Hi,
Pls try to solve the below problem
A farmer has a field that measures 1000 ft wide by 2000 ft long. There is an untillable strip 20 ft wide on the inside edge of the field. In addition, there is a 30 ft wide untillable strip bisects the field into two squares (approximate). Approximately what percentage of the field is tillable?
A) 98%
B) 93%
C) 91%
D) 90%
E) 88%
OA to follow.
Good One to Solve-Try it !
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Area tillable after 20 fit strip = (1000-2*20)*(2000-2*20) = 960*1960imhimanshu wrote: farmer has a field that measures 1000 ft wide by 2000 ft long. There is an untillable strip 20 ft wide on the inside edge of the field. In addition, there is a 30 ft wide untillable strip bisects the field into two squares (approximate). Approximately what percentage of the field is tillable?
A) 98%
B) 93%
C) 91%
D) 90%
E) 88%
OA to follow.
After the 30 fit wide strip the area is divided in two parts means it can be in following ways:
1000*(2000-30)
so the two blocks will be 1000*985 each
Out of these 20 feet will go out from each area,
i.e. (1000-40)*(985-40)
So total tillable area = 2*960*945
% = 2*960*945*100/1000*2000 = 96*945/1000 = 96*0.945
= 91%
Charged up again to beat the beast
Is it B?
Consider the farm to be,
A..B
|..|
|..|
C..D
My calculation: Total Area: 2000*1000 = 20, 00, 000.
The length: 2000 and Width 1000 (Given).
With 20ft non-tileable area in the border, the area it takes : 120000 ft. (Approx.) How?
Consider 20ft * (1000+1000) ft [AB and CD].
Similarly, 20ft * (2000+2000) ft [AC and BD].
With 30ft non-tileable area partitioning the farm, the area it takes: 30000 ft. (Approx.). Because, 30ft along a width of 1000ft.
Total non-tileable area = 15000 ft.
Percentage: 150000/20,00,000 = 15/200 = 7.5% (Approx)
93% B.
Consider the farm to be,
A..B
|..|
|..|
C..D
My calculation: Total Area: 2000*1000 = 20, 00, 000.
The length: 2000 and Width 1000 (Given).
With 20ft non-tileable area in the border, the area it takes : 120000 ft. (Approx.) How?
Consider 20ft * (1000+1000) ft [AB and CD].
Similarly, 20ft * (2000+2000) ft [AC and BD].
With 30ft non-tileable area partitioning the farm, the area it takes: 30000 ft. (Approx.). Because, 30ft along a width of 1000ft.
Total non-tileable area = 15000 ft.
Percentage: 150000/20,00,000 = 15/200 = 7.5% (Approx)
93% B.
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IMO B.imhimanshu wrote:Hi,
Pls try to solve the below problem
A farmer has a field that measures 1000 ft wide by 2000 ft long. There is an untillable strip 20 ft wide on the inside edge of the field. In addition, there is a 30 ft wide untillable strip bisects the field into two squares (approximate). Approximately what percentage of the field is tillable?
A) 98%
B) 93%
C) 91%
D) 90%
E) 88%
OA to follow.
real2008, the calculated answer is 92.64, which is approximately 93%
What we think, we become
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