BANK & INSURANCE (MENSURATION) PART 2

Area of the circular surface = (924/1.5) = 616 m²
Let the radius of the painted shape be ‘r’ metres

Therefore,
(22/7) × r × r = 616
r² = (616 × 7)/22 = 196
r = √196 = 14 (Since, radius cannot be negative)

Diameter of the circular surface = 2 × 14 = 28 metres

Length of each side of the square lawn = 28 metres

Perimeter of the square lawn = 28 × 4 = 112 metres

Cost of fencing the lawn = 112 × 2.5 = Rs. 280

Base area of the cylinder = π × radius² = (22/7) × radius² = 1386
So, radius² = 1386 ÷ 22 × 7 = 441
So, radius = √441 = 21 cm

Curved surface area of the cylinder = 2 × π × radius × height
= 2 × (22/7) × 21 × 10 = 1320 cm²

Let the length and breadth of the rectangular sheet be ‘l’ metres and ‘b’ metres, respectively.

According to question:
2 × (l + b) = 34
(l + b) = 17 … (i)

And, (l × b) = 70 … (ii)

After solving equation (i) and (ii), we get
(l + b)² = (l − b)² + 4lb

(17)² = (l − b)² + (4 × 70)
289 − 280 = (l − b)²
(l − b)² = 9

So, (l − b) = 3 … (iii)
(Since, length is greater than breadth)

Hence, option d.

Let the radius of the circular park be ‘r’ metres

Circumference of a circle = 2 × π × r, where ‘r’ = radius of the circle

ATQ:
66 = 2 × (22/7) × r
44r = 7 × 66
r = 10.5

Area of the circular park excluding the path = (22/7) × 10.5 × 10.5 = (693/2) m²

Radius of the circular park including the path = 10.5 + 3.5 = 14 m

Area of the circular park including the path = (22/7) × 14 × 14 = 616 m²

Area of the path = Area of the circular park including the path − Area of the circular park
Area of the path = 616 − (693/2) = 269.5 m²

Let the length, breadth and height of the room be ‘l’ metres, ‘b’ metres and ‘h’ metres, respectively.

According to question:
54 = 2 × (l + b)
Or, (l + b) = 27 … (eq. i)

And, (l × b) = 180

Now, (l + b)² = (l − b)² + 4lb

So, (27)² = (l − b)² + 4 × 180
729 − 720 = (l − b)²
(l − b)² = 9

So, (l − b) = 3 … (eq. ii)

Adding equation (i) and (ii), we get
l = 15 And, b = 27 − 15 = 12

So, h = 15 + 15 × 0.2 = 18

Area of four walls of the room = 2 × (l + b) × h
= (54 × 18) = 972 m²

Let the radius of the sphere be ‘R’ cm.

2304π = (4/3) × π × R³
1728 = R³

So, R = 12

So, length of edge of the cube = 12 × 2 = 24 cm
Required volume = 24³ = 13824 cm

Let the length of the rectangle be ‘L’ cm.

Area of rectangle = Length × breadth
192 = L × 12
L = 16

Length of diagonal of rectangle = √(L² + B²)
So, length of the diagonal = √(12² + 16²) = 20 cm

Required percentage = (20 − 16)/16 × 100 = 25%

Area of the track = length of track × breadth of track
= 75 × breadth of track = 4500

So, breadth of track = 4500 ÷ 75 = 60

So, perimeter of the track = (75 + 60) × 2 = 270 metres

So, speed of the car = (270/10) = 27 m/s

Relative speed of the car with respect to the person = 27 + 5 = 32 m/s

So, distance between the car and the person, 2 seconds before they cross each other = 32 × 2 = 64 metres

Hence, option e.

Height of the box = 1800 ÷ (15 × 10) = 12 metres

Total surface area of the box = 2 × {(length × breadth) + (breadth × height) + (length × height)}
= 2 × {(15 × 10) + (10 × 12) + (15 × 12)}
= 2 × {(150 + 120 + 180)} = 2 × 450 = 900 m²

So, total cost of painting the box = 900 × 6 = Rs. 5,400

πr² + lb = 2058
(22/7) × 21 × 21 + l × 24 = 2058
1386 + 24l = 2058
24l = 2058 − 1386
24l = 672, l = 28 cm

Perimeter of rectangle = 2 × (l + b) = 2 × (28 + 24)
⇒ 2 × 52 = 104 cm