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- Question
A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.
Options- A. 14
- B. 13
- C. 12
- D. 11
- Correct Answer
- 14
ExplanationDistance covered in one revolution = = 88m.
- 1. The difference between circumference and semidiameter of a circle is 37 cm. What is the diameter of the circle ?
Options- A. 18 cm
- B. 12 cm
- C. 16 cm
- D. 14 cm Discuss
- 2. A typist uses a sheet measuring 20cm by 30cm lengthwise.If a margin of 2 cm is left on each side and a 3 cm margin on top and bottom, then percent of the page used for typing is
Options- A. 64%
- B. 74%
- C. 84%
- D. 94% Discuss
- 3. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m.What is the area of the field?
Options- A. 47500
- B. 37500
- C. 57500
- D. 77500 Discuss
- 4. A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is?
Options- A. 10ft
- B. 20ft
- C. 30ft
- D. 40ft Discuss
- 5. The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is
Options- A. 240
- B. 140
- C. 40
- D. 340 Discuss
- 6. If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area ?
Options- A. 108 sq.cm
- B. 112 sq.cm
- C. 116 sq.cm
- D. 120 sq.cm Discuss
- 7. Find the area of the square whose side is equal to the diagonal of a rectangle of length 3 cm and breadth 4 cm.
Options- A. 25 sq.cm
- B. 16 sq.cm
- C. 9 sq.cm
- D. 4 sq.cm Discuss
- 8. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
Options- A. 22
- B. 44
- C. 66
- D. 88 Discuss
- 9. The breadth of a rectangular field is 75% of its length. If perimeter of the field is 1050 m, what will be th area of field ?
Options- A. 65700 sq.m
- B. 54500 sq.m
- C. 78700 sq.m
- D. 67500 sq.m Discuss
- 10. The length, breadth and height of a room are in the ratio 7:3:1. If the breadth and height are doubled while the length is halved, then by what percent the total area of the 4 walls of the room will be increased ?
Options- A. 90%
- B. 88%
- C. 85%
- D. 84% Discuss
Area problems
Search Results
Correct Answer: 14 cm
Explanation:
circumference of a circle = 2 r
=> 2 × 22/7 × r ? r = 37
=> 37/7 × r = 37
=> r = 7 cm.
Diameter D = 2r = 7×2 = 14 cm.
Correct Answer: 64%
Explanation:
Area of the sheet =
Area used for typing =
required % = =64%
Correct Answer: 37500
Explanation:
Let length = x meters, then breadth = 0.6x
Given that perimeter = 800 meters
=> 2[ x + 0.6x] = 800
=> x = 250 m
Length = 250m and breadth = 0.6 x 250 = 150m
Area = 250 x 150 = 37500 sq.m
Correct Answer: 20ft
Explanation:
let the radius of the pool be Rft
Radius of the pool including the wall = (R+4)ft
Area of the concrete wall = = sq feet
=> 8
Radius of the pool R = 20ft
Correct Answer: 140
Explanation:
distance covered in 1 revolution =
= 2 x (22/7) x 20 = 880/7 cm
required no of revolutions = 17600 x (7/880) = 140
Correct Answer: 120 sq.cm
Explanation:
The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle =1/2 x base x height => 1/2 x 24 x 10 = 120 sq.cm
Correct Answer: 25 sq.cm
Explanation:
Given length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Then, the diagonal of the rectangle
Then, it implies side of square = 5 cm
We know that Area of square = S x S = 5 x 5 = 25 sq.cm.
Correct Answer: 88
Explanation:
We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Correct Answer: 67500 sq.m
Explanation:
Let the length be 'l' and breadth be 'b'.
b = l × 3/4__________(a)
2(l+b) = 1050
l+b = 525___________(b)
From equations (a) and (b),
l = 300m, b = 225 m
Area = l × b
= 300 × 225
= 67500 sq.m.
Correct Answer: 90%
Explanation:
Let length, breadth and height of the room be 7, 3, 1 unit respectively.
Area of walls = 2(l+b)xh = 2(7+3)x1 = 20 sq. unit.
Now, length, breadth and height of room will become 3.5, 6 and 2 respectively.
Area of walls = 2(l+b)xh = 2(3.5+6)x2 = 38 sq. unit.
% Increase in the area of walls = (38-20)x100/20 = 90%.
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