Explore topic-wise InterviewSolutions in .

Given - 

ratio of the length and breadth = 7 : 4

Area of path = 416 m²

4 meter wide path

Formula used - 

If a path of x m wide run around a rectangular field having length and breadth l and b then,

area of path = 2x × (l + b + 2x)

Solution - 

Let the length and breadth be 7k and 4k.

⇒ 2 × 4 × (7k + 4k + 8) = 416

⇒ 11k = 44

⇒ k = 4 m

⇒ breadth of rectangle = 4k = 16 m

∴ breadth of rectangle = 16 m.

Given:

20% of the area of the road = 290.4 m²

Width of the road = 3 m

Formula used:

Area of a circle = 22/7 × r²

Where, r = radius of the circle.

Calculations:

Let us assume the radius of the park = r

Area of the park = 22/7 × r²

Area of the park with the road = 22/7 × (r + 3)²

ATQ,

22/7 × (r + 3)² - 22/7 × r² = 290.4 × 5

⇒ (2r + 3) × 3 = 462

⇒ r = 75.5

Diameter of the park = 151 m

Given:

Length of circular path = 64cm

Diameter of Circular garden = 10m = 1000 cm

Radius of Circular garden = 10/2 m = 500 cm

Quantity I:

Inner radius = 500 cm 

Outer radius =  564 cm

Area of path = Outer area - Inner area

⇒ 22/7 × (5642 - 5002)

⇒ 22/7 × (564 + 500)(564 - 500)

⇒ 22/7 × (1064)(64)

⇒ 21.4016 m2

⇒ 21 m2(approx.)

Given:

A path of uniform width runs round inside the rectangular field of 38 m long and 32 m wide. if the path occupies 600 m².

Concept used:

Area of rectangle = Length × Breadth

Calculation:

Let the width of path be x m.

Area of rectangular field = 38 × 32 = 1216 m²

Area of rectangular field without path = (38 - 2x) (32 - 2x)

⇒ 1216 - 64x - 76x + 4x²

⇒ 4x² - 140x + 1216

Area of path 

⇒ 1216 - (4x² - 140x + 1216)

⇒ 140x - 4x² 

The path of the rectangular field occupies is 600 m²

⇒ 140x - 4x² = 600

⇒ 4x² - 140x + 600

⇒ x² - 35x + 150

⇒ x² - 30x - 5x + 150

⇒ x(x - 30) - 5(x - 30)

⇒ (x - 5) (x - 30)

∴ x = 5 because \(x \ne 30\)

∴ x = 5m is the width of the path. 

Given: 

The cost of painting the lateral surface area of a cube at the rate of 0.15 rupee per cm² is 38.4 rupee

Concept: 

Lateral surface area = Cost of painting the lateral surface/cost of painting per cm² 

The lateral surface area of cube = 4 × (Side)²

The surface area of cube = 6 × (Side)²

Calculation: 

The lateral surface area of the cube = 38.4/0.15

⇒ 256 cm²

Now, 

⇒ 4 × (Side)² = 256

⇒ Side = (64)^1/2

⇒ Side = 8 cm

The surface area of the cube is 

⇒ 6 × 8²

⇒ 6 × 64

⇒ 384 cm²

∴ The required surface area of the cube is 384 cm².

Given:
Ratio of sides of the wooden box = 1 : 2 : 3
Volume of the wooden box = 750 cm³
Rate of painting = Rs. 21 per cm²

Concept used:
Total surface area of cuboidal box = 2 × (lb + bh + hl)
Volume of cuboidal box = l × b × h
Cost of painting the cuboidal box = Rate × Total surface area of box

Calculation:
Let sides of the box be a, 2a and 3a.
Volume of box = 750
⇒ a × 2a × 3a = 750
⇒ 6a³ = 750
⇒ a³ = 750/6
⇒ a³ = 125
⇒ a = 5
Surface area of box = 2(a × 2a + 2a × 3a + 3a × a)
⇒ 2(2a2 + 6a2 + 3a2)
⇒ 2 × 11a2
⇒ 22 × 5 × 5
⇒ 550 cm²
Cost of painting = 21 × 550
⇒ Rs. 11550
∴ The cost of painting the outer side of the wooden box is Rs. 11550.

Given:

Volume of the cylinder = 12320 cm3

Height of the cylinder = 20 cm

Formula used:

Total surface area of the cylinder = 2 π r(r + h)

Volume of the cylinder = π r2 h

Here, h and r is the height and radius respectively

Calculation:

Volume of the cylinder = π r2 h

⇒ 22/7 × r2 × 20 = 12320

⇒ r2 = 196

⇒ r = 14 cm

Now, Total surface area of the cylinder = 2 π r(r + h)

⇒ 2 × 22/7 × 14(14 + 20)

⇒ 2992 cm2

Cost of painting = 4 × 2992 = 11968

∴ The cost of painting is Rs 11968

Given

External radius = 10 cm

Internal diameter = 6 cm

Formula Used 

Volume of hollow sphere = 4/3 π(R₁³ - R₂³)

Calculation 

Volume of material = 4/3 ×  22/7 (10³ - 3³) cm³

⇒ 4/3 × 22/7 (1000-27) cm³

⇒4/3 ×  22/7 × 973 cm³  =  4077.33 ≈ 4077 cm³

∴ The required answered is 4077 cm³.

Given 

Dimension of rectangular sheet = 40 cm × 25 cm 

Formula Used 

Area of rectangle = length × breadth

Area of circular button = πr² 

Calculation;

Area of sheet = 40 × 25 cm²   = 1000 cm²

Area  of button =  3.14 × 1 × 1 cm² = 3.14 cm²

Area of 224 buttons = 224 × 3.14 cm²  =703.36

Area of sheet left out = (1000 - 703.36) cm²  = 296.64 cm²

∴ The required answer is None of these 

Given :

The surface area of two spheres are in ratio = 49 : 25

Formula used : 

\(Area of sphere = 4πr^2 \)

\(Volume \ of \ sphere = \frac{4}{3} π r^3\)

Calculation:

According to question 

⇒ \( 4πR^2 : 4πr^2 = 49 : 25\)

⇒ R : r = 7 : 5

⇒ \(Volume\ of \ sphere = \frac{4}{3}( π 7^3): \frac{4}{3} (π 5^3)\)

Hence R^{3 :} r ³ is 343 : 125.

Given:

The ratio of volume of two spheres = 1 ∶  8

Formula used:

The volume of sphere = (4/3) × π r³

The surface area of sphere = 4π r²

Calculations:

Let us take the radius of  1st sphere to be r1

Let us take the radius of the 2nd sphere to be r2

The volume of  1st sphere to be v₁ =  (4/3) × π r₁3

The volume of  1st sphere to be v1 =  (4/3) × π r₂3

The ratio of volume the two spheres = \(\)[ (4/3) × π r₁³] ÷ [(4/3) × π r23]

⇒ [ (4/3) × π r13] ÷ [(4/3) × π r23] = 1 ∶ 8

⇒ (r₁ ÷ r₂₎³ = 1/8

⇒ (r1 ÷ r2)3  = (1/2)³

⇒ r1 ÷ r2 = 1/2

The surface area of 1st sphere = 4π r₁2

The surface area of 2nd sphere = 4π r₂2

The ratio of surface area of two spheres = (4π r12) ÷ (4π r22)

⇒ (r₁ ÷ r₂)² = (1/2)²

⇒ 1/4 

∴ The ratio of surface area of two spheres = 1 ∶ 4

Given: 

Perimeter of quadrant is 11 feet 

Formula Used: 

Perimeter of quadrant = πR/2 + 2R 

Calculation: 

(22/7 × R)/2 + 2R = 11 

⇒ 50R = 11 × 7 × 2

⇒ R = 77/25 = 3.08 feet 

∴ The radius of circle is 3.08 feet.

Given:

Volume of sphere = 38808 cm³

Concept:

First calculate the radius of the sphere then proceed to find the surface area.

Formula used:

Volume of sphere = (4/3)πR³

Surface area of sphere = 4πR²

Calculation:

Volume of sphere = (4/3)πR³

⇒ 38808 = (4/3) πR³

⇒ 38808 = (4 × 22 × R³)/21

⇒ (38808 × 21)/88 = R³

⇒ R = 21 cm

Surface area of sphere = 4πR²

⇒ 4 × (22/7) × 21 × 21

⇒ 5544 cm²

∴ The surface area of the sphere is 5544 cm2. 

Given:

TSA = 27 cm²

Formula used:

TSA = 6a²   (where a is the side of the cube)

Diagonal =  length of the longest rod inside the cubical room = a√3

Calculations:

According to the question, 6 × a² = 27

⇒ a² = 27/6

⇒ a2 = 9/2

⇒ a = 3/√2 cm

⇒ Diagonal = (3/√2) × √3

⇒ Diagonal = 3√3/√2 cm

∴  The length of the longest rod inside the cubical room is 3√3/√2 cm.

Given 

Length the volume of a cuboid  = X 

Breath the volume of a cuboid  = Y 

Height the volume of a cuboid  = Z

Formula used 

Volume of cuboid = length × breath × height 

Calculation 

(X) × (Y) × (Z) = XYZ

∴ Volume of cuboid is XYZ.

Given:

⇒ Curved surface area of cylinder = 440

⇒ 440 = 2πrh

Then,

⇒ h = r + 3

Solving,

⇒ h = 10cm and r = 7cm

⇒ Radius of circle = 7cm

⇒ Circumference of a circle = 2π × radius

= 2 × 22/7 × 7

= 44 cm

∴ Circumference of a circle is 44cm.

Given :-

The radius of a circular cone is 6 cm and its height is 7 cm

Concept :-

Volume of cone = (1/3)πR²h

Calculation :-

⇒ Volume of cone = (1/3) × (22/7) × (6)² × 7

⇒ Volume of cone = (1/3) × (22/7) × 36 × 7

⇒ Volume of cone = 12 × 22

⇒ Volume of cone = 264 cm³

∴ Volume of cone is 264 cm³

Calculation:

Area of triangle = 1/2 × (Base × Height)

∴ The formula to find out the area of triangle is 1/2 × (Base × Height)

Given:

Total surface area of cube = 108 cm²

Formula used:

TSA of cube = 6a²

Volume of cube = a³

Calculation:

Let, the side of the cube be ‘a’

∵ TSA of cube = 6a²

⇒ 108 cm² = 6a²

⇒ a² = 108/6

⇒ a² = 18 cm²

⇒ a = √18 = 3√2 cm

∴ Volume of cube = a³

= (3√2)³

= 54√2 cm³

Given:

Diameter of wheel = 1.4 m

Radius of wheel = 1.4/2 = 0.7 m

No. of revolution in 60 second = 5

Concept:

Calculate the no. of revolution of the wheel in 6 second, and then multiply it with the circumference of the wheel.

Formula Used:

Circumference of circle = 2πr

Where,

r → Radius of circle

Calculation:

Number of revolutions in a second = 5/60 = 1/12

No of revolution in 6 second = 6/12 = 1/2

Distance covered by the wheel in 6 seconds = (1/2) × 2 × (22/7) × 0.7 = 2.2 m

∴ Distance covered by the wheel in 6 seconds = 2.2 m 

Given :

If a point lies inside a circle, what will be the number of tangents drawn from that point to the circle

Calculation :

No tangents line can be drawn through a point within a circle, since any such line must be a secant line.

However, two tangent lines can be drawn to a circle from a point P outside of the circle. 

∴ The required answer is 0

Given

Diameter of base of cylinder = 2 cm

Height of cylinder = 16 cm

Formula used

Volume of sphere = (4/3)πr³, where r is the radius of the sphere

Volume of Cylinder = πR²h, where R is the radius of circular base and h is the height of cylinder

Calculation

Radius of base of cylinder = 1 cm

Volume of cylinder = πR2h

⇒ π(1)²(16) = 16π

Volume of 12 spheres = 12 × (4/3)πr³

⇒ 16πr³

 Volume of cylinder = volume of 12 spheres 

⇒ 16π = 16πr³

⇒ r³ = 1

⇒ r = 1 cm

∴ Diameter = 2 × radius = 2 × 1 = 2 cm

GIVEN:

Perimeter of Square = 256 cm

Ratio of Radius to height = 2x : 5x

Radius of cylinder = 12.5% of side of the square.

FORMULA USED:

1) Side of square = perimeter of square/4

2) Volume of cylinder = πr²h cm³

VALUES USED:

⇒ 12.5% = 1/8

CALCULATION:

⇒ Side of square = 256/4 = 64 cm

⇒ Radius of cylinder = 12.5% of side

⇒ Radius of cylinder = 12.5% of 64 = 8 cm

⇒ Ratio of Radius to height = 2x : 5x

⇒ 2x = 8 cm

⇒ 5x = 20 cm

⇒ Volume of cylinder = 4022.85 cm³

(4) 10 × 5 = 50 sq m. 

6 × 8 = 48 sq m. 

50 + 48 = 98 sq m

Given :

Length, breadth and height of the tank are respectively 15 m, 28 dm and 6 m 

75% of the whole volume need to be taken out of the tank 

Outlet tap can pump out 7 litres of water per minute

Formula used :

The total volume of the cuboid = length × breadth × height

Total time taken to pump out water = Total volume/Water thrown out per unit time  

Calculations :

Volume of the cuboid = 15 ×  2.8 × 6    (1 dm = (1/10) m)

⇒ 252 m³ or 252000 litres                 (1 m3 = 1000 litres)

Now 75% of the total volume = 75% of 252 m³

⇒ 189 m³ or 189000 litres              

Total time taken to pump out 75% of the water = 189000/7 

⇒ 27000 minutes 

Total time in hours = (1/60) × 27000 

⇒ 450 hours 

∴ total time taken to pump out 75% of the water is 450 hours 

Given:

Length of the cuboid is 20 cm

Ratio of length, breadth and height of the cuboid is 4 : 2 : 3 respectively

Formula used:

Total surface area of cuboid = 2{(length × breadth) + (breadth × height) + (height × length)}

Calculation:

Ratio of length, breadth and height is given

So, length = 4x = 20 cm

⇒ x = 5 cm

So, breadth = 2x = 10cm

Height = 3x = 15 cm

Total surface area = 2{(length × breadth) + (breadth × height) + (height × length)}

⇒ 2{(20 × 10) + (10 × 15) + (15 × 20)}

⇒ 2(200 + 150 + 300)

⇒ 2 × 650 = 1300 cm²

∴ The total surface area of the cuboid is 1300 cm²

Given:

The length of the cuboid is 4/3 of its breadth

The height of the cuboid is 1/3^rd of its breadth

The height of the cuboid is 10 cm

Formula used:

The volume of cuboid = length × breadth × height

Calculation:

Breadth = height × 3

⇒ 10 × 3 = 30 cm

Length = 4/3 of its breadth

⇒ 30 × (4/3) = 40 cm

The volume of cuboid = length × breadth × height

⇒ 40 × 30 × 10 = 12000 cm³

∴ The volume of the cuboid is 12000 cm³

Given:

Radius of cylinder = 3 cm

Height of cylinder = 4 cm 

Formula Used:

Volume of cylinder = π (Radius)² Height

Volume of sphere = (4/3)π (Radius)³

Concept Used:

One figure is melted and another figure is formed from that in this case volumes of both the figures will be equal.

Calculation:

Volume of cylinder = π (3 cm)² (4 cm)

⇒ π × 9 × 4 cm³

⇒ 36π cm³

Volume of sphere = (4/3)π (Radius)³

⇒ (4/3)π (Radius)³ = 36π cm³

⇒ (Radius)³ = 27 cm³

⇒ Radius = 3 cm 

∴ Radius of the sphere is 3 cm.

Given:

The base of the triangle is 10 cm and the height of the triangle is 4 cm

Concept:

The area of the triangle = (1/2) × (Base) × (Height)

Calculation:

⇒ The area of the triangle = (1/2) × 10 × 4 = 20 sq.cm

∴ The required result will be 20 sq.cm

Given:

Side of the square = 10 cm

Formula:

Area of the triangle = ½ × Base × height

Calculation:

Height of the triangle = 10 cm

Base of the triangle = 10 cm

Given:

The number of rounds run by Manik = 8

Side of a square field = 350

Formula used:

The perimeter of the square = 4 × side

1 km = 1000 m

Calculations:

Each round = Perimeter of a square

⇒ 4 × 350 = 1400 m

8 rounds = 1400 × 8

⇒ 11200 m

11200 ÷ 1000 = 11 km and 200 meters    (∵ 1 km = 1000 meters)

∴ The required distance is 11 km and 20​0 meters

Concept

Number of edges of cube = 12

Volume of cube = (edge)³

Calculation

The Sum of edges = 24 cm

Length of each edge = 24/12 = 2 cm

Volume of cube = 2³ = 8 cu. cm

Given:

Area of a circle = 154 cm²

Area of another circle = 2 × 154 cm²

Formula used:

Area of circle = πr²

Calculation:

Let r be the radius of the circle and R is the radius of another circle.

Area of a circle = 154 cm²

Area of another circle = 2 × 154 cm²

The ratio of their areas = (πr²)/( πR²)

⇒ 154/(2 × 154) = r²/R²

⇒ 1/2 = r²/R²

⇒ r/R = 1/√2

∴ The ratio of their radius is 1 ∶ √2.

Calculation:

Area of triangle = (1/2) × (Base × Height)

∴ The required formula is (1/2) × (Base × Height)

Given:

Number of revolutions = 5000

Distance covered = 11 km = 1100000 cm

Formula used:

Circumference of circle = 2πr

Calculation:

Circumference of circle = 1100000/5000

⇒ 220 cm

According to the question,

⇒ 2πr = 220 cm

⇒ 2 × (22/7) × r = 220 cm

⇒ r/7 = 5 cm

⇒ r = 35 cm

Diameter = 2r

⇒ 2 × 35

⇒ 70 cm

∴ The diameter of circle is 70 cm

Given:

The base and height of the right-angled triangle are 12 m and 16m respectively

Formula used:

Area of the right-angled triangle = ½ × base × height

The volume of the prism = Area of the base × Height of the prism

The number of the edges of the prism = The number of the sides of the base × 3

Calculation:

The number of the edges of the prism = The number of the sides of the base × 3

⇒ 3 × 3 = 9

Area of the base = ½ × 12 × 16 = 96 m2

The volume of the prism = 96 × 9 = 864 m3

∴ The number of edges and its volume is 9 and 864 m3

Given:

Length of wire is 28 cm

A strait wire is converted into a square

Formula used:

Perimeter of square = 4 × side

Diagonal of square = √2 × side

Calculation:

When the wire is transformed into the square then the length of wire is equal to the perimeter of square

∵ Length of wire = Perimeter of square

∴ 4 × side = 28 cm

⇒ Side = 7 cm

Now, Diagonal of square = √2 × 7 = 7√2 cm

Given:

Area of semi-circle is 77 square cm

Formula used:

Area of semicircle = (πr²)/2

Perimeter of semicircle = (πr) + 2r

Calculation:

Area of semi-circle = 77 square cm

∴ 77 = (πr²)/2

⇒ r = 7 cm

Now, Perimeter of semi-circle = (πr) + 2r = (π × 7) + 2 × 7 = 36 cm

Given:

A garden diameter is 22 m 

A playground having radius of 4 m

Formula used:

Area of circle = πr²

where,

r is Radius of the circle

Calculations:

The radius of the garden is 22/2 = 11 m

According to the question, we have

The area of the garden is

⇒ Area = (22/7) × 11 × 11

⇒ Area = 380.28 m²

And, Radius of the playground is 4 m

The area of the playground is

⇒ Area = (22/7) × 4 × 4

⇒ Area = 50.28 m²

The remaining area is 

⇒ Remaining area = 380.28 - 50.28

⇒ Remaining area = 330 m²

∴ The remaining area of a circular field is 330m².

Given:

The ratio of diagonals of squares is 5 : 7.

Formula Used:

Area of triangle = ½ × (Diagonal)²

Calculations:

Let the diagonals of the square be 5x and 7x respectively.

Area of Square₁ = ½ × (5x)²                 

Area of Square₂ = ½ × (7x)²

The ratio of their areas = ½ × (5x)² : ½ × (7x)²

⇒ 25 : 49

∴ The ratio of areas of squares is 25 : 49. 

Given:

The total surface area of the regular tetrahedron = 96√3 cm2

Formula used:

The total surface area of the regular tetrahedron = √3 × a2

The height of the tetrahedron = √6/3 × a

Where a is the side of the regular tetrahedron

Concept used:

The tetrahedron is a generally triangular pyramid

Calculation:

Let the side of the regular tetrahedron be a cm

As Surface area of regular tetrahedron = √3 × a2

⇒ √3 × a2 = 96√3

⇒ a = √96 = 4√6 cm

Now,

The height of the tetrahedron = √6/3 × a

⇒ √6/3 × 4√6

⇒ (6 × 4)/3

⇒ 8 cm

∴ The height of the regular tetrahedron is 8 cm