Quantitative Aptitude PYQs

Shortcut Trick

Speed of 2^nd train = 300 ÷ 3 = 100 km/h.

Ratio of speeds (Train 1 : Train 2) = 5 : 8.

Since 8 units = 100 km/h ⇒ 1 unit = 12.5 km/h.

Speed of 1^st train = 5 units = 5 × 12.5 = 62.5 km/h.

∴ The correct answer is 62.5 km/h.

Alternate Method

Given: Ratio of speeds = 5 : 8 | Distance (Train 2) = 300 km | Time (Train 2) = 3 hours.

Formula Used: Speed = Distance ÷ Time

⇒ Speed of 2^nd train = Distance ÷ Time

⇒ Speed of 2^nd train = 300 ÷ 3 = 100 km/h.

⇒ Let the speed of the 1^st train be 5x and the 2^nd train be 8x.

⇒ 8x = 100

⇒ x = 100 ÷ 8 = 12.5

⇒ Speed of 1^st train = 5x = 5 × 12.5 = 62.5 km/h.

∴ The correct answer is 62.5 km/h.

Additional Information

Speed-Distance Relationship

When time is constant, Speed is directly proportional to Distance (S₁/S₂ = D₁/D₂).

Unit Conversion

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

Relative Speed

For two objects moving in the same direction, Relative Speed = S₁ − S₂. In opposite directions, Relative Speed = S₁ + S₂.

Shortcut Trick

Identify the mode by finding the observation with the highest frequency.

Number 2 appears 4 times, 3 appears 3 times, 4 appears 2 times, and 5 appears 1 time.

The value 2 occurs the maximum number of times.

∴ The correct answer is 2 hr.

Alternate Method

Given: Data points for 10 students = 2, 3, 4, 2, 5, 3, 2, 4, 2, 3.

Formula: Mode is the value that appears most frequently in a data set.

To find the mode, we construct a frequency table:

⇒ Data value 2: Frequency = 4

⇒ Data value 3: Frequency = 3

⇒ Data value 4: Frequency = 2

⇒ Data value 5: Frequency = 1

⇒ Comparing frequencies: 4 > 3 > 2 > 1

⇒ Since the value 2 hr occurs with the maximum frequency of 4.

∴ The correct answer is 2 hr.

Additional Information

Arithmetic Mean

The average of the data set calculated as (Sum of all observations) ÷ (Total number of observations).

Median

The middle value of a data set when arranged in ascending or descending order. For even n, it is the mean of (n/2)^th and (n/2 + 1)^th terms.

Empirical Relationship

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

Bimodal and Multimodal

A data set with two modes is called bimodal, and one with more than two modes is called multimodal.

Shortcut Trick

If two numbers x and y are given, their Third proportional is calculated as y² ÷ x.

Here, let x = a² − b² and y = a + b.

Third proportional = (a + b)² ÷ (a² − b²)

Applying identity a² − b² = (a − b)(a + b), we get (a + b)² ÷ ((a − b)(a + b)).

Simplified result = (a + b) / (a − b).

∴ The correct answer is ((a+b))/((a-b)).

Alternate Method

Given:

First term = a² − b²

Second term = a + b

Condition: a ≠ b

We know that if x, y, z are in Continued Proportion, then:

x ÷ y = y ÷ z

⇒ z = y² ÷ x

Substitute the given expressions:

⇒ z = (a + b)² ÷ (a² − b²)

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

⇒ z = ((a + b) × (a + b)) ÷ ((a + b) × (a − b))

Canceling the common term (a + b) from the numerator and denominator:

⇒ z = (a + b) ÷ (a − b)

∴ The correct answer is ((a+b))/((a-b)).

Additional Information

Mean Proportional

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

Fourth Proportional

If four quantities a, b, c, d are in proportion (a:b :: c:d), then the fourth proportional d is (b × c) ÷ a.

Continued Proportion

Three quantities a, b, c are in continued proportion if a/b = b/c, which implies b² = ac.

Shortcut Trick

To find the Ratio of Boys to Teachers, equate the common entity (Girls) in both ratios.

Boys : Girls = 5 : 7 and Girls : Teachers = 14 : 1

Multiply the first Ratio by 2 to make the Girls count equal to 14: (5 × 2) : (7 × 2) = 10 : 14.

Now, the combined Ratio of Boys : Girls : Teachers = 10 : 14 : 1.

Extracting the required terms, Boys : Teachers = 10 : 1.

∴ The correct answer is 10:1.

Alternate Method

Given:

⇒ Ratio of Boys to Girls = 5 : 7

⇒ Ratio of Girls to Teachers = 14 : 1

Formula Used:

⇒ If A:B and B:C are given, then A:C = (A/B) × (B/C)

Calculations:

⇒ Let Boys be B, Girls be G, and Teachers be T.

⇒ B / G = 5 / 7

⇒ G / T = 14 / 1

⇒ To find B / T, we multiply the two ratios: (B / G) × (G / T)

⇒ B / T = (5 / 7) × (14 / 1)

⇒ B / T = (5 × 14) / (7 × 1)

⇒ B / T = 70 / 7

⇒ B / T = 10 / 1

⇒ B : T = 10 : 1

∴ The correct answer is 10:1.

Additional Information

Compound Ratio

When two or more Ratios are multiplied together, the result is called a Compound Ratio. For ratios a:b and c:d, the Compound Ratio is (ac):(bd).

Duplicate Ratio

The Duplicate Ratio of any ratio a:b is calculated by squaring both terms, resulting in a² : b².

Proportion

A Proportion is an equation which states that two Ratios are equal (a/b = c/d), where the product of extremes equals the product of means (ad = bc).

Shortcut Trick

HCF = LCM of numbers ÷ LCM of ratio parts

LCM of ratio parts 10, 13, and 11 = 10 × 13 × 11 = 1430

HCF = 2860 ÷ 1430 = 2

∴ The correct answer is 2.

Alternate Method

Given:

Ratio of three numbers = 10 : 13 : 11

LCM of the numbers = 2860

Formula Used:

If numbers are in ratio a : b : c, then their LCM = HCF × LCM(a, b, c).

Calculations:

⇒ Let the HCF of the three numbers be x.

⇒ The numbers are 10x, 13x, and 11x.

⇒ Since 10, 13, and 11 have no common factors (they are co-prime), their LCM = 10 × 13 × 11 = 1430.

⇒ LCM of the numbers = x × 1430

⇒ 2860 = 1430x

⇒ x = 2860 ÷ 1430

⇒ x = 2

∴ The correct answer is 2.

Additional Information

Relationship for Two Numbers

For any two numbers a and b, Product of numbers (a × b) = HCF(a, b) × LCM(a, b).

HCF (Highest Common Factor)

The HCF is the largest positive integer that divides each of the given numbers without leaving a remainder.

LCM (Least Common Multiple)

The LCM is the smallest positive integer that is divisible by each of the given numbers.

Co-prime Numbers

Two or more numbers are said to be co-prime if their HCF is 1. Their LCM is simply the product of the numbers.

Shortcut Trick

The age difference between two people always remains constant over time.

Ratio of X and Y = 4 : 5 ⇒ Difference = 5 − 4 = 1 unit.

Actual difference = 3 years (given for 5 years later, but it is constant).

∴ 1 unit = 3 years.

Sum of present ages = (4 + 5) units = 9 units = 9 × 3 = 27 years.

∴ The correct answer is 27 years.

Alternate Method

Given: Present age ratio of X and Y = 4 : 5. Future age difference = 3 years.

Formula Used: Age = Ratio Part × Value of 1 Unit; Age difference is invariant.

⇒ Let the present ages of X and Y be 4k and 5k respectively.

⇒ Age of X after 5 years = (4k + 5) years.

⇒ Age of Y after 5 years = (5k + 5) years.

⇒ Difference = (5k + 5) − (4k + 5) = 3

⇒ k = 3.

⇒ X's present age = 4 × 3 = 12 years.

⇒ Y's present age = 5 × 3 = 15 years.

⇒ Sum of present ages = 12 + 15 = 27 years.

∴ The correct answer is 27 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals remains the same throughout their lives, regardless of how many years pass.

Sum of Ages in Future/Past

If the current sum of ages of 'n' people is S, then after 't' years, the sum of ages will be S + (n × t).

Ratio Proportionality

In age problems, if the ratio is a : b and the difference is 'd', the units can be directly mapped: (b − a) units = d.

Shortcut Trick

Ratio of Cost Price (CP) to Selling Price (SP) based on price: 25 : (32 × 0.8) = 25 : 25.6

Ratio based on Weight fraud: 800 : 1000 = 4 : 5

Combined Ratio (CP : SP) = (25 × 4) : (25.6 × 5) = 100 : 128

Profit % = 128 − 100 = 28%

∴ The correct answer is 28% profit.

Alternate Method

Given: CP = ₹25/kg, MP = ₹32/kg, Discount = 20%, False weight = 800 g instead of 1000 g.

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

⇒ Selling Price (SP) per kg claimed = 32 − (20% of 32)

⇒ SP = 32 − 6.4 = ₹25.6

⇒ Since the shopkeeper uses 800 g instead of 1 kg, his Actual CP is for 800 g.

⇒ Actual CP = (25 ÷ 1000) × 800 = ₹20

⇒ Profit = 25.6 − 20 = ₹5.6

⇒ Overall Profit % = (5.6 ÷ 20) × 100

⇒ Profit % = 0.28 × 100 = 28%

∴ The correct answer is 28% profit.

Additional Information

Dishonest Dealer Weight Gain

When a dealer sells at CP but uses less weight, Profit % = [Error ÷ (True Value − Error)] × 100.

Marked Price and Discount

Marked Price (MP) = CP × [100 + Markup%] ÷ 100. Selling Price (SP) = MP × [100 − Discount%] ÷ 100.

Successive Percentage Change

Overall change for two successive percentages a and b is [a + b + (a × b)/100] %. Use negative values for discounts or losses.

Shortcut Trick

To find the maximum possible length of each equal piece from ropes of lengths 54 m, 72 m, and 90 m, we directly calculate their Highest Common Factor (HCF).

Lengths of the ropes: 54 m, 72 m, 90 m

HCF(54, 72, 90) = 18

Thus, each rope can be cut into smaller pieces of exactly 18 m with no leftover rope.

∴ The correct answer is 18 m.

Alternate Method

Given:

Length of the first rope = 54 m

Length of the second rope = 72 m

Length of the third rope = 90 m

Formula Used:

Maximum possible length of each piece = HCF of (54, 72, 90)

54 m 18m 18m

Shortcut Trick

Price after first 15% discount = 2000 − 300 = ₹1700.

Required second discount amount = 1700 − 1530 = ₹170.

Second discount % = (170 ÷ 1700) × 100 = 10%.

∴ The correct answer is 10%.

Alternate Method

Given:

Marked Price (MP) of the jacket = ₹2,000

Final Selling Price (SP) = ₹1,530

First Discount (D₁) = 15%

Formula Used:

Intermediate Price = MP × (100 − D₁) / 100

D₂ % = [(Intermediate Price − Final SP) / Intermediate Price] × 100

Calculations:

⇒ First discount amount = 15% of 2000 = (15/100) × 2000 = ₹300

⇒ Price after first discount = 2000 − 300 = ₹1700

⇒ Let the second discount percentage be x%

⇒ Second discount amount = 1700 − 1530 = ₹170

⇒ x = (Second discount amount / Price after first discount) × 100

⇒ x = (170 / 1700) × 100 = 10%

∴ The correct answer is 10%.

Additional Information

Successive Discount Formula

The single equivalent discount for two successive discounts of a% and b% is given by (a + b − ab/100)%.

Selling Price and Marked Price Relation

SP = MP × (1 − d₁/100) × (1 − d₂/100), where d₁ and d₂ are the successive discount percentages.

Marked Price Calculation

Marked Price = [Cost Price × (100 + Profit%)] / (100 − Discount%).

Shortcut Trick

Simplify √20 as 2√5.

The expression becomes (√5 + 2√5)².

Add like surds: (3√5)².

Square the term: 3² × (√5)² = 9 × 5 = 45.

∴ The correct answer is 45.

Alternate Method

Given: The expression to simplify is (√5 + √20)².

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

⇒ Let a = √5 and b = √20.

⇒ (√5)² + (√20)² + 2 × √5 × √20

⇒ 5 + 20 + 2 × √(5 × 20)

⇒ 25 + 2 × √100

⇒ 25 + 2 × 10

⇒ 25 + 20

⇒ 45

∴ The correct answer is 45.

Additional Information

Algebraic Identities

The square of a sum is given by (a + b)² = a² + 2ab + b², while the square of a difference is (a − b)² = a² − 2ab + b².

Properties of Surds

For any positive real numbers x and y, the product rule states √x × √y = √(xy) and the power rule states (√x)² = x.

Simplifying Square Roots

A square root is simplified by extracting the largest perfect square factor, such as √20 = √(4 × 5) = 2√5.

Shortcut Trick

Let the speed of Train A be S_A. Length L_A = S_A × 24.

Speed of Train B = 54 × (5/18) = 15 m/s.

When crossing Train B in opposite direction: (L_A + 150) = (S_A + 15) × 20.

Substitute L_A: 24S_A + 150 = 20S_A + 300 ⇒ 4S_A = 150.

S_A = 37.5 m/s = 37.5 × (18/5) = 135 km/h.

L_A = 37.5 × 24 = 900 m.

∴ The correct answer is 900 m and 135 km/h.

Alternate Method

Given:

Time to pass fixed pole = 24 sec

Time to cross Train B = 20 sec

Length of Train B = 150 m

Speed of Train B = 54 km/h

Formula Used:

Distance = Speed × Time

Relative Speed (Opposite) = S₁ + S₂

Calculations:

⇒ Let Speed of Train A = v m/s and Length of Train A = L m.

⇒ From first condition: L = v × 24 → (i)

⇒ Speed of Train B in m/s = 54 × (5/18) = 15 m/s.

⇒ From second condition (crossing in opposite direction):

⇒ Total Distance = L + 150

⇒ Relative Speed = v + 15

⇒ (L + 150) = (v + 15) × 20 → (ii)

⇒ Substituting (i) in (ii): 24v + 150 = 20v + 300

⇒ 4v = 150 ⇒ v = 37.5 m/s

⇒ Speed of Train A in km/h = 37.5 × (18/5) = 135 km/h.

⇒ Length of Train A = 37.5 × 24 = 900 m.

∴ The correct answer is 900 m and 135 km/h.

Additional Information

Relative Speed Concept

For objects moving in the same direction, Relative Speed = S₁ − S₂. For opposite directions, Relative Speed = S₁ + S₂.

Unit Conversion

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

Train Crossing a Pole vs Train

When a train crosses a pole/man, distance = length of train. When it crosses another train/platform, distance = sum of both lengths.

Shortcut Trick

In a trapezium (where AB || DC), the length of the segment EF joining the midpoints of the diagonals is half the difference of the parallel sides.

Length of EF = 1/2 × |AB − DC|

⇒ EF = 1/2 × |21 − 11|

⇒ EF = 1/2 × 10 = 5 cm

∴ The correct answer is 5 cm.

Alternate Method

Given: AB = 21 cm, DC = 11 cm (Parallel sides), E and F are midpoints of AC and BD.

Formula Used: Length of segment joining midpoints of diagonals = (1/2) × (Difference of parallel sides).

Calculations:

⇒ Parallel side 1 (AB) = 21 cm

⇒ Parallel side 2 (DC) = 11 cm

⇒ Difference of parallel sides = 21 − 11 = 10 cm

⇒ Length of segment EF = (1/2) × 10

⇒ EF = 5 cm

∴ The correct answer is 5 cm.

Additional Information

Median of a Trapezium

The segment joining the midpoints of the non-parallel sides is called the median, and its length is the average of the parallel sides: (AB + DC) ÷ 2.

Midpoint Segment Property

In any quadrilateral, the segment joining the midpoints of the diagonals lies on the median segment joining the midpoints of the non-parallel sides.

Isosceles Trapezium

In an isosceles trapezium, the non-parallel sides (AD and BC) are equal, and the diagonals (AC and BD) are also equal in length.

Shortcut Trick

Profit of 20% means Selling Price is 120% of Cost Price (CP).

Given 120% = ₹1,200 ⇒ 1% = ₹10.

For a Loss of 10%, the required Selling Price is (100 − 10)% = 90%.

Required Price = 90% × 10 = ₹900.

∴ The correct answer is ₹900.

Alternate Method

Given:

Initial Selling Price (SP₁) = ₹1,200

Initial Profit% = 20%

Required Loss% = 10%

Formula Used:

Cost Price (CP) = [SP ÷ (100 + Profit%)] × 100

New Selling Price (SP₂) = CP × [(100 − Loss%) ÷ 100]

Calculations:

⇒ Cost Price (CP) = [1,200 ÷ (100 + 20)] × 100

⇒ CP = [1,200 ÷ 120] × 100

⇒ CP = 10 × 100 = ₹1,000

⇒ New Selling Price (SP₂) = 1,000 × [(100 − 10) ÷ 100]

⇒ SP₂ = 1,000 × (90 ÷ 100)

⇒ SP₂ = 1,000 × 0.9 = ₹900

∴ The correct answer is ₹900.

Additional Information

Profit and Loss Percentage

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

Marked Price and Discount

Marked Price (MP) is the label price. Discount is calculated on MP: Selling Price (SP) = MP − Discount.

Successive Profit/Loss

If an article is sold at a% profit and then b% profit, the Effective Profit % = [a + b + (a × b)/100]%.

Shortcut Trick

The filling rate of the second tap (19 hr) and the emptying rate of the third tap (19 hr) are identical.

When opened together, they neutralize each other's effect completely.

Net filling depends only on the first tap which fills the full cistern in 4 hours.

Time to fill one-fourth of the cistern = 4 hr × (1/4) = 1 hr.

∴ The correct answer is 1 hr.

Alternate Method

Given:

Time taken by Tap A (filling) = 4 hr

Time taken by Tap B (filling) = 19 hr

Time taken by Tap C (emptying) = 19 hr

Formula Used:

Efficiency = Total Work ÷ Time Taken

Time = Required Work ÷ Combined Efficiency

Calculations:

⇒ Total Efficiency of (A + B + C) = 19 + 4 − 4 = 19 units/hr

⇒ Required Work = 1/4 of total capacity = 76 ÷ 4 = 19 units

⇒ Time Taken = Required Work ÷ Total Efficiency

⇒ Time = 19 ÷ 19 = 1 hr

∴ The correct answer is 1 hr.

Additional Information

Inlet and Outlet Pipes

Inlet pipes fill the tank (positive efficiency), while outlet/leakage pipes empty the tank (negative efficiency).

Net Efficiency

If multiple pipes are open, the net efficiency is the algebraic sum of their individual efficiencies: E_net = E₁ + E₂ − E₃.

Fractional Work

The time taken to fill 1/n part of the tank is simply the time taken to fill the full tank divided by n.

Shortcut Trick

To find the fourth proportional (x) of three numbers a, b, and c, use the direct formula: x = (b × c) ÷ a.

Here, a = 37, b = 25, and c = 888.

x = (25 × 888) ÷ 37

x = 25 × 24 = 600.

∴ The correct answer is 600.

Alternate Method

Given: The numbers 37, 25, and 888 are the first three terms of a proportion, and x is the fourth proportional.

Formula Used: If four numbers a, b, c, d are in proportion (a : b :: c : d), then the product of extremes equals the product of means (a × d = b × c).

⇒ 37 ÷ 25 = 888 ÷ x

⇒ 37 × x = 25 × 888

⇒ x = (25 × 888) ÷ 37

⇒ x = 22200 ÷ 37

⇒ x = 600

∴ The correct answer is 600.

Additional Information

Fourth Proportional

For four non-zero quantities a, b, c, d, if a : b = c : d, then d is the fourth proportional to a, b, c.

Third Proportional

If a : b = b : c, then c is the third proportional to a and b, calculated as c = b² ÷ a.

Mean Proportional

The mean proportional (or geometric mean) between a and b is the value x such that a : x = x : b, which is x = √(a × b).

Shortcut Trick

Actual Increase = ₹220 − ₹200 = ₹20

Percentage Increase = (Increase ÷ Original Price) × 100

Percentage Increase = (20 ÷ 200) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Initial Price of the book = ₹200

Final Price of the book = ₹220

Formula Used:

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

Calculations:

⇒ Price Increase = ₹220 − ₹200 = ₹20

⇒ Percentage Increase = (20 ÷ 200) × 100

⇒ Percentage Increase = (1 ÷ 10) × 100

⇒ Percentage Increase = 10%

∴ The correct answer is 10%.

Additional Information

Percentage Change Formula

The formula for percentage change is [(New Value − Old Value) ÷ Old Value] × 100. If the result is positive, it is an increase; if negative, it is a decrease.

Fraction to Percentage

To convert any fraction into a percentage, multiply it by 100 and add the % symbol. For example, 1/5 × 100 = 20%.

Percentage Base Change

If a value increases from A to B, the base is A. If it decreases from B back to A, the base for calculation becomes B.

Shortcut Trick

Divisibility rule of 9 states that the sum of the digits must be divisible by 9.

Sum of digits of 5135n2 = 5 + 1 + 3 + 5 + n + 2 = 16 + n

The next higher multiple of 9 after 16 is 18.

⇒ 16 + n = 18 ⇒ n = 2

∴ The correct answer is 2.

Alternate Method

Given:

The number 5135n2 is completely divisible by 9.

Formula/Property Used:

A number is divisible by 9 if and only if the sum of its digits is a multiple of 9.

Calculation:

⇒ Sum of digits of the number = 5 + 1 + 3 + 5 + n + 2

⇒ Sum of digits = 16 + n

⇒ Since the number is divisible by 9, (16 + n) must be equal to 18, 27, 36, etc.

⇒ For the least value of n (where n is a single digit, 0 ≤ n ≤ 9):

⇒ 16 + n = 18

⇒ n = 18 − 16

⇒ n = 2

∴ The correct answer is 2.

Additional Information

Divisibility Rule of 3

A number is completely divisible by 3 if the sum of its digits is a multiple of 3.

Divisibility Rule of 11

A number is divisible by 11 if the difference between the sum of digits at odd places and even places is either 0 or a multiple of 11.

Divisibility Rule of 8

A number is divisible by 8 if the number formed by the last three digits is divisible by 8.

Shortcut Trick

The given ratio of three numbers is 3 : 2 : 13.

Since the numbers in the ratio (3, 2, and 13) are all prime numbers, their LCM is simply the product of these numbers multiplied by their HCF.

HCF = LCM ÷ (Product of Ratio parts)

HCF = 2574 ÷ (3 × 2 × 13) = 2574 ÷ 78

HCF = 33

∴ The correct answer is 33.

Alternate Method

Given: Ratio of numbers = 3 : 2 : 13 | LCM = 2574

Formula Used: LCM of (ax, bx, cx) = x × LCM(a, b, c), where x is the HCF.

⇒ Let the three numbers be 3x, 2x, and 13x, where x is the HCF.

⇒ LCM(3x, 2x, 13x) = x × LCM(3, 2, 13)

⇒ Since 3, 2, and 13 are co-prime, their LCM = 3 × 2 × 13 = 78

⇒ Total LCM = 78x

⇒ 78x = 2574

⇒ x = 2574 ÷ 78

⇒ x = 33

∴ The correct answer is 33.

Additional Information

Relationship for Two Numbers

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

Numbers are called co-prime if their Highest Common Factor (HCF) is 1. The LCM of co-prime numbers is always their product.

HCF and LCM of Fractions

HCF = (HCF of Numerators) ÷ (LCM of Denominators) and LCM = (LCM of Numerators) ÷ (HCF of Denominators).

Given:

Equation: 3⁻²² + 3⁻¹⁹ + 3⁻¹⁷ + 3⁻¹⁶ = k × 3⁻²⁴

Formula used:

The exponent rule for division: a^m ÷ aⁿ = a^{m − n}

Calculations:

k × 3⁻²⁴ = 3⁻²² + 3⁻¹⁹ + 3⁻¹⁷ + 3⁻¹⁶

⇒ k = (3⁻²² + 3⁻¹⁹ + 3⁻¹⁷ + 3⁻¹⁶) ÷ 3⁻²⁴

⇒ k = 3^{(−22 − (−24))} + 3^{(−19 − (−24))} + 3^{(−17 − (−24))} + 3^{(−16 − (−24))}

⇒ k = 3² + 3⁵ + 3⁷ + 3⁸

⇒ k = 9 + 243 + 2187 + 6561

⇒ k = 9000

∴ The correct answer is option (4).

Shortcut Trick

Find two Factors of the product 144 that add up to the sum 26.

List factors: 18 × 8 = 144

Check sum: 18 + 8 = 26 (Matches given sum)

Required Positive Difference = 18 − 8 = 10

∴ The correct answer is 10.

Alternate Method

Given: Sum (a + b) = 26 and Product (ab) = 144

Formula Used: (a − b)² = (a + b)² − 4ab

⇒ (a − b)² = (26)² − 4 × 144

⇒ (a − b)² = 676 − 576

⇒ (a − b)² = 100

⇒ a − b = √100

⇒ a − b = 10

∴ The correct answer is 10.

Additional Information

Quadratic Equation Construction

If sum S and product P of two numbers are known, the numbers are roots of x² − Sx + P = 0.

Algebraic Identities

Useful relation: (a + b)² − (a − b)² = 4ab.

Finding Individual Numbers

Once sum (S) and difference (D) are found, the numbers are (S+D)÷2 and (S−D)÷2.