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Least possible value

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by saidov.mikhail » Mon Jul 01, 2013 2:40 am
If y is an integer, then the least possible value of |23-5y| is

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Please help me with this one as I don't understand explanation from OG. (It's a question #50 from OG 12th Edition).

As I understood "least possible value" means the value that is the least. How to move further stepwise?

thanks,
misha
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by GMATGuruNY » Mon Jul 01, 2013 3:20 am
saidov.mikhail wrote:If y is an integer, then the least possible value of |23-5y| is

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
|a-b| = the DISTANCE between a and b.
Thus, |23-5y| = the distance between 23 and 5y.
To minimize this distance, the value of 5y must be AS CLOSE AS POSSIBLE to 23.
Options:
If y=4, then 5y = 20.
If y=5, then 5y = 25.
If y=6, then 5y = 30.
The LEAST possible distance between 23 and 5y will be yielded by the option in red.
If y=5, then |23-5y| = |23-25| = 2.

The correct answer is B.
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by freyesinsb » Mon Jul 01, 2013 8:00 am
Doesn't that make the answer E? If Y=5?
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by GMATGuruNY » Mon Jul 01, 2013 2:42 pm
freyesinsb wrote:Doesn't that make the answer E? If Y=5?
The question stem asks not for the value of y but for the least possible value of the following expression:
|23-5y|.
When y=5, we get the least possible value of |23-5y|:
|23-5y| = |23-5*5| = |23-25| = |-2| = 2.
Thus, the least possible value of |23-5y| is 2 (answer choice B).
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by Scott@TargetTestPrep » Wed May 20, 2015 4:51 am
saidov.mikhail wrote:If y is an integer, then the least possible value of |23-5y| is

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Please help me with this one as I don't understand explanation from OG. (It's a question #50 from OG 12th Edition).

As I understood "least possible value" means the value that is the least. How to move further stepwise?

thanks,
misha
Solution:

To solve this question, we must make sure we interpret it correctly. We are not finding the least possible value of y, but rather the least possible value of |23-5y| (the absolute value of 23 - 5y). Remember that the smallest value that can result from taking the absolute value is zero. Thus we need to make 23 - 5y as close to zero as possible.

We know that 5y is a multiple of 5, so let's first look at the multiples of 5 closest to 23. We have "20" and "25". Let's subtract both of these from 23 and see which one produces the smallest result. When 5y = 20, y is 4 and when 5y = 25, y is 5. Let's start with letting y = 4.

|23-5(4)|

|23-20|

|3| = 3

Next, let's let y equal 5.

|23-5(5)|

|23-25|

|-2| = 2

We see that the smallest possible value of |23-5y| is 2.

The answer is B

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