Solid Geometry

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Solid Geometry

by bharti.2010 » Thu Jan 05, 2012 7:54 am
A lady grows cabbages in her garden that is in the shape of a square. Each cabbage takes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year?

A. 11236
B. 11025
C. 14400
D. 12696
E. Cannot be determined

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by pemdas » Thu Jan 05, 2012 8:05 am
q,Q are outputs for diff. years q=r^2 and Q=R^2
r,R are diff. sides for two areas inside the bigger square (garden)
Q-q=211, find Q?

(R^2-r^2)=211

It's clear that the difference between two squares may mean that R and r may be assigned any values. The output Q will depend on the value of R, hence not clear.

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bharti.2010 wrote:A lady grows cabbages in her garden that is in the shape of a square. Each cabbage takes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year?

A. 11236
B. 11025
C. 14400
D. 12696
E. Cannot be determined
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by shankar.ashwin » Thu Jan 05, 2012 9:02 am
Imagine 2 concentric squares,

Given Area(bigger) - Area(smaller) = 211

We are asked Area(bigger) = ?

Looking at the answer choices, we know side(bigger) > 100

A is 106^2

now since units digit is 1, Area(smaller) should end in 5, consider 105

106^2 - 105^2 = 211 - Success

A IMO

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by GMATGuruNY » Thu Jan 05, 2012 10:11 am
bharti.2010 wrote:A lady grows cabbages in her garden that is in the shape of a square. Each cabbage takes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year?

A. 11236
B. 11025
C. 14400
D. 12696
E. Cannot be determined
Let T² = the area of the garden this year.
Let L² = the area of the garden last year.

Since each cabbage takes up 1 square foot, the area of the garden = the number of cabbages produced.
Since the number of cabbages produced increases by 211:
T² - L² = 211.
(T+L)(T-L) = 211.

211 is a prime number. It's only factors are 211 and 1.
Since the area = the number of cabbages, T and L -- the dimensions of the garden this year and last year -- are integers.
Thus, T+L = 211 and T-L = 1.

Adding the two equations:
(T+L) + (T-L) = 211+1.
2T = 212.
T = 106.

Thus, the area of the garden this year = 106² = 11236.

The correct answer is A.
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by santhoshsram » Thu Jan 05, 2012 2:00 pm
Forgive me if this turns out to be a silly question.

The question says the lady grows cabbages in a SQUARE garden and each cabbage takes 1 SQ FEET of area in the garden. The first question that came to my mind after reading this is, "Is the garden completely packed with end-to-end with cabbages?". If the answer is yes, then the number of cabbages = X*X (X=length of square). If the answer is no, for example if only 1/4th of the garden was filled with cabbages, we'd end up with choice E.

Is it ok to assume the garden is packed?
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by pemdas » Thu Jan 05, 2012 4:20 pm
bharti.2010 wrote:The shape of the area used for growing the cabbages has remained a square in both these years.
we are having squares for both years and not other shape quadrilaterals to answer your question about packing
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by santhoshsram » Thu Jan 05, 2012 5:10 pm
pemdas wrote:
bharti.2010 wrote:The shape of the area used for growing the cabbages has remained a square in both these years.
we are having squares for both years and not other shape quadrilaterals to answer your question about packing
I think, I didn't get my question through clearly last time. Let me try to do it with more clarity.

The question says the shape remained a square.

Case 1: Not fully packed.
=========================
Let us assume she grew 10 cabbages last year. This year the number of cabbages is 10 + 211 = 221. If the garden (SQUARE) originally had a side of say 100feet, its area would be have been 10,000 sqft. Enough to grow 10 cabbages last year AND enough to grow 221 cabbages this year. So in this case the garden was/is a square in both years, but it is the same square. And I just picked a conveniently large number 100, it can be 200, 300 or whatever.

See my point?

Unless we assume that both last year and this year the entire garden was packed with cabbages we won't be able solve it as mentioned. In other words, we are assuming there is no empty space in the garden. Is that ok? Because if we do not assume that, we can pick another answer choice.

I mean, just imagine a really really large square garden that is large enough to fit X cabbages as well as X+211 cabbages.
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by santhoshsram » Thu Jan 05, 2012 5:14 pm
If the choices did not have "Cannot be determined" then we could have possibly ruled out this situation.
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by GMATGuruNY » Thu Jan 05, 2012 5:31 pm
santhoshsram wrote:
pemdas wrote:
bharti.2010 wrote:The shape of the area used for growing the cabbages has remained a square in both these years.
we are having squares for both years and not other shape quadrilaterals to answer your question about packing
I think, I didn't get my question through clearly last time. Let me try to do it with more clarity.

The question says the shape remained a square.

Case 1: Not fully packed.
=========================
Let us assume she grew 10 cabbages last year. This year the number of cabbages is 10 + 211 = 221. If the garden (SQUARE) originally had a side of say 100feet, its area would be have been 10,000 sqft. Enough to grow 10 cabbages last year AND enough to grow 221 cabbages this year. So in this case the garden was/is a square in both years, but it is the same square. And I just picked a conveniently large number 100, it can be 200, 300 or whatever.

See my point?

Unless we assume that both last year and this year the entire garden was packed with cabbages we won't be able solve it as mentioned. In other words, we are assuming there is no empty space in the garden. Is that ok? Because if we do not assume that, we can pick another answer choice.

I mean, just imagine a really really large square garden that is large enough to fit X cabbages as well as X+211 cabbages.
You have a point. The intended meaning of each cabbage takes 1 square foot of area seems to be: The garden is divided into 1x1 squares, and each 1x1 square is occupied by exactly 1 cabbage. The GMAT would use wording less open to interpretation.
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by pemdas » Thu Jan 05, 2012 5:45 pm
santhoshsram wrote:And I just picked a conveniently large number 100, it can be 200, 300 or whatever.

See my point?

Unless we assume that both last year and this year the entire garden was packed with cabbages we won't be able solve it as mentioned. In other words, we are assuming there is no empty space in the garden. Is that ok? Because if we do not assume that, we can pick another answer choice.

I mean, just imagine a really really large square garden that is large enough to fit X cabbages as well as X+211 cabbages.
first of all, it's not as easy and simple as you imagine. Shankar also solved for answer and Mitch came with breaking factors of prime. I stopped at the eve and decided unclear.

There is one major constraint in this problem, and it's name is cabbage (one sq.foot area). Each cabbage is one, whole, unit and you could not count these as half cabbages or 1/3 cabbages. Therefore, even if theoretically you can have X+216 in the garden, practically you may have only Integer A^2- Integer B^2 = 216. This is resolved in expert's Mitch answer with great elegance. If you account for constraint cabbage is integer and then figure the way for T^2-L^2=216, you are done.

it's very precise and clear now, the answer is A.

looked up the last post by expert.
@Mitch, even if the question doesn't convey each cabbage occupies 1*1 sq.ft, it tells the dimensions of each cabbage, which is again whole unit (must be integer) and gathered together the cabbages form some area. This area will be construed by individual cabbages (say scattered within the square shaped garden) and aggregated will make some area. We are interested in area and it's composite sides. We are not measuring the areas of garden portions packed from end-to-end. We measure areas.
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by santhoshsram » Thu Jan 05, 2012 6:03 pm
pemdas wrote: There is one major constraint in this problem, and it's name is cabbage (one sq.foot area). Each cabbage is one, whole, unit and you could not count these as half cabbages or 1/3 cabbages. Therefore, even if theoretically you can have X+216 in the garden, practically you may have only Integer A^2- Integer B^2 = 216. This is resolved in expert's Mitch answer with great elegance. If you account for constraint cabbage is integer and then figure the way for T^2-L^2=216, you are done.

it's very precise and clear now, the answer is A.
I still don't think I put forth my idea clearly. As Mich pointed out two posts earlier, this question is open to interpretation. As Mich says, the garden is divided into 1x1 smaller squares. Each cabbage has a size of 1x1. The integer constraint would mean that each smaller square can have 1 and only 1 cabbage. But how about empty 1x1 squares with no cabbages? No? Why not?
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by pemdas » Thu Jan 05, 2012 6:07 pm
responding to pm:

the only way we cannot select answer A is when the aggregated area is not producing square
then 106^=11236 is also formed as 53*212. We may not take T^2-L^2=216 in such case. The original question says only area used is square and as you have doubt it could be packed not to the end of either side. We have rectangle then and may not take T^2 or L^2. Under such case, your argumentation is sensible and desire to pick E is understandable.
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