Quantitative Aptitude PYQs

Shortcut Trick

The ratio of diagonals is 3 : 4, so the ratio of half-diagonals is also 3 : 4.

In a rhombus, half-diagonals and the side form a right-angled triangle.

Using the Pythagorean triplet (3, 4, 5), the side corresponds to ratio 5.

Given side length 5 units → 12.5, so 1 unit → 2.5.

Shorter diagonal ratio is 3 × 2 = 6 units → 6 × 2.5 = 15 units.

∴ The correct answer is 15 units.

Alternate Method

Given:

Ratio of diagonals (d₁ : d₂) = 3 : 4

Length of each side (a) = 12.5 units

Formula Used:

In a rhombus, the relation between side and diagonals is: a² = (d₁/2)² + (d₂/2)²

Calculations:

⇒ Let the diagonals be d₁ = 3x and d₂ = 4x.

⇒ (12.5)² = (3x/2)² + (4x/2)²

⇒ 156.25 = (1.5x)² + (2x)²

⇒ 156.25 = 2.25x² + 4x²

⇒ 156.25 = 6.25x²

⇒ x² = 156.25 ÷ 6.25 = 25

⇒ x = √25 = 5

⇒ Length of shorter diagonal (d₁) = 3 × 5 = 15 units.

∴ The correct answer is 15 units.

Additional Information

Properties of Rhombus Diagonals

The diagonals of a rhombus bisect each other at right angles (90°).

Area of a Rhombus

The area can be calculated using diagonals: Area = 1/2 × d₁ × d₂.

Perimeter of a Rhombus

Since all sides are equal, the perimeter = 4 × side.

Given:

{18 + [24 ÷ (3 of 2)] × 4} − 10

Formula used:

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

Calculations:

{18 + [24 ÷ (3 of 2)] × 4} − 10

⇒ {18 + [24 ÷ (3 × 2)] × 4} − 10

⇒ {18 + [24 ÷ 6] × 4} − 10

⇒ {18 + 4 × 4} − 10

⇒ {18 + 16} − 10

⇒ 34 − 10

⇒ 24

∴ The correct answer is option (1).

Shortcut Trick

Given: 2 cot A = 4 → cot A = 2.

Using the Trigonometric Identity: cosec² A = 1 + cot² A.

Substituting the value: cosec² A = 1 + (2)² = 1 + 4 = 5.

Expression value: (5 + 1) ÷ (5 − 1) = 6 ÷ 4 = 1.5.

∴ The correct answer is 1.5.

Alternate Method

Given:

2 cot A = 4

⇒ cot A = 4 ÷ 2 = 2

Formula Used:

1. cot A = Base ÷ Perpendicular
2. Pythagoras Theorem: H² = P² + B²
3. cosec A = Hypotenuse ÷ Perpendicular

Calculations:

Let Perpendicular (P) = 1, then Base (B) = 2 (since cot A = B/P = 2/1).

⇒ H² = 1² + 2² = 1 + 4 = 5

⇒ H = √5

⇒ cosec A = H ÷ P = √5 ÷ 1 = √5

⇒ cosec² A = (√5)² = 5

Required Expression = (cosec² A + 1) ÷ (cosec² A − 1)

⇒ (5 + 1) ÷ (5 − 1)

⇒ 6 ÷ 4

⇒ 1.5

∴ The correct answer is 1.5.

Additional Information

Fundamental Trigonometric Identities

There are three main identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Ratios

Cosecant is the reciprocal of sine (cosec θ = 1/sin θ) and cotangent is the reciprocal of tangent (cot θ = 1/tan θ).

Pythagoras Theorem in Trigonometry

For any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (H² = P² + B²).

Shortcut Trick

Ratio of Passed students to Failed students = 11 : 2

Total units in the class = 11 + 2 = 13 units

Given, Failed students (2 units) = 44

⇒ 1 unit = 44 ÷ 2 = 22

Total students = 13 units × 22 = 286

∴ The correct answer is 286 students.

Alternate Method

Given:

Ratio (Passed : Failed) = 11 : 2

Number of Failed students = 44

Formula:

Total Students = Passed Students + Failed Students

⇒ Let the number of passed students be 11x and failed students be 2x.

⇒ According to the question, 2x = 44

⇒ x = 44 ÷ 2 = 22

⇒ Number of passed students = 11 × 22 = 242

⇒ Total number of students = Passed + Failed = 242 + 44 = 286

∴ The correct answer is 286 students.

Additional Information

Ratio and Total Value

If two quantities are in the ratio a:b, the total quantity can be represented as (a + b)k, where k is a constant multiplier.

Unitary Method in Ratios

To find the total, first divide the given numerical value by its corresponding ratio parts to find the value of 1 unit, then multiply by the total ratio parts.

Percentage Conversion

The fail percentage in this class is (Failed parts ÷ Total parts) × 100 = (2 ÷ 13) × 100 ≈ 15.38%.

Shortcut Trick

If the ratio of two numbers is a : b and their HCF is H, then their LCM is given by the formula:

LCM = H × a × b (provided a and b are coprime)

Given a = 3, b = 8, and H = 6.

⇒ LCM = 6 × 3 × 8 = 144.

∴ The correct answer is 144.

Alternate Method

Given:

Ratio of numbers = 3 : 8

Highest Common Factor (HCF) = 6

Formula Used:

Number = Ratio part × HCF

Product of two numbers = HCF × LCM

Calculations:

⇒ Let the two numbers be 3x and 8x.

⇒ The HCF of 3x and 8x is x (since 3 and 8 are coprime).

⇒ Given x = 6.

⇒ Number₁ = 3 × 6 = 18

⇒ Number₂ = 8 × 6 = 48

⇒ Using the product rule: 18 × 48 = 6 × LCM

⇒ LCM = (18 × 48) ÷ 6

⇒ LCM = 3 × 48 = 144

∴ The correct answer is 144.

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

Two numbers are called co-prime if their HCF is 1. If the terms of a ratio are co-prime, the HCF of the actual numbers is the common multiplier.

LCM of Ratios

If the ratio of two numbers is x : y and their HCF is H, then their LCM = H × x × y and the numbers are Hx and Hy.

Shortcut Trick

Area of Trapezium = Average of bases × Height

Average of bases = (8 + 12) ÷ 2 = 10 cm

Area = 10 × 5 = 50 cm²

Therefore, the Area of the trapezium is 50 cm².

∴ The correct answer is 50 cm².

Alternate Method

Given:

Length of first base (a) = 8 cm

Length of second base (b) = 12 cm

Height (h) = 5 cm

Formula Used:

Area of Trapezium = 1/2 × (a + b) × h

Calculations:

⇒ Area = 1/2 × (8 + 12) × 5

⇒ Area = 1/2 × 20 × 5

⇒ Area = 10 × 5

⇒ Area = 50 cm²

Thus, the area of the trapezium is 50 cm², which corresponds to Option (2).

∴ The correct answer is 50 cm².

Additional Information

Properties of a Trapezium

A quadrilateral with at least one pair of parallel sides. If the non-parallel sides are equal, it is called an isosceles trapezium.

Median of a Trapezium

The line segment joining the midpoints of the non-parallel sides is the median. Length = (Base₁ + Base₂) ÷ 2.

Area of a Rhombus

Area = 1/2 × d₁ × d₂, where d₁ and d₂ are the lengths of its diagonals.

Given:

Expression: ((19/18) ÷ (19/6)) ÷ ((5/18) × (16/15) + (4/6)) + (6/8) ÷ (24/25) of (25/8)

Formula used:

The BODMAS rule is used to solve this expression, which stands for Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Calculations:

((19/18) ÷ (19/6)) ÷ ((5/18) × (16/15) + (4/6)) + (6/8) ÷ (24/25) of (25/8)

⇒ (19/18 × 6/19) ÷ (8/27 + 4/6) + (6/8) ÷ (24/25 × 25/8)

⇒ (1/3) ÷ (8/27 + 2/3) + (6/8) ÷ 3

⇒ (1/3) ÷ (8/27 + 18/27) + (6/8 × 1/3)

⇒ (1/3) ÷ (26/27) + 1/4

⇒ (1/3 × 27/26) + 1/4

⇒ 9/26 + 1/4

⇒ (18 + 13)/52

⇒ 31/52

∴ The correct answer is option (3).

Shortcut Trick

Profit = 25% = 1/4 ⇒ Ratio of CP : SP = 4 : 5

Given SP = 5 units = ₹600 ⇒ 1 unit = ₹120

Total CP = 4 units = 4 × 120 = ₹480

Original Purchase Price = Total CP − Repair = 480 − 50 = ₹430

∴ The correct answer is ₹430.

Alternate Method

Given: Selling Price (SP) = ₹600, Profit Percentage = 25%, Repair Cost = ₹50.

Formula Used: Total Cost Price (CP) = Selling Price × [100 / (100 + Profit%)]

⇒ Total CP = 600 × [100 / (100 + 25)]

⇒ Total CP = 600 × (100 / 125)

⇒ Total CP = 600 × (4 / 5)

⇒ Total CP = 120 × 4 = ₹480

⇒ Original Purchase Price = Total CP − Repair Expenses

⇒ Original Purchase Price = 480 − 50 = ₹430

∴ The correct answer is ₹430.

Additional Information

Concept of Total Cost Price

The Total Cost Price includes the Original Purchase Price plus all overheads like repair, transportation, and taxes. Profit or Loss is always calculated on this final sum.

Profit Percentage Formula

Profit % = [(SP − Total CP) / Total CP] × 100. This measures the gain relative to the entire investment.

Effective Selling Price

If a profit of P% is made, then SP = Total CP × [(100 + P)/100]. Conversely, Total CP = SP × [100/(100 + P)].

Shortcut Trick

Average Speed = Total Distance ÷ Total Time

Total Distance = 120 + 180 + 100 = 400 km

Total Time = (120 ÷ 40) + (180 ÷ 60) + (100 ÷ 50) = 3 + 3 + 2 = 8 hours

Average Speed = 400 ÷ 8 = 50 km/h

∴ The correct answer is 50 km/h.

Alternate Method

Given:

Distance₁ = 120 km, Speed₁ = 40 km/h

Distance₂ = 180 km, Speed₂ = 60 km/h

Distance₃ = 100 km, Speed₃ = 50 km/h

Formula Used:

Time = Distance ÷ Speed

Average Speed = (Total Distance) ÷ (Total Time)

Calculations:

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

⇒ Time taken for second part (t₂) = 180 ÷ 60 = 3 hours

⇒ Time taken for third part (t₃) = 100 ÷ 50 = 2 hours

⇒ Total Time (T) = 3 + 3 + 2 = 8 hours

⇒ Total Distance (D) = 120 + 180 + 100 = 400 km

⇒ Average Speed = 400 ÷ 8 = 50 km/h

∴ The correct answer is 50 km/h.

Additional Information

Average Speed for Equal Distances

If a car travels two equal distances at speeds x and y, the Average Speed is given by the Harmonic Mean: 2xy ÷ (x + y).

Unit Conversion

To convert km/h to m/s, multiply the speed 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 the difference of their speeds. In opposite directions, it is the sum of their speeds.

Shortcut Trick

Find the scaling factor between the first two terms: 5.6 ÷ 0.8 = 7.

Apply the same scaling factor to the second term to find the third proportional (n).

n = 5.6 × 7 = 39.2.

∴ The correct answer is 39.2.

Alternate Method

Given: First term (a) = 0.8, Second term (b) = 5.6, Third proportional = n.

Formula Used: If a, b, and n are in continued proportion, then a : b :: b : n, which implies n = b² ÷ a.

⇒ a : b :: b : n

⇒ 0.8 : 5.6 :: 5.6 : n

⇒ 0.8 ÷ 5.6 = 5.6 ÷ n

⇒ n = (5.6 × 5.6) ÷ 0.8

⇒ n = 31.36 ÷ 0.8

⇒ n = 39.2

∴ The correct answer is 39.2.

Additional Information

Mean Proportional

The mean proportional between two numbers a and b is given by x = √(a × b). It is the value that satisfies a : x = x : b.

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d satisfies a : b = c : d. It is calculated as d = (b × c) ÷ a.

Continued Proportion

A series of numbers a, b, c, d... are in continued proportion if the ratio between successive terms is constant: a/b = b/c = c/d.

Shortcut Trick

Price after 1^st discount = 16000 × 0.85 = 13600

2^nd discount amount = 13600 − 12240 = 1360

N% = (1360 ÷ 13600) × 100 = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Marked Price (MP) = ₹16,000

First Discount = 15%

Second Discount = N%

Final Selling Price (SP) = ₹12,240

Formula Used:

SP = MP × (100 − d₁)/100 × (100 − d₂)/100

Calculations:

⇒ 12,240 = 16,000 × (100 − 15)/100 × (100 − N)/100

⇒ 12,240 = 16,000 × 85/100 × (100 − N)/100

⇒ 12,240 = 160 × 85 × (100 − N)/100

⇒ 12,240 = 13,600 × (100 − N)/100

⇒ 12,240 ÷ 136 = 100 − N

⇒ 90 = 100 − N

⇒ N = 100 − 90

⇒ N = 10%

∴ The correct answer is 10%.

Additional Information

Net Successive Discount Formula

The single equivalent discount for two successive discounts a% and b% is given by: [a + b − (a × b)/100]%.

Relationship between MP and CP

MP/CP = (100 + Profit%)/(100 − Discount%). This is useful when both discount and profit percentages are known.

Marked Price (List Price)

The price printed on an item is the Marked Price. Discounts are always calculated on this value to arrive at the Selling Price.

Shortcut Trick

Calculate Cost Price (CP) per unit: 1000 ÷ 20 = ₹50.

Given Selling Price (SP) per unit = ₹55.

Profit percentage = [(SP − CP) ÷ CP] × 100

Profit % = [(55 − 50) ÷ 50] × 100 = (5 ÷ 50) × 100 = 10%.

∴ The correct answer is 10%.

Alternate Method

Given:

Total Quantity of rice = 20 kg

Total Cost Price (CP) = ₹1,000

Selling Price (SP) per kg = ₹55

Formula Used:

Total SP = SP per unit × Total Quantity

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

Calculations:

⇒ Total Selling Price (SP) = 55 × 20 = ₹1,100

⇒ Total Profit = 1,100 − 1,000 = ₹100

⇒ Profit Percentage = (100 ÷ 1,000) × 100

⇒ Profit % = 10%

∴ The correct answer is 10%.

Additional Information

Loss Percentage

When CP is greater than SP, Loss % = [(CP − SP) ÷ CP] × 100.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). Discount % = (Discount ÷ MP) × 100.

Relation between CP and SP

SP = CP × [1 + (Profit % ÷ 100)] or SP = CP × [1 − (Loss % ÷ 100)].

Shortcut Trick

Mode is defined as the value that appears most frequently in a given data set.

By scanning the data: 12, 15, 18, 18, 20, 18, 22, 22, 25, 28, 22, 15, 18, 12

The number 18 is repeated 4 times, which is higher than any other number.

∴ The correct answer is 18.

Alternate Method

Given Data: 12, 15, 18, 18, 20, 18, 22, 22, 25, 28, 22, 15, 18, 12

Formula Used: Mode = Value with the maximum frequency.

Let us count the frequency of each distinct number in the dataset:

⇒ Number 12 occurs 2 times.

⇒ Number 15 occurs 2 times.

⇒ Number 18 occurs 4 times.

⇒ Number 20 occurs 1 time.

⇒ Number 22 occurs 3 times.

⇒ Number 25 occurs 1 time.

⇒ Number 28 occurs 1 time.

⇒ Comparing frequencies: 4 > 3 > 2 > 1.

⇒ Since 18 has the highest frequency of 4, it is the mode.

∴ The correct answer is 18.

Additional Information

Empirical Relationship

For a moderately skewed distribution, the relationship between mean, median, and mode is: Mode = 3 × Median − 2 × Mean.

Types of Mode

A distribution with one mode is Unimodal, two modes is Bimodal, and more than two modes is Multimodal.

Mean and Median

Mean is the average of all values (Sum ÷ Count), while Median is the middle value when data is arranged in ascending or descending order.

Shortcut Trick

For half-yearly compounding, the rate becomes 5% (10% ÷ 2) and the time periods become 3 (1.5 × 2).

Successive percentage for 3 periods of 5% = 15.7625%.

Compound Interest = 12,000 × 15.7625 / 100 = ₹1,891.50.

Total Amount = 12,000 + 1,891.50 = ₹13,891.50.

∴ The correct answer is ₹13,891.50.

Alternate Method

Given:

Principal (P) = ₹12,000

Annual Rate (R) = 10%

Time (T) = 1.5 years

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

Since the interest is compounded half-yearly:

⇒ Half-yearly rate (r) = 10% / 2 = 5%

⇒ Number of half-years (n) = 1.5 × 2 = 3

⇒ A = 12000 × (1 + 5/100)³

⇒ A = 12000 × (105/100)³

⇒ A = 12000 × (21/20) × (21/20) × (21/20)

⇒ A = 12000 × 9261 / 8000

⇒ A = 1.5 × 9261

⇒ A = ₹13,891.50

∴ The correct answer is ₹13,891.50.

Additional Information

Effective Rate for Half-Yearly Compounding

When interest is compounded half-yearly, the effective annual rate is slightly higher than the nominal rate. For a nominal rate R, the effective rate is [(1 + (R/2)/100)² − 1] × 100.

Difference between CI and SI

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

Compounding Frequencies

If compounded quarterly, the rate is divided by 4 (R/4) and the time is multiplied by 4 (4T). If compounded monthly, divide rate by 12 and multiply time by 12.

Shortcut Trick

Find two factors of the product 220 that sum to 31.

220 → 20 × 11

Check Sum: 20 + 11 = 31 (Matches given sum)

Required Difference: 20 − 11 = 9

∴ The correct answer is 9.

Alternate Method

Given:

Sum of numbers (x + y) = 31

Product of numbers (xy) = 220

Formula Used:

The algebraic identity relating sum, product, and difference is:

(x − y)² = (x + y)² − 4xy

Calculations:

⇒ (x − y)² = (31)² − 4 × (220)

⇒ (x − y)² = 961 − 880

⇒ (x − y)² = 81

⇒ (x − y) = √81

⇒ x − y = 9

∴ The correct answer is 9.

Additional Information

Quadratic Relation

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

Identity for Sum of Squares

The sum of the squares of the numbers is given by x² + y² = (x + y)² − 2xy.

Combined Identity

(x + y)² + (x − y)² = 2(x² + y²), which relates the sum, difference, and squares of two numbers.

Shortcut Trick

Total expenditure for each type is ₹4,800.

Difference in Price × Difference in Quantity = 40 × 10 = 400.

Product of the number of items and the price per item for biographies must be 4800.

Trial with factors: If Biography books = 30, Price = 4800 ÷ 30 = ₹160.

Then Mystery novels = 40, Price = 4800 ÷ 40 = ₹120. (Price difference 160 − 120 = ₹40, matches!)

∴ The correct answer is 30.

Alternate Method

Given:

Total spend on Biography books = ₹4,800

Total spend on Mystery novels = ₹4,800

Price difference = ₹40 per book

Quantity difference = 10 books

Formula Used:

Expenditure = Price per item × Quantity

Calculations:

Let the number of biography books be x.

Then, the number of mystery novels is x + 10.

⇒ 4800 ÷ x − 4800 ÷ (x + 10) = 40

⇒ 120 ÷ x − 120 ÷ (x + 10) = 1

⇒ 120 × [ (x + 10 − x) ÷ (x(x + 10)) ] = 1

⇒ 120 × 10 = x² + 10x

⇒ x² + 10x − 1200 = 0

⇒ x² + 40x − 30x − 1200 = 0

⇒ x(x + 40) − 30(x + 40) = 0

⇒ (x − 30)(x + 40) = 0

⇒ x = 30 (since quantity cannot be negative)

∴ The correct answer is 30.

Additional Information

Inverse Proportion Logic

When Total Expenditure is constant, the Price per unit is inversely proportional to the Quantity purchased (P ∝ 1/Q).

Quadratic Roots Property

For the equation ax² + bx + c = 0, if c is negative, one root will be positive and one will be negative. In physical contexts like quantity, only the positive root is valid.

Middle Term Splitting

To solve x² + 10x − 1200 = 0, find two numbers whose product is −1200 and sum is 10. The numbers are 40 and −30.

Shortcut Trick

Net percentage change formula for successive increases: x + y + (x × y)/100

Given: x = 15% and y = 15%

Net Increase = 15 + 15 + (15 × 15)/100

Net Increase = 30 + 225/100 = 30 + 2.25 = 32.25%

∴ The correct answer is 32.25%.

Alternate Method

Given:

First percentage increase = 15%

Second percentage increase = 15%

Calculations:

⇒ Let the initial price of the article be 100.

⇒ Price after first 15% increase = 100 + (15% of 100) = 115.

⇒ Price after second 15% increase = 115 + (15% of 115).

⇒ 15% of 115 = (15 × 115) / 100 = 1725 / 100 = 17.25.

⇒ Final Price = 115 + 17.25 = 132.25.

⇒ Net Percentage Increase = Final Price − Initial Price = 132.25 − 100 = 32.25%.

∴ The correct answer is 32.25%.

Additional Information

Successive Percentage Change

When a value is changed by x% and then the resulting value is changed by y%, the net change is given by: [x + y + (xy/100)]%.

Successive Discount Formula

If two successive discounts of x% and y% are given, the equivalent single discount is: [x + y − (xy/100)]%.

Fraction Multiplying Factor

For an increase of 15%, the multiplying factor is (1 + 15/100) = 1.15. For two such increases, Net Factor = 1.15 × 1.15 = 1.3225.

Shortcut Trick

Apply the successive percentage change formula: Net Change = a + b + (a × b / 100)

Given: a = +20 (increase) and b = −25 (decrease)

Net % Change = 20 − 25 + [20 × (−25) / 100]

Net % Change = −5 − 5 = −10%

Negative result indicates a percentage decrease.

∴ The correct answer is 10% decrease.

Alternate Method

Given:

First change = 20% increase

Second change = 25% decrease

⇒ Let the initial price of the trouser be 100 units.

⇒ Price after 20% increase = 100 + (20% of 100) = 100 + 20 = 120 units.

⇒ Price after 25% decrease = 120 − (25% of 120)

⇒ Price after decrease = 120 − (1/4 × 120) = 120 − 30 = 90 units.

⇒ Net change in price = Final Price − Initial Price = 90 − 100 = −10 units.

⇒ Net percentage change = (Net Change / Initial Price) × 100

⇒ Net percentage change = (−10 / 100) × 100 = −10%.

∴ The correct answer is 10% decrease.

Additional Information

Net Percentage Change Formula

For two successive changes of x% and y%, the net change is given by [x + y + (xy/100)]%. Always use a negative sign for a percentage decrease.

Fractional Multiplier Method

Convert percentages to fractions: 20% increase = 6/5 and 25% decrease = 3/4. Net effect = (6/5) × (3/4) = 18/20 = 9/10. A ratio of 10:9 implies a 1/10 reduction, which is 10%.

Successive vs. Simple Change

Successive changes are calculated on the updated value, not the original base, making them different from a single cumulative arithmetic sum of percentages.

Shortcut Trick

Ratio of Mohit : Rahul (Present) = 3 : 1 (Difference = 2 parts)

Ratio (After 12 years) = 2 : 1 → Multiply by 2 to equate difference → 4 : 2

Increase in units (4 − 3 or 2 − 1) = 1 unit → 12 years

Sum of present age ratios = 3 + 1 = 4 units

Sum of ages = 4 × 12 = 48 years

∴ The correct answer is 48 years.

Alternate Method

Given:

Present age of Mohit = 3 × Present age of Rahul

After 12 years, Mohit age = 2 × Rahul age

Let the present age of Rahul be x years.

⇒ Present age of Mohit = 3x years.

After 12 years:

⇒ Age of Rahul = x + 12

⇒ Age of Mohit = 3x + 12

As per the condition:

⇒ 3x + 12 = 2(x + 12)

⇒ 3x + 12 = 2x + 24

⇒ 3x − 2x = 24 − 12

⇒ x = 12

Present age of Rahul = 12 years.

Present age of Mohit = 3 × 12 = 36 years.

⇒ Sum of their present ages = 12 + 36 = 48 years.

Therefore, the correct answer corresponds to Option (3).

∴ The correct answer is 48 years.

Additional Information

Constant Age Difference

The mathematical difference between the ages of two individuals never changes with time. If A is 5 years older than B today, he will remain 5 years older forever.

Ratio Adjustment Rule

To solve age problems using ratios, ensure the difference between the ratio parts of the two people is the same for both time periods by multiplying the ratios with suitable integers.

Sum of Ages

If the sum of ages of 2 people is S, then after n years, the new sum will be S + 2n.

∴ The correct answer is 48 years.

Shortcut Trick

Apply the Ratio-Unitary method for direct variation:

Distance ratio = 120 : 225 ⇒ 8 : 15

Since Petrol ∝ Distance, the petrol ratio is also 8 : 15.

Given 8 units = 8 liters, then 15 units = 15 liters.

∴ The correct answer is 15 liters.

Alternate Method

Given:

Distance₁ = 120 km, Petrol₁ = 8 liters

Distance₂ = 225 km

Formula Used:

Petrol required = (Petrol₁ ÷ Distance₁) × Distance₂

Calculations:

⇒ Petrol required for 1 km = 8 ÷ 120

⇒ Petrol required for 1 km = 1/15 liters

⇒ Petrol required for 225 km = (1/15) × 225

⇒ Petrol required = 15 liters

∴ The correct answer is 15 liters.

Additional Information

Direct Proportion

Two quantities x and y are in direct proportion if the ratio x/y remains constant. In this case, Distance/Petrol = Constant.

Fuel Efficiency (Mileage)

Mileage is the distance covered per unit of fuel. Mileage = Total Distance ÷ Total Fuel. Here, 120 ÷ 8 = 15 km/liter.

Unitary Method

The method of finding the value of a single unit first and then multiplying to find the required value for multiple units.