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 Post subject: GMAT Symbols
PostPosted: Thu Aug 11, 2011 5:17 pm 
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Operation # is defined as: a # b = 4a² + 4b² + 8ab for all non-negative integers. What is the value of (a + b) + 3, when a # b = 100?
A. 5
B. 8
C. 10
D. 13
E. 17

(B) We know that a # b = 100 and a # b = 4a² + 4b² + 8ab. So
4a² + 4b² + 8ab = 100

We can see that 4a² + 4b² + 8ab is a well-known formula for (2a + 2b)². Therefore (2a + 2b)² = 100.
(2a + 2b) is non-negative number, since both a and b are non-negative numbers. So we can conclude that 2(a + b) = 10.

(a + b) + 3 = 10/2 + 3 = 8.

The correct answer is B.
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The only question I have about this problem is the transition from (2a + 2b)² = 100 to 2(a + b) = 10. Obviously the square root was taken but only of the (a + b)?


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 Post subject: Re: math (t.1, qt.5): algebra, sumbols
PostPosted: Thu Aug 11, 2011 5:21 pm 
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Joined: Sun May 30, 2010 2:23 am
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questioner wrote:
The only question I have about this problem is the transition from (2a + 2b)² = 100 to 2(a + b) = 10. Obviously the square root was taken but only of the (a + b)?
The square root was taken of the whole expression within the brackets, 2a + 2b.

So (2a + 2b)² = 100 yields 2a + 2b = 10. Then we just factor out 2: 2a + 2b = 2(a + b).


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 Post subject: Re: math (t.1, qt.5): algebra, sumbols
PostPosted: Thu Dec 08, 2011 2:46 pm 
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What happened to 8ab when you solved 4a² + 4b² + 8ab = 100 to (2a+ 2b)² = 100?
The 8ab was just dropped?


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 Post subject: Re: math (t.1, qt.5): algebra, sumbols
PostPosted: Thu Dec 08, 2011 2:50 pm 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
The 8ab was just dropped?
No, we've used the well-known formula x² + 2xy + y² = (x + y)².

In our case it is:
4a² + 8ab + 4b² = 100
(2a)² + 2 × (2a) × (2b) + (2b)² = 100
So we apply the formula for x = 2a and y = 2b.
(2a + 2b)² = 100


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