Area problems
- 1. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Options- A. 25% increase
- B. 50% increase
- C. 50% decrease
- D. 75% decrease Discuss
Correct Answer: 50% increase
Explanation:
Original area = xy.
| New length = | x | . |
| 2 |
New breadth = 3y.
| New area = | ❨ | x | x 3y | ❩ | = | 3 | xy. |
| 2 | 2 |
| ∴ Increase % = | ❨ | 1 | xy x | 1 | x 100 | ❩% | = 50%. |
| 2 | xy |
- 2. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:
Options- A. 10%
- B. 10.08%
- C. 20%
- D. 28% Discuss
Correct Answer: 28%
Explanation:
| Decrease in area |
|
||||||||||
|
|||||||||||
|
| ∴ Decrease % = | ❨ | 7 | xy x | 1 | x 100 | ❩% | = 28%. |
| 25 | xy |
- 3. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?
Options- A. 40
- B. 50
- C. 120
- D. Data inadequate
- E. None of these Discuss
Correct Answer: None of these
Explanation:
Then, length = (x + 20) metres.
| Perimeter = | ❨ | 5300 | ❩ | m = 200 m. |
| 26.50 |
∴ 2[(x + 20) + x] = 200
⟹ 2x + 20 = 100
⟹ 2x = 80
⟹ x = 40.
Hence, length = x + 20 = 60 m.
- 4. The diagonal of the floor of a rectangular closet is 7½ feet. The shorter side of the closet is 4½ feet. What is the area of the closet in square feet?
Options- A.
5 1 4 - B.
13 1 2 - C. 27
- D. 37 Discuss
Correct Answer: 27
Explanation:
| Other side | = |
|
ft | ||||||||||||
| = |
|
ft | |||||||||||||
| = |
|
ft | |||||||||||||
| = | 6 ft. |
∴ Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.
- 5. 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. 34
- B. 40
- C. 68
- D. 88 Discuss
Correct Answer: 88
Explanation:
So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
- 6. An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:
Options- A. 2%
- B. 2.02%
- C. 4%
- D. 4.04% Discuss
Correct Answer: 4.04%
Explanation:
∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
| ∴ Percentage error = | ❨ | 404 | x 100 | ❩% | = 4.04% |
| 100 x 100 |
- 7. A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?
Options- A. 20
- B. 24
- C. 30
- D. 33 Discuss
Correct Answer: 30
Explanation:
Then, AB + BC = 2x metres.
AC = √2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
| Saving % = | ❨ | 0.59x | x 100 | ❩% | = 30% (approx.) |
| 2x |
- 8. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
Options- A. 15360
- B. 153600
- C. 30720
- D. 307200 Discuss
Correct Answer: 153600
Explanation:
| Perimeter = Distance covered in 8 min. = | ❨ | 12000 | x 8 | ❩m = 1600 m. |
| 60 |
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 x 320) m2 = 153600 m2.
- 9. The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:
Options- A. 9 cm
- B. 18 cm
- C. 20 cm
- D. 41 cm Discuss
Correct Answer: 18 cm
Explanation:
Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81
⟹ (l + b) = 9.
∴ Perimeter = 2(l + b) = 18 cm.
- 10. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
Options- A. 2.91 m
- B. 3 m
- C. 5.82 m
- D. None of these Discuss
Correct Answer: 3 m
Explanation:
Area of the lawn = 2109 m2.
∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
⟹ x2 - 100x + 291 = 0
⟹ (x - 97)(x - 3) = 0
⟹ x = 3.
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