Quantitative Aptitude PYQs

Shortcut Trick

Total age 4 years ago = 44 − (2 × 4) = 36 years.

Ratio of Priya to Rohan 4 years ago = 2 :

1. Sum of ratio parts (2 + 1) = 3 units → 36 years ⇒ 1 unit = 12 years.

Priya age 4 years ago = 2 × 12 = 24 years.

Present age of Priya = 24 + 4 = 28 years.

∴ The correct answer is 28 years.

Alternate Method

Given: Sum of Priya (P) and Rohan (R) = 44 years. Condition: (P − 4) = 2 × (R − 4).

⇒ P + R = 44 ⇒ R = 44 − P

⇒ P − 4 = 2 × ((44 − P) − 4)

⇒ P − 4 = 2 × (40 − P)

⇒ P − 4 = 80 − 2P

⇒ 3P = 84

⇒ P = 28

∴ The correct answer is 28 years.

Additional Information

Age Difference Rule

The difference between the ages of two persons remains constant throughout their lives, regardless of the number of years passed.

Sum of Ages Change

If the sum of ages of 'n' people is S, then 't' years ago, the sum was S − (n × t), and 't' years hence, it will be S + (n × t).

Shortcut Trick

Given equation: 1/(1 − tanθ) + 1/(1 + tanθ) = 3

Simplify: 2/(1 − tan²θ) = 3 ⇒ 1 − tan²θ = 2/3 ⇒ tan²θ = 1/3

For acute angle θ, tanθ = 1/√3 ⇒ θ = 30°

Therefore, cosecθ = cosec 30° = 2.

∴ The correct answer is 2.

Alternate Method

Given:

1/(1 − tanθ) + 1/(1 + tanθ) = 3

Formula Used:

(a − b)(a + b) = a² − b²

1 + tan²θ = sec²θ

cosecθ = 1/sinθ

Calculations:

⇒ [(1 + tanθ) + (1 − tanθ)] / [(1 − tanθ)(1 + tanθ)] = 3

⇒ 2 / (1 − tan²θ) = 3

⇒ 2 = 3 − 3 tan²θ

⇒ 3 tan²θ = 1

⇒ tan²θ = 1/3

⇒ tanθ = 1/√3 (since θ is an acute angle, tanθ must be positive)

⇒ θ = 30°

⇒ cosecθ = cosec 30° = 2

∴ The correct answer is 2.

Additional Information

Trigonometric Identities

Recall that sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, and cosec²θ − cot²θ = 1.

Standard Values for 30°

For θ = 30°, sinθ = 1/2, cosθ = √3/2, tanθ = 1/√3, and cosecθ = 2.

Shortcut Trick

Let the Marked Price be 100%.

Given Discount = 50%, so Selling Price = 100% − 50% = 50%.

The extra amount without discount represents the discount value itself.

Since 50% discount = ₹8,453, then the amount paid (50%) is also ₹8,453.

∴ The correct answer is ₹8,453.

Alternate Method

Given: Discount % = 50%, Extra amount if no discount = ₹8,453.

Formulas: Discount = MP − SP | Discount % = (Discount ÷ MP) × 100 | SP = MP − Discount

⇒ Let the Marked Price (MP) be x.

⇒ If no discount is given, she pays the Marked Price. The extra amount = MP − SP = Discount.

⇒ Discount = 50% of x = ₹8,453.

⇒ 0.50 × x = 8,453 ⇒ x = 8,453 ÷ 0.50 = ₹16,906.

⇒ Amount paid (SP) = MP − Discount = 16,906 − 8,453 = ₹8,453.

∴ The correct answer is ₹8,453.

Additional Information

Marked Price & Discount

The Marked Price is the price listed on the label. Discount is always calculated on the Marked Price unless stated otherwise.

The 50% Rule

When a 50% discount is applied, the Selling Price and the Discount Value are always equal, as the price is split exactly in half.

Shortcut Trick

Rate of 5% in fraction = 1/20. For 2 years, the ratio of Principal to Amount is (20)² : (21)².

Ratio = 400 : 441

Given, 400 units = ₹4,400 ⇒ 1 unit = ₹11

Amount = 441 units × 11 = ₹4,851

∴ The correct answer is ₹4,851.

Alternate Method

Given:

Principal (P) = ₹4,400

Rate (R) = 5% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 4400 × (1 + 5/100)²

⇒ A = 4400 × (1 + 1/20)²

⇒ A = 4400 × (21/20) × (21/20)

⇒ A = 4400 × (441/400)

⇒ A = 11 × 441

⇒ A = ₹4,851

∴ The correct answer is ₹4,851.

Additional Information

Effective Interest Rate for 2 Years

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

Difference between CI and SI

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

Compounding Half-Yearly

When interest is compounded half-yearly, the rate is halved (R/2) and the time is doubled (2n).

Shortcut Trick

Wages are distributed in the ratio of the efficiencies of the workers when they work for the same duration.

Efficiency ratio A : B : C = 1/17 : 1/51 : 1/68 = 12 : 4 : 3

Total ratio parts = 12 + 4 + 3 = 19 units

B's share = (4 ÷ 19) × 5,985 = 4 × 315 = ₹1,260

∴ The correct answer is ₹1,260.

Alternate Method

Given: Time taken by A = 17 days, B = 51 days, C = 68 days. Total wage = ₹5,985.

Formula Used: Total Work = LCM of time; Efficiency = Total Work ÷ Time; Wage ∝ Efficiency.

⇒ Total efficiency = 12 + 4 + 3 = 19 units

⇒ 19 units = ₹5,985

⇒ 1 unit = 5,985 ÷ 19 = ₹315

⇒ B's wage = 4 units = 4 × 315 = ₹1,260

∴ The correct answer is ₹1,260.

Additional Information

Wage Distribution Rule

If workers work for the same amount of time, the wage is distributed in the ratio of their efficiencies. If time periods differ, wage ratio = (E₁×T₁) : (E₂×T₂).

Efficiency-Time Relationship

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

Work and Wages

Total Wage = (Individual Work Done ÷ Total Work) × Total Amount.

Shortcut Trick

Let Total Voters be 100%.

Voters who cast votes = 100% − 20% = 80%.

Winner gets 70% of 80%, Loser gets 30% of 80%.

Difference = (70% − 30%) of 80% = 40% of 80% = 32%.

32% = 23,200 ⇒ 1% = 725 ⇒ 100% = 72,500.

∴ The correct answer is 72,500.

Alternate Method

Given:

Non-voters = 20%

Winner share of cast votes = 70%

Winning margin = 23,200

Calculations:

⇒ Let total number of voters = x

⇒ Votes cast = 80% of x = 0.8x

⇒ Winner's votes = 70% of 0.8x = 0.56x

⇒ Loser's votes = (100% − 70%) of 0.8x = 30% of 0.8x = 0.24x

⇒ Margin = 0.56x − 0.24x = 0.32x

⇒ 0.32x = 23,200

⇒ x = 23,200 ÷ 0.32

⇒ x = 72,500

∴ The correct answer is 72,500.

Additional Information

Invalid Votes Concept

If some votes are invalid, subtract them from the cast votes before calculating the winner's share: Valid Votes = Cast Votes − Invalid Votes.

Percentage Change

Margin percentage based on total voters is (Winner% − Loser%) × (Cast Vote%).

Total Population vs Voters

Voters are usually a subset of the total population; questions may involve multiple stages of reduction from Total Population → Voter List → Casted Votes.

Shortcut Trick

Speed is distance per unit time.

Direct division: 120 ÷ 2 = 60 km/h.

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Distance = 120 km

Time = 2 hours

Formula Used:

Speed = Distance ÷ Time

Calculations:

⇒ Speed = 120 km ÷ 2 hours

⇒ Speed = 60 km/h

∴ The correct answer is 60 km/h.

Additional Information

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

Total distance traveled divided by the total time taken. For two equal distances at speeds x and y, Average Speed = 2xy ÷ (x + y).

Shortcut Trick

Let Cost Price (CP) be 100.

Marked Price (MP) = 100 + 25% of 100 = 125.

Selling Price (SP) = 100 + 15% of 100 = 115.

Discount = MP − SP = 125 − 115 = 10.

Discount % = (Discount ÷ MP) × 100 = (10 ÷ 125) × 100 = 8%.

∴ The correct answer is 8%.

Alternate Method

Given: Markup% = 25%, Profit% = 15%.
Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = CP × (100 + Profit%) ÷ 100 | Discount% = (MP − SP) ÷ MP × 100.

⇒ Assume CP = 100 units.
⇒ MP = 100 × (125 ÷ 100) = 125 units.
⇒ SP = 100 × (115 ÷ 100) = 115 units.
⇒ Discount Amount = MP − SP = 125 − 115 = 10 units.
⇒ Discount% = (10 ÷ 125) × 100 = 8%.

∴ The correct answer is 8%.

Additional Information

Marked Price & Discount

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

Relation between CP and MP

The direct ratio is MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%).

Shortcut Trick

Base circumference (2πr) = 176 cm ⇒ Radius (r) = 176 ÷ (2 × 22/7) = 28 cm.

Total Surface Area (TSA) = 2πr(h + r) = (Circumference) × (h + r).

TSA = 176 × (56 + 28) = 176 × 84.

Calculation: 176 × 84 = 14,784 cm².

∴ The correct answer is 14,784 cm².

Alternate Method

Given: Height (h) = 56 cm, Circumference = 176 cm.

Formula Used: Circumference = 2πr | TSA = 2πr(h + r).

⇒ 2 × (22/7) × r = 176

⇒ (44/7) × r = 176

⇒ r = 176 × (7/44) = 4 × 7 = 28 cm

⇒ TSA = 2 × (22/7) × 28 × (56 + 28)

⇒ TSA = 176 × 84

⇒ TSA = 14,784 cm²

∴ The correct answer is 14,784 cm².

Additional Information

Curved Surface Area of Cylinder

The area of the side surface only, given by CSA = 2πrh.

Volume of Cylinder

The total space occupied by the cylinder, calculated as V = πr²h.

Base Area of Cylinder

The area of one of the circular ends, given by Area = πr².

Shortcut Trick

For an equilateral triangle, the sum of perpendicular distances (p₁, p₂, p₃) from an interior point to the sides is equal to the height (h).

h = p₁ + p₂ + p₃ = 12.7 + 14.8 + 11.5 = 39 cm.

Area = h² / √3 = 39² / √3 = 1521 / √3 = 507√3 cm².

∴ The correct answer is 507√3 cm².

Alternate Method

Given: Perpendiculars p₁ = 12.7 cm, p₂ = 14.8 cm, p₃ = 11.5 cm.

Formula: Sum of perpendiculars (p₁+p₂+p₃) = (√3/2) × side (a).

⇒ Let the side of the equilateral triangle be 'a'.

⇒ Area of triangle = Sum of areas of three smaller triangles formed with the interior point.

⇒ (√3/4)a² = (1/2)×a×p₁ + (1/2)×a×p₂ + (1/2)×a×p₃

⇒ (√3/4)a² = (1/2)a (p₁ + p₂ + p₃)

⇒ (√3/2)a = 12.7 + 14.8 + 11.5 = 39

⇒ a = 78 / √3 = 26√3 cm

⇒ Area = (√3/4) × (26√3)²

⇒ Area = (√3/4) × 676 × 3 = 169 × 3 × √3 = 507√3 cm²

∴ The correct answer is 507√3 cm².

Additional Information

Viviani's Theorem

It states that the sum of the distances from any interior point to the sides of an equilateral triangle is constant and equal to the altitude (h) of the triangle.

Equilateral Triangle Formulas

Height (h) = (√3/2)a; Area = (√3/4)a²; In-radius (r) = a/(2√3); Circum-radius (R) = a/√3.

Area in terms of Height

If the height 'h' is known, the area of an equilateral triangle can be directly calculated as Area = h² / √3.

Shortcut Trick

4.2 / 6.3 = 2/3

x/2.1 + 2/3 = 4

x/2.1 = 4 − 2/3 = 10/3

x = (10/3) × 2.1 = (10/3) × (21/10) = 7

∴ Answer = 7

Alternate Method

Given:

x/2.1 + 4.2/6.3 = 4

Formula used:

Simplification of fractions

Calculation:

⇒ 4.2/6.3 = 42/63 = 2/3

⇒ Equation becomes:

x/2.1 + 2/3 = 4

⇒ x/2.1 = 4 − 2/3

⇒ = (12 − 2)/3 = 10/3

⇒ x = (10/3) × 2.1

⇒ = (10/3) × (21/10)

⇒ = 7

∴ The correct answer is 7.

Additional Information

Fraction Conversion

Convert decimals to fractions for easy solving

Equation Solving

Isolate variable step-by-step

Simplification

Reduce fractions before multiplying

Key Insight

Convert all terms into common format

Shortcut Trick

Difference in Distance = 15 − 10 = 5 km

Difference in Fare = ₹330 − ₹230 = ₹100

Per-kilometre Charge = ₹100 ÷ 5 = ₹20 per km

Base Fare = Total Fare − (Distance × Per-km Charge) = 230 − (10 × 20) = ₹30

∴ The correct answer is Base fare = ₹30; Per-kilometre Charge = ₹20.

Alternate Method

Given:

Let Base Fare be 'B' and Rate per km be 'R'.

⇒ Case 1: B + 10R = 230 — (i)

⇒ Case 2: B + 15R = 330 — (ii)

Calculations:

Subtracting equation (i) from (ii):

⇒ (B + 15R) − (B + 10R) = 330 − 230

⇒ 5R = 100

⇒ R = 20

Substitute R = 20 in equation (i):

⇒ B + 10(20) = 230

⇒ B + 200 = 230

⇒ B = 30

∴ The correct answer is Base fare = ₹30; Per-kilometre Charge = ₹20.

Additional Information

Linear Equation Modeling

This problem follows the format y = mx + c, where 'm' is the variable rate, 'c' is the fixed constant (base fare), and 'y' is the total cost.

Simultaneous Equations

To find two unknowns, we require at least two independent linear equations. The elimination method is usually the fastest way to solve for the variable rate.

Unitary Application

In fixed plus variable cost problems, the entire difference in total cost is attributed to the difference in variable quantity only.

Shortcut Trick

Combined efficiency of A, B and C = (Total Work ÷ 3) + (Total Work ÷ 10) + (Total Work ÷ 15)

Using LCM (3, 10, 15) = 30 as base work.

Total efficiency = 10 + 3 + 2 = 15 units/day.

Twice the work = 30 × 2 = 60 units.

Time taken = 60 ÷ 15 = 4 days.

∴ The correct answer is 4 days.

Alternate Method

Given: Time taken by A = 3 days, B = 10 days, C = 15 days.

Formula Used: Total Work = LCM of individual days; Efficiency = Total Work ÷ Time; Time = Work ÷ Efficiency.

⇒ Combined efficiency of (A + B + C) = 10 + 3 + 2 = 15 units/day.

⇒ Original work = 30 units.

⇒ Twice the work = 30 × 2 = 60 units.

⇒ Required time = 60 ÷ 15 = 4 days.

∴ The correct answer is 4 days.

Additional Information

Work and Efficiency Relationship

Efficiency is inversely proportional to the time taken to complete a task. If time ratio is x : y, then efficiency ratio is y :

x. Pipes and Cisterns

Similar logic applies where filling pipes have positive efficiency and leakage pipes have negative efficiency.

Work and Wages

Wages are distributed in the ratio of work done or the ratio of efficiencies when all persons work for the same duration.

Shortcut Trick
For any distribution following the Empirical Relationship, if the Mode and Median are equal, the Mean must also be equal to that same value.
Given: Mode = 46 and Median = 46.
Observation: Since Mode = Median, the distribution is considered symmetrical at this point, leading to Mean = 46.
∴ The correct answer is 46.
Alternate Method
Given:
⇒ Mode = 46
⇒ Median = 46
Formula Used:
The Empirical Relationship between the three measures of central tendency is:
⇒ Mode = 3 × Median − 2 × Mean

Calculations:
⇒ 46 = 3 × 46 − 2 × Mean
⇒ 46 = 138 − 2 × Mean
⇒ 2 × Mean = 138 − 46
⇒ 2 × Mean = 92
⇒ Mean = 92 ÷ 2
⇒ Mean = 46
∴ The correct answer is 46.
Additional Information
Empirical Formula
Karl Pearson's formula relates the three averages as: Mode = 3 Median − 2 Mean. It is applicable for moderately skewed distributions.
Symmetrical Distribution
In a perfectly Symmetrical Distribution (like the Normal Distribution), the values of Mean, Median, and Mode are all identical.
Skewed Distributions
For Positively Skewed data: Mean > Median > Mode. For Negatively Skewed data: Mode > Median > Mean.

Shortcut Trick

Apply the algebraic identity (a − b)² = a² − 2ab + b².

Square of the first term: (3m)² = 9m².

Square of the second term: (5n)² = 25n².

The middle term is −2 × (3m) × (5n) = −30mn.

Combine all parts: 9m² − 30mn + 25n².

∴ The correct answer is 9m² − 30mn + 25n².

Alternate Method

Given: (3m − 5n)²

Formula Used: (a − b)² = a² − 2ab + b²

⇒ Identifying values: a = 3m and b = 5n.

⇒ Expansion: (3m)² − 2 × (3m) × (5n) + (5n)²

⇒ (3 × 3 × m × m) − (2 × 3 × 5 × m × n) + (5 × 5 × n × n)

⇒ 9m² − 30mn + 25n²

⇒ This expression corresponds to Option 4.

∴ The correct answer is 9m² − 30mn + 25n².

Additional Information

Square of a Sum Identity

(a + b)² = a² + 2ab + b². The only difference is the sign of the middle term.

Difference of Two Squares

(a − b)(a + b) = a² − b². This is used for factorization and simplifying products.

Cubic expansion of Difference

(a − b)³ = a³ − 3a²b + 3ab² − b³. Note the alternating signs in the result.

Shortcut Trick

The angle formed by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the other two angles:

⇒ ∠STR = (∠P + ∠Q) ÷ 2

⇒ 17° = (∠P + ∠Q) ÷ 2 ⇒ ∠P + ∠Q = 34°

Given ratio ∠P : ∠Q = 8 : 9, the sum of ratio parts is 8 + 9 = 17 units.

Difference between ∠Q and ∠P = 9 − 8 = 1 unit = 34° × (1 ÷ 17) = 2°.

∴ The correct answer is 2°.

Alternate Method

Given:

⇒ In quadrilateral PQRS, the bisectors of ∠R and ∠S meet at point T.

⇒ ∠STR = 17°

⇒ Ratio of ∠P to ∠Q = 8 : 9

Formula Used:

⇒ Sum of interior angles of a quadrilateral = 360°

⇒ Sum of angles in a triangle = 180°

Calculations:

In triangle STR:

⇒ ∠STR + ∠TSR + ∠TRS = 180°

Since ST and RT are the angle bisectors of ∠S and ∠R respectively:

⇒ ∠TSR = ∠S ÷ 2

⇒ ∠TRS = ∠R ÷ 2

Substituting these values into the triangle angle sum equation:

⇒ 17° + (∠S ÷ 2) + (∠R ÷ 2) = 180°

⇒ (∠S + ∠R) ÷ 2 = 180° − 17°

⇒ (∠S + ∠R) ÷ 2 = 163°

⇒ ∠S + ∠R = 326°

In quadrilateral PQRS, the sum of all interior angles is 360°:

⇒ ∠P + ∠Q + ∠R + ∠S = 360°

⇒ ∠P + ∠Q + 326° = 360°

⇒ ∠P + ∠Q = 360° − 326°

⇒ ∠P + ∠Q = 34°

Given that the ratio of ∠P to ∠Q is 8 : 9, let ∠P = 8x and ∠Q = 9x:

⇒ 8x + 9x = 34°

⇒ 17x = 34°

⇒ x = 2°

The difference between the measures of ∠Q and ∠P is:

⇒ ∠Q − ∠P = 9x − 8x = x

⇒ x = 2°

∴ The correct answer is 2°.

Additional Information

Angle Bisector Theorem in Quadrilaterals

The angle formed by the bisectors of two adjacent angles of a quadrilateral is always equal to half the sum of the remaining two opposite angles.

Sum of Interior Angles of a Polygon

For any convex polygon with n sides, the sum of all interior angles is given by the formula (n − 2) × 180°.

Shortcut Trick

When distance is constant, Speed is inversely proportional to Time.

Speed Ratio (Usual : New) = 7 : 4 → Time Ratio (Usual : New) = 4 : 7.

Difference in time units = 7 − 4 = 3 units.

Given, 3 units = 12 minutes ⇒ 1 unit = 4 minutes.

Usual Time = 4 units × 4 minutes = 16 minutes.

∴ The correct answer is 16 min.

Alternate Method

Given: New Speed = 4/7 of Normal Speed, Delay = 12 minutes.

Formula Used: Usual Time = [Late time × Numerator] ÷ [Difference between Numerator and Denominator]

⇒ Let Usual Speed be S and Usual Time be T.

⇒ Distance D = S × T

⇒ New Speed = 4/7 S, New Time = T + 12

⇒ D = (4/7 S) × (T + 12)

⇒ S × T = 4/7 S × (T + 12)

⇒ 7T = 4T + 48

⇒ 3T = 48

⇒ T = 16 minutes.

∴ The correct answer is 16 min.

Additional Information

Proportionality Rule

If Speed becomes a/b of its original value, the Time taken becomes b/a of the original time.

Average Speed

Average speed for equal distances at different speeds x and y is given by 2xy ÷ (x + y).

Relative Speed

When two objects move in the same direction, relative speed is the difference of their speeds.

Shortcut Trick

3125 = 5⁵ ⇒ x − y = 5

2187 = 3⁷ ⇒ 3x + y = 7

Solve ⇒ x = 3, y = −2

19x − 18y = 19×3 − 18(−2)

= 57 + 36 = 93

∴ Answer = 93

Alternate Method

Given:

5^{x − y} = 3125

3^{3x + y} = 2187

Formula used:

a^m = aⁿ ⇒ m = n

Calculation:

⇒ 3125 = 5⁵

⇒ x − y = 5 ...(1)

⇒ 2187 = 3⁷

⇒ 3x + y = 7 ...(2)

⇒ Add (1) and (2)

⇒ 4x = 12 ⇒ x = 3

⇒ Substitute in (1)

⇒ 3 − y = 5 ⇒ y = −2

⇒ 19x − 18y = 19×3 − 18(−2)

⇒ = 57 + 36

⇒ = 93

∴ The correct answer is 93.

Additional Information

Exponent Rule

If bases are same, equate powers

Simultaneous Equations

Solve linear equations step-by-step

Substitution Method

Find one variable and substitute

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%

∴ The correct answer is 10% decrease.

Alternate Method

Given: First increase = 20%, Second decrease = 25%

Formula Used: Final Price = Initial Price × (100 + %change₁)/100 × (100 + %change₂)/100

⇒ Let the initial price be 100 units.

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

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

⇒ 120 − (25/100 × 120) = 120 − 30 = 90 units.

⇒ Net change = 100 − 90 = 10 units.

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

∴ The correct answer is 10% decrease.

Additional Information

Successive Percentage Change

When a value is changed by x% and then by y%, the effective change is (x + y + xy/100)%. Use positive for increase and negative for decrease.

Fractional Conversion

20% increase is a multiplier of 6/5 and 25% decrease is a multiplier of 3/4. Net multiplier = 6/5 × 3/4 = 18/20 = 9/10, which indicates a 1/10 or 10% decrease.

Product Stability

If an item price increases by x%, the consumption must decrease by [x/(100+x)] × 100 to keep the expenditure constant.

Shortcut Trick

Expenditure ratio is the product of Price ratio and Quantity ratio.

Expenditure (Jowar : Gram) = (1 × 4) : (5 × 3)

Expenditure (Jowar : Gram) = 4 : 15

∴ The correct answer is 4 : 15.

Alternate Method

Given:

Ratio of Market Prices (Jowar : Gram) = 1 : 5

Ratio of Quantities Consumed (Jowar : Gram) = 4 : 3

Formula Used:

Expenditure = Price × Quantity

Calculations:

⇒ Let the prices be 1x and 5x respectively.

⇒ Let the quantities be 4y and 3y respectively.

⇒ Expenditure on Jowar = 1x × 4y = 4xy

⇒ Expenditure on Gram = 5x × 3y = 15xy

⇒ Ratio of expenditure = 4xy : 15xy

⇒ Ratio = 4 : 15

∴ The correct answer is 4 : 15.

Additional Information

Ratio Multiplication

When two independent variables contribute to a third variable (like Price and Quantity to Expenditure), their respective ratios are multiplied component-wise.

Inverse Relationship

If expenditure is constant, price and quantity are inversely proportional (Price₁/Price₂ = Quantity₂/Quantity₁).

Percentage Change in Expenditure

Net % change in expenditure = a + b + (a × b)/100, where a and b are percentage changes in price and quantity.