40 = a + 3b
5 = |–a|
If a and b are both integers, What is the only possible value that b can take? Options:
A) 14
B) 15
C) 13
D) 12
E) 10
First understand the meaning of 5 = |-a|   1) when we write |-x| this means the same as |x|.   2) Mod is very easy concept if you solve mod question by considering as a distance. when a mod is written as |x-(a)| = b, this means the distance from point 'a' (both side left and right of 'a' on number line) is b (This is same as given case). |x-(a)| &lt; b means the distance is between the two extreme distance(left and right side of 'a' on number line, considering the max distance is 'b' from 'a’. |x-(a)| &gt; b means the distance is greater than the distance of 'b'..i.e the value of a could be anywhere more than 'b'.   Now come to the question. First its given|-a| = 5 ==&gt; |a| = 5 ===&gt; |a-0| = 5 ==&gt; the distance from zero is equal to 5. So the point will be -5 or 5, as distance from 0 to (-5) is 5 and distance from 0 to 5 is 5. Thus, the value of a will be either 5 or -5.   Now, lets move to equation a + 3b = 40 ==&gt; b = (40 - a)/3. According to question, b is an integer, hence to make b integer (40 - a) must be divisible by 3. Now remove the value which can’t make (40-a) divisible by 3 from the possible values of a (i.e. 5, -5). It will remain with -5 which makes the value of b = (40 - (-5))/3 = 45/3 ==&gt; b = 15...hence answer is 15.
<result>B</result>
40 = a + 3b
5 = |–a|
If a and b are both integers, What is the only possible value that b can take?