Quantitative Aptitude PYQs

Given:

0, 1, 2, 9

Formula Used:

Natural numbers start from 1 and include all positive integers.

Calculation:

⇒ Natural numbers: {1, 2, 3, 4, ...}

⇒ 0 is not part of the natural numbers.

∴ The correct answer is Option 1.

Given:

Natural numbers: {1, 2, 3, ...}

Integer numbers: {..., -3, -2, -1, 0, 1, 2, 3, ...}

Formula Used:

Natural numbers are positive whole numbers starting from 1.

Integers include all whole numbers, both positive and negative, including zero.

Calculation:

⇒ Number 0 is an integer but not a natural number.

⇒ Number 5 is both a natural number and an integer.

⇒ Number -5 is an integer but not a natural number.

⇒ None is not a valid option as there is a number common to both sets.

∴ The correct answer is 5.

Given:

Number of boxes = 200

Cost per box = 120

Extra expense = 1800

100 boxes sold at 150 each

100 boxes sold at 130 each

Formula Used:

Profit% = (Profit ÷ Total Cost Price) × 100

Calculation:

⇒ Cost of 200 boxes = 200 × 120 = 24000

⇒ Total cost price = 24000 + 1800 = 25800

⇒ Selling price of first 100 boxes = 100 × 150 = 15000

⇒ Selling price of remaining 100 boxes = 100 × 130 = 13000

⇒ Total selling price = 15000 + 13000 = 28000

⇒ Profit = 28000 − 25800 = 2200

⇒ Profit% = (2200 ÷ 25800) × 100

⇒ Profit% ≈ 8.53

∴ The correct answer is 8.53%.

Given:

\(\angle QOR = 110^\circ\)

O is the incenter of triangle PQR

Formula Used:

\(\angle BIC = 90^\circ + \frac{1}{2}\angle A\) (Angle at incenter)

Calculation:

⇒ \(\angle QOR = 90^\circ + \frac{1}{2}\angle QPR\)

⇒ \(110 = 90 + \frac{1}{2}\angle QPR\)

⇒ \(\frac{1}{2}\angle QPR = 20\)

⇒ \(\angle QPR = 40^\circ\)

∴ The correct answer is \(40^\circ\).

Given:

(tan20°)² / (cosec70°)² + (cot20°)² / (sec70°)² + 2tan15° tan45° tan75°

Formula Used:

cosecθ = 1 / sinθ

secθ = 1 / cosθ

sin(70°) = cos(20°)

cos(70°) = sin(20°)

tan45° = 1

tan15° × tan75° = 1

Calculation:

⇒ (tan20°)² / (cosec70°)² = tan²20° × sin²70° = tan²20° × cos²20° = sin²20°

⇒ (cot20°)² / (sec70°)² = cot²20° × cos²70° = cot²20° × sin²20° = cos²20°

⇒ sin²20° + cos²20° = 1

⇒ 2tan15° tan45° tan75° = 2 × 1 × 1 = 2

⇒ Total value = 1 + 2 = 3

∴ The correct answer is 3.

Given:

2 × sin q × cos(90° − q) = 1/2

0° < q < 90°

Formula Used:

cos(90° − θ) = sin θ

sin²θ + cos²θ = 1

sin(2θ) = 2sinθcosθ

Calculation:

⇒ 2 × sin q × cos(90° − q) = 2 × sin q × sin q = 1/2

⇒ 2sin²q = 1/2

⇒ sin²q = 1/4

⇒ sin q = 1/2

⇒ q = 30°

⇒ sin(2q) + cos(2q) = sin60° + cos60°

⇒ = √3/2 + 1/2

⇒ = (√3 + 1)/2

∴ The correct answer is (√3 + 1)/2.

Given:

Amount after 4 years = 9504

Amount after 9 years = 12384

Formula Used:

Simple Interest = Principal × Rate × Time ÷ 100

Amount = Principal + Simple Interest

Calculation:

⇒ Interest for 5 years = 12384 − 9504 = 2880

⇒ Interest per year = 2880 ÷ 5 = 576

⇒ Interest for 4 years = 576 × 4 = 2304

⇒ Principal = 9504 − 2304 = 7200

⇒ 576 = (7200 × R) ÷ 100

⇒ R = (576 × 100) ÷ 7200

⇒ R = 8

∴ The correct answer is 8%.

Given:

Amount after 8 years = 240% of principal

Principal = 3000

Time = 3 years

Formula Used:

Simple interest = Principal × Rate × Time ÷ 100

Calculation:

⇒ 240% = Principal + Interest

⇒ Interest in 8 years = 240% − 100% = 140%

⇒ Rate per year = 140 ÷ 8 = 17.5%

⇒ Simple interest = 3000 × 17.5 × 3 ÷ 100

⇒ Simple interest = 1575

∴ The correct answer is 1575.

Given:

70% of total money distributed between Priya and Meena

Ratio (Priya : Meena) = 5 : 9

Rates = 6% and 10%

Time = 4 years

Difference in interest = 2304

Formula Used:

Simple Interest = Principal × Rate × Time ÷ 100

Calculation:

⇒ Let total money = T

⇒ Distributed money = 70% of T = 0.7T

⇒ Priya share = 0.7T × 5 ÷ 14 = 0.25T

⇒ Meena share = 0.7T × 9 ÷ 14 = 0.45T

⇒ Priya interest = 0.25T × 6 × 4 ÷ 100 = 0.06T

⇒ Meena interest = 0.45T × 10 × 4 ÷ 100 = 0.18T

⇒ Difference = 0.18T − 0.06T = 0.12T

⇒ 0.12T = 2304

⇒ T = 2304 ÷ 0.12 = 19200

∴ The correct answer is 19200.

Given:

Length of stick = 48 cm

Angle with floor = 60°

Formula Used:

Height = Hypotenuse × sin θ

Calculation:

⇒ Height = 48 × sin 60°

⇒ Height = 48 × (√3 ÷ 2)

⇒ Height = 24√3

∴ The correct answer is 24√3 cm.

Shortcut Trick

In a 30°−60°−90° triangle, the side opposite to the 30° angle is always half of the hypotenuse.

Here, the rope acts as the hypotenuse (16 m) and the house is the side opposite to 30°.

Height of house = Hypotenuse ÷ 2

Height of house = 16 ÷ 2 = 8 metres.

∴ The correct answer is 8 metres.

Alternate Method

Given:

Length of the rope (Hypotenuse, AC) = 16 metres

Angle of elevation (θ) = 30°

Formula Used:

sin θ = Perpendicular (Height) ÷ Hypotenuse

Calculations:

⇒ sin 30° = Height ÷ 16

⇒ 1/2 = Height ÷ 16

⇒ Height = 16 × (1/2)

⇒ Height = 8 metres

∴ The correct answer is 8 metres.

Additional Information

Trigonometric Ratios

Standard values: sin 30° = 1/2, cos 30° = √3/2, and tan 30° = 1/√3.

30°−60°−90° Triangle Rule

The sides of such a triangle are in the ratio 1 : √3 : 2, where 1 is opposite to 30° and 2 is the hypotenuse.

Angle of Elevation

The angle formed by the line of sight with the horizontal when looking upwards at an object.

Given:

Pipe A fills the tank in 14 hours

Pipe B empties the tank in 8 hours

Initial water in tank = 1/4

Formula Used:

Total work = LCM of 14 and 8

Work per hour = Total work ÷ Time

Calculation:

⇒ LCM of 14 and 8 = 56

⇒ Total work = 56 units

⇒ Work by A in 1 hour = 56 ÷ 14 = 4

⇒ Work emptied by B in 1 hour = 56 ÷ 8 = 7

⇒ Net work per hour = 4 − 7 = −3

⇒ Water initially in tank = 56 ÷ 4 = 14 units

⇒ Time to empty 14 units = 14 ÷ 3

⇒ Time = 4 hour 40 min

∴ The correct answer is 4 hour 40 min.

Given:

4 women + 6 girls can do work in 12 days

24 women + 46 girls can do work in 2 days

Let:

Work done by 1 woman in 1 day = W

Work done by 1 girl in 1 day = G

Formula used:

Work = Rate × Time

Calculations:

From first condition:

⇒ (4W + 6G) × 12 = Total work

From second condition:

⇒ (24W + 46G) × 2 = Total work

Since work is same:

⇒ (4W + 6G) × 12 = (24W + 46G) × 2

⇒ 48W + 72G = 48W + 92G

⇒ 48W cancel

⇒ 72G = 92G

⇒ 92G − 72G = 0

⇒ 20G = 0 ⇒ G = 0

⇒ Put G = 0 in (4W + 6G) × 12 = Work

⇒ (4W) × 12 = Work

⇒ Work = 48W

⇒ From second: (24W) × 2 = 48W 

Now for 15 women + 20 girls:

⇒ Work per day = 15W + 20G

⇒ = 15W + 0 = 15W

⇒ Time = Total work ÷ Rate

⇒ Time = 48W ÷ 15W

⇒ Time = 48 ÷ 15 = 3.2 days

⇒ 0.2 day = 0.2 × 24 = 4.8 hours

⇒ 0.8 hour = 0.8 × 60 = 48 minutes

∴ Required time = 3 days 4 hours 48 minutes

Shortcut Trick

Height of taller pole (H) = Height of smaller pole (h) + distance × tan(θ)

H = 10 + 50 × tan(30º)

H = 10 + 50 × (1 / √3) ≈ 10 + 50 × 0.57735

H = 10 + 28.87 = 38.87 m

∴ The correct answer is 38.87 m.

Alternate Method

Given:

Height of smaller pole (AB) = 10 m

Distance between poles (BD) = 50 m

Angle of elevation from top of smaller pole (θ) = 30º

Let the taller pole be CD and the smaller pole be AB.

From the figure, AB = ED = 10 m and AE = BD = 50 m.

In right-angled triangle AEC:

⇒ tan(30º) = CE / AE

⇒ 1 / √3 = CE / 50

⇒ CE = 50 / √3

⇒ CE = 50 / 1.732

⇒ CE ≈ 28.87 m

Total height of the taller pole (CD) = CE + ED

⇒ CD = 28.87 + 10

⇒ CD = 38.87 m

∴ The correct answer is 38.87 m.

Additional Information

Angle of Elevation

The angle formed by the line of sight with the horizontal when an object is viewed above the horizontal level.

Trigonometric Ratios

In a right-angled triangle, tan θ = Perpendicular / Base. Common value: tan 30º = 1/√3 ≈ 0.577, tan 45º = 1, tan 60º = √3 ≈ 1.732.

Angle of Depression

The angle formed by the line of sight with the horizontal when an object is viewed below the horizontal level.

Shortcut Trick

The radius, distance from the center, and half of the chord always form a right-angled triangle.

Half of the chord = 8 ÷ 2 = 4 cm.

Using the standard Pythagorean triplet (3, 4, 5):

If the two sides are 3 cm and 4 cm, the hypotenuse (radius) must be 5 cm.

∴ The correct answer is 5 cm.

Alternate Method

Given: Length of chord (AB) = 8 cm, Distance from center (OM) = 3 cm.

Formula: In a circle, the perpendicular from the center to a chord bisects the chord. Radius² = (Half-chord)² + (Distance)².

⇒ Let O be the center and AB be the chord.

⇒ OM ⊥ AB, so AM = MB = AB ÷ 2

⇒ AM = 8 ÷ 2 = 4 cm

⇒ In right ΔOMA, OA² = OM² + AM²

⇒ Radius² = 3² + 4²

⇒ Radius² = 9 + 16 = 25

⇒ Radius = √25 = 5 cm

∴ The correct answer is 5 cm.

Additional Information

Perpendicular Bisector Property

The perpendicular drawn from the center of a circle to a chord bisects the chord into two equal segments.

Pythagorean Theorem

In any right-angled triangle, Hypotenuse² = Base² + Perpendicular². Common triplets include (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Diameter and Radius Relation

The diameter of a circle is the longest chord and its length is exactly 2 × radius.

Equal Chords Property

Chords of equal length are equidistant from the center of the circle.

Shortcut Trick

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

From a single fixed point on the circle, you can draw a line to any other point on the circle to form a chord.

Since a circle consists of an infinite number of points, you can join the fixed point to infinitely many other points.

∴ The correct answer is Infinite.

Alternate Method

Given: A circle with center O and a point P on the circumference.

Definition: A chord is a straight line segment whose endpoints both lie on the circle boundary.

⇒ Let P be a fixed point on the circle.

⇒ Let Q₁, Q₂, Q₃, ... be other distinct points on the circle.

⇒ By joining P to Q₁, we get chord PQ₁. By joining P to Q₂, we get chord PQ₂, and so on.

⇒ A circle is a set of infinitely many points in a plane that are at a given distance from a center.

⇒ Since there are infinite points available to be chosen as the second endpoint Q, the number of chords that can be drawn through P is infinite.

∴ The correct answer is Infinite.

Additional Information

Diameter as a Chord

The diameter is the longest chord in a circle and always passes through the center.

Chord Bisection

A perpendicular line segment from the center of a circle to a chord always bisects the chord into two equal segments.

Tangent vs Chord

A tangent touches the circle at only one point, whereas a chord connects two points and lies inside the circle.

Shortcut Trick

Ratio of length and breadth is 14 : 11 ⇒ 2 × (14x + 11x) = 150 ⇒ x = 3.

Area of Rectangle = (14 × 3) × (11 × 3) = 42 × 33 = 1386 cm².

Top surface of cylinder is a circle: (22/7) × R² = 1386 ⇒ R² = 441 ⇒ R = 21 cm.

Height H = R / 1.05 = 21 / 1.05 = 20 cm.

Curved Surface Area (CSA) = 2 × (22/7) × 21 × 20 = 2640 cm².

∴ The correct answer is 2640 cm².

Alternate Method

Given: Ratio of L : B = 14 : 11, Perimeter = 150 cm, R = 105% of H, Area of rectangle = Top surface area of cylinder.

Formula Used: Perimeter = 2(L + B) | Area of circle = πR² | CSA of cylinder = 2πRH

Step 1: Find dimensions of the rectangle

⇒ Let L = 14x and B = 11x.

⇒ Perimeter = 2(14x + 11x) = 150

⇒ 2(25x) = 150 ⇒ 50x = 150 ⇒ x = 3

⇒ L = 14 × 3 = 42 cm, B = 11 × 3 = 33 cm

⇒ Area of rectangle = 42 × 33 = 1386 cm²

Step 2: Find radius (R) and height (H) of the cylinder

⇒ Area of top surface (πR²) = 1386

⇒ (22/7) × R² = 1386 ⇒ R² = 1386 × (7/22) = 63 × 7 = 441

⇒ R = √441 = 21 cm

⇒ Since R = 105% of H ⇒ 21 = 1.05 × H

⇒ H = 21 / 1.05 = 20 cm

Step 3: Calculate Curved Surface Area (CSA)

⇒ CSA = 2 × (22/7) × 21 × 20

⇒ CSA = 2 × 22 × 3 × 20 = 2640 cm²

∴ The correct answer is 2640 cm².

Additional Information

Total Surface Area (TSA) of Cylinder

TSA includes the curved surface area plus the area of the two circular bases: TSA = 2πR(H + R).

Volume of Cylinder

The volume represents the space inside the cylinder: Volume = πR²H.

Rectangle Properties

Area = Length × Breadth and Diagonal = √(L² + B²).

Percentage to Fraction Conversion

105% can be written as 105/100 or 21/20 in simplified fraction form.

Shortcut Trick

In a circle where diameter BA is produced to meet tangent PC at point P:

Measure of ∠ACP = (90° − ∠BPC) ÷ 2

Given ∠BPC = 34°

∠ACP = (90° − 34°) ÷ 2 = 56° ÷ 2 = 28°

∴ The correct answer is 28°.

Alternate Method

Given: AB is the diameter of a circle with centre O. PT is a tangent at point C. BA produced meets PT at P. ∠BPC = 34°.

Formula Used: Sum of angles in a triangle = 180°; Angle in a semicircle = 90°; Radius is perpendicular to the tangent at the point of contact.

PAOBC34°?

⇒ In ▵OCP, ∠OCP = 90° (Since OC ⊥ PT)

⇒ ∠COP = 180° − (∠OCP + ∠BPC)

⇒ ∠COP = 180° − (90° + 34°) = 56°

⇒ ∠COA = ∠COP = 56°

⇒ In ▵OAC, OA = OC (Radii of the same circle)

⇒ ∠OCA = ∠OAC (Angles opposite to equal sides)

⇒ ∠OCA = (180° − ∠COA) ÷ 2

⇒ ∠OCA = (180° − 56°) ÷ 2 = 124° ÷ 2 = 62°

⇒ Now, ∠ACP = ∠OCP − ∠OCA

⇒ ∠ACP = 90° − 62° = 28°

∴ The correct answer is 28°.

Additional Information

Alternate Segment Theorem

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment (∠ACP = ∠ABC).

Angle in a Semicircle

Any angle subtended by the diameter at the circumference of the circle is always a right angle (90°).

Radius and Tangent Property

The radius of a circle is always perpendicular to the tangent at the point of contact (90° angle).

Given:

(√12 + √27) ÷ √3

Formula Used:

√(ab) = √a × √b

Calculation:

⇒ (√12 + √27) ÷ √3

⇒ (2√3 + 3√3) ÷ √3

⇒ 5√3 ÷ √3

⇒ 5

∴ The correct answer is 5.

Given:

(5 + 2√6)(5 − 2√6)

Formula Used:

(a + b)(a − b) = a² − b²

Calculation:

⇒ (5 + 2√6)(5 − 2√6)

⇒ 5² − (2√6)²

⇒ 25 − 4 × 6

⇒ 25 − 24

⇒ 1

∴ The correct answer is 1.