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

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Math Expert
Joined: 02 Sep 2009
Posts: 53066
Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat  [#permalink]

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23 Dec 2014, 08:49
00:00

Difficulty:

5% (low)

Question Stats:

88% (01:36) correct 12% (01:50) wrong based on 116 sessions

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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

Kudos for a correct solution.

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Posts: 75
Location: India
GMAT 1: 690 Q49 V34
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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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23 Dec 2014, 09:18
1
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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Manager
Joined: 21 Aug 2010
Posts: 174
Location: United States
GMAT 1: 700 Q49 V35
Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat  [#permalink]

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23 Dec 2014, 20:45
1
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]

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23 Dec 2014, 23:43
2
Quote:
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

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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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24 Dec 2014, 01:19
2

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]

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25 Dec 2014, 03:56
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.
Math Expert
Joined: 02 Sep 2009
Posts: 53066
Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat  [#permalink]

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08 Jan 2015, 06:56
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.

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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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28 Mar 2018, 09:30
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.)

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Re: Operation # is defined as: a # b = 4a^2 + 4b^2 + 8ab for all non-negat   [#permalink] 28 Mar 2018, 09:30
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