Quantitative Aptitude PYQs

Given:

Number of students = 15

Initial average = 64

Wrong marks = 58

Correct marks = 85

New student marks = 10% more than new average

Formula Used:

Average = Total ÷ Number

Total = Average × Number

Percentage increase = Value × (1 + Percentage/100)

Calculation:

⇒ Initial total = 64 × 15 = 960

⇒ Corrected total = 960 - 58 + 85 = 987

⇒ New average = 987 ÷ 15 = 65.8

⇒ New student marks = 65.8 × (1 + 10 ÷ 100) = 65.8 × 1.1 = 72.38

⇒ Final total = 987 + 72.38 = 1059.38

⇒ Final average = 1059.38 ÷ 16 = 66.21

∴ The correct answer is 66.21.

Given:

Total income = 160000

Mutual fund investment = 35%

Fixed deposit investment = 25%

Recurring deposit investment = 40%

Mutual fund growth = 20%

Fixed deposit growth = 10%

Recurring deposit growth = 15%

Formula Used:

Investment = Total × Percentage ÷ 100

Return = Investment × Growth ÷ 100

Total percentage return = (Total return ÷ Total investment) × 100

Calculation:

⇒ Mutual fund investment = 160000 × 35 ÷ 100 = 56000

⇒ Fixed deposit investment = 160000 × 25 ÷ 100 = 40000

⇒ Recurring deposit investment = 160000 × 40 ÷ 100 = 64000

⇒ Mutual fund return = 56000 × 20 ÷ 100 = 11200

⇒ Fixed deposit return = 40000 × 10 ÷ 100 = 4000

⇒ Recurring deposit return = 64000 × 15 ÷ 100 = 9600

⇒ Total return = 11200 + 4000 + 9600 = 24800

⇒ Total percentage return = 24800 ÷ 160000 × 100 = 15.5

∴ The correct answer is 15.5%.

Given:

Increase = 25%

Discount = 20%

Final price = 48000

Formula Used:

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

Final value after discount = New value × (1 - Percentage ÷ 100)

Calculation:

⇒ Let original price = x

⇒ After 25% increase = x × (1 + 25 ÷ 100) = x × 1.25

⇒ After 20% discount = x × 1.25 × (1 - 20 ÷ 100) = x × 1.25 × 0.8

⇒ x × 1.25 × 0.8 = 48000

⇒ x × 1 = 48000

⇒ x = 48000

∴ The correct answer is 48000.

Shortcut Trick

If a number ends in 2, its square must end in 2² = 4.

While all given options end in 4, the result of a square must be a perfect square.

By inspection, 12² = 144, while 124, 134, and 154 are not perfect squares.

∴ The correct answer is 144.

Alternate Method

Given: The unit digit of the original number is 2.

Formula: If the unit digit of a number 'n' is 'a', then the unit digit of n² is the unit digit of a².

⇒ Unit digit of the square = 2 × 2 = 4.

⇒ Let us test consecutive perfect squares ending in 4:

⇒ 2² = 4

⇒ 8² = 64

⇒ 12² = 144

⇒ 18² = 324

⇒ Among the choices provided (124, 144, 134, 154), only 144 is a perfect square.

∴ The correct answer is 144.

Additional Information

Unit Digit of Squares

The square of a number ending in 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 will always end in 0, 1, 4, 9, 6, 5, 6, 9, 4, or 1 respectively.

Properties of Perfect Squares

A perfect square never ends with the digits 2, 3, 7, or 8. Also, the number of zeros at the end of a perfect square is always even.

Digital Sum Property

The digital sum (sum of digits until a single digit is reached) of any perfect square is always 1, 4, 7, or 9.

Shortcut Trick

For any number ending in 25, the square root ends in 5.

Identify the remaining part of the number: 20.

Find two consecutive integers whose product is 20 ⇒ 4 × 5 = 20.

The smaller integer 4 is the tens digit of the root.

∴ The correct answer is 45.

Alternate Method

Given: Number = 2025

Formula Used: Prime Factorization Method for square root calculation.

Calculations:

⇒ 2025 = 5 × 405

⇒ 405 = 5 × 81

⇒ 81 = 3 × 3 × 3 × 3

⇒ 2025 = (5 × 5) × (3 × 3) × (3 × 3)

⇒ √2025 = √(5² × 3² × 3²)

⇒ √2025 = 5 × 3 × 3

⇒ √2025 = 45

∴ The correct answer is 45.

Additional Information

Square of Numbers Ending in 5

To square a number ending in 5 (say A5), the result is always [A × (A+1)] followed by 25. Example: 45² = [4 × 5]25 = 2025.

Perfect Square Identification

A perfect square never ends in the digits 2, 3, 7, or 8. Its digital sum (sum of digits reduced to a single digit) is always 1, 4, 7, or 9.

Estimation Method

Since 40² = 1600 and 50² = 2500, the square root of 2025 must lie between 40 and 50. Since it ends in 5, the only possible value is 45.

Given:

External diameter = 42 cm ⇒ R = 21 cm

Internal diameter = 36 cm ⇒ r = 18 cm

Formula used:

Area = π(R² − r²)

Calculations:

⇒ Area = 3.1416 × (21² − 18²)

⇒ = 3.1416 × (441 − 324)

⇒ = 3.1416 × 117

⇒ = 367.57 cm²

∴ Area = 367.57 cm².

Given:

(125x³ - 343y³) ÷ (5x - 7y) = Ax² - By² + Cxy

Formula Used:

a³ - b³ = (a - b)(a² + ab + b²)

Calculation:

⇒ 125x³ = (5x)³

⇒ 343y³ = (7y)³

⇒ (125x³ - 343y³) = (5x - 7y)(25x² + 35xy + 49y²)

⇒ (125x³ - 343y³) ÷ (5x - 7y) = 25x² + 35xy + 49y²

⇒ Compare with Ax² - By² + Cxy

⇒ A = 25

⇒ -B = 49 ⇒ B = -49

⇒ C = 35

⇒ 2A + B - C = 2 × 25 + (-49) - 35

⇒ 50 - 49 - 35 = -34

∴ The correct answer is -34.

Shortcut Trick

For an equation in the form x² + y² + z² = 2(ax + by + cz) − k, if k = a² + b² + c², then x = a, y = b, and z = c.

Comparing with the given equation: x = 5, y = 6, and z = 4.

Check k: 5² + 6² + 4² = 25 + 36 + 16 = 77 (Matches the constant 77).

Required value = 2(5 + 6) − 3(4) = 2(11) − 12 = 22 − 12 = 10.

∴ The correct answer is 10.

Alternate Method

Given Expression: x² + y² + z² = 2(5x + 6y + 4z) − 77

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

⇒ x² + y² + z² = 10x + 12y + 8z − 77

⇒ x² − 10x + y² − 12y + z² − 8z + 77 = 0

⇒ (x² − 2 × 5 × x + 25) + (y² − 2 × 6 × y + 36) + (z² − 2 × 4 × z + 16) = 0

⇒ (x − 5)² + (y − 6)² + (z − 4)² = 0

Since the sum of perfect squares of real numbers is zero, each term must be zero:

⇒ x − 5 = 0 ⇒ x = 5

⇒ y − 6 = 0 ⇒ y = 6

⇒ z − 4 = 0 ⇒ z = 4

Calculating the required value:

⇒ 2(x + y) − 3z = 2(5 + 6) − 3(4)

⇒ 2(11) − 12 = 22 − 12 = 10

∴ The correct answer is 10.

Additional Information

Perfect Square Identity

The identities (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b² are fundamental in solving quadratic variable equations.

Sum of Squares Principle

For any real numbers A, B, and C, if A² + B² + C² = 0, then A = 0, B = 0, and C = 0. This is used to extract values from combined square equations.

Coefficient Comparison Method

In competitive exams, if x² is on one side and 2(ax) is on the other, x usually equals a provided the constants balance out to complete the square.

Shortcut Trick

The Curved Surface Area (CSA) of a cone can be directly calculated using the base circumference (C) and slant height (l).

Formula: CSA = (1/2) × Circumference × Slant Height

⇒ CSA = (1/2) × 44 × 14

⇒ CSA = 22 × 14 = 308 cm²

∴ The correct answer is 308 cm².

Alternate Method

Given:

Base circumference of the cone = 44 cm

Slant height (l) of the cone = 14 cm

Formula Used:

1. Circumference = 2 × π × r
2. Curved Surface Area (CSA) = π × r × l

Calculations:

⇒ 2 × (22/7) × r = 44

⇒ (44/7) × r = 44

⇒ r = (44 × 7) / 44

⇒ r = 7 cm

Now, calculating the Curved Surface Area:

⇒ CSA = π × r × l

⇒ CSA = (22/7) × 7 × 14

⇒ CSA = 22 × 14

⇒ CSA = 308 cm²

∴ The correct answer is 308 cm².

Additional Information

Total Surface Area of a Cone

It is the sum of the curved surface area and the base area: TSA = πr(l + r).

Volume of a Cone

The space occupied by the cone is given by the formula: V = (1/3)πr²h, where h is the vertical height.

Slant Height Relation

The relationship between slant height (l), radius (r), and height (h) is: l² = r² + h² or l = √(r² + h²).

Shortcut Trick
The expression sec2A − tan2A is a fundamental trigonometric identity.
By identity, for any angle A (where defined), sec2A − tan2A = 1.
The value of sin A = 3/5 is given to define the angle, but the result of the identity remains constant regardless of the specific value of A.
∴ The correct answer is 1.

Alternate Method
Given: sin A = 3/5, where A is an acute angle.
Formula: sin A = Perpendicular (P) / Hypotenuse (H), sec A = H / Base (B), tan A = P / B.

⇒ Let Perpendicular (P) = 3 and Hypotenuse (H) = 5.
⇒ Base (B) = √(H2 − P2) = √(52 − 32) = √(25 − 9) = √16 = 4.
⇒ sec A = H / B = 5/4.
⇒ tan A = P / B = 3/4.
⇒ sec2A − tan2A = (5/4)2 − (3/4)2
⇒ sec2A − tan2A = 25/16 − 9/16 = 16/16 = 1.
∴ The correct answer is 1.

Additional Information
Trigonometric Identities
There are three primary Pythagorean identities: sin2θ + cos2θ = 1, 1 + tan2θ = sec2θ, and 1 + cot2θ = cosec2θ.
Acute Angle Ratios
For an acute angle (0° < A < 90°), all trigonometric ratios (sin, cos, tan, etc.) are positive.
Pythagorean Triplet
The numbers (3, 4, 5) form a Pythagorean triplet, where 32 + 42 = 52, often used in right-angled triangle problems.

Given:

cosx = 12 ÷ 13

Formula Used:

sin²x + cos²x = 1

(a + b)² + (a - b)² = 2(a² + b²)

Calculation:

⇒ sinx = 5 ÷ 13

⇒ (1 + sinx)² + (1 - sinx)² = 2(1² + sin²x)

⇒ = 2(1 + (5 ÷ 13)²)

⇒ = 2(1 + 25 ÷ 169)

⇒ = 2 × 194 ÷ 169 = 388 ÷ 169

⇒ cos²x = (12 ÷ 13)² = 144 ÷ 169

⇒ Required value = (388 ÷ 169) ÷ (144 ÷ 169)

⇒ = 388 ÷ 144

⇒ = 97 ÷ 36

∴ The correct answer is 97 ÷ 36.

Given:

Time₁ = 2

Time₂ = 5

Ratio of amounts = 132 ÷ 115

Formula Used:

Amount = Principal × (1 + Rate × Time ÷ 100)

Calculation:

⇒ Let principal = P, rate = R

⇒ Amount in 2 years = P × (1 + 2R ÷ 100)

⇒ Amount in 5 years = P × (1 + 5R ÷ 100)

⇒ P(1 + 5R ÷ 100) ÷ P(1 + 2R ÷ 100) = 132 ÷ 115

⇒ (1 + 5R ÷ 100) ÷ (1 + 2R ÷ 100) = 132 ÷ 115

⇒ 115(1 + 5R ÷ 100) = 132(1 + 2R ÷ 100)

⇒ 115 + 575R ÷ 100 = 132 + 264R ÷ 100

⇒ 575R ÷ 100 - 264R ÷ 100 = 132 - 115

⇒ 311R ÷ 100 = 17

⇒ 311R = 1700

⇒ R = 1700 ÷ 311

⇒ R ≈ 5.47

∴ The correct answer is 5.47%.

Given:

AB = 9

BC = 12

CD = 11

Formula Used:

In a tangential quadrilateral: AB + CD = BC + AD

Calculation:

⇒ AB + CD = BC + AD

⇒ 9 + 11 = 12 + AD

⇒ 20 = 12 + AD

⇒ AD = 20 - 12 = 8

∴ The correct answer is 8 cm.

Given:

A : B = 3 : 4

B : C = 5 : 9

Total profit = 71000

Formula Used:

Equalize common term in ratios

Share = Total profit × Individual ratio ÷ Sum of ratios

Calculation:

⇒ A : B = 3 : 4

⇒ B : C = 5 : 9

⇒ LCM of 4 and 5 = 20

⇒ A : B = 15 : 20

⇒ B : C = 20 : 36

⇒ A : B : C = 15 : 20 : 36

⇒ Total ratio = 15 + 20 + 36 = 71

⇒ Share of C = 71000 × 36 ÷ 71

⇒ = 1000 × 36 = 36000

∴ The correct answer is 36000.

Given:

4 boys + 6 girls → 8 days

3 boys + 7 girls → 10 days

Formula Used:

Work = Rate × Time

Total work is constant

Calculation:

⇒ Let 1 boy = b, 1 girl = g

⇒ (4b + 6g) × 8 = 1

⇒ 4b + 6g = 1 ÷ 8

⇒ (3b + 7g) × 10 = 1

⇒ 3b + 7g = 1 ÷ 10

⇒ Multiply first by 3: 12b + 18g = 3 ÷ 8

⇒ Multiply second by 4: 12b + 28g = 4 ÷ 10

⇒ Subtract: (12b + 28g) - (12b + 18g) = 4 ÷ 10 - 3 ÷ 8

⇒ 10g = 2 ÷ 5 - 3 ÷ 8

⇒ 10g = 16 ÷ 40 - 15 ÷ 40 = 1 ÷ 40

⇒ g = 1 ÷ 400

⇒ Rate of 10 girls = 10 × 1 ÷ 400 = 1 ÷ 40

⇒ Time = 1 ÷ (1 ÷ 40) = 40

∴ The correct answer is 40 days.

Given:

∠B = 90

AB = 12

AC = 13

Formula Used:

sinA = Opposite ÷ Hypotenuse

cosA = Adjacent ÷ Hypotenuse

In right triangle: Hypotenuse² = Base² + Perpendicular²

Calculation:

⇒ BC² = 13² - 12²

⇒ BC² = 169 - 144 = 25

⇒ BC = 5

⇒ sinA = BC ÷ AC = 5 ÷ 13

⇒ cosA = AB ÷ AC = 12 ÷ 13

⇒ sinA + cosA = 5 ÷ 13 + 12 ÷ 13

⇒ = 17 ÷ 13

∴ The correct answer is 17 ÷ 13.

Given:

secθ + tanθ = 5

Formula Used:

(secθ + tanθ)(secθ - tanθ) = 1

secθ = 1 ÷ cosθ

tanθ = sinθ ÷ cosθ

Calculation:

⇒ secθ - tanθ = 1 ÷ 5

⇒ Add: (secθ + tanθ) + (secθ - tanθ) = 5 + 1 ÷ 5

⇒ 2secθ = 26 ÷ 5

⇒ secθ = 13 ÷ 5

⇒ cosθ = 5 ÷ 13

⇒ tanθ = secθ - 1 ÷ 5 = 13 ÷ 5 - 1 ÷ 5 = 12 ÷ 5

⇒ sinθ = tanθ × cosθ

⇒ sinθ = (12 ÷ 5) × (5 ÷ 13) = 12 ÷ 13

∴ The correct answer is 12 ÷ 13.

Shortcut Trick

By definition, a Chord is a straight line segment whose endpoints both lie on a circular arc.

The longest chord of any circle is the one that passes through the center.

This specific central chord is uniquely defined as the Diameter of the circle.

∴ The correct answer is Diameter.

Alternate Method

Given: A chord that passes through the center of a circle.

Step-by-step Explanation:

⇒ A Chord is any line segment joining two points on the circumference of a circle.

⇒ If the endpoints of the chord are A and B, and the segment AB passes through Center O, then AO and OB are both radii.

⇒ Length of Chord AB = AO + OB = Radius + Radius.

⇒ Since 2 × Radius = Diameter, the chord passing through the center is the Diameter.

⇒ This is the longest chord possible in a circle.

∴ The correct answer is Diameter.

Additional Information

Radius

A line segment connecting the center of the circle to any point on its boundary; it is exactly half of the diameter (r = d ÷ 2).

Secant

A straight line that intersects a circle at two distinct points, extending infinitely in both directions.

Tangent

A straight line that touches the circle at exactly one point and is perpendicular to the radius at that point.

Circumference

The linear distance around the edge of the circle, calculated using the formula 2 × π × r or π × d.

Given:

Radius = 10

Chord₁ = 16

Chord₂ = 12

Formula Used:

Distance from center to chord = √(r² - (l ÷ 2)²)

Calculation:

⇒ Distance for chord 16 = √(10² - 8²)

⇒ = √(100 - 64) = √36 = 6

⇒ Distance for chord 12 = √(10² - 6²)

⇒ = √(100 - 36) = √64 = 8

⇒ Case 1 (same side) = |8 - 6| = 2

⇒ Case 2 (opposite sides) = 8 + 6 = 14

∴ The correct answer is 2 cm or 14 cm.

Given:

Distance = 180

Time = 6

Car speed increase = 20%

Bike speed increase = 25%

Required distance = 50% more than 180 = 270

Formula Used:

Speed = Distance ÷ Time

Time = Distance ÷ Speed

New speed = Original × (1 + Percentage ÷ 100)

Calculation:

⇒ Bus speed = 180 ÷ 6 = 30

⇒ Car speed = 30 × (1 + 20 ÷ 100) = 30 × 1.2 = 36

⇒ Bike speed = 36 × (1 + 25 ÷ 100) = 36 × 1.25 = 45

⇒ Time taken by car = 270 ÷ 36 = 7.5 hours

⇒ Time taken by bike = 270 ÷ 45 = 6 hours

⇒ Difference = 7.5 - 6 = 1.5 hours

⇒ Difference in minutes = 1.5 × 60 = 90

∴ The correct answer is 90 minutes.