Question

If \[2y\otimes y=3\,\text{mod 5}\] , then the value of y is,
(A) 2
(B) 0
(C) 4
(D) 1

Answer 1

Hint: We have the expression, \[2y\otimes y=3\,\text{mod 5}\] . Here, 3 is the remainder, 5 is the divisor and \[2y\otimes y\] is the dividend. We know the relation between dividend, divisor, quotient and remainder.
\[\text{Dividend=Divisior}\times \text{Quotient+Remainder}\] . Put 3 as remainder, 5 as divisor and \[2y\otimes y\] as dividend. Take quotient equal to 1 and solve it further.

Complete step-by-step answer:
According to the question, it is given that the expression is,
\[2y\otimes y=3\,\text{mod 5}\] …………………(1)
We have to get the value of y after solving this expression.
In the expression \[2y\otimes y=3\,\text{mod 5}\] , 3 is the remainder when \[2y\otimes y\] is divided by 5.
It means that 5 is the divisor and \[2y\otimes y\] is the dividend. So,
Divisor = 5 ………………………(2)
Remainder = 3 ………………………….(3)
Dividend = \[2y\otimes y\] ……………………..(4)
We know the relation between dividend, divisor, quotient and remainder.
\[\text{Dividend=Divisior}\times \text{Quotient+Remainder}\] ……………………………(5)
Now, using this relation and putting the value of dividend, divisor and remainder in equation (5), we get
\[2y\otimes y\text{=5}\times \text{Quotient+3}\]
\[\Rightarrow 2y\otimes y-3=5\times \text{Quotient}\] ……………………(6)
In equation (6), we can see that the remainder is 0. So, we can say that the expression \[2y\otimes y-3\] is divisible by 5.
Let the quotient be 1.
Now, putting quotient = 1 in equation (6), we get
\[\begin{align}
  & \Rightarrow 2y\otimes y-3=5\times \text{Quotient} \\
 & \Rightarrow 2y\times y-3=5\times 1 \\
\end{align}\]
Taking 3 to the RHS of the above equation, we get
\[\begin{align}
  & \Rightarrow 2{{y}^{2}}=3+5 \\
 & \Rightarrow 2{{y}^{2}}=8 \\
 & \Rightarrow {{y}^{2}}=4 \\
\end{align}\]
Now, applying square to both LHS and RHS of the above equation, we get
\[\begin{align}
  & \Rightarrow y=\sqrt{{{4}^{2}}} \\
 & \Rightarrow y=\pm 2 \\
\end{align}\]
So, the value of y can be +2 and -2.
Hence, the correct option is (A).


Note: In this question, one might get confused because we have taken quotient as 1 in the expression, \[2y\otimes y-3=5\times \text{Quotient}\] and in the question, the value of quotient is not given. So, here we have to consider the value of quotients. One can also think of taking quotients as -1. Id we do so, then
\[\begin{align}
  & \Rightarrow 2y\otimes y-3=5\times \text{Quotient} \\
 & \Rightarrow 2y\times y-3=5\times -1 \\
 & \Rightarrow 2{{y}^{2}}=3-5 \\
 & \Rightarrow 2{{y}^{2}}=-2 \\
 & \Rightarrow {{y}^{2}}=-1 \\
\end{align}\]
The solution of this equation is not possible. There is no solution to y if quotient is equal to -1.