Quantitative Aptitude PYQs

Given:

Diagonals are x and y.

Formula Used:

 Area of rhombus = ½ × (Product of lengths of its diagonals)

Calculation: 

Diagonal = x and y

∴ Area of rhombus = xy/2

Given:

Smallest side = 3x

Concept Used:

As one of the angles of the triangle is 90°, the triangle is a right angle triangle and the length of its three sides form a Pythagorean triple.

The Pythagorean triple with minimum values is (3, 4, 5),

Calculation:

Smallest side = 3x = 3

⇒ x = 1

Middle side = 4√y = 4

⇒ y = 1

Largest sides = 5³√z = 5

⇒ z = 1

∴ The minimum values of x, y and z are 1, 1, 1

Calculation:

A circle that passes through all the three vertices of the triangle is known as the circumcircle of the triangle.

As only one circle can be drawn by using three points, hence only one such circle is possible for a triangle.

∴ Maximum no. of circumcircles of a triangle = 1

Given: 

θ = 30°, radius = 6 cm

Formula used

If an arc subtends an angle of θ at the centre of circle, then,

Length of the arc = (θ/360°) × Circumference of circle = (θ/360°) × 2π × (radius)

Calculation

Length of arc = (30°/360°) × 2 × 3.14 × 6 = 3.14 cm

∴ Length of Arc is 3.14 cm

Given

Ladder reaches 4.8 m and 1.4m height on two walls in opposite sides of room.

Calculation

Let ‘x’ m and ‘y’ m be the distance of the foot of ladder from the two respective walls, as shown below,

Applying Pythagoras theorem,

⇒ 5² = 1.4² + x²

⇒ x² = 25 – 1.96 = 23.04

⇒ x = √23.04 = 4.8 m

Similarly,

⇒ 5² = 4.8² + y²

⇒ y² = 25 – 23.04 = 1.96

⇒ y = √1.96 = 1.4 m

∴ Breadth of room = x + y = 4.8 + 1.4 = 6.2 cm

Given

Radius of circular disk = 1 unit

Concept used

Diagonal of square = Diameter of circular disk

Calculation

The largest square plate can be cut from a circular disk as shown below,

Here, Diagonal of square plate = Diameter of circular disk = 2 × Radius of circular disk

⇒ Diagonal of largest square plate = 2 × 1 = 2 units

∵ Diagonal of square = √2 × Side of square

⇒ Side of largest square plate = 2/√2 = √2 units

∴ Area of largest square plate = (side)² = (√2)² = 2 sq. units

Concept Used:

If the centres of the three balls placed on the plane are joined, they will form an equilateral triangle with side equal to ‘2r’ units, as shown below

Height of triangular pyramid = √(2/3) × Base side

Calculation:

And, if the centre of the 4^th ball is joined to the three centres, a triangular pyramid will be formed with a base of an equilateral triangle

⇒ Base side of triangular pyramid = 2r units Height of triangular pyramid = √(2/3) × Base side

⇒ Height of triangular pyramid = 2√2r/√3 units

This is equal to the perpendicular distance of the centre of 4^th ball to the line joining the centres of the other balls

Thus, the distance of centre of 4^th ball from plane can be computed by adding the radius of the ball to this distance

∴ Distance of centre of 4^th ball from plane = 2√2r/√3 + r = (2√2 + √3)r/√3 units

Given:

The surface area of the sphere = p

The curved surface area of the cylinder = q

Formula used

Curved surface area of sphere = 4π × (radius)²

Curved surface area of cylinder = 2π × radius × height

Calculation

When a right circular cylinder just encloses a sphere as shown below, the radius of the sphere and cylinder are equal and the height of cylinder is equal to the diameter of the sphere

Let, radius of sphere = radius of cylinder = r

Here, height of cylinder = 2r

Now,

Curved surface area of sphere = 4π × (radius)²

⇒ p = 4πr²

Curved surface area of cylinder = 2π × radius × height

⇒ q = 2πr × 2r = 4πr²

∴ p = q

Given:

AD = DC = CA = 20 and BC = 12 units and ∠ABC = 90° 

Formula used

Area of equilateral triangle = (√3/4) × (side)²

Area of right angles triangle = ½ × base × height

Calculation

The given quadrilateral can be drawn as shown below,

∆ACD is an equilateral triangle, as AC = AD = DC = 20 units

⇒ Area of ∆ACD = (√3/4) × (side)² = (√3/4) × (20)² = 100√3 = 100 × 1.732 = 173.2 sq. units

In ∆ABC, AC = 20 units & BC = 12 units,

Applying Pythagoras theorem,

⇒ AB² = AC² – BC² = 400 – 144 = 256

⇒ AB = √256 = 16 cm

⇒ Area of ∆ABC = ½ × base × height = ½ × AB × BC = ½ × 16 × 12 = 96 sq. units

∴ Area of quadrilateral ABCD = 173.2 + 96 = 269.2 ≅ 269 sq. units

Given:

Diameter of circle = PS = 9 cm and

PQ = QR = RS = 3 cm

Formula used

Area of semi-circle =  \(\frac{1}{2}\)×π(Radius)²

Calculation:

According to the question, PS = 2 × 9 = 18 cm

⇒ PQ = QR = RS = 6 cm 

Area of semi-circle =  \(\frac{1}{2}\) × π(Radius)2

Area of the shaded region

A₁ = \(\frac{1}{2}π(9)^2-\frac{1}{2}π(6)^2\)

⇒ A₁ = \(\frac{π}{2}(9^2-6^2)\) 

⇒ A₁ = \(\frac{45π}{2}\) cm2    ----(1)

Area of the unshaded region

A₂ = \(\frac{1}{2}π(9)^2+\frac{1}{2}π(6)^2\)

⇒ A₂  = \(\frac{117π}{2}\)     -----(2)

∴ Required ratio = (45π/8)∶ (117π/8) = 5 ∶ 13

Given: 

Radius of each circle = 2 cm

Formula used

Area of equilateral triangle = (√3/4) × (side)²

Area of sector with angle θ = (θ/360°) × π × (radius)²

Calculation

Radius of each circle = 2 cm

The triangle ABC is an equilateral triangle with side equal to twice the radius of the circle

⇒ Side of ∆ABC = 2(2) = 4 cm

⇒ Area of ∆ABC = (√3/4) × (side)² = (√3/4) × (4)² = 4√3 cm²

Now, the triangle ABC cuts three sectors, one from each circle with angle 60°,

∵ Area of sector with angle θ = (θ/360°) × π × (radius)²

⇒ Area of three sectors = 3 × (60°/360°) × π × (2)² = 2π cm²

∴ Area of shaded region = Area of ∆ABC – Area of three sectors = (4√3 – 2π) cm²

Given:

AB = CD = 6√3 units & AD = BC = 6 unit

Formula used

Area of sector with angle θ = (θ/360°) × π × (radius)²

Calculation

Let O be the centre of the rectangle ABCD and also the centre of the circle, as shown below

In this figure AB = CD = 6√3 units & AD = BC = 6 units

Applying Pythagoras theorem in ∆ABC,

⇒ AC² = AB² + BC² = 108 + 36 = 144

⇒ AC = √144 = 12 cm

⇒ Radius of circle = OA = 12/2 = 6 cm

Also, OX = BC/2 = 6/2 = 3 cm

Considering ∆AXO,

⇒ cos ∠AOX = OX/OA = 3/6 = ½

⇒ ∠AOX = cos^{– 1} (1/2) = 60°

⇒ Central angle of sector = ∠AOB = 2(60°) = 120°

∵ Area of sector with angle θ = (θ/360°) × π × (radius)²

⇒ Area of sector with angle 120° = (120°/360°) × π × (6)² = 12π cm²

 Now,

Area of shaded region = Area of sector with angle 120° – Area of ∆AOB

∵ Area of ∆AOB = ½ × AB × OX = ½ × 6√3 × 3 = 9√3 cm²

∴ Area of shaded region = 12π – 9√3 = 3(4π – 3√3) cm²

Given

 \(\frac{y}{x} = 6\) and \(\frac{z}{x} = 5\)

Concept Used:

Sum of angles in a straight line = 180°

Calculation

y/x = 6

⇒ y = 6x

Also, z/x = 5

⇒ z = 5x

�� Sum of angles in a straight line = 180°

⇒ x + y + z = 180°

⇒ x + 6x + 5x = 180°

⇒ x = 180°/12 = 15°

∴ The value of x is 15°

Formula used:

Area of trapezium when parallel and non-parallel sides are given:

If a, b and c, d are the length of two parallel and non-parallel sides

of the trapezium then

Area of trapezium 

\(A = \frac{a+b}{k}{\sqrt{s(s-k)(s-c)(s-d)}}​​\)

Where, k = b - a, & \(s = \frac{k+c+d}{2}\)

Calculation:

a = 4, b = 7, c = 3 and d = 2

⇒ k = b - a = 7 - 4 

⇒ k = 3

\(s = \frac{3+3+2}{2} = 4\)

Hecne, are of trapezium 

\(A = \frac{a+b}{k}{\sqrt{s(s-k)(s-c)(s-d)}}\)

⇒ \(A = \frac{4+7}{3}{\sqrt{4(4-3)(4-3)(4-2)}}\)

⇒ \(A = \frac{22}{3}{\sqrt{2}}\)

Alternate Method

Formula used

Area of trapezium = ½ × Sum of lengths of parallel sides × Height

Calculation

Let DX be equal to ‘x’ cm in the trapezium ABCD, shown below

AB = 4 cm, BC = 3 cm, CD = 7 cm and DA = 2 cm

⇒ XY = AB = 4 cm

⇒ YC = 7 – 4 – x = (3 – x) cm

Using Pythagoras theorem, we can write,

⇒ AX = √(AD² – DX²) = √(4 – x²)

⇒ BY = √(BC² – YC²) = √[9 – (3 – x)²] = √(6x – x²)

But, height of trapezium = AX = BY

⇒ √(4 – x²) = √(6x – x²)

⇒ 4 – x² = 6x – x²

⇒ x = 4/6 = 2/3

⇒ AX = BY = √(4 – 4/9) = √(32/9) = 4√2/3 cm

Now,

Area of trapezium = ½ × Sum of lengths of parallel sides × Height

⇒ Area of trapezium ABCD = = ½ × (AB + CD) × AX

⇒ Area of trapezium ABCD = ½ × 11 × 4√2/3

∴ Area of trapezium ABCD = \(\frac{{22\sqrt 2 }}{3};c{m^2}\) cm²

Concept used

1. The opposite angles formed by two intersecting lines are congruent
2. The sum of angles of a triangle is 180°

Calculation

From the given figure, we can write

⇒ ∠BAD = 180° – (∠1 + ∠2)

⇒ ∠ABC = 180° – (∠3 + ∠4)

⇒ ∠BCD = 180° – (∠5 + ∠6)

⇒ ∠ADC = 180° – (∠7 + ∠8)

Adding all four equations, we get

⇒ ∠BAD + ∠ABC + ∠BCD + ∠ADC = 720° – (∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8)

∵ Sum of angles of quadrilateral is 360°,

⇒ 360° = 720° – (∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8)

∴ (∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8) = 720° – 360° = 360°

Alternate MethodThis is a star and mesh type of problem. The type of triangles is not mentioned.

Let all triangles be isosceles and the figure among triangles is square. Now, each triangle will be isosceles and right-angled.

In any triangle-

90 + 2∠1 = 180

∠1 = 45° 

Now, ∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = ∠7  = 45° 

∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 8 × 45° = 360° 

∴ The required sum is 360°.

Given:

∠AEF = 95°, ∠BHS = 110°, and ∠ABC = x

Concept used

1. External angle of a triangle is equal to the sum of its two opposite angles.
2. When two parallel lines intersect a single straight line, then the corresponding angles formed are equal.

Calculation

As we know, when two parallel lines intersect a single straight line, then the corresponding angles so formed are equal,

⇒ ∠EGH = ∠AEF = 95°

⇒ ∠BGH = 180° – 95° = 85°

Now, the external angle of a triangle is equal to the sum of its two opposite angles,

Considering ∆BGH,

⇒ ∠BHS = ∠BGH + ∠GBH

⇒ 110° = 85° + x

⇒ x = 110° – 85° = 25°

∴ The value of x is 25°

Given:

∠EAC = 40°

∠FDG = 55°

∠HAB = x°

Formula used:

Sum of angles in straight line = 180°

Calculation:

As we know, when two parallel lines are intersect another parallel straight lines, then the corresponding angles so formed are equal,

⇒ ∠CAB = ∠FDG = 55°

Now, sum of angles in straight line = 180°

⇒ ∠EAC + ∠CAB + ∠HAB = 180°

⇒ 40° + 55° + x° = 180°

⇒ x = 180° – 95° = 85°

∴ The value of x is 85°.

Given:

The distance between AC = 28 unit

Formula used

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

Calculation

⇒ Area of semi-circle with diameter AC = (π/8) × (28)² = 98π sq. units

∵ AF = FC = 28/2 = 14 units

⇒ Area of two semi-circles with diameter AF = 2 × (π/8) × (14)² = 49π sq. units

∴ Total area of three semi-circles = 98π + 49π = 147π = 147 × (22/7) = 462 sq. units

Given:

Length of rectangle = 8 cm

Breadth of ractangle = 6 cm

Formula used

Area of rectangle = Length × Breadth

Area of circle = (π/4) × (diameter)²

Calculation

Area of rectangle = 8 × 6 = 48 cm²

Also, Diameter of circle = Diagonal of rectangle

⇒ Diameter of circle = √(6² + 8²) = √100 = 10 cm

⇒ Area of circle = (3.14/4) × 100 = 78.5 cm²

∴ Area of shaded region = Area of circle – Area of rectangle = 78.5 – 48 = 30.5 cm²

Firstly, the opposite angles formed by the intersection of two lines are equal,

⇒ ∠AOD = ∠BOC

Hence, statement 2 is correct

 

Let ‘h’ units be the height XY of trapezium ABCD, and ‘a’ units and ‘b’ units be the lengths of the parallel sides AD and BC respectively

⇒ OX + OY = XY = h

As we know, the diagonal of a trapezium divides each other in proportion,

⇒ ∆AOD and ∆BOC are similar triangles, hence, their heights are in the same proportion as their sides

⇒ OX/OY = AD/BC = a/b

⇒ OX = ah/(a + b)

⇒ OY = bh(a + b)

⇒ Area (∆AOD) + Area (∆BOC) = [1/2 × a × ah/(a + b)] + [1/2 × b × bh/(a + b)]

⇒ Area (∆AOD) + Area (∆BOC) = h(a² + b²)/2(a + b)  ---- (1)

Now,

Area (∆AOB) + Area (∆DOC) = Area of trapezium ABCD – [Area (∆AOD) + Area (∆BOC)]

⇒ Area (∆AOB) + Area (∆DOC) = (a + b)h/2 – h(a² + b²)/2(a + b)

⇒ Area (∆AOB) + Area (∆DOC) = abh/(a + b)  ---- (2)

Comparing eq.

(1) and (2),

⇒ Area (∆AOD) + Area (∆BOC) ≠ Area (∆AOB) + Area (∆DOC)

Hence, statement 1 is not correct

 

Also, the sum of the lengths of the two sides of a triangle is greater than the length of its third side

In ∆ABD, AB + AD > BD

In ∆BCD, BC + CD > BD

In ∆ABC, AB + BC > AC

In ∆ADC, AD + CD > AC

Adding the four equations, we get,

⇒ 2(AB + BC + CD + AD) > 2(AC + BD)

⇒ AB + BC + CD + AD > AC + BD

Hence, statement 3 is correct

∴ Only statements 2 and 3 are correct