Quantitative Aptitude PYQs

Given:

First number = 2⁴ × 3² × 5³ × 7

Second number = 23 × 5 × 7 × 11

Formula used:

HCF = Product of common prime factors with the smallest powers

Calculation:

Prime factors of first number

⇒ 2⁴, 3², 5³, 7

Prime factors of second number

⇒ 23, 5, 7, 11

Common prime factors = 5 and 7

Smallest powers

⇒ 5¹, 7¹

⇒ HCF = 5 × 7

⇒ 35

∴ HCF = 35.

Given:

Initial population = 40000

First year increase = 20%

Second year decrease = 10%

Third year increase = 25%

Formula Used:

New value after increase = Original value × (1 + Percentage ÷ 100)

New value after decrease = Original value × (1 − Percentage ÷ 100)

Calculation:

⇒ Population after first year = 40000 × (1 + 20 ÷ 100)

⇒ Population = 40000 × 1.20

⇒ Population = 48000

⇒ Population after second year = 48000 × (1 − 10 ÷ 100)

⇒ Population = 48000 × 0.90

⇒ Population = 43200

⇒ Population after third year = 43200 × (1 + 25 ÷ 100)

⇒ Population = 43200 × 1.25

⇒ Population = 54000

∴ The correct answer is 54000.

Given:

Thief's speed = 20 km/h

Police car's speed = 30 km/h

Thief starts at 10:00 AM

Police car starts at 10:30 AM

Formula Used:

Distance = Speed × Time

Relative speed = Speed of police car - Speed of thief

Time = Distance ÷ Relative speed

Calculation:

⇒ Thief's head start time = 10:30 AM - 10:00 AM = 0.5 hours

⇒ Distance covered by thief in 0.5 hours = 20 × 0.5 = 10 km

⇒ Relative speed = 30 - 20 = 10 km/h

⇒ Time taken by police car to catch thief = 10 ÷ 10 = 1 hour

⇒ Police car catches thief at 10:30 AM + 1 hour = 11:30 AM

∴ The correct answer is 11:30 AM.

Shortcut Trick

Total work = LCM(15, 30) = 30 units.

Efficiency of A = 2 and B = 1 (Combined = 3 units/day).

Work done by (A+B) in 5 days = 3 × 5 = 15 units.

Remaining work = 30 − 15 = 15 units.

Time for B to finish remaining work = 15 ÷ 1 = 15 days.

∴ The correct answer is 15 days.

Alternate Method

Given: Time taken by A = 15 days and B = 30 days. They work together for 5 days.

Formula: Total Work = LCM of individual time; Efficiency = Total Work ÷ Time taken.

⇒ Combined efficiency of A + B = 2 + 1 = 3 units/day.

⇒ Work done by A and B together in 5 days = 5 × 3 = 15 units.

⇒ Remaining work = Total Work − Completed Work = 30 − 15 = 15 units.

⇒ Time taken by B to complete the remaining work = 15 ÷ 1 = 15 days.

∴ The correct answer is 15 days.

Additional Information

Relation between Time and Efficiency

Time and efficiency are inversely proportional to each other. If the ratio of time taken is x:y, then the ratio of efficiency is y:x.

Pipes and Cisterns

The same LCM method applies to pipes, where filling pipes have positive efficiency and outlet/leakage pipes have negative efficiency.

Work and Wages

Wages are distributed among workers based on the ratio of the actual work done by each person.

Given:

Selling price = 1800

Profit = 20%

Formula Used:

Selling price = Cost price × (1 + Profit ÷ 100)

Calculation:

⇒ 1800 = Cost price × (1 + 20 ÷ 100)

⇒ 1800 = Cost price × 1.20

⇒ Cost price = 1800 ÷ 1.20

⇒ Cost price = 1500

∴ The correct answer is 1500.

Shortcut Trick

For Simple Interest, use the ratio: (n₁ − 1) / T₁ = (n₂ − 1) / T₂

Here, n₁ = 2 (doubles), T₁ = 8 years, and n₂ = 5 (five times).

(2 − 1) / 8 = (5 − 1) / T₂

1 / 8 = 4 / T₂ ⇒ T₂ = 4 × 8 = 32 years.

∴ The correct answer is 32 years.

Alternate Method

Given: Sum doubles (A = 2P) in 8 years.

Formula: Simple Interest (SI) = (P × R × T) / 100

⇒ Case 1: SI₁ = Amount − Principal = 2P − P = P

⇒ P = (P × R × 8) / 100 ⇒ R = 100 / 8 = 12.5%

⇒ Case 2: New Amount = 5P, so SI₂ = 5P − P = 4P

⇒ 4P = (P × 12.5 × T₂) / 100

⇒ 4 = (12.5 × T₂) / 100

⇒ T₂ = (4 × 100) / 12.5 = 400 / 12.5 = 32 years

∴ The correct answer is 32 years.

Additional Information

Simple Interest Multiplier Rule

If a sum becomes n times in T years, the rate R is given by R = [(n − 1) × 100] / T.

Constant Interest Property

In Simple Interest, the interest earned for every equal interval of time remains the same.

Compound Interest Comparison

In Compound Interest, if a sum doubles in T years, it becomes n^k times in k × T years.

Shortcut Trick

Investment Scheme A : Scheme B = 150 : 100 = 3 : 2

Given Scheme A = ₹2,400 ⇒ 3 units = ₹2,400, so 2 units (Scheme B) = ₹1,600

Total SI = (2,400 × 4 × r/100) + (1,600 × 5 × (r+2)/100) = 1,920

96r + 80r + 160 = 1,920 ⇒ 176r = 1,760 ⇒ r = 10%

∴ The correct answer is 10%.

Alternate Method

Given:

Investment in Scheme A (P_A) = ₹2,400

Time for Scheme A (T_A) = 4 years, Rate (R_A) = r%

Investment in Scheme B (P_B): P_A is 50% more than P_B ⇒ 2,400 = 1.5 × P_B

Time for Scheme B (T_B) = 5 years, Rate (R_B) = (r + 2)%

Total Simple Interest = ₹1,920

Formula Used: Simple Interest (SI) = (P × R × T) / 100

⇒ P_B = 2,400 / 1.5 = ₹1,600

⇒ SI_A = (2,400 × r × 4) / 100 = 96r

⇒ SI_B = (1,600 × (r + 2) × 5) / 100 = 80(r + 2) = 80r + 160

⇒ Total SI = SI_A + SI_B = 1,920

⇒ 96r + 80r + 160 = 1,920

⇒ 176r = 1,920 − 160

⇒ 176r = 1,760

⇒ r = 1,760 / 176 = 10

∴ The correct answer is 10%.

Additional Information

Simple Interest (SI)

The interest is calculated only on the initial Principal amount for the entire duration. Formula: SI = (P × R × T) / 100.

Amount in SI

The total money accumulated after the time period. Amount = Principal + Simple Interest.

Effective Interest Rate

For Simple Interest, the total effective rate over T years is simply (R × T)% of the principal.

Given:

a + b + c = 12

ab + bc + ca = 40

Formula Used:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Calculation:

⇒ (12)² = a² + b² + c² + 2 × 40

⇒ 144 = a² + b² + c² + 80

⇒ a² + b² + c² = 144 − 80

⇒ a² + b² + c² = 64

∴ The correct answer is 64.

Shortcut Trick

A perfect square never ends with the unit digits 2, 3, 7, or 8.

Checking the given options by their unit digits:

(1) 245982 ends in 2 (Eliminated) | (2) 496838 ends in 8 (Eliminated)

(3) 178343 ends in 3 (Eliminated) | (4) 571536 ends in 6 (Possible)

∴ The correct answer is 571536.

Alternate Method

Given numbers: 245982, 496838, 178343, 571536.

Method Used: Unit digit verification and Digital Sum analysis (The sum of digits of a perfect square must be 1, 4, 7, or 9).

Analyze 571536:

⇒ Unit digit = 6. Property: If a square ends in 6, its tens digit must be odd.

⇒ Tens digit of 571536 is 3 (Odd), which satisfies the condition.

⇒ Digital sum of 571536 = 5 + 7 + 1 + 5 + 3 + 6 = 27.

⇒ Reduced digital sum = 2 + 7 = 9.

⇒ Since 9 is a valid digital sum for squares, 571536 is a potential perfect square (756² = 571536).

∴ The correct answer is 571536.

Additional Information

Unit Digit Rule

A perfect square always ends in 0, 1, 4, 5, 6, or 9. It never ends in 2, 3, 7, or 8.

Digital Sum Rule

The sum of the digits of a perfect square (root digit) is always 1, 4, 7, or 9. For example, 12² = 144 (1+4+4=9).

Last Two Digits Rule

The last two digits of any perfect square must be among the last two digits of the squares of numbers from 1 to 24.

Ending in 6 Rule

If a perfect square ends in 6, the digit in the tens place is always an odd number (e.g., 16, 36, 196, 256).

Shortcut Trick

Ratio of the milk solutions = 4 : 6 = 2 : 3.

Net fat percentage = [(Ratio₁ × Fat₁) + (Ratio₂ × Fat₂)] ÷ Total Ratio parts.

Net fat percentage = [(2 × 25) + (3 × 40)] ÷ (2 + 3) = (50 + 120) ÷ 5 = 170 ÷ 5 = 34%.

∴ The correct answer is 34%.

Alternate Method

Given:

Quantity of first solution = 4 liters with 25% fat.

Quantity of second solution = 6 liters with 40% fat.

Formula Used:

Total Fat Content = (Quantity₁ × Fat %₁) + (Quantity₂ × Fat %₂)

Final Fat Percentage = (Total Fat Content ÷ Total Quantity) × 100

⇒ Fat in 4 liters of solution = 4 × (25 ÷ 100) = 1.0 liter

⇒ Fat in 6 liters of solution = 6 × (40 ÷ 100) = 2.4 liters

⇒ Total quantity of final mixture = 4 + 6 = 10 liters

⇒ Total fat in final mixture = 1.0 + 2.4 = 3.4 liters

⇒ Final fat percentage = (3.4 ÷ 10) × 100 = 34%

∴ The correct answer is 34%.

Additional Information

Rule of Alligation

It states that the ratio of the quantities is the inverse ratio of the differences from the mean: (Quantity₁ ÷ Quantity₂) = (C₂ − M) ÷ (M − C₁), where M is the mean value.

Weighted Average

The combined average of two or more groups can be calculated using A_avg = (n₁a₁ + n₂a₂ + ...) ÷ (n₁ + n₂ + ...).

Percentage Concentration

In mixture problems, concentration refers to the amount of solute (like fat or salt) present in the total volume of the solution.

Given:

Increase in radius = 20%

Formula Used:

Volume of sphere = 4πr³ ÷ 3

Volume ∝ r³

Calculation:

⇒ New radius = r × (1 + 20 ÷ 100)

⇒ New radius = 1.2r

⇒ New volume ∝ (1.2r)³

⇒ New volume ∝ 1.2³ × r³

⇒ New volume ∝ 1.728r³

⇒ Increase factor = 1.728 − 1

⇒ Increase factor = 0.728

⇒ Percentage increase = 0.728 × 100

⇒ Percentage increase = 72.8%

∴ The correct answer is 72.8%.

Given:

Total surface area of hemisphere = 300π

Formula Used:

Total surface area of hemisphere = 3πr²

Volume of hemisphere = 2πr³ ÷ 3

Calculation:

⇒ 3πr² = 300π

⇒ r² = 300π ÷ 3π

⇒ r² = 100

⇒ r = 10

⇒ Volume = 2π × 10³ ÷ 3

⇒ Volume = 2π × 1000 ÷ 3

⇒ Volume = 2000π ÷ 3

∴ The correct answer is 2000π ÷ 3.

Given:

1 ÷ 2

0.55

3 ÷ 5

0.625

Formula Used:

Fraction to decimal = Numerator ÷ Denominator

Calculation:

⇒ 1 ÷ 2 = 0.5

⇒ 3 ÷ 5 = 0.6

⇒ 0.5 < 0.55 < 0.6 < 0.625

⇒ 1 ÷ 2 < 0.55 < 3 ÷ 5 < 0.625

∴ The correct answer is 1 ÷ 2, 0.55, 3 ÷ 5, 0.625.

Given:

Area of the circle = 144π cm²

Formula Used:

Area of a circle = πr²

Circumference of a circle = 2πr

Calculation:

⇒ πr² = 144π

⇒ r² = 144

⇒ r = √144

⇒ r = 12 cm

⇒ Circumference = 2πr

⇒ Circumference = 2π × 12

⇒ Circumference = 24π cm

⇒ Circumference ≈ 75.4 cm

∴ The correct answer is 75.4 cm.

Given:

27³ ÷ 3⁵ = 3ⁿ

Formula Used:

a^m ÷ aⁿ = a^{(m − n)}

27 = 3³

Calculation:

⇒ 27³ = (3³)³

⇒ 27³ = 3⁹

⇒ 3⁹ ÷ 3⁵

⇒ 3^{(9 − 5)}

⇒ 3⁴

⇒ n = 4

∴ The correct answer is 4.

Given:

x − y = 8

xy = 5

Formula Used:

x³ − y³ = (x − y)(x² + xy + y²)

(x − y)² = x² − 2xy + y²

Calculation:

⇒ x² − 2xy + y² = 8²

⇒ x² + y² = 64 + 2 × 5 = 74

⇒ x² + xy + y² = 74 + 5 = 79

⇒ x³ − y³ = 8 × 79 = 632

⇒ (x + y)² = x² + 2xy + y²

⇒ (x + y)² = 74 + 10 = 84

⇒ 8(x + y)² = 8 × 84 = 672

⇒ x³ − y³ + 8(x + y)² = 632 + 672 = 1304

∴ The correct answer is 1304.

Given:

\(\frac{(\sin(x) + \cos(x))^2}{1 + \sin(x)\cos(x)}\)

Formula Used:

\((a + b)^2 = a^2 + b^2 + 2ab\)

\(\sin^2(x) + \cos^2(x) = 1\)

Calculation:

⇒ \((\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x\)

⇒ \(\frac{1 + 2\sin x \cos x}{1 + \sin x \cos x}\)

∴ The simplified form is \(\frac{1 + 2\sin(x)\cos(x)}{1 + \sin(x)\cos(x)}\).

Given:

sin A = 5/13

Formula Used:

sin²A + cos²A = 1

tan A = sin A / cos A

Calculation:

⇒ sin²A = (5/13)² = 25/169

⇒ cos²A = 1 - sin²A = 1 - 25/169 = 144/169

⇒ cos A = √(144/169) = 12/13

⇒ tan A = sin A / cos A = (5/13) ÷ (12/13)

⇒ tan A = 5/12

∴ The correct answer is 5/12.

Given:

\(\tan A = \frac{1}{\sqrt{3}}\)

Formula Used:

\(2 \sin A \cos A = \sin 2A\)

\(\tan 30^\circ = \frac{1}{\sqrt{3}}\)

Calculation:

⇒ \(\tan A = \frac{1}{\sqrt{3}} \Rightarrow A = 30^\circ\)

⇒ \(2 \sin A \cos A = \sin 2A\)

⇒ \(\sin(2 \times 30^\circ) = \sin 60^\circ\)

⇒ \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)

∴ The correct answer is \(\frac{\sqrt{3}}{2}\).

Given:

5x − 2y = 10

Formula Used:

Equation of line: y = mx + c

Slope of perpendicular line = −1 ÷ m

Calculation:

⇒ 5x − 2y = 10

⇒ −2y = −5x + 10

⇒ y = (5/2)x − 5

⇒ Slope of given line m = 5/2

⇒ Slope of perpendicular line = −1 ÷ (5/2)

⇒ Slope = −2/5

∴ The correct answer is −2/5.