51 = 4a + 3b
5 > |–a|
If a and b are both integers, What is the highest values of ‘a’ that can be used to solve the above equation.	Choose the correct choice from the following answers
A) 1
B) 2
C) 3
D) -3
E) 0
Let us understand the meaning of 5 &gt; |-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. |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' - as per this scenario.....hence the value of 'a' must be between these two extremes. |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| &lt; 5 ==&gt; |a| &lt; 5 ===&gt; |a-0| &lt; 5 ==&gt; the distance from zero is less than 5. So the point will be -4,-3,-2,-1,0,1,2,3,4 as distance from 0 to these values is less than 5.   Now, lets move to equation 4a + 3b = 51 ==&gt; b = 17 – (4 *(a/3)). According to question, b is an integer, hence to make b integer a must be divisible by 3. Now remove the value which can’t make a divisible by 3 from the possible values of a. It will remain with -3 and 3. Out of these two values 3 is higher and it which makes the value of b = (17 - (4 * (3/3)) = 17 - 4 ==&gt; b = 13 (an integer). Hence the Answer is 3 (value 3).
<result>C</result>
51 = 4a + 3b
5 > |–a|
If a and b are both integers, What is the highest values of ‘a’ that can be used to solve the above equation.