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If a and b are integers and a = |b + 2| + |3 – b|, does a = 5?

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If a and b are integers and a = |b + 2| + |3 – b|, does a = 5? [#permalink]

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If a and b are integers and a = |b + 2| + |3 – b|, does a = 5?

(1) b < 3
(2) b > –2
[Reveal] Spoiler: OA
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Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5? [#permalink]

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New post 23 Feb 2017, 11:06
vikasp99 wrote:
If a and b are integers and a = |b + 2| + |3 – b|, does a = 5?

(1) b < 3
(2) b > –2

Dear vikasp99,

I'm happy to respond. :-) Wow, Kaplan wrote a wonderful problem here! This is great.

OK, before we dive into the DS, let's unpack the prompt a bit.

The nature of how an absolute value behaves changes when the argument (the expression inside the absolute value) changes to zero. As a general rule,
When u ≥ 0, |u| = u
When u < 0, |u| = -u

Thus,
When b ≥ -2, |b + 2| = b + 2
When b < -2, |b + 2| = -(b + 2) = -b - 2
When b ≤ 3, |3 – b| = 3 - b
When b > 3, |3 – b| = -(3 - b) = b - 3

We have three different regions of the number line to consider.
When b > 3, then
|b + 2| + |3 – b| = (b + 2) + (b - 3) = 2b - 1
When -2 ≤ b ≤ 3,
|b + 2| + |3 – b| = (b + 2) + (3 - b) = 5
When b < -2,
|b + 2| + |3 – b| = ( -b - 2) + (3 - b) = 1 - 2b

Thus, if b has a value anywhere in the region -2 ≤ b ≤ 3, the expression equals 5.

Thus, the individual statements are not sufficient, but combined, we have enough information to give a definitive "yes" to the prompt question. Together, the statements are sufficient.

OA = (C)

Does all this make sense?
Mike :-)
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Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5? [#permalink]

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New post 24 Feb 2017, 06:59
vikasp99 wrote:
If a and b are integers and a = |b + 2| + |3 – b|, does a = 5?

(1) b < 3
(2) b > –2



take 3 ranges
(1) b<-2 -----> -b-2+ 3-b =-2b+1---> value depends on b....
(2) -2<= b < 3------> b+2+3-b = 5
(3) b>=3--------> b+2-3+b = 2b-1----> value depends on b....

only satisfying range is -2<= b < 3

from options , we require both options

Ans C
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If a and b are integers and a = |b + 2| + |3 – b|, does a = 5? [#permalink]

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New post 24 Feb 2017, 12:31
I dont understand how you explain this without having statement numbers 1) and 2)..

Please be kind enough to explain me the explicit way to solve this.


I need your help also
mikemcgarry
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Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5? [#permalink]

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New post 24 Feb 2017, 14:33
nishanfaith wrote:
I dont understand how you explain this without having statement numbers 1) and 2)..

Please be kind enough to explain me the explicit way to solve this.

I need your help also
mikemcgarry

Dear nishanfaith,

I'm happy to respond. :-) First of all, my friend, I would urge you to read through my algebraic solution step-by-step, because if you understand that, then you really will understand this problem deeply. If you understand what I did there, you will understand what I concluded about the statements individually.

I will also show the basics of a picking numbers approach. Simply notice that at b = +3, the insides of one of the absolute values is zero, and at b = -2, the insides of the other absolute value is zero. Points that make the insides equal to zero tend to be "behavior change" points for absolute values.

Statement #1: b < 3
Try b = 0
a = |b + 2| + |3 – b| = |2| + |3| = 2 + 3 = 5
This choice produces a "yes" to the prompt question.
Try b = -10 (a point on the other side of the "behavior change" point at x = -2)
a = |b + 2| + |3 – b| = |-10 + 2| + |3 - (-10)| = |-8| + |13| = 8 + 21 = 21
This choice produces a "no" to the prompt question.
As soon as we have two different choices producing two different answers, we know that the statement is not sufficient.

Statement #2: b > –2
Again, use b = 0.
We know from the previous case that this makes a = 5.
This choice produces a "yes" to the prompt question.
Try b = +10 (a point on the other side of the "behavior change" point at x = +3)
a = |b + 2| + |3 – b| = |10 + 2| + |3 - 10| =|12| + |-7| = 12 + 7 = 19
This choice produces a "no" to the prompt question.
Again, we have two different choices producing two different answers, so we know that the statement is not sufficient.

The individual statements are insufficient, so we have to combine the statements.
Combined statements: -2 < b < 3
We already know that b = 0 means a = 5, which produces a "yes" to the prompt question.
Try b = +2
a = |b + 2| + |3 – b| = |2 + 2| + |3 – 2| = |4| + |1| = 4 + 1 = 5
That also produces a "yes" answer.
Try b = -1
a = |b + 2| + |3 – b| = |-1 + 2| + |3 – (-1)| = |1| + |4| = 1 + 4 = 5
That one also produces a "yes" answer.

OK, here's the issue. Picking numbers is an excellent way to demonstrate something is not sufficient, because if we can produce two different answers to the prompt question, we know it's not sufficient. Picking numbers categorically cannot demonstrate that something is sufficient. You see, there are an infinite number of fractions and decimals between -2 and +3, and of course we can't try every single case. We have shown that it produces the same answer for three cases, and that's certainly suggestive, but we can't be certain that there's not some decimal value somewhere in there that would produce a different answer. At this point, we would have to use either algebra (as I did above) or logic. I suggest that this is the best place to go back and look at the algebra above for what happens on this interval.

Does all this make sense?
Mike :-)
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Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5? [#permalink]

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Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5?   [#permalink] 22 Aug 2017, 09:18
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