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# A lady grows cabbages in her garden that is in the shape of

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A lady grows cabbages in her garden that is in the shape of [#permalink]

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11 Sep 2008, 22:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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11 Sep 2008, 22:34
A:

I dont know how to solve this in a cool way, but I discovered a pattern pretty quick:

If the side of last year's garden is 2 and this year's garden is 3 the difference in square feet is 5 (3*3 - 2*2).
If the side of last year's garden is 3 and this year's garden is 4 the difference in square feet is 7 (4*4 - 3*3).
If the side of last year's garden is 4 and this year's garden is 5 the difference in square feet is 9 (5*5 - 4*4).

You get the idea. So if the difference is 211, then 211/2=105, thus:
105*105 = last year's garden
106*106 = this year's garden = 11236

I wonder what's the mathematical way to solve it.
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12 Sep 2008, 03:37
I arrived at A using the same basic method as asdert.
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12 Sep 2008, 03:53
asdert wrote:
A:

I dont know how to solve this in a cool way, but I discovered a pattern pretty quick:

If the side of last year's garden is 2 and this year's garden is 3 the difference in square feet is 5 (3*3 - 2*2).
If the side of last year's garden is 3 and this year's garden is 4 the difference in square feet is 7 (4*4 - 3*3).
If the side of last year's garden is 4 and this year's garden is 5 the difference in square feet is 9 (5*5 - 4*4).

You get the idea. So if the difference is 211, then 211/2=105, thus:
105*105 = last year's garden
106*106 = this year's garden = 11236

I wonder what's the mathematical way to solve it.

I did not get your reasoning here Can you please elaborate ? ?
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12 Sep 2008, 07:15
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amitdgr wrote:
I did not get your reasoning here Can you please elaborate ? ?

Amit,

Since I couldn't write a proper equation at the beginning, I figured there had to be a pattern (GMAT wouldn't give you a problem you can't solve in less than 2-3 minutes)

So I wondered, what would be the difference if the initial were 2 sq ft and this years were 3 sq ft. After that, I wondered about a 3sq ft and a 4 sq ft respectively.

Results follow:

If the side of last year's garden is 2 and this year's garden is 3 the difference in square feet is 5 (3*3 - 2*2).
If the side of last year's garden is 3 and this year's garden is 4 the difference in square feet is 7 (4*4 - 3*3).
If the side of last year's garden is 4 and this year's garden is 5 the difference in square feet is 9 (5*5 - 4*4).

So I noticed a pattern: The difference is always odd, and it increments by 2 every number I add.

So there must be a number that when squared and added 211 yields the initial number squared.

When I solved I noticed a different pattern immediately:
If you divide the difference of the two squares by 2, it will give you the average of the numbers that you initially multiplied.

Let's take sides 2 and 3 and their squares: 4 and 9. ==>
Therefore, after you divide 5 (the result from 9-4) by 2, the result is 2.5, which is the average of 2 and 3.

If you repeat the process, you'll notice that this holds true.

Therefore:
211/2 = 105.5. So the initial numbers that were multiplied were 105 and 106, which are the sizes of last year's and this year's garden.

After that, you just have to square 106 to get A.

However, now that I'm thinking about it, probably the fastest way to solve would have been to write an equation after I found the relationship where x equals this year's garden:
$$x^2 - 211 = (x-1)^2$$
$$x^2 - 211 = x^2-2x+1$$
$$2x = 212$$
$$x = 106$$

I hope this helps and that my alternative approach doesn't confuse you.
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12 Sep 2008, 08:27
asdert wrote:
amitdgr wrote:
I did not get your reasoning here Can you please elaborate ? ?

Amit,

Since I couldn't write a proper equation at the beginning, I figured there had to be a pattern (GMAT wouldn't give you a problem you can't solve in less than 2-3 minutes)

So I wondered, what would be the difference if the initial were 2 sq ft and this years were 3 sq ft. After that, I wondered about a 3sq ft and a 4 sq ft respectively.

Results follow:

If the side of last year's garden is 2 and this year's garden is 3 the difference in square feet is 5 (3*3 - 2*2).
If the side of last year's garden is 3 and this year's garden is 4 the difference in square feet is 7 (4*4 - 3*3).
If the side of last year's garden is 4 and this year's garden is 5 the difference in square feet is 9 (5*5 - 4*4).

So I noticed a pattern: The difference is always odd, and it increments by 2 every number I add.

So there must be a number that when squared and added 211 yields the initial number squared.

When I solved I noticed a different pattern immediately:
If you divide the difference of the two squares by 2, it will give you the average of the numbers that you initially multiplied.

Let's take sides 2 and 3 and their squares: 4 and 9. ==>
Therefore, after you divide 5 (the result from 9-4) by 2, the result is 2.5, which is the average of 2 and 3.

If you repeat the process, you'll notice that this holds true.

Therefore:
211/2 = 105.5. So the initial numbers that were multiplied were 105 and 106, which are the sizes of last year's and this year's garden.

After that, you just have to square 106 to get A.

However, now that I'm thinking about it, probably the fastest way to solve would have been to write an equation after I found the relationship where x equals this year's garden:
$$x^2 - 211 = (x-1)^2$$
$$x^2 - 211 = x^2-2x+1$$
$$2x = 212$$
$$x = 106$$

I hope this helps and that my alternative approach doesn't confuse you.

Thanks for the clear explanation It helped
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12 Sep 2008, 21:39
good logic amitdgr !!
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13 Sep 2008, 01:34
x2suresh wrote:
good logic amitdgr !!

you meant asdert
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Re: cabbages   [#permalink] 13 Sep 2008, 01:34
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