consider the following series of arithmetic expressions. int a = 11; int b = 4; int c = 15…

consider the following series of arithmetic expressions. int a = 11; int b = 4; int c = 15; system.out.println(3 * a - 2 * b); system.out.println(2 * c - a % 11); what is the output of this program? 25 2 25 32 124 32
Answer
Explanation:
Step1: Calculate first expression
$3\times a - 2\times b=3\times11 - 2\times4=33 - 8 = 25$
Step2: Calculate second expression
Here, assuming % is the modulus operator. $2\times c - a%11=2\times15-11%11 = 30 - 0=30$. But it seems there might be a mis - typing in the code as the second expression in the code has an extra * symbol. If we assume the correct expression is $2\times c - a%11$, and if we consider the code as it is written and assume the * is a mistake and we just calculate $2\times c - a%11$ with $a = 11$, $c = 15$. If we assume the code is correct as written and we calculate step - by - step with the wrong syntax considered as a non - existent operation and just do arithmetic with the numbers in the right order for the second println as $2\times c - a$ (ignoring the %11 part which seems out of place with the wrong * there), we have $2\times15-11 = 30 - 11=19$. But if we assume the correct operation is $2\times c - a%11$, we get $2\times15-11%11=30 - 0 = 30$. Let's assume the correct operation is $2\times c - a%11$ and get the result as $30$. However, if we assume the code has a simple arithmetic operation $2\times c - a$ (ignoring the wrong % usage), we have $2\times15-11 = 29$. If we assume the correct operation is as it should be in Java with the modulus and correct order:
$2\times c - a%11=2\times15-11%11 = 30-0 = 30$. The most likely correct way in Java context for the second operation $2\times c - a%11$ gives $30$. But if we consider the code has a wrong symbol and we just do basic arithmetic $2\times c - a$ we get $2\times15 - 11=19$. Let's assume the correct Java - like operation for modulus and calculate $2\times c - a%11$.
$2\times15-11%11=30 - 0=30$.
Answer:
25 30