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A farmer has a rectangular field that measures 1000 ft wide

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u2lover
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A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   14 Jul 2006, 15:07
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A farmer has a rectangular field that measures 1000 ft wide by 2000 ft long. There is an untillable strip 20 ft wide on the inside edges 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%

math gurus, would you please find the fastest way to solve this... took me 5 min :shock:



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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   Updated on: 26 Jul 2006, 14:00
got D. 90%

also seek for the quickest way to solve it.

totoal = 2,000,000
tillable = 30x1000 + 2 (1000x20) + 4 (965x20) = 147,200
so % = 1- (147,200/2,000,000) = approx 93%

Originally posted by MA on 14 Jul 2006, 19:52.
Last edited by MA on 26 Jul 2006, 14:00, edited 1 time in total.
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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   22 Jul 2006, 20:02
OA is B... took me 5:19 to solve it :shock:
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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   22 Jul 2006, 23:16
(1960*960-960*30)/(1000*2000)
= [960 * 1930/(1000*2000)]* 100
= 48 * 193/100 = 9264/100 = 92.64
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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   22 Jul 2006, 23:33
2*((1960-30)/2*960))/(2000*1000) = .9264
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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   23 Jul 2006, 00:59
Go for untillible
all width of untillible 20+30+20 = 70

= [ (70*1000 + 1930*20*2) / 1000*2000 ] * 100
= (70*5+193*2)/10*10
= 736/100 = 7.36%

Approximately 7%
100-7 = 93%
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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   23 Jul 2006, 01:12
They key is to approximate the untillable area. Ignore the double counting because percentage-wise, it is negligible.

Perimeter of field*20 + width*30
6000*20+1000*30*=150,000

So tillable area is 150,000/2,000,000= 7.5% (slightly overstated)

Tillable area is thus very close to 92.5% (slightly understated)

Go for 93%. This can take less than one minute!
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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   26 Jul 2006, 04:10
B 93%

Area of land = 2000 *1000
Length of 1st square = 965
width = 960

Area of tillable land = 2* 960*965

Percentage = 2*960*965*100/2000*1000 ~ 93%
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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink] New post   27 Jul 2006, 23:33
Area of center strip = 30*1000 = 30,000
Area of border strips = Perimeter of field * 20 - 4(20^2) = 120000 - 400 or approx 120000

Total untillable = 120000 + 30000 = 150000

% untillable = 150000/(2000*1000) = 7.5%
% Tillable = 93 % (Approx)

Dint take me more than 2 min.



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Re: A farmer has a rectangular field that measures 1000 ft wide [#permalink]
27 Jul 2006, 23:33
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