Quantitative Aptitude PYQs

Shortcut Trick

Total Marked Price = 25 × 3,000 = ₹75,000

Successive Discount % = 10 + 5 − (10 × 5) ÷ 100 = 14.5%

Net Multiplier = 100% − 14.5% = 85.5% = 0.855

Final Amount = 75,000 × 0.855 = ₹64,125

∴ The correct answer is ₹64,125.

Alternate Method

Given:

Marked Price (MP) per item = ₹3,000

Quantity = 25 items (More than 20 items)

Discounts = 10% (primary) and an additional 5% (successive)

Calculations:

⇒ Total MP = 25 × 3,000 = ₹75,000

⇒ Since quantity is 25 (which is > 20), both 10% and 5% discounts apply.

⇒ Price after 10% Discount = 75,000 − (10% of 75,000)

⇒ Price after 10% Discount = 75,000 − 7,500 = ₹67,500

⇒ Price after Additional 5% Discount = 67,500 − (5% of 67,500)

⇒ Final Amount = 67,500 − 3,375

⇒ Final Amount = ₹64,125

∴ The correct answer is ₹64,125.

Additional Information

Successive Discount Formula

When two discounts d₁ and d₂ are applied successively, the single equivalent discount is calculated as (d₁ + d₂ − (d₁ × d₂) ÷ 100)%.

Net Selling Price Formula

SP = MP × (100 − d₁)/100 × (100 − d₂)/100. This is useful for calculating the final price directly after multiple discounts.

Marked Price vs. Selling Price

Marked Price is the price printed on the item, while Selling Price is the final amount paid after deducting all applicable discounts from the MP.

Shortcut Trick

Total Area = Number of Tiles × (Base × Height)

Total Area = 500 × (8 × 6)

Total Area = 500 × 48 = 24,000 cm²

∴ The correct answer is 24,000 cm².

Alternate Method

Given:

Total identical tiles = 500

Base (b) of each parallelogram-shaped tile = 8 cm

Height (h) of each parallelogram-shaped tile = 6 cm

Formula used:

Area of a parallelogram = Base × Height

Calculations:

⇒ Area of one tile = 8 cm × 6 cm

⇒ Area of one tile = 48 cm²

⇒ Total area of the floor = Number of tiles × Area of one tile

⇒ Total area = 500 × 48 cm²

⇒ Total area = 24,000 cm²

∴ The correct answer is 24,000 cm².

Additional Information

Area of Parallelogram

The area is determined by multiplying the length of the base by the perpendicular height (A = b × h).

Perimeter of Parallelogram

The perimeter is the total length of the boundary, calculated as P = 2 × (a + b), where 'a' and 'b' are adjacent sides.

Properties of Parallelogram

Opposite sides are parallel and equal, opposite angles are equal, and diagonals bisect each other.

Shortcut Trick

Total surface area of a sphere = 4πr².

If 25% is masked, then the Painted Area = 75% of Total Surface Area.

Painted Area = 75% × 4π(10)² = 0.75 × 400π = 300π cm².

∴ The correct answer is 300π cm².

Alternate Method

Given:

Radius of the sphere (r) = 10 cm

Masked percentage = 25%

Formula Used:

Total Surface Area (TSA) of a sphere = 4πr²

Painted Area = TSA − Masked Area = TSA × (100% − Masked %)

Calculations:

⇒ Total Surface Area (TSA) = 4 × π × (10)²

⇒ TSA = 4 × π × 100 = 400π cm²

⇒ Masked Area = 25% of 400π = 0.25 × 400π = 100π cm²

⇒ Painted Area = 400π − 100π

⇒ Painted Area = 300π cm²

∴ The correct answer is 300π cm².

Additional Information

Volume of a Sphere

The amount of space occupied by the sphere is given by Volume = (4/3)πr³.

Hemisphere Surface Area

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr².

Surface Area to Volume Ratio

For a sphere, the ratio is (4πr²) ÷ (4/3πr³) = 3 ÷ r.

Shortcut Trick

Total weighted parts = 3 + 2 + 5 = 10 units.

Total required weighted score = 68 × 10 = 680.

Contribution of Physics and Chemistry = (3 × 60) + (2 × 70) = 180 + 140 = 320.

Required contribution from Mathematics = 680 − 320 = 360.

Mathematics marks = 360 ÷ 5 = 72.

∴ The correct answer is 72.

Alternate Method

Given:

Ratio of weights (w₁ : w₂ : w₃) = 3 : 2 : 5

Marks in Physics (x₁) = 60, Marks in Chemistry (x₂) = 70

Weighted Average (A_w) = 68

Formula Used:

Weighted Average (A_w) = (w₁x₁ + w₂x₂ + w₃x₃) ÷ (w₁ + w₂ + w₃)

Calculations:

⇒ 68 = (3 × 60 + 2 × 70 + 5 × x₃) ÷ (3 + 2 + 5)

⇒ 68 = (180 + 140 + 5x₃) ÷ 10

⇒ 68 × 10 = 320 + 5x₃

⇒ 680 = 320 + 5x₃

⇒ 5x₃ = 680 − 320

⇒ 5x₃ = 360

⇒ x₃ = 360 ÷ 5

⇒ x₃ = 72

∴ The correct answer is 72.

Additional Information

Weighted Average Concept

Unlike a simple average where each value contributes equally, a weighted average assigns specific importance (weights) to each value. The sum of the products of weights and values is divided by the sum of the weights.

Net Deviation Method

The sum of weighted deviations from the weighted average is always zero. Σw_i(x_i − A_w) = 0. This can be used to find missing values by balancing positive and negative deviations.

Arithmetic Mean vs Weighted Mean

Arithmetic Mean is a special case of Weighted Mean where all weights are equal (e.g., w₁ = w₂ = w₃ = 1).

Shortcut Trick

Net percentage gap between 10% loss and 5% profit = 10% + 5% = 15%.

According to the question, this 15% difference corresponds to ₹450.

⇒ 15% = ₹450

⇒ 1% = ₹30

⇒ Cost Price (100%) = 30 × 100 = ₹3,000.

∴ The correct answer is ₹3,000.

Alternate Method

Given: Initial Loss = 10%, Final Profit = 5%, Price Increase = ₹450.

⇒ Let the Cost Price (CP) of the watch be ₹x.

⇒ Initial Selling Price (SP₁) = CP − 10% of CP = 0.90x

⇒ New Selling Price (SP₂) = CP + 5% of CP = 1.05x

⇒ According to the problem: SP₂ − SP₁ = 450

⇒ 1.05x − 0.90x = 450

⇒ 0.15x = 450

⇒ x = 450 ÷ 0.15

⇒ x = 45000 ÷ 15

⇒ x = 3000

∴ The correct answer is ₹3,000.

Additional Information

Profit and Loss Percentage

Profit % is always calculated on Cost Price unless stated otherwise. Formula: Profit% = (Profit ÷ CP) × 100.

Marked Price and Discount

Discount is the reduction offered on the Marked Price (MP). SP = MP − Discount.

Successive Percentage Shortcut

For two successive changes of a% and b%, net change = (a + b + ab/100)%.

Shortcut Trick

Average speed = Total distance ÷ Total time

Total distance = 120 + 120 + 100 = 340 km

Total time = (120 ÷ 40) + (120 ÷ 60) + (100 ÷ 80) = 3 + 2 + 1.25 = 6.25 hours

Average speed = 340 ÷ 6.25 = 54.4 km/h

∴ The correct answer is 54.4 km/h.

Alternate Method

Given:

First segment: Distance (d₁) = 120 km, Speed (s₁) = 40 km/h

Second segment: Distance (d₂) = 120 km, Speed (s₂) = 60 km/h

Third segment: Distance (d₃) = 100 km, Speed (s₃) = 80 km/h

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Time for first part (t₁) = 120 ÷ 40 = 3 hours

⇒ Time for second part (t₂) = 120 ÷ 60 = 2 hours

⇒ Time for third part (t₃) = 100 ÷ 80 = 1.25 hours

⇒ Total Distance = 120 + 120 + 100 = 340 km

⇒ Total Time = 3 + 2 + 1.25 = 6.25 hours

⇒ Average Speed = 340 ÷ 6.25 = 54.4 km/h

∴ The correct answer is 54.4 km/h.

Additional Information

Average Speed for Equal Distances

If a certain distance is covered at s₁ speed and the same distance is covered at s₂ speed, the average speed is (2 × s₁ × s₂) ÷ (s₁ + s₂).

Speed Unit Conversion

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

Relative Speed

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

Shortcut Trick

Use the direct distance formula: d = h(cot θ₁ − cot θ₂), where d is the distance moved toward the tower.

Here, d = 100 m, θ₁ = 30°, and θ₂ = 60°.

⇒ 100 = h(cot 30° − cot 60°)

⇒ 100 = h(√3 − 1/√3) ⇒ 100 = h(2/√3)

⇒ h = (100 × √3) / 2 = 50√3 m.

∴ The correct answer is 50√3 m.

Alternate Method

Given: Angle of elevation at point P = 30°, Angle at point Q = 60°, distance PQ = 100 m.

In right ▵ABQ:

⇒ tan 60° = AB/BQ

⇒ √3 = h/BQ ⇒ BQ = h/√3

In right ▵ABP:

⇒ tan 30° = AB/BP

⇒ 1/√3 = h/(BQ + 100)

⇒ BQ + 100 = h√3

Substituting the value of BQ:

⇒ h/√3 + 100 = h√3

⇒ 100 = h√3 − h/√3

⇒ 100 = (3h − h) / √3

⇒ 100 = 2h / √3

⇒ 2h = 100√3

⇒ h = 50√3 m.

∴ The correct answer is 50√3 m.

Additional Information

Trigonometric Ratios

In a right-angled triangle, tan θ = Perpendicular/Base and cot θ = Base/Perpendicular.

Special Angles

Standard values: tan 30° = 1/√3, tan 60° = √3, and tan 45° = 1.

Angle of Elevation

The angle formed by the line of sight with the horizontal when looking upward at an object.

Shortcut Trick

Effective Compound Interest rate for 2 years at 10% = 10 + 10 + (10 × 10)/100 = 21%.

Given 21% of the Principal (y) = ₹630.

Total Amount payable = Principal (100%) + Interest (21%) = 121%.

If 21% → ₹630, then 1% → ₹30.

121% → 121 × 30 = ₹3,630.

∴ The correct answer is ₹3,630.

Alternate Method

Given: Compound Interest (CI) = ₹630, Rate (R) = 10% p.a., Time (T) = 2 years, Principal = ₹y.

Formula: Amount (A) = P(1 + R/100)^T and CI = A − P

⇒ CI = y[(1 + 10/100)² − 1]

⇒ 630 = y[(11/10)² − 1]

⇒ 630 = y[121/100 − 1]

⇒ 630 = y[21/100]

⇒ y = (630 × 100) ÷ 21

⇒ y = 30 × 100 = ₹3,000

⇒ Total Amount (A) = Principal (y) + CI

⇒ Total Amount (A) = 3000 + 630 = ₹3,630

∴ The correct answer is ₹3,630.

Additional Information

Effective Rate of Interest

For a rate R% compounded annually for 2 years, the effective rate is R + R + (R×R)/100 %.

CI and SI Difference (2 Years)

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

Compound Interest Formula

Amount A = P(1 + r/n)^nt, where n is the number of times interest is compounded per year.

Given:

x = 119^0.49

y = 119^0.62

x^z = y⁶

Formula used:

(a^m)ⁿ = a^{m × n}

Calculation:

Substitute the values of x and y into the given relation:

⇒ (119^0.49)^z = (119^0.62)⁶

⇒ 119^{0.49 × z} = 119^{0.62 × 6}

Since the bases are the same, we can equate the exponents:

⇒ 0.49 × z = 0.62 × 6

⇒ 0.49 × z = 3.72

⇒ z = 3.72 ÷ 0.49

⇒ z ≈ 7.59

∴ The correct answer is option (2).

Shortcut Trick

The difference between a 2-digit number and its reverse is always 9 × (difference of digits).

Given difference = 27 ⇒ 9 × (Digit Difference) = 27 ⇒ Digit Difference = 3.

Sum of digits = 9 (Given). Let digits be x and y.

Larger digit = (9 + 3) ÷ 2 = 6 | Smaller digit = (9 − 3) ÷ 2 = 3.

Since the reversed number exceeds the original, the units digit must be larger ⇒ Number = 36.

∴ The correct answer is 36.

Alternate Method

Given:

Sum of the digits of a 2-digit number = 9

Difference between interchanged number and original number = 27

Formula Used:

A 2-digit number with tens digit x and units digit y is represented as: 10x + y.

Calculations:

⇒ Let the tens digit be x and the units digit be y.

⇒ Original number = 10x + y

⇒ x + y = 9 → (Equation 1)

⇒ Number obtained by interchanging the digits = 10y + x

⇒ (10y + x) − (10x + y) = 27

⇒ 9y − 9x = 27

⇒ y − x = 3 → (Equation 2)

⇒ Adding Equation 1 and Equation 2:

⇒ (x + y) + (y − x) = 9 + 3

⇒ 2y = 12 ⇒ y = 6

⇒ Substituting y = 6 in Equation 1: x + 6 = 9 ⇒ x = 3

⇒ Original Number = 10(3) + 6 = 36

∴ The correct answer is 36.

Additional Information

Number Reversal Property

The difference between a 2-digit number and its reverse is always a multiple of 9, specifically 9 × |tens digit − units digit|.

Sum of Reversed Numbers

The sum of a 2-digit number and its reverse is always a multiple of 11, specifically 11 × (tens digit + units digit).

Place Value Concept

In any n-digit number, the value is the sum of (digit × 10^position−1). For a 2-digit number, it is 10a + b.

Shortcut Trick

Use the successive percentage formula: x + y + (xy / 100)

Here, x = 18% and y = 22%

Total Increase = 18 + 22 + (18 × 22 / 100)

Total Increase = 40 + 396 / 100 = 40 + 3.96 = 43.96%

∴ The correct answer is 43.96%.

Alternate Method

Given: Increase in 2021 = 18%; Increase in 2022 = 22%.

Assume the initial number of electric vehicles was 100.

⇒ Number after 2021 increase = 100 + (18% of 100) = 118

⇒ Number after 2022 increase = 118 + (22% of 118)

⇒ 22% of 118 = (22 × 118) / 100 = 25.96

⇒ Final number of vehicles = 118 + 25.96 = 143.96

⇒ Total percentage increase = 143.96 − 100 = 43.96%

∴ The correct answer is 43.96%.

Additional Information

Successive Percentage Formula

For two successive changes of a% and b%, the net change is (a + b + ab/100)%. Use positive values for increase and negative values for decrease.

Multiplying Factor Method

Final Value = Initial Value × (1 + r₁/100) × (1 + r₂/100). For 18% and 22% increases, the factors are 1.18 and 1.22 respectively.

Three Successive Changes

If there is a third change c%, calculate the net of a and b first, then use that result as a new 'a' with the third value 'c'.

Shortcut Trick

Decrease in population = 8,000 − 7,200 = 800

Fractional decrease = 800 ÷ 8,000 = 1/10

Percentage Decrease = 1/10 × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Initial Population of the town = 8,000

Final Population after one year = 7,200

Formula Used: Percentage Decrease = [(Original Value − New Value) ÷ Original Value] × 100

⇒ Total Decrease = 8,000 − 7,200 = 800

⇒ Percentage Decrease = (800 ÷ 8,000) × 100

⇒ Percentage Decrease = (1 ÷ 10) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Percentage Change Rule

Percentage change is always calculated on the Initial Value (Base Value). Formula: [(Difference) ÷ (Initial Value)] × 100.

Successive Percentage Change

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

Population Growth Formula

If population grows at a constant rate r% annually, Population after n years = P × (1 + r/100)ⁿ.

Given:

((22/5))¹⁸ × ((5/22))⁸ × ((22/5))⁶ = ((5/22))^9m+4

Formula used:

(a/b)ⁿ = (b/a)⁻ⁿ

a^x × a^y = a^x+y

Calculation:

To find the value of m, we first equate the bases on the left-hand side to match the right-hand side base of (5/22).

Using the reciprocal rule (a/b)ⁿ = (b/a)⁻ⁿ:

⇒ ((5/22))⁻¹⁸ × ((5/22))⁸ × ((5/22))⁻⁶ = ((5/22))^9m+4

Using the law of exponents (a^x × a^y = a^x+y) to combine the terms:

⇒ ((5/22))^{(−18 + 8 − 6)} = ((5/22))^9m+4

⇒ ((5/22))⁻¹⁶ = ((5/22))^9m+4

Since the bases are now equal, we compare the exponents:

⇒ −16 = 9m + 4

⇒ 9m = −16 − 4

⇒ 9m = −20

⇒ m = −20/9

∴ The correct answer is option (4).

Shortcut Trick

Efficiency ratio of X and Y = 15 : 20 = 3 : 4.

Total work = 20 × 3 = 60 units.

Work done by X and Y in 5 days = 5 × (3 + 4) = 35 units.

Remaining work = 60 − 35 = 25 units.

Time taken by Y to finish = 25 ÷ 4 = 6.25 days.

∴ The correct answer is 6.25 days.

Alternate Method

Given:

Time taken by X = 20 days

Time taken by Y = 15 days

Time worked together = 5 days

Formula used:

Efficiency = Total Work ÷ Time Taken

Calculations:

⇒ Total efficiency of X and Y = 3 + 4 = 7 units/day

⇒ Work completed by X and Y in 5 days = 7 × 5 = 35 units

⇒ Remaining work after X leaves = 60 − 35 = 25 units

⇒ Time taken by Y to complete 25 units = 25 ÷ 4

⇒ Time = 6.25 days

∴ The correct answer is 6.25 days.

Additional Information

Pipes and Cisterns

This concept follows the same principles as time and work, where an inlet pipe has positive efficiency and an outlet pipe has negative efficiency.

Work and Wages

Wages are distributed among workers in the ratio of the total work done by each, which is (Efficiency × Time).

Alternate Day Work

In such problems, calculate the work done in one full cycle (e.g., 2 days) and then find the total cycles needed to approach the total work.

Shortcut Trick

For a number to be divisible by 11, the difference between the sum of digits at odd positions and even positions must be 0 or a multiple of 11.

Sum of digits at odd places (7+2+3+2) = 14

Sum of digits at even places (9+4+p+5) = 18 + p

Difference = (18 + p) − 14 = p + 4

For divisibility, p + 4 = 11 ⇒ p = 7.

∴ The correct answer is 7.

Alternate Method

Given: The 8-digit number is 79243p25 and it is divisible by 11.

Formula Used: A number is divisible by 11 if |(Sum of digits at odd places) − (Sum of digits at even places)| is 0, 11, 22, etc.

Calculations:

⇒ Sum of digits at odd places = 7 + 2 + 3 + 2 = 14

⇒ Sum of digits at even places = 9 + 4 + p + 5 = 18 + p

⇒ Difference = (Sum of even places) − (Sum of odd places)

⇒ Difference = (18 + p) − 14 = p + 4

⇒ For the number to be divisible by 11, (p + 4) must be a multiple of 11.

⇒ Since p is a single digit (0−9), the only possible multiple is 11.

⇒ p + 4 = 11

⇒ p = 11 − 4 = 7

∴ The correct answer is 7.

Additional Information

Divisibility by 3 and 9

A number is divisible by 3 (or 9) if the sum of all its digits is divisible by 3 (or 9).

Divisibility by 8

A number is divisible by 8 if its last three digits are divisible by 8.

Divisibility by 11 (Alternative)

Subtract the last digit from the remaining number. If the result is divisible by 11, the original number is too.

Shortcut Trick

Net change in sum = New value − Old value = 20 − 10 = +10

Net change in mean = Net change in sum ÷ Total count = 10 ÷ 5 = +2

Since the result is positive, the mean increases by the calculated value.

∴ The correct answer is It increases by 2.

Alternate Method

Given: Total numbers (n) = 5, Initial Arithmetic Mean = 20.

Formula Used: Total Sum = Mean × Number of observations

⇒ Initial total sum = 20 × 5 = 100

⇒ One number changes from 10 to 20.

⇒ New total sum = 100 − 10 + 20 = 110

⇒ New Arithmetic Mean = 110 ÷ 5 = 22

⇒ Difference in mean = 22 − 20 = +2

∴ The correct answer is It increases by 2.

Additional Information

Properties of Arithmetic Mean

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

Effect of Replacement

When a value x is replaced by y, the change in the mean is given by (y − x) ÷ n, where n is the total number of observations.

Weighted Average

The combined mean of two groups with sizes n₁, n₂ and means x₁, x₂ is (n₁x₁ + n₂x₂) ÷ (n₁ + n₂).

Shortcut Trick

Since the filling rate of the second tap and the emptying rate of the third tap are identical (both 17 hr), they cancel each other out.

Net filling rate = 1/4 + 1/17 − 1/17 = 1/4 of the cistern per hour.

Time to fill the full cistern = 4 hours.

Time to fill one-fourth of the cistern = 1/4 × 4 = 1 hour.

∴ The correct answer is 1 hr.

Alternate Method

Given: Filling time for Tap A = 4 hr, Tap B = 17 hr, and emptying time for Tap C = 17 hr.

Total capacity (Work) = LCM of (4, 17, 17) = 68 units.

⇒ Total net efficiency = 17 + 4 − 4 = 17 units/hr.

⇒ Required work (one-fourth of cistern) = 68 ÷ 4 = 17 units.

⇒ Time required = Required Work ÷ Net Efficiency

⇒ Time = 17 ÷ 17 = 1 hour.

∴ The correct answer is 1 hr.

Additional Information

Net Efficiency in Pipes and Cisterns

The net rate of filling is the sum of the rates of inlet pipes minus the sum of the rates of outlet pipes.

Relationship between Time and Work

Total Work = Efficiency × Time. For a cistern, work is represented by the volume or capacity, often calculated as the LCM of time intervals.

Concept of Negative Work

An emptying tap performs negative work. If a tap empties a tank in 'n' hours, its efficiency is taken as −1/n per hour.

Shortcut Trick

LCM of numbers = (LCM of ratio parts) × HCF

Ratio of numbers = 3 : 2 : 11

LCM of ratio parts (3, 2, 11) = 3 × 2 × 11 = 66 (since they are co-prime)

66 units = 3564

1 unit (HCF) = 3564 ÷ 66 = 54

∴ The correct answer is 54.

Alternate Method

Given:

Ratio of three numbers = 3 : 2 : 11

LCM of the numbers = 3564

Formula Used:

If the numbers are ax, bx, and cx, where x is the HCF, then their LCM = x × LCM(a, b, c).

Calculations:

Let the three numbers be 3x, 2x, and 11x, where x is the HCF.

⇒ LCM(3x, 2x, 11x) = 3564

⇒ x × LCM(3, 2, 11) = 3564

⇒ x × (3 × 2 × 11) = 3564

⇒ 66x = 3564

⇒ x = 3564 ÷ 66

⇒ x = 54

∴ The correct answer is 54.

Additional Information

HCF and LCM Relation

For any two numbers A and B, the product of the numbers is equal to the product of their HCF and LCM: A × B = HCF × LCM.

Co-prime Numbers

If a set of numbers are co-prime (meaning they have no common factors other than 1), their HCF is 1 and their LCM is the product of the numbers.

HCF and LCM Properties

The HCF of a group of numbers is always a factor of their LCM. HCF is the largest number that divides all given numbers exactly.

Shortcut Trick

Combine the ratios by equating the common term B: A : B : C = (15 × 14) : (17 × 14) : (17 × 19) = 210 : 238 : 323.

Total units = 210 + 238 + 323 = 771 units.

Given 771 units = 1542 runs ⇒ 1 unit = 2 runs.

A's score = 210 × 2 = 420.

∴ The correct answer is 420.

Alternate Method

Given:

Ratio A : B = 15 : 17

Ratio B : C = 14 : 19

Total runs scored = 1542

Formula Used:

If A : B = x : y and B : C = p : q, then A : B : C = (x × p) : (y × p) : (y × q)

Calculations:

⇒ To merge the ratios, find the LCM of B's values (17 and 14), which is 238.

⇒ A : B = (15 × 14) : (17 × 14) = 210 : 238

⇒ B : C = (14 × 17) : (19 × 17) = 238 : 323

⇒ Combined Ratio A : B : C = 210 : 238 : 323

⇒ Sum of parts = 210 + 238 + 323 = 771

⇒ Value of 1 part = 1542 ÷ 771 = 2

⇒ Runs scored by A = 210 × 2 = 420

∴ The correct answer is 420.

Additional Information

Merging Ratios

To combine ratios like A:B and B:C, the common variable (B) must have the same value in both. Multiply each ratio by the value of B from the other ratio.

Unitary Method in Ratios

Once the total number of parts is known, the value of one part is calculated as (Total Value ÷ Total Parts). Individual shares are then (Individual Parts × Value of 1 part).

Inverse Ratio

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

Shortcut Trick

Check the sum of the bases: 22 + (−16) + (−6) = 22 − 22 = 0.

Apply the algebraic identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.

Result = 3 × 22 × (−16) × (−6) = 66 × 96 = 6336.

∴ The correct answer is 6336.

Alternate Method

Given: The expression to evaluate is 22³ + (−16)³ + (−6)³.

Formula Used: If a + b + c = 0, then the identity a³ + b³ + c³ = 3abc holds true.

⇒ Assign values: a = 22, b = −16, and c = −6.

⇒ Verify the condition: a + b + c = 22 + (−16) + (−6) = 22 − 22 = 0.

⇒ Substitute into the identity: 22³ + (−16)³ + (−6)³ = 3 × 22 × (−16) × (−6).

⇒ Multiply the constants: 3 × 22 = 66 and (−16) × (−6) = 96.

⇒ Final calculation: 66 × 96 = 66 × (100 − 4) = 6600 − 264 = 6336.

∴ The correct answer is 6336.

Additional Information

General Algebraic Identity

The complete identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

Special Conditions

The expression a³ + b³ + c³ = 3abc occurs when either a + b + c = 0 or a = b = c.

Negative Base Powers

Since the exponent 3 is an odd number, the cube of a negative number remains negative, i.e., (−x)³ = −x³.