BANK & INSURANCE (MENSURATION) PART 1

Total Questions: 60

11. Sum of areas of a circle and a square is equal to 170 sq. cm. The diameter of a circle is 14 cm. What is the ratio of the circumference of circle to the perimeter of square?

Correct Answer: (d) 11 : 4
Solution:Circumference of circle = 44 cm
πr² + a² = 170
π7² + a² = 170
a² = 170 − 154
a = 4 cm
Perimeter of square = 16 cm
Required ratio = 44:16 = 11:4

12. The length and breadth of a rectangle are in the ratio of 7 : 6 and its area is 6048 m². Find the perimeter of the rectangle.

Correct Answer: (e) 312 m
Solution:

Let the length and breadth of the rectangle is 7x and 6x.
then, (7x × 6x) = 6048, or, x² = 144, x = 12 meter.
length = 84 and breadth = 72 meter.
Perimeter = 2 (84 + 72) = 312 meter.

13. A rectangular plot is ready to be paved with bricks. The plot is 32 m long and 18 m broad and the dimensions of the bricks are 24 cm by 10 cm. Find the number of bricks required to pave the plot?

Correct Answer: (e) 24000
Solution:

Number of bricks required = (area of plot)/(area of 1 brick) = 3200 × 1800 / (24 × 10) = 24000

14. The length of a rectangular field is 75% more than its breadth. If the total cost of fencing the field at the rate of Rs. 6 per metre was Rs. 264, what is the area of the field? (in square metres)

Correct Answer: (a) 112
Solution:

Let the breadth of rectangle be x m.
Then, length = x + 75x/100 = 7x/4
Perimeter of the rectangle = 264/6 = 44 m.
2(x + 7x/4) = 44
4x + 7x = 22 × 4
11x = 88
x = 8.
Length = (7x)/4 = 14 m
Thus, Area of rectangle = 8 × 14 = 112 sq m.

15. A cylindrical iron rod of length 80 cm and area of cross section 7546 cm² is melted down and casted into 220 cubes of equal size. Find the length of each side of the cube.

Correct Answer: (b) 14 cm
Solution:Let, each side cube be ‘x’ cm.
Volume of cylinder = 220 × volume of one cube
7546 × 80 = 220 × x³
x = 14
Therefore, each side of cube = 14 cm

16. A room is 12m long, 6m wide and 5m high. There are 2 doors with dimensions 2m × 1.5m; two windows of dimensions 1.5m × 1.8m and two more windows of 1m height and 0.5m breadth. Find the cost of applying wallpaper to cover the room at Rs. 10/m².

Correct Answer: (c) Rs. 1,676
Solution:

Area of 4 walls = 2(l + b)h = 2(12 + 6) × 5 = 180 m²
Area of 2 doors = 2 × 2 × 1.5 = 6 m²
Area of 4 windows = (2 × 1.5 × 1.8) + (2 × 1 × 0.5) = 6.4 m²
Area to be wallpapered = 180 − 6 − 6.4 = 167.6 m²
Cost = 167.6 × 10 = Rs. 1,676.

17. Mr. Vishal Kuthiala wanted to build an independent house. So he purchased a rectangular plot with dimensions 48 m × 40 m. He planned in such a way that there should be a path of uniform width inside and all around the plot of area 336 sq. m. What should be the width of the path?

Correct Answer: (a) 2 m
Solution:Let the width be x m.
(48 × 40) − (48 − 2x)(40 − 2x) = 336
1920 − (1920 − 176x + 4x²) = 336
4x² − 176x + 336 = 0
x² − 44x + 84 = 0
(x − 42)(x − 2) = 0
x = 2 (since x cannot be 42).

18. What would be the cost incurred in purchasing bricks of dimensions 4 cm × 9 cm × 16 cm to construct a wall of dimensions 6 m × 6 m × 8 m. 10% of the volume of the wall is occupied by cement and the remaining bricks and it is also given that 1000 bricks cost Rs. 500?

Correct Answer: (b) Rs. 2,25,000
Solution:Volume of wall = 600 cm × 600 cm × 800 cm = 288000000 cm³.
Volume taken by cement = 10% of 288000000,
Volume to be used by bricks = 288000000 − 28800000 = 259200000.
Number of bricks required = 259200000/(4 × 9 × 16) = 450000
Cost of purchasing bricks = (450000/1000) × 500 = Rs 225000.

19. The inner and outer radius of a circular track is 3:5. If the cost of carpeting the circular track is Rs 3660 at the rate of Rs 36.6 per sq m, then find the inner diameter.

Correct Answer: (e) 8.46 m
Solution:Area of the track = 3660/36.6 = 100;
π(R² − r²) = 100 ⇒ π(25x² − 9x²) = 100 ⇒ x² = (7 × 100)/(22 × 16) = 1.98 ⇒ x = 1.41
Inner diameter = 3 × 2 × 1.41 = 8.46 m

20. If the length of a rectangular plot is decreased by 50%, it becomes a square. It is also given that the ratio of area of the rectangle to its parameter is 10/3 m. The cost of fencing along the length (per meter) of the rectangle is 50% less than fencing the breadth (per meter). If the cost of fencing the breadth is Rs. 30 per m, find the total cost of fencing the plot.

Correct Answer: (d) Rs. 1200
Solution:Let the length be a and breadth be b.
a = 2b,
ab/2(a + b) = 10/3
a = 20 m, b = 10 m.
Cost of fencing along the length = Rs.15 per meter
Net cost of fencing along the total length = 40 × 15 = 600
Net cost of fencing along the total breadth = 20 × 30 = 600
Total cost = Rs.1200.