BANK & INSURANCE (MENSURATION) PART 3

Total Questions: 60

1. Tom can paint one side of a triangular wall of base 18 metres and height 25 metres in 18 days. Ram can paint 80% of one side of a rectangular wall of length ______ metres and breadth 25 metres in 28.8 days. There is a circular wall of diameter 28 metres. If Tom and Ram work together for 12 days on painting the circular wall, then they would have finished painting one side and painted 59 m² of the other side. (Note: Thickness of all the walls are negligible).

Ques: What is the value that will come in place of the blank?

Correct Answer: (c) 63 metres
Solution:

Area of one side of the triangular wall = (1/2) × base × height = (1/2) × 18 × 25 = 225 m²

So, painting efficiency of Tom = 225 ÷ 18 = 12.5 m²/day

Area of one side of the circular wall = π × radius² = (22/7) × (28/2) × (28/2) = 616 m²

So, total area painted by Ram and Tom = 616 + 59 = 675 m²

Out of which, area painted by Tom = 12 × 12.5 = 150 m²

So, area painted by Ram = 675 − 150 = 525 m²

So, painting efficiency of Ram = 525 ÷ 12 = 43.75 m²/day

So, area painted by Ram in 28.8 days = 43.75 × 28.8 = 1260 m² = 80% of the area of one side of the rectangular wall

So, 100% of area of one side of the rectangular wall = 1260 ÷ 0.8 = 1575 m²

And so, length of the rectangular wall = 1575 ÷ 25 = 63 metres

Hence, option c.

2. A well of radius 7 metres is dug out in the shape of a cylinder such that the curved surface area of the well (excluding embankment) is 1320 m². Out of total mud taken out in this process, 440 m³ of mud is thrown away and rest is used to make an embankment (of constant width) around the well such that the height of the embankment is 14 metres. What is the width (or thickness of the embankment)?

Correct Answer: (d) 5 metres
Solution:

Curved surface area of the well = 2 × π × radius × height
= 2 × (22/7) × 7 × height = 1320

So, height of the well = 1320 ÷ (2 × 22) = 30 metres

So, volume of the well = π × radius² × height
= (22/7) × 7² × 30 = 4620 m³ = volume of the mud dug out from the well

So, volume of mud used for the embankment = 4620 − 440 = 4180 m³

Let the radius of the well = ‘R₁’ metres
Let the radius of the embankment from the centre of the well = ‘R₂’ metres

Then, volume of the embankment = (π × R₂² × height of the embankment) − (π × R₁² × height of the embankment)

= π × (R₂² − R₁²) × 14
= (22/7) × (R₂² − 7²) × 14
= 44 × (R₂² − 49) = 4180

So, (R₂² − 49) = 4180 ÷ 44 = 95

So, R₂² = 95 + 49 = 144
So, R₂ = 12

So, width of the embankment = 12 − 7 = 5 metres

Hence, option d.

3. Achilles, Agatha, and Aleksanteri are racing around a circular track of ‘D’ metres. Achilles beats Agatha by 800 metres, and Aleksanteri by 1200 metres. Similarly, Agatha beats Aleksanteri by 600 metres. If the race lasts for ‘R’ rounds, then which of the following cannot be the value of ‘R’, given that length of track is an integer?

Correct Answer: (c) 18
Solution:

Length of race = D × R = DR metres

Distance covered by Agatha in the time that Achilles takes to cover DR metres = (DR − 800) metres

And, distance covered by Aleksanteri in the time that Achilles takes to cover DR metres = (DR − 1200) metres

And, distance covered by Aleksanteri in the time that Agatha takes to cover DR metres = (DR − 600) metres

ATQ:
[(DR − 800)/(DR − 1200)] = [DR/(DR − 600)]

(DR − 600) × (DR − 800) = DR × (DR − 1200)

DR² − 600DR − 800DR + 480000 = DR² − 1200DR

480000 = 200DR

So, DR = 2400

So, R = (2400/D)

Among given options, only 18 is not a factor of 2400.

So, all the other options are possible answers.

Hence, option c.

4. Curved surface area and volume of a right circular cylinder is 1056 cm² and 7392 cm³. Find the total surface area (in cm²) of a right circular cone whose radius is 8 cm and height is as same as the height of the room whose length is 20% more than the height of the cylinder and breadth is 20% less than radius of the cylinder. Area of four walls of the room is 768 cm².

Correct Answer: (d) 200π
Solution:

Let the radius and height of the cylinder be ‘r’ cm and ‘h’ cm, respectively.

So, 1056 = 2 × (22/7) × r × h ⇒ r = 168/h

And, 7392 = πr²h = (22/7) × (168/h)² × h

h = 12 and r = 14

Length of the room = (12 × 1.2) = 14.4 cm
And, breadth of the room = (14 × 0.8) = 11.2 cm

Let the height of the room be ‘H’ cm

According to the data given:
768 = 2 × (14.4 + 11.2) × H

 H = (384/25.6)
So, H = 15

So, height of the cone (H) = 15 cm

And, radius of the cone (r) = 8 cm (given)

So, slant height of the cone (l) = √[(15)² + (8)²]
= √(225 + 64) = √289 = 17 cm

So, total surface area of the cone = πr(l + r) = π × 8 × (17 + 8) = 200π cm²

Hence, option d.

5. Directions (5-6): Answer the question based on the information given below.

A boy has some materials such a cone, cylinder, sphere, hemisphere etc. He decided to make some structure by joining these materials. The radius of the cone is 50% more than the radius of the sphere of volume of 16384 cm³. The height of cylinder is twice the height of the cone. The curved surface area of the cone is 1800 cm². The total surface area of the hemisphere is 4356 cm² and the radius of hemisphere is 10% more than the radius of the cylinder. If cone is attached to one end of the cylinder and hemisphere is on the other end of cylinder with the centre of the circle of all the materials joined to form a structure lie on same axis and is wrapped with a gift cover which costs him Rs. 6 per sq. cm². [Use π = 3]

Ques: Find the cost of wrapping the whole structure thus formed.

Correct Answer: (a) Rs. 42984
Solution:

Let radius of the sphere is ‘r’ cm

Volume of sphere = (4/3) × π × r³
(4/3) × π × r³ = 16384

r³ = 4096 ⇒ r = 16 cm

So, radius of cone = 1.5 × 16 = 24 cm

Curved surface area of cone = 1800 cm²

Let slant height and height of the cone is ‘l’ and ‘h’ cm respectively.
π × r × l = 1800
3 × 24 × l = 1800
l = 1800/72 = 25 cm

So, height of cone = √(l² − r²)
h = √(25² − 24²) = √(625 − 576) = √49 = 7 cm

Now, height of cylinder = 2 × 7 = 14 cm
Let radius of the hemisphere be ‘R’ cm

So, 3 × π × R² = 4356
R² = 4356/9 = 484
R = 22 cm

So, radius of cylinder = 22/1.1 = 20 cm

Surface to be wrapped = curved surface area of cylinder + curved surface area of cone + curved surface area of the hemisphere + π × (24² − 20²) + π × (22² − 20²)

= 2 × π × 20 × 14 + 1800 + 2 × π × 22 × 22 + π × 176 + π × 84

= 1680 + 1800 + 2904 + 780
= 7164 cm²

Desired cost = 6 × 7164 = Rs. 42984

6. Find the total volume of the whole structure thus formed.

Correct Answer: (d) 42128 cm³
Solution:

Let radius of the sphere is ‘r’ cm

Volume of sphere = (4/3) × π × r³
(4/3) × π × r³ = 16384
r³ = 4096 ⇒ r = 16 cm

So, radius of cone = 1.5 × 16 = 24 cm
Curved surface area of cone = 1800 cm²

Let slant height and height of the cone is ‘l’ and ‘h’ cm respectively.
π × r × l = 1800
3 × 24 × l = 1800
l = 1800/72 = 25 cm

So, height of cone = √(l² − r²)
h = √(25² − 24²) = √(625 − 576) = √49 = 7 cm

Now, height of cylinder = 2 × 7 = 14 cm
Let radius of the hemisphere be ‘R’ cm

So, 3 × π × R² = 4356
R² = 4356/9 = 484 ⇒ R = 22 cm

So, radius of cylinder = 22/1.1 = 20 cm

Total volume of the structure thus formed =
= 1/3 × π × 24 × 24 × 7 + π × 20 × 20 × 14 + 2/3 × π × 22 × 22 × 22

= 4032 + 16800 + 21296 = 42128 cm³

Hence, option d.

7. The sum of the length and breadth of a rectangular field is 32 metres. Which of the following can be cost of fencing (in Rs.) at the rate of Rs. 4/metre, and cost of cultivating the field (in Rs.) at the rate of Rs. 2/m², if the length and breadth of the field are in the ratio of 5:3?

Correct Answer: (c) 256, 480
Solution:

Perimeter of the field = 32 × 2 = 64 metres

Therefore, cost of fencing the field = 64 × 4 = Rs. 256

Length of the field = 32 × (5/8) = 20 metres
Therefore, breadth of the field = 32 − 20 = 12 metres

Therefore, cost of cultivating the field = 12 × 20 × 2 = Rs. 480

Hence, option c.

8. A man walking diametrically across a semicircular playground takes 5 minutes less than if he had kept walking around the circular path. His speed is 50 meters per minute, then find the area of the square whose side is equal to the radius of the semicircle

Correct Answer: (b) 47961 sq.m
Solution:

Let the radius of the circle is r.

So, we can say,
Time taken by the person to go along the diameter = distance/speed = 2r/50 min

Time taken by the person to go along the semicircular path = distance/speed = πr/50 min

Given πr/50 − 2r/50 = 5 minutes
⇒ πr − 2r = 250
⇒ (π − 2)r = 250 ⇒ r = 250/(π − 2)
= 219 m (approximately)

Radius of the playground = 219 meters.
So, area of square = 219 × 219 = 47961 sq.m

9. Directions (9-10): Answer the questions based on the information given below.

A park is in the shape of rectangle surmounted by a semicircle along its length such that the mid-point of the length of the rectangular part coincides with the centre of the semi-circle. The length of the rectangular part is 20 m more than its breadth. A fence is build along the boundary of the park leaving a gap of 5 metres on each of the length of the rectangular part. The cost of fencing at the rate of Rs. 22.5 per metre is Rs. 7650. In the rectangular part, a concrete floor has to be made leaving a uniform gap of ‘y’ metres from all the sides. The area of the non-concrete rectangular part of the park is 1100 m².

Ques: What is the value of (x – y)?

Correct Answer: (a) 30
Solution:

Let the length of the rectangular part be ‘2x’ metres

Therefore, breadth of the rectangular part = (2x − 20) metres

Radius of the semi-circular part = 2x/2 = ‘x’ metres

Length of the park to be fenced =
(2x + 2x − 5 − 5 + 2x − 20 + 2x − 20) + [2 × (22/7) × x]/2

= ((78x − 350)/7) metres

((78x − 350)/7) = 7650/22.5

78x − 350 = 2380
78x = 2730
x = 35

Therefore, length of the rectangular part = 2x = 70 metres

Breadth of the rectangular part = (2x − 20) = 50 metres

Radius of the semi-circular part = x = 35 metres

Now,
Let the concrete floor is made by leaving a uniform gap of ‘y’ metres

Therefore, total area of the rectangular part = 70 × 50 = 3500 metres

Area of the concrete floor = (70 − 2y)(50 − 2y)
= (4y² − 240y + 3500)

4y² − 240y + 3500 = 3500 − 1100

4y² − 240y + 1100 = 0

4y² − 220y − 220y + 1100 = 0

4y(y − 5) − 220(y − 5) = 0

(4y − 220)(y − 5) = 0

y = 55, 5

But ‘y’ cannot be equal to 55 because breadth is of 50 metres only

Therefore, y = 5

(x − y) = 35 − 5 = 30

10. If whole park had been emptied then find difference between the cost of cultivating the semicircular part at the rate of Rs. 2.5/m² and the cost of cultivating the rectangular part at the rate of Rs. 4.5/m².

Correct Answer: (e) Rs. 10937.5
Solution:

Let the length of the rectangular part be ‘2x’ metres

Therefore, breadth of the rectangular part = (2x − 20) metres

Radius of the semi-circular part = 2x/2 = ‘x’ metres

Length of the park to be fenced = (2x + 2x − 5 − 5 + 2x − 20 + 2x − 20) + [2 × (22/7) × x]/2

= ((78x − 350)/7) metres

((78x − 350)/7) = 7650/22.5

78x − 350 = 2380
78x = 2730
x = 35

Therefore, length of the rectangular part = 2x = 70 metres

Breadth of the rectangular part = (2x − 20) = 50 metres

Radius of the semi-circular part = x = 35 metres

Now,
Let the concrete floor is made by leaving a uniform gap of ‘y’ metres
Therefore, total area of the rectangular part = 70 × 50 = 3500 metres

Area of the concrete floor = (70 − 2y)(50 − 2y) = (4y² − 240y + 3500)

4y² − 240y + 3500 = 3500 − 1100
4y² − 240y + 1100 = 0
4y² − 20y − 220y + 1100 = 0
4y(y − 5) − 220(y − 5) = 0
(4y − 220)(y − 5) = 0

y = 55, 5

But ‘y’ cannot be equal to 55 because breadth is of 50 metres only

Therefore, y = 5

Cost of cultivating the rectangular part = 3500 × 4.5 = Rs. 15750

Cost of cultivating the semicircular part = (1/2) × (22/7) × 35² × 2.5 = Rs. 4812.5

Required difference = 15750 − 4812.5 = Rs. 10937.50

Hence, option e.