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# 650 plus level question

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Manager
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650 plus level question [#permalink]  28 May 2011, 02:07
00:00

Difficulty:

45% (medium)

Question Stats:

55% (03:23) correct 45% (02:24) wrong based on 40 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?

A) 98%

B) 93%

C) 91%

D) 90%

E) 88%
[Reveal] Spoiler: OA
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Re: 650 plus level question [#permalink]  28 May 2011, 03:42
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:

%=100%*\frac{2*(1000-2*20)*(1000-20-\frac{30}{2})}{1000*2000} = 0.9264
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Manager
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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:

%=100%*\frac{2*(1000-2*20)*(1000-20-\frac{30}{2})}{1000*2000} = 0.9264

Good method ...fast and quick
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Manager
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Re: 650 plus level question [#permalink]  28 May 2011, 06:58
can u please explain the same with the help of diagram..
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Re: 650 plus level question [#permalink]  28 May 2011, 07:51
2
KUDOS
Expert's post

[Reveal] Spoiler:
Attachment:

114310.png [ 3.71 KiB | Viewed 2865 times ]

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Re: 650 plus level question [#permalink]  28 May 2011, 10:42
1
KUDOS
960*965*2/(1000*2000) = 93.04%
B.
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Re: 650 plus level question [#permalink]  28 May 2011, 12:26
okai .. its C

rectangle of 1000*2000

1000-2*20 = 960
(2000-2*20-30)/2 = 965
2 squares = 2*965*960/(1000*2000) = 93% approx.
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Re: 650 plus level question [#permalink]  29 May 2011, 01:42
3
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 2656 times ]

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Re: 650 plus level question [#permalink]  31 May 2011, 11:00
1
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

kudos from me
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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..
Manager
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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

must admit it took me 2 minutes+ to solve
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Re: 650 plus level question [#permalink]  21 Jun 2011, 12:37
Shalom!
Please tell me that this type of question is not in the 600 to 700 level question range on the GMAT.
Manager
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Re: 650 plus level question [#permalink]  21 Jun 2011, 12:40
Lol, I was REALLY thinking the same thing! I'm afraid it is
Manager
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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?
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Re: 650 plus level question [#permalink]  22 Jun 2011, 04:42
very good explanation
Re: 650 plus level question   [#permalink] 22 Jun 2011, 04:42
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