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- Question
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
- Correct Answer
- 3 m
ExplanationArea of the park = (60 x 40) m2 = 2400 m2.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.
Area problems
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- 1. 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:
√l2 + b2 = √41.Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81
⟹ (l + b) = 9.
∴ Perimeter = 2(l + b) = 18 cm.
- 2. 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.
- 3. 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:
Let the side of the square(ABCD) be x metres.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 - 4. 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:
100 cm is read as 102 cm.∴ 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 - 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:
We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
- 6. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
Options- A. 814
- B. 820
- C. 840
- D. 844 Discuss
Correct Answer: 814
Explanation:
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.Area of each tile = (41 x 41) cm2.
∴ Required number of tiles = ❨ 1517 x 902 ❩ = 814. 41 x 41 - 7. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?
Options- A. 16 cm
- B. 18 cm
- C. 24 cm
- D. Data inadequate
- E. None of these Discuss
Correct Answer: 18 cm
Explanation:
2(l + b) = 5 b 1 ⟹ 2l + 2b = 5b
⟹ 3b = 2l
b = 2 l 3 Then, Area = 216 cm2
⟹ l x b = 216
⟹l x 2 l = 216 3 ⟹ l2 = 324
⟹ l = 18 cm.
- 8. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
Options- A. 1520 m2
- B. 2420 m2
- C. 2480 m2
- D. 2520 m2 Discuss
Correct Answer: 2520 m2
Explanation:
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m2 = 2520 m2.
- 9. A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:
Options- A. Rs. 456
- B. Rs. 458
- C. Rs. 558
- D. Rs. 568 Discuss
Correct Answer: Rs. 558
Explanation:
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558. 100 - 10. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
Options- A. 40%
- B. 42%
- C. 44%
- D. 46% Discuss
Correct Answer: 44%
Explanation:
Let original length = x metres and original breadth = y metres.Original area = (xy) m2.
New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is
= (36/25)xy - xy
= xy(36/25 - 1)
= xy(11/25) or (11/25)xy
∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%. 25 xy
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