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.