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If a + b = 20, then (a + b/2) + (b + a/2) =

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Director
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If a + b = 20, then (a + b/2) + (b + a/2) = [#permalink] New post 06 Feb 2018, 21:05
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If a + b = 20, then (a + b/2) + (b + a/2) =

A) 10
B) 20
C) 30
D) 40
E) 50
[Reveal] Spoiler: OA
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Manager
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Re: If a + b = 20, then (a + b/2) + (b + a/2) = [#permalink] New post 06 Feb 2018, 22:16
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Notice that if you just add the terms you get a simple question:

(a + b/2) + (b + a/2) = 1.5a + 1.5b = ?

Then you can factor:

1.5a + 1.5b = 1.5(a + b)

and since we know a + b = 20, then 1.5(a + b) = 30
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Re: If a + b = 20, then (a + b/2) + (b + a/2) = [#permalink] New post 17 Feb 2018, 11:38
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Expert's post
amorphous wrote:
If a + b = 20, then (a + b/2) + (b + a/2) =

A) 10
B) 20
C) 30
D) 40
E) 50


Here's another approach:

GOAL: find the value of (a + b/2) + (b + a/2)

GIVEN: a + b = 20
Let's take some values of a and b that satisfy the equation a+b = 20, and plug them into our goal expression...

Well, a = 20 and b = 0 satisfy the equation a + b = 20

Now take (a + b/2) + (b + a/2), and replace a and b with 20 and b = 0
We get: (a + b/2) + (b + a/2) = (20 + 0/2) + (0 + 20/2)
= (20 + 0) + (0 + 10)
= 20 + 10
= 30

Answer: C

Cheers,
Brent
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Re: If a + b = 20, then (a + b/2) + (b + a/2) = [#permalink] New post 17 Feb 2018, 16:22
SherpaPrep wrote:
Notice that if you just add the terms you get a simple question:

(a + b/2) + (b + a/2) = 1.5a + 1.5b = ?

Then you can factor:

1.5a + 1.5b = 1.5(a + b)

and since we know a + b = 20, then 1.5(a + b) = 30


I just used random numbers to plug in and solve, not sure if it would work in any case however.
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Re: If a + b = 20, then (a + b/2) + (b + a/2) = [#permalink] New post 05 Mar 2018, 08:26
the most confident approach is the algebraic one -- the property does not change if you change the sides of the properties because all in all you are adding. So, a+b=20, 20+ a/2+b/2. Then the LCM of the numbers is 2 giving (40+a+b)/2=30
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Re: If a + b = 20, then (a + b/2) + (b + a/2) =   [#permalink] 05 Mar 2018, 08:26
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