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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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23 Feb 2017, 00:04
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67% (01:18) correct 33% (01:19) wrong based on 189 sessions

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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
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4667
Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5?  [#permalink]

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23 Feb 2017, 12:06
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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24 Feb 2017, 07:59
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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24 Feb 2017, 13:31
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.

mikemcgarry
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4667
Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5?  [#permalink]

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24 Feb 2017, 15:33
1
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.

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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22 Aug 2017, 10:18
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Clearly Ans is C
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Re: If a and b are integers and a = lb+2l + l3-bl, does a = 5?  [#permalink]

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19 May 2018, 11:42
We need to find if a=5.

1) b<3. Trying for all possible values of b, b=2, a=5; b=1,a=5; b=0,a=5; b=-1,a=5, b=-2,a=5, b=-3,a=7. Not Sufficient.

2) b>-2/ Trying for possible values of b, b=-1,a=5; b=0,a=5; b=5,a=9. Not Sufficient.

Combining a and b, we get -2<b<3 and a=5 for all possible values of b according to the combined limits. Hence, sufficient.
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If a and b are integers and a = lb+2l + l3-bl, does a = 5?  [#permalink]

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19 May 2018, 12:52
dyro1218 wrote:
If a and b are integers and a = lb+2l + l3-bl, does a = 5?

(1) b < 3
(2) b > -2

The answer Says "C" But statement (2) is also works as well by itself.

both are works too, However, C = represent "Both statements TOGETHER are sufficient, but Neither statement ALONE is sufficient. which tells you the Statement (2) isn't correct for a lone.

I don't know is it C is correct or B is correct?

Hey dyro1218

The reason statement 2 is not sufficient by itself is that we are given b > -2.

Case 1: If b=3, then a = l3+2l + l3-3l = 5 + 0 = 5
Case 2: If b=6, then a = l6+2l + l3-6l = 8 + 3 = 11

Since we can have both possibilities, statement 2 by itself is not enough.

Hope this helps you!
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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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19 May 2018, 20:02
Hi

|x-y| = Distance of x from y .
So, |b + 2| = distance of b from -2
and |3-b| = |b -3|= Distance of b from 3

a = |b + 2| + |3 – b| , means a is the sum of distance of b from -2 and 3

Now see the attached sketch, a = 5 only if b is between -2 & 3.
Attachment:

gmatbusters.jpeg [ 43.83 KiB | Viewed 365 times ]

St1) B<3. now b can be between -2 & 3 or b can be < -2
Not sufficient.
St 2) B>-2, b can be between -2 & 3 or b can be >3
Not Sufficient.

Combining both St1 & St2 , we get b is between -2 & 3. So a = 5.

SUFFICIENT

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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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20 May 2018, 02:47
Here is another explanation for the question
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WhatsApp Image 2018-05-20 at 15.15.27.jpeg [ 58.35 KiB | Viewed 312 times ]

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Re: If a and b are integers and a = |b + 2| + |3 – b|, does a = 5? &nbs [#permalink] 20 May 2018, 02:47
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