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Area - General Questions (2)

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  • Area - General Questions
9. 

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:

A. 10%
B. 10.08%
C. 20%
D. 28%

Answer: Option D

Explanation:

Let original length = x and original breadth = y.

Decrease in area
= xy - 1-sym-oparen-h1 80 x x 90 y 1-sym-cparen-h1
100 100
 
= 1-sym-oparen-h1 xy - 18 xy 1-sym-cparen-h1
25
 
= 7 xy.
25
 Decrease % = 1-sym-oparen-h1 7 xy x 1 x 100 1-sym-cparen-h1% = 28%.
25 xy
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10. 

The diagonal of the floor of a rectangular closet is 71-div-1by2feet. The shorter side of the closet is 41-div-1by2feet. What is the area of the closet in square feet?

A. 5 1/4
B. 13 1/2
C. 27
D. 37

Answer: Option C

Explanation:

Other side =√
1-sym-oparen-h1 15 1-sym-cparen-h1 2 - 1-sym-oparen-h1 9 1-sym-cparen-h1 2
2 2
ft
  =
√225 - 81
4 4
ft
  =
√144
4
ft
  = 6 ft.

 Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

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11. 

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?

A. 2.91 m
B. 3 m
C. 5.82 m
D. None of these

Answer: Option B

Explanation:

Area 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.

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12. 

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

A. 40%
B. 42%
C. 44%
D. 46%

Answer: Option C

Explanation:

Let original length = x metres and original breadth = y metres.

Original area = (xy) m2.

New length = 1-sym-oparen-h1 120 x 1-sym-cparen-h1m = 1-sym-oparen-h1 6 x 1-sym-cparen-h1m.
100 5
New breadth = 1-sym-oparen-h1 120 y 1-sym-cparen-h1m = 1-sym-oparen-h1 6 y 1-sym-cparen-h1m.
100 5
New Area = 1-sym-oparen-h1 6 x x 6 y 1-sym-cparen-h1m2 = 1-sym-oparen-h1 36 xy 1-sym-cparen-h1m2.
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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13. 

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?

A. 16 cm
B. 18 cm
C. 24 cm
D. Data inadequate
E. None of these

Answer: Option B

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.

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14. 

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:

A. 2%
B. 2.02%
C. 4%
D. 4.04%

Answer: Option D

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 = 1-sym-oparen-h1 404 x 100 1-sym-cparen-h1% = 4.04%
100 x 100
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15. 

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:

A. 15360
B. 153600
C. 30720
D. 307200

Answer: Option B

Explanation:

Perimeter = Distance covered in 8 min. = 1-sym-oparen-h1 12000 x 8 1-sym-cparen-h1m = 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.

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