Quantitative Aptitude PYQs

Given:

The radius of circum-circle of right-angled triangle = 10 cm

Length of altitude drawn to the hypotenuse = 8 cm

Concept:  In the case of a right triangle, the hypotenuse is a diameter

of the circumcircle, and its center is exactly at the midpoint of the

hypotenuse.

Formula used:

Area of a triangle(A) = 1/2 × base × height

Calculation:

Altitude to the hypotenuse BM = 8 cm

From the concept discussed above,

Radius of circle OC = OA = OB = 10 cm

Hence, hypotenuse AC = 2 × 10 = 20 cm

Therefore required area

⇒ A = 1/2 × 20 × 8

∴ Area of the triangle = 80 cm²

Given:

Sum of areas of two circles touching each other externally = 41π cm²

Distance between the centers = 9 cm

Formula used:

Area of a circle = πx²

Where x = radius of the circle

Calculation:

Let radii of two circles be R and r

According to the question:

R + r = 9 cm    ----(i)

Also, πR² + πr² = 41π

⇒ R² + r² = 41

We know that 

(R + r)² = R2 + r² + 2Rr

⇒ 81 = 41 + 2Rr

⇒ Rr = 20    ----(ii)

From (i) and (ii), we get

⇒ R(9 - R) = 20

⇒ R² - 9R + 20 = 0

⇒ (R - 5)(R - 4) = 0

⇒ R = 5, 4; r = 4, 5

Hence, their diameters will be 8 cm and 10 cm respectively

∴ Difference between their diameters = 2 cm

Given:

Area of square X = p

Area of square formed with side as diagonal of square X = q

Formula used:

Area of a square = a²

Diagonal of the square = √2a

a = side of the square

Calculation:

Let the side of the square X = a

⇒ p = a²

⇒ q = (√2a)² = 2a²

∴ p/q = 1/2

Given:

Area of the rhombus = 336 cm²

Length of one diagonal of the rhombus = 48 cm

Formula used:

Area of a rhombus = (product of its diagonals)/2

Diagonal of a rhombus bisect each other

Calculation:

Let other diagonal be d₂

According to the question:

336 = (48 × d₂)/2

⇒ d₂ = 14 cm

Now, from the figure,

AB² = AC² + BC²

⇒ AB² = 24² + 7²

⇒ AB = side of the rhombus = 25 cm

∴ Perimeter of the rhombus = 25 × 4 = 100 cm

Given:

Length of the minute hand of the clock = 21 cm

Formula used:

Area of a circle = πr²

Calculation:

Minute hand can be assumed as radius of the circle(clock)

Hence, area of the clock swept by minute hand = 22/7 × 21 × 21 = 1386 cm²

Now, In 60 minutes the area swept by minute hand = 1386 cm2

In 1 minute the area swept by minute hand = 1386/60 cm2

In 20 minutes the area swept by minute hand = 20 × 1386/60 cm2

Area will be 1/3^rd of the total area of the clock swept by minute hand

∴ Required area = 1386/3 = 462 cm²

Given:

Length of the room(l) = 21 m

Breadth of the room(b) = 16 m

Longest rod that can be placed in the room = 29 m

Formula used:

(Longest diagonal of a cuboid)² = l² + b² + h²

h = height of the diagonal

Calculation:

Longest rod that can be placed in the room can be called the longest diagonal of the room

According to the question:

29² = 21² + 16 ² + h²

h² = 144 

∴ Height of the room = 12 m

Given:

Internal radius of hemispherical bowl = 18 cm

Internal radius of bottle = 3 cm

Internal height of bottle = 4 cm

Formula used:

Volume of a hemisphere = (2/3) πR³

Volume of a cylinder = πr²h

R = radius of hemisphere, h = height of the cylinder

r = radius of the cylinder

Calculation:

Total volume of liquid in hemisphere = 2/3 × π ×18³ = 3888π cm³

Let the number of bottles be n

Then according to the question:

3888π = n × π × (3)² × 4

⇒ n = 108

∴ Total number of bottles = 108

Given:

Density of the metal = 7 g/cm³

Internal and external radii of the shell(spherical) = 3 cm and 6 cm respectively

Formula used:

Mass = Volume × Density

Volume of a sphere = (4/3) π(R³ - r³)

R = Internal radius, r = external radius

Calculation:

Volume of the shell = (4/3) × π × (6³ - 3³) = 252π cm³

Mass of the shell(M) = 252π × 7

⇒ M = 252 × (22/7) × 7 

∴ Mass of the shell = 5544 g

Given:

Height of the cone = 16 cm

Diameter of the hemisphere or cone = 14 cm

Formula used:

Volume of a cone = (1/3) πr²h

Volume of a hemisphere = (2/3) πr³

r = radius, h = height of the cone

Calculation:

Volume of the solid(V) = Volume of the cone + Volume of the hemisphere

⇒ V = [(1/3) × π × (7)2 × 16] + [(2/3) × π × (7)³]

⇒ V = 1540

∴ Volume of the solid = 1540 cm³

Given:

Volume of each cube = 343 cm³

Formula used:

Volume of a cube = a³

Total surface area of a cuboid(TSA) = 2(lb + bh + hl)

a = side of the cube, l = length of the cuboid

b = breadth of the cuboid, h = height of the cuboid

Calculation:

Let the side of each cube be a

According to the question:

a³ = 343

⇒ a = 7 cm

Total length of the cuboid = 7 + 7 + 7 = 21 cm

Breadth of the cuboid = 7 cm

Height of the cuboid = 7 cm

⇒ TSA of the cuboid = 2[(21 × 7) + (7 × 7) + (21 × 7)]

∴ Total surface area of the cuboid = 686 cm²

Given:

Side of the cubical block = 14 cm

The radius of the hemisphere = 7 cm

Formula used:

The surface area of a cube = 6a2

The curved surface area of a hemisphere = 2πr2

a = side of the cube

r = radius of the hemisphere

Calculation:

Total surface area of the solid = SA(cube) + CSA(hemisphere) - Area

of the base of the hemisphere 

⇒ SA = 6a² + 2πr² - πr2

⇒ SA = [6 × 14²] + [(22/7) × 7 × 7] 

∴ Total surface area of the solid = 1330 cm2

Given:

Diameter of the coin = 3.5 cm

Thickness(height) of the coin = 4 mm = 0.4 cm

Dimension of the cuboid = 21 cm × 11 cm × 7 cm

Formula used:

Volume of the cuboid = lbh

Volume of a cylinder = πr²h

l = length, b = breadth, h = height, r = radius 

Calculation:

Let n be the number of coins that will be melted to form the cuboid

Radius of the coin = 3.5/2 = 1.75 cm

Volume of one coin = (22/7) × (1.75)² × 0.4 = 3.85 cm³

Volume of the cuboid = 21 × 11 × 7 = 1617

Let n be the number of coins will be melted

According to the question

⇒ 3.85 n = 1617

⇒ n = 420

∴ Number of coins to be melted is 420.

Given:

Radii of the tub(frustum) are 105 cm and 42 cm respectively

Vertical height of the tub(h) = 16 cm

Formula used:

Slant height of a frustum(l) = \(\sqrt{h^2 + (R - r)^2}\)

R = radius of the larger end

r = radius of the smaller end

Calculation:

According to the question:

l = \(\sqrt{16^2 + (105 - 42)^2}\)

⇒ l = 65

∴ Slant height of the tub = 65 cm

Given:

ABC is a triangle, right-angled at B

AB = 8 cm, BC = 6 cm

It is made to revolve around BC to form a cone

Formula used:

Total surface area(TSA) of cone = πr(l + r)

r = radius of the cone

l = slant height  of the cone

Calculation:

AC² = AB² + BC²

⇒ AC² = 6² + 8²

⇒ AC = 10 cm

From the figure it can be deduced that after revolving,

Slant height (l) will be 10 cm and radius will be 8 cm

TSA = πr(l + r)

⇒ TSA = (22/7) × 8 × (10 + 8)

⇒ TSA = 452.16 ~ 452.57

∴ Total surface area of the cone = 452 cm²

Given:

There are two cylinders, placed one above another

Radius and height of the base cylinder are 7 cm and 20 cm respectively

Radius and height of the top cylinder are 3.5 cm and 10 cm respectively

Formula used:

Volume of a cylinder = πr²h

r = radius of the cylinder

h = height of the cylinder

Calculation:

Total volume of the figure(V) = Sum of volume of both cylinders

⇒ V = [(22/7) × 72 × 20] + [(22/7) × (3.5)² × 10]

⇒ V = 3080 + 385

⇒ V = 3465 cm³ = 0.003465 m³

If 1 m³ contains 10000 kg

Then, 0.003465 m³ contains 34.65 kg

∴ Mass of the rod = 34.65 kg

Given 

Radius of cone and hemisphere = x 

and Height of cone = 2x 

Formula Used 

Volume of cone = (1/3) πR²H

Volume of hemispher = (2/3) πR³  

Calculation 

The volume of the solid = Vol. of cone + Vol. of the hemisphere

⇒ (1/3) πx²(2x) + (2/3) π(x³) 

⇒ (4/3) πx³ = Volume of the sphere with radius x 

∴  The volume of the solid is equal to that of the volume of the sphere of radius x.

Given 

Radius of cone and hemisphere = x 

And Height of cone = 2x 

Formula Used:

Curved Surface area of cone = πrl

Total Surface area of cone = πrl + πr²

Curved surface area of hemisphere = 2πr² 

Calculation 

Total Surface area of solid = CSA of Cone + CSA of hemisphere

⇒ πrl + 2πr²

⇒ (22/7) × x × √5 x + 2 × (22/7) × x²

⇒ (22/7) × x² (√5 + 2)   

⇒ 13.31 x²

∴ The total surface area of the solid is 13.31 x²

Given  

Radius of cone and hemisphere = x 

and Height of cone = 2x 

Radius and height of cylinder x and 3x respectively. 

Formula Used 

Volume of cone = (1/3) πR2H

Volume of hemispher = (2/3) πR3  

Volume of cylinder = πR2H

Calculation 

Volume of water left in the cylinder = (Volume of cylinder) - Volume of (cone + hemisphere) 

⇒ πx².3x - [(1/3)πx².2x + (2/3)πx³]

⇒ πx2.3x - [(2πx³/3) + (2πx3/3)​]

⇒ (πx²) [3x - (2x/3 + 2x/3)]

⇒ (πx²)(5x/3) 

⇒ (22/7) × (5/3) × x³

⇒ 5.24 x³

∴ The approximate volume left in the cylinder is 5.24x³.

Given: 

Radius of circle = 2.1 cm

Formula Used: 

Area of sector = (θ/360) × πr²   

Area of triangle = (1/2) × base × height

Calculation:

Area of triangle = (1/2) × 2.1 × 2.1 sin120° = 1.9 cm²

Area of segment = Area of sector - Area of triangle corresponding to the sector 

Area of segment = [(120°/360°) × (22/7) × (2.1)²] - (1.9)

⇒ (1/3) × (22/7) × (2.1)² - 1.9

⇒ 4.62 cm² - 1.9 cm²

⇒ 2.71 cm²

∴ The approximate area of minor segment is 2.71 cm2

Given:

Radius of circle = 2.1 cm

Formula Used:

Area of sector = (θ/360) × πr2   

Area of triangle = (1/2) × base × height

Calculation:

Area of triangle = (1/2) × 2.1 × 2.1 sin120° = 1.9 cm2

Area of segment = Area of sector - Area of triangle corresponding to the sector 

Area of minor segment = [(120°/360°) × (22/7) × (2.1)2] - (1.9)

⇒ (1/3) × (22/7) × (2.1)2 - 1.9

⇒ 4.62 cm2 - 1.9 cm2

⇒ 2.71 cm2

Now, Area of major segment = Area of circle - Area of minor segment 

⇒ (22/7) × (2.1)² - (2.71)

⇒ 11.15 cm²

∴ The area of major segment = 11.15 cm²