Question

a,b,c are prime numbers, $x$ is an even number, $y$ is an odd number. Which of the following are is /are never true?
$\left(\text{I}\right)\text{ a + x = b}$
$\left(\text{II}\right)\text{ b + y = c}$
$\left(\text{III}\right)\text{ ab = c}$
$\left(\text{IV}\right)\text{ a + b = c}$
$\left(\text{A}\right)\text{ I and II}$
$\left(\text{B}\right)\text{ II and III}$
$\left(\text{C}\right)\text{ only III}$
$\left(\text{D}\right)\text{ III and IV}$

Answer 1

Hint: We have to check all the conditions from I to IV whether they are true or not by using the given data. We take some examples of the numbers and check the validation for each equation to get the required answer.

Complete step by step answer:
It is given that $a,b,c$ are prime numbers, $x$ is an even number, and $y$ is an odd number.
We have to solve that one by one,
The first condition $a+x=b$ is true.
We can prove it by giving an example like $3+4=7$ , where $a=3$, is a prime number and $x=4$ is an even number and $b=7$ is also a prime number.
The second condition $b+y=c$ is true.
It can also be proved by an example like $2+1=3$, where $b=2$ is a prime number and $y=1$ is an odd number and $c=3$ is also a prime number.
The third condition $ab=c$ is not true.
This can be proved by an example like $5\times 3=15$, where $a=5$ and $b=3$ both are prime numbers but $c=15$ is not a prime number.
The fourth condition is $a+b=c$ also true.
It is also proved by an example like $2+3=5$ where $2,3,5$ are prime numbers.
Hence all the parts are satisfied with the given data.
Finally, we get the third condition only never true.

Thus the correct option is C.

Note:
Prime numbers are those numbers that are only divisible by itself or $1$. For example $2,3,$ etc.
Odd numbers are those numbers that can never be divided by $2$. For example $9,15$ etc.
Even numbers are those numbers which are always divisible by $2$and have no remainder. For example $6,8$ etc.
It is to be kept in mind that when a prime number is multiplied by a prime number, it can never be a prime number.