Quantitative Aptitude PYQs

Let the side of the given cube be ‘a’ units.

⇒ Side of cube = CH = HG = FG = a units

In ∆CHG,

⇒ CG² = CH² + HG² = a² + a² = 2a²

Now, in ∆CGF,

⇒ CF² = CG² + FG² = 2a² + a² = 3a²

⇒ Diagonal of cube = CF = a√3 units

⇒ l = a√3

⇒ a = l/√3 units

∵ There are six faces is a cube,

∴ Total surface area of cube = 6a² = 6(l²/3) = 2l²

Let the perimeter of triangle, square and circle be equal to ‘x’ cm

Perimeter of equilateral triangle = 3(side) = x

⇒ Side of equilateral triangle = x/3

⇒ Area of equilateral triangle = (√3/4) × (side)² = (√3/4) × (x²/9) = x²/12√3

⇒ T ≅ 0.048x²     ...1)

Similarly,

Perimeter of square = 4(side) = x

⇒ Side of square = x/4

⇒ Area of square = (side)² = x²/16

⇒ S = 0.0625x²     ...2)

Also,

Perimeter of circle = 2π(radius) = x

⇒ Radius of circle = x/2π

⇒ Area of circle = π(radius)² = π × x²/4π² = x²/4π

⇒ C ≅ 0.0795x²     ...3)

∴ Comparing equation (1), (2) and (3), we can write, T < S < C

As we know, ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides,

We can write,

⇒ (Ratio of corresponding sides)² = Ratio of areas

⇒ (Ratio of corresponding sides)² = (7 – 4√3) / (7 + 4√3)

⇒ (Ratio of corresponding sides)² = [(7 – 4√3) / (7 + 4√3)] × [(7 – 4√3) / (7 – 4√3)]

⇒ (Ratio of corresponding sides)² = (7 – 4√3)² / [(7 + 4√3)(7 – 4√3)]

∵ (a + b)(a – b) = (a² – b²)

⇒ (Ratio of corresponding sides)² = (7 – 4√3)² / (49 – 48)

⇒ (Ratio of corresponding sides)² = (7 – 4√3)²

∴ Ratio of corresponding sides = (7 – 4√3)

Let AB be the chord and O be the centre of the circle, as shown below,

⇒ OA = OB = radius of circle = r (let)

⇒ AB = r√3

As we know, the perpendicular drawn from the centre of the circle to a chord bisects the chord and the two triangles so formed are congruent,

⇒ AE = BE = AB/2 = r√3/2

⇒ ∠AOE = ∠BOE

In ∆AOE,

⇒ sin ∠AOE = AE/OA

⇒ sin ∠AOE = (r√3/2) /r = √3/2

⇒ sin ∠AOE = sin 60°

⇒ ∠AOE = 60°

⇒ ∠AOB = 60° + 60° = 120°

Now, the angle subtended by a chord at the centre of the circle is equal to twice the angle subtended by the chord at some other point in the same segment of the circle,

⇒ ∠ACB = (∠AOB) /2 = 120°/2 = 60°

∵ ACBD is a cyclic quadrilateral, its opposite angles are supplementary

⇒ ∠ADB = 180° – ∠ACB = 180° – 60° = 120°

But,

Angle subtended by chord AB in minor arc = k(Angle subtended by chord AB in a major arc)

⇒ ∠ADB = k (∠ACB)

⇒ 120° = k (60°)

⇒ k = 2

Short trick∶

∠BDC = 90° – ∠A/2 = 90° – 25° = 65°

Detailed solution∶

In the figure given below,

∵ Sum of angles of triangle = 180°

⇒ ∠ABC + ∠ACB = 180° – 50° = 130°     ...1)

Also, sum of angles in straight line = 180°

⇒ ∠DBC = (180° – ∠ABC) /2

⇒ ∠DCB = (180° – ∠ACB) /2

Adding the two equations, we get,

⇒ ∠DBC + ∠DCB = 180° – (∠ABC + ∠ACB) /2

Substituting from (1),

⇒ ∠DBC + ∠DCB = 180° – 130°/2 = 180° – 65° = 115°

∴ ∠BDC = 180° – (∠DBC + ∠DCB) = 180° – 115° = 65°

∵ Volume of cone = (1/3)π × (base radius)² × height

⇒ Ratio of the volume of two cones = (Ratio of base radii)² × Ratio of heights

Given, ratio of base radii = 3 ∶ 1 and Ratio of heights = 1 ∶ 3

∴ Ratio of their volumes = (3/1)² × (1/3) = 9/3 = 3 ∶ 1

In the figure given below,

Given, ∠AOC = 5(∠AOD)

∵ Sum of angles in straight line is 180°

⇒ ∠AOC + ∠AOD = 180°

⇒ 6(∠AOD) = 180°

⇒ ∠AOD = 180°/6 = 30°

⇒ ∠BOC = ∠AOD = 30° (∵ opposite angles are equal)

⇒ ∠AOC = 5(30°) = 150°

⇒ ∠BOD = ∠AOC = 150° (∵ opposite angles are equal)

∴ The four angles at O are 30°, 30°, 150°, 150°

Let the co-ordinates of the points P be (x, y), A be (a, 0) and B be (0, b)

Given, PA² + PB² = k (constant)

⇒ (x – a)² + y² + x² + (y – b)² = k

⇒ x² – 2ax + a² + y² + x² + y² – 2by + b² = k

⇒ 2x² – 2ax + a² + 2y² – 2by + b² = k

⇒ x² – ax + a²/2 + y² – by + b²/2 = k/2

⇒ x² – ax + a²/4 + a²/4 + y² – by + b²/4 + b²/4 = k/2

⇒ (x – a/2)² + a²/4 + (y – b/2)² + b²/4 = k/2

⇒ (x – a/2)² + (y – b/2)² = k/2 – a²/4 – b²/4

⇒ (x – a/2)² + (y – b/2)² = c (constant)

This is the equation of a circle,

∴ The locus of the point P is a circle.

Concept:

Pythagoras theorem: For a right-angled triangle

Hypotenuse² = Base² + Perpendicular²

Calculation:

Let AB = 6 units, BC = 8 units in the triangle shown below,

Applying Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 36 + 64

⇒ AC = √100 = 10 cm

Now,

⇒ AC³ = (10)³ = 1000

⇒ AB³ + BC³ = 6³ + 8³

⇒ AB3 + BC3  = 728

∴ AC³ > AB³ + BC³

In the figure shown below, the shaded region is the region bounded externally by the square and internally by the circle,

Given, side of square = 2a cm

⇒ Area of square = (side)² = 4a² cm²

∵ The circle touches the four sides of the square,

⇒ Diameter of circle = 2a cm

⇒ Area of circle = (π/4) × (diameter)² = (π/4) × 4a² = πa² cm²

∴ Area of shaded region = Area of square – Area of circle = (4 – π)a² cm²

Given, AB = p cm and AC = q cm

∵ Area of right-angled triangle = 1/2 × Base × Height

⇒ Area of triangle ABC = pq/2

Applying Pythagoras theorem in ∆ABC,

⇒ BC² = AB² + AC²

⇒ BC² = (p² + q²)

∵ Area of semi-circle = (π/8) × (diameter)²

⇒ Area of semi-circle with diameter AB = (π/8) × (p)² = πp²/8 cm²

⇒ Area of semi-circle with diameter AC = (π/8) × (q)² = πq²/8 cm²

⇒ Area of semi-circle with diameter BC = (π/8) × (p² + q²) = π(p² + q²) /8 cm²

∴ Area of shaded region = (pq/2) + (πp²/8) + (πq²/8) – [π(p² + q²) /8] = (pq/2) cm²

Radius of circle = OP = OA = 6 cm

Given, AT = 4 cm

⇒ OT = 6 + 4 = 10 cm

∵ The radius from the centre of circle to the point of tangency is perpendicular to the tangent

⇒ ∠OPT = 90°

Applying Pythagoras theorem in ∆OPT,

⇒ OT² = OP² + PT²

⇒ 100 = 36 + PT²

⇒ PT = √64 = 8 cm

∴ Length of tangent PT = 8 cm

Diameter of circle = AD = 2(9) = 18 cm

⇒ AB = BC = CD = 18/3 = 6 cm

⇒ BD = BC + CD = 6 + 6 = 12 cm

∵ Area of semi-circle = (π/8) × (diameter)²

⇒ Area of semi-circle with diameter AD = (π/8) × (18)² = 81π/2 cm²

⇒ Area of semi-circle with diameter AB = (π/8) × (6)² = 9π/2 cm²

⇒ Area of semi-circle with diameter BD = (π/8) × (12)² = 18π cm²

∴ Area of shaded region = (81π/2) + (9π/2) – 18π = 45π – 18π = 27π cm²

As we know, the angles subtended by an arc in the same segment of a circle are equal

⇒ ∠BDC = ∠BAC = 30°

∵ Sum of angles of a triangle is 180°,

In triangle BCD,

∠BCD + ∠CBD + ∠CDB = 180°

∠BCD + 70° + 30° = 180°

∴ ∠BCD = 180° – 30° – 70° = 80°

As we know, an exterior angle of a triangle is equal to the sum of the two opposite interior angles,

⇒ ∠CEF = ∠EDC + ∠ECD

⇒ ∠CEF = 32° + 32° = 64°

Also, the angle subtended by an arc at the centre of the circle is twice the angle subtended at some other point in the same segment of the circle,

⇒ ∠COF = 2(∠FDC) = 2(∠EDC)

⇒ ∠COF = 2(32°) = 64°

Given, ΔABR ~ ΔPQR

Also, PQ = 3 cm, AB = 6 cm, BR = 8.2 cm and PR = 5.2 cm,

As we know, the ratio of corresponding sides of similar triangles are equal,

⇒ PQ/AB = PR/AR

⇒ 3/6 = 5.2/AR

⇒ AR = 2 × 5.2 = 10.4 cm

Similarly,

⇒ PQ/AB = QR/BR

⇒ 3/6 = QR/8.2

⇒ QR = 8.2/2 = 4.1 cm

∴ Lengths of QR and AR are 4.1 cm and 10.4 cm respectively

Concept:

Pythagoras theorem:

It states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. 

Calculation:

Given, AD = 9 cm and DC = 4 cm

⇒ AC = AD + DC = 9 + 4 = 13 cm

In ∆ABC and ∆ADB, side AB is common, ∠ABC = ∠ADB = 90°

Also, ∠BAC = ∠BAD (common angle)

⇒ ∆ABC and ∆ADB are similar triangles, hence, the ratio of their corresponding sides is equal,

⇒ AB/AD = AC/AB

⇒ AB² = AC × AD = 13 × 9 = 117

Now, in ∆ADB,

⇒ AB² = AD² + BD²

⇒ 117 = 81 + BD²

∴ BD = √36 = 6 cm

Let ‘r’ be the radius of the small circle and MN is the tangent to the two equal semi-circles, as shown in the figure given below,

⇒ ∠OMP = ∠OMQ = 90°

⇒ MN = Radius of bigger semi-circle = 108/2 = 54 cm

⇒ Radius of two equal semi-circles = PX = PM = QM = QY = 54/2 = 27 cm

⇒ Radius of small circle = OX = OY = ON = r

Considering ∆OMP,

⇒ OP = OX + PX = (27 + r) cm

⇒ OM = MN – ON = (54 – r) cm

Applying Pythagoras theorem,

⇒ OP² = OM² + PM²

⇒ (27 + r)² = (54 – r)² + (27)²

⇒ (27)² + r² + 54r = (54)² + r² – 108r + (27)²

⇒ 162r = (54)²

⇒ r = 54/3 = 18 cm

∵ Area of semi-circle = (π/2) × (radius)²

⇒ Area of bigger semi-circle = (π/2) × (54)² = 1458π cm²

⇒ Area of small semi-circles = (π/2) × (27)² = 729π/2 cm²

⇒ Area of small circle = π × (18)² = 324π cm²

∴ Area of shaded region = 1458π – 2(729π/2) – 324π = 1458π – 1053π = 405π cm²

Given, AB = BC = AC = 30 cm

Also, (XY + XP + YQ) = 40 cm

∵ XY is parallel to BC and remaining two sides are common,

⇒ ∆ABC and ∆AXY are similar triangles, i.e., ∆AXY is also an equilateral triangle

⇒ XY = AX = BY

Similarly, ∆BXP and ∆QYC are also similar to ∆ABC and are equilateral triangles,

⇒ XP = BX = BP

⇒ YQ = YC = QC

Now,

⇒ AB = AX + XB = XY + XP     ...1)

⇒ AC = AY + YC = XY + YQ     ...2)

⇒ BC = BP + PQ + QC

⇒ BC = XP + PQ + YQ     ...3)

Adding equation (1), (2) and (3),

⇒ AB + AC + BC = 2(XY + XP + YQ) + PQ

Substituting values,

⇒ 30 + 30 + 30 = 2(40) + PQ

∴ PQ = 90 – 80 = 10 cm

Area of square = (side)² = (4)² = 16 cm²

The sum of the area of four quadrants of a circle removed from the corners of the square is equal to the area of the circle of diameter 2 cm

⇒ Radius of circle = diameter/2 = 2/2 = 1 cm

⇒ Area of circle = π(radius)² = 22/7 cm²

∴ Area of shaded region = Area of square – 2(Area of circle) = 16 – 44/7 = 68/7 = \(9\frac{5}{7}\) cm²