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A farmer has a field that measures 1000 ft wide by 2000 ft l [#permalink]
28 May 2011, 02:07

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This post was BOOKMARKED

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Difficulty:

75% (hard)

Question Stats:

53% (03:28) correct
47% (03:13) wrong based on 87 sessions

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, and a 30 ft wide untillable strip bisects the field into two squares (approximate). Approximately what percentage of the field is tillable?

Re: 650 plus level question [#permalink]
28 May 2011, 03:42

2

This post received KUDOS

Expert's post

Actually you can solve the problem pretty fast by using following approach:

1. one shorter inside strip with width of 20 ft takes 20/2000 = 1% of field 2. There is 2 short strips, 2 long strips (twice as long as shorts ones) and one short but wider strip that equals 30/20 = 1.5 short strips. 3. Approximately we have 2 + 2*2 + 1.5 = 7.5 short strips --> ~ 7.5% or 92.5% 4. As we didn't take into account overlaps between strips it will be slightly higher than 92.5%.

Or you can use calculations but I think it will take more time:

Re: 650 plus level question [#permalink]
28 May 2011, 06:07

walker wrote:

Actually you can solve the problem pretty fast by using following approach:

1. one shorter inside strip with width of 20 ft takes 20/2000 = 1% of field 2. There is 2 short strips, 2 long strips (twice as long as shorts ones) and one short but wider strip that equals 30/20 = 1.5 short strips. 3. Approximately we have 2 + 2*2 + 1.5 = 7.5 short strips --> ~ 7.5% or 92.5% 4. As we didn't take into account overlaps between strips it will be slightly higher than 92.5%.

Or you can use calculations but I think it will take more time:

Re: 650 plus level question [#permalink]
29 May 2011, 01:42

3

This post received KUDOS

ruturaj wrote:

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, and 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%

Total Area = 1000*2000 Tillable Square's side horizontally = (2000-20-30-20)/2 = 1930/2 = 965 Tillable Square's side vertically = (1000-20-20) = 960 = 960

Consider it as 960: % = \frac{2*960*960}{1000*2000}*100=\frac{2*0.96*0.96*1}{2}*100=(0.96)^2*100=92.16 \approx 93%

Why approximated to 93 and not 91 because we shortened one side from 965 to 960. Thus, in reality the squares are bigger.

Ans: "B"

By the way, I looked up tillable after solving.

tillable: arable, cultivable, cultivatable

Attachment:

tillable_field.PNG [ 5.2 KiB | Viewed 3136 times ]

Re: 650 plus level question [#permalink]
31 May 2011, 11:00

1

This post received KUDOS

fluke wrote:

ruturaj wrote:

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, and 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%

Total Area = 1000*2000 Tillable Square's side horizontally = (2000-20-30-20)/2 = 1930/2 = 965 Tillable Square's side vertically = (1000-20-20) = 960 = 960

Consider it as 960: % = \frac{2*960*960}{1000*2000}*100=\frac{2*0.96*0.96*1}{2}*100=(0.96)^2*100=92.16 \approx 93%

Why approximated to 93 and not 91 because we shortened one side from 965 to 960. Thus, in reality the squares are bigger.

Ans: "B"

By the way, I looked up tillable after solving.

tillable: arable, cultivable, cultivatable

Attachment:

tillable_field.PNG

this was quite smart fluke

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Re: 650 plus level question [#permalink]
02 Jun 2011, 12:28

Guys, seriously.. If I would get this question on the actual exam I would seriously start crying or something.. Pfff it took me 30minutes to friggin understand what the question is about... Sigh..

Re: 650 plus level question [#permalink]
13 Jun 2011, 13:52

i just summed the differences:

20*1000+20*1980+20*980+20*1960 will be the frame. the bisector will be 30* 960 (which can be considered as 20*960 for approximation, remembering that the rounding can cost only 1/2 %)

20*(1980+1960+1000+980+2*960)/2000*1000=the rest is simple and got 7%(+-)which must be substracted from 100

Re: 650 plus level question [#permalink]
21 Jun 2011, 14:41

Shalom! I have a question. If I can expect to see this type of question in the 600 to 700 range then how do I prepare to calculate the answer without the use of a calculator?

Re: 650 plus level question [#permalink]
09 Jun 2014, 08:15

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