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Given:

9² + √4² − 10 × √81 − 15

Formula used:

BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Calculations:

9² + (√4)² − 10 × √81 − 15

⇒ 81 + 4 − 10 × 9 − 15

⇒ 81 + 4 − 90 − 15

⇒ 85 − 105

⇒ −20

∴ The correct answer is option (3).

Shortcut Trick

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

Efficiency: A = 3 units/day, B = 1 unit/day.

Work done by (A + B) in 2 days = (3 + 1) × 2 = 8 units.

Remaining work = 36 − 8 = 28 units.

Time taken by B to complete it = 28 ÷ 1 = 28 days.

∴ The correct answer is 28 days.

Alternate Method

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

Formula Used: Efficiency = Total Work ÷ Time; Time = Work ÷ Efficiency

⇒ Total efficiency of A and B = 3 + 1 = 4 units/day.

⇒ Work completed by both in 2 days = 4 × 2 = 8 units.

⇒ Remaining work to be done by B = 36 − 8 = 28 units.

⇒ Days taken by B = 28 ÷ 1 = 28 days.

∴ The correct answer is 28 days.

Additional Information

Work and Wages

Wages are always distributed in the ratio of work done by the individuals, which is usually the ratio of their efficiencies if they work for the same duration.

Pipes and Cisterns

The concept is identical to Time and Work, where an inlet pipe does positive work and an outlet pipe (leak) does negative work.

Combined Efficiency

If multiple people work together, their net efficiency is the sum of their individual efficiencies (E_total = E₁ + E₂ + ...).

Shortcut Trick

Remainder = 42

Divisor = 3 × 42 = 126

Quotient = 126 ÷ 9 = 14

Dividend = (126 × 14) + 42 = 1764 + 42 = 1806

∴ The correct answer is 1806.

Alternate Method

Given: The remainder is 42. The divisor is 9 times the quotient and 3 times the remainder.

Formula Used: Dividend = (Divisor × Quotient) + Remainder

⇒ Remainder = 42

⇒ Divisor = 3 × 42 = 126

⇒ Quotient = 126 ÷ 9 = 14

⇒ Dividend = (126 × 14) + 42

⇒ Dividend = 1764 + 42 = 1806

∴ The correct answer is 1806.

Additional Information

Division Algorithm

For any two integers a and b (b ≠ 0), there exist unique integers q and r such that a = bq + r, where 0 ≤ r < |b|.

Remainder Rule

The remainder must always be non-negative and strictly smaller than the divisor.

Verification of Division

To verify a division result, compute (Divisor × Quotient) + Remainder; the sum must equal the dividend.

Shortcut Trick

Ratio of quantities sold: 50 : (30% of 50) : (50 − 15) ⇒ 50 : 15 : 35 ⇒ 10 : 3 : 7.

Net Profit x% = [(10 × 37) + (3 × 26) − (7 × 24)] ÷ (10 + 3 + 7)

x% = [370 + 78 − 168] ÷ 20 ⇒ 280 ÷ 20 = 14%.

∴ The correct answer is 14%.

Alternate Method

Given: 50% goods at 37% profit, 30% of remaining at 26% profit, rest at 24% loss.

Assume total quantity = 100 units and Cost Price (CP) of each unit = ₹1.

⇒ Total Cost Price = ₹100.

⇒ Case 1: 50 units sold at 37% profit ⇒ Profit₁ = 50 × 37/100 = ₹18.5.

⇒ Remaining units = 100 − 50 = 50 units.

⇒ Case 2: 30% of 50 = 15 units sold at 26% profit ⇒ Profit₂ = 15 × 26/100 = ₹3.9.

⇒ Case 3: Remaining units = 50 − 15 = 35 units sold at 24% loss ⇒ Loss₃ = 35 × 24/100 = ₹8.4.

⇒ Total Net Profit = Profit₁ + Profit₂ − Loss₃

⇒ Total Net Profit = 18.5 + 3.9 − 8.4 = ₹14.

⇒ Overall Profit Percentage (x) = (Total Profit ÷ Total CP) × 100

⇒ x = (14 ÷ 100) × 100 = 14%.

∴ The correct answer is 14%.

Additional Information

Weighted Average Formula

Net Profit/Loss% = (n₁p₁ + n₂p₂ + ... + n_kp_k) ÷ (n₁ + n₂ + ... + n_k), where n is quantity and p is profit(+)/loss(−) percentage.

Net Profit Percentage

Overall Profit % = [(Total Selling Price − Total Cost Price) ÷ Total Cost Price] × 100.

Successive Percentage

If a value is changed by a% and then by b%, the net change is [a + b + (a × b)/100] %.

Shortcut Trick

For consecutive years in Compound Interest, the Amount of the previous year acts as the Principal for the subsequent year.

Interest for the 3^rd year = ₹28,800 − ₹24,000 = ₹4,800.

Rate of Interest (R) = (Interest ÷ Principal for the year) × 100

R = (4,800 ÷ 24,000) × 100 = 1/5 × 100 = 20%.

∴ The correct answer is 20%.

Alternate Method

Given:

Amount after 2 years (A₂) = ₹24,000

Amount after 3 years (A₃) = ₹28,800

Formula Used:

A = P(1 + R ÷ 100)ⁿ

Calculations:

⇒ A₂ = P(1 + R/100)² = 24000

⇒ A₃ = P(1 + R/100)³ = 28800

⇒ A₃ ÷ A₂ = [P(1 + R/100)³] ÷ [P(1 + R/100)²]

⇒ 28800 ÷ 24000 = (1 + R/100)

⇒ 288 ÷ 240 = 1 + R/100

⇒ 1.2 = 1 + R/100

⇒ R/100 = 1.2 − 1 = 0.2

⇒ R = 0.2 × 100 = 20%

∴ The correct answer is 20%.

Additional Information

Successive Growth in Compound Interest

When time periods are consecutive, the annual rate of interest can be calculated as [(A_n+1 − A_n) ÷ A_n] × 100.

Effective Interest Rate

If interest is compounded n times a year, the effective annual rate is given by R_eff = [1 + (R ÷ n)]ⁿ − 1.

Difference between SI and CI

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

Shortcut Trick

Identify the excess cost: 6 Naveen Pens cost ₹15 more each than Rajat Pens.

Total excess cost = 6 × ₹15 = ₹90.

Subtract excess from total: ₹540 − ₹90 = ₹450 (This represents the cost if all 10 pens were Rajat Pens).

Cost of one Rajat Pen = ₹450 ÷ 10 = ₹45.

∴ The correct answer is ₹45.

Alternate Method

Given:

Total pens = 4 Rajat + 6 Naveen

Total Cost Price = ₹540

Price difference = Naveen Pen − Rajat Pen = ₹15

Let the cost of one Rajat Pen be ₹R.

Then, the cost of one Naveen Pen = ₹(R + 15).

⇒ 4 × R + 6 × (R + 15) = 540

⇒ 4R + 6R + 90 = 540

⇒ 10R + 90 = 540

⇒ 10R = 540 − 90

⇒ 10R = 450

⇒ R = 450 ÷ 10

⇒ R = 45

∴ The correct answer is ₹45.

Additional Information

Linear Equation in One Variable

An equation of the form ax + b = c, where a, b, and c are constants. It is solved by isolating the variable x on one side.

Substitution Method

Expressing one variable in terms of another (e.g., N = R + 15) and substituting it into the main equation to reduce the number of unknowns.

Cost Price (CP)

The total amount paid by a buyer to acquire an item. Total CP = (Quantity 1 × Rate 1) + (Quantity 2 × Rate 2).

Shortcut Trick

Direct application of the Time-Speed-Distance relationship:

Time = Distance ÷ Speed

Time = 150 ÷ 50 = 3 hr

∴ The correct answer is 3 hr.

Alternate Method

Given:

Total Distance (D) = 150 km

Speed (S) = 50 km/h

Formula Used:

Time (T) = Distance ÷ Speed

Calculations:

⇒ Time taken = 150 ÷ 50 ⇒ Time taken = 3 hr

∴ The correct answer is 3 hr.

Additional Information

Fundamental Formulas

Distance = Speed × Time | Speed = Distance ÷ Time.

Unit Conversion

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

Average Speed

For a trip with different speeds over equal distances, Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Shortcut Trick

Arrange the monthly incomes in ascending order: 35, 37, 38, 39, 40, 41, 42, 44, 46, 50.

Count of observations (n) = 10 (which is even).

Median = Average of the middle two terms (5^th and 6^th terms).

Median = (40 + 41) ÷ 2 = 40.5.

∴ The correct answer is 40.5.

Alternate Method

Given: Monthly incomes of 10 employees (in ₹ thousands) are 35, 40, 38, 50, 46, 42, 39, 37, 41, and 44.

Formula Used:

For an even number of observations (n):

Median = [(n/2)^th observation + (n/2 + 1)^th observation] ÷ 2

⇒ Total observations (n) = 10

⇒ Arrange incomes in ascending order: 35, 37, 38, 39, 40, 41, 42, 44, 46, 50

⇒ n/2 = 10/2 = 5^th term

⇒ n/2 + 1 = 10/2 + 1 = 6^th term

⇒ 5^th term = 40 and 6^th term = 41

⇒ Median = (40 + 41) ÷ 2

⇒ Median = 81 ÷ 2 = 40.5

∴ The correct answer is 40.5.

Additional Information

Median for Odd Number of Terms

If n is odd, the Median is simply the middle-most value, calculated as the ((n + 1) ÷ 2)^th term after sorting.

Empirical Relationship

For moderately skewed distributions: Mode = 3 × Median − 2 × Mean.

Effect of Outliers

The Median is a measure of central tendency that is robust against outliers, unlike the arithmetic mean which is heavily influenced by extreme values.

Shortcut Trick

When a large sphere is melted into n identical small spheres, the radius ratio is R/r = n^1/3.

For n = 27, R/r = 27^1/3 = 3.

Ratio of Surface Area = R² : (n × r²) = (R/r)² : n.

Required Ratio = 3² : 27 = 9 : 27 = 1 : 3.

∴ The correct answer is 1:3.

Alternate Method

Given: Radius of the larger metallic sphere (R) = 6 units, Total number of smaller spheres (n) = 27.

Formula Used: Volume of sphere = (4/3)πr³; Surface area of sphere = 4πr².

⇒ Total Volume remains constant: Volume of Large Sphere = 27 × Volume of Small Sphere

⇒ (4/3)π × 6³ = 27 × (4/3)π × r³

⇒ 216 = 27 × r³

⇒ r³ = 216 / 27 = 8

⇒ r = √³8 = 2 units

⇒ Surface Area of Large Sphere (SA_L) = 4π × R² = 4π × 6² = 144π

⇒ Total Surface Area of 27 Small Spheres (SA_{S_total}) = 27 × (4π × r²)

⇒ SA_{S_total} = 27 × 4π × 2² = 27 × 16π = 432π

⇒ Required Ratio = SA_L : SA_{S_total} = 144π : 432π

⇒ Ratio = 144 / 432 = 1 / 3

∴ The correct answer is 1:3.

Additional Information

Volume Conservation

In melting and recasting problems, the initial volume of the object always equals the final total volume of all new objects formed.

Surface Area to Volume Relationship

If the radius of a sphere is scaled by a factor k, its surface area scales by k² and its volume scales by k³.

Surface Area of Hemisphere

For a sphere cut in half, the curved surface area is 2πr² and the total surface area is 3πr².

Shortcut Trick

Ratio 4 years ago (Teacher : Daughter) = 3 : 1 (Difference = 2 units)

Ratio 8 years hence (Teacher : Daughter) = 2 : 1 → 4 : 2 (Difference = 2 units)

Increase in ratio units = 4 − 3 = 1 unit ⇒ Total time gap = 4 + 8 = 12 years.

1 unit = 12 years ∴ Teacher's age 4 years ago = 3 × 12 = 36 years.

Present age of teacher = 36 + 4 = 40 years.

∴ The correct answer is 40 years.

Alternate Method

Given: 4 years ago, the teacher was 3 times her daughter's age. 8 years later, she will be twice her daughter's age.

Let the daughter's age 4 years ago be x.

Then, the teacher's age 4 years ago = 3x.

⇒ Present age of teacher = 3x + 4

⇒ Present age of daughter = x + 4

According to the second condition (8 years from now):

⇒ (3x + 4 + 8) = 2 × (x + 4 + 8)

⇒ 3x + 12 = 2 × (x + 12)

⇒ 3x + 12 = 2x + 24

⇒ 3x − 2x = 24 − 12

⇒ x = 12

Present age of teacher = 3 × 12 + 4 = 36 + 4 = 40 years.

∴ The correct answer is 40 years.

Additional Information

Constant Age Difference

The numerical difference between the ages of two individuals always remains constant, regardless of how many years pass.

Ratio Balancing Method

When solving with ratios, multiply the ratios by the difference of the other ratio's parts to make the unit difference uniform across time periods.

Linear Equation Application

For multiple conditions, use the present age as the base variable (x) to avoid confusion when moving backward or forward in time.