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Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat

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Bunuel
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Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   23 Dec 2014, 09:49
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Tough and Tricky questions: Word Problems.



Operation # is defined as: a # b = 4a^2 + 4b^2 + 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


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B
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   24 Dec 2014, 02:19
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Answer = B. 8

Given that \(4a^2 + 4b^2 + 8ab = 100\)

\(a^2 + 2ab + b^2 = 25\)

\((a+b)^2 = 25\)

a+b = 5

a+b+3 = 8
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   24 Dec 2014, 00:43
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Operation # is defined as: a # b = 4a^2 + 4b^2 + 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

1) 4a^2 + 4b^2 + 8ab -> (2a+2b)^2 = a # b

2) a # b = 100 = (2a+2b)^2
3) 10 = (2a+2b)
4) 5 = a+b this is the important info for our answer, because we dont need to know the exact values for a or b.
5) (a + b) + 3 = 5 + 3 = 8

The answer is 8, b!
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   23 Dec 2014, 10:18
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Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab
==> a # b = 4 * (a + b)^2
when a # b = 100
==> (a + b) = 5 (ignoring negative solution)

Thus (a + b) + 3 = 8, Answer B
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   23 Dec 2014, 21:45
Bunuel wrote:

Tough and Tricky questions: Word Problems.



Operation # is defined as: a # b = 4a^2 + 4b^2 + 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


Kudos for a correct solution.



4a^2 + 4b^2 + 8ab = 100 --->(1)
=>a^2 + b^2 + 2ab = 25 ---> Divide (1) by 4
=> (a + b) ^2 = 25
=> a+b = 5

=>(a + b) + 3 = 8

ANs B
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   25 Dec 2014, 04:56
a#b=100=4a*a +4b*b + 8ab
Dividing both sides by 4

a*a+ b*B+ 2ab=25 ==> (a+b)sq=25 ==> a+b=+-5. Taking a+b as -5 gives answer as 2 which is not an option. Taking a=B=5 gives the answer as 8.

hence option b.
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   08 Jan 2015, 07:56
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Bunuel wrote:

Tough and Tricky questions: Word Problems.



Operation # is defined as: a # b = 4a^2 + 4b^2 + 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


Kudos for a correct solution.


OFFICIAL SOLUTION:

(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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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   28 Mar 2018, 10:30
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Bunuel wrote:

Operation # is defined as: a # b = 4a^2 + 4b^2 + 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


a # b = 100

4a^2 + 4b^2 + 8ab = 100

a^2 + b^2 + 2ab = 25

(a + b)^2 = 25

a + b = 5

So (a + b) + 3 = 5 + 3 = 8.

(Note: a + b can’t be -5 since a # b is defined for non-negative integers only. If a + b = -5, then at least one of a and b has to be negative.)

Answer: B
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink] New post   05 Dec 2019, 09:16
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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat [#permalink]
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