks (100%) = 2.5 × 100 = 250

∴ The correct answer is 250.

Alternate Method

Given:

Case 1: Score = 84%, Result = Passed by 62 marks

Case 2: Score = 45%, Result = Failed by 35.5 marks

Formula Used:

Passing Marks = (Score% × Max Marks) − Pass Margin

Passing Marks = (Score% × Max Marks) + Fail Margin

Calculations:

⇒ Let Maximum Marks be x.

⇒ Passing Marks = 0.84x − 62

⇒ Passing Marks = 0.45x + 35.5

⇒ 0.84x − 62 = 0.45x + 35.5

⇒ 0.84x − 0.45x = 35.5 + 62

⇒ 0.39x = 97.5

⇒ x = 97.5 ÷ 0.39

⇒ x = 250

∴ The correct answer is 250.

Additional Information

Passing Percentage

The minimum score required to pass. Passing % = (Passing Marks ÷ Maximum Marks) × 100.

Percentage to Absolute Value

To find the total value when a percentage difference is known: Total = (Difference in values ÷ Difference in %) × 100.

Passing Marks Relation

Obtained Marks + Failing Marks = Passing Marks = Obtained Marks − Passing Margin.

Shortcut Trick

Sum of ages of husband and wife at marriage = 28 × 2 = 56 years.

Sum of ages 2 years later (when daughter was born) = 56 + (2 × 2) + 0 = 60 years.

Present total age of 3 family members = 24 × 3 = 72 years.

Increase in total age since daughter's birth = 72 − 60 = 12 years.

Years passed since daughter's birth = 12 ÷ 3 = 4 years.

Total years since marriage = 2 (before birth) + 4 (after birth) = 6 years.

∴ The correct answer is 6 years.

Alternate Method

Given:

Average age of husband and wife at marriage = 28 years

Time after marriage when daughter was born = 2 years

Present average age of 3 members = 24 years

Formula Used:

Total Age = Average Age × Number of persons

Calculations:

⇒ Sum of ages of husband (H) and wife (W) at marriage = 28 × 2 = 56

⇒ Let they got married 'x' years ago.

⇒ Present age of H + W = 56 + 2x

⇒ Present age of daughter (born 2 years after marriage) = x − 2

⇒ Present total sum = (56 + 2x) + (x − 2)

⇒ 24 × 3 = 54 + 3x

⇒ 72 = 54 + 3x

⇒ 3x = 72 − 54

⇒ 3x = 18

⇒ x = 6

∴ The correct answer is 6 years.

Additional Information

Average Age Concept

If the current average age of n people is A, then after k years, the average age will be A + k.

Sum of Ages

Total sum of ages increases by (Number of people × Number of years) over a period of time.

Birth in Average Problems

When a child is born, the number of members in the family increases, which immediately affects the denominator in the average calculation.

Shortcut Trick

Profit Ratio = Capital × Time

Ratio of (Anusha : Minal) = (55,000 × 12) : (Capital_M × 5) = 8 : 3

⇒ (11,000 × 12) : (Capital_M) = 8 : 3

⇒ 1,32,000 / Capital_M = 8 / 3

⇒ Capital_M = (1,32,000 × 3) / 8 = 16,500 × 3 = 49,500

∴ The correct answer is ₹49,500.

Alternate Method

Given:

Anusha's Capital (C_A) = ₹55,000

Anusha's Time (T_A) = 12 months

Minal's Time (T_M) = 12 − 7 = 5 months

Profit Ratio (P_A : P_M) = 8 : 3

Formula Used:

Ratio of Profits = (Capital₁ × Time₁) / (Capital₂ × Time₂)

Calculations:

⇒ Let Minal's capital be ₹x.

⇒ P_A / P_M = (C_A × T_A) / (C_M × T_M)

⇒ 8 / 3 = (55,000 × 12) / (x × 5)

⇒ 8 / 3 = 6,60,000 / 5x

⇒ 8 × 5x = 6,60,000 × 3

⇒ 40x = 19,80,000

⇒ x = 19,80,000 / 40

⇒ x = 49,500

∴ The correct answer is ₹49,500.

Additional Information

Partnership Basic Rule

Profit is shared in the ratio of the product of Investment and Time Period: P1 : P2 = C1 × T1 : C2 × T2.

Compound Partnership

When partners invest different amounts for different time periods, it is called a compound partnership.

Working Partner

A partner who manages the business is a working partner and often receives a salary or fixed percentage of profit before the remaining profit is divided.

Shortcut Trick

Net Efficiency of both pipes = 9 (Filling) − 4 (Emptying) = 5 units/min.

Total Capacity (LCM of 20, 45) = 180 units.

Work to fill one-fourth of the tank = 180 ÷ 4 = 45 units.

Time taken = 45 ÷ 5 = 9 min.

∴ The correct answer is 9 min.

Alternate Method

Given:

Time taken by Filling Pipe = 20 min

Time taken by Emptying Pipe = 45 min

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Total capacity of the tank = LCM(20, 45) = 180 units.

⇒ Net efficiency when both pipes work together = 9 − 4 = 5 units/min.

⇒ Required part of the tank to be filled = 1/4 × 180 = 45 units.

⇒ Time taken = Required Capacity ÷ Net Efficiency

⇒ Time taken = 45 ÷ 5

⇒ Time taken = 9 min.

∴ The correct answer is 9 min.

Additional Information

Pipes & Cisterns Basics

Filling pipes are considered as positive work (+), while emptying pipes (leaks) are considered as negative work (−).

Net Work Done

If pipe A fills in 'x' hours and pipe B empties in 'y' hours, net work in 1 hour = (1/x − 1/y).

Time and Efficiency Relationship

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

Shortcut Trick

Total divisors for any prime power pⁿ is (n + 1).

Option 1 & 3: p³ and q³ ⇒ 3 + 1 = 4 divisors.

Option 2: p × q ⇒ (1 + 1) × (1 + 1) = 4 divisors.

Option 4: p² ⇒ 2 + 1 = 3 divisors.

∴ The correct answer is p² for prime p.

Alternate Method

Given: Four expressions where p and q are prime numbers.

Formula Used: If a number is written in its prime factored form N = a^m × bⁿ, the total number of positive divisors is (m + 1) × (n + 1).

⇒ For p³: The divisors are 1, p, p², p³. Total = 4.

⇒ For p × q: The divisors are 1, p, q, pq. Total = 4.

⇒ For q³: Similar to p³, the divisors are 1, q, q², q³. Total = 4.

⇒ For p²: The divisors are 1, p, p². Total = 3.

⇒ Therefore, p² has 3 divisors, which is not equal to 4.

∴ The correct answer is p² for prime p.

Additional Information

Divisor Rule for Powers

If N = p^a, where p is prime, the number of factors is always (a + 1). For example, 2² = 4 has (2 + 1) = 3 factors (1, 2, 4).

Perfect Squares Property

Every perfect square has an odd number of distinct factors, while non-perfect squares always have an even number of factors.

Prime Product Property

The product of two distinct prime numbers (p × q) always results in exactly 4 divisors: {1, p, q, pq}.

Shortcut Trick

The mode is the value that appears most frequently in a data set.

Frequency of 55 → 4 times (at positions 1, 4, 11, 17)

Frequency of 44 → 3 times (at positions 5, 9, 13)

Frequency of 52, 45, and 54 → 2 times each

Frequency of 48, 50, 49, 43, and 47 → 1 time each

Highest frequency is 4, which belongs to the number 55.

∴ The correct answer is 55.

Alternate Method

Given data set: 55, 48, 50, 55, 44, 52, 45, 52, 44, 49, 55, 54, 44, 54, 43, 45, 55, 47

Formula: Mode = Observation with the maximum frequency.

Frequency Distribution Value: 55 Freq: 4 (MAX) Value: 44 Freq: 3 Value: 52/45 Freq: 2 Mode = 55 (Highest occurring value)

⇒ Counting the occurrences of each unique number in the list:

⇒ 43: 1 time | 44: 3 times | 45: 2 times | 47: 1 time | 48: 1 time

⇒ 49: 1 time | 50: 1 time | 52: 2 times | 54: 2 times | 55: 4 times

⇒ The value 55 occurs 4 times, which is more than any other value.

⇒ Therefore, Mode = 55.

∴ The correct answer is 55.

Additional Information

Empirical Formula

For a moderately skewed distribution, the relationship is: Mode = 3 × Median − 2 × Mean.

Types of Data Distributions

Unimodal has one mode, Bimodal has two modes, and Multimodal has more than two modes. If no value repeats, there is no mode.

Central Tendency Measures

Mean is the average, Median is the middle value after sorting, and Mode is the most frequent value.

Shortcut Trick

Total interest percentage for 2 years = Rate × Time = 5% × 2 = 10%.

Given that 10% of Principal = ₹720.

To find the sum of money (100%), we multiply by 10.

Principal = 720 × 10 = ₹7200.

∴ The correct answer is ₹7200.

Alternate Method

Given:

Simple Interest (SI) = ₹720

Time (T) = 2 years

Rate (R) = 5% per annum

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ 720 = (P × 5 × 2) ÷ 100

⇒ 720 = (P × 10) ÷ 100

⇒ 720 = P ÷ 10

⇒ P = 720 × 10

⇒ P = ₹7200

∴ The correct answer is ₹7200.

Additional Information

Amount in Simple Interest

The total Amount (A) is the sum of the Principal (P) and the Simple Interest (SI). Formula: A = P + SI = P(1 + RT/100).

Rate of Interest Calculation

If the sum, interest, and time are known, the Rate (R) can be calculated as R = (SI × 100) ÷ (P × T).

Constant Interest Property

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

Shortcut Trick

Total interest percentage for 9 months at 9% per annum = 9 × (9 ÷ 12) = 6.75%.

Simple Interest = 6.75% of ₹28,260.

⇒ 28260 × 0.0675 = 1907.55.

∴ The correct answer is ₹1907.55.

Alternate Method

Given:

Principal (P) = ₹28,260

Rate of interest (R) = 9% per annum

Time period (T) = 9 months

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ T in years = 9 ÷ 12 = 3/4 years

⇒ SI = (28260 × 9 × 3/4) ÷ 100

⇒ SI = (28260 × 27) ÷ 400

⇒ SI = 763020 ÷ 400

⇒ SI = 1907.55

∴ The correct answer is ₹1907.55.

Additional Information

Total Amount Formula

The total amount (A) to be returned after the time period is the sum of the Principal and the Interest: A = P + SI.

Time Conversion for Interest

When time is given in months, divide by 12 to convert to years. When given in days, divide by 365 (standard year) or 366 (leap year).

Effective Rate of Interest

Effective Rate = R × T. In this case, 9% × 0.75 = 6.75% is the total interest percentage applied to the principal.

Shortcut Trick

SP of Article X = 1300 + 734 = 2034.

Effective selling price factor for X = (1 − 0.84) × (1 − 0.25) = 0.16 × 0.75 = 0.12 of MP.

Selling price for Y = (1 − 0.87) = 0.13 of MP.

Since 12 units = 2034, then 13 units = (2034 ÷ 12) × 13 = 169.5 × 13 = 2203.5.

Profit on Article Y = 2203.5 − 1300 = 903.5.

Profit % = 903.5 ÷ 13 = 69.5%.

∴ The correct answer is 69.5%.

Alternate Method

Given:

Cost Price (CP) of Article X and Article Y = ₹1300 each.

Successive discounts on X = 84% and 25%.

Profit on X = ₹734.

Discount on Y = 87%.

Formula Used:

1. Selling Price (SP) = CP + Profit
2. SP = MP × (1 − d₁/100) × (1 − d₂/100)
3. Profit % = (Profit / CP) × 100

Calculations:

⇒ SP of Article X = 1300 + 734 = ₹2034.

⇒ Let the Marked Price (MP) be M.

⇒ M × (1 − 84/100) × (1 − 25/100) = 2034

⇒ M × 0.16 × 0.75 = 2034

⇒ M × 0.12 = 2034

⇒ M = 2034 / 0.12 = ₹16950.

⇒ SP of Article Y = MP × (1 − 87/100) = 16950 × 0.13

⇒ SP of Article Y = ₹2203.5.

⇒ Profit on Article Y = 2203.5 − 1300 = ₹903.5.

⇒ Profit % = (903.5 / 1300) × 100 = 69.5%.

∴ The correct answer is 69.5%.

Additional Information

Successive Discounts Formula

When two successive discounts of a% and b% are given, the single equivalent discount is calculated as (a + b − ab/100)%.

Markup and Discount Relation

Profit % can be related to markup and discount as: (100 + P)/(100 − D) = MP/CP, where P is profit % and D is discount %.

Successive Percentage Change

For any quantity changing by x% and then y% successively, the net percentage change is given by (x + y + xy/100)%.