Quantitative Aptitude PYQs

Shortcut Trick
Let the side of the square (and length of rectangle) be s.
New square area − Rectangle area = 3
(s − 4)2 − (6 × s) = 3
Testing option 3: Area of rectangle = 78 ⇒ 6 × s = 78 ⇒ s = 13.
Check: (13 − 4)2 − 78 = 92 − 78 = 81 − 78 = 3. (Satisfied)
∴ The correct answer is 78 cm2.
Alternate Method

Given:
Width of rectangle (w) = 6 cm
Length of rectangle (l) = Side of square (s)
New side of square = s − 4
Formula Used:
Area of Rectangle = length × width
Area of Square = (side)2
Calculations:
⇒ Area of Rectangle = 6 × s
⇒ Area of New Square = (s − 4)2
⇒ (s − 4)2 = 6s + 3
⇒ s2 − 8s + 16 = 6s + 3
⇒ s2 − 14s + 13 = 0
⇒ (s − 13)(s − 1) = 0
⇒ s = 13 (Since s − 4 must be positive, s > 4)
⇒ Area of rectangle = 6 × 13 = 78 cm2
∴ The correct answer is 78 cm2.
Additional Information
Area and Perimeter of Rectangle
Area = length × width; Perimeter = 2 × (length + width).
Area and Perimeter of Square
Area = side2; Perimeter = 4 × side; Diagonal = side × √2.
Quadratic Roots
For ax2 + bx + c = 0, roots are found by factorisation or x = [−b ± √(b2 − 4ac)] / 2a.

Shortcut Trick

Speed of Train A = 36 × (5/18) = 10 m/s. Length A = 10 × 20 = 200 m.

Length B = 200 + 100 = 300 m. Platform = 500 m.

Total Distance for B = 300 + 500 = 800 m. Speed B = 18 × (5/18) = 5 m/s.

Time taken by B = 800 ÷ 5 = 160 sec.

∴ The correct answer is 160 sec.

Alternate Method

Given: Speed of Train A = 36 km/hr, Time to cross pole = 20 sec, Length of Train B = Length A + 100 m, Speed of Train B = 18 km/hr, Platform length = 500 m.

Formula Used: Distance = Speed × Time; Time = Distance ÷ Speed; 1 km/hr = 5/18 m/s.

⇒ Speed of Train A in m/s = 36 × (5/18) = 10 m/s.

⇒ Length of Train A = Speed × Time = 10 × 20 = 200 m.

⇒ Length of Train B = 200 + 100 = 300 m.

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

⇒ Total Distance to be covered by B = Length of B + Length of Platform = 300 + 500 = 800 m.

⇒ Time taken by B = 800 ÷ 5 = 160 sec.

∴ The correct answer is 160 sec.

Additional Information

Unit Conversion

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

Train Crossing Concepts

When a train crosses a pole or stationary person, distance = length of train. When it crosses a platform or bridge, distance = length of train + length of platform.

Relative Speed

If two trains move in the same direction, relative speed = S₁ − S₂. If they move in opposite directions, relative speed = S₁ + S₂.

General solution:

Agent A

Number of health insurance sold to males and females = 150

Number of health insurance sold to males = 50

Number of health insurance sold to females = (150 - 50) = 100

Number of vehicle insurance sold to females = 85

Number of vehicle insurance sold to males = 85/5 ×  4 = 68

Agent B

Number of health insurance sold to males and females = 120

Number of health insurance sold to males = 50

Number of health insurance sold to females = (120 - 50) = 70

Number of vehicle insurance sold to females = 70

Number of vehicle insurance sold to males = 70/5 ×  4 = 56 

Agent C

Number of health insurance sold to males and females = 140

Number of health insurance sold to males = 50

Number of health insurance sold to females = (140 - 50) = 90

Number of vehicle insurance sold to females = 55

Number of vehicle insurance sold to males = 55/5 ×  4 = 44

Solution:

Number of health insurance sold by A = 150

Number of vehicle insurance sold by A = 153

If agent A sold '2y' more health and 'y' more vehicle insurance then

(150 + 2y) - (153 + y) = 27

-3 + y =  27

y = 30

General solution:

Agent A

Number of health insurance sold to males and females = 150

Number of health insurance sold to males = 50

Number of health insurance sold to females = (150 - 50) = 100

Number of vehicle insurance sold to females = 85

Number of vehicle insurance sold to males = 85/5 ×  4 = 68

Agent B

Number of health insurance sold to males and females = 120

Number of health insurance sold to males = 50

Number of health insurance sold to females = (120 - 50) = 70

Number of vehicle insurance sold to females = 70

Number of vehicle insurance sold to males = 70/5 ×  4 = 56 

Agent C

Number of health insurance sold to males and females = 140

Number of health insurance sold to males = 50

Number of health insurance sold to females = (140 - 50) = 90

Number of vehicle insurance sold to females = 55

Number of vehicle insurance sold to males = 55/5 ×  4 = 44

Solution:

Number of health and vehicle insurance sold by A = (100 + 50 + 68 + 85) = 303

Number of health and vehicle insurance sold by C = (50 + 90 + 44 + 55) = 239

Required ratio = (303: 239)

General solution:

Agent A

Number of health insurance sold to males and females = 150

Number of health insurance sold to males = 50

Number of health insurance sold to females = (150 - 50) = 100

Number of vehicle insurance sold to females = 85

Number of vehicle insurance sold to males = 85/5 ×  4 = 68

Agent B

Number of health insurance sold to males and females = 120

Number of health insurance sold to males = 50

Number of health insurance sold to females = (120 - 50) = 70

Number of vehicle insurance sold to females = 70

Number of vehicle insurance sold to males = 70/5 ×  4 = 56 

Agent C

Number of health insurance sold to males and females = 140

Number of health insurance sold to males = 50

Number of health insurance sold to females = (140 - 50) = 90

Number of vehicle insurance sold to females = 55

Number of vehicle insurance sold to males = 55/5 ×  4 = 44

Solution: 

Number of health insurance done by B = (50 + 70) = 120

Number of vehicle insurance done by C = (55 + 44) = 99

Difference between the total cost = 99 ×  2000 - 120 ×  1500 = (198000 - 180000) = 18000

General solution:

Agent A

Number of health insurance sold to males and females = 150

Number of health insurance sold to males = 50

Number of health insurance sold to females = (150 - 50) = 100

Number of vehicle insurance sold to females = 85

Number of vehicle insurance sold to males = 85/5 ×  4 = 68

Agent B

Number of health insurance sold to males and females = 120

Number of health insurance sold to males = 50

Number of health insurance sold to females = (120 - 50) = 70

Number of vehicle insurance sold to females = 70

Number of vehicle insurance sold to males = 70/5 ×  4 = 56 

Agent C

Number of health insurance sold to males and females = 140

Number of health insurance sold to males = 50

Number of health insurance sold to females = (140 - 50) = 90

Number of vehicle insurance sold to females = 55

Number of vehicle insurance sold to males = 55/5 ×  4 = 44

Solution:

Number of persons sold health insurance = 100

Total number of health insurance sold to females by all the agents = (100 + 70 + 90) = 260

Required percentage = 100/260 ×  100 = 38.46%

General solution:

Agent A

Number of health insurance sold to males and females = 150

Number of health insurance sold to males = 50

Number of health insurance sold to females = (150 - 50) = 100

Number of vehicle insurance sold to females = 85

Number of vehicle insurance sold to males = 85/5 ×  4 = 68

Agent B

Number of health insurance sold to males and females = 120

Number of health insurance sold to males = 50

Number of health insurance sold to females = (120 - 50) = 70

Number of vehicle insurance sold to females = 70

Number of vehicle insurance sold to males = 70/5 ×  4 = 56 

Agent C

Number of health insurance sold to males and females = 140

Number of health insurance sold to males = 50

Number of health insurance sold to females = (140 - 50) = 90

Number of vehicle insurance sold to females = 55

Number of vehicle insurance sold to males = 55/5 ×  4 = 44

Solution:

Vehicle insurance sold by B to all the people is = (56 + 70) = 126

21% = 126

100% = 600

Then the health insurance sold by all the people = (50 + 70) = 120

Required percentage = 120/600 × 100 = 20%

Shortcut Trick
Combined efficiency of A + B + C = 1/40, A = 1/20, B = 1/24.
Efficiency of Emptying Pipe C = (1/20 + 1/24) − 1/40 = (6 + 5 − 3)/120 = 8/120 = 1/15.
Time taken by C to empty the cistern = 15 hours.
Total Capacity = Emptying Rate × Time = 160 × 15 = 2400 liters.
∴ The correct answer is 2400 liters.
Alternate Method
Pipe Time (Hours) Total Work (LCM) Efficiency A (Inlet) 20 120 units +6 B (Inlet) 24 +5 A + B + C 40 +3
⇒ Total efficiency of A + B = 6 + 5 = 11 units/hr.
⇒ Efficiency of Pipe C = (A + B + C) − (A + B) = 3 − 11 = −8 units/hr.
⇒ Time taken by Pipe C alone to empty = 120 ÷ 8 = 15 hours.
⇒ Given: Pipe C empties 160 liters in 1 hour.
⇒ Total Capacity = 160 liters/hr × 15 hours = 2400 liters.
∴ The correct answer is 2400 liters.
Additional Information
Inlet and Outlet Pipes
Inlet pipes carry a positive efficiency (filling), while outlet/leakage pipes carry a negative efficiency (emptying).
Capacity Formula
Total Capacity = Rate of flow × Total time taken to fill or empty the tank.
Pipes & Cisterns Logic
If a tank is filled in 'x' hours and emptied in 'y' hours (y > x), the net part filled in 1 hour is (1/x − 1/y).

Shortcut Trick

Initial Ratio Alcohol : Water = 7 : 5 (Total 12 units)

Add 6L Alcohol and 5L Water → New Ratio = 34 : 25

Using Cross-Multiplication Method: (7 × 25) − (34 × 5) units = (34 × 5) − (25 × 6)

⇒ |175 − 170| units = |170 − 150| ⇒ 5 units = 20 litres ⇒ 1 unit = 4 litres

⇒ Initial Total Mixture = 12 units = 12 × 4 = 48 litres

∴ The correct answer is 48 litres.

Alternate Method

Given: Initial ratio = 7 : 5. Added Alcohol = 6L, Added Water = 5L. Final ratio = 34 : 25.

Let initial quantity of Alcohol be 7x and Water be 5x.

⇒ (7x + 6) / (5x + 5) = 34 / 25

⇒ 25 × (7x + 6) = 34 × (5x + 5)

⇒ 175x + 150 = 170x + 170

⇒ 175x − 170x = 170 − 150

⇒ 5x = 20 ⇒ x = 4

⇒ Initial Mixture = 7x + 5x = 12x = 12 × 4 = 48 litres.

∴ The correct answer is 48 litres.

Additional Information

Cross-Multiplication Technique

Useful for ratio problems where different constants are added/subtracted to parts: (a × d − b × c) units = (c × k₂ − d × k₁).

Mixture Replacement Formula

Final Quantity = Initial Quantity × (1 − Replacement/Total)ⁿ, where n is the number of times the process is repeated.

Alligation Rule

Used to find the ratio in which two mixtures of different concentrations are mixed to get a desired mean concentration.

The series follows the following pattern:

82 – 1 = 63

92 – 1 = 80

102 – 1 = 99

112 – 1 = 120

122 – 1 = 143

132 – 1 = 168

The series follows the following pattern:

7 × 1 + 1 = 8 

8 × 2 + 1 = 17 

17 × 3 + 1 = 52

52 × 4 + 1 = 209 

209 × 5 + 1 = 1046

The series follows the following pattern:

16 + 144 = 160

160 + 121 = 281

281 + 100 = 381

381 + 81 = 462

462 + 64 = 526

103 + 8³ = 615

615 + 7³ = 958

958 + 6³ = 1174

1174 + 5³ = 1299

1299 + 4³ = 1363

1363 + 3³ = 1390

Calculation:

The series follows the following pattern:

⇒ 7 × 1 = 7

⇒ 7 × 2 = 14

⇒ 14 × 3 = 42

⇒ 42 × 4 = 168

⇒ 168 × 5 = 840

∴ The missing term in the series is 42

Shortcut Trick

Total outcomes = 50 (numbers 1 to 50).

Multiples of 8 up to 50 = 50 ÷ 8 ≈ 6 (Specifically: 8, 16, 24, 32, 40, 48).

Probability = Favorable outcomes ÷ Total outcomes = 6 ÷ 50 = 3/25.

∴ The correct answer is 3/25.

Alternate Method

Given:

Total number of tickets = 50.

Condition: Drawn ticket must be a multiple of 8.

Formula Used:

Probability P(E) = n(E) ÷ n(S)

Calculations:

⇒ Sample space n(S) = {1, 2, 3, ..., 50} = 50.

⇒ Favorable outcomes E = {8, 16, 24, 32, 40, 48}.

⇒ Number of favorable outcomes n(E) = 6.

⇒ P(E) = 6 ÷ 50

⇒ P(E) = 3 ÷ 25

∴ The correct answer is 3/25.

Additional Information

Definition of Probability

Probability is the measure of likelihood of an event occurring, calculated as favorable outcomes divided by total possible outcomes. 0 ≤ P(E) ≤ 1.

Complementary Event

The probability of an event not occurring is P(E') = 1 − P(E). For this problem, probability of not picking a multiple of 8 is 1 − 3/25 = 22/25.

Multiples in a Range

The count of multiples of 'k' in the range 1 to 'n' is given by the greatest integer function [n/k]. Here, [50/8] = 6.

Shortcut Trick

Let the largest prime number in Set A be P and the largest odd number in Set B be Q.

Given: P + Q = 64 and P = 3Q − 4.

Substitute P: (3Q − 4) + Q = 64 ⇒ 4Q = 68 ⇒ Q = 17.

Then P = 64 − 17 = 47.

Set A consists of 3 consecutive prime numbers ending in 47: 41, 43, 47.

∴ The correct answer is 41.

Alternate Method

Given: Largest number of Set A = P, Largest number of Set B = Q.

Equations:

⇒ P + Q = 64 --- (i)

⇒ P = 3Q − 4 --- (ii)

Calculation:

⇒ Put (ii) in (i): (3Q − 4) + Q = 64

⇒ 4Q = 68

⇒ Q = 17 (Largest odd number of Set B)

⇒ P = 64 − 17 = 47 (Largest prime number of Set A)

⇒ Set A contains 3 consecutive primes: 41, 43, 47.

⇒ Smallest number = 41.

∴ The correct answer is 41.

Additional Information

Prime Numbers

A number greater than 1 with exactly two factors: 1 and itself. Consecutive primes after 40 are 41, 43, 47, 53, 59.

Consecutive Odd Numbers

Numbers of the form 2n + 1. The difference between two consecutive odd numbers is always 2.

Linear Equations

Solving systems of equations using substitution or elimination methods to find unknown variables.

Shortcut Trick

Total percentage spent = 40% + 20% + 10% = 70%

Remaining savings percentage = 100% − 70% = 30%

Given, 30% of income = ₹15,000

100% of income = (15,000 ÷ 30) × 100 = ₹50,000

∴ The correct answer is ₹50,000.

Alternate Method

Given:

Expenditure on household items = 40%

Expenditure on clothes = 20%

Investment in recurring deposits (RD) = 10%

Actual savings = ₹15,000

Calculations:

⇒ Total percentage utilized = 40% + 20% + 10% = 70%

⇒ Percentage saved = 100% − 70% = 30%

⇒ Let total income be X.

⇒ 30% of X = 15,000

⇒ (30 ÷ 100) × X = 15,000

⇒ X = (15,000 × 100) ÷ 30

⇒ X = 500 × 100

⇒ X = 50,000

∴ The correct answer is ₹50,000.

Additional Information

Successive Percentage Spending

If a person spends X% and then Y% of the remaining amount, the final value is calculated as Initial × (100−X)/100 × (100−Y)/100.

Income-Expenditure Relationship

Monthly Income = Total Expenditure + Savings. This basic identity is used to solve most percentage-based budget problems.

Fractional Conversions

Convert percentages to fractions to speed up calculations: 20% = 1/5, 40% = 2/5, 10% = 1/10.

Shortcut Trick
Evaluate values directly for given a = −3, b = −1, c = 6.
Quantity I = (−27 + 1 + 36) ÷ 4 = 10 ÷ 4 = 2.5
Quantity II = −3 ÷ (−27 + 1) = −3 ÷ −26 = 3/26 ≈ 0.115
Since 2.5 > 0.115, Quantity I > Quantity II.
∴ The correct answer is Quantity I > Quantity II.
Alternate Method
Given: a = −3, b = −1, c = 6
Quantity I Calculation:
⇒ a3 = (−3)3 = −27
⇒ b2 = (−1)2 = 1
⇒ c2 = (6)2 = 36
⇒ Quantity I = (−27 + 1 + 36) ÷ 4
⇒ Quantity I = 10 ÷ 4 = 2.5
Quantity II Calculation:
⇒ a / (a3 + b2) = −3 / [(−3)3 + (−1)2]
⇒ Quantity II = −3 / (−27 + 1)
⇒ Quantity II = −3 / −26 = 3/26 ≈ 0.115
Comparison:
⇒ 2.5 > 0.115
∴ The correct answer is Quantity I > Quantity II.
Additional Information
Power of Negative Numbers
A negative number raised to an odd power results in a negative value: (−x)n = −xn. A negative number raised to an even power results in a positive value: (−x)n = xn.
Comparison of Quantities
Always ensure both quantities are in the same form (decimal or fraction) before comparing. For fractions, cross-multiplication can be used to determine which is larger.

Perimeter of a square = 2 × 2 × (11 + 10) = 84 cm

⇒ Side of the square = 84/4 = 21 cm

Quantity A:

Circumference of the semicircle whose diameter is equal to the side of the square = 21 + 22/7 × 21/2

⇒ 21 + 33 = 54 cm

Quantity B:

Circumference of the circle which is inscribed inside the square = 2 × 22/7 × 21/2 = 66 cm

∴ Quantity A < Quantity B

Shortcut TrickQuantity A: Speed X = 420/7 = 60 kmph. Ratio 5:3 → Speed Y = (3/5)×60 = 36 kmph. Relative speed = 60 + 36 = 96 kmph. Distance in 3 hrs = 96×3 = 288 km. Initial gap = 100 km. Crossing occurs, then new distance = 288 − 100 = 188 km.Quantity B: Speed = 300/12 = 25 m/s = 25×18/5 = 90 kmph. Distance in 2 hrs = 90×2 = 180 km.Comparison: 188 km > 180 km ⇒ A > B.∴ The correct answer is Quantity A > Quantity B.Alternate MethodQuantity A:⇒ Speed of Train X = 420 km ÷ 7 hours = 60 kmph.⇒ Speed of Train Y = (60 ÷ 5) × 3 = 36 kmph.⇒ Relative Speed (Towards each other) = 60 + 36 = 96 kmph.⇒ Total Distance covered in 3 hours = 96 × 3 = 288 km.⇒ Final Distance between them = |Initial distance − Distance covered| = |100 − 288| = 188 km.Quantity B:⇒ Speed of Bus = 300 m ÷ 12 s = 25 m/s.⇒ Speed in kmph = 25 × (18 ÷ 5) = 90 kmph.⇒ Distance in 2 hours = 90 × 2 = 180 km.⇒ Comparing Quantity A (188 km) and Quantity B (180 km), A > B.∴ The correct answer is Quantity A > Quantity B.Additional InformationRelative SpeedWhen two objects move towards each other, relative speed is the sum of their individual speeds (S1 + S2). If moving in the same direction, it is the difference |S1 − S2|.Unit ConversionTo convert speed from m/s to kmph, multiply by 18/5. To convert kmph to m/s, multiply by 5/18.Distance After CrossingIf objects cover more distance than the initial gap, they cross each other. New distance = |Initial Gap − Relative Distance Covered|.